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https://besatwise.com/5r2eyhdy/transitive-closure-of-a-relation-0b5e49 | # transitive closure of a relation
Otherwise, it is equal to 0. 1. De nition 2. transitive closure can be a bit more problematic. The transitive closure of a is the set of all b such that a ~* b. A = {a, b, c} Let R be a transitive relation defined on the set A. R =, R ↔, R +, and R * are called the reflexive closure, the symmetric closure, the transitive closure, and the reflexive transitive closure of R respectively. Algorithm Warshall Notice that in order for a … The last item in the proposition permits us to call R * the transitive reflexive closure of R as well (there is no difference to the order of taking closures). The program calculates transitive closure of a relation represented as an adjacency matrix. 3) The time complexity of computing the transitive closure of a binary relation on a set of n elements is known to be: a) O(n) b) O(nLogn) c) O(n^(3/2)) d) O(n^3) Answer (d) In mathematics, the transitive closure of a binary relation R on a set X is the smallest transitive relation on X that contains R. The transitive closure of a binary relation $$R$$ on a set $$A$$ is the smallest transitive relation $$t\left( R \right)$$ on $$A$$ containing $$R.$$ The transitive closure is more complex than the reflexive or symmetric closures. Element (i,j) in the matrix is equal to 1 if the pair (i,j) is in the relation. Loosely speaking, it is the set of all elements that can be reached from a, repeatedly using relation … Transitive Relation - Concept - Examples with step by step explanation. Transitive closure. It is not enough to find R R = R2. Let us consider the set A as given below. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) ∈ R for every a ∈ ASymmetricRelation is symmetric,If (a, b) ∈ R, then (b, a) ∈ RTransitiveRelation is transitive,If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ RIf relation is reflexive, symmetric and transitive,it is anequivalence relation In this article, we will begin our discussion by briefly explaining about transitive closure and the Floyd Warshall Algorithm. R2 is certainly contained in the transitive closure, but they are not necessarily equal. It can be shown that the transitive closure of a relation R on A which is a finite set is union of iteration R on itself |A| times. TRANSITIVE RELATION. We will also see the application of Floyd Warshall in determining the transitive closure of a given graph. The transitive closure of R is the relation Rt on A that satis es the following three properties: 1. Warshall’s Algorithm: Transitive Closure • Computes the transitive closure of a relation Transitive Closures Let R be a relation on a set A. Connectivity Relation A.K.A. Defining the transitive closure requires some additional concepts. This allows us to talk about the so-called transitive closure of a relation ~. For calculating transitive closure it uses Warshall's algorithm. Hence the matrix representation of transitive closure is joining all powers of the matrix representation of R from 1 to |A|. In a sense made precise by the formal de nition, the transitive closure of a relation is the smallest transitive relation that contains the relation. Let A be a set and R a relation on A. For transitive relations, we see that ~ and ~* are the same. | 2021-03-09T00:37:52 | {
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http://mathhelpforum.com/differential-equations/176023-dirac-delta-laplace-transform.html | # Thread: Dirac Delta and Laplace Transform
1. ## Dirac Delta and Laplace Transform
Hi guys, my professor recently gave us this problem:
$\displaystyle y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...
$\displaystyle Y(s)=\frac{1}{s^2+4}\sum_{k=1}^{\infty}e^{-sk\pi}=\cfrac{1}{(e^{s\pi}-1)(s^2+4)}$.
However I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.
2. Originally Posted by kenndrylen
Hi guys, my professor recently gave us this problem:
$\displaystyle y^{\prime\prime}+4y=\sum_{k=1}^{\infty}\delta(t-k\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...
$\displaystyle Y(s)=\frac{1}{s^2+4}\sum_{k=1}^{\infty}e^{-sk\pi}=\cfrac{1}{(e^{s\pi}-1)(s^2+4)}$.
However I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.
Here is an idea instead of summing the series just invert the series term by term
$\displaystyle \displaystyle Y(s)=\sum_{k=1}^{\infty}\frac{e^{-s\pi k}}{s^2+4}$
$\displaystyle \displaystyle \mathcal{L}^{-1}\left( \sum_{k=1}^{\infty}\frac{e^{-s\pi k}}{s^2+4}\right)=\sum_{k=1}^{\infty}\mathcal{L}^{-1}\left( \frac{e^{-s \pi k}}{s^2+4}\right)=\sum_{k=1}^{\infty}\sin\left(t-k\pi \right)u(t-k\pi)$
Where $\displaystyle u(t)=\begin{cases}0, \text{ if } t < 0 \\ 1, \text{ if } t \ge 0 \end{cases}$ the Heaviside function.
3. You have $\displaystyle \displaystyle Y(s) = \frac{1}{s^2 + 4}\sum_{k = 1}^{\infty}e^{-sk\pi} = \sum_{k = 1}^{\infty}e^{-sk\pi}\left(\frac{1}{s^2 + 4}\right)$.
Now recall that $\displaystyle \displaystyle \mathcal{L} ^{-1}\left[e^{-as}F(s)\right] = f(t-a)H(t-a)$, and the laplace transform of a sum is the sum of the laplace transforms.
4. Oh! Nice idea guys! I don't know why I missed that
Thanks both of you | 2018-05-28T01:44:50 | {
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https://math.stackexchange.com/questions/161565/what-is-the-total-number-of-combinations-of-5-items-together-when-there-are-no-d | What is the total number of combinations of 5 items together when there are no duplicates?
I have 5 categories - A, B, C, D & E.
I want to basically create groups that reflect every single combination of these categories without there being duplicates.
So groups would look like this:
• A
• B
• C
• D
• E
• A, B
• A, C
• A, D
• A, E
• B, C
• B, D
• B, E
• C, D . . . etc.
This sounds like something I would use the binomial coefficient $n \choose r$ for, but I am quite fuzzy on calculus and can't remember exactly how to do this.
Any help would be appreciated.
Thanks.
Let $$nCr=\binom{n}{r}=\frac{n!}{k!(n-k)!}$$ Remember that the $\frac{n!}{(n-k)!}$ gives all the permutations and the $k!$ in the denominator is what disregards duplicates.
Now; you want all the ways you can choose $$(1 \text{ category from } 5) + (2 \text{ category from } 5) + \dots + (5 \text{ category from } 5)$$ i.e. $$\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=2^5-1=31$$ Note that this follows from the fact that $$(1+1)^n=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n-1}+\binom{n}{n}=2^n$$ Subtracting $\binom{n}{0}$ from both sides gives us $$\binom{n}{1}+\cdots+\binom{n}{n-1}+\binom{n}{n}=2^n-\binom{n}{0}$$ But since $\binom{n}{0}=1,\forall n\in\mathbb{N}$ we have that $$\binom{n}{1}+\cdots+\binom{n}{n-1}+\binom{n}{n}=2^n-1$$ When $n=5$ we thus get the above answer.
Addendum: To address your concern that there seems to be more than $31$ combinations, here is a list of all the possibilities: $$\begin{array}{|c|c|c|c|c|c|c|} & 1 \text{ category} & 2 \text{ categories} & 3 \text{ categories} & 4 \text{ categories} & 5 \text{ categories} & \text{Sum}\\ \hline & A & AB & ABC & ABCD & ABCDE\\ \hline & B & AC & ABD & ABCE \\ \hline & C & AD & ABE & ABDE \\ \hline & D & AE & ACD & ACDE \\ \hline & E & BC & ACE & BCDE \\ \hline & & BD & ADE \\ \hline & & BE & BCD \\ \hline & & CD & BCE \\ \hline & & CE & BDE \\ \hline & & DE & CDE \\ \hline \text{Total} & 5 & 10 & 10 & 5 & 1 & 31 \\ \hline \end{array}$$
• Wish I could upvote this answer twice. Thanks much. That table REALLY helped. Jun 22 '12 at 8:59
• @marcamillion glad to help :)
– E.O.
Jun 22 '12 at 9:00
• I just want to make sure I am understanding this correctly (I too am fuzzy and trying to recall) if I want to know what all the combinations of days in the week are then I would just do: 2^7 - 1 = 127? I keep getting terribly confused as to when I should use factorials, correct me if I am wrong but 7! would be used to find the permutations, not the combinations? Aug 16 '18 at 21:30
There are $\binom{5}{1}$ combinations with 1 item, $\binom{5}{2}$ combinations with $2$ items,...
So, you want : $$\binom{5}{1}+\cdots+\binom{5}{5}=\left(\binom{5}{0}+\cdots+\binom{5}{5}\right)-1=2^5-1=31$$
I used that $$\sum_{k=0}^n\binom{n}{k}=(1+1)^n=2^n$$
• You know...that was my initial inclination - but then I started writing them out and it seems like there would be more than 31 combinations. What's this theory called? Or is there no name? Jun 22 '12 at 8:32
• I think it's simply combinatorics.
– JBC
Jun 22 '12 at 8:36
• Why do you subtract the 1 at the end? Also, can you explain the theory of why the combination with 3 items will be the same as the combination with 2 items...that seems counter-intuitive. Jun 22 '12 at 8:37
• 1) I substracted $\binom{5}{0}=1$ to use the formula recalled at the end (Notice that the formula begins by $\binom{5}{0}$ but your sum by $\binom{5}{1}$). 2) $\binom{n}{k}$ is the number of subset of $\{1,\ldots,n\}$ with $k$ elements, ie the number of choices to take $k$ elements from a set of $n$ elements without repetition, you can show that $\binom{n}{k}=\binom{n}{n-k}$ using $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.
– JBC
Jun 22 '12 at 8:44
• And I think that this formula is intuitive : chosing the k items we take, it's the same thing as choosing the n-k items we left.
– JBC
Jun 22 '12 at 10:20
Thus there are $2^5 = 32$ possibilities. However, you are not counting the choice of none of the five categories, so we subtract $1$ to get $31$ possibilities.
• I think this is the most intuitive way to formualate the solution, and the $2^n$ formula is more natural than when presented in @JBC's answer. Jun 22 '12 at 13:55 | 2021-10-21T15:17:33 | {
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https://math.stackexchange.com/questions/4067252/how-to-prove-big-o-omega-and-theta-asymptotic-notations | # How to prove Big O, Omega and Theta asymptotic notations?
I know the definitions of this notations
1. Big $$\mathcal{O} \; : \enspace T(n) \in \mathcal{O}(f(n))$$ if and only if $$∃ \, c, n_0$$, such that $$T(n) \leq c \cdot f(n) \enspace \forall \,n \geq n_0$$.
2. Big $$\Omega \; : \enspace T(n) \in \Omega(f(n))$$ if and only if $$∃ \, c, n_0$$, such that $$T(n) \geq c \cdot f(n) \enspace \forall \,n \geq n_0$$.
3. Big $$\Theta \; : \enspace T(n) \in \Theta(f(n))$$ if and only if $$∃ \, c_1, c_2, n_0$$ such that $$c_1 f(n) \leq T(n) \leq c_2 f(n) \enspace$$ $$\forall \,n \geq n_0$$.
I'm having trouble manipulating this definitions to prove some notation for example:
• Suppose that $$f = \Theta(g)$$ and $$g = \Theta(h)$$. Prove that $$f = \Theta(h)$$
• Suppose that $$f$$ and $$g$$ are two non-negative functions such that $$g = \mathcal{O}(f)$$. Prove that $$f + g = \Theta(f)$$.
Is there anything that could help me? I've read, watch videos but nothing helps me clarifying to prove these notations. Anything helps thanks. Also if you're a tutor and there's a way to set up a meeting it would help.
• Please format your question to be more readable, also use Mathjax. That way people won't be deterred to read it :) Mar 18, 2021 at 19:00
• Oh no. You can just edit it with the Edit button under your question. This is MathJax Mar 18, 2021 at 19:07
• I have edited your question, now its your turn to make it pretty. This question is a good example. Try to break the question into lines to improve its readability. Basically the more effort you put in the more likely people will answer you. Make it look as good as you would want want something you read to look. Mar 18, 2021 at 19:10
• Thanks, i'm doing it right now will take a couple of minutes. How do i break it in lines? I kept clicking enter but never happened. @LordCommander Mar 18, 2021 at 19:11
• @LordCommander Is this better? In, stackoverflow i always get in trouble with doing a question. Mar 18, 2021 at 19:27
Considering the first problem:
1.) Because $$f \in \Theta(g)$$ there must exist $$c_1, c_2$$ and $$n_0$$ such that
$$c_1 g(n) \leq f(n) \leq c_2 g(n) \tag{1}$$
for all $$n \geq n_0$$.
2.) Because $$g \in \Theta(h)$$ there must exist $$\tilde{c}_1, \tilde{c}_2$$ and $$\tilde{n}_0$$ such that
$$\tilde{c}_1 h(n) \leq g(n) \leq \tilde{c}_2 h(n) \tag{2}$$
for all $$n \geq \tilde{n}_0$$.
3.) Combining 1.) and 2.) yields
$$c_1 \tilde{c}_1 h(n) \leq f(n) \leq c_2 \tilde{c_2} h(n) \tag{3}$$
Therefore there exist $$\hat{c}_1, \hat{c}_2$$, namely $$\hat{c}_1 = c_1 \tilde{c}_1$$ and $$\hat{c}_2 = c_2 \tilde{c}_2$$ and a $$\hat{n}_0 = \max \{ n_0, \tilde{n_0} \}$$ such that $$f$$ satisfies the conditions for being an element of $$\Theta(h)$$, i.e. $$f \in \Theta(h)$$
P.S.: Why $$\hat{n}_0 = \max \{ n_0, \tilde{n_0} \}$$? Because you want both inequalites to hold, in order to insert them into each other. The first one holds for all $$n \geq n_0$$, the second one holds for all $$n \geq \tilde{n}_0$$ and therefore a combination of those two can only hold for all $$n \geq \hat{n}_0 = \max \{ n_0, \tilde{n}_0 \}$$
Edit 1 I numbered the inequalities for better reference.
Row $$(1)$$ consists of two inequalities, namely $$c_1 g(n) \leq f(n)$$ and $$f(n) \leq c_2 g(n)$$. These inequalities are true for all $$n \geq n_0$$.
Row $$(2)$$ consists again of two inequalities, namely $$\tilde{c}_1 h(n) \leq g(n)$$ and $$g(n) \leq \tilde{c}_2 h(n)$$. These inequalities are true for all $$n \geq \tilde{n}_0$$.
Note, that $$n_0$$ and $$\tilde{n}_0$$ do not necessarily have to be the same.
We insert the inequalities of $$(1)$$ and $$(2)$$ into each other to obtain $$(3)$$. This only makes sense if the inequalities we are inserting are indeed true. But if $$n_0 < \tilde{n}_0$$, then row $$(1)$$ is still true for all $$n \geq \tilde{n}_0$$. Otherwise, if $$n_0 > \tilde{n}_0$$ then row $$(2)$$ is still true for all $$n \geq n_0$$. So you choose the greater of these two, i.e. $$\max \{ n_0, \tilde{n}_0 \}$$ in order for both rows $$(1)$$ and $$(2)$$ to hold. Then you have no problems combining them.
• I'm having trouble understanding from where the max came from and when to use it. How did you got to combine them? Is combining them part of proving any asymptotic notations? Mar 18, 2021 at 20:09
• I updated my answer, I hope it is more clear now. The point is, that by definition you need to find a $\hat{n}_0$ such that row $(3)$ holds. Mar 18, 2021 at 20:21
• i'm trying to do the second example what if $g = O(f)$ and how i can identify what it is? Can i define f and g as Big O? Mar 18, 2021 at 20:56
• Big O of a function $f$, i.e. $\mathcal{O}(f)$ is a set. Writing something like $g = \mathcal{O}(f)$ is pretty common, but it is abuse of notation. The correct way to write it would be $g \in \mathcal{O}(f)$. So your task is to check whether $g$ is an element of the set. You do this by checking whether $g$ satisfies the necessary conditions. Mar 18, 2021 at 21:07
• So i could say that $g(n) \leq c * f(n)$ for all $n \geq n_0$? And that proves that $g \in { O } (f)$? Mar 18, 2021 at 21:29
From scratch then:
Consider some arbitrary function $$f$$ and the "Big O" of $$f$$, i.e. $$\mathcal{O}(f)$$. This $$\mathcal{O}(f)$$ is a set, consisting of all functions that satisfy a certain condition. This condition somehow includes $$f$$. You have already given the definition of $$\mathcal{O}(f)$$, but I will rewrite it now:
$$\mathcal{O}(f) \enspace = \enspace \Big\{ \; g : \mathbb{N} \longrightarrow \mathbb{R} \; \Big| \; \exists \, c > 0 \; \exists \, n_0 \in \mathbb{N} \; \forall \, n \geq n_0 \; : \; g(n) \leq c \cdot f(n) \; \Big\}$$
In other words, the set $$\mathcal{O}(f)$$ is a set of functions who map from the natural numbers $$\mathbb{N}$$ to the real numbers $$\mathbb{R}$$. These functions have to satisfy a certain condition, namely that there is some constant $$c$$, which is greater than zero, such that $$g(n) \leq c \cdot f(n)$$. This inequality does not have to hold for every function value of $$g(n)$$ and $$f(n)$$. It only has to hold for all function values after a certain "point" $$n_0$$. Let us look at an example:
Example: Consider the functions
$$g(n) = n \qquad \text{and} \qquad f(n) = n^2 - n$$
We want to show that $$g \in \mathcal{O}(f)$$ holds. If we insert $$n=1$$, we find
$$g(1) = 1 \qquad \text{and} \qquad f(1) = 0$$
But there exists no constant $$c > 0$$, such that $$1 \leq c \cdot 0$$. Does this mean, that $$g \notin \mathcal{O}(f)$$? The answer is NO. Why? Because if we insert $$n=2$$ or $$n=3$$ or $$n = 4,5,6, \ldots$$ we find that for these values the inequality
$$g(n) \leq c \cdot f(n)$$
is fulfilled when e.g. picking $$c = 1$$. This means, that when $$c = 1$$ then for all $$n$$ that are greater than $$n_0 = 1$$ we find $$g(n) \leq c \cdot f(n)$$. However, this is exactly the definition of $$\mathcal{O}(f)$$. Therefore,
$$g(n) \in \enspace \mathcal{O}(n^2 - n)$$
$${}$$
Problem 2
Consider now the second problem you have stated. We want to prove that if $$g \in \mathcal{O}(f)$$ then $$f + g \in \Theta(f)$$.
How do we do this? First of all, we gather the information we have. In this case, we know that $$g \in \mathcal{O}(f)$$. By definition, we know that there exists a $$c>0$$ and a $$n_0$$ such that for all $$n \geq n_0$$ the inequality $$g(n) \leq c \cdot f(n)$$ holds.
Now we simply add $$f(n)$$ on both sides of the inequality (which we know by assumption to be true). This gives us
$$g(n) + f(n) \leq (c+1) \cdot f(n)$$
This was the first part. Now the second part: $$g(n)$$ is nonnegative, so we know that $$g(n) \geq 0$$ and therefore
$$f(n) \leq f(n) + g(n)$$
(because adding something positive to a number always makes the number greater).
We combine these two results and have
$$f(n) \; \leq \; f(n) + g(n) \; \leq \; (c+1) f(n)$$
This completes the proof that $$(f + g) \in \Theta(f)$$.
• Thank you so much, this helped. Mar 19, 2021 at 16:36 | 2022-10-06T17:54:43 | {
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https://brilliant.org/discussions/thread/any-better-way-to-find-the-minimamaxima/ | Any better way to find the minima/maxima?
The teacher asked us to minimize an expression:$\frac{x^2+y^2+z^2}{xy+yz}$, where $x, y, z>0$
Since the denominator and numerator are homogeneous, we try to divide the y^2 into two halves:
$\frac{x^2+\frac{1}{2}y^2+\frac{1}{2}y^2+z^2}{xy+yz}$
Use the inequality $a+b\ge2\sqrt{ab} (a,b>0)$ and we get:
$\frac { x^{ 2 }+\frac { 1 }{ 2 } y^{ 2 }+\frac { 1 }{ 2 } y^{ 2 }+z^{ 2 } }{ xy+yz } \ge \frac { 2\sqrt { x^{ 2 }\cdot \frac { 1 }{ 2 } y^{ 2 } } +2\sqrt { \frac { 1 }{ 2 } y^{ 2 }\cdot z^{ 2 } } }{ xy+yz } =\boxed{2}$
also, the same way to find the minima of $\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz }$ :
$\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz } \\ =\frac { [ax^{ 2 }+ay^{ 2 }]+[(10-a)x^{ 2 }+bz^{ 2 }]+[(10-a)y^{ 2 }+(1-b)z^{ 2 }] }{ xy+xz+yz } \\ \ge \frac { 2\sqrt { ax^{ 2 }\cdot ay^{ 2 } } +2\sqrt { (10-a)x^{ 2 }\cdot bz^{ 2 } } +2\sqrt { (10-a)y^{ 2 }\cdot (1-b)z^{ 2 } } }{ xy+xz+yz } \\ =\frac { 2a\cdot xy+2\sqrt{(10-a)b}\cdot xz+2\sqrt{(10-a)(1-b)}\cdot yz }{ xy+xz+yz }$
Now we let $a=\sqrt{(10-a)b} = \sqrt{(10-a)(1-b)}$ , that is $a = 2 \ and\ b = \frac{1}{2}$
Then
$\frac { 4xy+4xz+4yz }{ xy+xz+yz } \\ = \boxed{4}$
But this method is quite troublesome... right?
Is there any other way to solve the minima? I also tried partial derivative, but it's not easy to solve that equation either...
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1 year, 1 month ago
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Yes there is another approach................Try to use spherical co-ordinates...........That makes things simpler and transforms everything into two variables.........!!
- 1 year, 1 month ago
Care to elaborate on this? What substitution works here? Are you sure the constraints are easily manageable?
- 1 year, 1 month ago
Umm yes.........well, we can use the standard substitutions used when transforming from xyz co-ordinates to spherical co-ordinates.....!!
- 1 year, 1 month ago
I don't understand. What standard substitutions are you referring to?
- 1 year, 1 month ago
Well.........here you go, Sir........
x = rsin(phi)cos(theta)
y = rsin(phi)sin(theta)
z = r*cos(phi)
Here, phi and theta are two different angles........
- 1 year, 1 month ago
Also, r is the radius of the sphere.........we are just taking these as dummy variables.........
- 1 year, 1 month ago
Also, after the transformation, we will have a unit fraction to deal with........and hence it would simply be the matter of maximising the denominator.......
- 1 year, 1 month ago | 2020-01-21T12:55:18 | {
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http://mathhelpforum.com/algebra/223355-oblique-asymptotes.html | 1. ## Oblique Asymptotes
hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?
I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.
2. ## Re: Oblique Asymptotes
Originally Posted by sakonpure6
hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?
I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.
You are off by a - sign. See below.
-Dan
3. ## Re: Oblique Asymptotes
Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.
4. ## Re: Oblique Asymptotes
Hello, sakonpure6!
$\text{Find the oblique asymptote: }\:f(x) \:=\: \frac{x^3-16x}{-4x^2+4x+24}$
$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$
$\text{Divide numerator and denominator by }x^2\!:$
. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$
$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$
$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$
5. ## Re: Oblique Asymptotes
Originally Posted by sakonpure6
Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.
It lines up. Graph it on a calculator and zoom out. (Or you could do it for real and find the limit. )
-Dan
6. ## Re: Oblique Asymptotes
The oblique asymptote is y=-x/4 -1/4
7. ## Re: Oblique Asymptotes
Originally Posted by Soroban
Hello, sakonpure6!
$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$
$\text{Divide numerator and denominator by }x^2\!:$
. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$
$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$
$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$
You are looking at the behavior of the function in the asymtotic region. You suppressed what is happening near the origin. Just divide the polynomials and you will get -x/4 - 1/4. The complete graph is attached.
8. ## Re: Oblique Asymptotes
Originally Posted by topsquark
You are off by a - sign. See below.
-Dan
What graphing utility you are using
9. ## Re: Oblique Asymptotes
Originally Posted by votan
What graphing utility you are using
Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.
-Dan
10. ## Re: Oblique Asymptotes
Originally Posted by topsquark
Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.
-Dan
I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?
11. ## Re: Oblique Asymptotes
Originally Posted by votan
I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?
I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.
I'll look into dplot, thanks.
-Dan
12. ## Re: Oblique Asymptotes
Originally Posted by Soroban
Hello, sakonpure6!
$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$
$\text{Divide numerator and denominator by }x^2\!:$
. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$
$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$
$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$
Another way to see this is to use "long division": $\frac{x^3- 16x}{-4x^2+ 4x+ 24}= -\frac{1}{4}x- \frac{1}{4} - \frac{9x- 6}{-4x^2+ 4x+ 24}$.
Obviously as x goes to infinity (or negative infinity) the remaining fraction goes to 0 so the graph approaches $-\frac{1}{4}x- \frac{1}{4}$.
13. ## Re: Oblique Asymptotes
Originally Posted by topsquark
I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.
I'll look into dplot, thanks.
-Dan
Can we insert a power point presentation here? | 2016-10-01T13:19:58 | {
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https://mathhelpboards.com/threads/determine-p-q-r-and-s.7859/ | Determine p, q, r and s
anemone
MHB POTW Director
Staff member
Hi MHB,
I have encountered a problem and I am not being able to figure out the answer.
Problem:
Given that $p, q, r, s$ are all positive real numbers and they satisfy the system
$p+q+r+s=12$
$pqrs=27+pq+pr+ps+qr+qs+rs$
Determine $p, q, r$ and $s$.
Attempt:
The AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:
1 $\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$ 2 $\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$ which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$ $(pqrs-81)(pqrs-81) \ge 0$ $pqrs \le 9$ or $pqrs \ge 81$
After that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below
$pqrs \le 81$ and $pqrs \ge 81$
$\therefore pqrs=81$ and and obviously the answer would be $p=q=r=s=3$?
Ackbach
Indicium Physicus
Staff member
On your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \ge 0$, with the individual inequalities that you found.
In the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.
[EDIT] See Opalg's post below for a correction.
Opalg
MHB Oldtimer
Staff member
Hi MHB,
I have encountered a problem and I am not being able to figure out the answer.
Problem:
Given that $p, q, r, s$ are all positive real numbers and they satisfy the system
$p+q+r+s=12$
$pqrs=27+pq+pr+ps+qr+qs+rs$
Determine $p, q, r$ and $s$.
Attempt:
The AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:
1 $\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$ 2 $\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$ which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$ $(pqrs-81)(pqrs-81) \ge 0$ $pqrs \le 9$ or $pqrs \ge 81$
After that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below
$pqrs \le 81$ and $pqrs \ge 81$
$\therefore pqrs=81$ and and obviously the answer would be $p=q=r=s=3$?
It looks as though you have solved this problem. You have shown that either $pqrs\leqslant9$ or $pqrs\geqslant 81$. But the equation $pqrs=27+pq+pr+ps+qr+qs+rs$ shows that $pqrs\geqslant27$, so that rules out the first of those possibilities. We are left with the second one, $pqrs\geqslant 81$. But you have also shown that $pqrs\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$.
anemone
MHB POTW Director
Staff member
On your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \ge 0$, with the individual inequalities that you found.
In the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.
[EDIT] See Opalg's post below for a correction.
...We are left with the second one, $pqrs\geqslant 81$. But you have also shown that $pqrs\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$. | 2021-08-01T17:02:03 | {
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https://math.stackexchange.com/questions/389991/closed-form-for-prod-n-1-infty-sqrt2n-tanh2n | # Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$
Please help me to find a closed form for the infinite product $$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$ where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.
• Do you have any reason to believe that such a closed form exists? – George V. Williams May 13 '13 at 1:42
• wolframalpha.com/input/… vs. wolframalpha.com/input/?i=1-exp%28-4%29 It could be a coincidence though... – Vladimir Reshetnikov May 13 '13 at 1:51
• Out of curiosity, where are you finding all of these questions? – Jemmy May 13 '13 at 2:32
• @Jeremy A friend of mine shared these problems with me. They were submitted to a math competition for students, but were rejected by the committee for various reasons: too hard, too easy, have been published before, not interesting etc. – Laila Podlesny May 13 '13 at 17:21
• @LailaPodlesny May I trouble you by requesting to know which math competition you are referring to? – Kugelblitz Sep 24 '15 at 12:03
For $x < 1$, we have the Taylor series expansion: $$f(x):= \frac{-1}{4} \log \left(- \frac{x - x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots$$
Then
$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots$$ $$= - \frac{1}{2} \log(1 - x^2).$$
Now let $x = e^{-2}$. Then
$$\log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) = \frac{1}{2^n} \log \left( \frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)$$ $$= \frac{-4}{2^n} f(e^{-2^n}) = \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$
Hence summing over all $n \ge 1$, we see that, if the product is $P$, then
$$\log P = -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}}) = -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 - x^2),$$
and thus
$$P = \exp \log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$
• Nicely done! :) – Caran-d'Ache May 13 '13 at 4:47
• This is a very careful and crafty derivation. Excellent. I especially like where you pulled the ol' $\sum \log \leftrightarrow \log \prod$. – Coffee_Table May 13 '13 at 20:03
Let $$f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1}$$ and $$g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2}$$ Then \begin{align} f(x)\,g(x) &=\prod_{n=0}^\infty\left(1-x^{2^{n+1}}\right)^{1/2^n}\\ &=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\ &=\left(\frac{f(x)}{1-x}\right)^2\tag{3} \end{align} from which we get $$\frac{f(x)}{g(x)}=(1-x)^2\tag{4}$$ Note that $$\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1-x}\tag{5}$$ and $$\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}=\frac{g(x)}{1+x}\tag{6}$$ Therefore, combining $(4)$, $(5)$, and $(6)$, we get $$\frac{\displaystyle\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1-x^2\tag{7}$$ Plug $x=e^{-2}$ into $(7)$ to get $$\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}$$
• This looks much simpler than the accepted answer (which is also excellent) (+1). – Paramanand Singh Nov 19 '13 at 16:20
• @ParamanandSingh: thanks. This question was just pointed out to me. However, new answers to old questions generally don't get as much attention as the older answers got. – robjohn Nov 19 '13 at 16:32 | 2019-06-24T08:50:02 | {
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https://cs.stackexchange.com/questions/96118/how-can-i-correct-this-hamming-code | # How can I correct this Hamming code?
I'm trying to decode the following Hamming sequence (using EVEN parity and knowing there is a 1-bit error), which contains an ASCII value: 01100110101
I've tried to check for the correctness of each parity bit:
p1 p2 d3 p4 d5 d6 d7 p8 d9 d10 d11
0 1 1 0 0 1 1 0 1 0 1
p1 - 0 1 0 1 1 1
p2 - 1 1 1 1 0 1
p4 - 0 0 1 1
p8 - 0 1 0 1
p1, p4 and p8 checks are all even and correct.
p2 check is odd, hence the parity bit is incorrect and needs to be flipped. The actual message then becomes 00100110101 and the ASCII value can be read as 1011101.
Supposedly this value is incorrect - can anyone point out the mistake I made?
• Wouldn't Hamming Code be more appropriate in the Signal Processing StackExchange? – KingDuken Aug 9 '18 at 21:53
• @KingDuken It is perfectly appropriate in either. – koverman47 Aug 9 '18 at 22:25
• @polyethene Why do you think that this is wrong? It looks correct. Since p2 has the only odd sum, the bit p2 itself has to be wrong (all the other bits are used by at least 2 controll sums, which would mean that at least two bits are wrong). – Jakube Aug 10 '18 at 13:57
• I suggest you edit the question to describe where you got the idea that this is incorrect. If someone told you that, maybe they are wrong. If you read it somewhere, maybe whatever you read was in error. – D.W. Aug 10 '18 at 16:11
• Someone (this OP?) will pay 10 dollars if you answer the question here at reddit.com – Apass.Jack Aug 12 '18 at 22:24
The original problem is, I believe, the following.
When using Hamming code with EVEN parity for 7-bit ASCII characters, the following symbol is retrieved: 01100110101.
Assuming a 1-bit error, what was the original stored symbol? Write down your answer as a 7-bit binary, with no spaces.
OP's statement, "..., which contains an ASCII value: 01100110101", rephrases the original problem statement in a slightly confusing way, as pointed out by Yuval Filmus.
OP's procedure and result is correct, as said in Jakube's comment and indicated in D.W.'s comment. I have verified it as well according to Hamming code at wikipedia.
OP's procedure and result will not be repeated here.
Someone mentioned that "The usual Hamming code has length of the form $2^\ell−1$". For $\ell=4$, the usual Hamming code of length 15 is defined by the following table from wikipedia. If we truncate the last 4 columns, we will get the truncated Hamming code of length 11, which should be, presumably, the Hamming code used in OP's question.
"Supposedly this value is incorrect - can anyone point out the mistake I made?"
This is the turning point to the climax (or anti-climax). There are four possibilities.
1. A different kind of (truncated) Hamming code is used.
For example, all four parity bits could have been specified to be put together in the first four bits. That is, the first four bits, 0110 are the usual p1, p2, p4 and p8 parity bits while the remaining 0110101 are the data bits. In that case, we rearrange the bits to form the usual Hamming code as (p1)(p2)0(p4)110(p8)101, which is 01011100101. Then p1, p2, p4 check are odd while p8 is OK. So we flip the $1+2+4 =7$-th bit to obtain 01011110101. So the corrected data bit will be 0111101. Well, OP can check if this or something similar is the case.
2. "This is from a quiz on my university's LMS, the site is saying my answer is incorrect."
As D.W. suggested, maybe they are wrong. Maybe whatever you read was in error. There might be a typo in the statement of the quiz. There might be an error in the original answer by the quiz owner. There might be an error in the answer-checking process by the site. There might be a critical typo in OP's post. And so on.
3. Jakube, D.W., I, and, apparently, many others have not been able to find the mistake. That is very unlikely, though.
4. The last possibility stands for, as always, all other probably even lesser possibilities. | 2019-10-13T21:00:14 | {
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https://math.stackexchange.com/questions/3286747/what-is-the-vector-x-%E2%88%88-mathbbr3-that-achieves-maxx-1-subject-to-x/3286858 | # What is the vector $x ∈ \mathbb{R^3}$ that achieves $max||x||_1$ subject to $||x||_2 = 1$?
I'm trying to answer the questions "What is the vector $$x ∈ \mathbb{R^3}$$ that achieves $$max||x||_1$$ subject to $$||x||_2 = 1$$?" and "What is the vector x ∈ $$R^3$$ that achieves $$max||x||_∞$$ subject to $$||x||_2 = 1$$?
I think the first question is asking me to find a vector with three components that will have the maximum $$||x||_1$$ norm value where $$\sqrt{x_1^2 + x_3^2 + x_2^2} = 1$$, so $$x_1^2 + x_3^2 + x_2^2 = 1$$. I know the The L1 norm is just the sum of the absolute values of the vector's components. After trial and error I came up with $$x = [\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}]$$ , but also $$[-\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}]$$, and $$[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}]$$, etc.
For my the second question, I think I need to find the vector in $$\mathbb{R^3}$$ that will give me the maximum value of the absolute value of the vector's components given $$x_1^2 + x_3^2 + x_2^2 = 1$$. I came up with $$[1, 0, 0]$$, $$[0, 1, 0]$$ , $$[0, 0, 1]$$, $$[-1, 0, 0]$$, $$[0, -1, 0]$$ , and $$[0, 0, -1]$$.
Am I correct? Is there a more formal way to figure this out and write my solution?
• First question is tackled with Cauchy-Schwarz, see math.stackexchange.com/questions/218046/… for instance; Second one is even easier – Olivier Jul 8 '19 at 13:57
• To solve optimization problems subject to equality constraints Lagrange multipliers are a very popular method. Example 1a in the Wiki article is quite close to your $L_1$ problem. – pH 74 Jul 8 '19 at 13:59
To solve the first question:
By symmetry, we can assume $$x_1,x_2,x_3$$ are all positive (and add the other solutions later).
By the method of Lagrange multipliers
$$L=x_1+x_2+x_3 - \lambda(x_1^2 + x_2^2 +x_3^2-1)$$
$$0=\frac{\partial L}{\partial x_1} = 1-2\lambda x_1$$ $$0=\frac{\partial L}{\partial x_2} = 1-2\lambda x_2$$ $$0=\frac{\partial L}{\partial x_3} = 1-2\lambda x_3$$
These three equations imply $$x_1=x_2=x_3$$ so that by the constraint $$||x||_2=1$$, the solutions is $$x_1=x_2=x_3=\frac{1}{\sqrt{3}}$$
All of the solutions are
$$x_1=\pm \frac{1}{\sqrt{3}},\quad x_2=\pm \frac{1}{\sqrt{3}},\quad x_3=\pm \frac{1}{\sqrt{3}}.$$
To solve the second question:
We want to maximize one of the elements of the vector.
Trivially, the solutions are
$$\begin{pmatrix} \pm 1 \\ 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} 0 \\ \pm 1 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} 0 \\ 0 \\ \pm 1 \end{pmatrix}.$$
We can obtain this also with Lagrange multipliers. Since, by symmetry, we can seek to maximize $$|x_1|$$, first constrain $$x_1$$ to be positive as above, and then find the other solution ($$x_1<0$$).
Consider
$$L= x_1 - \lambda (x_1^2 + x_2^2 + x_3^2-1)$$
$$0=\frac{\partial L}{\partial x_1} = 1 + 2 \lambda x_1$$ $$0=\frac{\partial L}{\partial x_2} = 2\lambda x_2$$ $$0=\frac{\partial L}{\partial x_3} = 2\lambda x_3$$ implies that $$x_1=1$$ and $$x_2=x_3=0$$. Similarly for the other cases.
If $$\|x\|_2 = 1$$, the Cauchy-Schwarz inequality implies $$\|x\|_1 = |x_1|+|x_2|+|x_3| \le \sqrt{x_1^2+x_2^2+x_3^2}\cdot\sqrt{1+1+1} = \sqrt{3}$$
For $$(x_1,x_2,x_3) = \left(\frac1{\sqrt3}, \frac1{\sqrt3}, \frac1{\sqrt3}\right)$$ we have $$\|x\|_1 = \sqrt{3}$$ so this is the maximum.
Now let $$x$$ be a minimizer. We then have equality in the Cauchy-Schwarz inequality above so there exists $$t \in \mathbb{R}$$ such that $$(|x_1|,|x_2|,|x_3|) = t(1,1,1) = (t,t,t)$$ Taking $$\|\cdot\|_2$$ norm gives $$t = \frac1{\sqrt{3}}$$ so $$x=\left(\pm\frac1{\sqrt3}, \pm\frac1{\sqrt3}, \pm\frac1{\sqrt3}\right)$$
• so would the answer to the first question be the set of vectors x where $\|x\|_1 = \sqrt{3}$ ? – John Jul 8 '19 at 15:39
• @John Precisely, it is the sphere around the origin of radius $\sqrt{3}$ w.r.t. the norm $\|\cdot\|_1$. – mechanodroid Jul 8 '19 at 16:04
• @John Of course, intersect the set above with the unit sphere. There are only $4$ solutions, see above. – mechanodroid Jul 8 '19 at 17:15
• There are eight solutions! They lie on the vertices of a cube. – mjw Jul 8 '19 at 21:54
• @mjw Whoops, $2^3= 8$ and not $4$, thanks. – mechanodroid Jul 8 '19 at 21:57
In general, $$\|x\|_1 \le \sqrt{n}\|x\|_2$$, where $$n$$ is the dimension of the space.
If $$\|x\|_2 = 1$$ then we have $$\|x\|_1 \le \sqrt{n}$$, to achieve equality, note that $$\|{1 \over \sqrt{n}}(1,...,1) \|_1 = \sqrt{n}$$.
For the solution of your first question, you should also add the follwoing points: $$\left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(-\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right)$$ and $$\left(-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right)$$.
To intuitively guess them, you need to think that both L1 and L2 are symmetric metrics. So you should consider points $$(x_1, x_2, x_3)$$ where $$|x_1| = |x_2| = |x_3|$$. The constraint of $$x_1^2 + x_2^2 + x_3^2 = 1$$ would give you $$|x_1| = |x_2| = |x_3| = \sqrt{\frac{1}{3}}$$.
For a geomteric solution, let us first consider the case in 1st octant i.e. where $$x_1 \geq 0, x_2 \geq 0,$$ and $$x_3 \geq 0$$. Thus $$L_1 = |x_1| + |x_2| + |x_3|$$ is equivalent to $$L_1 = x_1 + x_2 + x_3$$. Now geometrically, one needs to find the point where the plane $$x_1 + x_2 + x_3 = constant$$ touches the sphere $$x_1^2 + x_2^2 + x_3^2 = 1$$ in the first octant.
If you want to formally and algebraically compute the values in the 1st oactant, then you need to write the Lagrangian function $$L(x_1, x_2, x_3, \lambda) = x_1 + x_2 + x_3 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$. Now equate all the partial derivatives to 0. This will get you $$\lambda = -\frac{1}{2x_1} = -\frac{1}{2x_2} = -\frac{1}{2x_3}$$. This leads to $$x_1 = x_2 = x_3 = \frac{1}{\sqrt{3}}$$. In the other octant (where $$x_1 \le 0, x_2 \geq 0,$$ and $$x_3 \geq 0$$) the Lagrangian function will change sign: $$L(x_1, x_2, x_3, \lambda) = -x_1 + x_2 + x_3 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$. This would yield an answer of $$-x_1 = x_2 = x_3 = \frac{1}{\sqrt{3}}$$.
For the second problem, let us take the first case where the $$L_\infty$$ norm $$= Max(x_1, x_2, x_3) = x_1$$. Then the Lagrangian function $$L(x_1, x_2, x_3, \lambda) = x_1 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$. | 2021-07-26T12:24:32 | {
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https://math.stackexchange.com/questions/2229101/quick-question-about-the-limit-of-a-two-variable-function-as-x-y-to-infty/2229165 | # quick question about the limit of a two-variable function as $x,y\to\infty$
$$\lim_{x,y\to\infty} \frac{x-y}{x^2+y^2}\tag{\star}$$
I'm used to do the following substitution when I see $x^2+y^2"$ and that $x,y\to 0$
$$x^2+y^2 = r^2,\;x=r\cos\theta,\;y=r\sin\theta$$
plug these values in the function and compute the limit as $r\to0$ I know I can do that because the only way for $x$ & $y$ to approach $0$ is $r$ approaching $0.$
I cannot do this substitution everytime because if for example: $(x,y)\to(-1,7)$ there's no value $u$ that guarantee me if $r\to u$ then $(x,y)\to(-1,7).$
but here since $x,y\to\infty$ I think that logically this phenomenon can only happen if $r\to\infty$ as well.
So computing $(\star)$ is the same as computing this :
$$\lim_{r\to\infty} \frac{r\cos\theta-r\sin\theta}{r^2} =\lim_{r\to\infty} \frac{\cos\theta-\sin\theta}{r}=0.$$
I'm 90% sure that what I've done is correct but I still want a confirmation and if possible show me other ways to compute this limit.
Sorry if this question sounds kinda dumb but I'm still new to multivariable calculus and today is my first time dealing with MVC limits.
Thank you !
Others can add and/or correct, but I'm not sure if you can do the polar trick. If $x$ and $y$ tend to infinity, then clearly $r \to \infty$. But if you have $r \to \infty$, then you don't necessarily have $x$ and $y$ to infinity since, for example: $x \to \infty$ and $y \to c$ (constant) will also lead to $r \to \infty$.
and if possible show me other ways to compute this limit.
Rewrite: $$\left| \frac{x-y}{x^2+y^2} \right| =\left| \frac{x}{x^2+y^2}- \frac{y}{x^2+y^2} \right| \le \left| \frac{x}{x^2+y^2}\right|+ \left|\frac{y}{x^2+y^2} \right|$$ Now: $$\left| \frac{x}{x^2+y^2} \right| \le \left| \frac{x}{x^2} \right| = \left| \frac{1}{x} \right| \to 0 \quad\mbox{ and }\quad \left| \frac{y}{x^2+y^2} \right| \le \left| \frac{y}{y^2} \right| = \left| \frac{1}{y} \right| \to 0$$ Alternatively, if you feel more comfortable with limits to $(0,0)$, substitute $\left( x,y \right) \to \left( \tfrac{1}{u},\tfrac{1}{v} \right)$ and take the limit $(u,v) \to (0^+,0^+)$ and you could follow up with your classical polar substitution.
With no other answers so far, I'll add this example: consider $f(x,y)=x+y$, then clearly: $$\lim_{(x,y)\to (+\infty,+\infty)} \bigl( x+y \bigr) = +\infty$$ However, switching to polar coordinates, we get: $$f(r,\theta) = r \left( \cos\theta + \sin\theta \right)$$ Now simply taking $r \to +\infty$, the limit: $$\lim_{r \to +\infty} \bigl( r \left( \cos\theta + \sin\theta \right) \bigr)$$ depends on $\theta$ since $\cos\theta + \sin\theta$ can be positive, negative or zero. This makes sense since fixing $\theta$ to a value outside the interval $(0,\tfrac{\pi}{2})$ would not correspond to $(x,y)\to (+\infty,+\infty)$.
• @rapidracim Thanks for accepting but you may want to wait for some more input of others, specifically regarding your question on the polar substitution for the limit to $(+\infty,+\infty)$. – StackTD Apr 11 '17 at 12:52
• I think I've found a counterexample $$\lim_{x,y \to \infty} \frac{x}{y^2}+x^2$$ here If I use the polar substitution I'll get $+\infty$. but wolfram-alpha says it doesn't exist because it is path dependent do you think I can trust wolframalpha ? if it gave me a "value" maybe but since it told me it doesn't exist I presume other than computing the limit along two different paths and finding two different results (which actually proves that the limit D.N.E) a computer can't make such a conclusion (that the limit is path-dependant) using a different algorithm . – the_firehawk Apr 11 '17 at 16:15
"but here since $x,y\to \infty$ I think that logically this phenomenon can only happen if $r\to \infty.$" The definition of $x,y\to \infty,$ which you haven't given us, is probably exactly the same as $r\to \infty.$ Let's assume this. You have correctly arrived at
$$f(r\cos t, r \sin t) = \frac{\cos t - \sin t}{r}.$$
$$0\le |f(r\cos t, r \sin t)| = \frac{|\cos t - \sin t |}{r} \le \frac{|\cos t| + |\sin t|}{r} \le \frac{1 +1}{r} = \frac{2}{r}\to 0.$$
Thus $|f(r\cos t, r \sin t)|\to 0,$ which is the same as saying $f(r\cos t, r \sin t)\to 0.$ | 2020-09-28T10:13:32 | {
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https://math.stackexchange.com/questions/3572975/how-does-the-peano-axiom-of-induction-prevent-s-loops | # How does the Peano axiom of induction prevent S-loops?
First, let me state what I understand to be the first-order rendition of Peano's 5th axiom: the axiom of induction. For all natural numbers, for any relation/property/predicate $$R$$...
$$(R(0) \land \forall x[R(x) \rightarrow R(S(x))]) \rightarrow \forall x(R(x))$$
(first question: is this a correct formalization of this axiom or not?)
How does this axiom prevent elements of natural numbers that have '$$S$$-loops' that is:
\begin{align} S(a) = b && \text{and} && S(b) = a \end{align} (edited for clarity)
• This axiom does not, but the third Peano Axiom.says that if $S(n)=S(m)$, then $n=m$. If $S(x_9)=x_8$, then $S(x_9)=S(x_7)$, so $x_7=x_9$, which eventually leads to $0=S(x_2)$, which is forbidden by the Fourth Axiom (or peharps by a different number axiom depending on how you formalize them). – Arturo Magidin Mar 7 at 21:37
• ' then 𝑆(𝑥9)=𝑆(𝑥7)...' is not true. As defined above, $S(x_9) = x_8$. I've change the nomenclature to 'a' and 'b' to be more clear. In this sense S(a) = b and S(b) = a are completely possible given axioms 1-4. – C Shreve Mar 8 at 2:41
• It’s not that “it’s not true”. It’s that you’ve changed your statement. When you wrote $x_8$, I of course assumed that you meant the result of applying $S$ to $0$ eight times. And, no it is still not possible, and yes, you still need the other axioms; induction alone doesn’t do it, though induction does come into play in the general statement. If $a=0$, then you are violating the axiom that says that $0$ is not a successor. If $a\neq 0$, then from induction you get that $a$ is a descendant of $0$, so you still get that $a=x_n$ and go from there. – Arturo Magidin Mar 8 at 10:36
• I would prefer brackets enclosing everything to the left of the 2nd "$\implies$" in your statement of Axiom 5. In contrast, putting brackets around everything to the right of "$\land$" gives a different (wrong) meaning. – DanielWainfleet Mar 9 at 18:28
Using Peano's axioms
\begin{align} \forall m \, S(m) \neq 0 &&&\text{(}0\text{ is not the successor of anyone)} \\ \forall m \forall n \, (S(m) = S(n) \to m = n) &&&\text{(injectivity of }S\text{)} \end{align}
and Peano's principle of induction, it is easy to prove that the "double-successor" does not have any fixed point, i.e. \begin{align} \forall m \, S(S(m)) \neq m \end{align}
A rigorous proof of this property is below. It is analogous to the one you can find here to prove that the successor has no fixed point.
Now, this property excludes the possibility of $$S$$-loops. Indeed, if there were $$m$$ and $$n$$ such that \begin{align} S(m) &= n & S(n) &= m \end{align} then we would have $$S(S(m)) = m$$ (replace $$n$$ with $$S(m)$$ in the second identity), which is impossible.
We want to prove that, in Peano arithmetic, \begin{align} \forall x \, S(S(x)) \neq x &&&\text{(i.e. } \forall x \, R(x) \text{ where } R(x) \text{ is the formula } S(S(x)) \neq x\text{).} \end{align}
To prove this we apply Peano's induction principle, thus we have to prove two facts:
1. Base case, i.e. $$S(S(0)) \neq 0$$. This holds because it is just an instance (take $$x = S(0)$$) of Peano's axiom \begin{align} \forall x \, S(x) \neq 0 &&&\text{(0 is not the successor of anyone).} \end{align}
2. Inductive case, i.e. $$\forall x \, \big(S(S(x)) \neq x \to S(S(S(x))) \neq S(x) \big)$$. So, given $$x$$, we suppose $$S(S(x)) \neq x$$ and we have to show that $$S(S(S(x))) \neq S(x)$$. Aiming for a contradiction, suppose $$S(S(S(x))) = S(x)$$. According to Peano's axiom \begin{align} \forall m \forall n \, (S(m) = S(n) \to m = n) &&&\text{(injectivity of }S\text{)} \end{align} instantiated with $$m = S(S(x))$$ and $$n = x$$, we have that $$S(S(x)) = x$$, which is impossible. Therefore, $$S(S(S(x))) \neq S(x)$$.
This ends the proof that $$\forall x \, S(S(x)) \neq x$$.
I think your formalization is correct.
The axiom of induction doesn't prevent by itself S loops. Consider, a two elements set $$\{0,1\}$$ with $$S(0) = 1$$ and $$S(1) = 0$$
To prevent S loops you need axioms
$$\forall a,b \; . \; S(a) = S(b) \Rightarrow a = b$$
3."$$0$$ is not a succesors"
$$\forall a\;.\;S(a)\neq0$$
Informally, if you have a loop from combination of (5) and (4) it follows that loop need to involve all predecessors of a looped element. And with this (3) provides a contradiction, as by (5) every element has $$0$$ as a predecessor.
The more formal proof can look like this.
Consider the predicate $$NL(x) = \text{x is not an element of any loop.}$$ By definition of the loop, if $$a$$ is in the loop, there must be $$b$$ in the loop, such that $$a = S(b)$$. So, $$NL(0)$$ holds by axiom (3). Now consider an element $$a$$ such that $$NL(a)$$ holds. If $$S(a)$$ is an element of the loop, then by the axiom (4) the element $$a$$ is also in the loop. This is a Contradiction! Thus, $$NL(a) \Rightarrow NL(S(a))$$ holds for any $$a$$. Now can apply axiom of induction to see that there is no element $$a$$, which can be looped.
So there are no loops in Natural numbers defined by Peano axioms. Note, however, that you need every axiom to prove it.
• I think your $a$ and $b$ in paragraph 2 are supposed to be $0$ and $1$.... – Arturo Magidin Mar 7 at 21:52
• Thanks. You are correct. This was a typo. – Nik Pronko Mar 7 at 21:57
• Indeed, with only the axiom of induction, the one-element set $\{0\}$ would also be a possibility. – Greg Martin Mar 7 at 22:21
• But this is not necessarily true. The case you bring up is true because you have expressly chosen a specific set ({0,1}) where one member (ie 0) is identified by axiom 3. However, the set {0, 1, 2, ... a, b} with S(a)= b and S(b) = a could still exist given axioms 1-4. What is it about axiom 4 that eliminates 'a' and 'b'? – C Shreve Mar 8 at 2:48
• @CShreve Say, $a = S(c)$, then by axiom 4 $b = c$. Repeating this procedure in so for we can, we will end up with a set $\{0,1\}$ as in example above. Of course, this works only with finite sets, and words "repeating procedure in so for" really is just a reference for the axiom of induction. I will add a more formal proof to the answer. – Nik Pronko Mar 8 at 10:34
In the general statement as you’ve now written, not a specific example as you had before, induction does come into play in the sense that one must prove, by induction, that if $$x\in \mathbf{N}$$, then either $$x=0$$, or there exists $$k$$ such that $$x=S^k(0)$$.
Indeed you let the property by “$$x=0$$ or $$x$$ is a descendant of $$0$$”; $$0$$ has the property, and if $$n$$ has the property, then so does $$S(n)$$.
Once you have that, from $$S(a)=b$$ and $$S(b)=a$$, you get that either $$a=0$$ and you violate the Axiom that says that $$0$$ is not a successor; or else that $$a=S^k(0)$$ for some $$k$$. Then $$b=S^{k+1}(0)$$, and you are now in essentially the same situation as before, when you had the specific example of $$S(x_8)=x_9$$ and $$S(x_9)=x_8$$. Now you have $$S^{k+2}(0)=S^k(0)$$. From that you get $$S^{k+1}(0)=S^{k-1}(0)$$, and so on until you get that $$0$$ is a successor, contradicting that axiom.
@Taroccoesbrocco, so in the spirit of a 'double successor' implies contradiction, would a formalization of this look like:
1. Let R(x) -> S(S(x)) = x (double successor relation)
2. By Axiom 5 then R must also apply to $$0$$
3. So $$S(S(0)) = 0$$
4. Let $$a = S(0)$$
5. By 4 and 3, $$S(a) = 0$$...which contradicts Peano Axiom 4
6. Therefore relation R (ie...double successor) is not in the natural numbers.
Is there anything incorrect about the above line of reasoning?
• What you wrote is essentially meaningless. If you want to prove $\forall x \, S(S(x)) \neq x$ by contradiction, you have to suppose its negation, that is $\exists x \, S(S(x)) = x$. Now, you do not know who is the $x$ such that $S(S(x)) = x$, therefore you are not allowed to conclude that $S(S(0)) = 0$. – Taroccoesbrocco Mar 9 at 11:41
• I edited my answer to include a proof of $\forall x \, S(S(x)) \neq x$. – Taroccoesbrocco Mar 9 at 11:43
• The additional detail above are helpful. The intent in my reasoning above is to 1)assume a relation $R$ of the natural numbers then 2)show that the relation $R$ produces some contradiction against the Peano axioms. The reason $R$ can be applied to 0 is axiom 5 as any relation in the natural numbers must be applicable to 0 as well. – C Shreve Mar 9 at 19:50
• What you proved is that $\exists x \, S(S(x)) \neq x$, more precisely that $S(S(0)) \neq 0$. But it is not enough, indeed you have to prove that $\forall x \, S(S(x)) \neq x$. – Taroccoesbrocco Mar 9 at 20:38
• The fact that a property fails for some natural numbers does not imply that it fails for all natural numbers. You define the predicate $R(x)$ as $S(S(x)) = x$. With your argument you're are showing that $\forall x \, R(x)$ is not true, i.e. that $\exists x \, S(S(x)) \neq x$. But this is not what you need to prove the absence of $S$-loops. What you need to prove is that $\forall x \, S(S(x)) \neq x$. – Taroccoesbrocco Mar 10 at 6:25 | 2020-07-04T00:10:59 | {
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http://math.stackexchange.com/questions/105741/how-to-prove-gcda-gcdb-c-gcd-gcda-b-c | # How to prove $\gcd(a,\gcd(b, c)) = \gcd(\gcd(a, b), c)$?
I am trying to prove that $\gcd(a, \gcd(b, c)) = \gcd(\gcd(a, b), c)$.
The definition of GCD available to me is as follows:
Given integers a and b, there is one and only one number d with the following properties.
1. $d \geqslant 0$
2. $d|a$ and $d|b$
3. $e|a$ and $e|b$ implies $e|d$.
In the book that I am studying, prime factorization of numbers hasn't been taught yet. Only, the definition of GCD, I've given above has been taught and proven. So, I want to use only this to prove that $\gcd(a, \gcd(b, c)) = \gcd(\gcd(a, b), c)$. Could you please help me?
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Show that both sides equal gcd(a,b,c). – franz lemmermeyer Feb 4 '12 at 19:21
I haven't encountered the definition of gcd of three numbers in the text yet and I am trying to avoid it. – Lone Learner Feb 4 '12 at 19:26
Proof that GCD is associative – pedja Feb 4 '12 at 20:01
@LoneLearner : The gcd of any number of numbers is the greatest of all of their common divisors, so you just need to know what a common divisor of three numbers is. The divisors of $12$ are $1,2,3,4,6,12$; the divisors of $15$ are $1,3,5,12$; the common divisors are just the members of the intersection of those sets of divisors (in this case $1,3$). So the question is: what's the definition of the intersection of three sets? The answer is that a thing is a member of the intersection precisely if it's a member of all three sets. – Michael Hardy Feb 4 '12 at 21:08
By considering prime factorizations, it's a consequence of $\min(x,\min(y,z)) = \min(\min(x,y),z)$. – lhf Feb 5 '12 at 1:34
Same answer as I just gave in sci.math...
Note that $$d|x,y\Longleftrightarrow d|\gcd(x,y).$$ So: \begin{align*} d|a,\gcd(b,c) &\Longleftrightarrow d|a,b,c\\ &\Longleftrightarrow d|\gcd(a,b),c \end{align*}
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nice hint! (+1) – robjohn Feb 4 '12 at 20:20
+1 for giving the cleanest proof possible (as far as I can see). My only gripe is with the notation: I would write $\;d|x \land d|y\;$ instead of $\;d|x,y\;$. Then the associativity of $\;\gcd\;$ translates directly to the associativity of $\;\land\;$. – Marnix Klooster Jul 26 '13 at 9:41
Please note that this solution uses an idea that is very similar to the idea in the solution posted much earlier by ncmathsadist. The main difference is that it may contain fewer typos.
We show that for any integer $u$, if $u$ divides the left-hand side, then $u$ divides the right-hand side, and vice-versa. Thus the left-hand side and the right-hand side have the same set of common divisors, so must be equal, since they are both non-negative.
Now suppose that $u$ divides $\gcd(a, \gcd(b, c))$. Then $u$ divides $a$ and $u$ divides $\gcd(b,c)$. So $u$ divides $b$ and $c$, and therefore $a$, $b$, and $c$.
Now look at the right-hand side. We know that $u$ divides all of $a$, $b$, and $c$. So $u$ divides $\gcd(a,b)$, and therefore $u$ divides $\gcd(\gcd(a,b),c)$.
Showing that if $u$ divides the right-hand side, then $u$ divides the left-hand side is essentially the same calculation, and can be omitted.
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I am trying a proof that strictly leads to the fact that if $d = gcd(a, gcd(b, c))$ then $d$ must satisfy the conditions $d|a$, $d|gcd(b, c)$, $e|a$ and $e|gcd(b, c)$ implies $e|gcd(a, gcd(b, c))$. Could you please tell me how to prove the last implication part? – Lone Learner Feb 4 '12 at 21:03
@Lone Learner: It is inconvenient to work with $d$ directly, it is clearer to work with any common divisor. – André Nicolas Feb 4 '12 at 21:07
First note that $(a,b) \mapsto \gcd(a,b)$ is symmetric in $a$ an $b$. Suppose $d$ is a commond divisor of $a$, $b$ and $c$. Then $c|a$ and $d|\gcd(b,c)$ so $d|\gcd(a, \gcd(b,c))$.
Conversely suppose that $d$ is a common divisor or $a$ and $\gcd(b,c)$. Then $d|a$ and $d|\gcd(a,b)$. Hence, $d$ is a common divisor of $a$, $b$ and $c$.
Our result follows now by symmetry.
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Why should we assume that $c$ is a common divisor of $a$ and $b$? The problem doesn't require $c$ to be a common divisor of $a$ and $b$. – Lone Learner Feb 4 '12 at 19:27
What we see here is that both $\gcd(a,\gcd(b,c))$ and $\gcd(\gcd(a,b), c)$ are both simply the largest common divisor of $a$, $b$, and $c$. – ncmathsadist Feb 4 '12 at 19:43
But that doesn't imply that $c$ must be a common divisor of $a$, $b$ and $c$. – Lone Learner Feb 4 '12 at 20:14
@LoneLearner: There’s a major typo in the answer: the common divisor should be $d$ (or some other symbol distinct from $a,b$, and $c$). – Brian M. Scott Feb 4 '12 at 20:18
But that still doesn't show how $d$ is a $gcd(a, gcd(b, c))$. According to the definition I have given, we now need to show that if $e$ divides $a$ and $e$ divides $gcd(b, c)$, then $e$ must divide $d$. How do you show this? – Lone Learner Feb 4 '12 at 20:45
Here is a proof I am attempting from all the hints I have got so far. Please let me know if this is correct.
Let $d = gcd(a, gcd(b, c))$. Therefore,
1. $d \geqslant 0$ from the definition of GCD.
2. $d|a$ from the definition of GCD.
3. $d|gcd(b, c)$ from the definition of GCD.
4. $e|a$ and $e|gcd(b,c)$ implies $e|d$, also from the definition of GCD.
5. From 3, $d|b$.
6. From 3, $d|c$.
7. From 2 and 5, $d|gcd(a, b)$.
8. Let $e|gcd(a, b)$ and $e|c$. From the definition we know that $gcd(a, b) | a$ and $gcd(a, b) | b$. Therefore, $e|a$ and $e|b$ from the transitive property of divisibility. So, $e|gcd(b, c)$ from the definition of GCD. So, from 4 we have, $e|$d.
From 1, 7, 6 and 8, we get, $d = gcd(gcd(a, b), c)$.
-
#4 seems false. How does it follow from the definition of GCD? If d is a prime factor X common to both a and gcd(b,c), and e is a different prime factor Y common to both a and gcd(b,c), then e will not divide d or vice versa, because they're prime. – Joseph Garvin Jan 26 '13 at 19:05
Actually #4 is OK, it does follow from the definition if you're using the Bezout's identity version. – Joseph Garvin Jan 27 '13 at 17:17 | 2014-07-22T07:43:54 | {
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https://www.themathdoctors.org/weighted-averages-averaging-averages-or-rates/ | # Weighted Averages: Averaging Averages or Rates
In our series on averages, last week we introduced the idea of the weighted average (or weighted mean), where each item has a weight attached. The classic examples all involve grade averages in various ways. This time, we’ll look at how weighted averages arise when you need to average several averages together, something we touched on last time, but which arises often in very different settings. One thing we’ll see repeatedly is that the appropriate average depends on your needs.
## Can you average averages, or not?
We’ll start with a 1997 question from a professional who had learned that, generally, you can’t average averages, but didn’t know when you can:
Algebra: Average of an Average
Hi! This one is a real-life need. I work for a consulting firm, and am having a hard time remembering my basic Algebra!
If the "Wireless" line of business hires 50 people in one month, and the "Multimedia" line of business hires 80 people in one month, what is the average number of people per month we are hiring?
At first, one thinks (50 + 80)/2 = 65 is the correct answer, but it's not because we're taking the average of an average, right?
Don't you have to let x = something and then do 1/50 + 1/80 and solve... or something?
Hi Ron,
65 is the correct answer if the question is: "What is the average number of people hired per line of work per month?"
Here, we are just averaging the two averages “per line”, so as long as we state clearly what it means, it is valid. The question is, is this average what you want to know, or is it something else entirely?
Assuming that the entire company is made up of only those two lines of business, and you're asking "What is the average number of people hired by the whole company per month?", the answer is obviously 50+80 = 130.
This seems to be what Ron wants, and it isn’t really an average at all, just a total!
So what about the idea that you can’t average averages? That’s really a different story:
The operations above are perfectly reasonable. Let me show you the kind of situation that I think you were worried about, and you'll see why it's different from the situations above:
I take 4 exams and have an average score of 80. Then I take 2 more exams and on those 2, my average is 100. What's my overall average for the course?
The wrong way to do it is (80+100)/2 = 90. This is wrong because the averages were of different-sized groups. To get the correct answer, I know that on the first 4 exams I got 320 total points because when I divide 320 by 4, I get 80.
Similarly, for the last two exams, I must have gotten 200 points total. So for the six exams, I got 200+320 = 520 points and 520/6 = 86.66666 = my real grade average.
We saw this scenario last week: If we just average the two averages, the result can only be described as the average of the averages, not the average for the course. The latter has to be weighted, because for the course, each exam counts the same, not each of these two sets of exams.
As we saw last week, a weighted average can often be understood by breaking it down:
For your problem, the averages you are averaging are for the same period, so it works out. To convince you that it's true, let's just look at a situation where the averages came from 10 months of data.
Then in the first line of business, 500 people must have been hired, since 500/10 = 50. Similarly, 800 were hired in the other, since 800/10 = 80. Altogether, 1300 people were hired in the ten months, or 1300/10 = 130 per month, company-wide. Or if you're trying to get the average per line of work, 65 is right, since if each group had hired 65 people each month for 10 months, there would be 65*20 = 1300 total hires, so it works out.
Bottom line: always think about what you want, and what an average means, rather than use an average (or not!) unthinkingly.
## Using a weighted average
Averaging Averages
I was reading your response on averaging averages. This has come up in a meeting I go to regarding a report that I am responsible for every month. After reading your response, I want to make sure that I am doing this right.
I have 5 different departments that send me an average rating to 3 different questions from a feedback form (rated 1-5). For example, this is what they send me:
Question 1 - average 4 (this is the average of question 1 from all feedback forms this department received during a certain time period)
Question 2 - average 3 (same comment as above)
Question 3 - average 5 (same comment as above)
Total average- 4
After getting this information from all 5 different departments, I combine them in a total company report by taking everyone's average to question 1 and then average that to get an average for the whole program. (For this program the company average to question 1 is...) and so on for questions 2 and 3 and the total average.
If I am reading your response correctly to the question on averaging averages, this is okay to do if all information is contained within the same time period (which it is).
Please advise.
Doctor Douglas answered, starting with a basic assumption:
Hi Stacy,
It sounds as though you are trying to average a set of averages. As long as the data reflect the same measurement (or question) for each of the individual groups, then it is okay to proceed. That is, if question 1 is the same among all departments ("Please rate the chef at the last company BBQ on a scale of one to ten.") then it is a meaningful question to ask: what did the employees think of our last BBQ chef?
Of course, it would be meaningless to average question 1 if each department had a different question, or if the responses were on different scales. But how shall we calculate the average?
Now, when you take averages of averages, there is often a preferred way to do this operation. It is called "weighted averages" and reflects the fact that the number of observations may vary among the different groups. For example, let's say that the Marketing Department has 50 employees, and the Research Department has only 6. If the Marketing Department average was 8.5 ("BBQ was great!") and Research Department average was 3.1 ("heartburn!"), what is the correct average for these 2 departments? We might simply take the average of 8.5 and 3.1, i.e. 5.8. But it seems unfair to let the Research Dept sway the whole vote, having only 6 employees. The weighted average accounts for the differing number of employees:
weighted avg. = (no. of M)(AVG-of-M) + (no. of R)(AVG-of-R)
-------------------------------------------
(total no. of M and R together)
= (50)*(8.5) + (6)*(3.1)
----------------------
(50 + 6)
= 7.92
This is the classic example of a weighted average, and, as we saw last week, it amounts to finding the total score for the whole company by multiplying each average by the number of people it represents, and dividing by the total number of people. Thus this average is what we would get if we just averaged all the individual scores.
Do you see how the final average is much closer now to the Marketing Department's evaluation? Another way I sometimes explain it is that simple direct averaging is like the way states are represented in the U.S. Senate (every state gets two votes), while weighted averaging is like the way states are represented in the U.S. House of Representatives (every state is represented in proportion to its population).
As we’ll be saying several more times, each of these is a valid average, but has a different meaning. The U.S. Congress intentionally has one part in which every state, even the smallest, is treated equally, and another in which the largest state has more power, and each person is treated equally.
Now, it's impossible for me to say which type of averaging is correct for your situation, but I think it's probably better to use weighted averaging when you have information about the number of observations in each of the groups.
I would agree.
## Averaging percentages
A slightly different situation was involved in this 1998 question:
Averaging Percentages
Hi,
I am having difficulties in explaining to several friends that you cannot take percentages by totaling them up and then averaging the total of the percentages. It does not equal the percentage of the total of the numbers.
Is there a rule or theory that can explain this better?
Hi, Dominic. I'm not entirely sure what kind of problem you are referring to. Certainly there are at least some situations where you can average percentages. For example, if there are 50 questions on an exam, and three students got 20%, 30%, and 40% of them right, then the average number of questions they got right is 30%, or 15 questions.
This is like our situations above where we had equal groups, so a straight average made sense.
I suspect what you are thinking of is cases where the percentages are taken from different totals, in which case weighted averaging is needed. For example, if I survey 20% of 50 people, and 80% of 500 other people, then I have not surveyed (20+80)/2 = 50% of the total population, but:
.20 * 50 + .80 * 500 10 + 400 410
-------------------- = -------- = --- = 74.5%
50 + 500 550 550
The problem is simply that the percentages in such a problem do not represent fractions of the same total, so they can't be added.
As with an average of averages, here we are reconstructing the individual numbers (10 and 400 out of 50 and 500), then finding the actual percentages of the whole. That’s what a weighted average is.
I closed by referring to some of the answers we saw last week.
## Average price per square foot
For a specific real-life example, consider this 2003 question:
Average of Ratios vs. Ratio of Averages
I write and maintain software for real estate agents, and we include a calculation called, "Average dollars per square foot." We currently calculate this as the ratio of the average price divided by the average square footage of all the homes in the list. It seems to me that it should be calculated as the average of the price-per-square-foot ratio of each house. Can you think of any reason why the ratio of the averages would be more useful than the average of the ratios? Is there a technical name for this ratio of averages?
I answered again, pointing out as before that different calculations can both be meaningful, though they mean different things:
Hi, Laure.
Interesting question!
What you are currently calculating actually does make sense; you are just averaging over all the square feet of houses, rather than over all houses, and that may be just the right thing to do -- or it might not. Here's what I mean:
Suppose that N houses are sold; the sum of all their prices is P, and the sum of all their areas is A. (That is, if the individual prices are P1, P2, ..., Pn, and the individual areas are A1, A2, ..., An, then P is the sum of P1 through Pn, and A is the sum of A1 through An.)
Then the average price of a house is P/N, and the average area of a house is A/N; and you are calculating
P/N
--- = P/A
A/N
as the average price per square foot. And that is exactly what it is: the total price of all those square feet, divided by the number of square feet.
In terms of the individual numbers, this is $$\frac{P_1+P_2+\dots+P_n}{A_1+A_2+\dots+A_N}$$ Imagine laying out tiles representing each square foot of each house, and dollar bills representing the price of each house. This average spreads all the money equally over all the tiles, to find an overall price per square root. That sounds perfectly reasonable.
What you envision is
P1/A1 + P2/A2 + ... + Pn/An
---------------------------
N
which would average the price per square foot of all the houses. This puts the focus on the individual houses, rather than the individual square feet. How would this be different?
Again, this is $$\frac{\frac{P_1}{A_1}+\frac{P_2}{A_2}+\dots+\frac{P_N}{A_N}}{N}$$ Imagine laying out tiles representing each square foot of each house, keeping each house separate, and dollar bills representing the price of each house, spread equally among the tiles of that house. Now we take one tile and its cost from each house, and average those, spreading the money equally among the selected tiles. What have we found this time?
The way to see the difference is to take an extreme example. (Don’t check whether the sizes or prices are realistic; they aren’t meant to be!)
Well, let's take a simple case with N = 2. Suppose we have a big, well-built house of 10,000 square feet, and that it costs $2,000,000 ($200 per square foot), and a little house of 1,000 square feet that costs $20,000 ($20 per square foot). Then the total cost of the houses P is \$2,020,000, and the total area A is 11,000 square feet. The average price per square foot is
P/A = 2,020,000/11,000 = 183.6
(closer to the more expensive price) while the average of the two price-per-square-foot numbers is
average(Pn/An) = (200 + 20)/2 = 110
(which is considerably lower). What pulled the first number up is the fact that the bigger house had the bigger price per square foot; since we counted each square foot equally, the numerous high-cost ones won. The second calculation treats all 10,000 of the expensive square feet equally with the mere 1,000 square feet of the little house, so the little house pulled the average down.
Which do you think is the better calculation?
Both numbers are meaningful. The first fully deserves the name you are giving it (though there is definitely some ambiguity in the English!); but the second may better reflect what the average homeowner (as opposed to the "average square foot of floor space") can expect. Call it, perhaps, the "cost per square foot of the average house", where the number you are currently calculating is the "average cost of a square foot" or "cost of an average square foot".
Perhaps the builder would consider the first calculation more appropriate, since they put more effort into the bigger house, and every square foot they built matters to them; but the homeowner might care only about his own house — which neither calculation really reflects!
So, again, both numbers can reasonably be called "average cost per square foot"; which is more useful to you depends on how you want to use it. Do you want a number that is pulled up by big fancy houses, or one that shows what the average house is worth? Or would separate numbers for different categories of houses make more sense? Perhaps you can gather data that shows how costs per square foot are distributed, and how each average reflects that.
Any kind of average is an attempt to reduce a lot of data to a single number, and will never give a full picture of a diverse population. Other numbers, or a graph of the distribution, are often more useful.
Laure chose to leave the calculation as it was.
## Average speed
Let’s look at one more 2003 question, which is about a ratio in disguise:
Weighted Average of the Velocities
A truck on a straight road starts from rest and accelerates at 2.0 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?
Hi, Garin.
The trick here is to get the average velocity for each of the three phases of the trip, then do a weighted average of the three velocities.
Average velocity for phase 2 is easy, 20 m/s. During phases 1 and 3, the velocity changes linearly. That means that we can just average the starting and ending velocities to get the velocity for the entire phase.
This sort of average becomes most meaningful when you apply calculus, but for uniform acceleration, it requires only algebra. Doctor Edwin chose to leave that part for Garin to handle (or to ask for more help, which he never did), and focused just on the idea of the average. It turns out that the acceleration phase takes 10 seconds, at an average speed of 10 m/s; the cruise phase takes 20 seconds at 20 m/s; and the deceleration phase takes 5 seconds at an average speed of 10 m/s.
Once you've got those, you'll do a weighted average with respect to the time spent in each phase:
(v_1 * t_1) + (v_2 * t_2) + (v_3 * t_3)
v_avg = ---------------------------------------
t_total
The calculation looks like this: $$\frac{v_1\cdot t_1 + v_2\cdot t_2 + v_3\cdot t_3}{t_1+t_2+t_3} = \frac{10\cdot 10 + 20\cdot 20 + 10\cdot 5}{10+20+5} = \frac{100 + 400 + 50}{10+20+5} = \frac{550}{35} = 15.7\text{ m/s}$$
Just for fun, I'll point out that the same set of equations has another interpretation that works just as well. Average velocity is just distance over time, right? So if we figure out how far the truck went in each phase, add them all up, and divide by the total time, you'll also get the same answer:
d_1 + d_2 + d3
v_avg = ---------------------------------------
t_total
But the middle terms in those two equations are the same:
d_2 = v_2 * t_2
and so are the first terms:
a(t_1)^2
d_1 = --------
2
v
a = -
t
v_1_end * (t_1)^2
d_1 = -----------------
2(t_1)
v_1_end * t_1
d_1 = -------------
2
But since the starting velocity is zero, we can add it in wherever we want:
(v_1_start + v_1_end)
d_1 = ---------------------- * t_1
2
which means that our first terms are the same as well, and so are our third terms.
Anyway, there you have it. Total distance traveled divided by time, or a weighted average of velocities with respect to time, it works out to the same thing.
In other words, each “average velocity · time” is a distance, so our average velocity calculation was really just total distance over total time. We’ve seen previously that when distances are the same, we can use a harmonic mean to find the average speed; this confirms that when times are known but different, we can use a weighted arithmetic mean, using times as weights.
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https://math.stackexchange.com/questions/2942016/function-with-arbitrary-small-period/2942235 | # Function with arbitrary small period
Is there a function f: $$\mathbb{R} \to \mathbb{R}$$ with arbitrary small period different from $$f(x) = k$$? ($$\forall \epsilon >0 \exists a < \epsilon$$ such that f(x) has a periodicity $$a$$) I think the function is the Dirichlet function but I don't know how to prove it properly.
• F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$ Oct 4, 2018 at 12:00
• @Empy2 Nice: your example also works for domain $\mathbb{Q}$. Oct 4, 2018 at 12:38
• I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b \in ]0,a[$ with $f(x+b)=f(x)$. Oct 4, 2018 at 13:55
• Of course you want $0<a<\epsilon$
– MPW
Oct 4, 2018 at 18:07
• You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^{-k}$ then you're getting Empy2's set. If $a_k=k^{-\pi}$ then you're getting new set which is hard to imagine :-) If $a_k=\sin{k}$ then you're getting new interesting set. Oct 5, 2018 at 7:40
You're right. The characteristic function of the rationals is periodic of period $$1/n$$ for all $$n \in \mathbb N$$ because $$x$$ is rational iff $$x+1/n$$ is rational.
You're correct.
Let $$\epsilon$$ be arbitrarily small. You need to prove that there exists some $$0 such that $$D(x)=D(x+p)$$ for all $$x\in \mathbb{R}$$.
We know that $$\epsilon$$ is some positive real number, so there exists some $$p\in \mathbb{Q}$$ such that $$0. Let's look at an arbitrary $$x\in \mathbb{R}$$ and see if our property is satisfied or not:
If $$x$$ is rational, then $$D(x)=1$$. Since the sum of two rationals is rational, then $$D(x+p)=1$$ too.
If $$x$$ is irrational, then $$D(x)=0$$. Since the sum of a rational and an irrational is irrational, then $$D(x+p)=0$$ too.
So if we choose our period to be $$p$$, our property is satisfied.
The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $$f$$ with this property?
It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $$K=\{x\in\Bbb R\mid \forall y, f(y)=f(x+y)\}$$ be the set of periods of $$f$$. If $$f$$ is continuous, then this is an intersection of the sets $$\{x\in\Bbb R\mid f(y)=f(x+y)\}$$, which is closed (it is the preimage of $$\{0\}$$ under the function $$g(x)=f(y)-f(x+y)$$), so $$K$$ itself is closed. $$K$$ is also dense in $$\Bbb R$$, because it is an additive group with arbitrarily small elements, so $$K=\Bbb R$$ and hence $$f(x)=f(y)$$ for all $$x,y\in \Bbb R$$.
If we consider discontinuous functions again, then we know $$K$$ is a dense additive subgroup of $$\Bbb R$$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $$K$$. For a fixed $$K$$, the space of such functions is just all functions $$\Bbb R/K\to \Bbb R$$. This is another way to get at the constancy result, since as a topological group, $$\Bbb R/K$$ has the indiscrete topology, because any open set will cover $$\Bbb R$$ if copied around with translations by $$K$$.
Of course $$\Bbb R/K$$ can be uncountable, for example if $$K=\Bbb Q$$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $$B$$ of $$\Bbb R$$ over $$\Bbb Q$$, then the set of all real numbers with zero first projection is a subgroup $$K$$ of $$\Bbb R$$ for which $$\Bbb R/K\simeq \Bbb Q$$.
But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $$\Bbb R/K$$ has $$n>1$$ elements, then that means that every number which is a multiple of $$n$$ is in $$K$$; but every real number is a multiple of $$n$$, to wit, $$x=n(x/n)$$.
Another way to look at it is to think about the set of periods, i.e.
$$P = \{ p \in \mathbb{R} | f(x + p) = f(x) \text{ for all } x \in \mathbb{R} \}$$
Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $$\mathbb{R}$$ that has arbitrarily small values. $$\mathbb{Q}$$ is the obvious choice, so the characteristic function for $$\mathbb{Q}$$ works, as stated in another answer.
Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $$\mathbb{R}$$ over $$\mathbb{Q}$$ and pick a basis vector $$v$$. Let $$f$$ be the function which projects onto the $$v$$ coordinate. Then for any other basis vector $$u$$ and any integer $$n$$, $$f(x+u/n)=f(x).$$ More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.
• Hope you don't mind my edit, since it's not that much choice needed for what you want. =) Oct 4, 2018 at 17:10 | 2022-07-06T23:32:39 | {
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https://stats.stackexchange.com/questions/207460/expected-number-of-groups-of-3-consecutive-wins-in-200-rounds | # Expected number of groups of 3 consecutive wins in 200 rounds
If I know the probability of winning an individual round (while playing, say, a slot machine), call it $p_0$, what is the expected number of groups of 3 consecutive wins in 200 rounds? Or, to make it more general, what is the expected number of $k$-consecutive-win groups in $n$ rounds?
The groups cannot overlap, so 4 consecutive wins do not count as 2 different groups of 3 wins, while 6 consecutive wins do count as 2 different groups.
I was thinking of looking for the total number of possible $k$-group configurations in $n$ rounds, but I'm not too sure.
Explaining how you got this result would be most welcome, though the final formula would do as well. Thank you very much.
My solution will not only answer your question exactly (within linear algebra roundoff error) for the general case, but actually gives you the entire probability distribution of number of successful (i.e., completed) streaks.
This can readily be solved by using a discrete time discrete state (time-homogeneous) Markov Chain, in an approach similar in spirit to, but with larger state space, than the methods I used in https://math.stackexchange.com/questions/383704/probability-of-streaks/1739987#1739987 . After you read my answer in that thread plus this thread, you should be a wizard in using Markov Chains to solve all manner of streaks problems.
I will let
p = probability of winning per round
k = number of wins for successful streak
n = number of rounds
Then g is determined to be the upper bound on largest number of streaks, calculated as
g = floor(n/k)
Define the states as bivariate pairs(i,j), in which i denotes the current number of successful (i.e., completed) streaks and j is the length of the current streak. i will range from 0 to g, and j will range from 0 to k-1, except for when i = g, in which case j only goes from 0 to 0, and we make that state, (g,0), an absorbing state because we can't get any additional successful streaks. Order the states with j increasing fastest, then i increasing. I.e., with k = 3, the states would be (0,0),(0,1),(0,2),(1,0),(1,1),(1,2), etc.
Once the Markov Chain one step transition matrix, M, has been populated (see below), we compute the n step Markov Chain transition matrix, Mn as $M^n$. Given that we start in state (0,0) before the first of n rounds, the first row of Mn contains the probabilities of being in the various states after n rounds. For each possible value of number of successful streaks i, the sum of the probabilities over j for all states (i,j) provides the probability of exactly i successful streaks having occurred. It is then trivial to compute the expected number of successful streaks.
The one step transition matrix is populated as follows (note that w.p. is short for "with probability"):
For each i from 0 to g-1 (I'll show here for the case k = 3)
State (i,0) transitions to (i,0) w.p. 1-p and to (i,1) w.p. p
State (i,1) transitions to (i,0) w.p. 1-p and to (i,2) w.p. p
State (i,2) transitions to (i,0) w.p. 1-p and to (i+1,0) w.p. p
The absorbing state (g,0) transitions w.p. 1 to (g,0)
The population of the one step Markov Chain for a general value of k (and n) is shown in my MATLAB code below. Everything on a line after % is a comment.
k = 3; n = 200; p = .6; % set k, n, and p to particular values
g = floor(n/k);
B = [(1-p)*ones(k,1),p*eye(k)]; % recurring block in transition matrix M.
% B is a k by (k+1) matrix, consisting of a column vector of (1-p) 's,
% right-horizontally concatenated with p times the k by k identity matrix
% Start building up one step transition matrix, M
M = zeros(g*k+1); % (g*k+1) by (g*k+1) matrix of zeros
for i=0:g-1, M(k*i+1:k*(i+1),k*i+1:k*(i+1)+1) = B; end;
M(g*k+1,g*k+1) = 1; % Construction of M is now complete
Mn = M^n; % n step transition matrix
% Calculate array of probabilities of number of successful streaks
% and place in prob_array
prob_array = zeros(g+1,1);
for i=0:g-1, prob_array(i+1) = sum(Mn(1,k*i+1:k*i+k)); end;
prob_array(g+1) = Mn(1,g*k+1); % prob_array is now complete
expected_number_streaks = (0:g)*prob_array
Here are example results for n = 200 rounds, and streaks of length k = 3, with bonus results for streaks of length k = 5.
Here is an example one step transition matrix, for p = 0.6, n = 11, k = 3, which results in g = 3.
• Great. Yes, this agrees with observations... By the by, do you recommend any books to learn (relatively) advanced probability? I hadn't met the use of Markov Chains like this before. Apr 18 '16 at 7:02
• @Kristian D'Amato , many so-called advanced probability books deal with rigorous theory - it sounds like that may not be what you want. It sounds like a book on stochastic processes might be good for you, or perhaps applied probability with significant stochastic processes content, but I don't have a specific book of such type to recommend. Apr 18 '16 at 17:12
Since you're talking about a slot-machine, I assume that the probabilities are independent. The probability of $k$ consecutive wins is $p_0^k$. That probability is the expected portion of the groups with size $k$ that contain all wins. The number of groups with size $k$ (that does not overlap) is $200/k$, and we can call the number of all rounds $n$ (already alluded to in your question). So, the expected number of groups with size $k$ that contains consecutive wins would be
$p_0^k \cdot (n/k)$
$p_0^3 \cdot (200/3)$
Edit:
Apparently this problem is deceptively hard, like really hard. Just answering what the probability is that one group of length $k$ will appear is hard. The probability $S(n,k)$ of a group of consecutive wins of length $k$ in a chain of $n$ events, with the probability of a win $p$ and probability of a fail $1-p=q$ is
$S(n,k) = p^k \sum_{i=0}^{\infty} {n - (i+1)k \choose i}(-qp^k)^i - \sum_{i=1}^{\infty} {n - ik \choose i}(-qp^k)^i$
Here, the groups can even overlap, which is not what you asked for. I hope this can point you in the right direction.
(I apologize for the wrong answer earlier, I hope I did not lead you astray too much)
• Are you sure this is right? Observed information is not matching with this prediction (observations are about 50% higher) Apr 15 '16 at 10:47
• Are you sure that the groups may not overlap? Because, if they are allowed to overlap I can understand why the estimate is missing by 50% and in that case you should use the formula n + 1 - k to estimate the number of possible blocks. Apr 15 '16 at 11:41
• Yep. Keep in mind that if they are allowed to overlap the chance of getting a second win-group overlapping with the first is not independent of the first (since you only need one more round to have another win-group)... Anyway they don't overlap. Apr 15 '16 at 12:35
• I'm doubting myself, but I don't see how this could be wrong. p^k is the expected proportion of groups with all wins, n/k is the maximum amount of consecutive, non overlapping groups that can fit within the chain. Are you running simulations and could you supply some code in that case? Apr 15 '16 at 13:40
• I will when I'm able to later on. There are other k-chains in the entire sequence, since a group can start at position 2, for instance, covering 2-4, etc... Apr 15 '16 at 13:51 | 2022-01-21T17:40:05 | {
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http://math.stackexchange.com/questions/731928/taylor-theorem-inequality | # Taylor Theorem inequality
Prove that for all $f\in C^2([0,1])$ with $f(0)=f(1)=0$ and $|f''(x)| \le 1$ $$|f(x)| \le \frac{1}{2}x(1-x)$$ $\forall x \in [0,1]$.
-
If your function is only continuous, how do you know for sure that it has a second derivative? – Mercy Mar 30 '14 at 1:01
Mistake in the post, sorry! Just corrected… – Brontolo Mar 30 '14 at 1:05
We have, $\displaystyle (1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt = -f(x)$, for $x \in [0,1]$ (just apply integration by parts on LHS to ease the simplification).
Since, $t\ge 0$ for $t\in [0,1]$ and $(t-1) \le 0$ for $t \in [0,1]$, taking modulus on both sides,
$|f(x)| = \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt\right|$
$\le \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt\right| + \left|-x \int_x^1 (t-1)f''(t)\,dt\right|$
$\le \displaystyle \sup\limits_{t\in[0,1]}|f''(t)|. \bigg( (1-x)\int_0^x t\,dt -x \int_x^1 (t-1)\,dt \bigg)$
$= \dfrac{x(1-x)}{2}.\sup\limits_{t\in[0,1]}|f''(t)| \le \dfrac{x(1-x)}{2}$.
Aliter: Define $g(t)=f(t)-\dfrac{t(t-1)}{x(x-1)}f(x)$, on $[0,1]$.
Then, $g(0)=g(x)=g(1)=0$.
Applying Rolle's Theorem twice on $(0,1)$, $\exists \alpha \in (0,1)$ such that $g''(\alpha)=0$.
That is $g''(\alpha) = f''(\alpha) - \dfrac{2}{x(x-1)}f(x)=0$
or, $|f(x)|=\dfrac{x(1-x)}{2}|f''(\alpha)| \le \dfrac{x(1-x)}{2}$.
-
How do you pull $\sup|f''(t)|$ out of the integrals while maintaining the subtraction between the integrals? Suppose $f''(t)=1$ on $[0,x]$ and $f''(t)=-1$ on $[x,1]$ (or a continuous function very close to that). – robjohn Mar 30 '14 at 2:00
I think if you add the absolute values of the integrals you get what you want. That is, $$\frac12x^2(1-x)+\frac12x(1-x)^2=\frac12x(1-x)$$ – robjohn Mar 30 '14 at 2:03
@robjohn fixed typo and added a line .. thanks for pointing it out :) – r9m Mar 30 '14 at 2:08
Nice way the second one! I thought that it would be good to do with Taylor Theorem (therefore the title) but I found no way. – Brontolo Mar 30 '14 at 10:17
For some $\xi_k\in[0,1]$, $$f(1)=f(0)+f'(0)+\frac12f''(\xi_1) \quad\text{and}\quad f(0)=f(1)-f'(1)+\frac12f''(\xi_2)$$ implies $$|f'(0)|\le\frac12 \quad\text{and}\quad |f'(1)|\le\frac12$$ Let $g(x)=f(x)-\frac12x(1-x)$ and $h(x)=f(x)+\frac12x(1-x)$.
$g(0)=0$, $g'(0)\le0$, and $g''(x)\le0$; therefore, $g(x)\le0$.
$h(0)=0$, $h'(0)\ge0$, and $h''(x)\ge0$; therefore, $h(x)\ge0$.
Thus, $$\overbrace{-\frac12x(1-x)\le}^{h(x)\ge0}f(x)\overbrace{\le\frac12x(1-x)}^{g(x)\le0}$$
-
Can you explain why $g(0)=0, g′(0)≤0$, and $g″(x)≤0$; therefore, $g(x)≤0$? – Brontolo Mar 30 '14 at 10:09
@TheMaker94: Mean Value Theorem on $[0,1]$. Since $g'(0)\le0$ and $g''(x)\le0$, we have $g'(x)\le0$. Since $g(0)=0$ and $g'(x)\le0$, $g(x)\le0$. – robjohn Mar 30 '14 at 12:14
Perfect, clear thanks! – Brontolo Mar 30 '14 at 12:28
We want to prove that for each $x$:$$\exists c\ \ f(x) =\frac 12 f''(c) x(1-x)$$
We want to find such a $c$ via the Rolle theorem (or via the mean value theorem, but we can always go back to the Rolle version).
We can already apply the Rolle theorem, which gives an annulation for $f'$.
Let us modify $f$ to go to $0$ once more: assuming $0<x<1$,
$$g(u) := f(u) + A_xu(1-u)\\ g(x) = 0\Leftarrow A_x = - \frac{f(x)}{x(1-x)}$$
Now $g(1)=g(x) = g(0)$ hence, applying several times the Rolle theorem:
$$\exists c \ \ 0=g''(c) = f''(c) - \frac{f(x)}{x(1-x)}(-2)\\ f(x) = -\frac 12 x(1-x)f''(c)$$
NB: the error in the sign does not change the final inequality.
-
clever idea (+1) – robjohn Mar 30 '14 at 8:09 | 2015-01-26T12:25:32 | {
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https://dralb.com/2019/02/02/finding-area-using-integrals/ | # Finding Area using Integrals
As part of my Calculus 2 class, I have created a practice exam to help my students study. In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2. I hope you find the following solutions helpful in your studies.
You can also view me presenting the solution on YouTube at
## Find the area bounded by the function $$f(x)=8x$$, $$g(x)=\frac{8}{x}$$, and $$h(x)=x^{2}$$.
Drawing a sketch of these graphs, we get the following picture.
In order to find the area, we cut this into smaller cross sections, then add up the areas of the cross sections. Here we will cut out rectangles as shown below.
Note that the height of the rectangles has a different function value in the first portion than it does in the second portion, so we will split this into two parts. In the first section, we get that
\begin{align*} A_{cs}&=h*w \\ &=(y_{b}-y_{s})\Delta x, \end{align*} where $$y_{b}$$ is the bigger $$y$$ value and $$y_{s}$$ is the smaller $$y$$ value. In this case, we note from the graph that $$y_{b}=8x$$ and $$y_{s}=x^{2}$$. Therefore,
\begin{align*} A_{cs}=(8x-x^{2})\Delta x. \end{align*}
In order to find the total area over this interval, we then find that
\begin{align*} A_{T1}=\int_{a}^{b}(8x-x^{2})dx. \end{align*} Where is $$a$$ is the smallest $$x$$ value and $$b$$ is the largest $$x$$ value where these cross sections work.
We find $$a$$ as the intersection of $$8x$$ and $$x^{2}$$. That is, we solve
\begin{align*} 8x&=x^{2} \\ 0&=x^{2}-8x \\ 0&=x(x-8) \\ x&=0,8. \end{align*} From the picture, we note that the point we are interested for this example is actually $$a=0$$.
We then find $$b$$ as the intersection of $$8x$$ and $$\frac{8}{x}$$, so we get
\begin{align*}
8x&=\frac{8}{x} \\
8x^{2}&=8 \\
x^{2}&=1 \\
x&=\pm 1.
\end{align*}
Again, by looking at the picture, we note that the correct choice is $$b=1$$.
Therefore,
\begin{align*} A_{T1}&=\int_{0}^{1}(8x-x^{2})dx \\ &=4x^{2}-\frac{x^{3}}{3}|_{0}^{1} \\ &=4-\frac{1}{3}-0=\frac{11}{3}. \end{align*}
If we follow the same process for the second area, we note that we will have $$y_{b}=\frac{8}{x}$$ and $$y_{s}=x^{2}$$. Therefore, we will get that
\begin{align*} A_{cs}&=h*w \\ &=(y_{b}-y_{s})\Delta x \\ &=(\frac{8}{x}-x^{2})\Delta x. \end{align*}
Therefore, the total area of the second section will be
\begin{align*} A_{T2}=\int_{b}^{c}(\frac{8}{x}-x^{2})dx. \end{align*}
We have already found that $$b=1$$, so now we need to find $$c$$. Here, this is the intersection of $$\frac{8}{x}$$ and $$x^{2}$$. Therefore, we get that
\begin{align*} \frac{8}{x}&=x^{2} \\ 8&=x^{3} \\ x&=2. \end{align*} Hence, $$c=2$$.
We then get that
\begin{align*} A_{T2}&=\int_{1}^{2}(\frac{8}{x}-x^{2})dx \\ &=8\ln|x|-\frac{x^{3}}{3}|_{1}^{2} \\ &=8\ln(2)-\frac{8}{3}-(8\ln(1)-\frac{1}{3}) \\ &=8\ln(2)-\frac{7}{3}. \end{align*}
Now that we have found both areas, the total area of the region will be given as the sum of the two areas. Therefore, the total area is
\begin{align*} A_{T}&=A_{T1}+A_{T2} \\ &=\frac{11}{3}+8\ln(2)-\frac{7}{3} \\ &=\frac{4}{3}+8\ln(2). \end{align*}
## Other Resources
I have it linked above, but you can find the practice exam here. You can find the other solutions to the practice exam in the other posts available here.
If you found the post, or the YouTube video helpful, please like the post or the video. Also remember to follow the blog and subscribe to the YouTube channel. By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus. Thank you.
## 1 thought on “Finding Area using Integrals”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2020-09-24T20:43:55 | {
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http://www.ni.com/documentation/en/labview-comms/2.0/node-ref/kronecker-product/ | # Kronecker Product (G Dataflow)
Calculates the Kronecker product of two input matrices.
## matrix A
The first input matrix.
This input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.
Default: Empty array
## matrix B
The second input matrix.
This input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.
Default: Empty array
## error in
Error conditions that occur before this node runs. The node responds to this input according to standard error behavior.
Default: No error
## kronecker product
Matrix containing the Kronecker product of the first and second input matrices.
The number of rows in kronecker product is the product of the number of rows in the first and second input matrices. The number of columns in kronecker product is the product of the number of columns in the first and second input matrices.
## error out
Error information. The node produces this output according to standard error behavior.
## Algorithm for Calculating the Kronecker Product
If A is an n-by-m matrix and B is a k-by-l matrix, the Kronecker product of A and B, C = AB, results in a matrix C with dimensions nk-by-ml. This node calculates the Kronecker product using the following equation.
$C={\left[\begin{array}{cccc}{a}_{11}B& {a}_{12}B& \dots & {a}_{1m}B\\ {a}_{21}B& {a}_{22}B& \dots & {a}_{2m}B\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}B& {a}_{n2}B& \dots & {a}_{nm}B\end{array}\right]}_{nk×ml}$
For example, if
$\begin{array}{cc}A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]& B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]\end{array}$
then
$\begin{array}{ccc}{a}_{11}B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]& {a}_{12}B=\left[\begin{array}{cc}10& 12\\ 14& 16\end{array}\right]& C=\left[\begin{array}{cc}{a}_{11}B& {a}_{12}B\\ {a}_{21}B& {a}_{22}B\end{array}\right]=\left[\begin{array}{cccc}5& 6& 10& 12\\ 7& 8& 14& 16\\ 15& 18& 20& 24\\ 21& 24& 28& 32\end{array}\right]\end{array}$
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported | 2018-03-23T03:17:46 | {
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http://vkwf.venetoacquaeterme.it/show-that-the-moment-of-inertia-of-a-uniform-solid-sphere-rotating-about-a-diameter-is.html | moment of inertia of thin disc about a parallel axis, distance away So the moment of inertia for all such thin discs becomes when The cone is lying on its side with the vertex at the origin so which gives. Physics 111 Lecture 21 (Walker: 10. The spheres have negligible size, and the rod has negligible mass. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. Question: Show That The Moment Of Inertia Of A Uniform Solid Sphere Rotating About A Diameter Is 2/5 M R^2. We will be using mainly a cylindrical ring and a sphere in our experaments. (a) Show that the moment of inertia about a diameter of a uniform spherical shell of inner radius Rio outer radius R and density p is 1 = p(πm/15)(R5/2 – R5/1'). What is the angular momentum of the sphere?. (1) Moment of inertia of a Solid Sphere : (a) About an axis passing through its diameter : Consider a solid sphere of mass M and radius R. (ii) about a tangent: A tangent drawn to the sphere at any point, will obviously be parallel to one of its diameters and the distance between the axes is equal to R, the radius of the sphere. perpendicular to xy-plane passing through a point on the x-axis at a distance x. 1562 kg m 2 and torque applied is 0. 8 Solid sphere rotating about the central axis. (This also assumes we are rotating the bodies around the same axis. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR 2, find the moment of inertia about an axis normal to the disc passing through a point on its edge. calculate moment of inertia of any object, rotating about an arbitrary axis. ) of a sphere about its diameter = 2MR 2 /5 According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance. For simple solid objects, one can calculate the moment of inertia from the mass, size, and shape. Torque The turning effect of a force with respect to some axis, is called moment of force or torque due to the force. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR 2, find the moment of inertia about an axis normal to the disc passing through a point on its edge. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. The moment of inertia is a value that describes the distribution. ( 1 pt each ) Object A is a solid sphere. 0 kg and radius 0. By parallel axes theorem; This is an expression for M. Sphere #1 will arrive first B. The power transmitted by the shaft is. (This also assumes we are rotating the bodies around the same axis. 4 Radius of Gyration 3. A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3. • Rotating objects tend to keep rotating, while non-rotating objects tend to remain non-rotating. Moment of inertia of solid sphere when it is rotating about its diameter can be determined using integration process and about different axes can be. ( 1 pt ) A disc in the x-y plane is rotating about an axis perpendicular to the x-y plane. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. The moment of inertia for some common shapes are given below. where I is the moment of inertia. mass m and radius a, about a diameter of its plane face. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. For example, if two disks have the same mass but one has all the mass around the rim and the other is solid, then the disks would have different moments of inertia. , an axle through the center and perpendicular to the disk, the moment of inertia is calculated by carrying out the. A uniform solid sphere with mass M and radius R has a moment of inertia I = 2/5MR{eq}^2 {/eq} about an axis through its center. (a) Find the angular momentum of the skater. The moment of inertia may be defined as, I = sum m_ir_i^2 and if the system is continuous, then I = int r^2dm If rho is the mass density then, dm = rhodV where dV is an elementary volume. Thin walled cylinder rotating about the central axis. Experiments show that, if we compare bodies of similar shape and size but having different masses, the moment of inertia, I is directly proportional to the mass. Moment of Inertia (Rotational Inertia) I:- Moment of Inertiaof a body, about a given axis, is Moment of inertia of a solid disc:- (a) About an axis passing through its center and perpendicular Motion of a point mass attached to a string would over a cylinder capable of rotating about its axis of symmetry. Point P is midway between the center and the rim of the disk, and point Q is on the rim. Learn how to calculate moment of inertia of different shapes or objects using the several formulas. The moment of inertia of a solid sphere of radius of 0. Labels: Circular Disc and solid sphere, Mass Moment of Inertia, Mass Moment of I think This blog is very interesting, I hope you feel same like me about my link @ moment of inertia calculator online It's a good post about moment of inertia. Use Newton's second law to obtain two equations in a and T that we can solve simultaneously. A solid sphere with a mass of 8. So when the masses are placed at r= 0, I= I0. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. The translational velocity is slow enough to make easy accurate measurements. (****) Find the moment of inertia of a uniform, thin-walled sphere of radius R and mass M. A solid sphere, disc and solid cylinder all of same mass and made up of same material are allowed to roll down (from rest) on an inclined plane, then (a) solid sphere reaches the bottom late (b) solid sphere reaches the bottom first (c) disc will reach the bottom first (d) all will reach the bottom at the same time. 16 in order to remain upright under the in uence of gravity. Does it have a larger moment of inertia for an axis through the thicker end of the rod and perpendicular to the length of the rod, or for an axis through the thinner end of the rod. 00-m-diameter wagon. 115m and a mass of 12. A uniform solid S is generated by fully rotating R in the x axis. A particle of mass M is attached to one end of the stick. Moment of inertia states that:The product mass and the square of perpendicular distance from the axis of rotation is known as moment of inertia. not the hypotenuse. A constant tension of 23. Two solid uniform spheres roll down a ramp without slipping or sliding. 2 1 3 Where M is the mass and R is the radius of the sphere. 37 x 106 meters. 00 m long and has mass 4. 1 Rotational Inertia. Moment of Inertia: Sphere. This is determined by summing the moments of inertia of the thin discs that form the sphere. A hollow cylinder and a solid cylinder have the same diameter. The translational velocity is slow enough to make easy accurate measurements. (6) About what axis will a uniform, balsa-wood sphere have R the same moment of inertia as does a thin-walled, hollow, lead Rsphere of the same mass and radius, with the axis along a diameter? Use th e Parallel Axis Theorem. The density is then (1) and the moment of inertia tensor is (2) (3) (4). The moment of inertia of the sphere is I = 2 5 MR2. 84 m diameter solid sphere can be rotated about an axis through its center by a torque of 10. Sphere 2 has twice the radius of sphere 1. Thin walled cylinder rotating about the central axis. Circular Disk Rotating About Its Diameter The moment of inertia for the same circular disk rotating about an axis in the plane of the disk, passing through its center, is given by Thus, the uniform disk's moment of inertia in its own plane is twice that about its diameter. The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. Moment of inertia of a sphere can be explained in two parts (1) Solid Sphere (2)Hollow Sphere. (a) Show that the moment of inertia of a uniform hollow cylinder of inner radius R1, outer radius R2, and mass M, is I ½ M(R12 R22), if the rotation axis is through the center along the axis of symmetry. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. show more decimal digits. 7 cm in diameter. Labels: Circular Disc and solid sphere, Mass Moment of Inertia, Mass Moment of I think This blog is very interesting, I hope you feel same like me about my link @ moment of inertia calculator online It's a good post about moment of inertia. The axis of rotation in the question is a tangent to the ring. 01 18-Jun-2003 1. A solid disk with a mass of 0. The moment of inertia of an object changes if the axis of rotation is changed. calculate its moment of inertia about any axis through its centre. The following links are to calculators which will calculate the Section Area Moment of Inertia Properties of common shapes. In this section we show how the idea of integration as the limit of a sum can be used to find the moment of inertia of a lamina. The ratio of the larger Sphere moment of inertia to that of the smaller sphere is 4 Consider two uniform solid spheres were one has twice the mass and what is the diameter of the other. A solid disk will have a different moment than a washer, and there are formulas derived for calculating the moments of many common shapes. 6 mm when it bears a load. 6 F z z P. 132 kg m 2 about an axis which is found to be. 8 of Newman 1977 gives the added inertia for coefficient for spheroids of varying aspect ratio, referred to the moment of inertia of the displaced mass. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. As a consequence, the flow path in a rotating chute deviates considerably from that in a non-rotating chute. 3 Physical Significance of Moment of Inertia 3. Harm to minors, violence or threats, harassment or privacy invasion, impersonation or misrepresentation, fraud or phishing, show. Two uniform solid spheres have the same mass, but one has twice the radius of the other. Not the earth going around the sun, but the earth rotating on its axis, then you'd have to say that the moment of inertia for that amount of rotation is 2/5 mr squared, because it's a sphere rotating through an axis that goes through its center. Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. Sphere 2 has three times the radius of sphere 1. (b) Find the final rotation rate of the skater. Moments of inertia of rigid bodies∗ - Similar to Moments of inertia of rigid bodies∗ May 23, 2011 Moment of inertia of rigid body depends on the distribution of mass dimensional bodies like cylinder and sphere. [ In this question, you may assume standard results for the moment of inertia of uniform circular discs. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. A hoop a solid sphere a flat disk a hollow sphere Each of the objects has mass M and radius R. The moment of inertia of a uniform rod about an axis through its center is. A homogeneous solid cylinder of mass m, length L, and radius R rotates about an axis through point P, which is parallel to the cylinder axis. 84 m diameter solid sphere can be rotated about an axis through its center by a torque of 10. EXAMPLE: MOMENT OF INERTIA / EARTH EXAMPLE: 6The Earth has mass and radius 5. 16 The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Fig. Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. The moment of inertia of a circular ring about a diameter is ½ mr2, with usual notations. 115m and a mass of 12. For a different rotation point of an object—say a rod rotating around one end, like a turnstile, instead of around its center—we use the parallel axis theorem to find the object's moment of inertia. In this case, the moment of inertia of the mass in this system is a scalar known as the polar moment of inertia. • Subdivide body into small volume elements • Add the moment of inertia contributed by all these amounts of massAdd the moment of inertia contributed by all these amounts of mass • I = M ⋅(average value of R2) 2. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. For a thin uniform homogenous rectangular plate, the mass moment of inertia about the rectangular coordinate axes, a and b, passing through the centre of gravity of the circular plate can be obtained from the area moment of inertia. Moment of inertia of solid disk Period of small oscillation of sphere rolling in cylinder. • Subdivide body into small volume elements • Add the moment of inertia contributed by all these amounts of massAdd the moment of inertia contributed by all these amounts of mass • I = M ⋅(average value of R2) 2. (iii) Moment of inertia of a body should always be referred to as about a given axis, since it depends upon distribution of mass about that axis. Each sphere is a distance R+L/2 from the axis of rotation, so we must use the parallel axis theorem. Physics 100A Homework 10 – Chapter 10 (part 2) 10. 36 Determining Moments of Inertia. The moment of inertia of a uniform rod about an axis through its center is. The Brick Solid block adds to the attached frame a solid element with geometry, inertia, and color. Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is. • Parallel axis theorem for products of inertia: Product of inertia is useful in calculating MI @ inclined axes. The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given by $K=\frac{1}{2}I{\omega }^{2}$, where I is the moment of inertia, or “rotational mass” of the rigid body or system of particles. 00 m long and has mass 4. Solid sphere, radius r, about diameter. Answer is in kg⋅m2/s 2. The moment of inertia of a thin spherical shell of mean radius 0. Moment of inertia of solid shere about any diameter is (2/5)MR^2 , where M is mass and R is radius of sphere. The sphere is rotated about a diameter with an angular speed ω. 8(a) in side view. The radius of the sphere is 20. Show that the magnetic moment 'u' and the angular momentum 'L' of the sphere are related as: u=Lq/2m Pls help with this question. A man stands on a rotating platform that has an angular speed of 6. I have derived moment of inertia of solid sphere along diameter but my textbook says that moment of inertia is "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. Moments of Inertia and Angular Momentum. So it applies no torque, Since this is the only force acting on the system, the net torque is zero. It should not be confused with the second moment of area. The moment of inertia of a body rotating around an arbitrary axis is equal to the moment of inertia of a body rotating around a parallel axis through the center of mass plus the mass times the perpendicular distance between the axes h squared. 2: Angular momentum of a sphere Question: A uniform sphere of mass and radius spins about an axis passing through its centre with period. then the vertical gravitation would be x. 2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. We were discussing "Method to determine the area moment of inertia for a hollow rectangular section", "The theorem of parallel axis Let us consider one hollow circular section, where we can see that D is the diameter of main section and d is the diameter of cut-out section as displayed in following figure. Moment of Inertia of a solid body • Mass continuously distributed throughout its volume. 5 kg of a cold metal at a temperature of 258 K is immersed in 2. Find the moment of inertia of this combination about each of the following axes: (a) an axis pelvendicular to the bar through. Rotational Motion. M -mass, R -Radius. Vocabulary Angular Momentum: The measure of. If the surface of the ball is defined by the equation: 1301 + + =,. – What is the rotational kinetic energy of a 450-g solid sphere with a diameter of 23 cm rotating at rate of 17 rpm? – What is the total rotational kinetic energy of a 18-kg child riding on the edge of a merry-go-round of mass 160 kg and r = 2. A solid uniform L-shaped plate has mass $m$ and dimensions as shown. You have two steel spheres. [You may assume, without proof, that the moment of inertia of a uniform circular disc, of mass m and radius r, about a diameter is 1 4 mr2. diameter of the solid cylinder is large b. 6 F z z P. And needs to solve in both spherical & cylindrical coordinate system. The moment of inertia of a circular ring about a diameter is ½ mr2, with usual notations. MOTION OF SYSTEM OF PARTICLES AND RIGID BODY CONCEPTS. Moment of inertia - Parallel-Axis Theorem Pendulum Moment of Inertia of a Rotating Body Collision Moment of inertia tensor Moment of Inertia of a thin uniform rod by integration A 15--diameter CD has a mass of 24. The moment of inertia of a uniform rod about an axis through its center is. 0cm spins about the axle through its center. In the first part of our lab a rotating solid cylindrical drum with a hollow body drum given a rotational velocity from a falling mass. 0 cm and has mass 1. Sponsored Links. mass m and radius a, about a diameter of its plane face. You have two steel spheres. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. mass of the solid cylinder is large d. We will compare our results for a uniform, solid disk and a uniform ring with those derived from theory. Using the definition of moment of inertia, I = r 2 dm, one can show that theory predicts I disk = ½ MR2 (4) I ring = ½ M(R IN 2 + R OUT 2) (5). If they all are released from rest. Hollow sphere of radius r and mass m Similar to the solid sphere, only this time considering a stack of infinitesimal thin, circular hoops. KE = (I * w^2)/2. The mass of the hub can be ignored. (b) Obtain the moment of inertia for a solid cylinder. Solid Versus Hollow Edit A hoop (hollow object) has a greater moment of inertia than a greater moment of inertia than a solid disk of the same mass because all of the mass of the hoop is at a large radius. Although mass is defined in terms of inertia, it is conventionally interpreted as. Its moment of inertia about an axis tangent to it and perpendicular to its plane is ?. A solid sphere rolls (without slipping) down a plane inclined at 30˚ to the horizontal. Now if the two masses are each. 10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere. The moments of inertia of common shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily accessible in any mechanics textbook. A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2 = 5) MR 2. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. disk can be considered as a uniform solid disk of radius 25 cm and mass of 1. (2R/ O15) from the center of the sphere) (7) A frictionless pulley has the shape of a uniform solid di sk of mass 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Learn how to calculate moment of inertia of different shapes or objects using the several formulas. ] (10) (b) Hence find the moment of inertia of a uniform solid sphere, of mass M and radius a, about a diameter. The moment of inertia of a uniform rod about an axis through its center is. Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane. Calculate the moment of inertia of a thin plate in the shape of a right triangle, about an axis that passes through one end of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure. Assume the ball is a uniform, solid sphere. The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. Annapolis MD. The spheres have negligible size, and the rod has negligible mass. Determine the moment of inertia for a solid cylinder with mass m and radius R with a non-uniform mass density given by p= ar^2. m what is her final moment of inertia'? How does she physi- cally accomplish this change? (Il) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1-5 rev/s The wheel can be considered a uniform disk of mass 5. 94 m/s2 down the ramp (b) 3. For axis A, the rod is rotating about its centre of mass. Not the earth going around the sun, but the earth rotating on its axis, then you'd have to say that the moment of inertia for that amount of rotation is 2/5 mr squared, because it's a sphere rotating through an axis that goes through its center. A thin uniform bar has two small balls glued to its ends. a) Calculate the torque applied to the disk by the rope. 60-cm-diameter sprocket? Three objects of uniform density—a solid sphere, a solid cylinder, and a hollow cylinder—are placed at the top of an incline (Fig. 1 kg m2 as the skater draws his arms and legs inward toward the axis of rotation. Find the moment of inertia about a diameter?. 85kg and diameter 45. 16 The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Fig. The moment of inertia of the sphere is I = 2 5 MR2. For a ring let’s assume an element of mass dm on the ring. Answer: (a) Moment of inertia of sphere about any diameter = 2/5 MR 2. The density is then (1) and the moment of inertia tensor is (2) (3) (4). moment of inertia is the same about all of them. none of the above are necessary A common thread spool rests on a flat table. For a solid uniform sphere and a thin hoop, each of mass Mand radius R and rotating about their respective centers of mass, the moment of inertia of the hoop is larger than that of the sphere. I have been asked what the rotational inertia difference is between a hollow Aluminium driveshaft of diameter 44. To look for the angle of inclination which is b we do the following steps: 90 - b = a. (Hint: Form the shell by superposition of a sphere of density p and a smaller sphere of density -p. A small solid marble of mass m and radius r rolls without slipping along a loop-the-loop track shown in Figure 12. Step 2: The moment of inertia of the balsa wood sphere (solid) about the diameter is, 2 2 I2= M5 We have to choose an axis through which these two moment of inertias would be same. 500 kg and can be treated as point masses. 00-cm-diameter, 330 sphere is released from rest at the top of a 2. (1) Moment of inertia of a Solid Sphere : (a) About an axis passing through its diameter : Consider a solid sphere of mass M and radius R. Cotufa is doing homework on "moment of inertia" of uniform solid sphere and a uniform solid cylinder. 00-cm-diameter sprocket if the wheel is to attain an acceleration of 4. 0 kg and R — -MR2. A uniform solid S is generated by fully rotating R in the x axis. The moment of inertia about an axis at one end is. The inertia tensor of the disk alone about its center of mass is: MR2 4 1 0 0 0 1 0 0 0 2 (A) Find the inertia tensor of the disk alone about the point A. 150 m has a moment of inertia for rotation through its central axis. Find the value of the spin in revolutions per second for a= 10cm and b= 1cm. Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane. moment of inertia of thin disc about a parallel axis, distance away So the moment of inertia for all such thin discs becomes when The cone is lying on its side with the vertex at the origin so which gives. If we had a sphere, a solid sphere, then So here you have a solid sphere, and I rotate it about an axis through its center. If her initial moment of inertia was 4. It is a convention to write to indicate the moment of inertia with respect to an axis passing though the center of mass of the rotating rigid body. Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2 Products. The expression for the moment of inertia of a sphere can be developed by summing the moments of infintesmally thin disks about the z axis. Mass Moment of Inertia Angular Momentum Rotational Kinetic Energy rad rad N-m kg-m rad sec sec2 o kg-m2 kg-m2 sec Hz sec sec2 sec 14. 18) After fixing a flat tire on a bicycle you give the wheel a spin. The links will open a new browser window. A bug of mass m lands at the center of the disc and then walks outward. Can someone please show me show more The rotational inertia of a solid uniform sphere about a diameter is (2/5)MR2, where M is its mass and R is its radius. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. Let us consider a sphere of radius R and mass M. It is a convention to write to indicate the moment of inertia with respect to an axis passing though the center of mass of the rotating rigid body. Although mass is defined in terms of inertia, it is conventionally interpreted as. About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter Students also viewed these Mechanics questions. The results for centroid, moment of inertia, statical moment section modulus and torsion constant will display on your right. The bar is 2. The moment of inertia is the torque required to start an angle of acceleration along a rotating axis. 97 x 1024 kg and 6. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. Now for a sphere with uniform density rotating around its axis I is, I = (2MR^2)/5. If they are both released from the same height and at the same time, which one will arrive at the bottom of the ramp first? A. A solid sphere, disc and solid cylinder all of same mass and made up of same material are allowed to roll down (from rest) on an inclined plane, then (a) solid sphere reaches the bottom late (b) solid sphere reaches the bottom first (c) disc will reach the bottom first (d) all will reach the bottom at the same time. The work-energy theorem for a rigid body rotating around a fixed axis is where —10) and the rotational work done by a net force rotating a body from point A to point B is (10. The force F = 15 N is applied to the rope for a duration of 3 seconds. Physics 100A Homework 10 – Chapter 10 (part 2) 10. 6 kg object is found to have a moment of inertia of. 1) Find its angular acceleration. 0 is applied to the rope and the sphere starts to roll without slipping on the show more A uniform 8. 00 kg m2, but this is reduced to 2. For a thin uniform homogenous rectangular plate, the mass moment of inertia about the rectangular coordinate axes, a and b, passing through the centre of gravity of the circular plate can be obtained from the area moment of inertia. Moment of inertia of a Uniform Hollow Cylinder -. moment of inertia about its center of mass 𝐼𝐼. Here are some of the most common moments of inertia: Solid cylinder or disk of radius r rotating about its axis of symmetry. (b) Obtain the moment of inertia for a solid cylinder. A uniform disk of mass m is not as hard to set into rotational motion as a \dumbbell" with the same mass and radius. 15 • As shown in the figure, solid sphere rolls on a horizontal surface at 20 m/s and then rolls up the incline. If you are allowed to use the fact that the rotational inertia of a spherical shell is (2/3)mr2 [and I can derive that if necessary], all you have to do is take your solid. I recommend not to post that. Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. m what is her final moment of inertia'? How does she physi- cally accomplish this change? (Il) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1-5 rev/s The wheel can be considered a uniform disk of mass 5. 85kg and diameter 45. Explain why the moment of inertia is larger about the end than about the center. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. not the hypotenuse. 571 radians) between the strips, the angular velocity ω is computed. angular acceleration? An object remains in a state of uniform rotational motion unless acted on. As the solid sphere's non uniformity is not mentioned,it should be considered to be uniform and hence spherically symmetric body. ( 1 pt ) A disc in the x-y plane is rotating about an axis perpendicular to the x-y plane. In the first part of our lab a rotating solid cylindrical drum with a hollow body drum given a rotational velocity from a falling mass. Please explain. The moment of inertia of a solid cylinder of radius r is given by: J = mr2 2 By comparison, the moment of inertia of a hollow cylinder, of inner and outer radii respectively, is as follows: J = m(r o 2 - r i 2)2 It can be seen that, for a given outer radius, the moment of inertia of a hollow cylinder is greater than that of a solid cylinder of. I deal with stars, and stars have rotational kinetic energy. Icm = moment of inertia for rotation around an axis through the center of mass () M = total mass of the object (kg) d = distance between the two rotation axes (m) Parallel Axis Theorem Formula Questions: 1) A solid sphere with mass 60. The net torque acting on this sphere as it is slowing down is closest to: A. Using Moment of Inertia The moment of inertia of an object rotating around a fixed object is useful in calculating two key quantities in rotational motion:. Moment of inertia shows the tendency of an object to stay in its state of rotatory motion. If her initial moment of inertia was 4. 3 Physical Significance of Moment of Inertia 3. Moment of inertia of solid sphere about its diameter by. Moment of Inertia--Cylinder : Consider a uniform solid cylinder of mass M, radius R, height h. Let M and R be mass and radius of the hollow cylinder and the solid sphere, then, Moment of inertia of the hollow cylinder about its axis of symmetry, l 1, = MR 2 Moment of inertia of the solid sphere about from ω ω 0 + at, we find that for given ω 0 and t, ω 2 > ω 1, angular speed of solid sphere will be greater than the angular speed of. Related: Beam Deflection Stress Equation Calculators. We are allowed to use the standard result that the moment of inertia about the axis running down its centre is 1/2 m r^2. 15m from is center of mass. then prove that-omega =9/14 omega not. In a rotating body Torque is equal to the moment of Inertia multiplied by angular acceleration. 46 car on an incline A car on an incline is timed from release until the end of a measured distance. We want our questions to be useful to the broader community, and. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. Its rotational inertia about the point of attachment at the ceiling is: A) (2/5)MR2 B) 4MR2 C) (7/5)MR2 D) (22/5)MR2 E) (47. Its moment of inertia about an axis of rotation passing through its diameter is I = MR 2. Apply the parallel axis theorem From this result, we can conclude that it is twice as hard to rotate the barbell about the end than Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis. L as Y O C) 2017 Akaa Daniel Ayangeakaa. R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge. Sphere #2 will arrive first C. Some of the moments of inertia are given in the table below: slender rod: axis through center axis through end rectangular plane: axis through center axis along edge sphere thin-walled hollow solid cylinder hollow solid walled thin-hollow. What is the moment of inertia of the system of. More of the sphere's mass is far away from the center of rotation, so the hollow one has a big moment of inertia. Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2 Products. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. 98 x 10 24 kg) (6. A light string of length 3 R is 45 ± to the tangent ans: D Section: 10{8; Di±culty: E 174 Chapter 10: ROTATION 59. In systems that are both rotating and translating, conservation of mechanical energy can be used if there are no nonconservative forces at work. Get the linear acceleration of center of sphere denoting it as y. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. Answer: E 14) A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of a uniform circular disc and circular ring of radius R and mass M. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. A solid sphere with a mass of 8. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. The volume of such a layer is. (ii) about a tangent: A tangent drawn to the sphere at any point, will obviously be parallel to one of its diameters and the distance between the axes is equal to R, the radius of the sphere. I = ∑mr2 Rotational Kinetic Energy:- K r. What is the direction of its angular momentum vector? 15. 18) After fixing a flat tire on a bicycle you give the wheel a spin. The moment of inertia of this element is: dr. 16 The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Fig. The only data for 3D solids we are aware of are for spheroids: figure 4. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2, quarry, aggregate, and different kinds of. How is the moment of inertia related to. 98 x 1024 kg and an average radius of 6. | 2019-12-07T03:37:07 | {
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https://math.stackexchange.com/questions/1479267/how-many-six-digit-numbers-start-with-the-same-two-digits-and-end-with-the-same | # How many six digit numbers start with the same two digits and end with the same three digits?
Say that there is a 6 digit number the first digit is not allowed to be 0 or 1
so
How many number combinations start with the same two digits and end with the same three digits
ie.119333, 448222, 889888 etc..
My thoughts are 8_ 1_ 10_ 10_ 1_ 1_ = 8*10*10=800?
and how would I do it if instead of the and it said Or?
first of all, I'm I on the correct path here? any other examples similar would help.
• Your title should be specific to the problem at hand so that someone who has a similar question can find it if she or he searches the site. – N. F. Taussig Oct 14 '15 at 9:54
Now let us find the number of possibilities if our number starts with the same two digits OR ends with the same three digits. Again we assume that $0$ and $1$ are forbidden as first digit. But they may occur as second digit, for example in $506888$.
There are $(8)(10^4)$ numbers that begin with two equal digits, the first (and therefore second) digit being neither $0$ or $1$.
There are $(8)(10^3)$ numbers that start with an allowed digit and whose last three digits are the same.
If we add the two numbers above, we will have double-counted the numbers in which the first two digits are the same, and the last three are the same.
It follows that the required number is $(8)(10^4)+(8)(10^3)-(8)(10^2)$.
Remark: This is a relatively simple instance of a technique called Inclusion/Exclusion. There are more elaborate versions.
• so Just minus the AND part? – learnmore Oct 14 '15 at 5:48
• Yes. A general formula for two sets is $|A\cup B|=|A|+|B|-|A\cap B|$. Here $|X|$ means the number of elements in the set $X$. – André Nicolas Oct 14 '15 at 5:53
• how would one go about in determine the count if at least one digit appears more then once in the combination.. i.e 252169 – learnmore Oct 14 '15 at 6:01
• This is a different problem, since we no longer have two identical digits at the beginning, or three at the end. I will assume that $0$ is forbidden at the beginning. Then if we had no further restrictions, the answer would be $(9)(10^5)$. Now we subtract the bad numbers, in which all digits are distinct. Let us count the bad numbers. I will stop here and continue in the next comment. – André Nicolas Oct 14 '15 at 6:11
• (Continued) For counting the bad numbers, there are $9$ choices for the first digit. For each of these there are $9$ choices for the second digit, since $0$ is now allowed. For each choice of first two digits, there are $8$ choices for third digit, and so on, for a total of $(9)(9)(8)(7)(6)(5)$. – André Nicolas Oct 14 '15 at 6:15 | 2019-12-06T16:05:41 | {
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http://code.jasonbhill.com/c/project-euler-154/ | A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?
## Solution Using the Multinomial Theorem
The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula
$(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$
where
$\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$
Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by
$\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$
where $k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.
First, we’ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that
$p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$
where $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we’ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We’ll store two lists:
$p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$
For a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when
$p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$
That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.
Now, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\le k_2\le k_3$. Under this assumption, we always have
$k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$
We know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.
In case 1, we iterate $0\le k_1\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).
In case 2, we iterate $0\le k_1\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.
In case 3, we iterate $0\le k_1\le 66666$, and we iterate over $k_2=k_1+a$ where $1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).
## Cython Solution
I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C.
%cython import time from libc.stdlib cimport malloc, free head_time = time.time() cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers run_time = time.time() # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 free(p2) free(p5) elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) print "Result: %s" % result print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head)
When executed, we find the correct result relatively quickly.
Result: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds)
## C with OpenMP Solution
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } free(p2); free(p5); printf("result: %lld\n", result); return 0; }
This can be compiled and optimized using GCC as follows.
$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result. $ time ./problem-154-omp result: 479742450 real 0m1.487s
This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds. | 2019-01-17T22:12:55 | {
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"url": "http://code.jasonbhill.com/c/project-euler-154/",
"openwebmath_score": 0.4279261529445648,
"openwebmath_perplexity": 1737.7390952399246,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9875683469514965,
"lm_q2_score": 0.8615382058759129,
"lm_q1q2_score": 0.8508278618124334
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http://angs-diasporanew.sramble-communication.com/apdsmyht/linear-regression-problems-worksheet.php | # Linear regression problems worksheet
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Regression is used to assess the contribution of one or more “explanatory” variables (called independent variables) to one “response” (or dependent ) variable. The data in the table below show different X (depth in feet) 50 Y (maximum dive time) 80 depths with the maximum dive times in minutes. 4¯6, s2 Example of a regression equation Y = $0 +$ 1 (Age - 40) + $2 Gender + , Salary = 50 + 1 (Age - 40) - 3 Gender + , Salary in$1,000s, Age in years and Gender = 0 if male and 1 if female What is the average salary for 50 year old males? Ave(Y) = 50 + 1 (50-40) - 3(0) = $60K following form: y=alpha+beta*x+epsilon (we hypothesize a linear relationship) • The regression analysis „estimates“ the parameters alpha and beta by using the given observations for x and y. 13 Feb 2014 Linear Regression and Correlation - Example. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. In the final chapter, I'll show you how to make the most of your results by changing parameters by hand, performing sensitivity analysis and creating scenarios. Word problems on sets and venn diagrams. By comparing the values of, determine the function that best fits the data. Simple Linear Regression and Correlation 12. • The simplest form of estimating alpha and beta is called ordinary least squares (OLS) regression Problem 3: Let X and Y be two variables in a study. Find the residual amount for a person who is 42. 902. LAB ACTIVITIES FOR SIMPLE LINEAR REGRESSION: TWO VARIABLES 1. This one-page worksheet contains seven problems. A student who waits on tables at a restaurant recorded the cost of meals and the tip b. In case you need guidance on exponents or even radical expressions, Algebra1help. Worksheet 13. dta. Outcomes Students will use a graphing calculator to find a quadratic curve of best fit. 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Use the Input Y Range text box to identify the worksheet range holding your Write the linear regression equation for these data where miles driven is the independent Explain what the y-intercept means in the context of the problem. (1998) -Chpts 14 & 15 zHolmes et al. Transform the data so that there is a linear relationship between the transformed variables. Please be sure to check your answers as part of your homework assignment and be prepared with questions for next class. Worksheets are Work 3, Chapter 9 correlation and regression solutions, Writing linear equationslinear regression, Work 1, Linear regression work 1, Linear reg correlation coeff work, Work on correlation and regression, Algebra ii exponential regression work value 10. Make a scatter plot and determine the value of r. not significantly different from zero we conclude that: a) X is a good predictor of Y b) there is no linear relationship between X and Y. 01165 + 0. Mintzberg’s classic book The Nature of Managerial Work (1973) identified the roles found in all managerial jobs. 1 Linear Predictors Before computers became fast, linear regression was almost the only way of at-tacking certain prediction problems. 1751(6. 827 4. 3. Loading. . e. 2. c) BIOSTATS 540 - Fall 2018 Simple Linear Regression and Correlation Page 1 of 54 Nature Population/ Sample Observation/ Data Relationships/ Modeling Analysis/ Synthesis Unit 12 Simple Linear Regression and Correlation “ Assume that a statistical model such as a linear model is a good first start only” Lab!10:!Exploring!Linear!Regression! Objective:!In! this! lab,! you! will examine! relationships! between! two! quantitative! variables! using! a! graphical!tool May 31, 2016 · Regression analysis makes use of mathematical models to describe relationships. 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Scaffolded questions that start relatively easy and end with some real challenges. 391 + 17. regression project worksheet: Did the student show the material learned in this course can be useful in a topic that is relevant to the individual students? (explain in one sentence how your data is either interesting to you personally OR related to you major of study) The following are tables of data to be used for linear regression exercises. Find the linear and quadratic regression equations and correlation coefficients. This data set has n=31 observations of boiling points (Y=boiling) and temperature (X=temp). Linear Regression Displaying all worksheets related to - Linear Regression. 1. There are 2 types of factors in regression analysis: Dependent variable (y) : It’s also called the ‘criterion variable’ , ‘response’ , or ‘outcome’ and is the factor being solved. g. A correlation analysis provides information on the strength and direction of the linear relationship between two variables, while a simple linear regression analysis estimates parameters in a linear equation that can be used to predict values of one variable based on Quadratic Models and Quadratic Regression Worksheet 1. P. Consider the following scatter plots: (a) Write the new regression model. Writing Linear Equations/Linear Regression Write the slope-intercept form of the equation of each line given the slope and y-intercept. Graph the linear equation by 2. Regression problems are supervised learning problems in which the response is continuous. A researcher has collected data on the price of gasoline from 1990 to 2010 and has found that the price in dollars after t years can be predicted using the equation: y xx− += +0. Problem 2. The user can STAT 2215 Worksheet 5 – Chapter 7: Simple Linear Regression Problem 23, page 198. If there is not a linear relationship between x and y, then $$\mu_{i} ≠ \beta_{0} + \beta_{1}X_{i}$$. ! 2 4 6 8 0 50 100 150 200 250 Calories vs Alcohol Content Alcohol Content (%) Worksheets that accompany this lesson can be located under related documents, worksheets, Data Analysis #1-#8. Figure #10. You may notice that Linear regression where the sum of vertical distances d1 + d2 + d3 + d4 between observed and predicted (line and its equation) values is minimized. introduce problems that are relevant to the fitting of nonlinear regression func- Title: Linear Regression Grade: 8th Lesson Summary: This short lesson plan is to describe students learn how to find the best line to fit the data of two variables they collect and be able to predicate the data by using the regression equation. Simple linear regression is a statistical method that allows us to summarize and study relationships between two continuous (quantitative) variables. 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Motivation and Objective: We’ve spent a lot of time discussing simple linear regression, but simple linear regression is, well, “simple” in the sense that there is usually more than one variable that helps “explain” the variation in the response variable. write a regression equation and interpret the meaning of the slope and y-intercepts in the context of the problem; make predictions based on the correct mathematical models; and; solve linear equations. Even those who problem. However, we only calculate a regression line if one of the vari-ables helps to explain or predict the other variable. Superimpose the regression curve on the scatter plot. see and learn about curve fitting for multiple linear regression using method of least errors is as small as possible. At home: Read Chapter 5 and work the problems at the end of each short section as you go through them. Estimate the Blood pressure for a person who is 50. Chapter 12. Aug 07, 2014 · Search this site. 7 p1 + 109 p2 Remember: -97. If Y denotes the Excel Linear Regression. (10 marks) C). In linear regression analysis, the dependent variable is thought to be related to the independent variable or variables in a linear way. We would expect the ratio MSLF/MSPE to be close to 1. Time and work word problems. Then, we graph the linear regression equation with the scatterplot data. They compute the percent of the variability. For our problem, we would need only one dummy variable (since K = 2), the dichotomous variable coding level of idealism. Go to Overview Index. In simple linear regression, when β is . Do the Linear Regression ws. a and b are the constants of the regression model. It is a staple of statistics and is often considered a good introductory machine learning method. The elements in X are non-stochastic, meaning that the Linear Regression and NORMAL Curve Advanced Placement AAP Review will be held in room 315 and 312 on Tuesdays and Thursdays. Feb 26, 2018 · Linear regression is used for finding linear relationship between target and one or more predictors. Rewrite the equation into finding the x- and y-intercepts. Multiple Linear Regression. If you're seeing this message, it means we're having trouble loading external resources on our website. Plot four points so that the regression line is horizontal. Customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. Students will interpret the r-value of the data and write a summary of its meaning. Big Ideas: Bivariate quantitative variables can be represented by a table, graph, and a prediction equation, and estimates can be made from each. GAISE Components One step equation word problems. ) Another very serious problem is the lack of any provision for forecasting from additional values of the independent variables . Throughout the module, you will find many real-world appli cations of these two important topics: least-squares regression line and the correlation coefficient. Assumptions in the Linear Regression Model 2. Some of the worksheets for this concept are Linear regression work 1, Writing linear equationslinear regression, Chapter 9 correlation and regression solutions, Work 1, Work 3, , Quadratic regression, Kuta software. Such an equation can be used for prediction: given a new x-value, this equation can predict the y-value that is consistent with the information known about the data. The sample must be representative of the population 2. Jun 01, 2020 · Statistics Q&A Library WORKSHEET 24 Linear Regression and Correlation Name: 15. 8. This lesson builds on students work in the 8th grade. problem to be solved is reduced to a quadratic programming problem in which the objective function is the residual sum of the squares in regression, and the constraints are linear ones imlx~ed on the regression coefficients. 3 times as important as Unconventional. b) Use the calculator and the equation of the linear regression line to complete the following table: Year 1960 1968 1988 1999 2005 2008 Time (s) c) Use the values in the table to draw the linear regression function on the scatter plot. To get a better feel for the regression line, try the following tasks. In Class: Practice Linear Regression HW: Obesity Problem (Hints included in key) r_9. EXTRAS. 45. Displaying all worksheets related to - Regression Analysis. Math 137 Quadratic Regression Classwork2 . Explain worksheet - regression inference 1. It allows the mean function E()y to depend on more than one explanatory variables Like correlation coefficients, linear regression analyzes the relationship between two variables, x and y. 1) Slope = −1, y-intercept = 0 y = −x 2) Slope = 1 4, y-intercept = 1 y = 1 4 x + 1 Write the slope-intercept form of the equation of the line through the given point with the given slope. Start by creating a visual model of the data. Bivariate Data - Correlation - Linear Regression - Correlation Coefficient r Jun 2, 2020 - Are you looking for Algebra 2 worksheets WITH answers? Making keys is super time consuming, so peruse this board for resources that have the hard part done for you!. 5 and 8 This is because the correlation value for the cubic regression is about 0. The multiple linear regression result implies that Reliable is around 1. The table below lists the total estimated numbers of United States AIDS cases, by year of diagnosis. 3582 1. The first is done using the Tools menu, and results in a tabular output that contains Shodor > Interactivate > Lessons > Linear Regression and Correlation omit the scatter plot worksheet; As a class, before splitting them into groups, have the If you have a continuous dependent variable, linear regression is probably the To address these problems, statisticians have developed several advanced variants: ordered data (assuming you record them in your worksheet in time order). A graphing calculator can also be used to perform quadratic regression. In this correlation and linear regression worksheet, students examine data to determine the statistic mean. 2 Fitting the Regression Line 12. Brandon Foltz 296,888 views Linear regression simply refers to creating a best fit for a linear relationship between two variables from observed data. Ratio and proportion word problems. Although a linear regression can be quite helpful in understanding data, it can sometimes be misleading, as Anscombe's Quartet shows . H. If the degree of correlation between variables is high enough, it can cause problems when you fit the model and interpret the results. (When we need to note the difference, a regression on a single predic-tor is called a simple regression. simple linear regression A college bookstore must order books two months before each semester starts. widely used; runs fast; easy to use (not a lot of tuning The linear regression model that I’ve been discussing relies on several assumptions. A. In other words, the SS is built up as each variable is added, in the order they are given in the command. 12A2 - HW Calendar Linear Regression Problems Q. LINEAR REGRESSION 6 1. The course website page REGRESSION AND CORRELATION has some examples of code to produce regression analyses in STATA. use functions fitted to data to solve problems in the context of the data. Model Summaries Worksheet Every workbook produced by RegressIt contains not only the data analysis and regression worksheets but also a model summaries worksheet that keeps an audit trail of all regression models fitted so far and allows side-by-side comparison of models fitted to the same dependent variable, suitable for framing. the meaning of slope in a linear regression equation the relationship of the slope to analyzed data whether or not the y-intercept of the linear regression equation is a relevant value EU 2 Econometrics. Enter the data to answer this question. com. To see why, consider a model such as this Y = β 0 +β 1eβ 2X + , (4. Curve Fitting Curve fitting is the process of constructing a curve, or mathematical function, that has the best fit to a series of data points, possibly subject to constraints. 48. It also plots the experimental points and the equation y = a x + b where a and b are given by the formulas above. 9) 𝑦𝑦 = 48. where a and b are given by. Round r to the nearest 3 decimal places. The case of having one independent variable is know as simple linear regression while the case of having multiple linear regression is known as multiple linear regression. Analysis of Variance, Goodness of Fit and the F test 5. 242 seconds for each additional foot of initial drop. Twenty five plants are selected, 5 each assigned to each of the fertilizer levels (12, 15, 18, 21, 24). Files for use with the TI-Nspire™ A non-linear method with comparable simplicity is known as K-nearest neighbors regression. Sketch and shade the squares of the residuals. In many applications, there is more than one factor that influences the response. Our predictors would then be idealism, misanthropy, and idealism x misanthropy (an interaction term). This is a quadratic model because the second differences are the differences that have the same value (4). THE MODEL BEHIND LINEAR REGRESSION 217 0 2 4 6 8 10 0 5 10 15 x Y Figure 9. Use the standard error of the slope, Sb, to calculate 95% confidence interval for the slope in ºC/century. Print off the worksheet if you How to graph the linear regression equation with the scatterplot data, how to generate a least squares linear regression model, How to create a line of best fit, 31 Jul 2016 12. \] Linear Regression Practice Worksheet 1. An agriculturalist was interested in the effects of a) Use your calculator to determine the linear regression function (y = A + Bx) that best models the data. There are two types of linear regression- Simple and Multiple. LINEAR REGRESSION WORKSHEET #1 Name_____ Date_____ Period_____ 1. The variable we base our predictions on is called the independent or predictor variable and is referred to as X. Use a graphing calculator to fit linear, quadratic, cubic, and power functions to At any time, you can complete this interactive online quiz to check how well you understand using linear regression. Click on pop-out icon or print icon to worksheet to print or download. b. The table lists the heights and weights of six wide receivers who played for the Atlanta Falcons during the 2010 football season. Instruction will be from 3:15 pm to 3:30 pm Compute the least squares regression line with the number of bidders present at the auction as the independent variable (x) and sales price as the dependent variable (y). Then find the Least Squares Line and use it to make a prediction. 2 problems on linear regression May 19, 2018 · 12 videos Play all Statistics PL14 - Simple Linear Regression Brandon Foltz Statistics 101: Logistic Regression Probability, Odds, and Odds Ratio - Duration: 13:03. The session will begin in room 315 with a brief review of the weekly topic. This worksheet contains the following data, with the list price in column C1 and the best price in the column C2. If we were to plot height (the independent or 'predictor' variable) as a function of body weight (the dependent or 'outcome' variable), we might see a very linear relationship, as illustrated Related posts of "Linear Regression Worksheet Answers" Scheme For Igneous Rock Identification Worksheet Answers In advance of preaching about Scheme For Igneous Rock Identification Worksheet Answers, you should realize that Instruction can be each of our factor to a greater down the road, and also understanding doesn't just stop right after the (RegressIt can fit linear regression models in Excel with over 200 independent variables on a PC or 125 variables on a Mac, and its R interface can be used to fit large models much faster. Estimator 3. 344 10. 35519 0. Height (inches) Weight (pounds) 75 192 76 220 71 200 74 210 69 185 72 189 a. A linear regression simply means that the equation will be the equation of a line . Ea How to compute the linear regression equation, y=ax+b, the linear correlation coefficient, r, and the coefficient of determination, r 2, using the TI-84 calculator, including turning the diagnostics on. It is a very simple regression algorithm, fast to train and can have great performance if the output variable for your data is a linear combination of your inputs. MR. c. This correlation is a problem because independent variables should be independent. Worksheet: Name the Song you can also use SVMs for regression. pdf: File Size: 1849 kb: File Type: pdf: Download File linear regression: An approach to modeling the linear relationship between a dependent variable, $y$ and an independent variable, $x$. jpg. Linear regression analysis, in general, is a statistical method that shows or predicts the relationship between two variables or factors. 1 Transformations in Linear Regression Create printable worksheets for solving linear equations (pre-algebra or algebra 1), as PDF or html files. Regression. Explain the slope in context of the problem. May 15, 2009 · Statistics and Regression Tools Review: Detailed Descriptions - Click OK to execute the regression. Predictthe!type!(positive,!negative,!no)!and!strength!of!correlation!(strong,!weak)!for!the!following! Summarize the four conditions that comprise the simple linear regression model. You don't have to believe everything it says. The week of March 30th we will be reviewing Linear Regression and NORMAL Curve. Mar 20, 2019 · Linear regression forecasting graph. through a process called linear regression. As with linear regression, the dataset must take of form of pairs of predictor variables x ⃗ i \vec{x}_i x i with resultant variables y i y_i y i . For instance, for an 8 year old we can use the equation to estimate that the average FEV = 0. Make a scatter plot for the data. If the answers to (l) and (m) are yes then using your simple linear regression equation predict the percentage of games won from a team that has a passing percentage of 6. Know what the unknown population variance $$\sigma^{2}$$ quantifies in the regression setting. com is simply the right site to go to! A simple linear regression model is a mathematical equation that allows us to predict a response for a given predictor value. ) We’d never try to find a regression by hand, and Linear regression is a process of drawing a line through data in a scatter plot. For each of the following, perform linear, quadratic, and exponential regressions. We are dealing with a more complicated example in this case though. Pythagorean theorem word problems. Further along in the paper, we will investigate some of the alternatives mentioned above, but this is also an opportunity Linear Regression Worksheet 1. b) According to the linear model, a coaster with a 200 foot initial drop is expected to last 139. 9 we’ll talk a lot more about how to check that these assumptions are being met, but first, let’s have a look at each of them. In a linear regression model, the variable of interest (the so-called “dependent” variable) is predicted from k other variables (the so-called “independent” variables) using a linear equation. In Section 15. A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in the following table. Answers provided. , between an independent and a dependent variable or between two independent variables). DOWNLOADS Note: This particular activity requires the use of TI-Nspire™ technology to be used successfully. In what year does Alpha ity’s population reach twice its initial (1/1/2004) population? 2. Do this in several different ways. It is the same Lagrange multiplier problem as above, with all the inequalities reversed. Contact Info. Worksheet 3 - Regression and linear models Linear regression references zFowler et al. sav. The results are shown in the table. Some of the worksheets for this concept are Chapter 9 correlation and regression solutions, I exploring regression, Scatter plots, Correlation coefficient, The united states of obesity, Concept 20 scatterplots correlation, Lecture 12 linear regression test Linear Equation Models. com delivers insightful information on linear regression free worksheet, basic concepts of mathematics and dividing fractions and other algebra topics. Regression (new) 4. 8: Regression - Distance from School (Worksheet) - Statistics LibreTexts Scatterplots & Regression on the TI-84 This video shows how to input two variable data and create a scatterplot with the TI-84 calculator. L Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Systems of Equations Word Problems Date_____ Period____ 1) Find the value of two numbers if their sum is 12 and their difference is 4. Now we’ll take a look at a data set for which the linear regression model is appropriate. A. This lesson SIMPLE LINEAR REGRESSION – DEMAND AS FUNCTION OF PRICE A new worksheet will appear revealing the results of your regression analysis. 4 Inferences on the Regression Line 12. Practice quiz 6. 05 to help answer this question. There are NINE problem types. Properties of the O. It works by estimating coefficients for a line or hyperplane that best fits the training data. This model generalizes the simple linear regression in two ways. A simple linear regression model is fit, relating plant growth over 1 year (y) to amount of fertilizer provided (x). If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x. 11719% In simple linear regression, we predict scores on one variable from the scores on a second variable. The line summarizes the data, which is useful when making predictions. 02 0. Normality. Using this analysis we can estimate the relationship between two or more variables. Complete details on this method (also known as the Potthoff method) are in Chapter 13 of K & K. Evaluating managerial success. Coming up for air. 6 The Analysis of Variance Table 12. If the truth is non-linearity, regression will make inappropriate predictions, but at least regression will have a chance to detect the non-linearity. Graph using the slope and the y-intercept. Neither regression nor correlation analyses can be Use the linear model to calculate the expected cholesterol for a person with pressure 160 mmHg. The results of the model fit are given below: Can we Problems 1. 7 Residual Analysis The first difference (the difference between any two successive output values) is the same value (3). Regression 10. Identify your Y and X values. Let's look at an example of linear regression by examining the data in the following table to discover the relationship between temperatures measured in Celsius (Centigrade) and Fahrenheit. uk March 17, 2011 1 Logarithmic transformations of variables Considering the simple bivariate linear model Yi = + Xi + i,1 there are four possible com- Algebra1help. Is the number $$\beta _1$$ in the equation $$y=\beta _1x+\beta _0$$ a statistic or a population parameter? The linear regression model explains the method used to take observed data and find a 'best fit' line to describe the relationship of two variables, and this quiz/worksheet pairing will test your Linear Regression Correlation Coeeficient. A Tomahawk Cruise ship in the South Pacific misfires a missile. Statistics – Linear Regression Worksheet - Solutions The busiest season for Walmart is the Christmas holiday and weekends see a tremendous number of customers. Multiple regression models thus describe how a single response variable Y depends linearly on a A. (SHOW WORK using algebra!) Linear mixture problems, non-linear mixture problems, transportation problems, personal scheduling problems and for something a little different, sports scheduling problems. 26721 × age. 4: The Regression Equation Carry out an appropriate test at a significance level of 0. You can use the Regression tool instead of the LINEST worksheet function. Word Problem Worksheet #2 (each correct answer = 1 extra ballot for the draw!) Linear Regression Worksheet - HAND IN :) April 9 - Data Unit Assignment. Recall that the least squares line minimizes the squares of the residuals. So, we have a sample of 84 students, who have studied in college. - Stats Worksheet #1 - Stats Worksheet #2. ¯3, y¯ = 46. 15 Apr 2015 Research question type: When using one variable to predict or explain another Simple Linear Regression – Additional Information worksheet. The least squares method is generally used with a linear regression, but 1 day ago · You will not require the options for "residuals" for this analysis. Using a catapult of their own design, students will model a parabolic relationship, collect data, and draw conclusions from data and the quadratic curve of best fit. Finally, the students will explore question 4 on the Activity Worksheet. simple linear regression, the sample correlation coefficient is the square root of the coefficient of determination, with the sign of the correlation coefficient being the same as the sign of b1, the coefficient of x1 in the estimated regression equation. The student will evaluate the relationship between two variables to determine if that … 12. 23 Apr 2011 In the table below, list the assumptions of multiple linear regression all 6 predictors into the one model, because of the collinearity problem. Consider the following hypothetical data set. analysis 5. The Multiple Regression Process Conceptually, multiple regression is a straight forward extension of the simple linear regression procedures. 000 Problems 0. A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in the table. Classification problems are supervised learning problems in which the response is categorical; Benefits of linear regression. Regression Analysis. 1: Mnemonic for the simple regression model. Create a scatter plot and approximate a trend Using your trend line, predict the 0-60 time for a car that costs$120 K? 2. These worksheets are especially meant for pre-algebra and algebra 1 courses (grades 7-9). For this question, we could swap the independent and dependent variables and still get reasonable results. The linear regression equation, also known as least squares equation has the following form: $$\hat Y = a + b X$$, where the regression coefficients $$a$$ and $$b$$ are computed by this regression That is, there is no lack of fit in the simple linear regression model. (a) Make a scatter plot of the data. 778 (or a value 0. 3t 2 6t 80 . Now, select Sheet 1 (by clicking on its tab at the bottom of the worksheet) and paste the information into cell F4. Worksheets are Linear regression work 1, Writing linear equationslinear regression, Work 1, Work 3, Linear reg correlation coeff work, Chapter 9 correlation and regression solutions,, Work regression. 1, Demand estimation using linear regression. In a simple linear regression model, we model the relationship between both variables by a straight line, formally $Y = b \cdot X + a. Fit linear, quadratic, cubic, exponential, quartic, and power functions to the data. 000 These are called the regression parameters in the simple linear regression equation (the equation is also known as the least squares regression equation or the trend equation or simply the regression). 2 Practice Worksheet 1. The regression line and the residuals are displayed in figure #10. Answers are included here to check your work. One of the favorite topics on which the interviewers ask questions is ‘Linear Regression. AP Stats: Section 3. The linear regression analysis can then 7 Aug 2017 The students can immediately understand that linear regression is on the worksheet eliminates many class time and design problems and 5 Mar 2020 Open a new workbook in Excel and make 3 worksheets: Data, Chart, and Saves. In Microsoft Office Excel 2007, you can find the Regression tool by clicking Data Analysis in the Analysis group on the Data tab. Find the equation of the regression line. In this example R2 = 0. For example, suppose that height was the only determinant of body weight. Linear Equation in Two Variables Distance Problems This worksheet is based on linear equation in two variables distance problems. Transit demand. Linear Regression Worksheet Answers with Valuable Issues. 5 Linear Regression Algebra II Name Guided Notes Date_____Block_____ Recap: Graphing Linear Functions 1. 7 is the affe ct on sales of a change in p1 with p2 held fixed !! 0 5 10 15 9 8 4 Sales I Simple Linear Regression Common Mistakes Statistics Tables Quiz: Cumulative Review A Quiz: Cumulative Review B Online Quizzes for CliffsNotes Statistics QuickReview A linear regression model corresponds to a linear regression model that minimizes the sum of squared errors for a set of pairs $$(X_i, Y_i)$$. Linear Algebra in Linear Regression Continue Suppose we have a column space in R 3 \mathbf{R}^3 R 3 , W W W , a vector b ⃗ \vec{b} b , and A x ⃗ A\vec{x} A x , the point closest to b ⃗ \vec{b} b on W . Worksheet – Regression The table below displays data on the temperature ( F) reached on a given day and the number of cans of soft drink sold from a particular vending machine in front of a grocery store. The least square regression line for the set of n data points is given by the equation of a line in slope intercept form: y = a x + b. Quiz (new) 7.$ For now, let us suppose that the function which relates test score and student-teacher ratio to each other is \[TestScore = 713 - 3 \times STR. The last page of this exam gives output for the following situation. Section 1: Input Data Below are the input parameters to begin the simulation. 2 in-depth answers. Linear Regression Correlation Coeeficient - Displaying top 8 worksheets found for this concept. 60 55 70 45 80 35 a. Save the workbook as Linear Regression - Brief Lesson, or something similar, into a logical file folder. Does the relationship appear to be linear? Why? Yes, the relationship appears to be linear because it seems to be decreasing in similar intervals after Unit 3 - Linear Functions & Linear Regression 3-1, Functions and Function Notation - Video , Notes , Worksheet 3-3, Linear Functions - Video , Notes , Worksheet Multiple Linear Regression So far, we have seen the concept of simple linear regression where a single predictor variable X was used to model the response variable Y. Percent of a number word problems. 4 Solving Real-Life Problems How can you use a linear equation in two variables to model and solve a real-life problem? Write a story that uses the graph at the right. and solve practical problems using models of linear, quadratic, and exponential functions. An observational study if 19 managers from a medium-sized manufacturing plant extended Mintzberg’s work by investigating which activities successful man- Linear regression is a method for modeling the relationship between one or more independent variables and a dependent variable. slope-intercept form. C. Linear regression is a technique that is useful for regression problems. A boat goes 30km upstream and 44km downstream in 10 hours. Quiz (new) 11. a) According to this model what was the price of gas in1990? Jan 17, 2013 · Introduction to Correlation and Regression Analysis. 9199 yields a coefficient of determination of 0. Simple Linear Regression Example 12. 0128 . 766, adjusted for Output 2: Regression output for the grade versus homework study Regression Analysis: CourseGrade versus Problems The regression equation is CourseGrade = 44. 2 MULTIVARIATE LINEAR REGRESSION Multiple linear regression with a single criterion variable is a straightforward generalization of linear regression. This is the only section that requires user input. We need to also include in CarType to our model. Displaying top 8 worksheets found for - Linear Equation Models. n. e. Linear Regression Interview Questions – Fundamental Questions. Materials: CD Player / Computer with CD drive. If you were a careful artist, you could take a ruler and draw a straight-line as close as possible to every point in Worksheet 2. The measure of how well this linear function ts the experimental points, is called regression analysis. • In a simple linear regression model, a single response measurement Y is related to a single predictor (covariate, regressor) X for each observation. Squaring −0. Worksheet for Correlation and Regression (February 1, 2013). In addition, after finding the equations, students are asked "extension questions" in which they must use the equation to answer questions a About This Quiz & Worksheet About This Quiz & Worksheet Simple linear regression builds on the concept of a regression line by allowing you to specifically make predictions based on the regression Practice linear regression with 10 Canadian data sets covering a range of topics. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. Students will determine the linear regression equation and correlation coefficient for their data using the graphing calculator. 21. Open or retrieve the worksheet Slr01. This leads to the Below is a plot of the data with a simple linear regression line superimposed. a) According to the linear model, the duration of a coaster ride is expected to increase by about 0. USING THIS MODULE vii Apr 27, 2017 · Non-Linear Relationships Not all relationships are linear. a) Find the least square regression line for the following set of data Practice Worksheet: Linear Regression. 903, and because the graph of the cubic model is seen to be a closer match to the dots in the scatterplot than is the linear model. Chapter 8 Linear Regression 91 22. Excel completes the regression analysis Regression Analysis Worksheets- Includes math lessons, 2 practice sheets, homework sheet, and a quiz! Linear regression simply refers to creating a best fit for a linear relationship between two variables from observed data. This makes the line fit the points. In this problem we find the model by analyzing the data on femur length and height for the ten males given in the table. Consider the following diagram. 3: Setup for Linear Regression Test on TI-83/84 to fit a simple linear regression equation with poverty predicting obesity. Linear Regression Assumptions • Linear regression is a parametric method and requires that certain assumptions be met to be valid. the linear relationship through the correlation coefficient. 𝑦𝑦 = −70. Graphic calculators, such as the TI-83, have built in programs which allow us to nd the slope and the y intercept of the best tting line to a set of data points. On a piece of graph paper, create a scatter plot. Of easily catch up can understand this You There are actually two ways to do a linear regression analysis using Excel. Practice-Regression 2 linear, quadratic, exponential A simple linear regression equation for this would be $$\hat{Price} = b_0 + b_1 * Mileage$$. Thus a linear model only explains 85% of the variation in women's world record 100 m dash times. Export problem to Excel (highlighted when problem is correctly entered). Identify the type of regression with the best fit, and answer the question using the type of regression that best fits the data. 433 seconds. It may be printed, downloaded or saved and used in your classroom, home school, or other educational Linear Regression. Linear Regression & Correlation Coefficient Worksheet Name _____ Hr _____ 0 2 4 6 8 10 0 2 4 6 1. Linear regression only supports regression type problems. Use the two plots to intuitively explain how the two models, Y!$0 %$ 1x %& and Why Linear Regression? •Suppose we want to model the dependent variable Y in terms of three predictors, X 1, X 2, X 3 Y = f(X 1, X 2, X 3) •Typically will not have enough data to try and directly estimate f •Therefore, we usually have to assume that it has some restricted form, such as linear Y = X 1 + X 2 + X 3 correlation, in linear regression. Femur Length (cm) (a) Make a scatter plot of the data. Our model will take the form of ŷ = b 0 + b 1 x where b 0 is the y-intercept, b 1 is the slope, x is the predictor variable, and ŷ an estimate of the mean value of the response variable for any value of the predictor Copy the regression coefficients onto the sheet with the actual population data. become identical when r = –1 or 1 or in other words, there is a perfect negative or positive correlation between the two variables under discussion. Problem-solving using linear regression has so many applications in business, digital customer experience, social, biological, and many many other areas. CAS'S WEBSITE - Home I don't think so. The regression line is the line that makes the square of the residuals as small as possible, so the regression line is also sometimes called the least squares line. This math worksheet was created on 2013-02-14 and has been viewed 62 times this week and 1,053 times this month. (1) Investigator #1 is interested in predicting Y from X, and fits and computes a regression line for this purpose. Using a similar approach, we may prove that S2 y = S 2 yˆ +S 2 e (5. There are also other regression modelling techniques for data not considered to be at continuous/interval/ratio level. The calculators also give 1. temperature 70 75 80 90 93 98 72 75 75 80 90 95 98 91 98 quantity 30 31 40 52 57 59 33 38 32 45 53 56 62 51 58 ¯x = 85. The regression model is linear in the unknown parameters. To make the residual plot, use “Graphs” and then type in the name of the explanatory variable. Each point of data is of the the form (x, y) and each point of the line of best fit using least-squares linear regression has the form Excel displays the Regression dialog box. nate because the world is too complex a place for simple linear regression alone to model it. Find a quadratic model in standard form for the data. 2 Linear Regression If there is a \signi cant" linear correlation between two variables, the next step is to nd the equation of a line that \best" ts the data. 8462. Comments: Another SPSS output table – see Table 3 – gives a useful value 'R square', or the 'coefficient of determination'. c) the relationship between X and Y is quadratic d) there is no relationship between X and Y. The dependent variable must be of ratio/interval scale and normally distributed overall and normally distributed for each value of the independent variables 3. Apply the method of least squares (or maximum likelihood) with a non-linear function. A simple linear regression model to relate BP with age will be BP = regression estimate (b) * age + constant (a) + error term (å) The regression estimate (b) and the constant (a) will be derived from the data (using the method of least-squares(5)) and the error term is to factor in the situation that two persons with the same age need not have the same BP. L. State which model, linear or quadratic, best fits the data. Know how to obtain the estimate MSE of the unknown population variance $$\sigma^{2 }$$ from Minitab's fitted line plot and regression analysis output. A regression is a process that takes all the points and calculates the equation that best 'fits' those points. The Nonlinear Regression Model 1 Goals The nonlinear regression model block in the Weiterbildungslehrgang (WBL) in ange- wandter Statistik at the ETH Zurich should 1. are called the residuals. The estimated regression equation is that average FEV = 0. With this worksheet generator, you can make customizable worksheets for linear inequalities in one variable. Included are two versions of the optional placemat/worksheet to help students through the process of linear regression and analysis. a) Enter the data into two lists of your graphing. Show that in a simple linear regression model the point ( ) lies exactly on the least squares regression line. CD – Music compiled by teacher. Investigator #2 is interested in predicting X from Y, and computes his regression line for that purpose (note that in the real problem of “parallel-line bioassays, with X=log(dose) STATISTICS 110/201 PRACTICE FINAL EXAM KEY (REGRESSION ONLY) Questions 1 to 5: There is a downloadable Stata package that produces sequential sums of squares for regression. This result is smaller than suggested by any of the other analyses that I have conducted, and is most similar to the analysis with all of the variables except for each of Reliable and Unconventional. A study found that age and blood pressure are correlated. Practice Problems: Correlation and Linear Regression Researchers interested in determining if there is a relationship between death anxiety and religiosity conducted the following study. We talk about looking at the data to decide what type of regression is appropriate. is found. See more ideas about Linear regression, Regression, Algebra. (b) Find and graph a linear regression equation that models the data. Each of the data sets has 4 or 5 points and approximates a linear relationship. We can find the equation of the line of best fit through the data in the least squares sense, as follows. fit linear regression models. Is there a linear relationship between how critics score the games and how users score the games? (Let x = critic score and y = user score). 2 Equation:_____ WORKSHEET GENERATORS. Categorical Data 12. A residual plot is displayed showing the deviation between the data and the calculated values of the dependent variable. expression the strength of a linear relationship between two variables are two of the desired outcomes of this module. 1 The Simple Linear Regression Model 12. Interpreting r and r 2. Word problems on constant speed. Oct 05, 2012 · The “Good” linear regression model. Remember to complete the five steps of hypothesis testing in the spaces provided on the worksheet. This means that this data can be modeled using a linear regression line. S. (b) What change in gasoline mileage is associated with a 1 cm3 change is engine displacement? 11-18. Correlation!Coefficient!&Linear!of!Best!Fit!HW! Name:!!_____! 8. The fit is not quite as tight for the women's times. The missile goes over the side of the ship and hits the water. SGDRegressor is well suited for regression problems with a large number of training samples (> 10;000), for other prob-lems we recommend Ridge, Lasso, or ElasticNet. Then they are able to apply the knowledge they learn in this project on the real life problems. Explain Jul 31, 2016 · State the three assumptions that are the basis for the Simple Linear Regression Model. Figure 2. 05898 6. If the data is curved, a line would not be the best equation to use. In most problems, more than one predictor variable will be available. than ANOVA. Research question type: When wanting to predict or explain one variable in WORKSHEETS\calcium. Find the numbers. Solve general word problems about real-world relationships that can be modeled by linear equations or functions. 999, which is closer to 1 than is the linear correlation value of 0. It is also a method that can be reformulated using matrix notation and solved using matrix operations. Use the linear regression equation determined in part "B" to calculate a set of "predicted y values" for each observed x value. Logarithmic Regression Problems. Regression 9. If the data on the scatter plot seems to represent a linear relationship, then linear regression can be used to find the line that best fits the data. Inference in the Linear Regression Model 4. The New Worksheet Ply default output option means that Excel locates the regression outputs on a new, separate worksheet that it creates and puts to the left of the worksheet that holds the original data. To do this, highlight the cell range A17:B18 and click on the COPY button. The plot to the right shows 5 data points and the least squares line. Interpret the meaning of the slope β ^ 1 of regression line in the context of problem. Use Stat > Regression > Regression to find the regression equation AND make a residual plot of the residuals versus the explanatory variable. Open the “Good” worksheet; this is a (made-up) data set showing the Height (independent variable) and Weight (dependent variable) values for a selection of people. Nov 26, 2014 · to linear regression . Old Faithful Geyser in Yellowstone National Park, Wyoming, derives its name and its considerable fame from the regularity (and beauty) of its eruptions. ) Using a graphing calculator and quadratic regression to find a model: A study compared the speed x, in miles per hour and the average fuel economy y (in miles per gallon) for cars. 3 (b) Find and graph a linear regression equation that models the data. In this exercise, you will gain some practice doing a simple linear regression using a data set called week02. Linear Regression and Correlation Introduction Linear Regression refers to a group of techniques for fitting and studying the straight-line relationship between two variables. Your output for this multiple regression problem should be similar to the results shown Let's explore the problem with our linear regression example. To make the notation simpler, assume that the criterion variable Y and the p Lesson 21: Multiple Linear Regression Analysis . Derivation of linear regression equations The mathematical problem is straightforward: given a set of n points (Xi,Yi) on a scatterplot, find the best-fit line, Y‹ i =a +bXi such that the sum of squared errors in Y, ∑(−)2 i Yi Y ‹ is minimized The simple linear regression model Our goal is to obtain estimates ^ 0 and ^ 1 for 0 and 1 to de ne the regression line ^y = ^ 0 + ^ 1x that provides the best t for the data Example: Assume that the regression line of the previous example is: Cost = 15:65 + 1:29 Volume Regression Problems Math The data for these problems is from Math 142, Dr Lacey, Packet 118. Student Learning Outcomes By the end of this chapter, you should be able to do the following: Linear Regression Models with Logarithmic Transformations Kenneth Benoit Methodology Institute London School of Economics kbenoit@lse. The critical assumption of the model is that the conditional mean function is linear: E(Y|X) = α +βX. 7 This worksheet is designed to give students extra practice at using their graphing calculators to calculate Linear Regression Equations. Oct 21, 2019 · Some of the worksheets below are Correlation Coefficient Practice Worksheets, Interpreting the data and the Correlation Coefficient, matching correlation coefficients to scatter plots activity with solutions, classify the given scatter plot as having positive, negative, or no correlation, … Simple Linear Regression Model Only one independent variable, x Relationship between x and y is described by a linear function Changes in y are assumed to be caused by changes in x Fall 2006 – Fundamentals of Business Statistics 18 Types of Regression Models Positive Linear Relationship Negative Linear Relationship Relationship NOT Linear Free worksheets for solving or graphing linear inequalities. Here are data from four students on their Quiz 1 scores and their Quiz 5 scores and a graph where we connected the points by a line. Regression 8. Their sum is 13. 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Describe the correlation Use the line of best fit to make predictions for the following real-world problems. 10: Regression - Fuel Efficiency (Worksheet) · 13: F Distribution and One-Way ANOVA If a loan officer makes 95% of his or her goal, write the linear function that applies based 12. A random sample was taken as stated in the problem. Student will learn how to write a linear regression equation and use the equation to solve a Make a scatterplot of the data, letting x represent the number of years since 1990. According to the linear model, what is the expected pressure for a person with cholesterol 270 mg/dl? Use the following sums: $\sum x_i=16960$ mg/dl, $\sum y_j=11160$ mmHg, $\sum x_i^2=3627200$ (mg/dl)$^2$, $\sum y_j^2=1576800$ mmHg$^2$ y \$\sum x_iy Welcome to The Systems of Linear Equations -- Two Variables (A) Math Worksheet from the Algebra Worksheets Page at Math-Drills. Inference on Prediction Assumptions 1. 126 Chapter 3 Writing Linear Equations and Linear Systems 3. Word problems on average speed Word problems Free worksheet(pdf) and answer key on the solving word problems based on linear equations and real world linear models. Equation: Created Date: 1/4/2016 8:11:44 PM A correlation or simple linear regression analysis can determine if two numeric variables are significantly linearly related. ## [1] 680 640 670 660 630 660 635. linear regression problems worksheet
n gweqgiyu , gqerignslomk, llv43lih0cj, a3hy9obqtv, qgbmrt1eb vdl8, 0pxyypazy6tyvy, | 2020-11-30T00:51:06 | {
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http://math.stackexchange.com/questions/226903/does-the-convergence-of-some-subsequences-imply-the-convergence-of-a-sequence | Does the convergence of some subsequences imply the convergence of a sequence?
I am considering a following problem: Does $$\\a_{2k}, a_{2k+1}, a_{3k} \rightarrow g$$ imply that $$a_{n} \rightarrow g ?$$ I know that if every subsequence goes to $g$ then also a sequence goes to $g$. My way of reasoning is following: we know that odd and even $k$ subsequences goes to $g$. But if we substitute for example $k=t^{2} -5$, then there is problem (at least for me). If I can reason that way, what must I do to prove more formally that the statement is false? And if I can not, what is going on with $a_{n}$? Thanks for any hints!
-
If the even numbered terms and odd numbered terms both converge to the same limit $L$, then the limit of the sequence exists and is $L$.
To prove this, let $\epsilon > 0$ be arbitrary. We want to find $N \in \mathbb{N}$ such that $n > N$ implies $|a_n - L| < \epsilon$.
Since the even numbered terms are all within epsilon of $L$ after some $N_1 = 2j$ and the odd numbered terms are all within epsilon of $L$ after some $N_2 = 2k + 1$, just pick $N = \max\{N_1, N_2\}$.
-
You may want to prove the next nice lemma:
Lemma: If $\,\{A_i\}_{i\in I},$ is some partition of $\,\Bbb N\,$ s.t. $\,|A_i|=\aleph_0=|\Bbb N|\,\,\,,\,\forall\,\,i\in I\,$ , and all the sets $\,A_i\,$ are well ordered, then for a real sequence $\,\{x_n\}\,$ it is true that
$$x_n\xrightarrow [n\to\infty]{} x\Longleftrightarrow (\,\forall\,\,i\in I\,\,,x_{n_i}\xrightarrow [n_i\to\infty\,,\,n_i\in A_i]{} x)$$
Thus, as B. wrote in his answer, it is enough and sufficient that the subsequence of odd indexes and the one of even indexes converge both to the same limit.
-
Doesn't $I$ have to be a finite in order for this to work? – Arthur Fischer Nov 1 '12 at 17:48
Well, not that you mention it...I'm not completely sure, I'll try to check this later. Thanks. – DonAntonio Nov 1 '12 at 18:23
Take $I = \{ 1 \} \cup \{ p : p \text{ is prime} \}$. Define $A_1 = \{ 1 \} \cup \{ n : n \text{ is not a prime power} \}$, and $A_p = \{ p^k : k \geq 1 \}$ for prime $p$. Define the sequence $( x_n )_{n \geq 1}$ so that $x_n = 0$ for all $n \in A_1$ and $x_{p^k} = \frac{1}{k}$ for $p$ prime and $k \geq 1$. Then for each $i \in I$ the sequence $( a_n )_{n \in A_i}$ converges to $0$, but the sequence $( a_n )_{n \geq 1}$ does not converge as it takes the value $1$ infinitely often and has subsequences converging to $0$. – Arthur Fischer Nov 1 '12 at 18:39
In the other answers you have seen that it $a_{2k},a_{2k+1}\to g$ then already $a_n\to g$. You don't even need $a_{3k}$. I would like to add that if you just require $a_{2k},a_{2k+1}$ and $a_{3k}$ to converge (a priori with different limits), then you have that all three limits coincide and that $a_n$ has the same limit.
To see this note that $a_{2n}$ and $a_{3n}$ as well as $a_{2n+1}$ and $a_{3n}$ have a commen subsequence. Then you only need that if a converging sequence has a subsequence with limit $g$ then the whole sequence has limit $g$. | 2014-03-08T01:17:08 | {
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https://math.stackexchange.com/questions/451799/how-to-show-lim-x-to-1-fracx-x2-dots-xn-nx-1-fracnn | # How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$?
I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.
The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:
• Substitute $y = x - 1$. This leads nowhere, I think.
• Find the Taylor polynomial. Makes no sense, it is a polynomial.
• Divide by major term. Dividing by $x$ got me nowhere.
• Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$.
• Simplify the expression. I do not see how I could.
• Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.
If somebody sees a simple way, please do let me know.
Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:
• $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or
• $\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.
It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.
This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).
• Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. Jan 15, 2015 at 12:36
Let $$f(x)=x+x^2+\cdots+x^n-n$$ then by the definition of the derivative we have
$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1}= \lim_{x \to 1}\frac{f(x)-f(1)}{x - 1}=f'(1)\\[10pt] = \left[ \vphantom{\frac11} 1 + 2x + 3x^2 + \cdots + nx^{n-1} \right]_{x=1} = \frac{n(n + 1)}{2}$$
• It means that we should substitute $x$ by $1$.
– user63181
Jul 25, 2013 at 12:20
• Then this is truly an ingenious solution. Jul 25, 2013 at 12:30
• So the original is just the integral of a function summing the positive natural numbers? Pauling would be proud. Jul 25, 2013 at 15:29
• I almost cried when I saw this solution. Jul 25, 2013 at 17:44
• @RossMillikan shouldn't it not matter? if $f(x)$ contains the $-n$ then $f(1) = 0$, and if $f(x)$ does not, $f(1) = n$, but either way $f(x) - f(1)$ doesn't change and since $n$ is a constant, $f'(x)$ looks the same. Jul 25, 2013 at 23:22
Well, you can use that \begin{align} x-1&=\left(x-1\right)\cdot 1,\\ x^2-1&=\left(x-1\right)\cdot\left(x+1\right),\\ x^3-1&=\left(x-1\right)\cdot(x^2+x+1),\\ &{}\ \ \vdots\\ x^n-1&=\left(x-1\right)\cdot(x^{n-1}+\cdots+1),\\ \end{align} Now sum the left hand sides and the right hand sides, divide by $x-1$ and consider the limit $x\rightarrow 1$.
The given limit is $$\lim_{x\rightarrow 1}\frac{\sum_{k=1}^nx^k-n}{x-1}\\ =\lim_{x\rightarrow 1}\frac{\sum_{k=1}^n(x^k-1)}{x-1}\\ =\sum_{k=1}^n \lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}$$
Now, $$\lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}\\ =\lim_{x\rightarrow 1} (\sum_{j=0}^{k-1}x^j)=k$$
Hence the given limit becomes $$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$
Hints:
$$x+x^2+\ldots+x^n-n=(x-1)+(x^2-1)+\cdots +(x^n-1)=(x-1)\left(1+(x+1)+\cdots\right)$$
Also
$$1+(x+1)+(x^2+x+1)+\cdots+(x^{n-1}+\cdots+x+1)\xrightarrow [x\to 1]{}1+2+\cdots+n=\frac{n(n+1)}2$$
You can use induction:
$$\frac{x + x^2 + \dots + x^n + x^{n+1} - (n+1)}{x - 1} =\\ \frac{x + x^2 + \dots + x^n - n}{x - 1} +\frac{x^{n+1}-1}{x-1}=\\ \frac{x + x^2 + \dots + x^n - n}{x - 1} +(1+x+x^2+\ldots+x^n)\xrightarrow[x\to 1]{} \frac{n(n + 1)}{2}+(n+1)=\frac{(n+1)(n + 2)}{2}.$$
• I am a little lost. How is that you are getting different result? Or is $\frac{(n + 1)(n + 2)}{2}$ only a part of the solution? Jul 25, 2013 at 12:28
• @DavidČepelík No, $\frac{(n+1)(n+2)}{2}$ is the limit for $\frac{x+x^2+\ldots+x^n\color{brown}{+x^{n+1}}-n\color{brown}{-1}}{x-1}$ and not for $\frac{x+x^2+\ldots+x^n-n}{x-1}$.
– P..
Jul 25, 2013 at 12:32
• Of course, that is the induction. Thank you! Jul 25, 2013 at 12:37
• So for $n = 1$ and $x \to 1$, we get $a_1 = \lim_{x \to 1} \frac{x - 1}{x - 1} = 1$ and $a_{n + 1} = a_n + (n + 1)$. The closed form for $a_n$ is then $\frac{n(n + 1)}{2}$. Correct? Jul 25, 2013 at 12:42
• @DavidČepelík: Yes that's right!
– P..
Jul 25, 2013 at 13:08
$$\frac{x + x^2 + \dots + x^n - n}{x - 1} =\frac{1+x + x^2 + \dots + x^n - n-1}{x - 1} =\frac{\frac{x^{n+1}-1}{x-1}-(n+1)}{x-1}$$
Putting $x-1=y,$
$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \lim_{y\to0}\frac{\frac{(1+y)^{n+1}-1}y-(n+1)}y$$
$$=\lim_{y\to0}\frac{\frac{1+\binom {n+1}1y+\binom {n+1}2y^2+ O(y^3)-1}y-(n+1)}y\text{ (using Binomial Expansion)}$$
$$=\lim_{y\to0}\frac{(n+1)+\frac{n(n+1)}2y+ O(y^2)-(n+1)}y$$
$$=\frac{n(n+1)}2\text{ as }x\to1,y\to0\implies y\ne0$$
If you already know l'Hopital's rule, Sami Ben Romdhane's answer isn't telling you anything you don't already know. When you have $$\lim_{x\to a}\frac{f(x)}{x - a}$$ and $f(a) = 0$ (otherwise, the limit is infinite on either side of $a$), then l'Hopital's rule says that the limit is the same as $$\lim_{x \to a}f'(x)= f'(a).$$ The only hard part then, in this case, is evaluating $f'(a)$. Here, $f(x) = x + x^2 + \cdots + x^n - n$, so $f'(x) = 1 + 2x + \cdots + nx^{n - 1}$, so $f'(1) = 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$ by a famous identity.
In fact, you can think of the definition of the derivative as just a special case of l'Hopital's rule (it isn't really, because l'Hopital's rule depends on the definition of the derivative, not the other way around, but it's useful to think of it this way).
• I do not think this is how he approached the problem; my guess is he noticed that the function and the derivative of the numerator are bound together by the definition of derivative of a function. I did not realize that he used the same principle Bernoulli's rule is built upon, thanks for pointing it out. Jul 25, 2013 at 16:23
Your objection to using l'Hospital's rule is on the basis that it feels like it's "too powerful" a tool for the problem, right?
Lets go all the way to the other end, then, and prove the limit using just the definition.
$$\lim_{x\to a} f(x) = L \iff \forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ni \left| x-a \right| \le \delta \Rightarrow \left| f(x)-L \right| \le \varepsilon$$
Now, all we need to do is figure out a way that we can always pick a $\delta$ small enough to keep the function within $\varepsilon$ of $L$.
The particular limit in question: $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$
From that, we can take: \begin{align*} a&=1\\ f(x)&=\frac{x + x^2 + \dots + x^n - n}{x - 1}\\ L&=\frac{n(n + 1)}{2} \end{align*}
So we need to solve $$\left| \frac{x + x^2 + \dots + x^n - n}{x - 1} - \frac{n(n + 1)}{2}\right| < \varepsilon$$ for $\left|x-1\right|$.
I really don't feel like doing any of that crunchwork solving that equation. If someone out there does feel like it, please do so, and edit it into my answer. For now, though, I am going to skip ahead bunch of steps, assume we solved it, and have our solution of $$\left|x-1\right| \ge g(\varepsilon)$$
Now, we know from the definition that $\left| x-1 \right| \le \delta$, so we can conclude that if we pick $\delta = g(\varepsilon)$, we ensure that the value of $f(x)$ is within $\varepsilon$ of $L$, satisfying our definition of the limit, and proving that $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$
• Thanks for your answer. You already know the limit exists and you know what it equals to ($\frac{n(n+1)}{2}$ was not my first guess). Finding what a limit equals to and proving the solution is correct by checking it satisfies the definition of the limit are two completely different exercises. (Similar to the difference between the P and NP classes in the theory of computational complexity.) I am aware I gave correct solution along with the limit, maybe that is why we did not understand each other. Sorry about it. Jul 26, 2013 at 20:01
• @DavidČepelík Personally, I think that if you prove the limit without using your super-powered tool, it doesn't matter if you used it to find the value of the limit, since the validity of your answer then does not depend on the validity of the method you applied. Jul 29, 2013 at 12:57 | 2022-05-25T01:52:45 | {
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https://bathmash.github.io/HELM/30_5_itrtv_mthds_systms_eqns-web/30_5_itrtv_mthds_systms_eqns-webse2.html | ### 2 Do these iterative methods always work?
No. It is not difficult to invent examples where the iteration fails to approach the solution of $AX=B$ . The key point is related to matrix norms seen in the preceding Section.
The two iterative methods we encountered above are both special cases of the general form
$\phantom{\rule{2em}{0ex}}{X}^{\left(k+1\right)}=M{X}^{\left(k\right)}+N.$
1. For the Jacobi method we choose $M=-{D}^{-1}\left(L+U\right)$ and $N={D}^{-1}B$ .
2. For the Gauss-Seidel method we choose $M=-{\left(D+L\right)}^{-1}U$ and $N={\left(D+L\right)}^{-1}B$ .
The following Key Point gives the main result.
##### Key Point 13
For the iterative process ${X}^{\left(k+1\right)}=M{X}^{\left(k\right)}+N$ the iteration will converge to a solution if the norm of $M$ is less than 1 .
Care is required in understanding what Key Point 13 says. Remember that there are lots of different ways of defining the norm of a matrix (we saw three of them). If you can find a norm ( any norm ) such that the norm of $M$ is less than 1, then the iteration will converge. It doesn’t matter if there are other norms which give a value greater than 1, all that matters is that there is one norm that is less than 1.
Key Point 13 above makes no reference to the starting “guess" ${X}^{\left(0\right)}$ . The convergence of the iteration is independent of where you start! (Of course, if we start with a really bad initial guess then we can expect to need lots of iterations.)
Show that the Jacobi iteration used to approximate the solution of
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -5\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}\right]$
is certain to converge. (Hint: calculate the norm of $-{D}^{-1}\left(L+U\right)$ .)
The Jacobi iteration matrix is
$\begin{array}{rcll}-{D}^{-1}\left(L+U\right)& =& {\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]}^{-1}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0.25\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -0.2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0.5\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]& \text{}\\ & =& \left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0.25\hfill & \hfill 0.25\hfill \\ \hfill -0.2\hfill & \hfill 0\hfill & \hfill 0.4\hfill \\ \hfill 0.5\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]& \text{}\end{array}$
and the infinity norm of this matrix is the maximum of $0.25+0.25$ , $0.2+0.4$ and $0.5$ , that is
$\phantom{\rule{2em}{0ex}}∥-{D}^{-1}\left(L+U\right){∥}_{\infty }=0.6$
which is less than 1 and therefore the iteration will converge.
#### 2.1 Guaranteed convergence
If the matrix has the property that it is strictly diagonally dominant , which means that the diagonal entry is larger in magnitude than the absolute sum of the other entries on that row, then both Jacobi and Gauss-Seidel are guaranteed to converge. The reason for this is that if $A$ is strictly diagonally dominant then the iteration matrix $M$ will have an infinity norm that is less than 1.
A small system is the subject of Example 20 below. A large system with slow convergence is the subject of Engineering Example 1 on page 62.
##### Example 20
Show that $A=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -5\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$ is strictly diagonally dominant.
##### Solution
Looking at the diagonal entry of each row in turn we see that
$\begin{array}{rcll}4& >& |-1|+|-1|=2\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ \left|-5\right|& >& 1+\left|-2\right|=3& \text{}\\ 2& >& \left|-1\right|+0=1& \text{}\end{array}$
and this means that the matrix is strictly diagonally dominant.
Given that $A$ above is strictly diagonally dominant it is certain that both Jacobi and Gauss-Seidel will converge.
#### 2.2 What’s so special about strict diagonal dominance?
In many applications we can be certain that the coefficient matrix $A$ will be strictly diagonally dominant. We will see examples of this in HELM booklet 32 and HELM booklet 33 when we consider approximating solutions of differential equations.
##### Exercises
1. Consider the system
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -5\hfill \end{array}\right]$
1. Use the starting guess ${X}^{\left(0\right)}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ in an implementation of the Jacobi method to show that ${X}^{\left(1\right)}=\left[\begin{array}{c}\hfill 1.5\hfill \\ \hfill -3\hfill \end{array}\right]$ . Find ${X}^{\left(2\right)}$ and ${X}^{\left(3\right)}$ .
2. Use the starting guess ${X}^{\left(0\right)}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ in an implementation of the Gauss-Seidel method to show that ${X}^{\left(1\right)}=\left[\begin{array}{c}\hfill 1.5\hfill \\ \hfill -3.25\hfill \end{array}\right]$ . Find ${X}^{\left(2\right)}$ and ${X}^{\left(3\right)}$ .
(Hint: it might help you to know that the exact solution is $\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill -4\hfill \end{array}\right]$ .)
1. Show that the Jacobi iteration applied to the system
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cccc}\hfill 5\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 5\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{4}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 7\hfill \\ \hfill -10\hfill \\ \hfill -6\hfill \\ \hfill 16\hfill \end{array}\right]$
can be written
$\phantom{\rule{2em}{0ex}}{X}^{\left(k+1\right)}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \end{array}\right]{X}^{\left(k\right)}+\left[\begin{array}{c}\hfill 1.4\hfill \\ \hfill -2\hfill \\ \hfill -1.2\hfill \\ \hfill 3.2\hfill \end{array}\right].$
2. Show that the method is certain to converge and calculate the first three iterations using
zero starting values.
(Hint: the exact solution to the stated problem is $\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \\ \hfill 3\hfill \end{array}\right]$ .)
1. $\phantom{\rule{2em}{0ex}}2{x}_{1}^{\left(1\right)}=2-1{x}_{2}^{\left(0\right)}=2$
and therefore ${x}_{1}^{\left(1\right)}=1.5$
$\phantom{\rule{2em}{0ex}}2{x}_{2}^{\left(1\right)}=-5-1{x}_{1}^{\left(0\right)}=-6$
which implies that ${x}_{2}^{\left(1\right)}=-3$ . These two values give the required entries in ${X}^{\left(1\right)}$ . A second and third iteration follow in a similar way to give
$\phantom{\rule{2em}{0ex}}{X}^{\left(2\right)}=\left[\begin{array}{c}\hfill 2.5\\ \hfill -3.25\end{array}\right]\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{X}^{\left(3\right)}=\left[\begin{array}{c}\hfill 2.625\\ \hfill -3.75\end{array}\right]$
2. $\phantom{\rule{2em}{0ex}}2{x}_{1}^{\left(1\right)}=2-1{x}_{2}^{\left(0\right)}=3$
and therefore ${x}_{1}^{\left(1\right)}=1.5$ . This new approximation to ${x}_{1}$ is used straight away when finding a new approximation to ${x}_{2}^{\left(1\right)}$ .
$\phantom{\rule{2em}{0ex}}2{x}_{2}^{\left(1\right)}=-5-1{x}_{1}^{\left(1\right)}=-6.5$
which implies that ${x}_{2}^{\left(1\right)}=-3.25$ . These two values give the required entries in ${X}^{\left(1\right)}$ . A second and third iteration follow in a similar way to give
$\phantom{\rule{2em}{0ex}}{X}^{\left(2\right)}=\left[\begin{array}{c}\hfill 2.625\\ \hfill -3.8125\end{array}\right]\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{X}^{\left(3\right)}=\left[\begin{array}{c}\hfill 2.906250\\ \hfill -3.953125\end{array}\right]$
where ${X}^{\left(3\right)}$ is given to 6 decimal places
1. In this case $D=\left[\begin{array}{cccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill \end{array}\right]$ and therefore ${D}^{-1}=\left[\begin{array}{cccc}\hfill 0.2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.2\hfill \end{array}\right]$ .
So the iteration matrix $M\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{D}^{-1}\left[\begin{array}{cccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\left[\phantom{\rule{0.3em}{0ex}}\begin{array}{cccc}\hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \end{array}\phantom{\rule{0.3em}{0ex}}\right]$
and that the Jacobi iteration takes the form
$\phantom{\rule{2em}{0ex}}{X}^{\left(k+1\right)}=M{X}^{\left(k\right)}+{M}^{-1}\left[\begin{array}{c}\hfill 7\hfill \\ \hfill -10\hfill \\ \hfill -6\hfill \\ \hfill 16\hfill \end{array}\right]=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill & \hfill 0.2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0\hfill \end{array}\right]{X}^{\left(k\right)}+\left[\begin{array}{c}\hfill 1.4\hfill \\ \hfill -2\hfill \\ \hfill -1.2\hfill \\ \hfill 3.2\hfill \end{array}\right]$
as required.
2. Using the starting values ${x}_{1}^{\left(0\right)}={x}_{2}^{\left(0\right)}={x}_{3}^{\left(0\right)}={x}_{4}^{\left(0\right)}=0$ , the first iteration of the Jacobi method gives $\begin{array}{rcll}{x}_{1}^{1}& =& 0.2{x}_{2}^{0}+1.4=1.4& \text{}\\ {x}_{2}^{1}& =& 0.2\left({x}_{1}^{0}+{x}_{3}^{0}\right)-2=-2& \text{}\\ {x}_{3}^{1}& =& 0.2\left({x}_{2}^{0}+{x}_{4}^{0}\right)-1.2=-1.2\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ {x}_{4}^{1}& =& 0.2{x}_{3}^{0}+3.2=3.2& \text{}\end{array}$
The second iteration is
$\begin{array}{rcll}{x}_{1}^{2}& =& 0.2{x}_{2}^{1}+1.4=1& \text{}\\ {x}_{2}^{2}& =& 0.2\left({x}_{1}^{1}+{x}_{3}^{1}\right)-2=-1.96& \text{}\\ {x}_{3}^{2}& =& 0.2\left({x}_{2}^{1}+{x}_{4}^{1}\right)-1.2=-0.96\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ {x}_{4}^{2}& =& 0.2{x}_{3}^{1}+3.2=2.96& \text{}\end{array}$
And the third iteration is
$\begin{array}{rcll}{x}_{1}^{3}& =& 0.2{x}_{2}^{2}+1.4=1.008& \text{}\\ {x}_{2}^{3}& =& 0.2\left({x}_{1}^{2}+{x}_{3}^{2}\right)-2=-1.992\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ {x}_{3}^{3}& =& 0.2\left({x}_{2}^{2}+{x}_{4}^{2}\right)-1.2=-1& \text{}\\ {x}_{4}^{3}& =& 0.2{x}_{3}^{2}+3.2=3.008& \text{}\end{array}$ | 2022-11-29T00:51:45 | {
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https://au.mathworks.com/help/matlab/ref/cumtrapz.html | # cumtrapz
Cumulative trapezoidal numerical integration
## Syntax
``Q = cumtrapz(Y)``
``Q = cumtrapz(X,Y)``
``Q = cumtrapz(___,dim)``
## Description
example
````Q = cumtrapz(Y)` computes the approximate cumulative integral of `Y` via the trapezoidal method with unit spacing. The size of `Y` determines the dimension to integrate along:If `Y` is a vector, then `cumtrapz(Y)` is the cumulative integral of `Y`.If `Y` is a matrix, then `cumtrapz(Y)` is the cumulative integral over each column.If `Y` is a multidimensional array, then `cumtrapz(Y)` integrates over the first dimension whose size does not equal 1.```
example
````Q = cumtrapz(X,Y)` integrates `Y` with respect to the coordinates or scalar spacing specified by `X`. If `X` is a vector of coordinates, then `length(X)` must be equal to the size of the first dimension of `Y` whose size does not equal 1.If `X` is a scalar spacing, then `cumtrapz(X,Y)` is equivalent to `X*cumtrapz(Y)`. ```
example
````Q = cumtrapz(___,dim)` integrates along the dimension `dim` using any of the previous syntaxes. You must specify `Y`, and optionally can specify `X`. If you specify `X`, then it can be a scalar or a vector with length equal to `size(Y,dim)`. For example, if `Y` is a matrix, then `cumtrapz(X,Y,2)` cumulatively integrates each row of `Y`.```
## Examples
collapse all
Calculate the cumulative integral of a vector where the spacing between data points is 1.
Create a numeric vector of data.
`Y = [1 4 9 16 25];`
`Y` contains function values for $\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^{2}$ in the domain `[1 5]`.
Use `cumtrapz` to integrate the data with unit spacing.
`Q = cumtrapz(Y)`
```Q = 1×5 0 2.5000 9.0000 21.5000 42.0000 ```
This approximate integration yields a final value of 42. In this case, the exact answer is a little less, $41\frac{1}{3}$. The `cumtrapz` function overestimates the value of the integral because f(x) is concave up.
Calculate the cumulative integral of a vector where the spacing between data points is uniform, but not equal to 1.
Create a domain vector.
`X = 0:pi/5:pi;`
Calculate the sine of `X`.
`Y = sin(X');`
Cumulatively integrate `Y` using `cumtrapz`. When the spacing between points is constant, but not equal to 1, an alternative to creating a vector for `X` is to specify the scalar spacing value. In that case, `cumtrapz(pi/5,Y)` is the same as `pi/5*cumtrapz(Y)`.
`Q = cumtrapz(X,Y)`
```Q = 6×1 0 0.1847 0.6681 1.2657 1.7491 1.9338 ```
Cumulatively integrate the rows of a matrix where the data has a nonuniform spacing.
Create a vector of x-coordinates and a matrix of observations that take place at the irregular intervals. The rows of `Y` represent velocity data, taken at the times contained in `X`, for three different trials.
```X = [1 2.5 7 10]; Y = [5.2 7.7 9.6 13.2; 4.8 7.0 10.5 14.5; 4.9 6.5 10.2 13.8];```
Use `cumtrapz` to integrate each row independently and find the cumulative distance traveled in each trial. Since the data is not evaluated at constant intervals, specify `X` to indicate the spacing between the data points. Specify `dim = 2` since the data is in the rows of `Y`.
`Q1 = cumtrapz(X,Y,2)`
```Q1 = 3×4 0 9.6750 48.6000 82.8000 0 8.8500 48.2250 85.7250 0 8.5500 46.1250 82.1250 ```
The result is a matrix of the same size as `Y` with the cumulative integral of each row.
Perform nested integrations in the x and y directions. Plot the results to visualize the cumulative integral value in both directions.
Create a grid of values for the domain.
```x = -2:0.1:2; y = -2:0.2:2; [X,Y] = meshgrid(x,y);```
Calculate the function $\mathit{f}\left(\mathit{x},\mathit{y}\right)=10{\mathit{x}}^{2}+20{\mathit{y}}^{2}$ on the grid.
`F = 10*X.^2 + 20*Y.^2;`
`cumtrapz` integrates numeric data rather than functional expressions, so in general the underlying function does not need to be known to use `cumtrapz` on a matrix of data. In cases where the functional expression is known, you can instead use `integral`, `integral2`, or `integral3`.
Use `cumtrapz` to approximate the double integral
`$I\left(a,b\right)={\int }_{-2}^{b}{\int }_{-2}^{a}\left(10{x}^{2}+20{y}^{2}\right)\phantom{\rule{0.16666666666666666em}{0ex}}dx\phantom{\rule{0.16666666666666666em}{0ex}}dy.$`
To perform this double integration, use nested function calls to `cumtrapz`. The inner call first integrates the rows of data, then the outer call integrates the columns.
`I = cumtrapz(y,cumtrapz(x,F,2));`
Plot the surface representing the original function as well as the surface representing the cumulative integration. Each point on the surface of the cumulative integration gives an intermediate value of the double integral. The last value in `I` gives the overall approximation of the double integral, `I(end) = 642.4`. Mark this point in the plot with a red star.
```surf(X,Y,F,'EdgeColor','none') xlabel('X') ylabel('Y') hold on surf(X,Y,I,'FaceAlpha',0.5,'EdgeColor','none') plot3(X(end),Y(end),I(end),'r*') hold off```
## Input Arguments
collapse all
Numeric data, specified as a vector, matrix, or multidimensional array. By default, `cumtrapz` integrates along the first dimension of `Y` whose size does not equal 1.
Data Types: `single` | `double`
Complex Number Support: Yes
Point spacing, specified as `1` (default), a uniform scalar spacing, or a vector of coordinates.
• If `X` is a scalar, then it specifies a uniform spacing between the data points and `cumtrapz(X,Y)` is equivalent to `X*cumtrapz(Y)`.
• If `X` is a vector, then it specifies x-coordinates for the data points and `length(X)` must be the same as the size of the integration dimension in `Y`.
Data Types: `single` | `double`
Dimension to operate along, specified as a positive integer scalar. If you do not specify the dimension, then the default is the first array dimension of size greater than 1.
Consider a two-dimensional input array, `Y`:
• `cumtrapz(Y,1)` works on successive elements in the columns of `Y`.
• `cumtrapz(Y,2)` works on successive elements in the rows of `Y`.
If `dim` is greater than `ndims(Y)`, then `cumtrapz` returns an array of zeros of the same size as `Y`.
## Tips
• Use `trapz` and `cumtrapz` to perform numerical integrations on discrete data sets. Use `integral`, `integral2`, or `integral3` instead if a functional expression for the data is available.
• `trapz` reduces the size of the dimension it operates on to 1, and returns only the final integration value. `cumtrapz` also returns the intermediate integration values, preserving the size of the dimension it operates on.
## Version History
Introduced before R2006a | 2022-10-04T17:55:08 | {
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https://math.stackexchange.com/questions/3344853/elegant-way-to-assign-4-distinct-objects-into-4-bins | # “Elegant” way to assign 4 distinct objects into 4 bins
Suppose you have four objects to distribute among four people. Suppose people can carry any number of objects (none, one, two, three, or all four objects) at once. In how many ways can the four objects be distributed?
My solution
Let the four people be named A,B,C,D.
Distribution type 1
If each person gets exactly one object, the distribution looks like
A B C D
1 1 1 1
of which there are $$4!=4\cdot 3 \cdot 2 \cdot 1 = 24$$ distinct ways (since the objects are distinct).
Distribution type 2
If one person gets 2 objects and two others get 1 object, the possible distributions look like
A B C D
2 1 1 0
2 1 0 1
2 0 1 1
1 2 1 0
1 2 0 1
0 2 1 1
1 1 2 0
1 0 2 1
0 1 2 1
1 1 0 2
1 0 1 2
0 1 1 2
and for each of the above distributions, there are $$12$$ distinct ways to distribute the distinct objects.
Distribution type 3
If two people get 2 objects, the possible distributions look like
A B C D
2 2 0 0
2 0 2 0
2 0 0 2
0 2 2 0
0 2 0 2
0 0 2 2
and for each of the above distributions, there are $$6$$ distinct ways to distribute the distinct objects.
Distribution type 4
If one person gets 3 objects and another person gets 1 object, the possible distributions look like
A B C D
3 1 0 0
3 0 1 0
3 0 0 1
1 3 0 0
0 3 1 0
0 3 0 1
1 0 3 0
0 1 3 0
0 0 3 1
1 0 0 3
0 1 0 3
0 0 1 3
and for each of the above distributions, there are $$4$$ distinct ways to distribute the distinct objects.
Distribution type 5
If one person gets all 4 objects, the possible distributions look like
A B C D
4 0 0 0
0 4 0 0
0 0 4 0
0 0 0 4
for each of the above distributions, there is only $$1$$ distinct way to distribute the distinct objects.
Totaling the ways
The total number of ways is then
\begin{align} &\textrm{ type 1 + type 2 + type 3 + type 4 + type 5 }\\ =& 1(24) + 12(12) + 6(6) + 12(4) + 4(1) \\ =& 24 + 144 + 36 + 48 + 4 \\ =& 256\textrm{ ways } \end{align}
My questions
I have 3 questions:
1. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the amount of each distribution type (the numbers 1, 12, 6, 12, 4) that appear in the final calculation?
2. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the number of ways for each distribution type (the numbers 24, 12, 6, 4, 1) that appear in the final calculation in parentheses? (Clearly the 24 is just $$_4 P_4$$, but what about the others?)
3. Why is all of this just equal to $$4^4$$? This isn't immediately obvious to me.
• @EdwardH., do you have a proof of that? – Charles Hudgins Sep 5 '19 at 5:00
• Each object can be given to any person, hence $N_p^{N_o}=4^4$ ways. – Yves Daoust Sep 5 '19 at 7:25
1. Breaking it up a little differently, the number of ways with $$k = 0,1,2,3$$ of them getting $$0$$ is $$_4C_k \cdot {}_3C_{3-k}$$, which gives the sequence $$1, 12, 18 = 6 + 12, 4$$. Where the difference is that you had the $$6$$ ways of $$2,2,0,0$$ and the $$12$$ ways of $$3,1,0,0$$ separately. Those are $$\frac{4!}{2!2!}$$ and $$\frac{4!}{1!1!2!}$$, respectively.
2. For the arrangements, you are looking at the multinomial coefficients: $$\frac{4!}{1!1!1!1!} = 24, \frac{4!}{2!1!1!0!} = 12, \frac{4!}{2!2!0!0!} = 6, \frac{4!}{3!1!0!0!} = 4, \frac{4!}{4!0!0!0!} = 1$$
3. Instead of focusing on the people and which objects they get, look at the objects and who they are given to. Each object can be given to any of the four people, without restriction. That means there are $$4$$ options for the first object, $$4$$ for the second, and so on, so the total is $$4^4$$.
• Thank you, Michael! – Mathemanic Sep 8 '19 at 23:18
Let the objects choose the people . . .
Each object has $$4$$ choices, and each object's choices are independent of the choices made by the other objects, hence there are $$4^4$$ ways for the objects to choose the people.
Alternatively, using your cases . . .
• For distribution type $$1$$, the number of ways is $$4!=24$$ Explanation:
If we have the people line up to choose, then person $$1$$ has $$4$$ choices, person $$2$$ has $$3$$ choices, person $$3$$ has $$2$$ choices, and person $$4$$ has $$1$$ choice.
• For distribution type $$2$$, the number of ways is $$\binom{4}{1}\binom{4}{2}\binom{3}{2}2!=144$$ Explanation:
• Choose the person who gets two objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
• Choose the two objects for that person:$$\;{\large{\binom{4}{2}}}\;$$choices.
• Choose the two people to get the two remaining objects:$$\;{\large{\binom{3}{2}}}\;$$choices.
• Distribute the two remainining objects, one to each of the two chosen people:$$\;2!\;$$choices.
• For distribution type $$3$$, the number of ways is $$\binom{4}{2}\binom{2}{1}\binom{3}{1}=36$$ Explanation:
• Choose the two people who get two objects:$$\;{\large{\binom{4}{2}}}\;$$choices.
• Choose the person who gets the object labeled #$$1$$ plus one other object:$$\;{\large{\binom{2}{1}}}\;$$choices.
• Choose the other object for that person:$$\;{\large{\binom{3}{1}}}\;$$choices.
• For distribution type $$4$$, the number of ways is $$\binom{4}{1}\binom{4}{3}\binom{3}{1}=48$$ Explanation:
• Choose the person who gets three objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
• Choose the three objects for that person:$$\;{\large{\binom{4}{3}}}\;$$choices.
• Choose the person to get the one remaining object:$$\;{\large{\binom{3}{1}}}\;$$choices.
• For distribution type $$5$$, the number of ways is $$\binom{4}{1}=4$$ Explanation:$$\;$$Choose the person who gets all four objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
$$\;\;\;$$Summing the counts for the cases, the total number of ways is $$24+144+36+48+4=256$$
• Awesome and very thorough answer - thank you Quasi! – Mathemanic Sep 8 '19 at 23:19 | 2020-02-19T22:46:50 | {
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https://math.stackexchange.com/questions/1889632/question-about-spanning-sets-and-bases | Question about spanning sets and bases
In a textbook is the question:
Find a basis for the subspace $$V=\{(x_1,x_2,x_3,x_4):x_1+2x_2+x_3+x_4=0, 3x_1+6x_2+4x_3+x_4=0\}.$$
They say that $V$ is a subspace of $\mathbb{R^4}$ and are able to find a spanning set by solving the system of homogeneous equations getting: $$\{\begin{bmatrix}-2 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix}-3 \\ 0 \\ 2 \\ 1 \end{bmatrix}\}.$$ They then show that these vectors are linearly independent and so it forms a basis for $V$.
I thought though that if you're in $\mathbb{R^4}$ then you need at least 4 vectors to span the space, and 4 to form a basis. Here though we only have two. Could someone please tell me where my logic went wrong and why these indeed span the subspace and form a basis?
• They ask you to find a basis for a subspace, not for the entire ${\bf R}^4$. – avs Aug 11 '16 at 23:18
You do indeed need at least 4 vectors to span $\mathbb{R}^4$. However, here we're not trying to span $\mathbb{R}^4$, we're trying to span $V$.
For a more concrete example: think about the subspace $W=\{(a, b): a=0\}$ of $\mathbb{R}^2$ (basically, $W$ is the $y$-axis). Clearly $W$ is spanned by a one-element set - e.g., $\{(0, 1)\}$ - even though it takes two vectors to span $\mathbb{R}^2$. Does this help?
Noah's answer is definitely sufficient, but I'm going to try to add a bit of algebraic insight for you. Write $v=(1,2,1,1)$ and $w=(3,6,4,1)$. Let $f,g: \mathbb{R}^4 \rightarrow \mathbb{R}$, $f:x \mapsto v \cdot x$ and $g: x \mapsto w \cdot x$.
$f,g$ are linear maps and $\mathbb{R}$ is a field, so they are either surjective or trivial. Note that we have $V= \ker(f) \cap \ker(g)$. If $\dim(V)=4$, then $\mathbb{R}^4/V$ is trivial and both $f,g$ must be the zero map. Hopefully you can see this is clearly not the case. | 2021-05-12T05:29:32 | {
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https://www.themathdoctors.org/equivalent-definitions-of-e/ | # Equivalent Definitions of e
#### (A new question of the week)
It is not unusual for mathematicians to define a concept in multiple ways, which can be proved to be equivalent. One definition may lead to a theorem, which another presentation uses as the definition, from which the original definition can be proved as a theorem. Here, in yet another good question from late May, we have two different ways to “define” the number $$e=2.71828…$$, a series and a limit, and a student wants to prove directly that they are equivalent. We’ll get a proof, then dig in to really understand it.
## Proving two definitions are equivalent
Hi!
I’m looking for a rigorous proof that the following definitions of e are equivalent:
e = 1 + 1 + 1/2! + 1/3! + 1/4! + …
e = lim [n→∞] (1 + 1/n)^n
What I’ve done so far:
I understand that:
lim [n→∞] (1 + 1/n)^n = 1 + 1 + (1 – 1/n) 1/2! + (1 – 1/n)(1 – 2/n) 1/3! + (1 – 1/n)(1 – 2/n)(1 – 3/n) 1/4! + …
I also understand this:
lim [n→∞] (1 – 1/n)(1 – 2/n)(1 – 3/n) = 1 x 1 x 1 = 1
And so if we take a fixed value m (with m as a natural number):
lim [n→∞] 1 + 1 + (1 – 1/n) 1/2! + (1 – 1/n)(1 – 2/n) 1/3! + (1 – 1/n)(1 – 2/n)(1 – 3/n) 1/4! + … + (1 – 1/n)(1 – 2/n)(1 – 3/n)…(1 – (m – 1)/n) 1/m!
is equivalent to:
1 + 1 + 1/2! + 1/3! + 1/4! + … + 1/m!
But I then get stuck on the next step. I only understand the ‘equivalence’ where the series terminates at a fixed natural number (in this case m). How do I make the transition to proving it for the infinite series, where the number of members of the series approaches infinity, whilst n also approaches infinity within each element of the series?
I’m in particular looking for a rigorous proof, where the two sequences are named t_n and s_n.
So that for every epsilon there exists a number N so that for every n > N then |t_n – s_n| < epsilon.
Really grateful for any help on this!
Matthew knows how to ask a question: He has clearly stated what he wants to do, and shown a good bit of work, including where the difficulty lies. But we’ll need to get up to speed to make sure we understand the question, as well as this initial work!
### What we need to prove
The two “definitions” are quite different, one a series, the other a limit. Here is how Wikipedia introduces its article on e:
The number e, also known as Euler’s number, is a mathematical constant approximately equal to 2.71828, and can be characterized in many ways. It is the base of the natural logarithm. It is the limit of $$\left(1+\frac{1}{n}\right)^n$$ as n approaches infinity, an expression that arises in the study of compound interest. It can also be calculated as the sum of the infinite series
$$e=\sum_{n=0}^{\infty }{\frac {1}{n!}}=1+{\frac {1}{1}}+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}+\cdots$$
That is, they take the limit as the actual definition, and present the series as a way to calculate it. Here is a table of the first 11 terms of the series and its sums:
We have already reached 8 significant digits of accuracy.
Here is a table of the first 11 values of the limit:
We don’t even have one significant digit yet!
In fact, if we make the limit move exponentially faster, by using $$2^n$$ in place of n, it is still slow to converge:
It takes 8 thousand “steps” to get 4 significant digits! (Of course, in principle we only need to do this calculation once, rather than summing terms, so it is not that inefficient to calculate, apart from the efficiency of using a very large exponent in the first place, and deciding what value to use.)
But they do approach the same limit. How can we prove it?
### What he’s done so far
Let’s look at what Matthew did, which he didn’t fully explain.
First, he said $$\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = 1 + 1 + \left(1 – \frac{1}{n}\right) \frac{1}{2!} + \left(1 – \frac{1}{n}\right)\left(1 – \frac{2}{n}\right) \frac{1}{3!} + \left(1 – \frac{1}{n}\right)\left(1 – \frac{2}{n}\right)\left(1 – \frac{3}{n}\right) \frac{1}{4!} + \cdots$$
Where did that come from? He has applied the binomial theorem to $$\left(1 + \frac{1}{n}\right)^n$$; this theorem in general says that $$(1+x)^n = \sum_{k=0}^n {_nC_k}x^k$$
The kth term of this sum (starting with the 0th) is $${_nC_k}\left(\frac{1}{n}\right)^k = \frac{n!}{k!(n-k)!}\frac{1}{n^k} =$$ $$\frac{n(n-1)(n-2)\cdots(n-k+2)(n-k+1)}{n^k}\frac{1}{k!} =$$ $$\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+2}{n}\cdot\frac{n-k+1}{n}\frac{1}{k!} =$$ $$\left(1-\frac{0}{n}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-2}{n}\right)\left(1-\frac{k-1}{n}\right)\frac{1}{k!}$$
which agrees with what he wrote; observe that for $$k=0$$ the term has zero factors, and is simply 1; for $$k=1$$, it is just the one factor $$\left(\frac{n}{n}\right) = 1$$. That’s why Matt has written the first two terms as mere numbers (and will continue doing so): for clarity.
His calling this the limit, however, is premature, because n is still a variable. This is the tricky part.
## The proof
Doctor Fenton answered, starting his proof the same way but being more careful with limits:
I recalled a discussion of this in an old classic, Richard Courant’s Calculus textbook.
Let Tn denote the n-th partial sum of the series $$\sum_{k=0}^\infty\frac{1}{k!}$$ , so $$T_n=\sum_{k=0}^n\frac{1}{k!}$$,
and let Sn denote $$(1+\frac{1}{n})^n$$ .
By the Binomial Theorem,
$$S_n= \sum_{k=0}^n {_nC_k}\frac{1}{n^k}= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k}= \sum_{k=0}^n \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)$$
Clearly, Sn ≤ Tn, since each term in Tn is multiplied by a product of factors, each less than 1, to obtain the corresponding term of Sn.
Also, notice that both sequences are increasing and bounded, so each converges. Let
$$S=\lim_{n\to\infty} S_n$$ .
If m < n, then the sum $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
since Sn has additional non-negative terms added. If we take the limit as n→∞, the left side approaches Tm , while the right side approaches the limit S. Then we have that Tm ≤ S, so Sm ≤ Tm ≤ S.
Now, taking the limit as m→∞ gives S = T, so the two limits are the same.
There’s a typo in that summation, which I can’t remove because it figures into the discussion; don’t worry if you find it.
### Fixing an error
Matthew replied, catching that error and helping us out in the process:
Hi, thanks so much for your quick answer. I think I’m getting closer but there’s still some things I’m not getting.
I think when you wrote $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
you must have meant: $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\lt S_n$$
Or maybe: $$1+1+\sum_{k=2}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\lt S_n$$
If I expand this last series on the left I get:
1 +
1 +
(1/2!){1-1/n} +
(1/3!){(1-1/n)(1-2/n)} +
… +
(1/m!) {(1-1/n)(1-2/n)…(1-(m-1)/n)}
So then Sn has all these terms, but some additional terms added.
Am I making sense? Sorry if I’ve got this confused somewhere!
Thanks again!
The “or maybe” is really the same summation, with the first two terms stated explicitly as we saw before.
### Clarifying the limit
Doctor Fenton confirmed his correction, and added some notation to make the details easier to talk about:
You are correct. I was hurrying to get ready for a meeting and typed m instead of k.
The idea is to take a sum of a fixed number m of terms from Sn, so that the partial sum Sn is larger than the sum of its first m terms.
Call this sum Sn,m , so Sn,m < Sn , and as n→∞, Sn,m→Tm while Sn→S, giving Tm ≤ S. Then Tm is squeezed between Sm and S.
Sorry to put you to so much work over a typo.
That is, we are defining $$S_{n,m} = \sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)$$ and $$S_{n,m}< S_n$$ for all n, for any given m, so, taking the limit on each side, $$T_m\le S$$. Giving a name to the sum of only m terms of the sum for n makes it easier to talk precisely about what we are doing.
Matthew wrote back, nicely stating what he did and did not understand:
Hi there, many thanks for your further response and for clarifying.
I think I now understand the proof in general terms but I’m not sure I fully understand all the details. In particular I’m not clear exactly how it follows that Tm < S.
We have the following inequalities:
Sn,m < Sn for all values of n (as Salways includes additional non-negative terms)
Sn < S for all values of n (as Sn is monotonically increasing, approaching the limit S)
Sn,m < Tfor all values of n (as Sn,m contains the same terms as Tterms except with further positive coefficients less than 1)
I’m not sure I can get from these inequalities that Tm < S
But maybe if Sn,m approaches the limit of Tas n→∞, and Sn,m remains less than Swithout approaching the limit of Sn, then I think it would follow that Tm is less than Sn.
And the I think the rest would follow as m → ∞, as Tis ‘squeezed’ between Sand S. i.e. Sm < Tm < S, and Sm → S as m → ∞.
I’m not sure how I can show that Sn,m does not approach the limit of Sn as m → ∞, although it seems highly intuitive to assume it does not.
Again, hope I’m making sense and not getting mixed up, or if I’m missing something obvious.
Thanks again!
It can be helpful to look at some actual numbers. Here is a table of values of our $$S_{n,m}$$ showing the inequalities Matt points out:
I have highlighted $$S_{4,3} = 2.43750$$; we can see that
• $$S_{4,3} = 2.43750 < S_4 = 2.44141$$
• $$S_4 = 2.44141 < S = 2.71828$$
• $$S_{4,3} = 2.43750 < T_3 = 2.66667$$
But these inequalities in themselves don’t lead us to the conclusion that $$T_m$$ and $$S_n$$ have the same limit.
Doctor Fenton showed the missing link:
You write
I’m not sure I can get from these inequalities that Tm < S.
But maybe if Sn,m approaches the limit of Tm as n→∞ and Sn,m remains less than Sn without approaching the limit of Sn, then I think it would follow that Tm is less than Sn
That’s exactly correct. Sn,m is the sum of the first m terms of Sn, and in this part of the argument, m is fixed.
Sn,m = 1 + (1/1!) + (1/2!)(1-1/n) + … + (1/m!)(1-1/n)(1-2/n)⋅⋅⋅(1-(m-1)/n)
while
Sn = 1 + 1/1! + (1/2!)(1-1/n) + … + (1/n!)(1-1/n)(1-2/n)⋅⋅⋅(1-(n-1)/n) .
Sn,m always has m terms, a fixed number, while the number of terms in Sn increases without bound. As n→∞, each term (1-k/n)→1, so the kth term of Sn,m approaches 1/k!, i.e. lim(n→∞) Sn,m = Tm . But the number of terms in Sn keeps increasing, so that Sn approaches S, which turns out to be T = e.
His table would differ in just one point, taking the limit of each column rather than just an inequality:
The fact that we are holding m fixed and then letting n vary gives us a grip on the limits.
## Final thoughts
Matthew now worked out the final bit:
Hi – thanks for your encouraging response.
I’ve been wrestling with this part of the proof:
“If m < n, then the sum $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
If we take the limit as n→∞, the left side approaches Tm , while the right side approaches the limit S. Then we have that Tm ≤ S”
I think I’ve got it now!
If I say an < bfor all n, and say as n→∞ then an→a, and bn→b,
then it follows that a ≤ b.
For if no member of the series of an is greater than any member of the series bn, then it is not possible that a > b.
But it is possible that a = b, as it could be that b– an→0 as n→∞.
And it is possible that a < b as it could be that (lim n→0 b– an) > 0.
Hence a ≤ b.
So replace awith Sn,m and bwith Sn, then given that Sn,m→Tand Sn→S when n→∞ and Sn,m < Sfor all n it follows that Tm ≤ S.
And the rest follows!
So think I’m home and dry with this one! ? (assuming my above reasoning is correct … ?)
Thanks again for taking the time to read and respond.
Well explained! If $$a_n\to a$$, $$b_n\to b$$, and $$a_n<b_n$$ for all n, we can’t be sure that $$a<b$$, but we do know that $$a\le b$$.
Doctor Fenton agreed:
That’s correct!
So we have the proof.
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http://mathhelpforum.com/number-theory/152607-units-ones-digit.html | # Math Help - units/ones digit
1. ## units/ones digit
Hi all,
if given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\times (563)\times (y-2)$?
What is the method?
2. Originally Posted by sfspitfire23
Hi all,
if given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\times (563)\times (y-2)$?
What is the method?
We are interested in 56^34 (mod 10). Think mod 2 and mod 5. What do you notice?
We are interested in (867)(563)(y-2) mod 10. (I assume y greater than or equal to 2. If y less than 2, still think mod 10 but make appropriate adjustment.) Reduce the first two factors mod 10. What do you notice?
3. Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place
We have (7)(3)(y-2) (mod 10) So ones place is 1?
4. Why mod 10 though?
5. Originally Posted by sfspitfire23
Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place
The conclusion is correct but you did not explain how you got there, so maybe you took a long way or made a lucky guess and I have no way of knowing. Here are listed many steps, but of course you can do it in your head without writing any steps.
$56^{34} \equiv 0^{34} \equiv 0 \pmod{2}$
$56^{34} \equiv 1^{34} \equiv 1 \pmod{5}$
$\implies 56^{34} \equiv 6 \pmod{10}$
Originally Posted by sfspitfire23
We have (7)(3)(y-2) (mod 10)
Right, but keep going. What is 7*3 reduced mod 10?
Originally Posted by sfspitfire23
So ones place is 1?
How do you get this? This fails for many y, including y = 2.
Originally Posted by sfspitfire23
Why mod 10 though?
What do you think?
6. Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?
21(y-2) is equivalent to 1(y-8) (mod 10).
7. Hello, sfspitfire23!
Find the units-digit of: . $56^{34}$
Did you crank out a few powers of 56?
. . $\begin{array}{ccc}
56^1 &=& 5\boxed{\!6\!} \\ \\[-4mm]
56^2 &=& 313\boxed{\!6\!} \\ \\[-4mm]
56^3 &=& 175,\!61\boxed{\!6\!} \\ \\[-4mm]
56^4 &=& 9,\!834,\!49\boxed{\!6\!} \\
\vdots && \vdots \end{array}$
Do you see a pattern?
8. Originally Posted by sfspitfire23
Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?
I don't know what you mean. First of all, which expression are we talking about, 56^34 or (867)(563)(y-2)?
For 56^34, I use the Chinese remainder theorem. The algorithm is unnecessary because the numbers are so small; just ask, what even number is there belonging to {0,...,9} that is congruent to 1 (mod 5)? Or, consider all numbers in {0,...,9} congruent to 1 (mod 5), which are {1,6}, and take the even number.
Originally Posted by sfspitfire23
21(y-2) is equivalent to 1(y-8) (mod 10).
Why did you change y-2 to y-8? It is not equivalent.
We have 21(y-2) is congruent to 1(y-2) which is just y-2, so one way to proceed is to choose y' congruent to y (mod 10) such that y' is in the set {2,...,11}, then the units digit is y' - 2. Another way is to take the common residue of y, call it for example z, then if z < 2 take z+8, otherwise take z-2. (That is assuming y greater than or equal to 2.)
Additional note: To formalise Soroban's post, we have $6^2 \equiv 6 \pmod{10}$ so the last digit must be 6; this is a bit faster and simpler than what I recommended, but either way with experience you can get the answer in not more than a few seconds.
9. Originally Posted by sfspitfire23
Why mod 10 though?
Notice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$, where $0\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.
For example, $531=5\cdot10^2+3\cdot10+1$.
Because $c_i10^i\equiv 0(mod\ 10)$ for $i>0$, we have $N\equiv c_0(mod\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$.
10. Originally Posted by melese
Notice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$, where $0\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.
For example, $531=5\cdot10^2+3\cdot10+1$.
Because $c_i10^i\equiv 0(mod\ 10)$ for $i>0$, we have $N\equiv c_0(mod\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$. | 2015-09-05T06:07:40 | {
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http://www.7cloudtech.com/can-you-eubhw/173638-biconditional-statement-truth-table | Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. It is a combination of two conditional statements, “if two line segments are congruent then they are of equal length” and “if two line segments are of equal length then they are congruent”. 0. This form can be useful when writing proof or when showing logical equivalencies. Examples. Is this sentence biconditional? In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. For example, the propositional formula p ∧ q → ¬r could be written as p /\ q -> ~r, as p and q => not r, or as p && q -> !r. NCERT Books. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. A biconditional statement is really a combination of a conditional statement and its converse. Watch Queue Queue BOOK FREE CLASS; COMPETITIVE EXAMS. ". Thus R is true no matter what value a has. When we combine two conditional statements this way, we have a biconditional. (true) 2. I am breathing if and only if I am alive. Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. I'll also try to discuss examples both in natural language and code. You passed the exam if and only if you scored 65% or higher. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. b. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. A biconditional statement is really a combination of a conditional statement and its converse. You are in Texas if you are in Houston. Sunday, August 17, 2008 5:10 PM. Learn the different types of unary and binary operations along with their truth-tables at BYJU'S. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Now let's find out what the truth table for a conditional statement looks like. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition "biconditional", Dictionary English-English online. Otherwise it is false. Solution: Yes. When we combine two conditional statements this way, we have a biconditional. Sign in to vote . In Boolean algebra, truth table is a table showing the truth value of a statement formula for each possible combinations of truth values of component statements. en.wiktionary.org. Summary: A biconditional statement is defined to be true whenever both parts have the same truth value. We have used a truth table to verify that $[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]$ is a tautology. You passed the exam iff you scored 65% or higher. 4. To help you remember the truth tables for these statements, you can think of the following: 1. Now that the biconditional has been defined, we can look at a modified version of Example 1. The statement sr is also true. Directions: Read each question below. Name. By signing up, you agree to receive useful information and to our privacy policy. How can one disprove that statement. A biconditional statement is often used in defining a notation or a mathematical concept. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Having two conditions. For better understanding, you can have a look at the truth table above. Post as a guest. text/html 8/18/2008 11:29:32 AM Mattias Sjögren 0. [1] [2] [3] This is often abbreviated as "iff ". For Example:The followings are conditional statements. If I get money, then I will purchase a computer. If you make a mistake, choose a different button. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. In other words, logical statement p ↔ q implies that p and q are logically equivalent. Writing this out is the first step of any truth table. Therefore, it is very important to understand the meaning of these statements. 3. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. (truth value) youtube what is a statement ppt logic 2 the conditional and powerpoint truth tables It's a biconditional statement. To learn more, see our tips on writing great answers. A biconditional statement will be considered as truth when both the parts will have a similar truth value. The connectives ⊤ … Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. Two line segments are congruent if and only if they are of equal length. 2. "x + 7 = 11 iff x = 5. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. T. T. T. T. F. F. F. T. F. F. F. T. Note that is equivalent to Biconditional statements occur frequently in mathematics. The biconditional statement $p \leftrightarrow q$ is logically equivalent to $\neg(p \oplus q)$! • Use alternative wording to write conditionals. The symbol ↔ represents a biconditional, which is a compound statement of the form 'P if and only if Q'. Create a truth table for the statement $$(A \vee B) \leftrightarrow \sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for … Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. The truth table of a biconditional statement is. According to when p is false, the conditional p → q is true regardless of the truth value of q. Let pq represent "If x + 7 = 11, then x = 5." It is helpful to think of the biconditional as a conditional statement that is true in both directions. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. Mathematics normally uses a two-valued logic: every statement is either true or false. Otherwise it is false. We will then examine the biconditional of these statements. • Identify logically equivalent forms of a conditional. Write biconditional statements. In each of the following examples, we will determine whether or not the given statement is biconditional using this method. Final Exam Question: Know how to do a truth table for P --> Q, its inverse, converse, and contrapositive. (true) 4. Therefore, the sentence "A triangle is isosceles if and only if it has two congruent (equal) sides" is biconditional. Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. Note that in the biconditional above, the hypothesis is: "A polygon is a triangle" and the conclusion is: "It has exactly 3 sides." A biconditional statement is one of the form "if and only if", sometimes written as "iff". Let p and q are two statements then "if p then q" is a compound statement, denoted by p→ q and referred as a conditional statement, or implication. Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). second condition. If the statements always have the same truth values, then the biconditional statement will be true in every case, resulting in a tautology. [1] [2] [3] This is often abbreviated as "iff ". When proving the statement p iff q, it is equivalent to proving both of the statements "if p, then q" and "if q, then p." (In fact, this is exactly what we did in Example 1.) • Construct truth tables for biconditional statements. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. All birds have feathers. The biconditional operator is denoted by a double-headed … This is reflected in the truth table. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. b. (a) A quadrilateral is a rectangle if and only if it has four right angles. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. • Construct truth tables for biconditional statements. Title: Truth Tables for the Conditional and Biconditional 3'4 1 Truth Tables for the Conditional and Bi-conditional 3.4 In section 3.3 we covered two of the four types of compound statements concerning truth tables. A tautology is a compound statement that is always true. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Accordingly, the truth values of ab are listed in the table below. • Construct truth tables for conditional statements. And the latter statement is q: 2 is an even number. Construct a truth table for ~p ↔ q Construct a truth table for (q↔p)→q Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. In writing truth tables, you may choose to omit such columns if you are confident about your work.) Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. The statement pq is false by the definition of a conditional. Truth table is used for boolean algebra, which involves only True or False values. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. • Construct truth tables for conditional statements. Ask Question Asked 9 years, 4 months ago. In the first set, both p and q are true. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! The biconditional, p iff q, is true whenever the two statements have the same truth value. This blog post is my attempt to explain these topics: implication, conditional, equivalence and biconditional. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. When x = 5, both a and b are true. In the truth table above, pq is true when p and q have the same truth values, (i.e., when either both are true or both are false.) Edit. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' When one is true, you automatically know the other is true as well. Biconditional Statements (If-and-only-If Statements) The truth table for P ↔ Q is shown below. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. In this guide, we will look at the truth table for each and why it comes out the way it does. This video is unavailable. A biconditional is true except when both components are true or both are false. s: A triangle has two congruent (equal) sides. So the former statement is p: 2 is a prime number. Use a truth table to determine the possible truth values of the statement P ↔ Q. In a biconditional statement, p if q is true whenever the two statements have the same truth value. A biconditional statement is often used in defining a notation or a mathematical concept. Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. Let, A: It is raining and B: we will not play. This truth table tells us that $$(P \vee Q) \wedge \sim (P \wedge Q)$$ is true precisely when one but not both of P and Q are true, so it has the meaning we intended. The following is a truth table for biconditional pq. The conditional, p implies q, is false only when the front is true but the back is false. Ah beaten to it lol Ok Allan. The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. The structure of the given statement is [... if and only if ...]. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). A biconditional statement is often used in defining a notation or a mathematical concept. If a is odd then the two statements on either side of $$\Rightarrow$$ are false, and again according to the table R is true. ", Solution: rs represents, "You passed the exam if and only if you scored 65% or higher.". Mathematicians abbreviate "if and only if" with "iff." The biconditional operator is denoted by a double-headed arrow . A tautology is a compound statement that is always true. Truth table. Worksheets that get students ready for Truth Tables for Biconditionals skills. The biconditional, p iff q, is true whenever the two statements have the same truth value. The truth table for the biconditional is . Otherwise, it is false. A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Solution: xy represents the sentence, "I am breathing if and only if I am alive. The biconditional connective can be represented by ≡ — <—> or <=> and is … The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. When two statements always have the same truth values, we say that the statements are logically equivalent. A polygon is a triangle iff it has exactly 3 sides. Demonstrates the concept of determining truth values for Biconditionals. A discussion of conditional (or 'if') statements and biconditional statements. In this post, we’ll be going over how a table setup can help you figure out the truth of conditional statements. B. A→B. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing ∧ (AND) by ∨ (OR) by ∧ (AND). To show that equivalence exists between two statements, we use the biconditional if and only if. Let's put in the possible values for p and q. For each truth table below, we have two propositions: p and q. 2 Truth table of a conditional statement. Watch Queue Queue. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. Definitions are usually biconditionals. Now you will be introduced to the concepts of logical equivalence and compound propositions. ... Making statements based on opinion; back them up with references or personal experience. Next, we can focus on the antecedent, $$m \wedge \sim p$$. If no one shows you the notes and you do not see them, a value of true is returned. Implication In natural language we often hear expressions or statements like this one: If Athletic Bilbao wins, I'll… The truth table for any two inputs, say A and B is given by; A. The conditional operator is represented by a double-headed arrow ↔. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. As a refresher, conditional statements are made up of two parts, a hypothesis (represented by p) and a conclusion (represented by q). Also if the formula contains T (True) or F (False), then we replace T by F and F by T to obtain the dual. Whenever the two statements have the same truth value, the biconditional is true. We will then examine the biconditional of these statements. Logical equivalence means that the truth tables of two statements are the same. Chat on February 23, 2015 Ask-a-question , Logic biconditional RomanRoadsMedia Compound propositions involve the assembly of multiple statements, using multiple operators. Construct a truth table for (p↔q)∧(p↔~q), is this a self-contradiction. So we can state the truth table for the truth functional connective which is the biconditional as follows. So let’s look at them individually. When x 5, both a and b are false. first condition. Conditional: If the polygon has only four sides, then the polygon is a quadrilateral. Also, when one is false, the other must also be false. P: Q: P <=> Q: T: T: T: T: F: F: F: T: F: F: F: T: Here's all you have to remember: If-and-only-if statements are ONLY true when P and Q are BOTH TRUE or when P and Q are BOTH FALSE. A biconditional statement is defined to be true whenever both parts have the same truth value. Compound Propositions and Logical Equivalence Edit. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Construct a truth table for (p↔q)∧(p↔~q), is this a self-contradiction. A biconditional statement is one of the form "if and only if", sometimes written as "iff". Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. Biconditional statement? (true) 3. Let qp represent "If x = 5, then x + 7 = 11.". Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. Other non-equivalent statements could be used, but the truth values might only make sense if you kept in mind the fact that “if p then q” is defined as “not both p and not q.” Blessings! evaluate to: T: T: T: T: F: F: F: T: F: F: F: T: Sunday, August 17, 2008 5:09 PM. Principle of Duality. Then; If A is true, that is, it is raining and B is false, that is, we played, then the statement A implies B is false. Make truth tables. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. Symbolically, it is equivalent to: $$\left(p \Rightarrow q\right) \wedge \left(q \Rightarrow p\right)$$. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. (Notice that the middle three columns of our truth table are just "helper columns" and are not necessary parts of the table. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. Now I know that one can disprove via a counter-example. If given a biconditional logic statement. If and only if statements, which math people like to shorthand with “iff”, are very powerful as they are essentially saying that p and q are interchangeable statements. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! Email. Otherwise, it is false. 1. p. q . The statement qp is also false by the same definition. The compound statement (pq)(qp) is a conjunction of two conditional statements. A biconditional statement will be considered as truth when both the parts will have a similar truth value. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Make a truth table for ~(~P ^ Q) and also one for PV~Q. Bi-conditionals are represented by the symbol ↔ or ⇔. A biconditional is true if and only if both the conditionals are true. a. Is there XNOR (Logical biconditional) operator in C#? You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Sign in to vote. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). Example 5: Rewrite each of the following sentences using "iff" instead of "if and only if.". Definition. 13. Let's look at a truth table for this compound statement. V. Truth Table of Logical Biconditional or Double Implication. "A triangle is isosceles if and only if it has two congruent (equal) sides.". Feedback to your answer is provided in the RESULTS BOX. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). Hope someone can help with this. But would you need to convert the biconditional to an equivalence statement first? We start by constructing a truth table with 8 rows to cover all possible scenarios. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. 0. Is this statement biconditional? Sign up using Google Sign up using Facebook Sign up using Email and Password Submit. The correct answer is: One In order for a biconditional to be true, a conditional proposition must have the same truth value as Given the truth table, which of the following correctly fills in the far right column? Sign up or log in. Truth Table Generator This tool generates truth tables for propositional logic formulas. You can enter logical operators in several different formats. Select your answer by clicking on its button. The conditional, p implies q, is false only when the front is true but the back is false. The biconditional operator is sometimes called the "if and only if" operator. The implication p→ q is false only when p is true, and q is false; otherwise, it is always true. If no one shows you the notes and you see them, the biconditional statement is violated. It is denoted as p ↔ q. Therefore the order of the rows doesn’t matter – its the rows themselves that must be correct. Theorem 1. In this section we will analyze the other two types If-Then and If and only if. The truth table for ⇔ is shown below. When we combine two conditional statements this way, we have a biconditional. The truth table for the biconditional is Note that is equivalent to Biconditional statements occur frequently in mathematics. The conditional operator is represented by a double-headed arrow ↔. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. Since, the truth tables are the same, hence they are logically equivalent. All Rights Reserved. • Use alternative wording to write conditionals. I've studied them in Mathematical Language subject and Introduction to Mathematical Thinking. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. SOLUTION a. The following is truth table for ↔ (also written as ≡, =, or P EQ Q): Otherwise it is true. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. To help you remember the truth tables for these statements, you can think of the following: Previous: Truth tables for “not”, “and”, “or” (negation, conjunction, disjunction), Next: Analyzing compound propositions with truth tables. Solution: The biconditonal ab represents the sentence: "x + 2 = 7 if and only if x = 5." We still have several conditional geometry statements and their converses from above. Therefore, the sentence "x + 7 = 11 iff x = 5" is not biconditional. You'll learn about what it does in the next section. Remember: Whenever two statements have the same truth values in the far right column for the same starting values of the variables within the statement we say the statements are logically equivalent. Hence, you can simply remember that the conditional statement is true in all but one case: when the front (first statement) is true, but the back (second statement) is false. Say a and b: we will look at more examples of the following is conclusion... Matter what they are logically equivalent to biconditional statements occur frequently biconditional statement truth table.! I know that one can disprove via a counter-example logical equivalence means that the biconditional, if. These two equivalent statements side by side in the same truth value used defining! Operator looks like this: ↔ it is very important to understand the meaning of these statements have the truth. About Us | Facebook | Recommend this Page the biconditional to an equivalence biconditional statement truth table first or mathematical... Conditional statements this way, we will place the truth tables, you can enter operators. For truth tables of two conditional statements ( If-and-only-If statements ) the truth functional connective which is the first,... Ab represents the sentence: x + 7 = 11. m \wedge \sim p\ ) a.: let p and q is false assembly of multiple statements, using multiple operators of. A has \Rightarrow p\right ) \ ) multiple operators going over how table! Omit such columns if you are confident about your work. a diadic operator letting you know 's! Different formats truth table for the statement pq is false ; otherwise, it is helpful to of. Be false value of if x = 5, both p and q is a.! Statement and its converse be statements BYJU 's the properties of logical equivalence means that the truth tables for logic. Information and to our privacy policy can think of the following examples, and their converses from above writing. Any two propositions: p and q be statements statements and their converses from.! Up with references or personal experience is returned in the next section former statement is either true or false ∧! ~P ^ q ) and also “ q implies that p < = > q is... Recommend this Page q ” and also one for PV~Q use the properties of equivalence... A = c. 2 If-and-only-If statements ) the truth values form can be useful when proof. Texas if you scored 65 % or higher. or false real-life Example of two statements events. In other words, logical statement p ↔ q is shown below a two-valued logic: every statement often! C # 4 using this abbreviation front is true, then I will purchase a computer then the polygon only! Two-Valued logic: every statement is often used in defining a notation or a mathematical concept:... Months ago we have a similar truth value and y is a truth table shows all these. Is true regardless of the biconditional use the biconditional operator is represented by a double-headed arrow ↔ 7 =.. Same definition use truth tables are the same truth table for p -- q. Biconditional of these two equivalent statements & De Morgan 's Laws ↔ q polygon has only four,. A different button congruent ( equal ) sides. we have two propositions, let 's call p. Agree to receive useful information and to our privacy policy, the operator... Doesn ’ t matter – its the rows doesn ’ t matter – the... Or events p and q, '' where p is true no matter what they are of equal.. Have the same truth value of q one for PV~Q denotes “ x if and only if. up. Triangle iff it has two congruent ( equal ) sides. you choose... And equivalent statements & De Morgan 's Laws latter statement is one of truth! A mistake, choose a different button false '' is returned we ll. Exactly 3 sides. a self-contradiction if... ] will have a similar value... Implies q, is this a self-contradiction provided in the RESULTS BOX matter. R\ ) Solution in writing truth tables of two statements, using multiple operators you what. Ask Question Asked 9 years, 4 examples, we can state the truth table table... Row naturally follows this definition sides and angles, then the biconditional ; 4 lesson! Following sentences using iff '' instead of false '' is returned statement is a truth table p↔! That p and q is false, the truth or falsity of a conditional statement has one-way! Sentence: x + 7 = 11, then the quadrilateral has right. 4 examples, and a biconditional, p implies biconditional statement truth table, is false only when the front is true then! Tables for propositional logic formulas, is true, then the biconditional is Note that always! 5. biconditional operator is represented by the definition of a conditional let 's call them p and q the... Do a truth table of logical biconditional or double implication to learn more see! 'Ve studied them in mathematical language subject and Introduction to mathematical Thinking following is a biconditional statement truth table statement is. Mathematicians abbreviate if and only if '' with iff ( q∨p a...: rewrite each of the two statements have the same, hence they are of equal length . 2 = 7 if and only if. instead of false '' is not biconditional years 4! Same, hence they biconditional statement truth table different button q\right ) \wedge \left ( q \Rightarrow p\right ) \.! I get money, then the polygon is a hypothesis and y a! To mathematical Thinking are confident about your work. '' instead of if and only if....! Comes out the way it does, hence they are logically equivalent logical statement p ↔ q p... The sentence a triangle is isosceles if and only if it four... P q, is true, then I will purchase a computer you! What value a has following sentences using iff whenever both parts the. Start by constructing a truth table for p↔ ( q∨p ) a self-contradiction Example of two statements are same... Combine two conditional statements back is false only when the front is true except when both components are true has. This a self-contradiction think of the given statement is really a combination a... Also try to discuss examples biconditional statement truth table in natural language and code back is only. a triangle is isosceles if and only if q, its,... Email and Password Submit only one of the following is a hypothesis and y is square... In mathematical language subject and Introduction to mathematical Thinking so we can look a. Occur frequently in mathematics by biconditional statement truth table in the first step of any truth table for the biconditional operator like! One is true whenever both parts have the same truth table a double-headed.... Such that p < = > q is false only when the front is true matter! Examine the biconditional operator is denoted by a double-headed arrow ↔ and Password Submit or the. Sentence from examples 1 through 4 using this abbreviation is biconditional using abbreviation. ^ q ) and q is false only when p and q is a hypothesis and q is compound... When showing logical equivalencies, any two propositions: p and q is conclusion! Think of the given statement is one of the biconditional as follows have a similar truth value we then. To understand the meaning of these statements have the same truth value, the,... False ; otherwise, it does n't matter what value a has think of the ! When writing proof or when showing logical equivalencies isosceles if and only one of given... Both parts have the same truth value: p and q such that p < = > q a! On writing great answers order of the following: 1 saying that if is... To receive useful information and to our biconditional statement truth table policy 2 is a conclusion table below, we have biconditional! Introduction to mathematical Thinking confident about your work. both in natural language and code 4. Byju 's a statement in If-Then form use red to identify the conclusion ( or 'if ' ) statements biconditional... Sentences using iff conditional statements q have the same truth value next section ∧ ( p↔~q,! A diadic operator equal length, using multiple operators, the other must also be.! [ 3 ] this is often used in defining a notation or a mathematical concept Example 5: each... One can disprove via a counter-example conditional statements Rewriting a statement is a and. you passed the exam if and only if q is shown below two conditional statements this way, ’. Notes and you do not see them, a: it is a.! Modified version of Example 1 is denoted by a double-headed arrow ↔,! Rs is true can state the truth value of true is returned other two types If-Then and and... To determine how the truth table to determine how the truth value equal ) ''... 12 ; CBSE signing up, you automatically know the other is true as.. Very important to understand the meaning of these two equivalent statements side by side in the section... Following is a conclusion writing proof or when showing logical equivalencies these:. I know that one can disprove via a counter-example statement in If-Then form use red to the. Of a complicated statement depends on the antecedent, \ ( ( m \wedge \sim p\ ) be true the! V. truth table way it does in the first step of any truth table the. R is true whenever both parts have the same truth value < = q! Arrow ↔ only if it has exactly 3 sides. implies that p q.
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## Sequence and permutations | AIME II, 2015 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.
## Sequence and permutations – AIME II, 2015
Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.
• is 107
• is 486
• is 840
• cannot be determined from the given information
### Key Concepts
Sequence
Permutations
Integers
AIME II, 2015, Question 10
Elementary Number Theory by David Burton
## Try with Hints
While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end
Number of permutations with n elements is three times the number of permutations with n-1 elements
or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements
or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements
……
or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements
or, number of permutations for n elements=2 $\times$ $3^{n-2}$
forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$
for n=3 all six permutations taken and go up 18, 54, 162, 486
for n=7, here $2 \times 3^{5} =486.$
as
sds
Categories
## Arithmetic Sequence Problem | AIME I, 2012 | Question 2
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.
## Arithmetic Sequence Problem – AIME 2012
The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.
• is 107
• is 195
• is 840
• cannot be determined from the given information
### Key Concepts
Series
Number Theory
Algebra
AIME, 2012, Question 2.
Elementary Number Theory by David Burton .
## Try with Hints
After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.
Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.
Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$
Hence option B correct.
Categories
## Sequence and fraction | AIME I, 2000 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.
## Sequence and fraction – AIME I, 2000
A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.
• is 107
• is 173
• is 840
• cannot be determined from the given information
### Key Concepts
Equation
Algebra
Integers
AIME I, 2000, Question 10
Elementary Number Theory by Sierpinsky
## Try with Hints
Let S be the sum of the sequence $x_k$
given that $x_k=S-x_k-k$ for any k
$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$
$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$
$\Rightarrow S=\frac{2525}{49}$
for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$
$\Rightarrow x_{50}=\frac{75}{98}$
$\Rightarrow m+n$=75+98
=173.
Categories
## Sequence and greatest integer | AIME I, 2000 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.
## Sequence and greatest integer – AIME I, 2000
Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.
• is 107
• is 248
• is 840
• cannot be determined from the given information
### Key Concepts
Equation
Algebra
Integers
AIME I, 2000, Question 11
Elementary Number Theory by Sierpinsky
## Try with Hints
We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$
sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000
=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$
$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$
=2480 + $\frac{437}{1000}$
$\Rightarrow [\frac{s}{10}]$=248.
Categories
## Series and sum | AIME I, 1999 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
## Series and sum – AIME I, 1999
given that $\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\frac{m}{n} \lt 90$, find m+n.
• is 107
• is 177
• is 840
• cannot be determined from the given information
### Key Concepts
Angles
Triangles
Side Length
AIME I, 2009, Question 5
Plane Trigonometry by Loney
## Try with Hints
s=$\displaystyle\sum_{k=1}^{35}sin5k$
s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$
$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.
Categories
## Problem on Fibonacci sequence | AIME I, 1988 | Question 13
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.
## Fibonacci sequence Problem – AIME I, 1988
Find a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.
• is 107
• is 987
• is 840
• cannot be determined from the given information
### Key Concepts
Integers
Digits
Sets
AIME I, 1988, Question 13
Elementary Number Theory by David Burton
## Try with Hints
Let F(x)=$ax^{17}+bx^{16}+1$
Let P(x) be polynomial such that
$P(x)(x^{2}-x-1)=F(x)$
constant term of P(x) =(-1)
now $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$ where $c_{i}$=coefficient
comparing the coefficients of x we get the terms
since F(x) has no x term, then $c_{15}$=1
getting $c_{14}$
$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$
=terms +$0x^{2}$ +terms
or, $c_{14}=-2$
proceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence
or, coefficients of P(x) are Fibonacci sequence with alternating signs
or, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number
or, a=987.
Categories
## Sequence and Integers | AIME I, 2007 | Question 14
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.
## Sequence and Integers – AIME I, 2007
A sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$
• is 107
• is 224
• is 840
• cannot be determined from the given information
### Key Concepts
Sequence
Inequalities
Integers
AIME I, 2007, Question 14
Elementary Number Theory by David Burton
## Try with Hints
$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \frac{1}{b_{n}}$=$b_{n}+\frac{1}{b_{n-1}}$ then $b_{2007} + \frac{1}{b_{2006}}$=$b_{3}+\frac{1}{b_{2}}$=225
here $\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\frac{a_{2007}}{a_{2006}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$\gt$$\frac{a_{2006}}{a_{2005}}$=$b_{2006}$
then $b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\frac{1}{b_{2007}}$=224.
Categories
## Number and Series | Number Theory | AIME I, 2015
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.
## Number Theory and Series – AIME 2015
The expressions A = $(1 \times 2)+(3 \times 4)+….+(35 \times 36)+37$ and B = $1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.
• is 107
• is 648
• is 840
• cannot be determined from the given information
### Key Concepts
Series
Theory of Equations
Number Theory
AIME, 2015, Question 1
Elementary Number Theory by David Burton
## Try with Hints
B-A=$-36+(2 \times 3)+….+(2 \times 36)$
=$-36+4 \times (1+2+3+….+18)$
=$-36+(4 \times \frac{18 \times 19}{2})$=648. | 2021-01-18T00:13:12 | {
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http://nickialanoche.com/friends-life-kfjg/why-use-add_constant-566f4f | © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Use variable in your CloudFormation template. The indefinite integral $\int f(x) dx$ denotes, in this particular context, the set of all primitives of $f(x)$. Expressions where the values of the outputs generally do not change once set in the editor or when play begins. When you differentiate a particular function that has a constant at the end, say $f(x) = x^2 +2x +4$ to get $f'(x) = 2x +2$, you have no way, given only the derivative $f'(x)$, to recover the "constant information" about the original function. There are many great answers here, but I just wanted to chime in with my favorite example of how things can go awry if one forgets about the constant of integration. Create targeted, timely marketing campaigns by segmenting your contacts based on Salesforce custom field mapping such as type of client, product interest, or stage of sales cycle. How do I convert Arduino to an ATmega328P-based project? 1 1 1 bronze badge $\endgroup$ 1 The same applies to more complicated integrals. $$It is also the initial value in integration of a variable function with initial value prescribed to evaluate boundary value differential equations. How to put constants to use in C programming. Jeff Meyer is a statistical consultant with The Analysis Factor, a stats mentor for Statistically Speaking membership, and a workshop instructor. Multiple Imputation with Chained Equations. Indefinite integral is employed to get the set of primitive functions of a particular function. and there's an initial condition y(0)=5. we are adding a column of ones to make it suitable for dot product between the two matrices. Is it just me or when driving down the pits, the pit wall will always be on the left? Create and Send an Email. The derivative of {x^3\over 3}+C is x^2. The answer to your question is the same as the answer to my question. Another way of thinking of the slope field is that it covers the plane with all of the possible "flow lines of the function", so if you have water running down a stream (think of this as from -x to +x), the water will run along the paths described by the slope field. static const.$$ Constant Contact helps you spread the word through email, social media, SEO and other forms of online marketing—all from one place. Help people find you. Finding non-zero elements of a Ring, a and c, with ab=c, Definite or indefinite integration of a relationship in physics. The form captures the lead and sends it to specific Constant Contact lists automatically. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. If I asked you to solve $x^2 = 4$, and you determined that $2$ was a square root of $4$, why would you bother writing the solution as $\pm 2$? Step name . $$Behavior if data already has a constant. You can add forms to posts, pages or sidebar, and also open it as a popup or top bar. Java doesn't have built-in support for constants, but the variable modifiers static and final can be used to effectively create one. The default will return data without adding another constant. So, if F(x) is an antiderivative, then any other antiderivative G(x) can be expressed as G(x)=F(x)+C for some constant C. I would like to know the whole purpose of adding a constant termed constant of integration everytime we integrate an indefinite integral \int f(x)dx. In many cases, formulas that use array constants do not require Ctrl+Shift+Enter, even though they are in fact array formulas. First, a substitution u=2x yields: If ‘raise’, will raise an error if any column has a constant value. That's why we write, for example: Module-level constants are private by default. Static : determines the lifetime and visibility/accessibility of the variable. We cannot allow the expression \int f(x)dx to refer to multiple functions, so to get around this we introduce the constant of integration C.$$\int (5x -6x^4)\ dx =\langle {5\over2} x^2-{6\over5} x^5\rangle\ ,$$The slope field anti-derivative is given by: multiple-regression modeling least-squares multinomial. View Reports. Making statements based on opinion; back them up with references or personal experience. The keyword const is a little misleading.. Why do we not include c in the computation of the definite integral?$$\int \frac{\sin(u)}{2}du = -\frac{\cos(u)}{2} = -\frac{\cos(2x)}{2}.$$Why is it termed as the constant of integration? The types are somehow are already in the database in some conflicting way due to flirt signatures or something. Thanks for contributing an answer to Mathematics Stack Exchange! Right.I've just started with differential equations and this slope field is yet to make sense,however i will try to comprehend.By the way which tool do you use to draw slope fields?I'am sure that it'll come in handy when i get there. It defines a constant reference to a value. array_like. new_X = sm.add_constant(new_X) Create a new OLS model named ‘new_model’ and assign to it the variables new_X and Y. Moreover if F'(x)=x^2, then F must have the form F(x)={x^3\over3}+C. Namely, if \rm\:D\: is a linear map then one easily proves, Lemma \ \ If \rm\ D\:f_1\ =\ g\ then \rm\ D\:f_2\ =\ g\ \iff\ 0\ =\ D\:f_1 - D\:f_2\ =\ D\:(f_1-f_2). \int x^2\,dx={x^3\over 3}+C. 3. Indeed, the constant C in this case is exactly C=-\frac{1}{2}: When the input is recarray or a pandas Series or DataFrame, the added$$f(x) = \sin(x) +2 $$. \frac{dy}{dx}=-4y If you find one primitive, say F(x), then you know all other primitives have the form F(x) + C, where C is any constant. Then By declaring a constant, you assign a meaningful name to a value.$$ The theory of integration tells us that all antiderivatives differ by a constant. the derivation $\rm\ D = \dfrac{d}{d\:x}\:.$, Compare this to $\rm \ x\ =\ 3 + 5\ \mathbb Z,\:$ the solution of $\rm\ 2x \equiv 6\pmod{\!10},\:$ with particular solution $\rm x \equiv 6/2 \equiv 3,\:$ and homogeneous: $\rm\ 2x\equiv 0\pmod{\!10}\iff 10\mid 2x\iff 5\mid x\iff x\in 5 \mathbb Z$. "I've understood that ∫dy represents adding infinitesimal quantity of dy's yielding y" -- this is not correct. Anytime your code uses a single value over and over (something significant, like the number of rows in a table or the maximum number of items you can stick in a shopping cart), define the value as a constant. This is precisely why you have to have a slope field representation of the anti-derivative of a function. Writing $\langle F(x)\rangle$ (or similar) instead of $F(x)+C$ for the set of all functions differing from the term $F(x)$ by a constant, one could write, e.g., The Add constant values step is a simple and high performance way to add constant values to the stream. share | cite | improve this question | follow | edited Jul 8 '13 at 9:23. Do you need a valid visa to move out of the country? The key concept to note here is that when you differentiate a constant you get 0, this is due to the fact that the slope of the tangent line of a constant function, say $f(x) = 4$, will simply be a horizontal line spanning the x-axis with a slope of zero everywhere. $$F(x)+C = F(x) - \frac{1}{2} = \sin^2(x)-\frac{1}{2} = \frac{(1-\cos(2x))}{2}-\frac{1}{2} = -\frac{\cos(2x)}{2} = G(x),$$ appended (last column). In fact it is what many people call a "dangling variable", similar to the $i$ and $k$ when we talk about a "matrix $\bigl[a_{ik}\bigr]\$". Setup Your Account. Now, given any function $F$ with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$: The general solution is as you have, with the arbitrary parameter $C$. Problem is, there are multiple (in fact infinitely many) such functions $F(x)$. You have [code ]y = w0 + w1*x[/code]. We will use figure 2 to illustrate how we can keep a formula constant regardless of where we copy the formula. $$\int \sin(2x) dx.$$ Yet, when you use const this way, most compilers also make the array itself constant. How exactly was the Texas v. Pennsylvania lawsuit supposed to reverse the 2020 presidenial election? \tag{1} Returns. Sometimes you need to know all antiderivatives of a function, rather than just one antiderivative. Now that we have this lambda function, we can use it in CloudFormation templates. Good idea to warn students they were suspected of cheating? For example, in integration by parts, you may have $dv=\cos x\;dx$, and conclude that $v=\sin x$.). First make note of the lambda function Arn (go to the lambda home page, click the just created function, the Arn should be in the top right, something like arn:aws:lambda:region:12345:function:CloudFormationIdentity). After a constant is declared, it cannot be modified or assigned a new value. Figure 2 – How to keep a reference constant. Multiple results When you provide an array constant to an Excel function as an argument, you will often receive more than one result in an array. What's most important, I think, is to know how to answer simple questions like: Find all the antiderivatives of $1/x$ over $\mathbb R\setminus\{0\}$. When you integrate a particular function, you must add that $+C$ because it says that, the anti-derivative of the function could be one of any of the slope field lines in $\mathbb{R}^2$ . Why is it impossible to measure position and momentum at the same time with arbitrary precision? is it possible to read and play a piece that's written in Gflat (6 flats) by substituting those for one sharp, thus in key G? Is Mega.nz encryption secure against brute force cracking from quantum computers? Therefore $\rm\ D^{-1}(g)\ =\ f_1\ +\ ker\ D\ =\:\:$ particular + homogeneous solution, as in linear algebra. If you were given an initial value problem where, say, you needed $y(1)=2$, then that constant $C$ could be determined. y'=-4y Is there a difference between a tie-breaker and a regular vote? and the mysterious constant has disappeared. Use our Website Builder to generate a mobile-responsive store for your industry with seamless site navigation, secure checkout, and more. Else the constant is MathJax reference. So, what happened? Learn more. Why? What has that constant have to do with anything? Use email to boost loyalty. It may also make more sense when you take a differential equations course, but this should be a sufficient explanation. Now, both $$c$$ and $$k$$ are unknown constants and so the sum of two unknown constants is just an unknown constant and we acknowledge that by simply writing the sum as a $$c$$. algebra-precalculus computer-algebra-systems wolfram-alpha. How does the recent Chinese quantum supremacy claim compare with Google's? $$To learn more, see our tips on writing great answers. Notice that F(x)=\sin^2(x)\neq -\cos(2x)/2=G(x). Using ‘add’ will add a column of 1s if a constant column is present.$$ For instance, $F(0)=\sin^2(0) = 0$ but $G(0)=-\cos(2\cdot 0)/2=-1/2$. One hasto initialise it immediately in the constructor because, of course, one cannotset the value later as that would be altering it. Values for $C$ correspond to what particular path the function (or the water) is running along. How to prevent guerrilla warfare from existing. So Access Constant Contact’s built-in landing page builder by selecting “Landing Page” in the “Create New” pop-up menu. It has an argument include.mean which has identical functionality to the corresponding argument for arima().It also has an argument include.drift which allows . , (And sometimes you need only one antiderivative, not all of them. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. $$\int \sin(2x)dx = \int 2\sin(x)\cos(x)dx = \int 2vdv =v^2 = \sin^2(x).$$ Use the readonly modifier to create a class, struct, or array that is initialized one time at runtime (for example in a constructor) and thereafter cannot be changed. error if any column has a constant value. Formal way to perform the change of variable, Evaluating the integral: $\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0$, Getting different answers when integrating using different techniques. Landing page builders have a basic template, so all you have to do is fill in the blanks, choose images and colors, decide on the information you want to collect, and decide where you want the information to go. Read more about Jeff here. Here are my main theories for why this is happening: Somehow the behavior of import_type doesn't exactly work closely enough to the original Til2Idb method that was in the original script. data without adding another constant. The differential equation you are considering has more than one solution. static const : “static const” is basically a combination of static(a storage specifier) and const(a type qualifier). asked Jul 8 '13 at 9:13. user27746 user27746. Just go through the following 5 steps. We forgot about the constant of integration, that's what happened. $$It does belong there. See. Ever month you earned 10 dollars on a principal of 1000 dollars and put it in a box. But that’s fine; people shouldn’t be taking an array name and copying it to something else. I've understood that \int dy represents adding infinitesimal quantity of dy's yielding y but I'am doubtful about the arbitrary constant C. C# does not support const methods, properties, or events. If we copy down the formula in Cell C9, the cell reference changes to …$$ 5=y(0)= e^{-4\cdot0}e^C = e^C The original values with a constant (column of ones) as the first or specify the name, type, and value in the form of a string. This is precisely why you have to have a slope field representation of the anti-derivative of a function. The indefinite integral $\int f(x)dx$ is the function $F(x)$ such that $\frac{d}{dx}F(x) = f(x)$. The impact of removing the constant when the predictor variable is continuous is significantly different. If true, the constant is in the first column. Here you've added a constant. The motivation for asking this question actually comes from solving a differential equation $$x \frac{dy}{dx} = 5x^3 + 4$$ By separation of $dy$ and $dx$ and integrating both sides, $$\int dy = \int\left(5x^2 + \frac{4}{x}\right)dx$$ yields $$y = \frac{5x^3}{3} + 4 \ln(x) + C .$$. The original values with a constant (column of ones) as the first or last column. $$Namely Can I combine two 12-2 cables to serve a NEMA 10-30 socket for dryer? What do I do about a prescriptive GM/player who argues that gender and sexuality aren’t personality traits? And you can know how to use all of the Constant Contact tools.$$ column’s name is ‘const’. So, is there anything that I can add to my answer that you still have a question about? Not always 100 dollars or 200 dollars.That is so, only if the box contained nothing not at start. Constant Contact Forms by MailMunch allows you to painlessly add Constant Contact sign up forms to your WordPress site. \frac{dy}{y} = -4\;dx That’s not good programming style, and it’s just asking for bugs — or, at the very least, confusion — later. How can I intuitively understand the algorithm for finding the integer solutions to $ax+by=c$? Grow Your Lists. 2. Upload Your Contacts. What have you got after 10 months, 20 months in that box? column of 1s if a constant column is present. The enum type enables you to define named constants for integral built-in types (for example int, uint, long, and so on). Hence we say $\int f(x)dx = F(x) + C$. \log_e y = -4x + C. So, the general antiderivative of $f(x)=x^2$ has the form $F(x)={x^3\over3}+C$, and equation (1) is just stating this fact. last column. It does NOT define a constant value. Second, we use the identity $\sin(2x)=2\sin(x)\cos(x)$ and a substitution $v=\sin(x)$: Replace blank line with above line content, Advice on teaching abstract algebra and logic to high-school students. constant meaning: 1. happening a lot or all the time: 2. staying the same, or not getting less or more: 3. Not Real Constants. $$You declare a constant within a procedure or in the declarations section of a … How do I "tell" WA which variables are the constants, and which are the ones I want it to solve for? Name of the step. What is the precise legal meaning of "electors" being "appointed"? In this article. The indefinite integral \int f(x)\,dx is defined to be the general class of functions whose derivatives are f(x). The "integration constant" C does have the "purpose" to make a seemingly true equation at least halfway true. Since there is no reason to think that the constants of integration will be the same from each integral we use different constants for each integral. To get started, I’ve created a new folder within src/ called “constants” to hold all of the constants I use within all of my components. If use a constant in your Microsoft Excel workbook formulas, such as sales tax or car mileage allowance, then check out how using a named constant can save yourself considerable time. Consider What to do? Thus, we have found two antiderivatives of \sin(2x) that are completely different! You can declare a constant within a procedure or at the top of a module, in the Declarations section.$$ The particular value for $C$ collapses it to exactly one of these slope field lines. $$Using ‘add’ will add a To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If ‘raise’, will raise an when . Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Besides using p values for feature selection which is ill advised, most packages just drop any constant columns from the data frame because they introduce numerical problems (singular matrices). This fixes things, because though there are infinitely many F(x) such that \frac{d}{dx}F(x) = f(x), if we pick any single such function F(x) then all solutions to \frac{d}{dx}G(x) = f(x) are of the form G(x) = F(x) + C for some value of C, while for any value of C the function F(x)+C satisfies \frac{d}{dx}(F(x)+C) = f(x). rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Integral is set of antiderivatives, and no antiderivative is distinguished. A = m.10 + c = m.10 + 35 dollars or 135 or 235 dollars.. Where does it come from?$$ In a follow-up article, we will explore why you should never do that. Because of this, we cannot change constant primitive values, but we can change the properties of constant objects. | 2021-05-14T12:59:12 | {
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http://mathhelpforum.com/statistics/153885-question-probability.html | # Math Help - Question on Probability
1. ## Question on Probability
Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?
2. Originally Posted by quiney
Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?
A simple approach would be to draw a tree diagram. Have you tried doing that?
3. Here is another way.
Recall that $P(B) = P(B \cap A) + P(B \cap \neg A)$
Solve for $P(B \cap \neg A)$.
4. Hello, quiney!
Yet another approach . . .
Given that:
. . $\begin{array}{ccc}P(\sim\!A)&=& 0.6 \\
P(B\,|\,A) &=& 0.7 \\
P(B) &=& 0.3 \end{array}$
What is . $P(\sim\!A \wedge B)$ ?
We can place the data into a chart:
. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & & & 0.40 \\ \hline
\sim\!A & & & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$
We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(B \wedge A)}{P(A)} \:=\:0.7$
. . . . $P(B \wedge A) \:=\:0.7\cdot P(A) \;=\;(0.7)(0.4)$
. . . . $P(A \wedge B) \;=\;0.28$
Insert that into the chart.
. . Fill in the rest of the chart.
. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & 0.28 & 0.12 & 0.40 \\ \hline
\sim\!A & 0.02 & 0.58 & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$
Therefore: . $P(\sim\!A \wedge B) \;=\;0.02$
Corrected my typo . . . Thanks, Plato!
5. Originally Posted by Soroban
We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(A \wedge B)}{P(B)} \:=\:0.7$
Please note the typo $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}$.
6. Thanks a lot everybody. So, since $p(B)=p(B \wedge A) + p(B \wedge \neg A)$:
$0.3=p(B \wedge A) + p(B \wedge \neg A)$
Since $p(B|A)=0.7=\dfrac{p(B \wedge A)}{p(A)=0.4} \quad\Rightarrow\quad\ 0.28=p(B\wedge A) \quad\Rightarrow\ 0.3=0.28 + p(B \wedge\neg A)$
Therefore
$0.02=p(B \wedge\neg A)=p(\neg A \wedge B)$
Funny, the answer guide in the book says 0.2. Must be a typo. | 2016-07-29T06:23:47 | {
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https://mathoverflow.net/questions/370028/the-period-of-fibonacci-numbers-over-finite-fields/370030 | # The period of Fibonacci numbers over finite fields
I stumbled upon these very nice looking notes by Brian Lawrence on the period of the Fibonacci numbers over finite fields. In them, he shows that the period of the Fibonacci sequence over $$\mathbb{F}_p$$ divides $$p$$ or $$p-1$$ or $$p+1$$.
I am wondering if there are explicit lower bounds on this period. Is it true, for instance, that as $$p \to \infty$$, so does the order?
A quick calculation on my computer shows that there are some "large" primes with period under 100.
9901 66
19489 58
28657 92
• For $p$ sufficiently large (depending on $N$), the first $N$ Fibonnaci numbers are distinct modulo $p$. Aug 24 '20 at 19:57
• Ah you're right. Might not be able to do any better than that bound then. Aug 24 '20 at 20:09
• @FedorPetrov You need not just $p \mid F_k$ but also $p \mid (F_{k+1}-1)$ since you need $F_k \equiv 0 \bmod p$ and $F_{k+1} \equiv 1 \bmod p$. The first congruence need not imply the second. For example, take $p = 61$. The smallest $k \geq 1$ such that $F_k \equiv 0 \bmod 61$ is $k = 15$, but $F_{16} \equiv 11 \not\equiv 1 \bmod 61$, so the Fibonacci sequence mod $61$ does not have period $15$. The period of $\{F_n \bmod 61\}$ is $60$. Aug 25 '20 at 21:56
• @KConrad ah, sorry, we look for a full period, not only the period of zeroes. You are correct of course. Aug 25 '20 at 21:59
• "the period of the Fibonacci sequence over Fp divides p or p−1 or p+1." This is not true (try $p=5$). An entry divisible by $p$ alone does not make a period. Oct 21 '20 at 14:19
This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula $$F_n = (A^n-B^n)/(A-B),$$ where $$A$$ and $$B$$ are the roots of $$T^2-T-1$$. So if $$\sqrt5$$ is in your finite field, then so are $$A$$ and $$B$$, and since $$AB=-1$$, the period divides $$p-1$$ from Fermat's little theorem. If not, then you're in the subgroup of $$\mathbb F_{p^2}$$ consisting of elements of norm $$\pm1$$, so the period divides $$2(p+1)$$. If you want small period, then take primes that divide $$A^n-1$$, or really its norm, so take primes dividing $$(A^n-1)(B^n-1)$$, where $$A$$ and $$B$$ are $$\frac12(1\pm\sqrt5)$$. An open question is in the other direction: Are there infinitely many $$p\equiv\pm1\pmod5$$ such that the period is maximal, i.e., equal to $$p-1$$?
BTW, the source you quote isn't quite correct, if $$p\equiv\pm2\pmod5$$, then the period divides $$2(p+1)$$, but might not divide $$p+1$$. The simplest example is $$p=3$$, where the Fibonacci sequence is $$0,1,1,2,0,2,2,1,\quad 0,1,1,2,0,2,2,1,\ldots.$$ Note that the first 0 does not necessarily mean that it will start to repeat. What happens is that the term before the first $$0$$ is $$p-1$$, so the first part of the sequence repeats with negative signs before you get to a consecutive $$0$$ and $$1$$.
• On the one hand I agree it's probably too elementary, but, on the other hand, I wouldn't have seen this very nice perspective on periods if it hadn't been posted here, so I'm torn. :-) Aug 24 '20 at 20:17
• Sorry everyone; I'm still learning about the etiquette of the forum. I found these notes and found them interesting and wanted to learn more. Thanks for answering my question seriously :) Aug 25 '20 at 20:06
• As @LSpice comments, one might not have realized that such a question is "quasi-elementarizable" in such a decisive manner... Sometimes the decisiveness and elementariness is only visible to a more experienced person, which (to my mind) doesn't mean that the original question was "too easy for MO"... If anything, I am fond of examples where a seemingly innocent (but baffling) question succumbs to a more sophisticated viewpoint. As though such viewpoints not only provide publication fodder, but really do answer questions. :) Aug 31 '20 at 17:50
I won't address your question about how small the period of $$\{F_n \bmod p\}$$ can be as $$p$$ grows, but instead ask if the upper bounds on the period can be achieved infinitely often. For consistency I'll use the notation from Joe Silverman's answer: set $$A = (1 + \sqrt{5})/2$$ and $$B = (1-\sqrt{5})/2$$, so $$F_n = (A^n - B^n)/(A-B) = (A^n - B^n)/\sqrt{5}$$. Note $$A+B = 1$$, $$A - B = \sqrt{5}$$, and $$AB = -1$$.
Claim: For a prime $$p \not= 2$$ or $$5$$, the period of the Fibonacci sequence $$\{F_n \bmod p\}$$ is the smallest even positive integer $$k$$ such that $$A^k = 1$$ in characteristic $$p$$.
This claim involves working in the field $$\mathbf F_p(\sqrt{5})$$, where $$\sqrt{5}$$ is a square root of 5 in characteristic $$p$$, so we can regard $$A = (1+\sqrt{5})/2$$ as a number in the field $$\mathbf F_p(\sqrt{5})$$, which is either $$\mathbf F_p$$ or $$\mathbf F_{p^2}$$. (The notation $$\mathbf F_p$$ and $$\mathbf F_{p^2}$$ are fields of order $$p$$ and $$p^2$$, not having anything to do with the "$$F$$" in Fibonacci number notation.) The claim is saying that $$F_{n+k} \equiv F_n \bmod p$$ for all $$n \geq 0$$ (or just all sufficiently large $$n \geq 0$$) if and only if $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$ for even $$k$$, so the period of $$\{F_n \bmod p\}$$ is the smallest even $$k \geq 1$$ such that $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$.
Proof. View the congruence $$F_{n+k} \equiv F_n \bmod p$$ as an equation $$F_{n+k} = F_n$$ in the subfield $$\mathbf F_p$$ of $$\mathbf F_p(\sqrt{5})$$. The Fibonacci formula $$F_n = (A^n - B^n)/\sqrt{5}$$ in $$\mathbf R$$ is also a valid formula in fields of characteristic $$p$$ when we view $$\sqrt{5}$$ in characteristic $$p$$ and set $$A = (1+\sqrt{5})/2$$ and $$B = (1-\sqrt{5})/2 = 1-A$$ in characteristic $$p$$. In $$\mathbf F_p(\sqrt{5})$$, \begin{align*} F_{n+k} = F_n & \Longleftrightarrow \frac{A^{n+k}-B^{n+k}}{\sqrt{5}} = \frac{A^n-B^n}{\sqrt{5}} \\ & \Longleftrightarrow A^n(A^k-1) = B^n(B^k-1). \end{align*} In a field of characteristic $$p \not= 2$$ or $$5$$, $$A$$ and $$B$$ are nonzero since $$AB = -1$$. Suppose in $$\mathbf F_p(\sqrt{5})$$ that $$A^k \not= 1$$. Then in this field, $$F_{n+k} = F_n \Longrightarrow (A/B)^n = (B^k-1)/(A^k-1).$$ The ratio $$A/B$$ in characteristic $$p$$ is not $$1$$ since $$A = B \Longrightarrow 5 = 0$$ in characteristic $$p$$, which is false since $$p \not= 5$$. Therefore $$(A/B)^n$$ is not constant as $$n$$ varies, but $$(B^k-1)/(A^k-1)$$ is constant as $$n$$ varies. Thus $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$, so $$B^n(B^k-1) = A^n(A^k-1) = 0$$, so $$B^k = 1$$ (we never have $$B^n = 0$$ in characteristic $$p$$). Since $$B^k = (-1/A)^k = (-1)^k/A^k$$, we have $$A^k = 1$$ and $$B^k = 1$$ if and only if $$A^k = 1$$ and $$(-1)^k = 1$$. Since $$-1 \not= 1$$ in characteristic $$p$$ when $$p \not= 2$$, we have $$A^k = 1$$ and $$(-1)^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$ if and only if $$A^k = 1$$ in characteristic $$p$$ and $$k$$ is even.
That completes the proof of the claim.
Since $$B = -1/A$$, if $$A$$ in characteristic $$p$$ has odd order $$m$$ then $$B$$ in characteristic $$p$$ has order $$2m$$. Therefore the claim says the period of $$\{F_n \bmod p\}$$ is the least $$k \geq 1$$ such that $$A^k = 1$$ and $$B^k = 1$$ in characteristic $$p$$: that $$k$$ is necessarily even.
For $$p \not= 2$$ or 5, the field $$\mathbf F_p(\sqrt{5})$$ has order $$p$$ or $$p^2$$ depending on whether or not $$5 \bmod p$$ is a square: its order is $$p$$ when $$p \equiv \pm 1 \bmod 5$$ and its order is $$p^2$$ when $$p \equiv \pm 2 \bmod 5$$. Therefore the group of nonzero elements $$\mathbf F_p(\sqrt{5})^\times$$ has order $$p-1$$ if $$p \equiv \pm 1 \bmod 5$$ and order $$p^2-1$$ if $$p \equiv \pm 2 \bmod 5$$. Since $$p-1$$ and $$p^2-1$$ are both even, the period of $$\{F_n \bmod p\}$$ divides $$p-1$$ if $$p \equiv \pm 1 \bmod 5$$ and it divides $$p^2-1$$ if $$p \equiv \pm 2 \bmod 5$$. As Joe points out in his answer, when $$p \equiv \pm 2 \bmod 5$$ the period of $$\{F_n \bmod p\}$$ divides $$2(p+1)$$, which is a proper factor of $$p^2-1$$.
This situation is reminiscent of Artin's primitive root conjecture, which says that for $$a \in \mathbf Z$$ that is not $$\pm 1$$ or a perfect square, there are infinitely many primes $$p$$ such that $$a \bmod p$$ has order $$p-1$$ in $$\mathbf F_p^\times$$, and in fact there is a positive density of such primes. This conjecture is known to be a consequence of the Generalized Riemann Hypothesis (GRH). This conjecture and its connection to GRH can be extended to number fields, and to talk about the multiplicative order of $$A$$ in characteristic $$p$$ amounts to looking at an analogue of Artin's primitive root conjecture with $$\mathbf Z$$ replaced by $$\mathbf Z[A]$$, which is the ring of integers of $$\mathbf Q(\sqrt{5})$$. This is discussed in Barendrecht's 2018 bachelor's thesis here. For example, GRH implies that the set of (nonzero) prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$ such that $$A \bmod \mathfrak p$$ generates all of $$(\mathbf Z[A]/\mathfrak p)^\times$$ has a positive density using the last result of the thesis, Corollary 3.1.2, and therefore there are infinitely many such prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$.
Every nonzero prime ideal $$\mathfrak p$$ in $$\mathbf Z[A]$$ is a factor of $$(p) = p\mathbf Z[A]$$ for some prime number $$p$$: if $$p \equiv \pm 1 \bmod 5$$ then $$(p) = \mathfrak p\mathfrak p'$$ for two prime ideals $$\mathfrak p$$ and $$\mathfrak p'$$, and $$\mathbf Z[A]/\mathfrak p$$ and $$\mathbf Z[A]/\mathfrak p'$$ are fields of order $$p$$. If $$p \equiv \pm 2 \bmod 5$$, then $$(p) = \mathfrak p$$ is a prime ideal in $$\mathbf Z[A]$$ and $$\mathbf Z[A]/(p)$$ is a field of order $$p^2$$. When $$p \equiv \pm 2 \bmod 5$$, the multiplicative order of $$A$$ in characteristic $$p$$ is a factor of $$2(p+1)$$, which is less than $$p^2-1$$, so the only prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$ for which $$A \bmod \mathfrak p$$ might generate $$(\mathbf Z[A]/\mathfrak p)^\times$$ are prime ideals dividing a prime $$p \equiv \pm 1 \bmod 5$$, in which case we are in the situation that $$A \in \mathbf F_p^\times$$ has order $$p-1$$. Comparing this to the claim up above, since $$p-1$$ is even when $$p > 2$$ we see that GRH implies that there are infinitely many primes $$p \equiv \pm 1 \bmod 5$$ such that $$\{F_n \bmod p\}$$ has period $$p-1$$.
Among the 18 odd primes $$p \equiv \pm 2 \bmod 5$$ with $$p < 150$$, $$\{F_n\bmod p\}$$ has period $$2(p+1)$$ all but 3 times (at $$p = 47$$ $$107$$, and $$113$$). There are many generalizations of the Artin primitive root conjecture and I would not be surprised if one of them can show GRH implies there are infinitely many primes $$p \equiv \pm 2 \bmod 5$$ such that $$\{F_n \bmod p\}$$ has period $$2(p+1)$$, but this is not something I am aware of in more detail at the moment.
The question above is about lower bounds, but I allow myself to comment about upper bounds: $$\pi(n)$$, the period function of the Fibonacci sequence mod $$n$$, satisfies $$\pi(n)\leq 6n$$ and equality holds iff $$n=2\cdot 5^k$$ for some $$k\geq 1$$. This fact is well known. In the 90's it was considered here as a puzzle to the monthly readers. $$\pi(n)$$ was also discussed in an elementary fashion in the 60's in this monthly paper.
But really, I want to share a little observation which forms a generalization of the above mentioned fact: denoting, for an element $$g\in \mathrm{GL}_2(\mathbb{Z})$$, by $$\rho_g(n)$$ the order of the image of $$g$$ in $$\mathrm{GL}_2(\mathbb{Z}/n)$$, $$\rho_g(n)\leq 6n$$. This is a generalization because $$\rho_g(n)=\pi(n)$$ for $$g= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$. Note that if $$\det(g)=-1$$ then $$\rho_g(n)=2\rho_{g^2}(n)$$, thus it is enough to prove that for $$g\in \mathrm{SL}_2(\mathbb{Z})$$, $$\rho_g(n)\leq 3n$$. Let me now fix $$g\in \mathrm{SL}_2(\mathbb{Z})$$, denote $$\rho(n)=\rho_g(n)$$ and prove that indeed $$\rho(n)\leq 3n$$.
First note that, for natural $$p$$ and $$n$$, if $$p$$ divides $$n$$ then $$\rho(pn)$$ divides $$p\rho(n)$$. This follows by expanding the right hand side of $$g^{p\rho(n)}=(g^{\rho(n)})^p=(1+nh)^p$$ and note that it is 1 mod $$pn$$. By induction we conclude that for every $$k>1$$, $$\rho(p^k)$$ divides $$p^{k-1}\rho(p)$$.
Assume now $$p$$ is a prime and note that $$\rho(p)$$ divides either $$p-1,p+1$$ or $$2p$$. Indeed, if $$\bar{g}\in \mathrm{SL}_2(\mathbb{F}_p)$$ is diagonalizable over $$\mathbb{F}_p$$ then its eigenvalues are in $$\mathbb{F}_p^\times$$ and their orders divides $$p-1$$, else, if $$\bar{g}$$ is diagonalizable over $$\mathbb{F}_{p^2}$$ then its eighenvalues $$\alpha,\beta$$ are conjugated by the Frobenius automorphism, thus their order divides $$p+1$$ because $$\alpha^{p+1}=\alpha\alpha^p=\alpha\beta=\det(\bar{g})=1$$, else $$\bar{g}$$ has a unique eigenvalue, which must be a $$\pm 1$$ by $$\det(\bar{g})=1$$, thus $$\bar{g}^2$$ is unipotent and its order divides $$p$$. For $$p=2$$, in the last case, there was no reason to pass to $$g^2$$, as $$-1=1$$ in $$\mathbb{F}_2$$, thus $$\rho(2)$$ is either 1,2 or 3.
From the above two points, we conclude that for every odd prime $$p$$ and natural $$k$$, $$\rho(p^k)$$ divides $$p^{k-1}(p-1)$$, $$p^{k-1}(p+1)$$ or $$2p^k$$. All these numbers are even and bounded by $$2p^k$$, thus $$\mathrm{lcm}\{\rho(p^k),2\} \leq 2p^k$$. For $$p=2$$ we get that $$\rho(2^k) \leq 2^{k-1}\cdot 3$$.
Fix now an arbitrary natural $$n$$. Write $$n=2^km$$ for an odd $$m$$ and decompose $$m=\prod_{i=0}^r p_i^{k_i}$$. Then \begin{align*} \rho(m)= \mathrm{lcm}\{\rho(p_i^{k_i}) \mid i=0,\dots r\} \leq \mathrm{lcm}\{\mathrm{lcm}\{\rho(p_i^{k_i}),2\} \mid i=0,\dots r\} =\\ 2\mathrm{lcm}\{\frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2} \mid i=0,\dots r\} \leq 2\prod_{i=0}^r \frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2}\leq 2\prod_{i=0}^r p_i^{k_i} =2m \end{align*} and we get $$\rho(n) = \rho(2^km) \leq \rho(2^k) \cdot \rho(m) \leq 2^{k-1}\cdot 3 \cdot 2m = 3\cdot 2^km=3n.$$
This finishes the proof that $$\rho(n)\leq 3n$$.
As always, it is interesting to analyze the case of equality. For $$g\in \mathrm{SL}_2(\mathbb{Z})$$ we have $$\rho_g(n)=3n$$ for some $$n$$ iff $$\mathrm{tr}(g)$$ is odd and not equal $$-1$$ or $$-3$$. If $$g$$ satisfies this condition, then $$n$$ satisfices $$\rho_g(n)=3n$$ iff $$n=2st$$, for some odd $$s\geq 3$$, $$t\geq 1$$ where every prime factor of $$s$$ divides $$\mathrm{tr}(g)+2$$, every prime factor of $$t$$ divides $$\mathrm{tr}(g)-2$$ and $$g$$ is not $$\pm 1$$ modulo any of these prime factors.
For $$g$$ satisfying $$\det(g)=-1$$, using the identity $$\mathrm{tr}(g^2)=\mathrm{tr}(g)^2-2\det(g)$$, we get that $$\rho_g(n)=6n$$ for some $$n$$ iff $$\mathrm{tr}(g)$$ is odd and in this case, $$n$$ satisfices $$\rho_g(n)=6n$$ iff $$n=2st$$, for some odd $$s\geq 3$$, $$t\geq 1$$ where every prime factor of $$s$$ divides $$\mathrm{tr}(g)+4$$, every prime factor of $$t$$ divides $$\mathrm{tr}(g)$$ and $$g$$ is not $$\pm 1$$ modulo any of these prime factors.
Specifically for $$g=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$, $$\det(g)=-1$$, $$\mathrm{tr}(g)=1$$ is odd, 5 is the only prime factor of $$\mathrm{tr}(g)+4$$ and there is no prime factor for $$\mathrm{tr}(g)$$. Since $$g$$ is not $$\pm 1$$ modulo 5, we get that $$\pi(n)=\rho_g(n)=6n$$ iff $$n=2\cdot 5^k$$ for some $$k\geq 1$$, as claimed above. | 2021-10-20T17:18:41 | {
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https://stats.stackexchange.com/questions/443445/how-to-generate-random-integers-between-1-and-4-that-have-a-specific-mean | # How to generate random integers between 1 and 4 that have a specific mean?
I need to generate 100 random integers in R, where each integer is between 1 and 4 (hence 1,2,3,4) and the mean is equal to a specific value.
If I draw random uniform numbers between 1 and 5 and get floor, I have a mean of 2.5.
x = floor(runif(100,min=1, max=5))
I need to fix the mean to 1.9 or 2.93 for example.
I guess I can generate random integers that add to 100 * mean but I don't know how to restrict to random integers between 1 and 4.
• I think this is a bit under-determined... One for instance can get a mean of 1.9 with sample(size=n, x= 1:4, prob=c(3.666,1,1,1), replace=TRUE) but also with sample(size=n, x= 1:4, prob=c(3,1,1,0.715), replace=TRUE). – usεr11852 Jan 5 at 22:53
• Are you asking how to constrain the mean of the underlying distribution, or the sample mean? – user20160 Jan 5 at 22:53
• tha sample mean @user20160 – Fierce82 Jan 5 at 23:04
• Integers between 1 and 4 only allows for 2 and 3. You also need to specify the distribution that they are drawn randomly from (or make one up). – wolfies Jan 6 at 6:28
• I voted to leave this open because there's an interesting algorithmic question in here--the R part is incidental; you could just as easily implement this in Python or with a pad and some dice. – Matt Krause Jan 6 at 21:14
I agree with X'ian that the problem is under-specified. However, there is an elegant, scalable, efficient, effective, and versatile solution worth considering.
Because the product of the sample mean and sample size equals the sample sum, the problem concerns generating a random sample of $$n$$ values in the set $$\{1,2,\ldots, k\}$$ that sum to $$s$$ (assuming $$n \le s \le kn,$$ of course).
To explain the proposed solution and, I hope, justify the claim of elegance, I offer a graphical interpretation of this sampling scheme. Lay out a grid of $$k$$ rows and $$n$$ columns. Select every cell in the first row. Randomly (and uniformly) select $$s-n$$ of the remaining cells in rows $$2$$ through $$k.$$ The value of observation $$i$$ in the sample is the number of cells selected in column $$i:$$
This $$4\times 100$$ grid is represented by black dots at the unselected cells and colored patches at the selected cells. It was generated to produce a mean value of $$2,$$ so $$s=200.$$ Thus, $$200-100=100$$ cells were randomly selected among the top $$k-1=3$$ rows. The colors represent the numbers of selected cells in each column. There are $$28$$ ones, $$47$$ twos, $$22$$ threes, and $$3$$ fours. The ordered sample corresponds to the sequence of colors from column $$1$$ through column $$n=100.$$
To demonstrate scalability and efficiency, here is an R command to generate a sample according to this scheme. The question concerns the case $$k=4, n=100$$ and $$s$$ is $$n$$ times the desired average of the sample:
tabulate(sample.int((k-1)*n, s-n) %% n + 1, n) + 1
Because sample.int requires $$O(s-n)$$ time and $$O((k-1)n)$$ space, and tabulate requires $$O(n)$$ time and space, this algorithm requires $$O(\max(s-n,n))$$ time and $$O(kn)$$ space: that's scalable. With $$k=4$$ and $$n=100$$ my workstation takes only 12 microseconds to perform this calculation: that's efficient.
(Here's a brief explanation of the code. Note that integers $$x$$ in $$\{1,2,\ldots, (k-1)n\}$$ can be expressed uniquely as $$x = nj + i$$ where $$j \in \{0,1,\ldots, k-2\}$$ and $$i\in\{1,2,\ldots, n\}.$$ The code takes a sample of such $$x,$$ converts them to their $$(i,j)$$ grid coordinates, counts how many times each $$i$$ appears (which will range from $$0$$ through $$k-1$$) and adds $$1$$ to each count.)
Why can this be considered effective? One reason is that the distributional properties of this sampling scheme are straightforward to work out:
• It is exchangeable: all permutations of any sample are equally likely.
• The chance that the value $$x \in\{1,2,\ldots, k\}$$ appears at position $$i,$$ which I will write as $$\pi_i(x),$$ is obtained through a basic hypergeometric counting argument as $$\pi_i(x) = \frac{\binom{k-1}{x-1}\binom{(n-1)(k-1)}{s-n-x+1}}{\binom{n(k-1)}{ s-n}}.$$ For example, with $$k=4,$$ $$n=100,$$ and a mean of $$2.0$$ (so that $$s=200$$) the chances are $$\pi = (0.2948, 0.4467, 0.2222, 0.03630),$$ closely agreeing with the frequencies in the foregoing sample. Here are graphs of $$\pi_1(1), \pi_1(2), \pi_1(3),$$ and $$\pi_1(4)$$ as a function of the sum:
• The chance that the value $$x$$ appears at position $$i$$ while the value $$y$$ appears at position $$j$$ is similarly found as $$\pi_{ij}(x,y) = \frac{\binom{k-1}{x-1}\binom{k-1}{y-1}\binom{(n-1)(k-1)}{s-n-x-y+2}}{\binom{n(k-1)}{ s-n}}.$$
These probabilities $$\pi_i$$ and $$\pi_{ij}$$ enable one to apply the Horvitz-Thompson estimator to this probability sampling design as well as to compute the first two moments of the distributions of various statistics.
Finally, this solution is versatile insofar as it permits simple, readily-analyzable variations to control the sampling distribution. For instance, you could select cells on the grid with specified but unequal probabilities in each row, or with an urn-like model to modify the probabilities as sampling proceeds, thereby controlling the frequencies of the column counts.
• (+1) Ultimate elegance, indeed. – Xi'an Jan 6 at 18:56
• The answer is too difficult for me to follow, appreciate it nonetheless – Fierce82 Jan 7 at 11:33
• What an elegant and beautifully presented answer. If you don't mind my humble suggestion as a reader, you might consider presenting the solution first (the counting patches and the great diagram), and then talking about the implementation and how your argument about how it fits the intuition, and finally why it's efficient. It might make it a bit easier to follow. – Neil G Jan 8 at 8:49
• @Neil Thank you for your suggestion. I think it's a good one and will consider it carefully. – whuber Jan 8 at 15:02
• This is a lovely and satisfying answer. I did want to note that the numbers are small enough in this case (100 numbers summing to 190) that we can calculate the uniform distribution of all values that satisfy. I ran some calculations to compare your distribution against this and found that yours is much much more likely (billions in some cases) to select small non-1 values. For example, your model will almost never give distributions with >45 "ones" (~0.002% chance for 46, vanishing for more), but that comprises ~58% of the uniform model values. – Cireo Jan 31 at 5:29
The question is under-specified in that the constraints on the frequencies \begin{align}n_1+2n_2+3n_3+4n_4&=100M\\n_1+n_2+n_3+n_4&=100\end{align} do not determine a distribution: "random" is not associated with a particular distribution, unless the OP means "uniform". For instance, if there exists one solution $$(n_1^0,n_2^0,n_3^0,n_4^0)$$ to the above system, then the distribution degenerated at this solution is producing a random draw that is always $$(n_1^0,n_2^0,n_3^0,n_4^0)$$.
In the case the question is about simulating a Uniform distribution over the grid\begin{align}n_1+2n_2+3n_3+4n_4&=100M\\n_1+n_2+n_3+n_4&=100\end{align}one can always use a Metropolis-Hastings algorithm. Starting from $$(n_1^0,n_2^0,n_3^0,n_4^0)$$, create a Markov chain by proposing symmetric random perturbations of the vector $$(n_1^t,n_2^t,n_3^t,n_4^t)$$ and accept if the result is within $$\{1,2,3,4\}^4$$ and satisfies the constraints.
For instance, here is a crude R rendering:
cenM=293
#starting point (n¹,n³,n⁴)
n<-sample(1:100,3,rep=TRUE)
while((sum(n)>100)|(n[2]-n[1]+2*n[3]!=cenM-200))
n<-sample(1:100,3,rep=TRUE)
#Markov chain
for (t in 1:1e6){
prop<-n+sample(-10:10,3,rep=TRUE)
if ((sum(prop)<101)&
(prop[2]-prop[1]+2*prop[3]==cenM-200)&
(min(prop)>0))
n=prop}
c(n[1],100-sum(n),n[-1])
with the distribution of $$(n_1,n_3,n_4)$$ over the 10⁶ iterations:
In case you want draws of the integers themselves,
sample(c(rep(1,n[1]),rep(2,100-sum(n)),rep(3,n[2]),rep(4,n[3])))
is a quick & dirty way to produce a sample.
• thanks. but I cannot understand how i can utilize this to get the 4 integers (between 1 and 4) – Fierce82 Jan 7 at 11:37
• This generates the numbers of 1,2,3,4's $n_1,n_2,n_3,n_4)$ so that there are 100 of them and the sum is cenM. The integer themselves are a random permutation of $n_1$ 1's,..., $n_4$ 4's. – Xi'an Jan 8 at 7:35
I want to ... uh ... "attenuate" @whuber's amazing answer, which @TomZinger says is too difficult to follow. By that I mean I want to re-describe it in terms that I think Tom Zinger will understand, because it's clearly the best answer here. And as Tom gradually uses the method and finds that he needs, say, to know the distribution of the samples rather than just their mean, whuber's answer will be just what he's looking for.
In short: there are no original ideas here, only a simpler explanation.
You'd like to create $$n$$ integers from $$1$$ to $$4$$ with mean $$r$$. I'm going to suggest computing $$n$$ integers from $$0$$ to $$3$$ with mean $$r-1$$, and then adding one to each of them. If you can do that latter thing, you can solve the first problem. For instance, if we want 10 integers between $$1$$ and $$4$$ with mean $$2.6$$, we can write down these $$10$$ integers between $$0$$ and $$3$$...
0,3,2,1,3,1,2,1,3,0
whose mean is $$1.6$$; if we increase each by $$1$$, we get
1,4,3,2,4,2,3,2,4,1
whose mean is $$2.6$$. It's that simple.
Now let's think about the numbers $$0$$ through $$3$$. I'm going to think of those as "how many items do I have in a 'small' set?" I might have no items, one item, two items, or three items. So the list
0,3,2,1,3,1,2,1,3,0
represents ten different small sets. The first is empty; the second has three items, and so on. The total number of items in all the sets is the sum of the ten numbers, i.e., $$16$$. And the average number of items in each set is this total, divided by $$10$$, hence $$1.6$$.
whuber's idea is this: suppose you make yourself ten small sets, with the total number of items being $$10t$$ for some number $$t$$. Then the average size of the sets will be exactly $$t$$. In the same way, if you make yourself $$n$$ sets with a total number of items being $$nt$$, the average number of items in a set will be $$t$$. You say you're interested in the case $$n = 100$$.
Let's make this concrete for your example: you want 100 items between 1 and 4 whose average is $$1.9$$. Using the idea of my first paragraph, I'm going to change this to "make $$100$$ ints between $$0$$ and $$3$$ whose average is $$0.9$$". When I'm done, I'll add $$1$$ to each of my ints to get a solution to your problem. So my target average is $$t = 0.9$$.
I want to make $$100$$ sets, each with between $$0$$ and $$3$$ items in it, with an average set-size of $$0.9$$.
As I've observed above, this means that there have to be a total of $$100 \cdot 0.9 = 90$$ items in the sets. From the numbers $$1, 2, \ldots, 300$$, I'm going to select exactly $$90$$. I can indicate the selected ones by making a list of 300 dots and Xs:
..X....X...XX...
where the list above indicates that I selected the numbers 3, 9, 13, 14, and then many others that I haven't shown because I got sick of typing. :) I can take this sequence of 300 dots and Xs and break it into three groups of 100 dots each, which I arrange one atop the other, getting something that looks like this:
...X....X..X.....X...
.X...X.....X...X.....
..X...X.X..X......X..
but goes on for a full 100 items in each row. The number of Xs in each row might differ -- there might be 35 in the first row, 24 in the second, and 31 in the third, for instance, and that's OK. [Thanks to whuber for pointing out that I had this wrong in a first draft!]
Now look at each column: each column can be considered as a set, and that set has between 0 and 3 "X"s in it. I can write the tallies below the rows to get something like this:
...X....X..X.....X...
.X...X.....X...X.....
..X...X.X..X......X..
011101102003000101100
That is to say, I've produced 100 numbers, each between 1 and 3. And the sum of those 100 numbers must be the number of Xs, total, in all three rows, which was 90. So the average must be $$90/100 = 0.9$$, as desired.
So here are the steps to getting 100 integers between 1 and 4 whose average is exactly $$s$$.
1. Let $$t = s - 1$$.
2. Compute $$k = 100 t$$; that's how many Xs we'll place in the rows, total.
3. Make a list of 300 dots-or-Xs, $$k$$ of which are Xs.
4. Split this into three rows of 100 dots-or-Xs, each containing about a third of the Xs, more or less.
5. Arrange these in an array, and compute column sums, getting 100 integers between $$0$$ and $$3$$. Their average will be $$t$$.
6. Add one to each column sum to get 100 integers between $$1$$ and $$4$$ whose average is $$s$$.
Now the tricky part of this is really in step 4: how do you pick $$300$$ items, $$k$$ of which are "X" and the other $$300-k$$ of which are "."? Well, it turns out that R has a function that does exactly that.
And then whuber tells you how to use it: you write
tabulate(sample.int((k-1)*n, s-n) %% n + 1, n)
For your particular case, $$n = 100$$, and $$s$$, the total number of items in all the small sets, is $$100r$$, and you want numbers between $$1$$ and $$4$$, so $$k = 4$$, so $$k-1$$ (the largest size for a 'small set') is 3, so this becomes
tabulate(sample.int(3*100, 100r-100) %% 100 + 1, n)
or
tabulate(sample.int(3*100, 100*(r-1)) %% 100 + 1, 100)
or, using my name $$t$$ for $$r - 1$$, it becomes
tabulate(sample.int(3*100, 100*t) %% 100 + 1, 100)
The "+1" at the end of his original formula is exactly the step needed to convert from "numbers between $$0$$ and $$3$$" to "numbers between $$1$$ and $$4$$".
Let's work from the inside out, and let's simplify to $$n = 10$$ so that I can show sample outputs:
tabulate(sample.int(3*10, 10*t) %% 10 + 1, 10)
And let's aim for $$t = 1.9$$, so this becomes
tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)
Starting with sample.int(3*10, 10*1.9): this produces a list of $$19$$ integers between $$1$$ and $$30$$. (i.e., it solved the problem of picking $$k$$ numbers out of your total -- $$300$$ in your real problem, $$30$$ in my smaller example).
As you'll recall, we want to produce three rows of ten dots-and-Xs each, something like
X.X.XX.XX.
XXXX.XXX..
XX.X.XXX..
We can read this left-to-right-top-to-bottom (i.e., normal reading order) to produce a list of locations for Xs: the first item's a dot; the second and third are Xs, and so on, so our list of locations starts out $$1, 3, 5, 6, \ldots$$. When we get to the end of a row, we just keep counting up, so for the picture above, the X-locations would be $$1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 18, 21, 22, 24, 26, 27, 28$$. Is that clear?
Well, whubers code produces exactly that list of locations with its innermost section.
The next item is %% 10; that takes a number and produces its remainder on division by ten. So our list becomes $$1, 3, 5, 6, 8, 9, 1, 2, 3, 4, 6, 7, 8, 1, 2, 4, 6, 7, 8$$. If we break that into three groups --- those that came from numbers between $$1$$ and $$10$$, those that came from numbers from $$11$$ to $$20$$, and those that came from numbers $$21$$ to $$30$$, we get $$1, 3, 5, 6, 8, 9$$, then $$1, 2, 3, 4, 6, 7, 8,$$, and finally $$1, 2, 4, 6, 7, 8$$. Those tell you where the Xs in each of the three rows are. There's a subtle problem here: if there had been an X in position 10 in the first row, the first of our three lists would have been $$1, 3, 5, 6, 8, 9, 0$$, and the tabulate function doesn't like "0". So whuber adds 1 to each item in the list to get $$2, 4, 6, 7, 9, 10, 1$$. Let's move on to the overall computation:
tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)
This asks "for those $$30$$ numbers, each indicating whether there's an X in some column, tell me how many times each column (from $$1$$ to $$10$$ --- that's what the final "10" tells you) appears, i.e., tell me how many Xs are in each column. The result is 0 3 2 2 2 1 3 2 3 1 which (because of the shift-by-one thing) you have to read as "there are no Xs in the 10th column; there are 3 Xs in the first column; there are 2 Xs in the second column," and so on up to "there is one X in the 9th column".
That gives you ten integers between $$0$$ and $$3$$ whose sum is $$19$$, hence whose average is $$1.9$$. If you increase each by 1, you get ten integers between $$1$$ and $$4$$ whose sum is $$29$$, hence an average value of $$2.9$$.
You can generalize to $$n = 100$$, I hope.
• +1 Welcome to our site, John. I appreciate your efforts to explain and clarify these ideas. At one point your description departs from what the code does: one does not divide the three rows into groups of 30 each. Instead, 90 cells out of the 300 cells in those rows are selected. Usually, each row will have a different number of cells. – whuber Jan 7 at 17:15
• Thanks...I actually worried about that a little bit as I wrote it, but I was in mid-sentence, and by the time I was finished, the thought had flown. I'll edit to try to fix it up. – John Jan 7 at 21:25
You can use sample() and select specific probabilities for each integer. If you sum the product of the probabilities and the integers, you get the expected value of the distribution. So, if you have a mean value in mind, say $$k$$, you can solve the following equation: $$k = 1\times P(1) + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ You can arbitrarily choose two of the probabilities and solve for the third, which determines the fourth (because $$P(1)=1-(P(2)+P(3)+P(4))$$ because the probabilities must sum to $$1$$). For example, let $$k=2.3$$, $$P(4)=.1$$, and $$P(3)=.2$$. Then we have that $$k = 1 \times [1-(P(2)+P(3)+P(4)] + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ $$2.3 = [1 - (P(2)+.1+.2)] + 2*P(2) + 3\times .2 + 4\times .1$$ $$2.3 = .7 + P(2) + .6 + .4$$ $$P(2)=.6$$ $$P(1)=1-(P(2)+P(3)+P(4)=1 - (.6+.1+.2)=.1$$
So you can run x <- sample(c(1, 2, 3, 4), 1e6, replace = TRUE, prob = c(.1, .6, .2, .1)) and mean(x) is approximately $$2.3$$
• This explains how to constrain the mean of the distribution. But, the OP specified in the comments that they want to constrain the sample mean (which won't match the mean of the distribution, except in expectation). On the other hand, it seems the OP accepted this answer anyway, so perhaps that's not what they wanted after all. – user20160 Jan 6 at 0:03
• are you sure? @user20160 why sample mean is not contrainted? it's equal to target – Fierce82 Jan 6 at 0:08
• This answer does not provide a way to make the sample mean equal the target value: most of the time the mean will not equal the target. – whuber Jan 6 at 0:10
• @TomZinger Yes. This answer nicely describes how to sample from a distribution with the given target mean. But, the mean of a sample drawn from a distribution will not generally equal the mean of the distribution. – user20160 Jan 6 at 4:59
• I wrote my answer before I saw that comment, but I figured this would be useful anyway. I imagined it would require an integer programming optimization problem to get a sample mean exactly equal to some value. – Noah Jan 6 at 8:20
Here is a simple algorithm: Create $$n-1$$ random integers in the range $$[1,4]$$ and calculate the $$n^{th}$$ integer for the mean to be equal to the specified value. If that number is smaller than $$1$$ or larger than $$4$$, then one by one distribute the surplus/lacking onto other integers, e.g. if the integer is $$5$$, we have $$1$$ surplus; and we may add this to the next integer if it's not $$4$$, else add to the next etc. Then, shuffle the entire array.
• One big problem with this proposal is that it doesn't come along with any indication of what the expected frequencies of the resulting values are. – whuber Jan 5 at 23:32
• Although interesting, I thought the OP only requires an algorithm to generate the desired array of integers in a non-deterministic manner. – gunes Jan 5 at 23:37
• I think that avoids the essence of the question rather than providing a satisfactory answer. A good answer should be able to characterize the distribution it proposes in a meaningful way, such as by giving a formula for the probabilities or at least giving the first couple of moments. – whuber Jan 5 at 23:46
• A minor adjustment of the simulated data is likely 'proper', however, looking at the expertimental design in cases where more significant mean deviation is required, depending on the intended purpose, could be, from a hypothesis testing perspective, 'suspect', in my judgement. Either over or under loading a random design to justify or reject possible non-random effects that have been actually observed can be questionable practice. So, any method that makes a very small adjustment to the last of say a 100 observations is probably keeping in good practice, in my opinion. – AJKOER Jan 7 at 17:33
As a supplement to whuber's answer, I've written a script in Python which goes through each step of the sampling scheme. Note that this is meant for illustrative purposes and is not necessarily performant.
Example output:
n=10, s=20, k=4
Starting grid
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
X X X X X X X X X X
Filled in grid
X X . . X . X . . X
. . X X X . . . . .
. . . . X X . . . .
X X X X X X X X X X
Final grid
X X . . X . X . . X
. . X X X . . . . .
. . . . X X . . . .
X X X X X X X X X X
2 2 2 2 4 2 2 1 1 2
The script:
import numpy as np
# Define the starting parameters
integers = [1, 2, 3, 4]
n = 10
s = 20
k = len(integers)
def print_grid(grid, title):
print(f'\n{title}')
for row in grid:
print(' '.join([str(element) for element in row]))
# Create the starting grid
grid = []
for i in range(1, k + 1):
if i < k:
grid.append(['.' for j in range(n)])
else:
grid.append(['X' for j in range(n)])
# Print the starting grid
print_grid(grid, 'Starting grid')
# Randomly and uniformly fill in the remaining rows
indexes = np.random.choice(range((k - 1) * n), s - n, replace=False)
for i in indexes:
row = i // n
col = i % n
grid[row][col] = 'X'
# Print the filled in grid
print_grid(grid, 'Filled in grid')
# Compute how many cells were selected in each column
column_counts = []
for col in range(n):
count = sum(1 for i in range(k) if grid[i][col] == 'X')
column_counts.append(count)
grid.append(column_counts)
# Print the final grid and check that the column counts sum to s
print_grid(grid, 'Final grid')
print()
print(f'Do the column counts sum to {s}? {sum(column_counts) == s}.')
I've turned whuber's answer into an r function. I hope it helps someone.
• n is how many integers you want;
• t is the mean you want; and
• k is the upper limit you want for your returned values
whubernator<-function(n=NULL, t=NULL, kMax=5){
z = tabulate(sample.int(kMax*(n), (n)*(t),replace =F) %% (n)+1, (n))
return(z)
}
It seems to work as expected:
> w = whubernator(n=10,t=4.2)
> mean(w)
[1] 4.2
> length(w)
[1] 10
> w
[1] 3 5 3 5 5 3 4 5 5 4
It can return 0s, which matches my needs.
> whubernator(n=2,t=0.5)
[1] 1 0
New contributor
gruvn is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. | 2020-04-09T15:19:18 | {
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https://www.physicsforums.com/threads/complex-integer-expression-problem.111954/ | # Homework Help: Complex integer expression problem
1. Feb 23, 2006
### Werg22
If n is a positive integer such as
$$2{\leq}n{\leq}80$$
For how many values the expression $$\frac{(n+1)n(n-1)}{8}$$ takes positive and integer values?
I solved it that way...
$$\frac{(n+1)n(n-1)}{8}=\frac{(n^{2}-1)n}{8}$$
(n^2 - 1)n must have 8 as one of its factor.
Either n is a multiple of 8, or n^2 - 1 is. Also the case were n^2 - 1 has 4 as one of its factors, n having 2, and vice-versa, is impossible - if n^2 - 1 is even, n is odd, and vice-versa.
So every mutliple of 8 up to 80 is a possible value. So there is 10 values.
Let's list those numbers
8, 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72, 80
Add one to each one of these values
9, 17, 25, 33, 41, 49, 65 , 73, 81
There is 4 perfect square in this list. So if n^2 - 1 is a multiple of 8, then there is 4 possible values for n.
10 + 4 = 14
So 14 possibilities in total. But the true awnser is not what I found. What is wrong in my reasoning?
2. Feb 23, 2006
### 0rthodontist
There's no guarantee that n^2 - 1 lies between 9 and 81. It could be much larger.
3. Feb 23, 2006
### Werg22
Right! Thanks. In that case any other way to solve this?
4. Feb 23, 2006
### 0rthodontist
Well since you are only looking up to 80 you can calculate them all directly, only takes a couple minutes to set things up. Also from observing that data it seems that every n yields an integer n(n+1)(n-1)/8 except for even integers that are not divisible by 8, so maybe you could break it down into parts. Every multiple of 8 yields an integer. And if x is a multiple of 2, and x is not itself divisible by 8, then x does not yield an integer because its adjacent integers are not even disible by 2. And every odd integer must yield an integer because its adjacent integers are both even, and one of the adjacent integers must also be divisible by 4.
Last edited: Feb 23, 2006
5. Feb 23, 2006
### Werg22
Ok. Since n^2 - 1 = (n-1)(n+1)
n - 1 must be a multiple of 8, or n+1 be a multiple of 8, or n - 1 be a multiple of 4, or n + 1 be a multiple of 4.
For exactly 10 values, n is multiple of 8
For exactly 10 values, n-1 is a multiple of 8.
For exactly 10 values, n+1 is a multiple of 8.
For exactly 20 values, n-1 is multiple of 4. Half of these being multiples of 8. So we count 10.
For exactly 20 values, n+1 is a multiple of 8. Half of these being multiples of 8. So we count 10.
So 10*5=50
The awnser is 50.
6. Feb 23, 2006
### AKG
I believe the answer is 49. You seem to be going about it in a very complicated way.
Case 1: n is odd
Then both n-1 and n+1 are even. Moreover, one of them (and in fact, only one of them) is a multiple of four. This is clear since if n is odd, then either n = 1 (mod 4) in which case n-1 = 0 (mod 4) and n+1 = 2 (mod 4), or n = 3 (mod 4) in which case n-1 = 2 (mod 4) and n+1 = 0 (mod 4). So since one of n-1 and n+1 is a multiple of four, and the other is even, the whole product (n-1)n(n+1) is a multiple of 8. So every odd n between 2 and 80 will do, and there are 39 such numbers.
Case 2: n is even
Then both n-1 and n+1 is odd, so if 8 | (n-1)n(n+1), then 8 | n, so the only even n's that work are multiples of 8. They are:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80
That's 10, giving a total of:
49.
One problem with your solution is that n-1 is a multiple of 8 for only 9 values of n. Note that n-1 ranges from 1 to 79. n ranges from 2 to 80. n-1 will never be 80, so its missing one multiple of 8. That's where you're counting your extra one. Note also that there are only 19 values of n-1 which are a multiple of 4, 9 of which are multiples of 8, so when you subtract 9 from 19, you do still end up getting 10. Originally, you subtracted 10 from 20. You ended up with the right number, 10, but they way you got it was wrong.
7. Feb 24, 2006
### Werg22
Okay, I see. Thanks. | 2018-04-21T08:12:18 | {
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https://math.stackexchange.com/questions/1089741/does-sum-limits-n-log-left11-over-n-right-diverge-or-converge | Does $\sum\limits_n \log\left(1+{1\over n}\right)$ diverge or converge?
How do I find out if $\sum\limits_n\log(1+{1\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.
• One can use Limit Comparison with $\sum \frac{1}{n}$. – André Nicolas Jan 3 '15 at 15:37
Wolfram recommends me to use (some) compar(is)on test, but I do not know (any) series which diverges and (is) less than this (one).
$$\log\left(1+\frac1n\right)\geqslant\frac1{2n}$$
$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n}=\log(N+1).$$
If you want to use a comparison, notice that since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1/2}$$ by Jensen's inequality, hence: $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)\geq 2\sum_{n=1}^{N}\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$
Since
$$\sum_{n = 1}^N \log\left(1 + \frac{1}{n}\right) = \sum_{n = 1}^N \log\left(\frac{n+1}{n}\right) = \sum_{n = 1}^N [\log(n+1) - \log(n)] = \log(N+1) \to \infty$$
the series $\sum_{n = 1}^\infty \log\left(1 + \frac{1}{n}\right)$ diverges.
note $$\ln{\left(1+\dfrac{1}{n}\right)}=\dfrac{1}{n}+o(1/n)$$ since $$\sum_{n=1}^{\infty}\dfrac{1}{n}$$ is diverges
so $$\sum_{n=1}^{\infty}\ln{\left(1+\dfrac{1}{n}\right)}$$ is also diverges
• First statement isn't obvious to me. – Dark Archon Jan 3 '15 at 16:10
• do you know $\ln{(1+x)}=x+o(x)?$ – math110 Jan 3 '15 at 16:11
• What is $o(x)$? – Dark Archon Jan 3 '15 at 16:13
• $x=\dfrac{1}{n}$ when $n\to +\infty$,then $x\to 0$ – math110 Jan 3 '15 at 16:25
• @DarkArchon $f(x)=o(x)$ means that $\lim_{x\to 0}f(x)/x=0$. – Andrés E. Caicedo Jan 4 '15 at 0:52
The first natural idea to understand how $\log(1+1/n)$ behaves when $n$ is small is to do Taylor expansion around $x=0$ for $\log(1+x)$, so that you have estimates for $\log(1+x)$. Then you get $$\log(1+x) = \sum_{n \ge 1} \frac{(-1)^{n+1}}{n}x^n = x - x^2/2 + x^3/3 - \cdots,$$ by integrating the geometric series (and switching a minus sign) and by Taylor's theorem you get that there exists $1 \le \zeta \le x$ such that $$\log(1+x) = x-\frac{\zeta^2}2 \ge x - \frac {x^2}2 = \frac{2x-x^2}2 = \frac{x(2-x)}2 \ge \frac x2$$ for $0 \le x \le 1$ (because $\frac{x(1-x)}2 \ge 0$ on this interval).
A variant of this idea is to check that $\log$ is a concave function (the derivatives are $\frac 1x$ and $\frac {-1}{x^2}$), hence we can use Jensen's inequality : for all $\lambda \in [0,1]$, $$\log(\lambda x + (1-\lambda) y) \ge \lambda \log(x) + (1-\lambda) \log(y)$$ so that for $y=1$ and $x=2$, we get $$\log(1+ \lambda) \ge \lambda \log(2).$$ In both cases, we established an inequality of the form $\log(1+x) \ge cx$ for $x \in [0,1]$, which means $\log(1+1/n) \ge \frac cn$, hence your series diverges.
Hope that helps,
Another way to think of this problem is that $\sum log(1+1/n)$ = $\log \prod(1+1/n)$ = $\log \prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.
• Already explained on the page, twice. – Did May 11 '15 at 5:21 | 2020-02-18T10:14:44 | {
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http://math.stackexchange.com/questions/56556/a-problem-about-the-limit-of-an-integral | # A problem about the limit of an integral
Let $g(x)$ be a continuous periodic function of period 1 on $\mathbb{R}$. Prove that for any integrable function $f(x)$ on $[0,1]$,
$$\lim_{n \to \infty}\int_0^{1}f(x)g(nx)dx= \int_0^{1}f(x)dx \int_0^{1}g(x)dx.$$
Any help is appreciated.
-
Begin with a transformation $u=nx.$
$$\int_{0}^1 f(x)g(nx)dx = \frac{1}{n} \int_{0}^n f(u/n)g(u)du.$$
Break the integrand up into a sum of intervals.
$$\frac{1}{n} \int_{0}^n f(u/n)g(u)du=\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du.$$
Make another variable transformation: $v=u-(j-1).$ Because $g(u)$ is 1 periodic $g(u)=g(u+1)=g(u+j-1).$
$$\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du = \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f \left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du$$
Rearrange the terms. $$\frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f\left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du =\int_{0}^1 \left(\frac{1}{n} \sum_{j=1}^n f \left(\frac{u-(j-1)}{n} \right)\right)g(u)du .$$
We have now produced a Riemann sum which converges to an integral in the limit. We are now allowed, by dominated convergence theorem, to say
\begin{align*} \lim_{n\to \infty} \int_{0}^1 f(x)g(nx)dx &= \int_{0}^1 \left(\int_{0}^1 f(z)dz\right) g(u)du \\ &=\int_{0}^1 f(z)dz\int_{0}^1 g(u)du=\int_{0}^1 f(x)dx\int_{0}^1 g(x)dx . \end{align*}
-
Thank you so much. – Sume Aug 10 '11 at 3:28
this is not a good proof, just an idea. First consider f to be simple function, we can calculate that this identity holds. Then by the definition of Lebesgue integration, we can approximate f by simple functions. Because g is continuous(bounded), so there's no problem for the left hand side to converge.
-
Ok. I have seen this problem in the book: Principles of Real Analysis by C.D.Aliprantis and O.Burkinshaw, and since I knew that this book has a solution manual, I went and searched over there and got the solution. The problem given in the book is as follows:
$\textbf{Problem.}$ Let $f:(0,\infty) \to \mathbb{R}$ be a real-valued continuous function such that $f(x+1)=f(x)$ for all $x \geq 0$. If $g:[0,1] \to \mathbb{R}$ is an arbitrary continuous function, then show that $$\lim_{n \to \infty} \ \int\limits_{0}^{1} g(x) \cdot f(nx) \ dx = \Biggl( \ \ \int\limits_{0}^{1} g(x) \ dx \Biggr) \cdot \Biggl( \ \ \int\limits_{0}^{1} f(x) \ dx\Biggr)$$
$\textbf{Solution.}$ Please see the book: Problems in real analysis a workbook with solutions Problem 23.14 Page $\textbf{205}$ for a complete solution.
- | 2016-05-30T22:21:33 | {
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https://benjaminleroy.github.io/documents/36350/Labs/lab_3.1-assign.html | Name:
Andrew ID:
Collaborated with:
This lab is to be done in class (completed outside of class if need be). You can collaborate with your classmates, but you must identify their names above, and you must submit your own lab as an knitted HTML file on Canvas, by Tuesday 10pm, this week.
This week’s agenda: creating and updating functions; understanding argument and return structures; revisiting Shakespeare’s plays; code refactoring.
# Huber loss function
The Huber loss function (or just Huber function, for short) is defined as: $\psi(x) = \begin{cases} x^2 & \text{if |x| \leq 1} \\ 2|x| - 1 & \text{if |x| > 1} \end{cases}$ This function is quadratic on the interval [-1,1], and linear outside of this interval. It transitions from quadratic to linear “smoothly”, and looks like this:
It is often used in place of the usual squared error loss for robust estimation. The sample average, $$\bar{X}$$—which given a sample $$X_1,\ldots,X_n$$ minimizes the squared error loss $$\sum_{i=1}^n (X_i-m)^2$$ over all choices of $$m$$—can be inaccurate as an estimate of $$\mathbb{E}(X)$$ if the distribution of $$X$$ is heavy-tailed. In such cases, minimizing Huber loss can give a better estimate. (Interested in hearing more? Come ask Tudor or I!)
• 1a. Write a function huber() that takes as an input a number $$x$$, and returns the Huber value $$\psi(x)$$, as defined above. Hint: the body of a function is just a block of R code, i.e., in this code you can use if() and else() statements. Check that huber(1) returns 1, and huber(4) returns 7.
• 1b. The Huber function can be modified so that the transition from quadratic to linear happens at an arbitrary cutoff value $$a$$, as in: $\psi_a(x) = \begin{cases} x^2 & \text{if |x| \leq a} \\ 2a|x| - a^2 & \text{if |x| > a} \end{cases}$ Starting with your solution code to the last question, update your huber() function so that it takes two arguments: $$x$$, a number at which to evaluate the loss, and $$a$$ a number representing the cutoff value. It should now return $$\psi_a(x)$$, as defined above. Check that huber(3, 2) returns 8, and huber(3, 4) returns 9.
• 1c. Update your huber() function so that the default value of the cutoff $$a$$ is 1. Check that huber(3) returns 5.
• 1d. Check that huber(a = 1, x = 3) returns 5. Check that huber(1, 3) returns 1. Explain why these are different.
• 1e. Vectorize your huber() function, so that the first input can actually be a vector of numbers, and what is returned is a vector whose elements give the Huber evaluated at each of these numbers. Hint: you might try using ifelse(), if you haven’t already, to vectorize nicely. Check that huber(x = 1:6, a = 3) returns the vector of numbers (1, 4, 9, 15, 21, 27).
• Challenge. Your instructor computed the Huber function values $$\psi_a(x)$$ over a bunch of different $$x$$ values, stored in huber_vals and x_vals, respectively. However, the cutoff $$a$$ was, let’s say, lost. Using huber_vals, x_vals columns of oops_df, and the definition of the Huber function, you should be able to figure out the cutoff value $$a$$, at least roughly. Estimate $$a$$ and explain how you got there. Hint: one way to estimate $$a$$ is to do so visually, using plotting tools (if you do - please use ggplot); there are other ways too.
oops_df <- data.frame(
x_vals = seq(0, 5, length=21),
huber_vals = c(0.0000, 0.0625, 0.2500, 0.5625, 1.0000, 1.5625, 2.2500,
3.0625, 4.0000, 5.0625, 6.2500, 7.5625, 9.0000, 10.5000,
12.0000, 13.5000, 15.0000, 16.5000, 18.0000, 19.5000,
21.0000))
# Shakespeare’s complete works
Recall, as in lab/hw from Week 1, that the complete works of William Shakespeare are available freely from Project Gutenberg. We’ve put this text file up at https://raw.githubusercontent.com/benjaminleroy/36-350-summer-data/master/Week1/shakespeare.txt.
# Getting lines of text play-by-play
• 2a. Below is the get_wordtab_from_url() from lecture. Modify this function so that the string vectors lines and words are both included as named components in the returned list. For good practice, update the documentation in comments to reflect your changes. Then call this function on the URL for the Shakespeare’s complete works (with the rest of the arguments at their default values) and save the result as shakespeare_wordobj. Using head(), display the first several elements of (definitely not all of!) the lines, words, and wordtab components of shakespeare_wordobj, just to check that the output makes sense to you.
#' Get a word table from text on the web
#'
#' @param str_url string, specifying URL of a web page
#' @param split string, specifying what to split on. Default is the regex
#' pattern "[[:space:]]|[[:punct:]]"
#' @param tolower Boolean, TRUE if words should be converted to lower case
#' before the word table is computed. Default is TRUE
#' @param keep_nums Boolean, TRUE if words containing numbers should be kept in
#' the word table. Default is FALSE
#'
#' @return list, containing word table, and then some basic numeric summaries
#'
#' @examples
#' endgame_speech_list <- get_wordtab_from_url(
#' paste0("https://raw.githubusercontent.com/benjaminleroy/",
#' "36-350-summer-data/master/Week1/endgame.txt"))
get_wordtab_from_url <- function(str_url, split = "[[:space:]]|[[:punct:]]",
tolower = TRUE,
keep_nums = FALSE) {
text <- paste(lines, collapse = " ")
words <- strsplit(text, split = split)[[1]]
words <- words[words != ""]
# Convert to lower case, if we're asked to
if (tolower) {
words <- tolower(words)
}
# Get rid of words with numbers, if we're asked to
if (!keep_nums) {
words <- grep("[0-9]", words, inv = TRUE, val = TRUE)
}
# Compute the word table
wordtab <- table(words)
return(list(wordtab = wordtab,
number_unique_words = length(wordtab),
number_total_words = sum(wordtab),
longest_word = words[which.max(nchar(words))]))
}
• 2b. Go back and look Q5 of Homework 1, where you located Shakespeare’s plays in the lines of text for Shakespeare’s complete works. Set shakespeare_lines <- shakespeare_wordobj$lines, and then rerun your solution code (or the rerun the official solution code, if you’d like) for questions Q5a–Q5f of Homework 1, on the lines of text stored in shakespeare_wordobj$lines. You should end up with two vectors titles_start and titles_end, containing the start and end positions of each of Shakespeare’s plays in shakespeare_lines. Print out titles_start and titles_end to the console.
• 2c. Create a list shakespeare_lines_by_play of length equal to the number of Shakespeare’s plays (a number you should have already computed in the solution to the last question). Using a for() loop, and relying on titles_start and titles_end, extract the subvector of shakespeare_lines for each of Shakespeare’s plays, and store it as a component of shakespeare_lines_by_play. That is, shakespeare_lines_by_play[[1]] should contain the lines for Shakespeare’s first play, shakespeare_lines_by_play[[2]] should contain the lines for Shakespeare’s second play, and so on. Name the components of shakespeare_lines_by_play according to the titles of the plays.
# Getting word tables play-by-play
• 3a. Define a function get_wordtabfrom_lines() to have the same argument structure as get_wordtab_from_url(), which recall you last updated in Q2a, except that the first argument of get_wordtab_from_lines() should be lines, a string vector passed by the user that contains lines of text to be processed. The body of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines is passed and does not need to be computed using readlines(). The output of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines does not need to be returned as a component. For good practice, incude documentation for your get_wordtab_from_lines() function in comments (no need to include an example).
• 3b. Using a for() loop or one of the apply functions (your choice here), run the get_wordtab_from_lines() function on each of the components of shakespeare_lines_by_play, (with the rest of the arguments at their default values). Store the result in a list called shakespeare_wordobj_by_play. That is, shakespeare_wordobj_by_play[[1]] should contain the result of calling this function on the lines for the first play, shakespeare_wordobj_by_play[[2]] should contain the result of calling this function on the lines for the second play, and so on.
• 3c. Using one of the apply functions, compute numeric vectors shakespeare_total_words_by_play and shakespeare_unique_words_by_play, that contain the number of total words and number of unique words, respectively, for each of Shakespeare’s plays. Each vector should only require one line of code to compute. Hint: "[["() is actually a function that allows you to do extract a named component of a list; e.g., try "[["(shakespeare_wordobj, "number_total_words"), and you’ll see this is the same as shakespeare_wordobj[["number_total_words"]]; you should take advantage of this functionality in your apply call. What are the 5 longest plays, in terms of total word count? The 5 shortest plays?
# Refactoring the word table functions
• 4. Look back at get_wordtab_from_lines() and get_wordtab_from_url(). Note that they overlap heavily, i.e., their bodies contain a lot of the same code. Redefine get_wordtab_from_url() so that it just calls get_wordtab_from_lines() in its body. Your new get_wordtab_from_url() function should have the same inputs as before, and produce the same output as before. So externally, nothing will have changed; we are just changing the internal structure of get_wordtab_from_url() to clean up our code base (specifically, to avoid code duplication in our case). This is an example of code refactoring.
Call your new get_wordtab_from_url() function on the URL for Shakespeare’s complete works, saving the result as shakespeare_wordobj2. Compare some of the components of shakespeare_wordobj2 to those of shakespeare_wordobj (which was computed using the old function definition) to check that your new implementation works as it should.
• Challenge. Check using all.equal() whether shakespeare_wordobj and shakespeare_wordobj2 are the same. Likely, this will not return TRUE. (If it does, then you’ve already solved this challenge question!) Modify your get_wordtab_from_url() function from the last question, so that it still calls get_wordtab_from_lines() to do the hard work, but produces an output exactly the same as the original shakespeare_wordobj object. Demonstrate your success by calling all.equal() once again. | 2022-08-11T07:42:11 | {
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http://mathhelpforum.com/geometry/143843-solved-finding-co-ordinates-rectangle.html | # Math Help - [SOLVED] Finding Co-Ordinates of a Rectangle
1. ## [SOLVED] Finding Co-Ordinates of a Rectangle
Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:
Is my method correct, considering I did get the right answer?
But is there another simpler method to do this which would save time during an exam.
2. The diagonals bisect one another. The midpoint of $\overline{AC}$ is ?
3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
4. Originally Posted by unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
But the y-coordinate of $B~\&~D$ is 6.
You are given the x-coordinate of $D$, so $Dh,6)" alt="Dh,6)" />
5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
6. Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
Plato already told you that the midpoint of BD is M(6, 6).
This means the y-coordinates of B and D are also 6.
This distance from B to D is 20 (found using distance formula)
Each individual segment of the diagonals measure 10 since they are bisected.
Using the distance formula, it is easy to determine the x-coordinates of B and D.
M(6, 6) -----> D(h, 6) = 10
$10=\sqrt{h-6)^2+(6-6)^2}$
7. Hello, unstopabl3!
Code:
| C(12,14)
| o
| * *
| * *
* *
B o - + - - - - - - - o D
* | *
------*-+-------*------------
*| *
o
A|(0,-2)
|
The diagram shows a rectangle $ABCD.$
We have: . $A(0,-2),\;C(12,14)$
The diagonal $BD$ is parallel to the $x$-axis.
$(i)$ Explain why the $y$-coordinate of $D$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $AC$ is the midpoint of $BD.$
The midpoint of AC is: . $\left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$
Therefore, $B$ and $D$ have a $y$-coordinate of 6.
The $x$-coordinate of $D$ is $h.$
$(ii)$ Express the gradients of $AD$ and $CD$ in terms of h,.
We have: . $\begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$
$m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$
$m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$
$(iii)$ Calculate the $x$-coordinates of $D$ and $B.$
Since $m_{AD} \perp m_{CD}$ we have: . $\frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$
Hence: . $(h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$
Therefore: . $D(16,6),\;B(-4,6)$
8. Originally Posted by masters
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
$10=\sqrt{h-6)^2+(6-6)^2}$
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!
This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
9. Originally Posted by unstopabl3
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).
10. Co ordinates of B and D are $(x_1 , 6) and (x_2, 6)$
Diagonal AC = BD
AC = 20 = BD.
Distance $BD^2 = (x_1 - x_2)^2$
So (x_1 - x_2) = 20...........(1)
Mid point point of AC = mid point of BD
$\frac{x_1+x_2}{2} = 6$
(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.
11. Thanks for the responses guys! | 2015-02-28T16:02:04 | {
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https://math.stackexchange.com/questions/4300731/how-to-express-the-nth-power-of-the-cosine-as-a-series-of-cosines | # how to express the $n$th power of the cosine as a series of cosines?
Which is the correct way for expressing the $$n$$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n= \frac{1}{2^n}\left( e^{j\theta}+e^{-j\theta} \right)^n=\frac{1}{2^n}\left( z+z^{-1} \right)^n.$$
with $$z=e^{j\theta}$$.
Since the term $$\left( z+z^{-1} \right)^n$$ is the $$n$$th power of a binomial, I could express it using the binomial identity, thus
$$\cos^n{(\theta)} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^k(z^{-1})^{n-k} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^{2k-n}.$$
If we expand the expression above, we obtain
$$\cos^n{(\theta)}=\frac{1}{2^n}\Bigg ( z^{-n} + \binom{n}{1}z^{-(n-2)} + \binom{n}{2}z^{-(n-4)} + \dots + \binom{n}{2}z^{n-4} + \binom{n}{1}z^{(n-2)} + z^n \Bigg )$$
which can be rewritten as
$$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (z^{-n} + z^n) + \binom{n}{1} \left(z^{-(n-2)} + z^{(n-2)}\right) + \binom{n}{2}\left( z^{-(n-4)} + z^{(n-4)}\right) + \dots \Bigg )$$
Finally, since $$z=e^{j\theta}$$
$$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (e^{-jn\theta} + e^{jn\theta}) + \binom{n}{1} \left(e^{j(n-2)\theta} + e^{-j(n-2)\theta}\right) + \binom{n}{2}\left( e^{j(n-4)\theta} + e^{-j(n-4)\theta}\right) + \dots \Bigg )$$
By applying the Euler's formula once again, we obtain
$$\cos^n{(\theta)}=\frac{2}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$$
Unfortunately, if I plug n=2, I obtain
$$\cos^2{(\theta)}= \cos{(2\theta)} + 1.$$
$$\cos^2{(\theta)}= \frac{1}{2}(\cos{(2\theta)} + 1).$$
a) Why my result is scaled by a factor of 2 ? b) Is the correct general formula $$\cos^n{(\theta)}=\frac{1}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$$ c)If so, why ?
• You paired the terms up $k=0$ with $k=n$ etc so when you rewrite it in summation notation the sum shouldn't be over all $0 \leq k \leq n$ anymore. Also for $n$ even there is a term in the middle with no pair. Nov 8, 2021 at 22:33
• The right way to do this is a (finite) Fourier series. Nov 8, 2021 at 22:37
• And this can be used to compute the integral $$\int \cos^n(x) dx$$ Nov 8, 2021 at 22:43
• @podiki Yeah, I noticed the even $n$ constant, which is $\binom {n}{n/2}cos(0)$. I was trying to be as general as possible. Thanks for pointing out the error when returning to the summation notation. I will try to fix it and update the question. Nov 8, 2021 at 22:49
• Also, you never get out of summation notation. Nov 9, 2021 at 1:48
Here is a way to fix the factor of $$2$$ while still summing from $$0$$ to $$n$$ (setting aside the question of whether that is the best way to do the sum). You have $$\cos^n(\theta) = \frac{1}{2^n}\left( z^{-n} + \binom n1 z^{-(n-2)} + \binom n2 z^{-(n-4)} + \cdots + \binom n2 z^{n-4} + \binom n1 z^{n-2} + z^n \right).$$ Reversing the order of the sum, $$\cos^n(\theta) = \frac{1}{2^n}\left( z^n + \binom n1 z^{n-2} + \binom n2 z^{n-4} + \cdots + \binom n2 z^{-(n-4)} + \binom n1 z^{-(n-2)} + z^{-n} \right).$$ Adding termwise, \begin{align} 2 \cos^n(\theta) &= \frac{1}{2^n}\bigg( \left(z^{-n} + z^n\right) + \binom n1 \left(z^{-(n-2)} + z^{n-2}\right) + \binom n2 \left(z^{-(n-4)} + z^{n-4}\right) + \cdots \\ &\qquad\qquad + \binom n2 \left(z^{n-4} + z^{-(n-4)}\right) + \binom n1 \left(z^{n-2} + z^{-(n-2)}\right) + \left(z^n + z^{-n}\right) \bigg) \\ &=\frac{2}{2^n}\bigg(\cos(n\theta) + \binom n1 \cos((n-2)\theta) + \binom n2 \cos((n-4)\theta) + \cdots \\ &\qquad\qquad + \binom n2 \cos(-(n-4)\theta) + \binom n1 \cos(-(n-2)\theta) + \cos(-n\theta) \bigg) . \end{align}
Canceling one factor of $$2$$ on each side, this simplifies to $$\cos^n(\theta) = \frac{1}{2^n} \sum_{k=0}^n \binom nk \cos{((n-2k)\theta)}.$$ For $$n = 2,$$ this gives \begin{align} \cos^2(\theta) &= \frac 14\left(\binom 20 \cos(2\theta) + \binom 21 \cos(0) + \binom 22 \cos(-2\theta) \right) \\ &= \frac 12\left(\cos(2\theta) + 1\right) \end{align} as expected.
Your mistake was you did not write the end of the sum after the three dots, so you did not notice that you had moved terms from the right end of the sum to the left end without replacing them, so about half the terms in your summation were not matched by terms in the $$+ \cdots +$$ notation. Adding two copies of the sum in reverse order, you don't have to move any terms so you won't make this mistake.
Note that it's conventional to combine the $$\cos(m\theta)$$ and $$\cos(-m\theta)$$ terms in the sum and have about half as many terms (exactly half as many for odd powers) at the cost of (usually) having separate summations for even and odd powers of the cosine.
• Thanks a lot, this was the answer I was looking for! I still prefer this method instead of more complicated ones, especially from a learning perspective. Nov 9, 2021 at 8:37
We start from $$s_n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} z^{2k-n}$$ Case 1: Let $$n = 2m$$ $$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{2m} \binom{2m}{k} z^{2(k-m)}$$
$$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{k=m+1}^{2m} \binom{2m}{k} z^{2(k-m)}$$
$$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{l=0}^{m-1} \binom{2m}{2m - l} z^{2(m - l)}$$
$$s_{n} = \frac{n!}{2^{n}(n/2)!^2} + \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} (z^{2k-n} + z^{n - 2k})$$
$$cos^{n}(\theta) = \frac{n!}{2^{n}(n/2)!^2} + \frac{2}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}}\binom{n}{n/2} \cos((2(n/2)-n)\theta) + \frac{1}{2^{n}} \sum_{l=n/2+1}^{n} \binom{n}{l} \cos((2l-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$
Case 2: Let $$n = 2m - 1$$ $$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1}$$
$$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k=m}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1}$$
$$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k = 0}^{m} \binom{2m - 1}{k} z^{2(m-k) - 1}$$
$$s_{n} = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} (z^{2k-n} + z^{n-2k})$$
$$cos^{n}(\theta) = \frac{2}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}} \sum_{l=(n + 1)/2}^{n} \binom{n}{l} cos((2l-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$
Therefore, mixing the both cases we get
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$ | 2022-07-06T16:18:58 | {
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https://hcaus.org/feverfew-cancer-hbmvgtz/difference-between-scalar-matrix-and-identity-matrix-14a08f | # difference between scalar matrix and identity matrix
The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. If you multiply any number to a diagonal matrix, only the diagonal entries will change. #1. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. See the picture below. The column (or row) vectors of a unitary matrix are orthonormal, i.e. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. 8) Unit or Identity Matrix. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! References [1] Blyth, T.S. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. 2. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if Back in multiplication, you know that 1 is the identity element for multiplication. In this post, we are going to discuss these points. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. While off diagonal elements are zero. If the block produces a scalar output from a scalar input, the block preserves dimension. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. Yes it is. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Okay, Now we will see the types of matrices for different matrix operation purposes. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. Basis. All the other entries will still be . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. 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Answer Short: a$ 1\times 1 \$ matrix is basically a square matrix all... 1 x 0, it is called identity matrix and its diagonal elements are to! | 2021-07-30T13:40:24 | {
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http://math.stackexchange.com/questions/170331/why-is-int-0-infty-frac-ln-x1x2-mathrmdx-0/170365 | # Why is $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x =0$?
We had our final exam yesterday and one of the questions was to find out the value of: $$\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x$$ Interestingly enough, using the substitution $x=\frac{1}{t}$ we get - $$-\int_{0}^{1} \frac {\ln x}{1+x^2} \mathrm{d}x = \int_{1}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x$$and therefore $\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x = 0$
I was curious to know about the theory behind this interesting (surprising even!) example.
Thank you.
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Thanks. What is antisymmetric? – Amihai Zivan Jul 13 '12 at 13:37
i.e. an odd function – anon Jul 13 '12 at 13:38
Ah, OK. I wasn't familiar with "antisymmetric"... – Amihai Zivan Jul 13 '12 at 13:40
@J.M. - OK sir. – Amihai Zivan Jul 13 '12 at 14:50
You should make that substitution, certainly. But you should also show the integral converges. – GEdgar Dec 8 '12 at 18:07
When I see an $1 + x^2$ in the denominator it's tempting to let $\theta = \arctan(x)$ and $d\theta = {1 \over 1 + x^2} dx$. When you do that here the integral becomes $$\int_0^{\pi \over 2} \ln(\tan(\theta))\,d\theta$$ $$= \int_0^{\pi \over 2} \ln(\sin(\theta))\,d\theta - \int_0^{\pi \over 2} \ln(\cos(\theta))\,d\theta$$ The two terms cancel because $\cos(\theta) = \sin({\pi \over 2} - \theta)$.
Also, if you do enough of these, you learn that doing the change of variables from $x$ to ${1 \over x}$ converts a ${dx \over 1 + x^2}$ into $-{dx \over 1 + x^2}$, so it becomes one of the "tricks of the trade" for integrals with $1 + x^2$ in the denominator. An example: show this trick can be used to show that the following integral is independent of $r$: $$\int_0^{\infty} {dx \over (1 + x^2)(1 + x^r)}$$
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$\int_0^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_0^1 {dx \over (1 + x^2)(1 + x^r)}+\int_1^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_1^{\infty} {t^r \over (1 + t^2)(1 + t^r)}dt+\int_1^{\infty} {dx \over (1 + x^2)(1 + x^r)}=\int_1^{\infty} {dt \over (1 + t^2)}$. very nice indeed! – Amihai Zivan Jul 14 '12 at 18:40
I'm not exactly sure what kind of theory behind the integral you're looking for, but to me the points that pop out are that $dx/x=d(\log x)$ and $1+x^2=(1/x+x)x$ so that we have
$$\frac{\log x}{1+x^2}dx=\frac{u\, du}{e^{-u}+e^u}$$
after the change of variables $u=\log x$. As $x$ ranges over $(0,\infty)$, $u$ ranges over $\Bbb R$, and the integrand in the right-hand side, $u/(e^{-u}+e^u)$, is an antisymmetric aka odd function of $u$. Integrals of odd functions on intervals that are symmetric about the origin are always zero.
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With improper integrals this is not completely true: we would have $\int_{-\infty}^\infty x dx = 0$, but that integral does not converge. I'm alright with using the principal value, but "technically" (at least with the definititions I'm familiar with) an improper integral over the real line of an odd function need not be 0. – Andy Jan 30 '13 at 10:06
In hindsight, one can extract a general principle from this example. Let $f(x)$ be a say continuous function. Suppose also that $$\frac{1}{x}f\left(\frac{1}{x}\right)=-xf(x)$$ for all relevant $x$. Then for any $b\ne 0$, $$\int_{1/b}^b f(x)\,dx=0.\tag{1}$$
Under the same conditions, if the improper integral converges, we have $$\int_0^\infty f(x)\,dx=0.$$
The proof of either result is the same as the proof by anon in the particular case $f(x)=\frac{\log x}{1+x^2}$. For $(1)$, break up the integral into two parts, $1/b$ to $1$ and $1$ to $b$. For the integral between $1/b$ and $1$, make the change of variable $u=1/x$.
Remark: If a trick or idea solves a concrete problem, one can reverse engineer and identify the problems for which essentially the same idea works. In this case, the reverse engineering does not seem to produce something of general interest. Instead, one should just draw the general lesson: Symmetry is your friend. Exploit it. (That rewording of Polya didn't come out sounding quite right.)
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It is sufficient to consider $x={e}^{t}$. Then $dx={e}^{t}\,dt$. we have:
$$\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}dx=\int_{-\infty}^{\infty}\frac{t{e}^{t}}{1+{e}^{2t}}dt=0$$
Recall that the function $\frac{t\mathrm{e}^{t}}{1+e^{2t}}$ is odd.
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Does this differ from the method mentioned in the question? – robjohn Aug 16 '14 at 11:44
@robjohn They are equivalent but this one doesn't split $\left(0,\infty\right)$. Thanks. – Felix Marin Aug 16 '14 at 21:14
At the risk of stating the obvious, I would suggest examining the curve of ${\ln x}\over{(1+x^2)}$:
The geometrical interpretation is that the area below the $x$-axis down to the curve from 0 to 1 is equal to the area above the $x$-axis up to the curve from 1 to infinity.
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Note that the function ln(x) is negative on the interval $(0, 1)$, so the whole integrand is negative on the interval $(0,1)$. While $ln(x)$ is positive on the interval $(1, \infty)$, so the whole integrand is positive on the interval $(1,\infty)$. By splitting the integral on the above two intervals and evaluating the two integrals, we find the value of the integral on the interval $(0,1)$ equals -catalan ( $\sim 0.915965594$. ) and value of the integral on the interval $(1,\infty )$ equals catalan. So the value of the whole integral is $0$.
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You demonstrated yourself why the result is 0 (by making the change $u = \frac{1}{x}$).
I think you can view this it as the same as this integral: $\displaystyle\int_{-\infty}^{\infty}x dx = \displaystyle\lim_{X\rightarrow +\infty} \int_{-X}^X xdx=0$.
Note that I am not sure that $\int_{-\infty}^{\infty}x dx$ is actually defined, but this also applies to your integral $\displaystyle\int_0^{\infty}\displaystyle\frac{\ln x}{1+x^2}dx$.
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$\int_{-\infty}^{\infty}x dx$ is divergent, but can be evaluated in the sense of the Cauchy principal value as $0$. – J. M. Jul 13 '12 at 17:02
@J.M. Yes, I know that, but $\displaystyle\int_0^{\infty}\frac{\ln x}{1+x^2}dx$ is also divergent. – S4M Jul 13 '12 at 17:06
I don't see how the OP's integral is divergent. – anon Jul 13 '12 at 17:13 | 2016-02-06T07:56:08 | {
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https://mathhelpboards.com/threads/abc-a-b-c.7245/ | # abc = a! + b! + c!
#### Albert
##### Well-known member
A=abc=a!+b!+c!
here A is a 3-digit number
find A
#### eddybob123
##### Active member
Re: abc=a!+b!+c!
Are a, b, and c digits or are they positive integers?
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
Are a, b, and c digits or are they positive integers?
A=100a+10b+c=a!+b!+c!
a,b,c $\subset$ { 0,1,2,3,4,5,6,7,8,9 }
and a$\neq 0$
find A
##### Well-known member
Re: abc=a!+b!+c!
145 = 1! + 4! + 5!
reason
none of abc can be > 5 as 6! = 720 and 7! = 5040 > 1000
one of them that is b or c= 5 ( a cannot be 5 as 5! = 120 and 5! + 4! + 3! < 200)
so a = 1, b= 5, c = ? or a = 1, b = ? , c = 5 ( it has to be < 5)
if a = 1 , b = 5 we get 1 + 120 + c ! > 150 and < 160
so c! > 29 so there is no c
if a = 1, c = 5 we get 1 + 120 + b! = 105 + 10 b
so b = 4
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
This is a very nice problem, Albert.
Actually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)
The number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).
Balarka
.
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
This is a very nice problem, Albert.
Actually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)
The number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).
Balarka
.
Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term
I'd prefer not telling that, that'd make things easier.
A hint may suffice, for the sake of keeping this problem fair enough :
The next number is not too large.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
I'd prefer not telling that, that'd make things easier.
A hint may suffice, for the sake of keeping this problem fair enough :
The next number is not too large.
The next one is
$$\displaystyle 40585 = 4!+0!+5!+8!+5!$$
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Yes! nice, Zaid!
These are called factorions base 10. See, A014080.
Balarka
.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Yes! nice, Zaid!
These are called factorions base 10. See, A014080.
Balarka
.
Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.
#### Albert
##### Well-known member
Re: abc=a!+b!+c!
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.
ans :1, 2, 145, 40585
I wrote a program (using Excel) and found no answer for 4 digits number
and the only five digits number is 40585
the first person proved this (if using computer not allowed) must be very smart
#### mathbalarka
##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!
Proving finiteness of the sequence is not hard. Note that any n-digit factorion has an upper bound $$\displaystyle n 9!$$ and a lower one $$\displaystyle 10^{(n-1)}$$. The first to exceed this bound is n = 7, Implying that the largest factorion is at most of 7 digits. | 2021-07-30T23:58:12 | {
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http://math.stackexchange.com/questions/187102/what-is-the-name-of-v-alpha | # What is the name of $V_\alpha$?
In the Von Neumann cumulative hierarchy, $V:=\bigcup_\alpha(V_\alpha)$ is called the universe. Is there a name for the individual levels $V_\alpha$?
Just as one can say "The closure of $A$ is defined as $$cl(A):={...}"$$
I would like to be able to say "The _______ of $\alpha$ is defined as $$V_\alpha:={...}"$$
-
The $\alpha$th level of the cumulative hierarchy?... The family of all sets of rank less than $\alpha$?... I don't recall hearing a zippy name applied to these families. – Arthur Fischer Aug 26 '12 at 14:15
Well "the $\alpha$th level ..." is more creative than anything I was able to come up with just now. – Travis Bemrose Aug 26 '12 at 14:16
The sets $V_\alpha$ are usually referred to either as levels of the cumulative hierarchy (as mentioned in the comments) or as rank initial segments of V. I don't know how to fill in the blank in "the __ of $\alpha$", but one could say "the rank initial segment of V determined by the ordinal $\alpha$" or simply "the rank initial segment $V_\alpha$ of $V$."
It's not good to call it just the $\alpha$th level of $V$ without specifying which hierarchy; that could cause confusion when $V=L$, because $V_\alpha \ne L_\alpha$ in general.
-
Welcome Trevor! – Asaf Karagila Sep 3 '12 at 21:48
I'd distinguish between "level $\alpha$ of the cumulative hierarchy" ($V_\alpha$) and "level $\alpha$ of the constructible hierarchy" ($L_\alpha$). – Carl Mummert Sep 3 '12 at 21:51
Thanks, Asaf! (Also, I will now edit my answer to reflect Carl's comment.) – Trevor Wilson Sep 3 '12 at 21:58
To give a minor modification to Trevor's answer,
You can say that $V_\alpha$ is "the set of set with von Neumann rank $<\alpha$" (or $\leq\alpha$, depending on your definition of rank).
If whatever you write will be read by people familiar with set theory (sans your teachers, of course, in this case go with the above suggestion) then using $V_\alpha$ is sufficient. This is such a common notation that you sometimes see things like $V_\alpha^M$ as the $V_\alpha$ set of elements from $M$ (and sometimes you see $M_\alpha$ instead).
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https://vueling-va.com/crunchy-synonym-uhvijsa/nvuf7.php?7b20a3=function-rules-algebra | Functions are usually represented by a function rule where you express the dependent variable, y, in terms of the independent variable, x. y = 2.50 â
x You can represent your function by making it into a graph. Here are a couple of the more traditional: (a) A function is a rule (or set of rules) by which each input results in exactly one output. Get access to hundreds of video examples and practice problems with your subscription! Click here for more information on our Algebra Class e-courses. Function Rules DRAFT. Below is the table of contents for the Functions Unit. Edit. I have several lessons planned to help you understand Algebra functions. Infinitely Many. In algebra several identities to find the x values by using this we can easily find the algebraic expression of the particular function. In the definitions we used $$\left[ {} \right]$$ for the function evaluation instead of the standard $$\left( {} \right)$$ to avoid confusion with too many sets of parenthesis, but the evaluation will work the same. College Algebra - Concepts Through Functions Function Notation. The following laws are true for all a , b , c {\displaystyle a,b,c} whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions. A rational function will be zero at a particular value of $$x$$ only if the numerator is zero at that $$x$$ and the denominator isnât zero at that $$x$$. Take a look at an example that is not considered a 2518 times. A function is when one variable or term depends on another according to a rule. Find the function rule for the function table. In algebra, in order to learn how to find a rule with one and two steps, we need to use function machines. A function is a relationship between two variables. If you can solve these problems with no help, you must be a genius! introduced to this term called a "function". Interpreting function notation. The easiest way to make a graph is to begin by making a table containing inputs and their corresponding outputs. Improve your math knowledge with free questions in "Evaluate a function" and thousands of other math skills. Finally, function composition is really nothing more than function evaluation. function. Therefore, this equation can be The first variable determines the value of the second variable. functions - but never called them functions. Click here to view all function lessons. when x = 5, y = 11. creature in Algebra land, a function is really just an equation with a What in the world is a We will only use it to inform you about new math lessons. The function is quadratic Since all, the function is quadratic and follows the form. For example, the function f (x) = x 2 + 2x â 3 has f (3) = 12 and f (â4) = 5. Yes, I know that these formal definitions only make it more confusing. No other number will correspond with 3, when using this calculates the answer to be 7. The equation y = 2x+1 is a function because every time that you box performs the calculation and out pops the answer. Our function is . Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. (Notice how our equation has 2 variables (x and y) When we input 3, the function box then substitutes 3 ⦠After you finish this lesson, view all of our Algebra 1 lessons and practice problems. 2 years ago. This is the currently selected item. being the center of the function box. One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. considered functions. Need More Help With Your Algebra Studies? A function is any rule to assign a value (for example) to a variable "y", depending on the value of variable "x". Let's take a look at an example with an actual equation. When we input 4 for x, we must take the square root of both sides in order to solve for y. Be sure to label your graph. 66% average accuracy. The goal is use the equation y = mx + b. fancy name and fancy notation. We had what was known as In this lesson, you will learn to write a function rule using information given in a table. So, what kinds of functions will you study? No other number can correspond with 5, when This means that the If it crosses more than once it is still a valid curve, but is not a function.. Some teachers now call it a "Function Box" and When x = 3, y = 7 Imagine the equation substitute 3 for x, you will get an answer of 7. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright © 2008-2019. (2*3 +1 = 7). Let's take a look at this another way. Functions and equations. This topic covers: - Evaluating functions - Domain & range of functions - Graphical features of functions - Average rate of change of functions - Function combination and composition - Function transformations (shift, reflect, stretch) - Piecewise functions - Inverse functions - Two-variable functions The function rule of algebra may be form of f(x), p(x),⦠to find the x value of the algebra functions. If you are given a table, usually you have to carefully examine the table to see what the function rule is. However, there is a nice fact about rational functions that we can use here. All right reserved. ... Algebra. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Register for our FREE Pre-Algebra Refresher course. variable y = 7. Boolean algebra also deals with functions which have their values in the set {0, 1}. function? A function rule such as cost = p + 0.08p is an equation that describes a functional relationship. Everything you need to prepare for an important exam! You put a number in, the function Function Rules based on Graphs In the last two Concepts, you learned how to graph a function from a table and from a function rule. Practice: Function rules from equations. 9th - 11th grade. In its most general form, algebra is the study of mathematical symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics. send us a message to give us more detail! For rational functions this may seem like a mess to deal with. Example of Graphing a Function Rule. If you are nervous, Algebra Class offers many lessons on understanding functions. Free functions calculator - explore function domain, range, intercepts, extreme points and asymptotes step-by-step. You will find more examples as you study the Play this game to review Algebra I. DEFINITION: A function can be defined in a variety of ways. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz Factoring Trinomials Quiz Solving Absolute Value Equations Quiz Order of Operations QuizTypes of angles quiz. Function rule in algebra means that we have to perform the arithmetic operation of two functions. Therefore, this does not satisfy the definition for a A sequence of bits is a commonly used for such functions. Next lesson. this is why: Here's a picture of an algebra function box. We have more than one value for y. Hopefully with these two examples, you now understand the difference Logarithm quotient rule Combining rules 3 and 4, we can multiply the denominator of the bottom fraction with the numerator of the upper fraction, which gives the combined numerator, and cancels the denominator of the lower fraction; we can then multiply the denominator of the upper fraction with the numerator of the lower fraction, to give the combined denominator and cancel the denominator of the upper fraction. Swipe through the slideshow below to ⦠We have the values of x as . Equations vs. functions. of functions in Algebra 1. every time. Your email is safe with us. Another common example is the subsets of a set E : to a subset F of E, one can define the indicator function that takes the value 1 on F, and 0 outside F. See: Logarithm rules Logarithm product rule. Video-Lesson Transcript Example 1. Boolean algebra allows the rules used in the algebra of numbers to be applied to logic. lessons in this chapter. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. labeled a function. The logarithm of the multiplication of x and y is the sum of logarithm of x and logarithm of y. log b (x â y) = log b (x) + log b (y). Function Rules DRAFT. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! There is a special relationship between the two variables of the function where each value in the input applies to only ⦠Math Algebra 1 Functions Functions and equations. Using (1,6) and (2,10), m = (10 - 6) / (2 - 1) = 4 / 1 = 4, Top-notch introduction to physics. A function may be thought of as a rule which takes each member x of a set and assigns, or maps it to the same value y known at its image.. x â Function â y. Click on the Math Gifs Algebra equation. Algebra 1, by James Schultz, Paul Kennedy, Wade Ellis Jr, and Kathleen Hollowelly. ewhitehurst8. On a graph, the idea of single valued means that no vertical line ever crosses more than one value.. lesson that interests you, or follow them in order for a complete study Learn More at mathantics.com Visit http://www.mathantics.com for more Free math videos and additional subscription based content! 3=81 a0 =1 If n,m 2 N, then an m = m p an =(m p a)n ax = 1 ax The rules above were designed so that the following most important rule not represent a function. It includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Edit. It seems like all equations would be When you input 5, you should get 11 because (2*5+1 = 1), so Algebra 2, by James Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy. Click here for more information on our affordable subscription options. We end up with y = 2 or -2. Letâs pick the x values then solve for its corresponding y values. Laws and Rules of Boolean algebra with Tutorial, Number System, Gray code, Boolean algebra and logic gates, Canonical and standard form, Simplification of Boolean function etc. In Algebra 1, we will study linear functions (much like linear equations) and quadratic I always go back to my elementary years when we learned about function: "the value of the first variable corresponds to one and only one value for the second value". functions. On this site, I recommend only one product that I use and love and that is Mathway If you make a purchase on this site, I may receive a small commission at no cost to you. Mathematics. And a further qualifier is that a function may have just one output value for every input value in its domain. A function(or a mapping) is a relation in which each element of the domain is associated with one and only one element of the range.Different types of functions explored here:inverse,composite,one-one,many-one,two-many.Worked examples and illustrations. If p is the price you pay for an item and 0.08 is the sales tax, the function rule above is the cost of the item. Save. These basic functions ⦠Obtaining a function from an equation. You are now deeper in your Algebra journey and you've just been Copyright © 2009-2020 | Karin Hutchinson | ALL RIGHTS RESERVED. Let us do this for example #3. (Notice how our equation has 2 variables (x and y). All weâre really doing is plugging the second function listed into the first function listed. an "in and out box". Improve your math knowledge with free questions in "Function transformation rules" and thousands of other math skills. When we input 3, the function box then substitutes 3 for x and If you input another number such as 5, you will get a different Vertical Line Test. Although it may seem at first like a function is some foreign The value of the first variable corresponds to one and only one value for the second variable. y (2 and -2). It seems pretty easy, right? 5. These coordinates would look like this: and . substituting into this equation. Instructor: Dr.Jo Steig . Here we have the equation: y = 2x+1 in the algebra function box. Basic-mathematics.com. We cannot say that the equation x = y2 represents a Represent combinational logic circuits free math videos and additional subscription based content recommendedscientific Notation QuizGraphing Slope QuizAdding Subtracting... Really doing is plugging the second variable page:: Disclaimer:: policy... Study linear functions ( much like linear equations ) and quadratic functions Properties..., the idea of single valued means that no vertical line ever crosses more than one value for input... Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy ) quadratic. A different output in function box performs the calculation and out pops the answer more! 3 for x and y ) about new math lessons in order learn. Relation that is defined using various mathematical operators set { 0, 1 } allows. Was known as an in and out pops the answer to be applied to.! About rational functions this may seem like a mess to deal with it to inform you about new math.. To be applied to logic allows the rules used in the set { 0 1! An equation that describes a functional relationship of Operations QuizTypes of angles Quiz inform you new. A function box '' and thousands of other math skills you put number. Quiz solving Absolute value equations Quiz order of Operations QuizTypes of angles Quiz { 0 1... Polynomials rational Expressions Sequences Power Sums Induction Logical Sets equation being the function rules algebra of the second variable quadratic! For the functions Unit to perform the arithmetic operation of two functions of video examples practice! With y = 2x+1 in the algebra function box Hutchinson | all RIGHTS.... Called them functions to it, let 's take a look at an with. Properties Partial Fractions Polynomials rational Expressions Sequences Power Sums Induction Logical Sets:! Will learn to write a function rule in algebra, a function is quadratic and follows the.! To my elementary years when we learned about functions - but never called them functions x... 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED on another according to a rule with one and one. Look at an example that is defined using various mathematical operators types of functions algebra. To find the x values by using this equation can be labeled a function is quadratic follows... About rational functions this may seem like a mess to deal with DonateFacebook:! Rules used in the algebra of numbers to be applied to logic linear functions ( much linear. And you 've just been introduced to this term called a function transformation rules '' and this why... Will study linear functions ( much like linear equations ) and quadratic...., y = 2x+1 in the algebra function box performs the calculation and out the. Rule, multiplication function rule in algebra means that we can use here usually you have perform! But is not considered a function rule such as groups, rings, and fields Subtracting! Can correspond with 5, you must be a genius to carefully examine the table to see the. A function '' at this another way Slope QuizAdding and Subtracting Matrices Factoring... Definitions only make it more confusing new math lessons, Area of irregular shapesMath problem solver we have equation... A picture of an algebra function box '' pins, Copyright © 2008-2019 's a! A genius considered functions '' and thousands of other math skills learn to write a function,! Wade Ellis Jr, Kathleen Hollowelly, and Kathleen Hollowelly, and even the math involved in baseball! Two steps, we will study linear functions ( much like linear equations ) and quadratic functions nervous, Class... Information given in a variety of ways 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED so what... To begin by making a table containing inputs and their corresponding outputs is using... Depends on another according to a rule or relation that is not considered a function is when one variable term! By using this we can easily find the x values then solve for.! But never called them functions your algebra journey and you 've just been introduced this... Lessons on understanding functions functions - but never called them functions it more confusing mathematical operators that we the., view all of our algebra Class e-courses function can be labeled a function . X = 3, the idea of single valued means that no vertical line ever more. The algebra function box like a mess to deal with 's answer that question: is... Line ever crosses more than once function rules algebra is still a valid curve, is... The goal is use the equation being the center of the first function listed into the first variable the! Algebra Class e-courses use it to inform you about new math lessons:. More you can read Injective, Surjective and Bijective, the idea of single valued that... Quiz order of Operations QuizTypes of angles Quiz boolean algebra allows the rules in... Used for such functions, division function rule, division function rule, division function rule.... Kathleen Hollowelly, and Paul Kennedy combinational logic circuits are given a table containing inputs their. You, or follow them in order to learn function rules algebra to find the Algebraic expression of the first function.! Kathleen Hollowelly, and Kathleen Hollowelly offers many lessons on understanding functions these formal definitions make. It, let 's answer that question: what is a function are! Means that no vertical line ever crosses more than once it is still a curve! Number can correspond with 3, when substituting into this equation also deals with functions which have their values the! Only make it more confusing out box '' 2 variables ( x calculates! Nervous, algebra Class e-courses making a table containing inputs and their corresponding outputs Power Sums Induction Logical Sets is! Question: what is a function rule such as groups, rings, and Kathleen,. After you finish this lesson, view all of our algebra 1 a... Mx + b we end up with y = 2x+1 in the set { 0, 1.! Injective, Surjective and Bijective multiplication function rule, multiplication function rule is then substitutes 3 for,., the function is quadratic Since all, the function box have equation... Equation that describes a functional relationship © 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED about -! Commonly used for such functions steps, we need to use function machines study of abstractions such groups... Linear equations ) and quadratic functions what is a rule or that. This another way to be 7 box performs the calculation and out the... To the study of functions will you study other number can correspond with 5, when substituting into equation. Different output you progress into algebra 2, you will learn to write function. A function box functions will you study the lessons in this lesson, function rules algebra will to. Tough algebra Word Problems.If you can read Injective, Surjective and Bijective Visit http: for... Quadratic Since all, the function box performs the calculation and out pops the answer to be applied logic... Would be considered functions is when one variable or term depends on another according a... Set { 0, 1 } everything from elementary equation solving to study! To be 7 graph is to begin by making a table, usually you have to examine... On another according to a rule function is quadratic Since all, the function box of equations System of System! Factoring Trinomials Quiz solving Absolute value equations Quiz order of Operations QuizTypes angles. You study on understanding functions letâs pick the x values by using equation... That question: what is a function boolean Expressions which are used to represent combinational logic.... Corresponds to one and two steps, we need to use function.! You need to use function machines follow them in order for a study... Y ) the center of the first variable corresponds to one and only one value find the x then. Important exam is quadratic Since all, the idea of single valued means that no vertical line crosses! You will get a different output combinational logic circuits their values in algebra... + b and additional subscription based content value for every input value its... Inputs and their corresponding outputs fact about rational functions this may seem like a mess deal... Evaluate a function is when one variable or function rules algebra depends on another to... Out more you can solve these problems with no help, you will be studying exponential functions equations... 2 or -2 Expressions Sequences Power Sums Induction Logical Sets to the study of such. I have several lessons planned to help you understand algebra functions = 2 or -2 the... Resource to a rule with one and two steps, we must the! Here for more information on our algebra Class e-courses be 7 more at mathantics.com Visit http: //www.mathantics.com for free... Of function rules algebra QuizTypes of angles Quiz information given in a variety of ways ) quadratic. To logic of both sides in order to solve for its corresponding y values function box particular.! Root of both sides in order to learn how to find the Algebraic expression of the box! Their values in the set { 0, 1 } be 7 you understand algebra...., the idea of single valued means that no vertical line ever more... | 2021-01-19T19:15:55 | {
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http://bgvl.chicweek.it/proof-by-contradiction-examples-and-solutions.html | # Proof By Contradiction Examples And Solutions
• Example Every prime number a irrational numbers. Proofs, the essence of Mathematics - tiful proofs, simple proofs, engaging facts. In particular m2 is even, which implies m is even, say m = 2x for x 2Z. – The allowable combinations, have a maximum value of 24. n odd ⇒ n2 odd 2. Use the method of proof by contradiction to prove the following statements. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Print Proof by Contradiction: Definition & Examples Worksheet 1. It will actually take two lectures to get all the way through this. In Class IX, you were introduced to the idea of proofs, and you actually proved many statements, especially in geometry. In these cases, when you assume the contrary, you negate the original. Euclid famously proved that there are an infinite number of prime numbers this way. Show that all cube numbers are multiples of 9. Both of these methods are called constructive proofs of existence. Similarly, Math 96 will also require you to write proofs in your homework solutions. Instructions You can write a propositional formula using the above keyboard. The "proof" by josgarithmetic" is wrong starting from his second line. Discrete Math Lecture 03: Methods of Proof 1. The Law is an essay written by the Frédéric Bastiat in 1850. 21 To Prove That V2 = 2/3 Is Irrational. Example: Prove that if. p 2 = a b 2 = a2 b2 2b2 = a2 This means a2 is even, which implies that a is even since. Textbook solution for Elementary Linear Algebra (MindTap Course List) 8th Edition Ron Larson Chapter A Problem 17E. In this case and and so we have found an example where but and thus disproving the statement. Is l Dillig, CS243: Discrete Structures Mathematical. The establishment of a fact by the use of evidence. Proof: By contradiction; assume that there is a rational number r and an irrational number s where the number r + s is rational. Solution: Suppose p 2 is rational, say p 2 = n=m for n;m 2Z. Another example, let n be an integer then ‘n is even’ and ‘n is odd’ is a contradiction Suppose we want to prove a proposition P then the procedure for proof by contradiction is as follows: 1. Let k be any even integer. Cube(b) ∧ a = b 2. The proof will use the following definitions. Then Therefore a 2must be even. By contradiction. is an integer and. Anything that can make a person believe that a fact or proposition is true or false. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Tindle, who. If a is even then amust be even (an. As a first example of proof by contradiction, consider the following theorem:. First and foremost, the proof is an argument. BaseCase:Whenn = 1 wehave111 − 6 = 5 whichisdivisibleby5. If a direct proof is straightforward then this is to be preferred – a direct proof usually provides more insight into the mathematical structure at hand. is a proposition that is always. Theorem: Greedy algorithm’s solution is optimal. Most of the proofs I think of should be accessible to a middle grade school student. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. Suppose there is some irrational number p such that -p is rational. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Proofs by contradiction are useful for showing that something is impossible and for proving the converse of already proven results. 2 Selected Homework Solutions 10. Name the left column Statements. statement q is true. Proof (by contradiction): Suppose greedy not optimal. Then use the. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. ó Solution: Assume that n is odd. I Observe that any rational number r can be written as p 2 pr 2 I We already proved p 2 is irrational. Considerthe number M = N + 1. This is also known as proof by assuming the opposite. Typically, one shows that if a solution to a problem existed, which in some. No possible constant value for x exists to make this a true equation. That is, suppose there is an. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Compared to a proof of contradiction you have the advantage that the goal is clear. 4 Proof by contradiction The idea of contradiction method is by showing is a contradiction of the statement , that is a tautology. Proof: This is easy to prove by induction. Proof by contradiction is often used when you wish to prove the impossibility of something. This also applies ifthe result is goingto beproven using mathematical induction. fn and fn+1 that have a common divisor d, where d is greater than 1. A z° x° y° 100° B O Solution Theorem 1 gives that z = y = 50 The value of x can be found by observing either of the following. To prove p, assume ¬p and derive a contradiction such as p ∧ ¬p. Give a proof by contradiction: prove that the square root of 2 is irrational. Valid Argument: 1. If x 2A B then x 2A (and not in B). ] Assume, to the contrary, that ∃ an integer n such that n 2 is odd and n is even. Direct Proof: Assume p, and then use the rules of inference, axioms, de - nitions, and logical equivalences to prove q. Use proof by contradiction to show that if n2 is an even integer then n is also an even integer. Inequalities 10 7. A logical contradiction is the conjunction of a statement S and its denial not-S. Any proof does. Solution manual for Analysis with an Introduction to Proof 5th Edition by Lay. If x 2A B then x 2A (and not in B). We have to prove 3 is irrational Let us assume the opposite, i. These solutions use information found in pages 154 - 160 of the textbook. The (Pedagogically) First Induction Proof 4 3. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. x = √(2k) –Not clear that sqrt(2k) is an even integer, or even an integer J Proof by contrapositive –prove that if x is odd then x2is odd. Therefore, 1 is the largest integer. This would mean that we can have at most 9 7 = 63 days we could have chosen. Examples of Proof by Contradiction. 6 [A level only] (a) Prove that the square root of 2 is irrational. I so p 1. Then there exists unmatched college c and unmatched student s. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Indirect Proof or Proof by Contradiction: Assume pand :qand derive a contradiction r^:r. Thus, the proposition is true. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. Compared to a proof of contradiction you have the advantage that the goal is clear. Smolka and J. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. •Alternatively, using contradiction, prove that it is not possible for such a thing notto exist. Then there exists integers. Figure 1 Solution: Proof. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. (It looks like that list omits the proof by the rational root theorem. The (Pedagogically) First Induction Proof 4 3. Again, you will want to go back to the de nition of \perfect square" as I have done in the two examples in the notes. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m' > m • m and w are not blocking Case #1 and #2 exhaust space. The preceding examples give situations in which proof by contradiction might be useful:. ExampleProve by contradiction that there is no greatest integer. -p = mln, where m and n are both integers and n # 0 3. This proof works by assuming the negation of the thing you want to prove and then finding some reason that this is absurd, namely by deducing a contradiction, like a statement and its opposite. 1, 2017W1 midterm 1): You're given an SMP instance where two men have the same preference list. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. ” I could go on, obviously, with countless examples of these kinds of posts on social media, not to mention news stories about biological men competing in women's athletics and young children—surprisingly young children—transitioning into another gender. Combining Proofs, cont. (b) Assume for a contradiction that the square root of 3 is rational, i. Proof by Contradiction. 21 To Prove That V2 = 2/3 Is Irrational. Let x be an integer. Induction step: Assume the theorem holds for n billiard balls. So, to prove "If P, Then Q" by the method of contrapositive means to prove "If Not Q, Then Not P". Each person is a vertex, and a handshake with another person is an edge to that person. In particular, the. Algebraic Examples Algebraic examples are often easier to follow at first than geometric. Solution: To prove this claim by contradiction, we will assumethat the negation is true; i. Another important method of proof is proof by contradiction. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. Thus, the proposition is true. Suppose p 2 is rational. Logical Form: 8n: n2 even =)neven. They clearly need to be proven carefully, and the cleverness of the methods of proof developed in earlier modules is clearly displayed in this module. Proof by contradiction: example Theorem: There are infinitely many primes. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. Since the nal is open-book, this doesn't make sense any more, so all the proofs will be of things you haven't done before. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. We shall show that you cannot draw a regular hexagon on a square lattice. Adding these together we get (2n+ 1) + 2m = 2n+ 2m+ 1 = 2(n+ m) + 1 which is of the form 2( integer ) + 1, which is an odd number. And you also want an explanation of what a proof by contradiction is, which also seems to be way too elementary. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. (It looks like that list omits the proof by the rational root theorem. , n and m have no prime factors in common. We assume 푝푝 ∧¬푞푞 , then show that this leads to a contradiction. Example of a Proof by Contradiction Theorem 4. Let ; then satisfies the following equation: Clearly, By Lemma 13, is the solution of , which is a contradiction. [1 mark] Assume positive integer solutions. Figure 1 Solution: Proof. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. The word Proof is italicized and there is some extra spacing, also a special symbol is used to mark the end of the proof. Let me show you another example where a contrapositive proof is so much easier to carry out. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. I Consider number q =(p1 p2 pk)+1: I q cannot be one of the primes as it is larger than any pi. In a non-constructive proof, one proves the statement using an indirect proof such as a proof by contradiction. And except for the beginning and end, to solve an indirect proof, you use the same techniques and theorems that you would use on regular proofs. Most of the proofs I think of should be accessible to a middle grade school student. A feasible solution is a solution that satis es the property P. ó Solution: Assume that n is odd. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. 1 Proving Negative Statements youtu. Use rules of inference, axioms, and logical equivalences to show that q must also be true. We must derive a contradiction. So, 0 = (x + y) (x y) = 2y. Then use the. Proofs and refutations: standard techniques for constructing proofs; counter-examples. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Discrete Mathematics This is a basic course for undergraduate students. geometry_terms_and_proof_by_contradiction. Hints and partial solutions are provided. Also I think it might help for you to study a few example proofs for greedy algorithms. So as an example, let's have the statements, P(n), sum of k is 1 to n of K which is basically sum one plus two plus three plus etc. is even, then. To prove: If x 2 is even, then x is even. Math 109, fall 2017 (Ioana), midterm 1 sample solutions October 26, 2017 For some problems, several sample proofs are given here. A Famous Contradiction Example. I don't understand this contradiction stuff. Proof by contradiction: Assume negation of what you are trying to prove (p q). Wrtiten response: Well done. # to derive a contradiction Then there is a finite list, p 1;:::;p k of elements of P. What makes it different is the way it begins and ends. We're not done with them! Also remember how to prove existence theorems (using an example) and disprove universal statements (using a counterexample). The Proof Page presents supplementary material (lecture notes, problem sets, and solution sets) to assist students moving academically and intellectually from "how to" mathematics, e. So a2 is a multiple of 3, and so must be a. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q ‘t‘ + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. Prove that if aand bare real numbers with aa. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. Suppose that a + br is rational. Properies of the modulus of the complex numbers. What's our proposition? Prove the following statement by contradiction: There is no integer solution to the equation x 2 - 5 = 0. There is no greatest even integer. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that. Complex numbers tutorial. # at most n elements in the list Then I can take the product p0= p 1 p k. Show that the conclusion Q(x) is true by using definitions, previously established results, and the rules for logical inference. Then derive the [Prove] statement using logic (known theorems, laws, etc. Proposition: Each natural number n>1isaprimeorproductofprimes. Proof: Form the contrapositive of the given statement. This is a well-written text, that can be readily used for introduction to proofs and logic course at the undergraduate level. If the following statement is true, give a proof. When trying to construct a proof it is sometimes useful to assume the opposite of the thing you are trying to prove, with a view to obtaining a contradiction. This is one of the base methods of reasoning. geometry_terms_and_proof_by_contradiction. 1 The method In proof by contradiction, we show that a claim P is true by showing that its negation ¬P leads to a contradiction. Advanced/wacky examples: This pdf has some great examples in Section 6(page 4) — they show how induction can be applied to all kinds of different mathematical problems. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Factor the right-hand side of this equation as 2 = (a−b)(a+b). A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. The difference between the two is that a proof by contradiction can be devised, but a proof by contrapositive. Henselian Valued Stable Fields I. Related Links. Solution: Suppose √2 is rational. In a proof by contradiction, we start with the supposition that the implication is false, and use this assumption to derive a contradiction. ] Suppose not. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Reach a contradiction. $\endgroup$ - D. You can use one of the above methods (direct proof, proof by contraposition or contradiction) to solve the p→q and q→p part. Wrtiten response: Well done. Therefore y = 0, contradicting that it is positive. Exam focused Online Study Pack. P and (not P) is a contradiction For example xx22−=1 0 Zero and 1 0 Not Equal to Zero[ ] −≠[ ] is a contradiction. One standard way of doing this is to make the first line “Suppose for the sake of contradiction that it is not true that (2 is irrational. In particular, the. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Hint: There are four parts to the proof. That is, suppose there is an integer n. Valid Argument: 1. a proof by contradiction. – If we had 5 pennies, we could replace them with a nickel. Proof: Suppose A. Lay Lee University. • Proof by contradiction • To prove that P is true, it is sufficient to prove that “not P implies Q” when Q is clearly false. SOLUTION Let x represent the length of the third side. To show that d : R !R de ned by d(x;y) = jx yjis a metric, for. Problem: Given that a, b, and c are odd integers, prove that equation ax 2 + bx + c = 0 can not have a rational root. [Hint: Assume that r = a/b is a root, where a and b are integers and a/b is in lowest terms. Because r + s is rational, we can write it as p / q for some integers p and q where q ≠ 0. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. The idea is to assume the hypothesis, then assume the. Similarly, Math 96 will also require you to write proofs in your homework solutions. Contraposition: Contradiction:. Students often find this emphasis difficult and new. We have a contradiction. Example Questions. 9 = 362,880. Example: Use proof by contradiction to prove that p 2 is irrational. We argue by contradiction. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. " Problem 2. p is the pumping length given by the PL. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. One example of a proof by contradiction is the proof that √2 is an irrational number: Assume that √2 is a rational number, meaning that there exists a pair of integers whose ratio is √2. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). p= -mln, where -m and n are both integers and n ± 0 4. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. Since n is odd, n = 2k + 1 for some integer k. Assume $$n$$ is a multiple of 3. Quiz SAP - C_THR84_2005 –Efficient Exam Syllabus, Because it can help you prepare for the C_THR84_2005 Exam Content exam, We have strong IT masters team to study the previous test to complete the C_THR84_2005 new dumps to follow the exam center's change and demand, SAP C_THR84_2005 Exam Syllabus PDF version: can be read under the Adobe reader, or many other free readers, including OpenOffice. The negative of an integer is. Typo: The hypothesis that r is not equal to 0 in the Example is not necessary (I was confusing this statement with a similar statement about the product, rx). Example 1: irrational. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Also, r = br b. Contradiction. The material in discrete mathematics is pervasive in the areas of data structures and. The first guy appears to misunderstand what proof by contradiction is. Then there exists integers aand bwith √2 = a/b, where b≠ 0 and aand b have no common factors (see Chapter 4). Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. (exercise) * Method of Proof by Contradiction Suppose the statement to be proved is false. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. For example, if every point lies on the x = y line, then each point would dominate all points below it, giving us n C 2 edges. bwhere b≠ 0. Solutions to propositional logic proof exercises October 6, 2016 1 Exercises 1. (I will not read any work on this question sheet). But, from the parity property, we know that an integer is not odd if, and only if, it is. Proof by contradiction examples Example: Proof that p 2 is irrational. A proof by contradiction in this case has the logical form ¬P ¬P→ (R∧ ¬R) ∴ R∧ ¬R 2. The max-flow, min-cut theorem Theorem: In any basic network , the value of the maximum flow is equal to the capacity of the minimum cut. Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. If we give a direct proof of ¬q → ¬p then we have a proof of p → q. have no common factors (see Chapter 4). 4 The number 3 is irrational. Thus, one might prove that the negation 8x2S;˘P(x) is false by deriving a contradiction. Which proof technique? Direct proof –express x2 as 2k for some k, i. Non-linear examples exhibit a few other quirks, and we will demonstrate them below also. In general, then, try to be specific when doing an existence proof, but if you cannot, it may still be possible to construct an example using some other existence result or another technique of proof. Main proof There is no generic radical root formula that applies universally to all quintic. Such proofs can be reviewed at the Proofs tutorial. both r and :r for some proposition r. by axiom that a number can be even or odd but not both at a time) so we can write n such as: n = 2k ; k is any integer (by definition of an even number). But every number's square is nonnegative, so y2 0, a contradiction. Mathematics cannot be a spectator sport. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q ‘t‘ + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). In order to illustrate this type of proof we assume that we know: 1. The number 2 is a prime number. Wiles's proof of Fermat's Last Theorem is a proof by British mathematician Andrew Wiles of a special case of the modularity theorem for elliptic curves. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Solution: Perform a proof by contradiction. Suppose that a + br is rational. No possible constant value for x exists to make this a true equation. • Direct proof • Contrapositive • Proof by contradiction • Proof by cases 3. Relation between Proof by Contradiction and Proof by Contraposition. We learn how to do it with a couple of worked examples. I pdivides both x= 1 2 k and q,and divides I =)pj x q 1. can be proved by showing that its contrapositive ¬ q → ¬ p. Proof \by contradiction": Suppose n < m vectors did span Rm, then there would be a pivot in every row, thus at least m pivots. Then there exists integers. Why can't we prove B is not true by finding a counter example?. 6: Let A, B, and X be sets. I just meant like it can be really useful in certain problems without explicitly asking for a contradiction proof. is an integer and. Thus x2 + 1 < 0 is false for all x ∈ S, and so the implication is true. 4, namely that for any integer. ICS 141: Discrete Mathematics I - Fall 2011 7-8 Indirect Proof Example: University of Hawaii Proof by Contraposition ! Theorem: (For all integers n) If 3n + 2 is odd, then n is odd. Giving a counter example 3+5=8 is even is not a proof by contradiction. some typical examples where you are expected to use proof by contradiction and I try below to cover all the possible situations I can think of. n odd ⇒ n2 odd 2. As a first example of proof by contradiction, consider the following theorem:. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. By contradiction. Proofs Proofs Proofs by Contradiction De nition Proof by Contradiction: A form of proof that establishes the truth or validity of a proposition by rst assuming that the opposite proposition is true, and then shows that such an assumption leads to a contradiction. I pdivides both x= 1 2 k and q ,and divides I =)pj q x 1. Finding a contradiction means that your assumption is false and therefore the statement is true. Suppose that x is a positive real number with. • Example Every prime number a irrational numbers. If it were rational, it could be expressed as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Hence, n2 = 4k2 +4k. Solutions 1. Smolka and J. One of the basic techniques is proof by contradiction. Proof: (direct proof) Assume that n is an even integer. But this is a contradiction, since the empty set cannot contain any elements, y or. Obtain an. If this is the case, we can factor the left side: x 2 - y 2 = (x-y)(x+y) = 1. If all our steps were correct and the result is false, our initial assumption must have been wrong. I don't understand this contradiction stuff. The main idea is to assume that the statement we want to prove is false, which leads us to contradiction. Proof (By contradiction) Suppose this language is context-free; then it has a context-free grammar. That is, suppose there exists a real number r such that r3 is irrational and r is rational. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Thus, 3n + 2 is even. If (A [B) X and (X B) (X A), then A B. If 3 - n2, then 3 - n. STEP 2 Reason logically until you reach a contradiction. Example 1 Find the value of each of the pronumerals in the diagram. Ex: p∧~p Claim:Suppose c is a contradiction. If 3jn then n = 3a for some a 2Z. The proof is by contradiction. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. John Smith is a man. The text covers topics one would expect to see in first course on logic and proofs, including proofs by contradiction and proof by induction. a proof by contradiction assumes that p is false and derives a contradiction, i. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. The proves the contrapositive of the original proposition,. You assume the opposite is true at the beginning only to end up to see the original assumption is not true. This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. If you can do that, that example is called a. The empty set is a subset of A, hence it is an element of the power set of A. Solutions are included. Let x be an odd integer. In this case and and so we have found an example where but and thus disproving the statement. Now this is a contradiction since the left hand side is odd, but the right side is even. Proof by Contradiction. 12, and if we can go by all the previous chapters, this will be our template for the exercises. This shows the negation is false, and hence that the original proposition is true. If f(2) = 8, explain why f(3) > 6. Trans and non-binary men belong. A First Example: Proof by Contradiction Proposition: There are no natural number solutions to the equation x2 y2 = 1. Therefore, the reasoning of the ontological argument dodges the parody, its reasoning is not parallel to the parody argument, and it cannot be used to prove the existence of a lost island. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. In an indirect geometric proof, you assume the opposite of what needs to be proven is true. Discrete Mathematics This is a basic course for undergraduate students. Obtain an. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. The argument is valid so the conclusion must be true if the premises are true. That would mean that there are two even numbers out there in the world somewhere that'll give us an odd number when we add them. (This is an indirect form of proof. The number 2 is a prime number. A Famous Contradiction Example. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. (Contrapositive) Let integer n be given. As a first example of proof by contradiction, consider the following theorem: Theorem 1. Example2 1. Proof by Contradiction: (AKA reductio ad absurdum). This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. Mathematical Proof/Methods of Proof/Proof by Induction. That is how Mathematical Induction works. The establishment of a fact by the use of evidence. ExampleProve by contradiction that there is no greatest integer. Section 4-7 : The Mean Value Theorem. In a proof by contradiction, or indirect proof, you show that if a proposition were false, then some false fact would be true. (i)Direct proof: we assume A is true. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. For every even integer n, N ≥ n. •Proof : Assume that the statement is false. ]! Then, by definition of rational, r = a/b and s = c/d for some integers a, b, c, and d with b ≠ 0 and d ≠ 0. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Wyke, or Mr. Solution LetP(n) bethemathematicalstatement 11n −6 isdivisibleby5. GS results in a stable matching (i. Proof: Form the contrapositive of the given statement. 8, 1113, Sofia, Bulgaria Communicated by Walter Feit Received November 10, 1997. Thursday 2/16/17. The Second Edition features new chapters on nested quantifiers and proof by cases, and the number of exercises has been doubled with answers to odd-numbered exercises provided. Prove the following statement by contradiction: The sum of two even numbers is always even. Then there exists unmatched college c and unmatched student s. Rajoub: "For the first time in the history of the conflict – the contradiction between [the occupation and] the interests of the world, international law, and its values have reached a peak, and we must not undermine this clash through the wrong actions – whether in word or in actions that deviate from the consensus. a method of disproving a. Give a proof by contradiction: prove that the square root of 2 is irrational. It contrasts what Bastiat considered as the proper function of the Law and the perversion of the Law. Such examples are called counter examples. (I will not read any work on this question sheet). It is usually not as neat as a two-column proof but is far easier to organize. Proof by contradiction means you assume the premise and the opposite of the conclusion and then derive some contradiction. to an equation then there is another integral solution that is smaller in some way. Prove that the sum of irrational and rational number is irrational using proof by contradiction. If x ∈ A ∩ B, then x ∈ A and x ∈ B by definition, so in particular x ∈ A. Proof by Contradiction. O is the centre of the circle and ∠AOB = 100. 1 √2 is an irrational number. (non-constructive proof) •Show that a player in a game has a winning strategy without actually sayingwhat it is! •Famous proof: There exist irrational x, y such that xyis rational Villanova CSC 1300 -Dr Papalaskari. You can put this solution on YOUR website!. Below are several more examples of this proof strategy. This is the way most people learn a new language | learn to say a. Proof: (direct proof) Assume that n is an even integer. An example is "Prove that the product of two nonzero real numbers is nonzero. So this is a valuable technique which you should use sparingly. Example of a Proof by Contradiction Theorem 4. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Solution: Suppose √2 is rational. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. Recall that for two integers x and y, we say x divides y if there exists an integer z such that xz = y. Wrtiten response: Well done. If all our steps were correct and the result is false, our initial assumption must have been wrong. Imagine, then, the thrill of being able to prove something in mathematics. We give the proof by contradiction. To prove a theorem of the form A IF AND ONLY IF B , you first prove IF A THEN B , then you prove IF B THEN A , and that's enough to complete the proof. The concept of proof by contradiction is to assume that P is false. By the closure property, we know b is an integer, so we see that 3jn2. If you can do that, you have used mathematical induction to prove that the property P is true for any element, and therefore every element, in the infinite set. Also see the Mathematical Association of America Math DL review (of the 1st edition) and the Amazon reviews. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. In the second group the proofs will be selected mainly for their charm. Since and are distinct points in the Hausdorff space , there must be disjoint, open neighborhoods such that and. This A Level Maths video takes you through a new method of proof called proof by contradiction. Proof by contradiction: Assume P(x) is true but Q(x) is false. Prove that if a square matrix A is a zero divisor (that is AB=0 for some non-zero matrix B) then det(A)=0. Selected Homework Solutions - Math 574, Frank Thorne 1. Use rules of inference, axioms, and logical equivalences to show that q must also be true. Example 3: Prove the following statement by contraposition: For all integers n, if n 2 is odd, then n is odd. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Prove: do not bisect each other. Proof by Contradiction is another important proof technique. Instead of it, we use proof by contradiction. Example of a Proof by Contradiction Theorem 4. Justify all of your decisions as clearly as possible. Proving "If A, then B" by contradiction Given the assumptions in A, show that B must be true because it cannot possibly be false. As a first example of proof by contradiction, consider the following theorem:. Then there exists integers. A proof of a very general idea could be preceded by an example in a specific context. Direct Proof: Assume that p is true. (Proof by Contradiction. Run M on hPi. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. So, we will discuss these methods in this lesson extensively. 2 Selected Homework Solutions 10. Solution: By contradiction. Proof by Contradiction Example: Use a proof by contradiction to give a proof that √2 is irrational. Introduction 1 2. Squaring both sides we get 2 = n 2=m2, so m2 = 2n. The Pigeonhole Principle 1 Pigeonhole Principle: Simple form Theorem 1. Statement: If A, then B Inverse: If B, then A Converse: If not A, then not B Contrapositive: If not B, then not A Which of these are logically equivalent? RTP: If A, then B Method: Assume not B Carry out logical, deductive steps Reach the conclusion not A Example:. The statement P1 says that x1 = 1 < 4, which is true. Then we have to find a statement R so that ¬P→ (R∧ ¬R) - a contradiction. I can use. If ¬P leads to a contradiction, then. Proof by Contradiction is another important proof technique. A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Since nm + 2n + 2m is odd, nm + 2n + 2m = 2k + 1 for some integer k. Similarly, since ris rational, we can express it as a/ bfor some integers aand. Finding a contradiction means that your assumption is false and therefore the statement is true. Final: Solutions ECS20 (Fall 2014) December 16, 2014 Part I: Proofs 1) Let a and b be two real numbers with a 0 and b 0. These notes explain these basic proof methods, as well as how to use definitions of new concepts in proofs. The proof is a sequence of mathematical statements, a path from some basic truth to the desired outcome. To write a two-column proof: Make a two-column form like this. Example: Prove that if you pick 22 days from the calendar, at least. Show that if n=k is true then n=k+1 is also true; How to Do it. (b) Prove that the square root of 3 is irrational. _____ Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1. 1 The number √3 is irrational. 3 Review the proof techniques on page 116−−118 Here is a result that is proved by three different proof techniques. Indirect Proof: Assume what you need to prove is false, and then […]. If (A [B) X and (X B) (X A), then A B. 2 Proof By Contradiction A proof is a sequence S 1;:::;S n of statements where every statement is either. Example of proof by contradiction. Eureka step and the eventual solution (Zeitz, 1999). Proof by mathematical induction. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. To change the symbol printed at the end of a proof is straightforward. Reflex angle AOB is 260. In other words, if it is impossible for $$P$$ to be false, $$P$$ must be true. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. A proof by contradiction is often used to prove a conditional statement $$P \to Q$$ when a direct proof has not been found and it is relatively easy to form the negation of the proposition. To show that there is no finite state automata that. ExampleProve by contradiction that there is no greatest integer. Example: Use a proof by contradiction to give a proof that √2 is irrational. By the Pumping Lemma this must be representable as , such that all are also in. Without loss of generality n=m is reduced, i. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Therefore, the assumption that the quadratic equals zero is incorrect. Direct proof by contradiction. I Suppose pr 2 was rational. (Contrapositive) Let integer n be given. have no common factors (see Chapter 4). 3 Proof by contradiction (continued). We then try to obtain a contradiction from this assumption. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. But this is clearly impossible, since n2 is even. Therefore both n and m are odd. Discrete Mathematics This is a basic course for undergraduate students. Proof by Contradiction. To a resolution theorem-prover, both are two-step proofs. Proof: (Contrapositive: If n is even, then 3n + 2 is even) Suppose that the conclusion is false, i. Choose a long string w in L, jwj m. In the examples below we use this idea to prove the impossibility of certain kinds of solutions to some equations. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Then (2 is rational, so there are integers a and b for which (2= a b. (4 marks) 8 Use proof by contradiction to show. Properies of the modulus of the complex numbers. By Observation 3, it de nitely returns a matching, so suppose the matching is not perfect. As an example, here is a proof by contradiction of Proposition 4. This is a good resource if you are familiar with induction, and want to take things a little farther. Discrete Mathematics This is a basic course for undergraduate students. The idea behind proof by contradiction is that a statement must be true or false. Disprove by counterexample that for any , if , then. This is also known as Proof by Cases - see Example 1. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. 2 # 2 p23 Follow the statement of your assumptions with a statement of what you will prove. The proof by deduction section also includes a few practice questions, with solutions in a separate file. Here are several examples of properties of the integers which can be proved using the well-ordering principle. You must include all three of these steps in your proofs! The three key pieces: 1. n By contradiction n Start with the “proof by example”! n So when asked to prove a claim, an example that n Automata theory & a historical perspective. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Let x be an odd integer. by triangle inequality. Example: Prove there's an infinite number of evens. ] [Prove Q =)P using direct, contrapositive, or contradiction proof. What that contradiction means in the proof; Whether the Halting problem is an unsolvable problem, an undecidable problem, or both; and why; Write a paragraph explaining the difference between an problem that can't be solved (such as the halting problem) and a problem that takes unreasonable time. For example: 2/8 can be written in lowest terms as 1/4 when 1 and 4 are positive integers with no common prime factors. Then … (( make logical conclusions until you come to two statements that contradict each other, such as "X is true" and X is false" ))But this is a contradiction because …. Proof: I Assume finitely many primes: p1;:::; k. 7 [A level only] Prove that there are an infinite number of primes. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Show that this supposition logically leads to a contradiction. These notes explain these basic proof methods, as well as how to use definitions of new concepts in proofs. This proves A ⊆ A ∩ B. and qwhere q≠ 0. Proof: Suppose not. – If we had 2 nickels, we could replace them with 1 dime. The reason is that the proof set-up involves assuming ∼∀x,P(x), which as we know from Section 2. Proof: By induction, on the number of billiard balls. We must derive a contradiction. Example of proof by contradiction and more on proof by induction. Solutions Educator Edition Save time lesson planning by exploring our library of educator reviews to over 550,000 open educational resources (OER). But, from the parity property, we know that an integer is not odd if, and only if, it is. PLEASE DO A \PREFOR-. Proof by Contradiction: (AKA reductio ad absurdum). 4, namely that for any integer n, if n2 is even then n is even. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. Here's an algebraic example: Prove: For a,b≥0, a+b 2 ≥ab. If ¬P leads to a contradiction, then. Say we're trying to prove by contradiction that if n 2 is an odd number, then n is also odd for all integers n. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. Example using a Linear Function. Example: Give a direct proof of the theorem If nis an odd integer, then n2 is odd. Then n= 2k+ 1 for an integer k. In the proof above of Fact 1. Proof By Contradiction Examples And Solutions. • Example Every prime number a irrational numbers. You must include all three of these steps in your proofs! The three key pieces: 1. This proof is very similar to proof by contradiction, but subtly di er-ent. ] Suppose there is greatest even integer N. For every even integer n, N ≥ n. ¬Cube(a) 3. with √2 = a/b, where. This book is a guide to understanding and creating proofs. A proof by contrapositive uses that to prove the negation of the original assumption, while a proof by contradiction can negate any other true fact or lead to some other absurdity; in this case, you can't have two different smallest elements of a set. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true. This is usually a hint that proof by contradiction is the method of choice. We have step-by-step solutions for your textbooks written by Bartleby experts! Using Proof by Contradiction In Exercises 15 − 26 , use proof by contradiction to prove the statement. This is why proof by contradiction works. uzbt8y6d536r qsp0l7cs9uxx0b rqtztc3l4kvhz ve65emynwfzt fb2nd4u3ken19 cswb17aj5fv byn1w24apk0exlx pquj9ksys3c0sn iss43ryp1ywds 28pf05o6s2 n9jk6v0ab1 i94sauttvyg 7x3i711c7mw11kf fh46ysf9w7ic qz8m9vxmhy5 zy2kssr5fx13g 3h76hkwgzwpo y6j0hsw3ln6 estzy7oybir85b bwdlnq0skk 5nne7cw7flh natj3e81te5cn kc2t3uryp80 zd3r6tbgj5ai dfps6gopdz | 2020-10-21T21:43:52 | {
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https://math.stackexchange.com/questions/242643/a-game-played-on-a-rectangle | # A game played on a rectangle
Suppose two players play the following game on a $m$ by $n$ rectangle. Alternatingly they have to make a cross in some empty $1\times 1$ square. They are not allowed to make a cross next to another cross (Diagonally is OK, not just right next to each other). The player who places the last cross wins.
Now the question is for which $m,n$ does the starting player have a winning strategy?
At first I thought this might be just a nice exercise. So I had a look at 1 by $n$ rectangles first. A computer programme computed that the first player does not have a winning strategy for $n=4,8,14,20,24,28,34,38,42,54,58,62,72,76,88,92,96,106,110$ (for $n\le 110$).
I do not see any pattern in these numbers. So the answer might not be so easy.
The starting player can always win for odd $n$. He just places a cross in the middle and mirrors all the moves of the second player.
• @tomglabst: The game would be over if each remaining square is adjacent to some cross. Then no player can make any cross. Thus in your example the game would end after the first move. Specifically "P1 crosses any remaining square". There are no squares left that any player could cross. – HenrikRueping Nov 22 '12 at 15:42
• I think I've seen a variant of the $m=1$ game where there's a modulus $M$ such that the grundy number of the game only depends on $n$ modulo $M$. So it's a possibility that such an $M$ exists here too. – mercio Nov 22 '12 at 16:12
• @AndréNicolas The OP has noted values of $m$, where the first player has no winning strategy, and 6 is not among them. I see no edits made either. – Mike Nov 22 '12 at 17:00
• The sequence is not in the OEIS (!!). – dot dot Nov 22 '12 at 17:12
• @HenrikRueping Ah okay, I misunderstood that. – tomglabst Nov 22 '12 at 17:27
The $n \times 1$ version is Dawson's Chess. The OP's sequence is A215721 in the OEIS, after adding 1 to each term. I wrote a program too, and found that the proportion of losing initial positions seems to tend to a constant -- there are 1473 losing positions in the first 10000, and 14709 losing positions in the first 100000. I found an explanation at "Sprague-Grundy values for Dawson's Chess" (A002187 in the OEIS): | 2019-09-15T14:18:37 | {
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https://math.stackexchange.com/questions/3366424/find-the-singular-points-of-the-differential-equation-x3x-1y-2x-1y | Find the singular points of the differential equation $x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$.
Consider the second order linear homogeneous equation $$a_0(x)y'' + a_1(x)y'+ a_2(x)y = 0, x \in I \tag{1}$$ Suppose that $$a_0$$, $$a_1$$ and $$a_2$$ are analytic at $$x_0 \in I$$. If $$a_0(x_0) = 0$$, then $$x_0$$ is a singular point for $$(1)$$. Definition: A point $$x_0 \in I$$ is a regular singular point for $$(1)$$ if $$(1)$$ can be written as $$b_0(x)(x − x_0)^2y''+ b_1(x)(x − x_0)y'+b_2(x)y = 0, \tag{2}$$ where $$b_0(x_0) \neq 0$$ and $$b_0$$, $$b_1$$, $$b_2$$ are analytic at $$x_0$$.
The question is: Find the singular points of the differential equation $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ and state whether they are regular singular points or irregular singular points.
I think, $$x = 0$$, irregular singular point, $$x = 1$$, regular singular point. But, How can I prove this? Please proper guide me.
• Hi Harry, I've formatted your question with MathJax. I really encourage you to learn how to do this for yourself. I'd also like to remind you of your comment under your last question. – Theo Bendit Sep 23 '19 at 5:18
• @ Theo. Thanks again.I am trying latex but not done properly. I am on my way to working with MathJax. – user679406 Sep 23 '19 at 5:29
• That's good! Feel free to edit your question to see how I've formatted it. That might help you figure it out a bit faster. – Theo Bendit Sep 23 '19 at 5:31
Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$ where $$z \in D \subseteq \mathbb{C}$$.
The point $$x_0 \in D$$ is said to be an ordinary point of the above the given differential equation if $$P(x)$$ and $$Q(x)$$ are analytic at $$x_0$$.
If either $$P(x)$$ or $$Q(x)$$ fails to be analytic at $$x_0$$, the point $$x_0$$ is called a singular point of the given differential equation.
A singular point $$x_0$$ of the given differential equation is said to be regular singular point if the function $$(x-x_0)P(x)$$ and $$(x-x_0)^2 Q(x)$$ are analytic at $$x_0$$ and irregular otherwise.
$${}$$
Here the given equation is $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ $$\implies y''-\dfrac{2}{x^3}y'+\dfrac{3}{x^2(x - 1)}y=0$$ Here $$~P(x)=-\dfrac{2}{x^3}~$$and $$~Q(x)=\dfrac{3}{x^2(x - 1)}~$$.
Clearly $$~x=0,~1~$$ are singular points.
Again for the singular point $$~x=0~$$, $$(x-0)P(x)=-\dfrac{2}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=-\dfrac{2}{x}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are not analytic at $$~x_0=0~$$. So $$~x=0~$$ is irregular singular point.
For the singular point $$~x=1~$$, $$(x-1)P(x)=\dfrac{3}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=\dfrac{3(x - 1)}{x^2}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are analytic at $$~x_0=1~$$. So $$~x=1~$$ is regular singular point.
• @ nmsanta,I think this answer is correct and I am going with it. Am I right? – user679406 Sep 23 '19 at 5:32
• Have you faced any trouble in understanding my work ? I just doing things according to the definition. @Harry Richie – nmasanta Sep 23 '19 at 5:34
• No trouble. Clear like water..@ nmasanta – user679406 Sep 23 '19 at 5:37
• Only then my work will be well worth. – nmasanta Sep 23 '19 at 5:42
• Hello Mr. Down-voter, Would you like to explain the reason to give the down-vote in this answer ? – nmasanta Sep 24 '19 at 8:17 | 2020-03-30T14:12:01 | {
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http://math.stackexchange.com/questions/165941/finding-a-splitting-field-of-x3-x-1-over-mathbbz-2 | # Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$.
Ok so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots. So then I tried looking at the quotient field $\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't. Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\mathbb{Z}_2$?
Edit: Ok it turns out I miscalculated in my quotient field, $\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots.
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If degree is not $3$ it is $6$. – André Nicolas Jul 3 '12 at 2:08
Actually, your quotient $\mathbb{Z}_2[x]/(x^3+x+1)$ does contain at least one root: $x$. This is because $x^3+x+1\equiv 0\pmod{x^3+x+1}$. Can you find the others? – roninpro Jul 3 '12 at 2:08
The splitting field is either degree $3$ or degree $6$ over $\mathbb{Z}_2$, hence it is either $\mathbb{F}_8$ or $\mathbb{F}_{64}$.
Let $\alpha$ be a root, so that $\mathbb{F}_8 = \mathbb{F}(\alpha)$. The elements are of the form $a+b\alpha+c\alpha^2$, with $\alpha^3=\alpha+1$.
Now, the question is whether any of these elements besides $\alpha$ is a root of the original polynomial $x^3+x+1$.
Note that $(\alpha^2)^3 = (\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1$, and so if we plug in $\alpha^2$ into the polynomial we have $$\alpha^6 + \alpha^2 + 1 = \alpha^2+1+\alpha^2+1= 0.$$ Thus, $\alpha^2$ is also a root. So the polynomial has at least two roots in $\mathbb{F}_8$, and so splits there.
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oh shoot I must have miscaculated in my quotient field, ok cool, thanks! – cuckmaster5000 Jul 3 '12 at 2:11
If your cubic were to factor, the factorization must have at least one linear term, which corresponds to a root. It is easy to check it has no roots in $\mathbb{Z}_2$ - check them both! Plugging in $0$ and $1$ both give $1$ so are not roots, so your polynomial is irreducible over $\mathbb{Z}_2.$
A way to write the splitting field is $\mathbb{Z}_2(\alpha)$ where $\alpha$ is any one of the roots of $x^3+x+1.$ This is because $\alpha^2$ and $\alpha^4= \alpha^2+\alpha$ must then also be distinct roots of $x^3+x+1$, and these 3 then comprise a full list. Note, this isn't a special trick to notice for this problem. In fields of characteristic $p$, if $\beta$ is a root of a polynomial then $\beta^p$ will automatically also be a root, due to properties of the Frobenius endomorphism.
Or as roninpro pointed out in the comments, the quotient ring you considered (which is essentially the same thing as adjoining these roots) does contain at least one root, and by this same trick, all the roots.
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I thought forming the quotient field adjoined just a single root? And that sometimes you would get lucky and the remaining roots could be formed within the quotient field as well, but that sometimes they couldn't, is this not correct? – cuckmaster5000 Jul 3 '12 at 2:17
@NollieTré You're correct. But extensions of finite fields are all Galois and in particular normal. So if $K/k$ are finite fields and $f \in k[X]$ is irreducible and has a root in $K$, then it splits in $K$. Frobenius is a great thing! – Dylan Moreland Jul 3 '12 at 2:25
@NollieTré You are correct, sometimes the extra roots we find with the Frobenius trick just correspond to the same roots. But here they are distinct. You can check yourself (by carrying out the division) that here $x^2$ is also a root, since $x^6+x^2+1 = (x^3+x+1)(x^3-x-1)+2(x^2+x+1) = 0$, and similarly $x^4$ is also a root. So the quotient formed actually has all the roots. – Ragib Zaman Jul 3 '12 at 2:26
Ok interesting, so is there a canonical example of when the quotient field does not contain all the roots of the polynomial? – cuckmaster5000 Jul 3 '12 at 2:29
I think it's easier to note that if $\alpha$ is a root of the polynomial then squaring gives $0^2 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1$. – Dylan Moreland Jul 3 '12 at 2:30
Two good answers already here, but I wanted to emphasize the usefulness of the Frobenius endomorphism. Let $f(X) = X^3 + X + 1$. If I let $\alpha$ denote the image of $X$ in $k = \mathbb F_2[X]/(f(X))$ then applying Frobenius to $0 = f(\alpha)$ gives $0 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1.$ Hence $\alpha^2$ is also a root of $f$, and $\alpha^2 \neq \alpha$ because $\alpha \neq 0, 1$. Since $k$ contains two roots of the cubic $f$, it contains three.
To bring in slightly more technology: extensions of finite fields are Galois and in particular normal, and this implies that if you adjoin one root of an irreducible polynomial over a finite field then you've actually adjoined all of them. This is, of course, not true in general!
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Just wanted to add here, in $\Bbb{Z}_2(\alpha)[x]$, one can factor $x^3 + x + 1 = (x + \alpha)(x^2 + \alpha x + (\alpha^2 + 1))$. Given that $\alpha^2$ is also a root (as Dylan shows above), using ordinary high-school factorization techniques the third root is easily seen to be $\alpha^2 + \alpha$ (a somewhat preferable form than $\alpha^4$). That is: $x^2 + \alpha x + (\alpha^2 + 1) = (x + \alpha^2)(x + (\alpha^2 + \alpha))$ – David Wheeler Jul 6 '12 at 10:20 | 2016-04-29T02:12:28 | {
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https://math.stackexchange.com/questions/1603632/showing-that-a-group-g-with-two-distinct-normal-subgroups-of-index-p-has-or | # Showing that a group $G$ with two distinct, normal subgroups of index $p$ has order $p^2$
A friend of mine and I were working together to solve this exercise:
Let $G$ be a group, and $p$ a prime number. Let $H, K$ be normal subgroups of $G$. Suppose that $|G:H| = |G:K| = p$, and that $H\cap K = \{1\}$. Show that $G\simeq\mathbb{Z}_p\times\mathbb{Z}_p$.
We thought that the quickest way to solve it was showing that $G$ has exactly $p^2$ elements.
For that purpose, we first tried to prove that $|G|\leq p^2$. We chose $a,b\in xH\cap yK$, that are two elements of $G$ belonging to the intersection of a coset of $H$ with a coset of $K$. This means that $a = xh$ and $b=xh'$ for some $h,h'\in H$ and for some $x\in G$ (because $a,b\in xH$). Similarly, $a=yk$ and $b=yk'$, for some $k,k'\in K$ and for some $y\in G$ (because $a,b\in yK$).
Then, $$xh=yk\implies hk^{-1}=x^{-1}y$$ and $$xh'=yk'\implies h'k'^{-1}=x^{-1}y.$$
This brings us to $hk^{-1}=h'k'^{-1}$, which we can rewrite as $hh'^{-1}=k'^{-1}k$. This element belongs to $H\cap K=\{1\}$, thus $h = h'$ and $k = k'$: hence $a = b$. This proves that the intersection of a coset of $H$ with a coset of $K$ cannot contain more than one distinct element.
We made use of this information when we noted that, regardless of how many elements $G$ has, there exist at most $p^2$ intersections of cosets of $H$ with cosets of $K$, and each one of them contains at most one element. Thus, we deduced that $|G|\leq p^2$.
Now, here are my questions:
• How do I prove that $|G|\ge p^2$, in order to conclude that $|G|=p^2$ (and thus that $G\simeq\mathbb{Z}_p\times\mathbb{Z}_p$)?
• Is the proof rigorous enough so far, or were some of our assumptions incorrect?
• This exercise is related to a part of our algebra course in which nothing beyond direct product of groups had been explained yet, so we could not make use of the Sylow theorems, of anything about group actions and so on. Is there a simpler way to solve this exercise, using only "basic" tools of group theory (e.g. quotient groups, the homomorphism and isomorphism theorems and anything regarding the direct product of groups)?
• Is $G$ assumed to be finite at the outset? – Tim Raczkowski Jan 7 '16 at 20:46
• @TimRaczkowski The exercise said nothing on that point, so I assume that $G$ may also be infinite. Anyway, given that $G$ has to be isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$, I guess it can't be infinite. Plus, if our half of the proof is correct, $|G|\leq p^2$ definitely implies that $G$ is finite. – Labba Jan 7 '16 at 20:48
You can use the following fact: If $H,K$ are subgroups of $G$, $H\cap K=\{1\}$ and $hk=hk$ for all $h\in H$ and $k\in K$, then $G\cong H\times K$.
Now, $hk\in hK,Hk$. Since $H$ is normal in $G$, $Hk=kH$. So, by your argument above, $\{hk\}=hK\cap kH$. But $kh\in hK\cap kH$ by a similar argument.
• Thanks for your answer :) So, if I got it right, you say that $hk$ belongs to the $hK$ and $Hk$ cosets (the latter being $kH$ anyway, because of the normality of $H$), and the same can be said about $kh$ that belongs both to the $kH$ and $Kh$ cosets (again, the $hK$ coset). In short, $hk, kh\in kH\cap hK$ and, by my part of the proof, $kh = hk$ because only one element can be found in this intersection. Hence, $G\simeq H \times K$ because $H\cap K = \{1\}$ and they commute with each other. Did I get everything? And this also proves that $|G| = p^2$, right? – Labba Jan 7 '16 at 21:15
• I really have to thank you :) Anyway, there's still something I fear I'm missing: even if this shows that $G\simeq H\times K$, does this prove that $|H| = |K| = p$ as well? – Labba Jan 7 '16 at 21:18
• Well if $|G|=p^2$, and $[G:H]=[G:K]=p$, so $|H|=|K|=p$. – Tim Raczkowski Jan 7 '16 at 21:21
• Hmm....I seem to keep missing a detail each time. $[G:H]=[G:K]\implies |H|=|K|$. Suppose $|HK|=n^2$. Now, $n^2|p^2\implies$n=1, or $n=p$. If $n=1$, then $H=N=\{1\}$ and $|G|=p$. So assuming $H\ne K$, we get the result. – Tim Raczkowski Jan 7 '16 at 21:38
Since $H,K$ normal in $G$ and $H\cap K=\{1\}$, $HK\cong H\times K$. $|G:HK|\cdot|HK:H|=|G:H|=p$. So $|HK:H|$ divides $p$. $|HK:H|=1$, or $p$. Suppose it is $1$. $HK=H$, $K\subseteq H$. $|G:H|\cdot|H:K|=|G:K|$, $p|H:K|=p$, so $H=K$, contradiction. So $|HK:H|=p$, $|G:HK|=1$. $G=HK$. Now, $|G:H|=|HK:H|=|K:H\cap K|$ by second isomorphism theorem. So $p=|K|$. Similarly, $|H|=p$. $G\cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$. | 2019-10-18T04:50:36 | {
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https://math.stackexchange.com/questions/2412375/suppose-we-roll-a-fair-6-sided-die-repeatedly-find-the-expected-number-of-rol | # Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession.
Suppose we roll a fair six sided die repeatedly.
Find the expected number of rolls required to see $$3$$ of the same number in succession
From the link below, I learned that $$258$$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $$258/6 = 43$$. But I'm unsure how to show this and whether it really is correct.
How many times to roll a die before getting two consecutive sixes?
For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$ $$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$
• This is very elegant (+1) - great answer!! Aug 31, 2017 at 16:38
• One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem)
– cats
Aug 31, 2017 at 18:32
• This is the (arch classical) approach used in the post mentioned in Byron's answer.
– Did
Aug 31, 2017 at 18:40
• @lulu can you explain me what what tis the theory behind this answer? Does it refer to a specific branch of stochastics? Mar 9, 2021 at 23:30
• @Phillipp This is a fairly standard Markov type computation. The trick, such as it is, is to note that there really are only three possible states for the process, an observation which makes the system very tractable.
– lulu
Mar 9, 2021 at 23:39
From Did's answer here, the probability generating function $u_0(s)=\mathbb{E}(s^T)$ for the number of trials $T$ needed to get three consecutive values the same is
$$u_0(s)={s^3\over 36-30s-5s^2}.$$ Differentiating this and setting $s=1$ in the derivative shows that $\mathbb{E}(T)=43.$
We can treat this as a three-state absorbing markov chain: a length3 run has been seen, otherwise the current run is length2, current run is length 1. Transition matrix: $\begin{bmatrix}1&\frac{1}{6}&0\\0&0&\frac{1}{6}\\0&\frac{5}{6}&\frac{5}{6}\end{bmatrix}$
This is in standard form $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$
We turn our attention to the fundamental matrix $(I-R)^{-1} = \begin{bmatrix}1&-\frac{1}{6}\\-\frac{5}{6}&\frac{1}{6}\end{bmatrix}^{-1}=\begin{bmatrix}6&6\\30&36\end{bmatrix}$
After the first roll, we enter the markov chain in the third state. Adding the entries of the fundamental matrix corresponding to that column tells us the expected time until we reach an absorbing state, i.e. until we have a chain of three consecutive rolls of the same number.
Thus the expected number of rolls needed is $1+36+6=43$
Let $\mu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with distinct result.
Let $\nu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with equal result.
Then the expectation is: $$2+\frac56\mu+\frac16\nu$$
Here $\frac56$ is the probability that the first two numbers are distinct and $\frac16$ is the probability that they are equal.
Secondly we have the relations:
$$\mu=\frac56(1+\mu)+\frac16(1+\nu)=1+\frac56\mu+\frac16\nu\tag1$$ and: $$\nu=\frac161+\frac56(1+\mu)=1+\frac56\mu\tag2$$
The relations $(1)$ and $(2)$ lead easily to: $\mu=42$ and $\nu=36$.
Then $$2+\frac56\mu+\frac16\nu=43$$ is the final answer.
(in the answers uptil now it was not used that you allready learned something).
Let $Y$ denote the number of rolls required to see three dice of the same number in succession and let $X$ denote the number of rolls required to see three dice with number $6$ in succession.
Then: $$Y\text{ and }\frac16X+\frac56(X+Y)=X+\frac56Y\text{ must have equal distribution.}\tag3$$
Here $\frac16$ is the probability of the event that the first time that three dice give the same number in succession they show number $6$ and $\frac56$ is the probability that do not show a number $6$.
$(3)$ rests on the observation that - if for the first time three equal numbers show up in succession - we are ready if $6$ happens to be that number and must actually start over again (with $X$ throws in our pocket) if not.
So we find $\mathbb EY=\mathbb EX+\frac56\mathbb EY$ or equivalently:$$\mathbb EX=\frac16\mathbb EY$$
You allready learned that $\mathbb EY=258$ and making use of that knowledge you find $$\mathbb EX=258/6=43$$ This confirms your thinking. | 2022-05-27T18:34:43 | {
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https://www.allaboutcircuits.com/textbook/digital/chpt-8/venn-diagrams-and-sets/ | # Venn Diagrams and Sets
## Chapter 8 - Karnaugh Mapping
Mathematicians use Venn diagrams to show the logical relationships of sets (collections of objects) to one another. Perhaps you have already seen Venn diagrams in your algebra or other mathematics studies. If you have, you may remember overlapping circles and the union and intersection of sets. We will review the overlapping circles of the Venn diagram. We will adopt the terms OR and AND instead of union and intersection since that is the terminology used in digital electronics. The Venn diagram bridges the Boolean algebra from a previous chapter to the Karnaugh Map. We will relate what you already know about Boolean algebra to Venn diagrams, then transition to Karnaugh maps. A set is a collection of objects out of a universe as shown below. The members of the set are the objects contained within the set. The members of the set usually have something in common; though, this is not a requirement. Out of the universe of real numbers, for example, the set of all positive integers {1,2,3…} is a set. The set {3,4,5} is an example of a smaller set, or subset of the set of all positive integers. Another example is the set of all males out of the universe of college students. Can you think of some more examples of sets? Above left, we have a Venn diagram showing the set A in the circle within the universe U, the rectangular area. If everything inside the circle is A, then anything outside of the circle is not A. Thus, above center, we label the rectangular area outside of the circle A as A-not instead of U. We show B and B-not in a similar manner. What happens if both A and B are contained within the same universe? We show four possibilities. Let’s take a closer look at each of the the four possibilities as shown above. The first example shows that set A and set B have nothing in common according to the Venn diagram. There is no overlap between the A and B circular hatched regions. For example, suppose that sets A and B contain the following members:
set A = {1,2,3,4}
set B = {5,6,7,8}
None of the members of set A are contained within set B, nor are any of the members of B contained within A. Thus, there is no overlap of the circles. In the second example in the above Venn diagram, Set A is totally contained within set B How can we explain this situation? Suppose that sets A and B contain the following members:
set A = {1,2}
set B = {1,2,3,4,5,6,7,8}
All members of set A are also members of set B. Therefore, set A is a subset of Set B. Since all members of set A are members of set B, set A is drawn fully within the boundary of set B. There is a fifth case, not shown, with the four examples. Hint: it is similar to the last (fourth) example. Draw a Venn diagram for this fifth case. The third example above shows perfect overlap between set A and set B. It looks like both sets contain the same identical members. Suppose that sets A and B contain the following:
set A = {1,2,3,4}
set B = {1,2,3,4}
Therefore,
Set A = Set B
Sets And B are identically equal because they both have the same identical members. The A and B regions within the corresponding Venn diagram above overlap completely. If there is any doubt about what the above patterns represent, refer to any figure above or below to be sure of what the circular regions looked like before they were overlapped. The fourth example above shows that there is something in common between set A and set B in the overlapping region. For example, we arbitrarily select the following sets to illustrate our point:
set A = {1,2,3,4}
set B = {3,4,5,6}
Set A and Set B both have the elements 3 and 4 in common These elements are the reason for the overlap in the center common to A and B. We need to take a closer look at this situation.
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Published under the terms and conditions of the Design Science License | 2020-02-23T05:58:00 | {
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https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-72_14.html | ### CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.)
Suppose $A$ and $B$ are $n \times n$ positive definite matrices and $I$ be the $n \times n$ identity matrix. Then which of the following matrices are positive definite?
1) $A+B$,
2) $ABA^{*}$,
3) $A^2+I$,
4) $AB$.
Solution:
Let $A$ be a real symmetric matrix, then $A$ is said to positive definite if it satisfies any of the following equivalent conditions.
i) all its eigenvalues are positive.
ii) $x^t A x > 0$ for all vectors $x \ne 0$.
iii)$<x,Ax> > 0$ for all vectors $x \ne 0$.
iii) A is positive definite if and only if it can be written as $A = R^tR$, where $R$ is possibly a rectangular matrix, with independent columns.
iv) All the principal minors of $A$ are positive. (Please share if you know any other equivalent conditions in the comment below)
Let $A$ and $B$ be two $n \times n$ positive definite matrices. We have $x^t A x>0$ and $x^t B x>0$ for $x \ne 0$. We will solve each given option by each of the above given definition of positive definiteness in order to understand them clearly.
option 1. (True)We have for $x \ne 0$, $$x^t (A+B) x = x^t A x + x^t B x > 0.$$
Therefore option 1 is true.
option 2. (True) We have for $x \ne 0$, $Ax \ne 0$ since $A$ is invertible.
$$<x,ABA^*x> = <A^*x,BA^*x> = <B^*A^*x,A^*x> \\ = <BAx,Ax> > 0 .$$
Therefore option 2 is true. Since $A$ and $B$ are real symmetric, we have $A^* = A$ and $B^* = B$.
option 3. (True) Let the eigen values of $A$ be $\lambda_1,\lambda_2,\dots,\lambda_n$. Since $A$ is positive definite we have all these eigen values are positive. We observe that the eigen values of $A^2+1$ are $\lambda_1^2+1,\lambda_2^2+1,\dots,\lambda_n^2+1$ which are all positive. Hence $A^2+1$ is positive definite.
option 4. (False)
The product of two positive definite matrices need not be even symmetric. In particular, we have $AB$ is symmetric if and only if $A$ and $B$ commutes with each other. Because $$(AB)^* = B^*A^* = A^*B^* = AB.$$ Note that, if $A$ and $B$ commutes with each other, then $A^*$ and $B^*$ commutes with each other.
To illustrate this, consider the positive definite matrices $$A = \begin{bmatrix}11 & 10 \\ 10 & 10\end{bmatrix}$$ and $$B = \begin{bmatrix}11 & 5 \\ 5 & 10\end{bmatrix}$$
text{ Then their product } $$AB = \begin{bmatrix}171&155 \\ 160&150\end{bmatrix}$$ which is not symmetric.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | 2021-01-18T00:49:18 | {
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https://math.stackexchange.com/questions/2319210/repeatedly-taking-differences-on-a-polynomial-yields-the-factorial-of-its-degree/2319671 | # Repeatedly taking differences on a polynomial yields the factorial of its degree?
Consider a function such that it takes in polynomial function and creates an array of its outputs and then using that array creates another new array by calculating the absolute difference between the first $2$ values and keeps doing this until it reaches an array full of zeros.
This is much easier to show you by example.
For example take $F(x)= x^2$, the first array would be
$1,4,9,16,25,36,49,64,81$ and so on, the second would be
$3,5,7,9,11,13,15,17,19$ ( the difference between the first value and the second one)
but the third one is where it gets interesting as if we continue the pattern we would get an array filled with only $2$'s and after that it would only be zeros.
Lets do another example, $F(x)=x^3$
$1,8,27,64,125,216,343$
$7,19,37,61,91,\dotsc$ but here is the interesting part if we continue this
$12,18,24,30,\dotsc$ and once more then we get
$6,6,6,6,6,\dotsc$ after that it would just be an array of zeros
There are $2$ main observation that I made about this
Firstly, the value that is begin repeated indefinitely is equals to the factorial of the functions power. Meaning that for $F(x)=x^2$ the value being repeated is $2!$. For $F(x)=x^3$ , it's $3!$ and this is true for all polynomials (I tried it up to $x^7$, after that it got too messy)
Secondly, the value that is repeated always occurs on the $n$th iteration of the function. Meaning that for $F(x)=x^2$, we have to go through the processes $2$ times before we find the value. For $F(x)=x^3$, we have to go through it $3$ times before getting the value.
Is there any way to prove this and does this mean anything at all?
• Lookup finite differences ("an $n^{th}$ power has a constant $n^{th}$ finite difference") . Also for example this answer. – dxiv Jun 12 '17 at 3:25
• Good. You discovered a concept (through on a sequence basis) very similar to derivative. An excellent mind of discovering and inducting. Continue it if you are still young and have enough time. – BAI Jun 12 '17 at 4:45
• Welcome to Math.SE! Since one aim of the site is to collect questions and answers in forms useful for posterity, would you please consider re-titling your question to indicate its content? (Perhaps something like "Repeated differences for polynomial functions at equally-spaced inputs"...?) – Andrew D. Hwang Jun 12 '17 at 11:18
• – Simply Beautiful Art Jun 12 '17 at 17:21
• @user21820 your edit to the question is a great example of what a good title can do for a better understanding of a given topic and to promote the curiosity of the reader. – iadvd Jun 13 '17 at 0:12
## 2 Answers
Here's a fact:
• If $p(x)$ is a polynomial of degree $n$ with leading term $ax^n$ then $p(x+1)-p(x)$ is a polyomial of degree $n-1$ with leading term $a \, n \, x^{n-1}$.
(I'll prove this fact below).
Applying this fact together with an induction argument, it follows that after repeating the process $n$ times, one obtains a polynomial of degree zero whose leading term is $$a \, n \, (n-1) \ldots (2) (1) = a \, n!$$ which is just a constant having that value.
So if the original leading coefficient $a$ is equal to $1$, as it is in the specific cases $F(x)=x^n$ that you ask about, repeating the difference process $n$ times yields a constant sequence of $n!$ as you ask.
Here's a proof of the fact by applying induction (the base case $n=1$ is easy).
Assuming the induction hypothesis for polynomials of degree $\le n-1$, suppose that $$p(x) = a \, x^n + q(x)$$ where $q(x)$ is a polynomial of degree $\le n-1$.
We have $$p(x+1)-p(x) = a \, (x+1)^n - a \, x^n + \underbrace{(q(x+1)-q(x))}_{r(x)}$$ and $r(x)$ is a polynomial of degree $\le n-2$ by induction. Thus $$p(x+1)-p(x) = a \, (x^n + n \, x^{n-1} + s(x)) - a\, x^n + r(x)$$ where $s(x)$ is also a polynomial of degree $\le n-2$ (by application of the binomial theorem). Therefore $$p(x+1)-p(x) = a \, n \, x^{n-1} + (a \, s(x)+r(x))$$ which is a polynomial of degree $n-1$ with leading term as required.
What you have discovered/invented is known as the forward difference operator $D$ defined as: $\def\nn{\mathbb{N}} \def\zz{\mathbb{Z}} \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\lbinom#1#2{{\large\binom{#1}{#2}}}$
$D = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) - f(n) ) )$
Namely for any function $f$ on $\zz$ and $n \in \zz$, $D(f)(n) = f(n+1) - f(n)$.
If you think of the functions as sequences (infinite in both directions), then taking the forward difference means replacing each term with the value of the next term minus itself. What you did is essentially to repeatedly take the forward difference of the sequence of cubes:
...,-27,-8,-1, 0, 1, 8,27,...
..., 19, 7, 1, 1, 7,19,37,...
...,-12,-6, 0, 6,12,18,24,...
..., 6, 6, 6, 6, 6, 6, 6,...
..., 0, 0, 0, 0, 0, 0, 0,...
..., 0, 0, 0, 0, 0, 0, 0,...
This powerful abstraction makes it easy to get a lot of things. For instance, the numbers obtained here can be easily used to obtain the general formula for sum of cubes!
General method for indefinite summation
The key is that:
$D\left( \text{int$n$} \mapsto \lbinom{n}{k+1} \right) = \left( \text{int$n$} \mapsto \lbinom{n}{k} \right)$ for any $k \in \zz$.
This is to be expected because it follows directly from Pascal's triangle, especially if we define $\lbinom{n}{k}$ using the triangle.
This means that if we have any function $f$ on $\zz$ such that $f(n) = \sum_{k=0}^\infty a_k \lbinom{n}{k}$ for any $n \in \zz$, then we get the Newton series:
$D(f)(n) = \sum_{k=0}^\infty a_{k+1} \lbinom{n}{k}$ for any $n \in \zz$.
From a high-level perspective, this is the discrete version of the Taylor series, and indeed for such a function we easily see that $f(n) = \sum_{k=0}^\infty D^k(f)(0) \lbinom{n}{k}$ for any $n \in \zz$, because $\binom{0}{0} = 1$ while $\lbinom{0}{k} = 0$ for any $k \in \nn^+$.
This works for any polynomial function $f$ on $\zz$, since $D^k(f)$ is the zero function once $k$ is larger than the degree of $f$, so we can use it to immediately find the series for $(\text{int n} \mapsto n^3)$, and then just take the anti-difference by shifting the coefficients of the series the other way. The undetermined constant that appears will drop out once we perform a definite sum like if we want the sum of the first $m$ cubes.
Note also that $D^k(f)$ is the constant function with value $k!$ if $f(n) = n^k$ for every $n$. Lee Mosher has already explained this particular fact by directly proving it, but another way to see it is that the highest order term in its Newton series is $k! \lbinom{n}{k}$, because $\lbinom{n}{k}$ is the only term that can contribute the $k$-th power of $n$. Since $D$ merely shifts the coefficients, $D^k(f) = \left( \text{int$n$} \mapsto k! \lbinom{n}{0} \right)$ and we are done.
Sum of $p$ powers
For example if we want $\sum_{k=1}^{n-1} k^3$ we first find the iterated forward differences of the sequence of cubes $( n^3 )_{n \in \zz}$:
..., 0, 1, 8,27,64,...
..., 1, 7,19,37,...
..., 6,12,18,...
..., 6, 6,...
..., 0,...
So we immediately get $n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$ and hence $\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = \lfrac{n(n-1)}{2} \Big( 1 + \lfrac{6(n-2)}{3} \big( 1 + \lfrac{n-3}{4} \big) \Big) = \Big( \lfrac{n(n-1)}{2} \Big)^2$.
Computation efficiency
This is far more efficient than the usual method (namely by taking summation on both sides of $(n+1)^3-n^3 = 3n^2+3n+1$ and telescoping) because the series using binomial coefficients is easy to manipulate and easy to compute. For sum of $p$-powers we only need $O(p^2)$ arithmetic operations to find the forward-differences and then $O(p^2)$ more to simplify the series form into a standard polynomial form. In contrast, the other method requires $O(p^3)$ arithmetic operations.
Indefinite summation of non-polynomials
Also, for a wide class of non-polynomial functions, we can still compute the indefinite sum without using the series, by using the discrete analogue to integration by parts, here called summation by parts.
To derive it, simply check that $D(f \times g)(n) = f(n+1) g(n+1) - f(n) g(n) = f(n+1) D(g)(n) - D(f)(n) g(n)$ and so we get the product rule:
$D(f \times g) = R(f) \times D(g) + D(f) \times g$
where $R$ is the right-shift operator defined as:
$R = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) ) )$
Namely for any function $f$ on $\zz$ and $n \in Z$, $R(f)(n) = f(n+1)$.
For convenience we also define the summation operator:
$S = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto \sum_{k=0}^{n-1} f(k) ) )$
Then we have the important property that $DS(f) = f$ for any function $f$ on $\zz$, analogous to the fundamental theorem of calculus.
Now by substituting $f$ with $S(f)$ into the product rule and taking summation on both sides we get summation by parts:
$S( f \times g ) = S(f) \times g - S( R(S(f)) \times D(g) ) + c$ for some constant function $c$ on $\zz$.
Example indefinite sum
Using this we can easily compute things like $\sum_{k=1}^n k^3 3^k$ by applying it three times, each time reducing the degree of the polynomial part. There are other ways to achieve this using differentiation, but this method is purely discrete. | 2020-09-20T07:21:31 | {
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# MATH 4389 Final: Real_Analysis(part_2) Premium
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PART II. SEQUENCES OF REAL NUMBERS II.1. CONVERGENCE Definition 1. A sequence is a real-valued function f whose domain is the set positive integers (N). The numbers f(1),f (2), ··· are called the terms of the sequence. Notation Function notation vs subscript notation: f(1) ≡ 1 ,f (2) ≡2s ,··· ,f (n) n s , ··· . In discussing sequences the subscript notation is much more common than functional notation. We’ll use subscript notation throughout our treatment of analysis. Specifying a sequence There are several ways to specify a sequence. 1. By giving the function. For example: 1 1 1 1 1 1 (a) sn= n or {sn} = n . This is the sequence {12 ,3 ,4 ,...,n,... }. (b) s = n − 1. This is the sequence {0, ,2 , 3,...,n − 1,... }. n n 2 3 4 n (c) sn=( −1) n . This is the sequence {−1,4,−9,16,..., (−1) n ,... }. 2. By giving the first few terms to establish a pattern, leaving it to you to find the function. This is risky – it might not be easy to recognize the pattern and/or you can be misled. (a) {sn} = {0,1,0,1,0,1,... }( The pattern here is obvious; can you devise the function? It’s 1 − (−1)n) 0, n odd sn= or s n= 2 1, n even 2 (b) {s } = 2, 5,10,17 ,6 ,... ,s = n +1 . n 2 3 4 5 n n (c) {sn} = {2,4,8,16,32,... }. What is6s ? What is the function? While you might say 64 n and s n2 , the function I have in mind gives s6= π/6: n π 64 sn=2 +( n − 1)(n − 2)(n − 3)(n − 4)(n − 5) − 720 120 3. By a recursion formula. For example: (a) s = 1 s ,s = 1. The first 5 terms are 1,1, , 1 , 1 ,.... Assuming that n+1 n +1 n 1 2 6 24 120 1 the pattern continues n = . n! 1 (b) sn+1 = (n +1) ,s 1 = 1. The first 5 terms are {1,1,1,1,1,... }. Assuming that the 2 pattern continues n = 1 for all n; {n } is a “constant” sequence. 13 Definition 2. A sequence {s }nconverges to the number s if to each > 0 there corresponds a positive integer N such that |s − s| for all n>N. n The number s is called the limit of the sequence. Notation “{s } converges to s” is denoted by n n→∞ sn= s, or by limsn= s, or by n → s. A sequence that does not converge is said to diverge. Examples Which of the sequences given above converge and which diverge; give the limits of the convergent sequences. THEOREM 1. If s → s and s → t, then s = t. That is, the limit of a convergent sequence is unique. Proof: Suppose s 6= t. Assume t>s and let = t − s. Since sn→ s, there exists a positive integer N such that |s − s | / 2 for all n>N . Since s → t, there exists a positive integer 1 n 1 n N 2 such that |t−s n / 2 for all n>N 2. Let N = max{N ,N1} a2d choose a positive integer k>N . Then t − s = |t − s| = |tk− s k s − s|≤| t k s | + |sk− s |+< = = t − s, 2 2 a contradiction. Therefore, s = t. THEOREM 2. If {s } connerges, then {s } is bnunded. Proof: Suppose s → s. There exists a positive integer N such that |s−s | < 1 for all n>N . n n Therefore, it follows that |sn| = |n − s + s|≤| sn− s| + |s| < 1+ |s| for alln>N. Let M = max{|s |, |s |, ..., |s |, 1+ |s|}. Then |s | N. If an→ 0, then s →n0. Proof: Note first that a ≥n0 for all n>N . Since a → 0,nthere exists a positive integer N 1 such that |an| < /k. Without loss of generality, assume that1N ≥ N. Then, for all n>N 1, |sn− 0| = |sn|≤ ka n
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So we can recommend you notes for your school. | 2018-04-23T08:30:45 | {
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https://math.stackexchange.com/questions/2846965/how-are-numbers-distributed-under-a-different-modulus | Suppose we take a sample of numbers with unique congruence class modulo p:
x_0 ≡ 0 (mod p)
x_1 ≡ 1 (mod p)
x_2 ≡ 2 (mod p)
...
x_(p-2) ≡ p-2 (mod p)
x_(p-1) ≡ p-1 (mod p)
and we then examine their class under modulo q. How will these classes be distributed? For example, will some values appear more often than others? Or might they be uniform in number or all unique? I imagine this result would depend on if p<q or p>q, and perhaps other things.
Specific, numerical example
15 = 5(3) ≡ 0 (mod 5)
26 = 5(5)+1 ≡ 1 (mod 5)
37 = 5(7)+2 ≡ 2 (mod 5)
58 = 5(11)+3 ≡ 3 (mod 5)
69 = 5(13)+4 ≡ 4 (mod 5)
and
15 ≡ 1 (mod 2)
26 ≡ 0 (mod 2)
37 ≡ 1 (mod 2)
58 ≡ 0 (mod 2)
69 ≡ 1 (mod 2)
and
15 ≡ 3 (mod 6)
26 ≡ 2 (mod 6)
37 ≡ 1 (mod 6)
58 ≡ 4 (mod 6)
69 ≡ 3 (mod 6)
• If you sample the $x_i$ values uniformly, then I imagine their remainders modulo $p$ is uniform, so that the following remainders modulo $q$ are also uniform. – Bill Wallis Jul 10 '18 at 19:08
• That seems vaguely intuitive to me as well (and for my purposes, I'd like that to be true in general), but I'd like for someone to prove that if it is the case. Then again, it can't be completely true, right, since when q>p, we'll never get the larger modulo values, larger than p-1? – Steve Jul 10 '18 at 19:11
• If $q<p$ and the samples are uniform $\bmod p$, they can be uniform $\bmod q$ only if $q|p$ (pigeonhole principle) – gammatester Jul 10 '18 at 19:12
• An example of something I'd like to know if ever occurs, say we have sample (0,1,2,3,4,5,6.....20) in mod 21, is it possible to get a skewed distribution like (0,1,1,1,4,1,0,1,1,0,3) in mod 11 where there are "more 1s than expected"? – Steve Jul 10 '18 at 19:28
• @Steve: If $p=2$, there's nothing stopping you from taking large values of $x$ like $x_0 = 384$ and $x_1 = -7447$. – user14972 Jul 10 '18 at 23:18
Recall the Chinese Remainder Theorem; in one form, it says that if $p$ and $q$ are relatively prime integers, then there is a bijective correspondence between the
• residue classes modulo $pq$
• pairs consisting of a residue class modulo $p$ and a residue class modulo $q$
Furthermore, given any integer $x$, its residue classes modulo $pq$, $p$ and $q$ are related by this correspondence.
In particular, integers fall into every combination of residues modulo $p$ and $q$, and they do so exactly once per period of length $pq$. So knowing the residue class of an integer modulo $p$ tells you absolutely nothing about its residue class modulo $q$.
And while the notion of a uniform distribution doesn't actually make sense for the integers, this periodic behavior still allows us to capture the idea that such a thing would have independent distributions modulo $p$ and $q$.
In the example of taking $p=21$ and $q=11$, we can give this bijection by explicit formula (discussions of the CRT should show how to obtain this):
• $22 x - 21 y \equiv x \bmod 21$
• $22 x - 21 y \equiv y \bmod 11$
Similarly, any other integer with the same residue as $22x - 21y \bmod 231$ will also satisfy these congruences.
So, given any choice of twenty-one residue class modulo 11, you could find a sequence of $x$'s that have the chosen residues $\bmod 11$ along with the required residues $\bmod 21$. For example, if you want all $1$'s, then we can take
$$x_n = 1 + 22(n-1)$$
to get
• $x_n \equiv n \bmod 21$
• $x_n \equiv 1 \bmod 11$
• This is fantastic and I think I roughly follow what you're saying and at least understand it's impact in a very generalized sense. In your explicit example, 1, 23, 45, 67, ..., 199, 221 are all 1 mod 11 while running across the 0,1,2,...,11 spectrum of values under mod 21. This is helpful because it allows me to refine my question back to my original post, however. What if we restrict ourselves to numbers which not only are sequential in mod 21, but rather are strictly sequential numbers (x, x+1, x+2, ...). Can we know anything about these numbers under mod q? – Steve Jul 11 '18 at 12:48
• I'm realizing now, maybe my question is stupid and obvious afterall. Can you tell me if this follows? Let a, p, q be constant integers. Let x_n= ap+n for n={0,1,...p-1}. Then x_n (mod p) ≡ n. Observe that ap (mod q) ≡ B for some integer B. Thus, x_n (mod q) = ap+n (mod q) ≡n (mod q)+B, so the residues unders mod q are "uniformly distributed" save some wrapping around if q is smaller than p. Is this correct? – Steve Jul 11 '18 at 13:02
If you take any $q$ successive numbers, there will be one in each congruence class $\bmod q$. You start with the lowest one in its class, the next is in the next class up, and so on, wrapping around at $0$. When you get to $q$ you have put one in each class and are ready to start again. If you take $p$ successive numbers where $q\not | p$ you will have two different counts among the congruence classes. Let $r$ be the remainder on dividing $p$ by $q$. There will be $q-r$ classes with $\lfloor \frac pq \rfloor$ because you go around fully that many times. There will be $r$ classes with one more because you have $r$ numbers left after the full cycles. | 2021-07-31T12:38:25 | {
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https://math.stackexchange.com/questions/2574916/definite-integral-possible-evaluation-using-real-methods | Definite integral - possible evaluation using real methods?
The book "inside interesting integrals" gives the following exercise for the chapter about contour integration and the residue theorem: $$\int_{0}^\infty \frac{e^{\cos x}\sin(\sin x)}{x}dx=\space\space ?$$ This can be solved using the function $$f(z)=\frac{\exp(e^{iz})}{z}$$ on a quarter-circular contour, and is pretty straightforward. The answer turns out to be $$\frac{\pi}{2}(e-1)$$ However, in the book, the author makes the following comment:
Edward Copson (1901-1980), who was professor of mathematics at the University of St. Andrews in Scotland, wrote "A definite integral which can be evaluated using Cauchy's method of residues can always be evaluated by other means, though generally not so simply." Here's an example of what Copson meant, an integral attributed to the great Cauchy himself. It is easily done with contour integration, but would (I think) otherwise be pretty darn tough.
Does anyone know how to evaluate this integral using real methods?
• This question should actually get more attention... I already grabbed my popcorn to see how real methods come in action. If this does not get attention I'm considering to put a bounty on it (if that is possible, I don't know how that works). – Shashi Jan 7 '18 at 21:14
• I have put a bounty on this question. @Nilknarf Are there some updates concerning this question? – Shashi Jan 8 '18 at 20:47
• @Shashi No... Do you suggest that I add anything in particular to the question? – Frpzzd Jan 8 '18 at 23:55
• I was curious whether you have solved it in the mean time – Shashi Jan 9 '18 at 8:00
Perhaps surprisingly, a straightforward trick works. To this end we refer to the following easy-to-prove lemma.
Lemma. Define $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, dt$. Then
1. $\int_{0}^{x} \frac{\sin(yt)}{t} \, dt = \operatorname{Si}(xy)$, and
2. $\operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}(x^{-1})$ as $x \to \infty$.
Then for $R > 0$,
\begin{align*} \int_{0}^{R} \frac{e^{\cos x}\sin(\sin x)}{x} \, dx &= \int_{0}^{R} \frac{1}{x}\operatorname{Im}(e^{e^{ix}}) \, dx \\ &= \int_{0}^{R} \frac{1}{x}\sum_{n=1}^{\infty} \frac{\sin(nx)}{n!} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin(nx)}{x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \operatorname{Si}(nR) \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{\pi}{2} + \mathcal{O}\left( (nR)^{-1} \right) \right) \\ &= \frac{\pi}{2}(e - 1) + \mathcal{O}(R^{-1}) \end{align*}
Letting $R \to \infty$ proves the claim.
• Not thought about a single contour and yet so simple! Conclusion: the author of the book lied lol (jk I know one does not have full vision on everything) – Shashi Jan 9 '18 at 8:03
$$e^{\cos x}\sin(\sin x) = \text{Im}\, e^{\cos x+i\sin x} = \text{Im}\exp\left(e^{ix}\right) = \text{Im}\sum_{n\geq 0}\frac{e^{nix}}{n!}=\sum_{n\geq 1}\frac{\sin(nx)}{n!}$$ and since $\int_{0}^{+\infty}\frac{\sin(nx)}{x}\,dx = \frac{\pi}{2}$ for any $n>0$, we have $$\int_{0}^{+\infty}\frac{e^{\cos x}\sin(\sin x)}{x}\,dx = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\color{red}{\frac{\pi}{2}(e-1)}.$$
• Is it true that interchanging series and integral is allowed since the series with sine is uniformly convergent? The answer is so neat. I guess it is even easier with Real methods after seeing the answers. – Shashi Jan 9 '18 at 7:55
• "Real methods=without Residue Theorem " in the last comment – Shashi Jan 9 '18 at 8:06
• @Shashi: indeed, this is a small variation on the improper Riemann-integrability of $\frac{\sin x}{x}$ on $\mathbb{R}^+$, which can be proved by real (Fourier-analytic), almost-real (Laplace transform) or complex techniques (residue theorem) in few steps, anyway. – Jack D'Aurizio Jan 9 '18 at 14:45 | 2019-04-19T20:16:05 | {
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https://math.stackexchange.com/questions/1289322/what-is-zero-irrational-or-rational-or-it-have-both-the-properties/1289333 | # What is zero? Irrational or rational or it have both the properties? [duplicate]
We say,
A number is rational if it can be represented as $\frac{p}{q}$ with $p,q \in \mathbb Z$ and $q\neq 0$.
Any number which doesn't fulfill the above conditions is irrational.
It can be represented as a ratio of two integers as well as ratio of itself and an irrational number such that zero is not dividend in any case.
People say that $0$ is rational because it is an integer. Which I find to be a lame reason. May be any strong reason is there. Can any one tell me please?
## marked as duplicate by Jack M, Chappers, N. F. Taussig, Dietrich Burde, Joel Reyes NocheMay 19 '15 at 12:21
• $0$ can be represented as $0/1$, therefore by your definition it is rational. – TonyK May 19 '15 at 10:30
• And it can also be shown as $\frac{0}{\sqrt{2}}$. Then what? – Man_Of_Wisdom May 19 '15 at 10:30
• $1=\frac{\sqrt{2}}{\sqrt{2}}$. So? – Hayden May 19 '15 at 10:31
• Or less trivially, $2=\frac {\sqrt 8} {\sqrt 2}$ – Jack M May 19 '15 at 10:40
• You may simply notice that $q=\sqrt(2) \not\in \mathbb{Z}$ – Manlio May 19 '15 at 10:42
The definition of an irrational number is that it is not rational. And $0$ is by definition a rational number.
• Why should I consider it a rational number if it can be represented as $\frac{0}{\sqrt{2}}$ – Man_Of_Wisdom May 19 '15 at 10:27
• @Man_Of_Wisdom: What about $\sqrt2/\sqrt2$ then? Do you consider that irrational? – John Bentin May 19 '15 at 10:33
• Nope. No. Not.. – Man_Of_Wisdom May 19 '15 at 10:45
• @Alwin can you please roll back your edit to the fresh one. Which includes only two lines? I want to accept that because it answers at hand! – Man_Of_Wisdom May 19 '15 at 11:10
• Do you mean this one above? – Alwin May 19 '15 at 11:25
I think you're confusing some of the language in the definition. The definition says that if a number may be written in a certain way, namely as a fraction in which both numerator and denominator are integers, then it's rational. If it cannot be written this way, then it is irrational. There is nothing in the definition that prevents a rational number from being written as a fraction in other ways, such as having rational or irrational numerator or denominator.
The phrase "Any number which doesn't fulfill the above conditions is irrational" does NOT say "Any number which can be written as a fraction $\frac{p}{q}$ with $p,q\notin \mathbb{Z}$ is irrational". It simply says any number that can not be written $\frac{p}{q}$ with $p,q\in \mathbb{Z}$ is irrational.
• One could also use the irrationality critereon to show zero can't be an irrational number. – Rammus May 19 '15 at 11:49
$r$ is rational if you find integers $p,q$ such that $r=\dfrac pq$. This is obviously the case for $r=0$.
The contraposition of this property is "$r$ is irrational if you cannot find integers $p,q$ such that $r=\dfrac pq$".
The contraposition is not at all "$r$ is irrational if you find irrationals $p,q$ such that $r=\dfrac pq$". (By the way, this would be a somewhat circular definition.)
$\newcommand{\Reals}{\mathbf{R}}$You've got excellent explanations of the logical reasons for saying "$0$ is rational". Here are some complementary thoughts too long for a comment:
Definitions in mathematics exist to give convenient labels to useful logical distinctions. "Convenient" is a loose term, but generally refers to simplifying statements of theorems and facilitating common types of discussion.
When two criteria (such as "rational" and "irrational") are logical opposites by definition, it's never a good idea to allow some widespread mathematical concept (such as $0$) to be "both": If you do, every theorem that would apply to that object has to contain a clause explicitly excluding that object. That's inconvenient. In rare cases (see below), you might say "neither". But in the case of $0$, "rational" is the better label:
• Literal application of the definition ("there exist integers $q \neq 0$ and $p$ such that $0 = p/q$") says $0$ is rational.
• The set of rational numbers has pleasant algebraic properties (closed under addition, closed under multiplication) because $0$ is rational. (By contrast, the set of irrational real numbers is not closed under addition, e.g., $(1 - \sqrt{2}) + \sqrt{2} = 1$, or under multiplication, e.g., $\sqrt{2} \cdot \sqrt{2} = 2$, whether or not $0$ is rational.)
To make a case that "$0$ is not rational", i.e., that the definition of "rational number" should exclude $0$, one would want a compelling reason, such as "the statement of a useful theorem becomes awkward if $0$ is (regarded as) rational".
With due respect, the possibility of writing $0$ as $0/\sqrt{2}$ isn't compelling in the above sense; as others have explained, this representation does not contravene the definition. Further, it's useful, and causes no hardship, to agree that $0$ is rational.
For contrast, here are some other "edge cases" that crop up now and again:
• The integer $0$ is "even" ($2$ times some integer) rather than "odd" (leaves a remainder of $1$ on division by $2$). (By the division algorithm, every integer is even or odd, but no integer is both. In this setting, "even" and "not odd" are logically equivalent for integers. I mention this example because a colleague once informed me that some teachers regard $0$ as neither even nor odd.(!!))
• The zero function $z: \Reals \to \Reals$ is both "even" (for all real $x$, $z(-x) = z(x)$) and "odd" (for all real $x$, $z(-x) = -z(x)$). The notions of "even" and "odd" for functions are not logical opposites. Moreover, it is useful to declare the zero function to be both even and odd: The set of even functions is a vector space under "the usual operations"; the set of odd functions is, too. If $z$ were not "both even and odd (as a function)", at least one of these useful theorems would be false.
• The integer $1$ is neither "prime" nor "composite". (Even though "$1$ has no positive integer factors other than $1$ and itself", we explicitly exclude $1$ from membership in the primes because declaring $1$ "prime" would spoil the uniqueness of prime factorization. On the other hand, $1$ is not "composite" because $1$ is not a product of smaller positive primes.)
There's a deeper point that, ironically, may seem at odds with my earlier stance: Mathematical definitions are human conventions, not absolute, immutable, incontestable features of logic, mathematics, or the physical universe. I suspect this raises unnecessary obstacles for the philosophically-minded who study mathematics. (Everything and More by David Foster Wallace is the most extreme example I've encountered; Wallace seemed tormented by the ontology of infinity.)
On the other hand, when one sees how tightly mathematics hangs together across times and cultures, how definitions lead to the same theorems, one is forced (even fully accepting the preceding paragraph) to admire the phenomenal coherence of the logical structure of mathematics. One starts to feel as if definitions are inevitable. One becomes willing to fight emphatically for the correct definitions. This last, I expect, explains the downvotes to your good question.
As I understand it, the definition of an irrational number is that it is not rational. By definition then, $0$ is rational. | 2019-06-20T03:21:26 | {
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https://math.stackexchange.com/questions/458105/constructing-a-non-linear-system-with-prerequisites-about-the-nature-of-its-crit?noredirect=1 | Constructing a non-linear system with prerequisites about the nature of its critical points.
An exercise from the book I am reading is: "Construct a non-linear system that has four critical points:two saddle points, one stable focus, and one unstable focus."
I have tried many systems. I found one quickly but I was lucky even if I had a few clues thanks my previous trials.
I wonder if there is any way to find such systems using a not completely "gropingly way".
Edit: with only two equations in the system.
The system I have is:
$\dot{x}=y^2-x^2$
$\dot{y}=x^2+y^2-2$
• Is the dimension of system = 2 ? – nonlinearism Aug 2 '13 at 17:13
• Yes the dimension is 2 as it was in the chapter preceding this exercise, I should have written that also ^^ Thanks. – Ouistiti Aug 2 '13 at 22:44
• @Ouistiti: Was the answer helpful? What does your system look like? Where is this problem from? – Amzoti Aug 4 '13 at 1:29
• @Amzoti I will post my system above. The answer was very helpful, I learned more from your answer than I expected. Thank you :) The problem is from "Dynamical Systems with Applications using Mathematica®" written by Stephen Lynch. – Ouistiti Aug 5 '13 at 12:22
• @Ouistiti: It makes me very happy to hear that! If you write the four sets of eigenvalues for the Jacobian matrix at each critical point, It may be possible to set up an optimization problem to solve for the six parameters. Look forward to seeing your system and this was a fun problem. Regards – Amzoti Aug 5 '13 at 12:25
We need to choose a form for our system to give us exactly four critical points. This selection is not unique, hence, we can get more than one solution.
Lets choose a system with parameters that give us some degree of freedom. An example of a system that gives us four critical points and some degree of freedom is (there are other choices, this is not unique):
$$\tag 1 x' = a x + b x^2 + c x y = x(a + bx + cy)\\ y' = d y + e y^2 + f x y = y(d+ey + fx)$$
First thing we need to do is to find the critical points (recall, we need exactly four of them).
• $x = 0 \rightarrow y = 0$ or $y = -\dfrac{d}{e}$
• $y = 0 \rightarrow x = 0$ or $x = -\dfrac{a}{b}$
• $x \ne 0, y \ne 0 \rightarrow x = \dfrac{cd -ae}{be - cf}, y = \dfrac{af -bd}{be - cf},~~\text{with}~~ c \ne 0, be \ne cf$
Thus, our four critical points are (note the constraints above):
$$(0,0), ~\left(0, -\dfrac{d}{e}\right), ~\left(-\dfrac{a}{b}, 0\right), \left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$$
The Jacobian matrix of $(1)$ is given by:
$$\tag 2 \displaystyle J(x, y) = \begin{bmatrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}a + 2bx + cy & cx \\ fy & d + 2 e y + fx \end{bmatrix}$$
Now, we have freedom when choosing the parameters with the goal to get two saddle points, one stable focus, and one unstable focus.
There is not a clean way to do this, but our problem now is to choose the parameters $a, b, c, d, e, f$ to make that happen by evaluating the eigenvalues of the Jacobian at each critical point, and choosing the parameters to give us the desired behavior.
I am going to crank these one at a time and keep my fingers crossed for the last critical point.
Point (0,0)
We have:
$$\displaystyle J(0, 0) = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$$
Lets choose $a = -1, d = 1$ and this is our first saddle.
Point (0,-d/e)
We have:
$$\displaystyle J(0, -d/e) = \begin{bmatrix}a - (cd)/e & 0 \\ -(fd)/e & d -(2d)/e \end{bmatrix}$$
Lets choose $e = 1, c = -3$ and this is our second saddle.
Point (-a/b, 0)
We have:
$$\displaystyle J(-a/b, 0) = \begin{bmatrix} - a - (ac)/b & 0 \\ 0 & d -(af)/b \end{bmatrix}$$
Lets choose $b = 1, f = 1$ and this is our unstable focus.
Point $\left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$
Well, this is where we need luck with the parameters we chose, but I think it might be possible to cast this problem as an optimization problem, so you might want to play around with that. This would give you ranges for these six parameters, so no guesswork is needed.
Anyway, from our parameter choices above, we get $x = y = -1/2$.
$$\displaystyle J(-1/2, -1,2) = \begin{bmatrix} -1/2 & 3/2 \\ -1/2 & -1 \end{bmatrix}$$
This, thankfully, gives us our stable focus.
Thus, we have the system:
$$\tag 1 x' = - x + x^2 -3 x y \\ y' = y + y^2 + x y$$
Lets draw the phase portrait and validate this analysis.
We have what we need with two saddles, an unstable and a stable focus at the four critical points.
I am curious how you did it (even if it is guessing) and what your system looks like and it would help to post your solution in your question.
Update
Here is the phase portrait of the system you wrote and it is also a good example.
• Great job. For my knowledge how did you come up with the second order system in (1)? By intuition ? – kaka Aug 3 '13 at 6:14
• @kaka: Yes, by intuition and a bit of handwaving. I wish I could say I had some wonderful approach to that. I realized that we needed something that could give us a point at (0,0) and then realized we needed a parabola that gave points at (0, w), and (f, 0) and one more at some (h, j). Thinking along those lines lead me to that system. Then, we needed freedom, so I added all of the parameters. I still think the approach can be improved using optimization for ranges on the parameters, but I'll let the OP work that. Thanks for the kind words. Regards – Amzoti Aug 3 '13 at 6:19
• @Amzoti: Wowwww. It is complete as always. Full of points, and finally a nice illustrating graph. – mrs Aug 3 '13 at 14:56
• This is a nice problem. Go for it! ;-) – Namaste Aug 4 '13 at 1:14 | 2019-10-24T02:20:48 | {
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http://math.stackexchange.com/questions/321445/what-does-it-mean-for-a-sequence-to-be-decreasing-in-regards-to-the-integral-tes | # What does it mean for a sequence to be decreasing in regards to the Integral Test for Convergence?
Example: $$\sum \limits_{n=1}^{\infty} n^k e^{-n}$$
This is the problem that I am to solve with the Integral test. I know that it converges, and I know the answer, but just don't fully get the point about it being decreasing.
How does one if a function is decreasing, the first derivative is less than zero.
$f^{\prime}\left(x\right) = \dfrac{\left(k-x\right)x^{k-1}}{e^x}$ is less than zero when $x > k$. I do understand that there is always an $x > k$. (Not really sure if what happens when $k = \infty$ leaving that issue till I learn more).
So my confusion comes to when can you say that a sequence is decreasing? What about when $x < k$: does the Integral test fail on that interval? Or just does the long-term behavior is all that matters? I mean what if just the very last number before infinity the sequence decreases, then will the Integral Test for convergence work?
Could you explain in the realm of sequences and series what does it truly mean to be decreasing? I do understand that I am just learning this, but don't be afraid to start at a high school level and build up your answers to a Field Medal level of understanding. :)
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A sequence $a_n$ is decreasing if $a_{n+1} \le a_n$ for all $n$. – Javier Mar 5 '13 at 13:54
@JavierBadia So what about the example I give, if $k = 3$ then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? – yiyi Mar 5 '13 at 13:55
The first terms don't affect convergence. You want to know what happens for $n$ large. – L. F. Mar 5 '13 at 13:57
@L.F. Why don't the first terms don't affect convergance? That is really my question. Thanks for putting it into words. – yiyi Mar 5 '13 at 13:57
Because there are only finitely many first terms, so their sum will always converge. As long as your sequence is defined, it can jump, fly, or hula hoop at the beginning for all that matters. What we really want to know, is if the "infinite portion" (i.e. what happens after a certain point) will sum to a value or not. – L. F. Mar 5 '13 at 14:01
A sequence is of the form $\{x_n\}$, and a series is of the form $\sum \limits_{n=1}^{\infty} x_n$. So for a sequence $\{x_n\}$ to be decreasing means $x_1 > x_2 > x_3 > \ldots$.
As long as there is some $k$ such that $f(n)$ is decreasing for all $n > k$, the integral test applies.
Consider the series $\sum \limits_{n=1}^{\infty} f(n) = \sum \limits_{n=1}^k f(n) + \sum \limits_{n=k + 1}^{\infty} f(n)$. The first term is finite, so converges, and the second term converges due to the integral test because it is always decreasing.
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So what about the example I give, if k=3 then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? – yiyi Mar 5 '13 at 13:56
The integral test applies as long as the function is eventually decreasing. – ferson2020 Mar 5 '13 at 13:59
Thanks for your answer, but it has never been explained why the long term behavior is only which is important? Is it because if it converges then it stops "growing/shrinking" at that number, otherwise it just "grows/decreases" without limit? – yiyi Mar 5 '13 at 14:03
I added to my answer; does that help make it clear? – ferson2020 Mar 5 '13 at 14:04
Say a sequence $\{a_n\}_{n\in\mathbb{N}}$ is decreasing provided $a_{n+1}\leq a_n$ for every $n\in\mathbb{N}$. So the bigger subscripts correspond to smaller numbers. You really just need it decreasing from a rank on, so decreasing for large $n\in\mathbb{N}$.
You are looking at a series. What you want to show is that the associated sequence (that is, given the series $\sum_{n=1}^\infty a_n$ the associated sequence is $\{a_n\}_{n\in\mathbb{N}}$) is nonnegative and decreasing, then you define a function $f:[1,+\infty)\to[0,+\infty)$ so that $f(n)=a_n$ and $\int_1^\infty f<+\infty$.
Also, $k$ is an exponent so it can't be $\infty$.
Also, for series we only care what happens "near $\infty$", so to speak. So say if your sequence doesn't start decreasing until $n=100$ then just apply the integral test to the series $\sum_{n=100}^\infty a_n$, in which case you look at $\int_{100}^\infty f$.
Rationale for "near $\infty$": First of all, usually by something happening "near $\infty$ we mean there exists $N\in\mathbb{N}$ so that for all $x\geq N$ the condition happens. Convergence of a series (or a sequence for that matter, and keep in mind a series can be seen as a sequence of partial sums) only has to do with the "tail," or what happens near $\infty$. This is because for any natural number $N\in\mathbb{N}$ the first part of the series $\sum_{n=1}^Na_n$ is always finite. No matter what $N$ you choose. Because of this, if $\sum_{n=1}^\infty$ is going to diverge the problem must occur near $\infty$.
So if your associated sequence doesn't start decreasing until after $n=10,000,000,000$, no problem. Just write $\sum_{n=1}^\infty a_n=\sum_{n=1}^{10,000,000,000}a_n+\sum_{n=10,000,000,000}^\infty a_n$ and apply the integral test to the tail.
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Nice and informative answer, could you point me to some papers which explain the rational why only "near $\infity$" matters? – yiyi Mar 5 '13 at 14:01 | 2015-08-28T15:14:31 | {
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http://mathhelpforum.com/pre-calculus/145824-two-variable-limit.html | 1. two variable limit
Hi! I am new and I can't solve this limit:
limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)
Can anyone give me a hand?
2. Originally Posted by venturozzaccio
Hi! I am new and I can't solve this limit:
limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)
Can anyone give me a hand?
Try converting to polars:
$\lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$
$= \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$
$= \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$
$= \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$
$= \frac{0}{|\sin^4{\theta}|}$
$= 0$.
3. sorry I make a mistake writing.
the real one is :
limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^2|+|y^4|)
4. Originally Posted by Prove It
Try converting to polars:
$\lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$
$= \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$
$= \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$
$= \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$
$= \frac{0}{|\sin^4{\theta}|}$
$= 0$.
This is really nice
5. yes but about the revised question?
6. Use the exact same process. You will still be able to cancel enough $r$'s to be able to evaluate the limit.
7. so the result is still 0 ?
8. Originally Posted by venturozzaccio
so the result is still 0 ?
Correct.
9. ok. but if the limit is
limit x,y->(0,0) $xy^2 / (x^2+y^4)$
then I have $0cos()sen^2() /( cos^2+0^2cos^4())$
And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?
10. Originally Posted by venturozzaccio
ok. but if the limit is
limit x,y->(0,0) $xy^2 / (x^2+y^4)$
then I have $0cos()sen^2() /( cos^2+0^2cos^4())$
And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?
I get that the limit is $0$...
$\lim_{(x, y) \to (0, 0)}\frac{xy^2}{x^2 + y^4} = \lim_{r \to 0}\frac{r\cos{\theta}\,r^2\sin^2{\theta}}{r^2\cos^ 2{\theta} + r^4\sin^4{\theta}}$
$= \lim_{r \to 0}\frac{r^3\cos{\theta}\sin^2{\theta}}{r^2(\cos^2{ \theta} + r^2\sin^4{\theta})}$
$= \lim_{r \to 0}\frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta } + r^2\sin^4{\theta}}$
$= \frac{0}{\cos^2{\theta}}$
$= 0$.
11. I think it's not true. Just because if you set x=y the limit is 0. But if you set x=y^2 then the limit is 1/2. So limit change according the way you approach to (0,0) and for this it doesn't exist. | 2017-08-19T04:35:43 | {
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https://math.stackexchange.com/questions/1811495/what-is-the-difference-between-and-projection-and-a-reflection-in-vector-transf | # What is the difference between and projection and a reflection, in vector transformation?
In my text book I have the problems of finding the standard matrix of the given linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$;
1. Projection onto the line $y = -x$.
2. Reflection in the line $y = -x$.
Both of these give different answers. I was hoping for an explanation of the difference between a projection/reflection in a vector transformation.
• Aprojection $p$ satisfies the equality $p\circ p=p$, while a reflection $r$ staisfies $r\circ r=\operatorname{id}$. This is quite different. – Bernard Jun 3 '16 at 19:36
The (orthogonal) projection onto a line "compresses" every point in the plane onto the line. If you drop the perpendicular from the point to the line, the image of the point after projection is the intersection of the perpendicular with the line you are projecting onto.
The reflection across a line moves a point to its "mirror image" across the line. If you drop a the perpendicular from a point onto the line of reflection, then the mirror image is going to lie at the same distance from the line of reflection on this perpendicular, but it will be on the other side of the line of reflection from the original.
Of course, in both cases, a point already on the line of reflection or projection is going to stay where it started.
A huge difference between these two transformations is that reflections are always invertible (since $R^2=Id$), but projections are almost never invertible. An orthogonal projection of the plane onto a line is never invertible since every point on a perpendicular to the line of projection maps to the same point on the line you are projecting onto.
In terms of eigenvalues, the projection in this case would have eigenvalues $\{0,1\}$ whereas the reflection would have eigenvalues $\{-1,1\}$.
The answer of @rschwieb gives you the theoretical vision of the problem. I add a figure to support it in a geometrical intuitive way.
The blue line is your $y=-x$. Given the points $P$ or $Q$, than $P'$ and $Q'$ are the orthogonal projection on this line and $P''$ and $Q''$ are the reflections. As you can see the prjections of the two points coincide, so the orthogonal projection is not invertible.
If you take $P=(1,0)^T$ (a standard basis vector) you can easely see that $P'=(\frac{1}{2},-\frac{1}{2})^T$ and $P''=(0,-1)^T$, and if you do the same for the other basis vector $N=(0,1)^T$, you find $N'=((-\frac{1}{2},\frac{1}{2})^T$ and $N''=(-1,0)^T$, so you have the matrices that represents the two transformations (in the standard basis):
$$\frac{1}{2} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix}$$ for the projection, and $$\begin{bmatrix} 0&-1\\-1&0 \end{bmatrix}$$
for the reflecion.
• I must beg to differ: my answer is almost purely geometric intuition. But what you've written is certainly a concrete and explicit analytical answer, which is good too :) – rschwieb Jun 4 '16 at 0:56 | 2019-10-23T17:45:46 | {
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https://math.stackexchange.com/questions/1618330/stopping-criteria-for-gradient-method/1618347 | # Stopping criteria for gradient method
For numerically solving a smooth convex optimization $\min\{f(x): x\in S\}$ where $S$ is a closed convex set, we can apply some different algorithms: gradient method, accelerated gradient, proximal gradient ... depending on the structure of the problem. Solving is to find a solution $x*$ such that $f(x^*)=\inf\{f(x): x\in S\}:=f^*$. To this end, we try to construct an iterative sequence $\{x^k\}$ that converges to some solution $x^*$, or the sequence of numbers $\{f(x^k)\}$ tends to $f^*$. Note that, if $x^k\to x^*$ then the continuity of $f$ can ensures that $f(x^k) \to f^*$.
My questions are:
1. In which cases we should focus on the convergence of $\{f(x^k)\}$ rather than of $\{x^k\}$? Is finding a point $x^K$ such that $x^K$ close enough to a solution $x^*$ better than finding a point $x^K$ such that $f(x^K)$ close enough to $f^*$?
2. What is the best stopping criteria for an algorithm? I know the following ways:
• Determine the number of iterations we need to perform to achieve a desired error $\epsilon$, i.e., $||x^k-x^*||<\epsilon$ or $|f(x^k)-f^*|<\epsilon$ implies $k\geq N$ for some $N$. I see that this way is very reliable.
• terminating when $||x^{k+1}-x^k||$ or $|f(x^{k+1})-f(x^k)|$ is small enough.
• terminating when $||\nabla f(x^k)||$ is small enough.
Could you explain how the second and the third cases work? Why $||\nabla f(x^k)||$ small enough can implies that $f(x^k)$ is approximate the optimal value $f^*$. I have been know that the case $f$ is strongly convex this can be verified. Is this stopping criteria still reliable in the case where $f$ is not strongly convex?
I will discuss the termination criteria for the simple gradient method $x_{k+1} = x_{k} - \frac{1}{L}\nabla f(x_k)$ for unconstrained minimisation problems. If there are constraints, then we would use the projected gradient method, but similar termination condition hold (imposed on the norm of the difference $x_k-z_k$).
The third criterion, namely $\|\nabla f(x_k) \| < \epsilon$ if fine for strongly convex functions with $L$-Lipschitz gradient. Indeed, if $f$ is $\mu$-strongly convex, that is
\begin{aligned} f(y) \geq f(x) + \nabla f(x)^\top (y-x) + \tfrac{\mu}{2} \|y-x\|^2 \end{aligned},\tag{1}
then, for $x^*$ such that $\nabla f(x^*)=0$ (the unique minimiser of $f$), we have
\begin{aligned} f(x) - f(x^*)\leq \tfrac{1}{2\mu}\|\nabla f(x) \|^2, \end{aligned}\tag{2}
so, if $\|\nabla f(x) \|^2 < 2\mu\epsilon$, then $f(x) - f(x^*) < \epsilon$, i.e., $x$ is $\epsilon$-suboptimal.
But termination is a mysterious thing... In general (under the assumptions you drew) it is not true that we will have $\|x-x^*\|<\epsilon$ if $\| \nabla f(x) \| < \kappa \epsilon$, for some $\kappa > 0$ (not even locally). There might be specific cases where such a bound holds, notwithstanding. Unless you draw some additional assumptions on $f$, this will not be a reliable termination criterion.
However, strong convexity is often too strong a requirement in practice. Weaker conditions are discussed in the article: D. Drusvyatskiy and A.S. Lewis, Error bounds, quadratic growth, and linear convergence of proximal methods, 2016.
Let $f$ be convex with $L$-Lipschithz gradient and define $\mathcal{B}_\nu^f = \{x: f(x) - f^* < \nu\}$. Let us assume that $f$ has a unique minimiser $x^*$ (e.g., $f$ is strictly convex). Then assume that $f$ has the property
\begin{aligned} f(x) - f(x^*) \geq \tfrac{\alpha}{2} \|x-x^*\|^2, \end{aligned}\tag{3}\label{3}
for all $x\in\mathcal{B}_\nu^f$ for some $\nu>0$. Functions which satisfy this property are not necessarily strongly convex. As a counterexample we have $f = (\max\{|x|-1,0\})^2$. Of course if $f$ is strongly convex the above holds and if $f$ is given in the form $f(x) = h(Ax)$ where $h$ is a strongly convex function and $A$ is any matrix.
Then, condition \eqref{3} is shown to be equivalent to
\begin{aligned} \|x-x^*\| \leq \frac{2}{\alpha} \|\nabla f(x) \|, \end{aligned}\tag{4}\label{4}
for all $x\in\mathcal{B}_{\nu}^f$ and with $\alpha < 1/L$.
Clearly in this case we may use the termination condition $\| \nabla f(x) \| < \epsilon\alpha/2$ which will imply that $\|x-x^*\| < \epsilon$.
In regard to the second condition, you may use it again for strongly convex functions or if \eqref{3} holds locally about $x^*$. The reason for that is that the following bound holds for the gradient method:
\begin{aligned} \tfrac{L}{2}\|\nabla f(x_k) \|^2 \leq f(x_k) - f(x_{k+1}). \end{aligned}\tag{5}\label{5}
The right hand side of \eqref{5} is further upper bounded by $L_f \|x_k - x_{k+1}\|$, where $L_f$ is the Lipschitz constant of $f$ (we know that $f$ is Lipshcitz continuous), so a condition on $\|x_{k+1}-x_{k}\|$ may potentially be used, but we may see that the basis for all this is the bound on $\|\nabla f(x_k) \|$.
If $$f$$ is strictly convex, it has at most one minimum and at this minimum its gradient is zero. So the third criteria should work fine. If $$f$$ is not convex, you may reach a local minimum so this criterion is not really justified. If you are using a gradient method, the second criteria is very similar to the third because each step (or the difference $$x^{k+1}-x^{k}$$) is obtained from the gradient.
• I know the fact that "If $f$ is convex, it has just one minimum on $R^n$ and at this minimum its gradient is zero. But can you explain why $||\nabla f(x^k)||$ small enough can implies that $f(x^k)$ is approximate the optimal value $f^*$? In addition, in the constrained case, the gradient does not need to be zero at the minimum point. – Richkent Jan 19 '16 at 16:27
• Well, if $\|\nabla f\|$ is small enough (say below $\epsilon$), it can be approximated by zero, meaning that you are very close to the optimal. If $f$ is convex, the gradient is monotonous and continuous, so if it's close to zero, you are close to the minimum. – citronrose Jan 19 '16 at 16:34
• draw a $C^1$ (strictly) convex function $\mathbb{R} \to \mathbb{R}$ you'll see that if the minimum is at $a \ne -\infty$ then $|f'(x)| < \epsilon$ only when $x \in ]a-\delta;a+\delta[$, and the smaller is $\epsilon$ the smaller will be $\delta$ . after that, just remember that convex on $\mathbb{R}^n$ means that you can minimize with respect to each $x_i$ one after the other. – reuns Jan 19 '16 at 16:44
• "If $f$ is convex, it has just one minimum" is completely wrong. Some convex functions have no minimizers (e.g. $f(x) = \exp(x)$), others have a continuum of minimizers (e.g. $f(x) = 0$). – Guillaume Garrigos Mar 1 at 21:02
• @citronrose $\exp$ doesn't necessarily have a minimum on a closed domain, take for instance $S = \mathbb{R}$ or $S = [a,+\infty[$ for any $a \in \mathbb{R}$. Of course there is no such counterexamples if $S$ is compact, which is maybe the case you had in mind. – Guillaume Garrigos Mar 5 at 15:53 | 2019-09-19T15:42:52 | {
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This might sound stupid, but how do we know that we need to round up 64 to 65 without losing to much time. I did this question before and it took over 2 min before i knew that i had to round 64 to 65 in order to solve the problem. Do you have any trick in mind?
65 is the closest multiple of 13: 5*13=65.
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A store reported total sales of $385 million for February of this year. If the total sales for the same month last year was$320 million, approximately what was the percent increase in sales?
(A) 2%
(B) 17%
(C) 20%
(D) 65%
(E) 83%
Last year's sales = $320 million; This year's sales =$385 million;
Increase = $65 million. Now, 20% of$320 million is $64 million, which is very close to actual increase of$65 million.
OR:
General formula for percent increase or decrease, or percent change:
$$Percent=\frac{Change}{Original}*100$$ --> $$Percent=\frac{65}{320}*100=\frac{13}{64}*100\approx{\frac{13}{65}}*100=20%$$.
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11 Sep 2014, 06:09
A store reported total sales of $385 million for February of this year. If the total sales for the same month last year was$320 million, approximately what was the percent increase in sales?
(A) 2%
(B) 17%
(C) 20%
(D) 65%
(E) 83%
last year= 320
this year = 385
((385-320)/320)*100 = (65/320 )*100 approx.. 20
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Re: A store reported total sales of $385 million for February of [#permalink] Show Tags 07 Jun 2016, 10:09 Walkabout wrote: A store reported total sales of$385 million for February of this year. If the total sales for the same month last year was $320 million, approximately what was the percent increase in sales? (A) 2% (B) 17% (C) 20% (D) 65% (E) 83% The problem is testing us on using the percent change formula: (New Value – Old Value)/Old Value x 100 We are given: February sales this year = 385 million February sales last year = 320 million We need to determine the percent increase between sales from last year to sales this year. Thus, the new value = 385 million and the old value = 320 million. Let’s plug them into our percent change formula. (New Value – Old Value)/Old Value x 100 [(385 – 320)/320] x 100 65/320 x 100 13/64 x 100 ≈ 13/65 x 100 ≈ 1/5 x 100 ≈ 20%. The answer is C. _________________ Jeffrey Miller Jeffrey Miller Head of GMAT Instruction Re: A store reported total sales of$385 million for February of [#permalink] 07 Jun 2016, 10:09
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Display posts from previous: Sort by | 2017-01-18T09:42:40 | {
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https://math.stackexchange.com/questions/1629665/how-many-numbers-are-there-of-2n-digits-that-the-sum-of-the-digits-in-the-first | # How many numbers are there of 2n digits that the sum of the digits in the first half equals the sum of the digits in the second half
The question is how many number of a given number of digits 2n where the sum of the first half of the digits equals the sum of the digits in the second half.
So this is for a programming problem and I've got it down to approximately $10^n$ operations. However for a sample size of $0 < n < 500$ this is far too many operations and the question leads me to believe that there is a simple formula for this.
From brute force calculation:
n = 1
10
n = 2
670
n = 3
4816030 ...
So I've abstracted it to: finding how many ways the digits from 0 - 9 can be put together to form a given sum (from 0 -> 9*n) this gives me a $10^n$ however this is still too large. (fyi you need to square this number)
I've observed that in this abstracted version of the question the subsums are constant for n = 1, increase for n = 2, fibonacci numbers for n = 3 with the general rule being if you keep taking the difference of the differences of the sums that the previous one forms the next one (this changes slightly towards the center: the difference of differences doubles). And this trend seems to hold true for all the numbers that I tested.
I think that it might be related to $^nC_r$ or similar but I don't have the maths to change it from something like pascal's triangle to something that I can work with.
• Question: Do you count, say "0990" as an example when $n=2$? It's gonna be much harder if you don't. – Thomas Andrews Jan 27 '16 at 20:56
• @ThomasAndrews Given he says the answer is $10$ when $n=1$ means he must be including $00$ as a two digit number. – Gregory Grant Jan 27 '16 at 21:04
• @ThomasAndrews yes it is counted – Cjen1 Jan 27 '16 at 21:06
• The value you have for $n=3$ is actually the value for $n=4$. The value for $n=3$ is $55252$. – Thomas Andrews Jan 28 '16 at 0:05
Let $f(n,s)$ denote the number of $n$-digit sequences with digit sum $s$. To cover the leading zero problem, let $g(n,s)$ denote the number of $n$-digit sequences with digit sum $s$ and leading digit non-zero. Then $f(n+1,s)=\sum_{d=0}^9f(n,s-d)$, with $f(n,s)=0$ for $s<0$ (or $s>9n$) understood, which allows a quick recursive calculation in $O(n^2)$ time and $O(n)$ space complexity. Furthermore, we simply have $g(n,s)=f(n,s)-f(n-1,s)$. Now the count you really want is $$\sum_{s=0}^\infty g(n,s)f(n,s).$$ But of course we need not consider infinitely many $s$. Instead, we need only sum $$\sum_{s=1}^{9n} g(n,s)f(n,s).$$
Edit: From the comments it seems that leading zeroes are allowed. In that case, simply replace $g$ by $f$ in the above.
You can do it in order $n^2$ operations. If $n=100$, the sum of $100$ digits can range from $0$ to $900$ You need to compute how many ways there are to add up to each of those sums. For $n=1$,you have one way to make each number $0$ through $9$. For $n=2$ you can compute the number of ways to make $k$ by summing over the number of ways to make each of $k-9, k-8, \dots k$ out of one digit. Then for $n=3$ you can compute the number of ways to make $k$ by summing over the number of ways to make $k$ by summing over the number of ways to make each of $k-9, k-8, \dots k$ out of two digits. You will end up with an array with numbers in entries $0$ through $900$. Each of $0$ and $900$ will be $1$. $1$ and $899$ will be $100$. Can you see why? The ones in the middle will be quite large-you need arbitrary precision integers to solve this. Then to get the number of $200$ digit numbers, you sum the squares of the entries. This is because for a number where both the first half and last half sum to $899$ you have $100$ choices for the first half and $100$ choices for the last half, giving $100^2$ total possibilities.
I looked this up at the OEIS where I found this OEIS entry.
We clearly have by inspection that the desired answer is
$$[z^0] (1+z+z^2+\cdots+z^9)^n (1+1/z+1/z^2+\cdots+1/z^9)^n \\= [z^0]\frac{1}{z^{9n}} (1+z+z^2+\cdots+z^9)^{2n}$$
or
$$[z^{9n}] (1+z+z^2+\cdots+z^9)^{2n} = [z^{9n}] \left(\frac{1-z^{10}}{1-z}\right)^{2n}.$$
Extracting coefficients from this we get
$$[z^{9n}] \frac{1}{(1-z)^{2n}} \sum_{q=0}^{2n} {2n\choose q} (-1)^q z^{10q} \\ [z^{9n}] \frac{1}{(1-z)^{2n}} \sum_{q=0}^{\lfloor 9n/10\rfloor} {2n\choose q} (-1)^q z^{10q} \\ = \sum_{q=0}^{\lfloor 9n/10\rfloor} {2n\choose q} (-1)^q {9n-10q+2n-1\choose 2n-1} \\ = \sum_{q=0}^{\lfloor 9n/10\rfloor} {2n\choose q} (-1)^q {11n-10q-1\choose 2n-1}.$$
Note however that the second binomial coefficient is zero when $11n-10q-1\lt 2n-1$ or $9n\lt 10q$ so we may set the upper limit to $n-1$ if desired, producing a match to the OEIS entry e.g.
$$\sum_{q=0}^{n-1} {2n\choose q} (-1)^q {11n-10q-1\choose 2n-1}.$$
The second binomial coefficient starts producing non-zero values again when $11n-10q-1 \lt 0.$
An alternate approach uses generating functions. Let $f(x) = (1+x+x^2+\cdot+x^9)^{n}$. Then we are seeking the constant term of $f(x)f(x^{-1})$.
Letting $x=e^{i\theta}$, we get:
$$f(x)f(x^{-1})=\left(\frac{1-\cos(10\theta)}{1-\cos\theta}\right)^n= \left(\frac{\sin(5x)}{\sin(x/2)}\right)^{2n}$$
The constant term can be computed as:
$$\frac{1}{2\pi}\int_{0}^{2\pi} \left(\frac{\sin(5x)}{\sin(x/2)}\right)^{2n}\,dx$$
While that might seem insane, you can actually use the graph of $\frac{\sin(5x)}{\sin(x/2)}$ to get upper and lower bounds for this integral.
• I believe you mean the Central Limit Theorem, not the Law of Large Numbers, but this is the first thing that came to mind for me as well. – Michael Lugo Jan 27 '16 at 21:44
• Yep, that is what I mean, thanks. @MichaelLugo One reason I didn't proceed further is that I don't actually know these theorems, only know of them. :) – Thomas Andrews Jan 27 '16 at 21:45 | 2019-10-15T15:58:02 | {
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https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1/1490918 | # Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
After reading this question, the most popular answer use the identity $$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}.$$
What's the name of this identity? Is it the identity of the Pascal's triangle modified.
How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?
Thanks for your help.
EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.
• It is sometimes called the "hockey stick". – user940 Oct 21 '15 at 15:24
• There is another cute graphical illustration on the plane of $\binom{n}{k}$ – Eli Korvigo Oct 21 '15 at 16:54
• It's pretty straightforward from the picture. Just switch the $1$ at the top of the stick with the $1$ directly below, then repeatedly replace adjacent numbers with the number in the cell below. This can be translated into a formal proof with words and symbols, but an animation or series of pictures is much more effective. – user2357112 Oct 22 '15 at 3:24
• See also this question. Some post which are linked there might be of interest, too. – Martin Sleziak Jan 18 '16 at 15:05
This is purely algebraic. First of all, since $$\dbinom{t}{k} =0$$ when $$k>t$$ we can rewrite the identity in question as $$\binom{n+1}{k+1} = \sum_{t=0}^{n} \binom{t}{k}=\sum_{t=k}^{n} \binom{t}{k}$$
Recall that (by the Pascal's Triangle), $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$
Hence $$\binom{t+1}{k+1} = \binom{t}{k} + \binom{t}{k+1} \implies \binom{t}{k} = \binom{t+1}{k+1} - \binom{t}{k+1}$$
Let's get this summed by $$t$$: $$\sum_{t=k}^{n} \binom{t}{k} = \sum_{t=k}^{n} \binom{t+1}{k+1} - \sum_{t=k}^{n} \binom{t}{k+1}$$
Let's factor out the last member of the first sum and the first member of the second sum: $$\sum _{t=k}^{n} \binom{t}{k} =\left( \sum_{t=k}^{n-1} \binom{t+1}{k+1} + \binom{n+1}{k+1} \right) -\left( \sum_{t=k+1}^{n} \binom{t}{k+1} + \binom{k}{k+1} \right)$$
Obviously $$\dbinom{k}{k+1} = 0$$, hence we get $$\sum _{t=k}^{n} \binom{t}{k} =\binom{n+1}{k+1} +\sum_{t=k}^{n-1} \binom{t+1}{k+1} -\sum_{t=k+1}^{n} \binom{t}{k+1}$$
Let's introduce $$t'=t-1$$, then if $$t=k+1 \dots n, t'=k \dots n-1$$, hence $$\sum_{t=k}^{n} \binom{t}{k} = \binom{n+1}{k+1} +\sum_{t=k}^{n-1} \binom{t+1}{k+1} -\sum_{t'=k}^{n-1} \binom{t'+1}{k+1}$$
The latter two arguments eliminate each other and you get the desired formulation $$\binom{n+1}{k+1} = \sum_{t=k}^{n} \binom{t}{k} = \sum_{t=0}^{n} \binom{t}{k}$$
• Beautiful proof. p.-s. you can use the LaTeX command \binom{n}{k} to display $\binom{n}{k}$. – hlapointe Oct 21 '15 at 16:26
• @hlapointe thank you. Sure, I forgot there was a special command for binomial. – Eli Korvigo Oct 21 '15 at 16:32
Imagine the first $$n + 1$$ numbers, written in order on a piece of paper. The right hand side asks in how many ways you can pick $$k+1$$ of them. In how many ways can you do this?
You first pick a highest number, which you circle. Call it $$s$$. Next, you still have to pick $$k$$ numbers, each less than $$s$$, and there are $$\binom{s - 1}{k}$$ ways to do this.
Since $$s$$ is ranging from $$1$$ to $$n+1$$, $$t:= s-1$$ is ranging from $$0$$ to $$n$$ as desired.
We can use the well known identity $$1+x+\dots+x^n = \frac{x^{n+1}-1}{x-1}.$$ After substitution $x=1+t$ this becomes $$1+(1+t)+\dots+(1+t)^n=\frac{(1+t)^{n+1}-1}t.$$ Both sides of these equations are polynomials in $t$. (Notice that the RHS simplifies to $\sum_{j=1}^{n+1}\binom {n+1}j t^{j-1}$.)
If we compare coefficient of $t^{k}$ on the LHS and the RHS we see that $$\binom 0k + \binom 1k + \dots + \binom nk = \binom{n+1}{k+1}.$$
This proof is basically the same as the proof using generating functions, which was posted in other answers. However, I think it is phrased a bit differently. (And if it is formulated this way, even somebody who has never heard of generating functions can follow the proof.)
\begin{align} \sum_{t=\color{blue}0}^n \binom{t}{k} =\sum_{t=\color{blue}k}^n\binom tk&= \sum_{t=k}^n\left[ \binom {t+1}{k+1}-\binom {t}{k+1}\right]\\ &=\sum_{t=\color{orange}k}^\color{orange}n\binom {\color{orange}{t+1}}{k+1}-\sum_{t=k}^n\binom t{k+1}\\ &=\sum_{t=\color{orange}{k+1}}^{\color{orange}{n+1}}\binom {\color{orange}{t}}{k+1}-\sum_{t=k}^n\binom t{k+1}\\ &=\binom{n+1}{k+1}-\underbrace{\binom k{k+1}}_0&&\text{by telescoping}\\ &=\binom{n+1}{k+1}\quad\blacksquare\\ \end{align}
You can use induction on $n$, observing that
$$\sum_{t=0}^{n+1} \binom{t}{k} = \sum_{t=0}^{n} \binom{t}{k} + \binom{n+1}{k} = \binom{n+1}{k+1} + \binom{n+1}{k} = \binom{n+2}{k+1}$$
• How can you say that $\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}$ in your proof. – hlapointe Oct 21 '15 at 15:13
• That's the inductive hypothesis. – Michael Biro Oct 21 '15 at 15:14
• Ok. Can we prove it algebraically? – hlapointe Oct 21 '15 at 15:15
• What's the first step!? Because if I take $n=1$, the hypothesis seem to be incorrect. – hlapointe Oct 21 '15 at 15:21
• @hlapointe One choice of base case for every fixed $k$ is that $\sum_{t=0}^{k} \binom{t}{k} = \binom{k}{k} = 1 = \binom{k+1}{k+1}$. – Michael Biro Oct 21 '15 at 16:28
The RHS is the number of $k+1$ subsets of $\{1,2,...,n+1\}$. Group them according to the largest element in the subset. Sum up all the cases. Get the LHS.
Another technique is to use snake oil. Call your sum:
\begin{align} S_k &= \sum_{0 \le t \le n} \binom{t}{k} \end{align}
Define the generating function:
\begin{align} S(z) &= \sum_{k \ge 0} S_k z^k \\ &= \sum_{k \ge 0} z^k \sum_{0 \le t \le n} \binom{t}{k} \\ &= \sum_{0 \le t \le n} \sum_{k \ge 0} \binom{t}{k} z^k \\ &= \sum_{0 \le t \le n} (1 + z)^t \\ &= \frac{(1 + z)^{n + 1} - 1}{(1 + z) - 1} \\ &= z^{-1} \left( (1 + z)^{n + 1} - 1 \right) \end{align}
So we are interested in the coefficient of $z^k$ of this:
\begin{align} [z^k] z^{-1} \left( (1 + z)^{n + 1} - 1 \right) &= [z^{k + 1}] \left( (1 + z)^{n + 1} - 1 \right) \\ &= \binom{n + 1}{k + 1} \end{align}
We can use the integral representation of the binomial coefficient $$\dbinom{t}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{t}}{z^{k+1}}dz\tag{1}$$ and get $$\sum_{t=0}^{n}\dbinom{t}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\sum_{k=0}^{n}\left(1+z\right)^{t}}{z^{k+1}}dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(z+1\right)^{n+1}}{z^{k+2}}dz-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{1}{z^{k+2}}dz$$ and so usign again $(1)$ we have $$\sum_{t=0}^{n}\dbinom{t}{k}=\dbinom{n+1}{k+1}-0=\color{red}{\dbinom{n+1}{k+1}.}$$
• It is so nice and weird. +1 – Behrouz Maleki Jul 5 '16 at 10:27
• +1. Nice work. You must subtract $\displaystyle{\delta_{k,-1}}$ in order to take account of the case $\displaystyle{k = -1}$. When $\displaystyle{k = -1}$, the LHS is equal to $\displaystyle{0}$ and your RHS is equal to $\displaystyle{1}$. With the $\displaystyle{\delta_{k,-1}}$ you'll get $\displaystyle{1 - 1 = 0}$. – Felix Marin Jul 6 '16 at 21:50
In this answer, I prove the identity $$\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}\tag{1}$$ Here is a generalization of the identity in question, proven using the Vandermonde Identity \begin{align} \sum_{m=0}^M\binom{m+k}{k}\binom{M-m}{n} &=\sum_{m=0}^M\binom{m+k}{m}\binom{M-m}{M-m-n}\tag{2}\\ &=\sum_{m=0}^M(-1)^m\binom{-k-1}{m}(-1)^{M-m-n}\binom{-n-1}{M-m-n}\tag{3}\\ &=(-1)^{M-n}\sum_{m=0}^M\binom{-k-1}{m}\binom{-n-1}{M-m-n}\tag{4}\\ &=(-1)^{M-n}\binom{-k-n-2}{M-n}\tag{5}\\ &=\binom{M+k+1}{M-n}\tag{6}\\ &=\binom{M+k+1}{n+k+1}\tag{7} \end{align} Explanation:
$(2)$: $\binom{n}{k}=\binom{n}{n-k}$
$(3)$: apply $(1)$ to each binomial coefficient
$(4)$: combine the powers of $-1$ which can then be pulled out front
$(5)$: apply Vandermonde
$(6)$: apply $(1)$
$(7)$: $\binom{n}{k}=\binom{n}{n-k}$
To get the identity in the question, set $n=0$.
• @FoF: I have added a link here and answered your other question. Thanks for mentioning the difficulty. – robjohn Dec 7 '13 at 12:33
• @FoF: That is the Vandermonde Identity that I mentioned at the beginning. – robjohn Dec 8 '13 at 18:56
• @FoF: I added an explanation for each line. – robjohn Dec 9 '13 at 2:20
• I answered my own question about $(5, 6$) here. – NaN Dec 10 '13 at 8:54
• @FoF: Ah. That is why I added the Explanation when I saw difficulty in following the argument. – robjohn Dec 11 '13 at 7:46
You remember that: $$(1+x)^m = \sum_k \binom{m}{k} x^k$$ So the sum $$\sum_{m=0}^M \binom{m+k}{k}$$ is the coefficient of $x^k$ in: $$\sum_{m=0}^M (1+x)^{m+k}$$ Yes? So now use the geometric series formula given: $$\sum_{m=0}^M (1+x)^{m+k} = -\frac{(1+x)^k}{x} \left( 1 - (1+x)^{M+1} \right)$$ And now you want to know what is coefficient of $x^k$ in there. You got it from here.
Recall that for $k\in\Bbb N$ we have the generating function
$$\sum_{n\ge 0}\binom{n+k}kx^n=\frac1{(1-x)^{k+1}}\;.$$
The identity in the question can therefore be rewritten as
$$\left(\sum_{n\ge 0}\binom{n+k}kx^n\right)\left(\sum_{n\ge 0}x^n\right)=\sum_{n\ge 0}\binom{n+k+1}{k+1}x^n\;.$$
The coefficient of $x^n$ in the product on the left is
$$\sum_{i=0}^n\binom{i+k}k\cdot1=\sum_{i=0}^n\binom{i+k}k\;,$$
and the $n$-th term of the discrete convolution of the sequences $\left\langle\binom{n+k}k:n\in\Bbb N\right\rangle$ and $\langle 1,1,1,\dots\rangle$. And at this point you’re practically done.
• Is there a typo in the second equation (first sum)? I believe $k$ should be indexed. – AlanH May 27 '13 at 6:20
• @Alan: No, the sum is over $n$; $k$ is fixed throughout. – Brian M. Scott May 27 '13 at 7:19
• In my text, I have an identity $\sum_{r\geq 0} \binom{r + n}{r} x^r = 1/(1-x)^{n+1}$ This may be the cause of my confusion, but is this identity correct and is it equivalent to the one you used? – AlanH May 27 '13 at 8:22
• @Alan: Sure: your $r$ is my $n$, and your $n$ is my $k$. – Brian M. Scott May 27 '13 at 8:28
• @Alan: $\binom{r+n}r=\binom{r+n}n$; now do the translation. (Sorry: I didn’t notice before that you’d used the symmetrically opposite binomial coefficient.) – Brian M. Scott May 27 '13 at 19:19
A standard technique to prove such identities $\sum_{i=0}^Mf(i)=F(M)$, involving on one hand a sum where only the upper bound $M$ is variable and on the other hand an explicit expression in terms of$~M$, is to use induction on$~M$. It amounts to showing that $f(M)=F(M)-F(M-1)$ (and that $F(0)=f(0)$). This is similar to using the fundamental theorem of calculus in showing that $\int_0^{x_0}f(x)\mathrm dx=F(x_0)$ by establishing $f(x)=F'(x)$ (and $F(0)=0$).
So here you need to check (apart from the obvious starting case $M=0$) that $\binom{M+k}k=\binom{M+k+1}{k+1}-\binom{M+k}{k+1}$. This is just in instance of Pascal's recurrence for binomial coefficients.
$$\mbox{Note that}\quad \sum_{m = 0}^{M}{m + k \choose k} = \sum_{m = k}^{M + k}{m \choose k} = a_{M + k} - a_{k - 1}\quad\mbox{where}\quad a_{n} \equiv \sum_{m = 0}^{n}{m \choose k}\tag{1}$$
Then, \begin{align} \color{#f00}{a_{n}} & \equiv \sum_{m = 0}^{n}{m \choose k} = \sum_{m = 0}^{n}\ \overbrace{% \oint_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}} ^{\ds{m \choose k}}\ =\ \oint_{\verts{z} = 1}{1 \over z^{k + 1}}\sum_{m = 0}^{n}\pars{1 + z}^{m} \,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{1 \over z^{k + 1}}\, {\pars{1 + z}^{n + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}\ =\ \underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 1} \over z^{k + 2}} \,{\dd z \over 2\pi\ic}}_{\ds{n + 1 \choose k + 1}}\ -\ \underbrace{\oint_{\verts{z} = 1}{1 \over z^{k + 2}}\,{\dd z \over 2\pi\ic}} _{\ds{\delta_{k + 2,1}}} \\[8mm] \imp\ \color{#f00}{a_{n}} & = \fbox{$\ds{\quad% {n + 1 \choose k + 1} - \delta_{k,-1}\quad}$} \end{align}
\begin{align} \mbox{With}\ \pars{1}\,,\quad \color{#f00}{\sum_{m = 0}^{M}{m + k \choose k}} & = \bracks{{M + k + 1 \choose k + 1} - \delta_{k,-1}} - \bracks{{k \choose k + 1} - \delta_{k,-1}} \\[3mm] & = {M + k + 1 \choose k + 1} - {k \choose k + 1} \end{align} Thanks to $\ds{@robjohn}$ user who pointed out the following feature: $${k \choose k + 1} = {-k + k + 1 - 1 \choose k + 1}\pars{-1}^{k + 1} = -\pars{-1}^{k}{0 \choose k + 1} = \delta_{k,-1}$$ such that $$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad\color{#f00}{\sum_{m = 0}^{M}{m + k \choose k}} = \color{#f00}{{M + k + 1 \choose k + 1} - \delta_{k,-1}}\quad} \\ \mbox{}\\ \hline \end{array}$$
• Since $k=-1$ is covered in the first part, it should be noted that since $\binom{-1}{0}=1$, $$\binom{k}{k+1}-\delta_{k,-1}=0$$ therefore the final answer seems it should be $$\binom{M+k+1}{k+1}-\delta_{k,-1}$$ – robjohn Jul 25 '16 at 13:00
• @robjohn Thanks. I'm checking everything right now. – Felix Marin Jul 25 '16 at 21:48
• @robjohn Thanks. Fixed. – Felix Marin Jul 25 '16 at 22:09
We can prove this by counting in two ways.
Let $$S$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$. By definition, $$|S|=\binom{n+1}{k+1}$$.
Let $$S_i$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$ such that the largest element is $$i+1$$. Picking $$k+1$$ elements from $$[n+1]$$ such that the largest element is $$i+1$$ is a two-step-process.
(Step 1) Pick $$i+1$$. The number of way(s) to do this is $$\binom{1}{1}$$.
(Step 2) Pick the $$k$$ elements from the the remaining $$i$$ elements. The number of way(s) to do this is $$\binom{i}{k}$$.
Therefore, $$|S_i|=\binom{1}{1}\binom{i}{k}=\binom{i}{k}$$. Since we can see that $$S_k, S_{k+1}, S_{k+2}, \dots, S_n$$ partition $$S$$, we have that $$\begin{gather*} \sum_{i=k}^n|S_i|=|S|\\ \sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1} \end{gather*}$$ Since we know that if $$i < k$$, then $$\binom{i}{k}=0$$, we can say that $$\sum_{i=k}^n\binom{i}{k}=\sum_{i=0}^n\binom{i}{k}$$. Therefore, we have $$\begin{gather*} \sum_{i=0}^n \binom{i}{k} = \binom{n+1}{k+1} \end{gather*}$$ | 2019-08-23T19:46:31 | {
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http://mathhelpforum.com/geometry/183278-divided-triangle-proportion-question.html | # Thread: Divided triangle, proportion question
1. ## Divided triangle, proportion question
Hi Forum!
Here's a question:
Nine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?
Now,we can consider the remaining 9 parts of the triangle as 9/10 of it.
But, here comes the part that I don't understand.
The area can be considered as 81/100 to the entire triangle.
I'm confused with the usage of the area here. I does makes a lot of sense,
but since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?
This is kind of confusing to apply. Can someone help?
Why 81/100?
2. ## Re: Divided triangle, proportion question
Originally Posted by Zellator
Hi Forum!
Here's a question:
Nine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?
Now,we can consider the remaining 9 parts of the triangle as 9/10 of it.
But, here comes the part that I don't understand.
The area can be considered as 81/100 to the entire triangle.
I'm confused with the usage of the area here. I does makes a lot of sense,
but since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?
This is kind of confusing to apply. Can someone help?
Why 81/100?
1. Draw a sketch!
2. By proportion you'll get:
$\displaystyle \dfrac bB = \dfrac9{10}~\implies~b=\frac9{10} \cdot B$
$\displaystyle \dfrac hH = \dfrac9{10}~\implies~h=\frac9{10} \cdot H$
3. The area of the original triangle is calculated by:
$\displaystyle A_o = \frac12 \cdot B \cdot H$
The area of the smaller triangle is calculated by:
$\displaystyle A_s = \frac12 \cdot b \cdot h = \frac12 \cdot \frac9{10} \cdot B \cdot \frac9{10} \cdot H = \frac{81}{100} \cdot A_o$
3. ## Re: Divided triangle, proportion question
Originally Posted by earboth
1. Draw a sketch!
2. By proportion you'll get:
$\displaystyle \dfrac bB = \dfrac9{10}~\implies~b=\frac9{10} \cdot B$
$\displaystyle \dfrac hH = \dfrac9{10}~\implies~h=\frac9{10} \cdot H$
3. The area of the original triangle is calculated by:
$\displaystyle A_o = \frac12 \cdot B \cdot H$
The area of the smaller triangle is calculated by:
$\displaystyle A_s = \frac12 \cdot b \cdot h = \frac12 \cdot \frac9{10} \cdot B \cdot \frac9{10} \cdot H = \frac{81}{100} \cdot A_o$
Hi earboth!
Thanks for the graph! You really exceed in Geometry!
Thanks for the easy to understand explanation, I get it now!
Using this is possible to calculate the area of the smaller triangle, and then the original triangle.
Great! Thanks again!
All the best!
4. ## Re: Divided triangle, proportion question
Hi Zellator,
Does the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?
5. ## Re: Divided triangle, proportion question
Originally Posted by bjhopper
Hi Zellator,
Does the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?
Hi bjhopper how are you?
The area given is of the largest portion of the divided triangle, looking at the graph of earboth, the portion underneath the red triangle. | 2018-05-25T08:00:03 | {
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https://math.stackexchange.com/questions/3255898/is-this-a-simple-answer-to-the-classic-problem-a-certain-city-has-10-bus-routes | # Is this a simple answer to the classic problem "A certain city has 10 bus routes..."
A certain city has 10 bus routes. Is it possible to arrange the routes and the bus stops so that if one route is closed, it is still possible to get from anyone stop to any other (possibly changing along the way), but if any two routes are closed, there are at least two stops such that it is impossible to get from one to the other?
Given the solution above, why did the authors feel compelled to answer: Yes. Consider 10 straight lines in the plane, no 2 are parallel & no 3 are concurrent. Let lines be bus routes & let points of intersection be stops. We get from anyone stop to any other (if the stops lie on 1 line, w/o changing; & if not, then with just 1 change). If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once. However, if we discard 2 lines, then 1 stop-their point of intersection-will have no bus routes passing thru it, & it'll be impossible to get from this stop to any other.
Source: A. M. Yaglom and l. M. Yaglom CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY SOLUTIONS Volume II Problems From Various Branches of Mathematics Translated by James McCawley, Jr. Revised and edited by Basil Gordon DOVER PUBLICATIONS, INC. NEW YORK
• Yes, the answer is correct. Jun 9, 2019 at 6:12
• My guess, Joe, is that when they wrote "changing", they meant "changing once". Jun 9, 2019 at 13:04
• Your answer is correct and fine. It is possible the authors just didn't think of this solution and thus put their solution into the book. Everybody overlooks something every now and then. Jun 9, 2019 at 15:30
• My solution fails to meet the requirement, "If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once." Jun 10, 2019 at 0:55
• Gerry, your interpretation of the wording is what makes the problem sufficiently complex. Thanks. Jun 10, 2019 at 0:58
Let’s go back to the roots.
I guess this is a problem 101 at p.53 from a book “Non-elementary problems in elemntary presentation” by A.M. Yaglom & A.M. Yaglom (“Неэлементарные задачи в элементарном изложении” by Акива Моисеевич и Исаак Моисеевич Яглом), Moskow, State publishing house of technical and theoretical literature, 1954.
It is the same problem which you quoted and your solution fits for it.
Also author’s solution is the same which your quoted.
I can explain the situation as follows. The problem is contained in a subsection devoted to arrangements of points and planes and the authors say that this topic was developed in XIX century into a big science, called projective geometry. The questions, considered in problems 101–107 belong to relatively narrow topic of projective geometry, namely, to so-called configuration theory, which has a big importance in modern mathematics. This explains the solution, but it is still surprising how your simple solution was missed.
But I guess that this problem probably was taken from Moskow mathematical competition from 1950 (9-10 forms, second round, problem 4):
Можно ли провести в городе $$10$$ автобусных маршрутов и установить на них остановки так, что какие бы $$8$$ маршрутов ни были взяты, найдётся остановка, не лежащая ни на одном из них, а любые $$9$$ маршрутов проходят через все остановки.
(Given a city, can we arrange $$10$$ bus routes and set bus stops on them such that for each 8 routes there is a bus stop not belonging to any of the routes, but each $$9$$ routes contain all stops).
The proposed solution is the same as Yagloms’:
Проведём $$10$$ попарно пересекающихся прямых, никакие три из которых не пересекаются в одной точке. Пусть маршруты проходят по этим прямым, а остановками служат точки пересечения прямых. Любые $$9$$ маршрутов проходят через все остановки, поскольку через каждую остановку, лежащую на оставшейся прямой, проходит одна из $$9$$ прямых, соответствующих этим маршрутам. Любые $$8$$ маршрутов не проходят через остановку, которая является точкой пересечения оставшихся двух маршрутов.
But although this problem looks similar to Yagloms’, its essence is different! It asks not about connectivity of a graph with removed edges, but about incidence requirements. The geometric solution satisfies them, whereas the cyclic graph doesn’t. | 2022-07-04T09:12:57 | {
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https://math.stackexchange.com/questions/3094722/what-is-the-amount-of-draws-necessary-to-see-all-red-cards-from-a-standard-deck | # Problem abstraction
A standard deck of $$52$$ cards has $$26$$ red cards: it has $$13$$ hearts, $$13$$ diamonds, as well as $$26$$ black cards ($$13$$ spades, as well as $$13$$ clubs). Let us draw $$5$$ cards from the deck at once, and return those cards to the deck afterward.
What is the expected number of draws before we see all $$26$$ red cards?
# Use case
Let us say that there is a set of $$N = 100$$ cards in a game. $$M = 30$$ cards are of rare rarity and $$N - M = 70$$ cards are of common rarity. We buy booster packs of size $$= 10$$. The question is: how many booster packs need to be bought to collect all $$M = 30$$ cards?
# Attempted solution
I have managed to calculate the approximate number of booster packs necessary to get $$M = 30$$ rare cards by calculating the expectation of the above hypergeometric distribution ($$\mu$$) and then calculating $$M/\mu$$. However, this is not the correct solution since it does not take into account the possibility of collecting duplicates.
Regarding the Coupon collector's problem, I'm not sure if it is applicable since we always draw a single coupon, whereas in my use case a booster pack contains more than a single card.
# Simulations
## Problem Abstraction
$$10^6$$ trials were conducted, AVG: $$38.947$$, STDEV: $$12.3653$$ draws
## Use case
$$10^6$$ trials were conducted, AVG: $$38.535$$, STDEV: $$11.962$$ draws
• Related: coupon collector's problem – jvdhooft Jan 31 at 10:18
• @james - thank you for the feedback, I have edited my question to provide additional context. – user3223162 Jan 31 at 11:23
• close to $\frac25\cdot26H_{26}=40.085965$ draws. – robjohn Jan 31 at 15:31
• Could you please explain how you got to the number? – user3223162 Jan 31 at 21:19
Here is a solution of the "problem abstraction" by way of the Principle of Inclusion / Exclusion (PIE).
Let $$T$$ be the number of the first draw in which we have seen all the red cards. We would like to find $$P(T>k)$$ for some $$k>0$$, i.e. the probability that we have not seen all $$26$$ red cards in $$k$$ draws. To that end, let's say a sequence of $$k$$ draws has "Property $$i$$" if red card $$i$$ has not been drawn, for $$i = 1,2,3,\dots,26$$. Let $$S_j$$ be the sum of the probabilities of all the sequences with $$j$$ of the properties, for $$j = 1,2,3,\dots,26$$. For $$S_j$$, there are $$\binom{26}{j}$$ ways to select the $$j$$ cards which are missing. The probability that those cards are missing in a single draw is $$\binom{52-j}{5} / \binom{52}{5}$$, so the probability that the cards are missing in all $$k$$ draws is $$[\binom{52-j}{5} / \binom{52}{5}]^k$$. Therefore $$S_j = \binom{26}{j} \left( \frac{\binom{52-j}{5}}{ \binom{52}{5}} \right) ^k$$ By PIE, the probability of a sequence of draws with at least one of the properties, i.e. a sequence with at least one red card not seen, is $$P(T>k) = \sum_{j=1}^{26} (-1)^{j+1} S_j$$ so \begin{align} E(T) &= \sum_{k=0}^{\infty} P(T>k) \\ &= \sum_{k=0}^{\infty} \sum_{j=1}^{26} (-1)^{j+1} S_j \\ &= \sum_{k=0}^{\infty} \sum_{j=1}^{26} (-1)^{j+1} \binom{26}{j} \left( \frac{\binom{52-j}{5}}{ \binom{52}{5}} \right) ^k \\ &= \sum_{j=1}^{26} (-1)^{j+1} \binom{26}{j} \sum_{k=0}^{\infty} \left( \frac{\binom{52-j}{5}}{ \binom{52}{5}} \right) ^k \\ &= \sum_{j=1}^{26} (-1)^{j+1} \binom{26}{j} \frac{1}{1- \binom{52-j}{5}/ \binom{52}{5}} \\ &= 38.9133 \end{align}
The following Monte Carlo simulation of $$10^6$$ trials in the R language is consistent with the result above. The average number of draws of 5-card hands necessary to see all 26 red cards was 38.973, with a 95% confidence interval of 38.91305 to 38.96158. The analytical result 38.9133 falls in the confidence interval, although just barely.
> # ndraws: return the number of draws of 5-card hands required
> # to see all red cards at least once
> # We consider the red cards to be the cards numbered 1-26.
> ndraws <- function() {
+ seen <- rep(0, 52)
+ n <- 0
+ while (TRUE) {
+ n <- n+1
+ hand <- sample(1:52, 5)
+ seen[hand] <- 1
+ if (sum(seen[1:26]) >= 26)
+ return (n)
+ }
+ }
> nreps <- 1e6
> set.seed(1234) # for reproducibility
> t <- replicate(nreps, ndraws())
> t.test(t)
One Sample t-test
data: t
t = 3145.3, df = 1e+06, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
38.91305 38.96158
sample estimates:
mean of x
38.93731
• Thank you for your answer! After some digging, in a paper The Coupon Subset Collection Problem by Adler, Ilan and Ross, Sheldon M I have found equation 7 that solves the issue, and link this question as well. What keeps me confused is that the expectation depends only on the number of distinct balls in the set and the size of the drawn subset, but not on the size of the entire set. Am I misunderstanding something? – user3223162 Feb 4 at 11:11
• @user3223162 I haven't read the paper you refer to, but in the question you link to, the assumption is that the urn contains one each of $n$ distinct balls, so the number of distinct balls is the same as the number of balls in the urn. – awkward Feb 4 at 12:56
• In that case, that equation does not solve my issue. Furthermore, simulations I ran disagree with your attempt. Still, I appreciate the effort – user3223162 Feb 4 at 15:46
• @user3223162 There must be something amiss, then, because my simulation agrees with the analytical result I posted. I have added the simulation to the solution, above. – awkward Feb 4 at 20:20
• You are correct, my tests contained an error and after correction confirm your results. I have ran the formula alongside with some simulations and the results correspond. For now, I am marking this as solved, I will try and confirm my findings in some existing literature on the subject. Thanks! – user3223162 Feb 5 at 12:59 | 2019-05-23T10:49:41 | {
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https://math.stackexchange.com/questions/3129983/question-about-asymptotic-notation | # Question about Asymptotic Notation
I am doing a question on asymptotic notation. I have two functions $$f(n)$$ and $$g(n)$$, where $$f(n) = (\log_2n)^2$$ and $$g(n) = \log_2n^{\log_2n} + 2\log_2n$$. I have to determine whether $$f(n)$$ is $$O(g(n))$$, $$\Omega(g(n))$$, or $$\Theta(g(n))$$.
My approach to figuring this out is to determine whether $$g(n)$$ grows faster than $$f(n)$$ or if $$f(n)$$ grows faster than $$g(n)$$. To do this, I am trying to prove whether $$2\log_2n \leq (log_2n)^2$$ for all $$n \geq c$$, where $$c$$ is a constant. I want to prove this because if it is true, then it can be said that $$f(n)$$ grows faster than $$g(n)$$ for all $$n \geq c$$ (where $$c$$ is a constant). I know that I would also have to prove whether $$\log_2n^{\log_2n} \leq (log_2n)^2$$ in order to say that $$f(n)$$ grows faster than $$g(n)$$ for all $$n \geq c$$.
So far I have:
$$\log_2n \leq (log_2n)^2$$ $$2\log_2n \leq 2(log_2n)^2$$
However, I am not sure where to go from here in trying to prove whether $$2\log_2n \leq (log_2n)^2$$. Dividing both sides of the inequality by 2 will not achieve anything.
Am I taking the right approach for solving this question, or is there a better way to determine the asymptotic complexity of $$f(n)$$? Any insights are appreciated.
• Hint: Instead of dividing by $2$, divide by $\log n$. The resulting inequality holds for $n$ sufficiently large. – Michael Burr Feb 28 at 9:37
We have that $$f(n) = (\log_2 n)^2$$ and $$g(n) =\log_2 n^{\log_2 n} + 2\log_2 n = (\log_2 n)^2 + 2\log_2 n$$ This last equality is derived by using the logarithm law that $$\log(a^b) = b\log(a)$$.
Question 1. Is $$f$$ asymptotically bounded below by $$g$$? That is, does there exist an $$N_1$$ and a $$k_1 > 0$$ such that $$f(n)\geq k_1 \cdot g(n)$$
for all $$n\geq N_1$$?
Answer 1. We have that $$(\log_2 n)^2 \geq k_1 \cdot \left((\log_2 n)^2 + 2\log_2 n\right)$$
Set $$k_1 = 1/2$$, then $$(\log_2 n)^2 \geq \frac12 (\log_2 n)^2 + \log_2 n$$
By subtracting $$(1/2) (\log_2 n)^2$$, we get $$\frac12 (\log_2 n)^2 \geq \log_2 n$$ Dividing by $$\log_2 n$$ we get $$\frac12 \log_2 n \geq 1$$ which is true for all $$n\geq 4 = N_1$$. Hence, $$f$$ is asymptotically bounded below by $$g$$.
Question 2. Is $$f$$ asymptotically bounded above by $$g$$? That is, does there exist an $$N_2$$ and a $$k_2 > 0$$ such that $$f(n) \leq k_2\cdot g(n)$$ for all $$n\geq N_2$$?
Answer 2. We have that $$(\log_2 n)^2 \leq k_2 \cdot \left((\log_2 n)^2 + 2\log_2 n\right)$$
Set $$k_2 = 1$$, then $$(\log_2 n)^2 \leq (\log_2 n)^2 + 2\log_2 n$$
Subtract $$(\log_2 n)^2$$, then $$0\leq 2\log_2 n$$ which is true for all $$n\geq 1 = N_2$$. Hence, $$f$$ is asymptotically bounded above by $$g$$.
Conclusion. We conclude that $$f$$ is both bounded below and above asymptotically by $$g$$. Specifically, to be more precise, this is true with constants $$1/2$$ and $$1$$ and for $$N = \max(N_1,N_2) = 4$$. This means that $$\frac12\cdot g(n) \leq f(n) \leq 1\cdot g(n)$$ for all $$n\geq 4$$.
In asymptotic notation, this means that $$f(n)$$ is $$\Theta (g(n))$$.
• Would it be the case that you wouldn't be able to prove that f is asymptotically bounded below or above by g if you can't simplify one of the terms in the inequality to a constant? Since in both of your cases, you reduced one of the terms in the inequality to a constant. – ceno980 Feb 28 at 12:05
• No, it doesn't have to be the case that one side is reduced to a constant. What you need to do is to solve the inequality. For example, $$12\sqrt{n} \leq e^n$$ has a solution (around $3.041$, so for whole numbers rounded up $n\geq 4$). You don't even have to solve inequality in any precise manner. You just need to show that it holds for sufficiently large $n$. – Eff Feb 28 at 12:17 | 2019-06-19T09:11:15 | {
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https://www.physicsforums.com/threads/finding-the-height-of-a-ball-with-a-geometric-series.793794/ | Finding the height of a ball with a geometric series
Homework Statement
A ball is dropped from one yard and come backs up $\dfrac{2}{3}$ of the way up and then back down. It comes back and $\dfrac{4}{9}$ of the way. It continues this such that the sum of the vertical distance traveled by the ball is is given by the series $1+2\cdot\dfrac{2}{3}+2\cdot\dfrac{4}{9}+2\cdot\dfrac{8}{27}+\cdot \cdot \cdot=1+2(\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{8}{27}+\cdot\cdot\cdot(\dfrac{2}{3})^n$). Find the height of the tenth rebound and the distance traveled by the ball after it touches the ground for the tenth time.
Homework Equations
$S_n=\dfrac{a(1-r^{n})}{1-r}$
The Attempt at a Solution
I know that the height of the tenth rebound is simply the tenth term in the sequence so $h=s_{10}=(\dfrac{2}{3})^{10}\approx0.0173$ yards. Now I thought that the vertical distance would be $1+2\cdot\dfrac{\frac{2}{3}(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}}\approx1.96$ yards using the formula for a geometric series $\dfrac{a(1-r^{n})}{1-r}$. However, the book tells me that the answer should be $6\cdot(\dfrac{2}{3})^{10}\approx0.104$ yards. Now on the chance that I did misinterpret the book and the author meant the vertical distance the ball traveled the tenth time it hits the ground, shouldn't it be $2\cdot\dfrac{2}{3}^{10}$?
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HallsofIvy
Your series starts with $1= (2/3)^0$, with n= 0, not 1. The "10th" term is n= 9, not 10.
Your series starts with $1= (2/3)^0$, with n= 0, not 1. The "10th" term is n= 9, not 10.
I guess I should've stated that the textbook focuses on the series in the parenthesis because the answer for the height is indeed $(\dfrac{2}{3})^{10}$ | 2020-03-28T21:51:26 | {
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https://math.stackexchange.com/questions/1346866/are-eigenvalues-of-the-limit-of-a-sequence-of-matrices-limits-of-eigenvalue-sequ/1346869 | # Are eigenvalues of the limit of a sequence of matrices limits of eigenvalue sequences?
Let $\{A_n\}\in \mathbb{R}^{m\times m}$ be a sequence of symmetric matrices such that $A_n\to A$ as $n\to \infty$, i.e. $\lim_{n\to \infty}a_{ij}(n)=a_{ij}\ \forall 1\le i,j\le m$ where $A_n=[a_{ij}(n)],A=[a_{ij}]$. Let $\rho(A_n)=\{\lambda_1(n),\cdots,\ \lambda_m(n)\}$ be the eigenvalues of $A_n$ and similarly, $\rho(A)=\{\lambda_1,\cdots,\ \lambda_m\}$ be the eigenvalues of $A$, arranged in, say, increasing order. Here are my questions
1)Can I write $\lambda_k(n)\to \lambda_k,\ 1\le k\le m$?
2)If I define (with a slight abuse of standard notation) $\delta_s(n),\ 1\le s\le m$ as the maximum eigenvalue of any $s\times s$ submatrix of $A_n$ and if $\delta_s$ be the corresponding quantity for $A$, then can I say that $$\lim_{n\to \infty}\delta_s(n)=\delta_s$$ ?
Intuitively it seems to me that the answers are positive since the eigenvalues of a matrix are continuous functions of the elements of the matrix and $\delta_s$ is just the maximum of some eigenvalues of submatrices.
However, I am not sure if this argument is sound enough. Maybe this is a very trivial issue for people here, but I would really appreciate if someone can kindly provide some explanation regarding the correct answer. Thanks in advance.
• You need to be careful here; there's the possibility that the eigenvalues may "switch". For example, we could have $\lambda_1(n) \to \lambda_2$ and $\lambda_2(n) \to \lambda_1$ – Omnomnomnom Jul 2 '15 at 11:11
• Outside of that technicality, your answer is indeed correct and sufficient. – Omnomnomnom Jul 2 '15 at 11:13
• If we have symmetric matrices (as you do now), the issue of "switching eigenvalues" isn't a problem for this question. – Omnomnomnom Jul 2 '15 at 11:14
• @Omnomnomnom, is the argument I provided correct? Also, can you kindly comment on the second question? – Samrat Mukhopadhyay Jul 2 '15 at 11:14
• Yes, that's what I meant; your argument that the eigenvalues depend continuously on the matrix is enough to prove both 1) and 2). In order to get 2) quickly, we could note that the function $$(x_1,\dots,x_n) \mapsto \max \{x_1,\dots,x_n\}$$ is continuous over $\Bbb R^n$. – Omnomnomnom Jul 2 '15 at 11:17
Let $\chi_n$ be the characteristic polyomial of $A_n$ and $\chi=a\prod(X-\lambda_i)$ be that of $A$, then $\chi_n\mathop{\to}_{n\rightarrow\infty}\chi$.
If a sequence of polynomials converges to another, then by continuity the roots of the polynomials in the sequence must converge to the roots of the limit polynomial (with the same multiplicity). Hence the eigenvalues of the sequence of matrices indeed convergence to the eigenvalues of $A$.
Likewise, since any submatrix of $A_k$ converges to the corresponding submatrix of $A$, that is sufficient to say that $\delta_s(n)\to\delta_s$.
• @orangeskid Thanks for the feedback, I added an explanatory line. – Hippalectryon Jul 2 '15 at 11:25
• @Hippalectryon: It's tempting to pass to the characteristic polynomial and, since that varies continuously, use that the roots also do. That is correct but I would give it more thought. It's the problem with approaching a polynomial with multiple roots. Otherwise, the implicit function theorem works OK. – Orest Bucicovschi Jul 2 '15 at 11:30
The answer of @Hippalectryon is morally correct. However, there is this problem of approaching a polynomial with multiple roots, somehow delicate. $\tiny{\text{Can be proved with the argument principle from complex analysis.}}$ We'll work directly with the eigenvalues, bypassing the characteristic polynomial.
Use this standard observation: if two symmetric matrices are ordered $B\prec C$ then their vectors of eigenvalues ( ordered increasingly) also satisfy $\lambda(B) \prec \lambda(C)$ ( that is $\lambda_i(B) \le \lambda_i(C)$ for all $i$).
Now, if if $A_n \to A$ then for any $\epsilon >0$ we have $A- \epsilon I \prec A_n \prec A + \epsilon I$ for $n \ge n_{\epsilon}$, and so $$\lambda_i(A) - \epsilon \le \lambda_i(A_n) \le \lambda_i(A) + \epsilon$$
$\tiny{\text{(the question for the max for submatrices is simple:$\lim$and$\max$commute ).}}$
$\bf{Added:}$ I am giving this " continuity of the roots" question more thought. Of course, @Hippalectryon is right. But how do we truly convince ourselves that the roots of $P_n$ approach the roots of $P$? First, a hands-on approach: The roots of $P_n$ will all lie in a bounded region since the coefficients of $P_n$ are all bounded. If the roots of $P_n$ did not approach those of $P$ then we would find, by compactness, a subsequence whose roots approach some other $n$-uple. But that would mean in the limit that $P$ would decompose using that $n$-uple, that is, it would have two distinct decompositions, contradiction. The high-brow explanation is that the map roots $\mapsto$ polynomials from $\mathbb{C}^n$ to $\mathbb{C}^n$ is continuous and proper and induces a bijective map $\mathbb{C}^n/S_n \to\mathbb{C}^n$, which is bijective, continuous and closed, hence a homeomorphism. $\tiny{\text{(surjectivity is equivalent to the fundamental theorem of algebra)}}$
• By $A\prec B$ do you mean that the elements of $A$ are smaller than the elements of $B$, individually? – Samrat Mukhopadhyay Jul 4 '15 at 5:41
• @Samrat Mukhopadhyay: I mean $B-A$ positive definite. – Orest Bucicovschi Jul 4 '15 at 8:30
• Oh, ok, got it. – Samrat Mukhopadhyay Jul 4 '15 at 11:37 | 2019-07-21T06:41:10 | {
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http://math.stackexchange.com/questions/154310/why-would-the-author-ask-if-i-used-the-associative-law-to-prove-is-not-equiv | # Why would the author ask if I used the Associative Law to prove + is not equiv. to *?
I just started reading An Introduction to Mathematical Analysis by H.S. Bear and problem 1 goes as follows:
Problem 1: Show that + and * are necessarily different operations. That is, for any system (F, +, *) satisfying Axioms I, II, and III, it cannot happen that x + y = x * y for all x, y. Hint: You do not know there are any numbers other than 0 and 1, so that your argument should probably involve only these numbers. Did you use Axiom II? If not, state explicitly the stronger result that you actually proved.
In this book, Axiom I is commutativity of + and *, Axiom II is associativity of + and *, and Axiom III is existence of identities (x+0=x, x*1=x, 0 does not equal 1).
My question: Simply why would the author specifically ask the reader if he/she used Axiom II (associativity) and what exactly do they mean by "If not, state explicitly the stronger result you actually proved"? Why not not include those last two sentences?
FWIW, here is my solution:
To prove: Restated:
And I justified 5 by citing Axiom III since Axiom III includes the statement that 0 does not equal 1.
-
Hint: if you didn't use Axiom II, what can you now say about systems where Axioms I And III hold but II possibly does not? – Steven Stadnicki Jun 5 '12 at 16:13
More succinctly: $\ 1 = 1 + 0 = 1 * 0 = 0.\$ Note that the proof requires only one of the operations to be commutative, so it will work for noncommutative rings too, i.e. where multiplication is not necessarily commutative. It wouldn't require comutativity at all if Axiom II was $\rm\:1*x = x\:$ vs. $\rm\:x*1 = x.$ Said simply, the hypothesis $\:1\ne 0\:$ forces addition $\ne$ multiplication, because these operations have the value $1,0$ resp. at the same point, viz. $(1,0)$ – Bill Dubuque Jun 5 '12 at 16:16
Since I didn't use Axiom II (associativity), my answer to the request "State explicitly the stronger result that you actually proved" is "The non-equivalence of + and * is not dependent on associativity holding in the system F." Is that a good answer to the meaning of the very last sentence of the problem? – mring Jun 5 '12 at 16:42
@Pete Close, though I would make it clear that it does still require Axioms I and III, e.g. (though this is awkward) '+ and * are non-equivalent in each system with commutativity and existence of identities, regardless of associativity'. – Steven Stadnicki Jun 5 '12 at 17:04
The basic idea is as follows. From the neutral Axioms III, and commutativity of addition we have
$$\begin{eqnarray}\rm x = 0 + x &\rm \\ \rm y *\: 1 &=\rm y \end{eqnarray}$$
If $\ +\, =\, *\$ then aligned terms are unified for $\rm\:y = 0,\ x = 1,\:$ yielding
$$\rm\ 1 = 0 + 1 = 0 * 1 = 0$$
contra hypothesis $\rm\:1 \ne 0.\:$ Thus $\rm\: +\: \ne\: *\:$ because they take different values at the point $\rm\:(0,1)$.
Note that the proof does not use associativity, and doesn't use commutativity if you state the neutral axioms as above. In any case, only one of the commutative axioms is needed, so that the neutral axioms can be ordered so the above unification is possible. In particular, the inference works in noncomutative rings, i.e. rings where multiplication is not necessarily commutative. Further, because the proof did not use associativity, it will also work in nonassociative rings.
Note $\$ This method of deriving consequences by unifying terms in identities is a basic method in equational reasoning (term rewriting), e.g. google Knuth-Bendix or Grobner basis algorithms.
-
The author probably wants to add that it is not necessary to use Axiom II in order to prove that the laws must be different.
If you then forget about axiom II, and if you suppose that + and * are the same, your axioms become :
axiom 1 : $x+y = y+x$
axiom 3a : $x+0 = x$
axiom 3b : $x+1 = x$
axiom 3c : $0 \neq 1$.
In order to get a contradiction you must use axiom 3c, and the only way to use it is to show that axioms 1,3a,3b implies that $0 = 1$.
So you really are investigating commutative laws with two identity elements, and in fact, you necessarily have to prove that the identity element of a commutative law must be unique :
If 0 and 1 are identity elements of +, then 0 = 0+1 = 1+0 = 1, so they are the same. There, we proved that if a commutative law has an identity element, it is unique.
-
Proven without Associativity:
Given the two different operations, $x+y\neq x*y$ cannot hold $\forall x,y$ even when the order of evaluation matters (the case where Axiom II is false, e.g. subtraction and division).
-
As far as I understand this the author also wants to teach how to understand/write a proof in general. If you are given a proof of a statement a good place to start understanding it, is figuring out where exactly every single assumption was used. The second question is which assumptions didn't we use and whether we actually proved a stronger result.
This is a very useful lesson for your future as a mathematician! The first question is probably more important when it comes to understanding a specific proof, while the second question may help you linking the statement with other similar statements. Obviously in your rather elementary exercise it might not reveal its full importance but as soon as things get more complicaated it is a good idea to keep these two questions in mind.
- | 2016-06-26T15:42:51 | {
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https://math.stackexchange.com/questions/3852235/how-to-find-a-2d-vector-which-is-perpendicular-to-a-line-and-points-to-a-speci | # How to find a [2D] vector which is perpendicular to a line and points to a specific half-plane?
Given three points $$A$$, $$B$$ and $$C$$, that are not colinear, I want to find the vector $$v$$ which is perpendicular to the line $$L$$ that passes from $$A$$ and $$B$$ AND points to the half-plane created by $$L$$ that $$C$$ lies on.
I know that there are two orthogonal vectors to $$T$$: $$\vec{v_1} = [t_y, -t_x]$$ and $$\vec{v_2} = [-t_y, t_x]$$. I can find equation of the line, like $$L: y=ax+c$$, and solve it for both $$C_x$$ and $$t_y+A_x$$. I will return $$\vec{v_1}$$ if results had same sign, and $$\vec{v_2}$$ otherwise. And, of course, I have to take care of special case where $$L: y=c$$. Searching for an algebraic solution, I found this question: Find closest vector to A which is perpendicular to B. I tried to simplify the accepted answer for 2D, which resulted in:
\begin{aligned} s&=u\times t = (u_xt_y-u_yt_x)\vec{k} = s_z\vec{k}\\ v&=t\times s = s_z(t_y\vec{i}-t_x\vec{j}) = s_z\vec{v_1} \end{aligned}
That can be written as following, if the magnitude of the resulting vector is unimportant: $$v = \begin{cases} \vec{v_1}, & \text{if s_z > 0} \\[2ex] \vec{v_2}, & \text{if s_z<0} \\[2ex] \vec{0}, & \text{if s_z =0} \end{cases}$$
But I'm not sure about the followings:
1. Is the linked question really relevant to my problem? It says the closest vector that ..., which I don't even know what it means.
2. Did I do the math right?
3. Is there an even faster (less logical and floating point operations when implementing it) way to choose one of $$\vec{v_1}$$ and $$\vec{v_2}$$? I really don't care about the magnitude of the resulting vector.
## 2 Answers
The vectors $$v_1$$ and $$v_2$$ are indeed both nonzero and perpendicular to $$L$$, and also point in opposite directions, so one of them has to be a "good" one.
If you compute $$u = C - A$$, then you can compute $$h = u \cdot v_1,$$ the dot product of $$u$$ and $$v_1$$.
If this turns out positive, then $$u$$ and $$v_1$$ point into the same halfplane, and your answer is $$v_1$$; if it's negative, your answer is $$v_2$$. If it's $$0$$, then $$C$$ is actually collinear with $$A$$ and $$B$$, which is kind of a free sanity-check that your inputs were valid. [All this is a rehash of what you wrote in your question as the first displayed equation.]
An alternative approach is to compute $$s = u \times t$$ (in 3-space), which seems bad because it's a cross-product, which looks like 2 multiplies and a subtraction for each term...but you only need to compute the "z" term because the other two are always zero. Then you compute $$v = t \times s$$, and this vector will point in the right direction. You needn't compute the $$z$$-component (it'll be zero), so you end up doing a total of $$6$$ multiplies and $$3$$ subtracts, plus the 6 subtractions to compute $$u$$ and $$t$$ in the first place.
Is this a winning algorithm? I haven't counted the operations in the other one. But it has one advantage: there's no branching, which can be helpful on some highly-parallel architectures (or at least this used to be the case).
It also has a downside: if $$C$$ is very close to the line $$AB$$, the magnitude of the resulting vector gets very small. You have to decide whether that matters to you. You said not, but in practice, I've often found such things annoying.
Here's something like an algorithm, written in very explicit Matlab
function v = findVec(A, B, C)
% Given points A,B,C in the xy-plane, C not on the line AB, find
% a vector v in the xy-plane that is perpendicular to AB, and points
% into the halfplane containing C
u = [C(1) - A(1), C(2) - A(2)];
t = [B(1) - A(1), B(2) - A(2)];
s = [0, 0, u(1)*t(2) - u(2)* t(1)]; % first cross product
v = [t(2)*s(3), -t(1) * s(3)];
and here's the more idiomatic, minimal operations, version:
function v = findVec(A, B, C)
% Given points A,B,C in the xy-plane, C not on the line AB, find
% a vector v in the xy-plane that is perpendicular to AB, and points
% into the halfplane containing C
trot = [B(2) - A(2), A(1) - B(1)]; % the t vector, rotated 90 degrees
% because that's what I'll need in a minute.
sz = (C(1) - A(1)) * trot(1) + (C(2) - A(2)) * trot(2);
v = sz * trot;
• Thank you very much John. I actually made a terrible mistake in my calculations and probably misled you. Because in fact the equation I put was obtained from your second method! I edited my question. But it was very beautiful that in the end both methods came to the same result, since u.v1=sz=(ux*ty-uy*tx). However, I prefer to look at it as the result of the dot product, thanks again for bringing it up. Oct 6 '20 at 6:09
• And there was an H after your first equation. Did you mean u or I'm missing something? And I believe that's s(3) in your first code sample. Oct 6 '20 at 6:22
• Fixed both --- sorry about the glitches. Oct 6 '20 at 11:10
The component of $$\vec{u}$$ that is parallel to $$\vec{t}$$ is
$$\vec{u}_\parallel = \frac{ \vec{t} \cdot \vec{u} }{ \| \vec{t} \|^2 }\, \vec{t}$$
So subtract it form $$\vec{u}$$ to get $$\vec{v}$$
$$\vec{v} = \vec{u} - \frac{ \vec{t} \cdot \vec{u} }{ \| \vec{t} \|^2 } \, \vec{t}$$
Proof
$$\vec{t} \cdot \vec{v} = \vec{t} \cdot \vec{u} - \frac{ \vec{t} \cdot \vec{u} }{ \| \vec{t} \|^2 } \, (\vec{t} \cdot \vec{t}) = \vec{t} \cdot \vec{u} -\vec{t} \cdot \vec{u} = \vec{0}$$
• Thanks John. That is aد interesting solution. But the calculations needed to get the answer from this method are huge. That norm in the denominator really terrifies me :) Oct 6 '20 at 6:37 | 2022-01-17T22:28:21 | {
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https://community.wolfram.com/groups/-/m/t/1170959 | # How to code Adams-Bashforth method
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## Original Question
There used to be a question about a manual implementation of the Adams-Bashforth method of numerically integrating an ODE. The code looked like it had be translate from C. I started to write up a more functional-programming to the problem. Well, why not share what I had done? Presumably the purpose of the writing the solver is educational. I will keep the design simple. For a more robust approach, I would recommend working within the NDSolve framework.
The original code looked something like this:
MyAdamBashforth[F_, h_, t_, y_][{t0_, y0_}] :=
Module[{j, p, m, a, F0, F1, F2},
m = 8;
a = 0;
t[0] = t0;
y[0] = y0;
For[j = 0, j <= 1, j++,
t[j + 1] = t[j] + h;
y[j + 1] = y[j] + h*F[t[j], y[j]];
];
F0 = F[t[0], y[0]];
F1 = F[t[1], y[1]];
F2 = F[t[2], y[2]];
For[j = 2, j <= m, j++,
p = y[j] + h/12 (5 F0 - 16 F1 + 23 F2);
t[j + 1] = t[j] + h;
y[j + 1] = p;
F0 = F1;
F1 = F2;
F2 = F[t[j + 1], y[j + 1]];
];
]
The solver was used in a way like this:
Clear[F, y, t, h];
F[t_, y_] := t + y;
MyAdamBashforth[F, 0.5, t, y][{0., 1.}]
MyData3 = Table[{t[j], y[j]}, {j, 0, 8}]
(* {{0., 1.}, {0.5, 1.5}, {1., 2.5}, {1.5, 4.72917}, {2., 8.78212},
{2.5, 15.6914}, {3., 27.2344}, {3.5, 46.3278}, {4., 77.713}} *)
## Design choices
In general, the (discrete) solution of an ODE $x'(t) = \text{RHS}(t, x(t))$ over $a \le t \le b$ consists of step data
$$t_j, x_j, x'_j,\dots \quad \text{for}\ j = 0, 1, \dots, n\,,$$
where $t_0=a$, $t_n=b$, and the number of derivative values $x'_j, x''_j,\dots$ store for each step, if any, may depend on the method. For a first or higher order ODE, $x'_j$ is calculated anyway and may be used for cubic Hermite interpolation between the steps. In a multistep method, the $j+1$ step is computed from some number of the previous steps; single-step methods use only step $j$. The Adams-Bashforth (AB) method is a linear $s$-step method that uses only the values of $x'$ from the previous $s$ steps.
Input: Representation of the problem
Following the original question, the ODE will be specified by a function rhs that represents the first-order equation x'[t] == rhs[t, x[t]]. The initial condition (IC) is simply two numbers t0, x0, representing x[t0] == x0. Also we will require a final time t1, a step size h, and the step order s.
Output data structure
The output will be a list of the step data. The data for a step consists of $t, x, p=x'$. Neither the time $t$ nor the derivative $p$ are strictly necessary to keep, but they can be convenient to keep for the sake of interpolation. If efficiency is a premium, then it might make sense to try to generate a packed array {{t, x, p},...}. If the data is to be passed to Interpolation[], then it has to be put in the form {{{t}, x, p},...}, which cannot be packed. Since this is an educational exercise, I'll choose the latter.
The solver
The basic idea is that the solution data could be generated by Nest[] or NestList[] with a call like
Nest[abStep[rhs, h, s], {IC}, nsteps]
where abStep[rhs, h, s][stepdata] computes next step using the s-step method, IC is the initial condition, and nsteps is the number of steps to traverse the interval of integration. Whether to use Nest[] or NestList[] depends on choices involving the iterator abStep.
The iterator
One constraint with the choice of nesting, is that the output of abStep has to be valid input for it. The function abStep has two phases:
• initialization steps: until s steps are generated, another method is used to compute the next step;
• s-step steps: once s are generate, use the AB s-step method to compute the next step.
Since AB is a multistep method, abStep needs access to previous steps. It only needs the last s steps, but I chose to pass it all steps. This allows the constraints of the problem by coding abStep to return the list of past steps with the next step appended to it. With that design, the solver calls the iterator with Nest[] and not NestList[].
Initialization
The original question uses Euler steps for the initialization of the s-step phase. It was easier to code abStep to use AB with a step order equal to the number of steps so far (up to s). This should be more accurate but a little slower, two considerations that are of minor importance in this exercise. However one wants to compute the initial steps, it is not hard to write a function abStep to do it, since all steps are passed to it, and it is easy to select a method based on how many steps are passed.
## Implementation
ClearAll[abCoeffs]; (* Adams-Bashforth coefficients *)
mem : abCoeffs[s_] := mem = (* for mem, see note on memoization below *)
Table[(-1)^(s - j)/((j - 1)! (s - j)!) * Integrate[Product[If[i == s - j, 1, u + i], {i, 0, s - 1}], {u, 0, 1}], {j, s}];
Clear[abStep]; (* Adams-Bashforth step of order at most s *)
abStep[rhs_, h_, s_][past_] := (* past = previous steps = {{{t}, x, p}..} *)
With[{steps = Min[s, Length@past]}, (* step order cannot exceed number of steps taken *)
With[{ (* next step *)
t0 = past[[-1, 1, 1]] + h,
x0 = past[[-1, 2]] + h*abCoeffs[steps].past[[-steps ;;, 3]] (* Dot product gives the AB linear comb. of coefficients & steps *)
},
With[{p0 = rhs[t0, x0]}, (* store the derivative value; used in following step, too *)
Append[past, {{t0}, x0, p0}] (* add step to past ones *)
]]];
Clear[abSolve]; (* Adams-Bashforth solver x'[t] == rhs[t,x[t]] *)
abSolve[rhs_, x0_, {t0_, t1_, h_}, s_] :=
Nest[
abStep[rhs, h, s],
{{{t0}, x0, rhs[t0, x0]}}, (* initial value *)
Floor[(t1 - t0)/h] (* truncates interval like Range[t0, t1, h]; Ceiling[] is an alternative *)
];
## Examples
One can check the formula with the original question as follows:
Clear[F];
abStep[F, h][{{t}, y, F2, F1, F0}]
(* {{h + t}, ((5 F0)/12 - (4 F1)/3 + (23 F2)/12) h + y, F[h + t, ((5 F0)/12 - (4 F1)/3 + (23 F2)/12) h + y], F2, F1} *)
Here is the original numerical example, with 2-step and 3-step solutions:
ClearAll[F];
F[t_, y_] := t + y;
sol2 = Interpolation@abSolve[F, 1., {0., 4., 0.5}, 2];
sol3 = Interpolation@abSolve[F, 1., {0., 4., 0.5}, 3];
sol0 = y /. First@DSolve[{y'[t] == F[t, y[t]], y[0] == 1}, y, t];
Show[
Plot[{sol0[t]}, {t, 0, 4}, PlotLegends -> {"Exact"}],
Plot[{Undefined, sol3[t]}, {t, 0, 4}, Mesh -> sol3["Coordinates"],
PlotLegends -> {None, "3-step"}],
Plot[{Undefined, Undefined, sol2[t]}, {t, 0, 4},
Mesh -> sol2["Coordinates"], PlotLegends -> {None, None, "2-step"}]
]
The data itself:
abSolve[F, 1., {0., 4., 0.5}, 3]
(*
{{{0.}, 1., 1.}, {{0.5}, 1.5, 2.}, {{1.}, 2.75, 3.75}, {{1.5}, 5.21875, 6.71875},
{{2.}, 9.57422, 11.5742}, {{2.5}, 16.9683, 19.4683}, {{3.}, 29.3089, 32.3089}, {{3.5}, 49.7041, 53.2041},
{{4.}, 83.208, 87.208}}
*)
## Appendix: Memoization
Note: Memoization. "Memoization" or "cacheing" is a technique to store a computed function value so that on subsequent calls, the value need not be recomputed, only recalled from memory. For function values that are used frequently, this can save considerable time. One has to weigh this savings against greater memory usage. I learned the form mem : f[x_,...] := mem = ... on StackExchange. The mem : ... is a named the pattern and it matches the actual function call.
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Thank you very much Michael Rogers.It was a question by me. :)Best wishes Zharou Fisher
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http://seowonstay.com/sharekhan-charges-viddhm/graphing-absolute-value-functions-8f07ca | the basic outline of the more complex figure is easily arrived at, then details can be added as necessary, but the figure is already recognizable for what it is. This video looks at graphing simple absolute value functions by hand. See y = 2|x| - 3.There's |x|.So draw y = 2x - 3at x > 0. This vertex is also the lowest point on the graph. Definition of the Absolute Value. Yes, they always intersect the vertical axis. 0. Save. He posts almost all his work as that user. 0. example. Graphing absolute value function given below. If we plot these points on the graph sheet, we will get a graph as given below. Have you ever tried to draw a picture of a rabbit, or cat, or animal? The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Next, a student will read the objective, "SWBAT graph absolute value functions on a coordinate plane". This leads to two different equations we can solve independently. Share practice link. There is an easy procedure you can follow to graph absolute value equations: Plot the value of x such that the expression inside the absolute value bars is 0. If we couldn’t observe the stretch of the function from the graphs, could we. These can be achieved by first starting with the parent absolute value function, then shifting the graph according to function transformations, flip graph if necessary and even may have to compress or decompress the graph. Learn vocabulary, terms, and more with flashcards, games, and other study tools. (c) The absolute value function intersects the horizontal axis at two points. Definition of percentage and definition of decimal, conversion of percentage to decimal, and... Robert Langlands: Celebrating the Mathematician Who Reinvented Math! Calculate |f(x)| and f(|x|): Quadratic Equation Note: Adding a positive number after the x inside the parentheses shifts the graph left, adding a negative (or subtracting) shifts the graph right. 0% average accuracy. DRAFT. a) (-9, 0) \quad\quad\quad b) (9, 0) \quad\quad\quad c) (0, -9) Play. Describe the Transformations using the correct terminology. Each coordinate pair represents the vertex of the graph of an absolute value function. Learn about the Life of Katherine Johnson, her education, her work, her notable contributions to... Graphical presentation of data is much easier to understand than numbers. Mathematics. 10th grade . The following steps will be useful to graph absolute value functions. 4 minutes ago. For example, |-7|=7 , |1|=1 ,and |-4|=4 ; the pattern is quite clear: the absolute value of a negative number is positive.In this section we consider the graph of f(x) and discuss how to graph |f(x)| and f(|x|) .. No, they do not always intersect the horizontal axis. Learn about the History of Fermat, his biography, his contributions to mathematics. This is an important function transformation. Each coordinate pair represents the vertex of the graph of an absolute value function. This blog deals with the three most common means, arithmetic mean, geometric mean and harmonic... How to convert units of Length, Area and Volume? Different types, Formulae, and Properties. These steps should be kept in mind in graphing absolute value function. Vertical and Horizontal Function Transformations. Because of this "minus", the positive values provided by the absolute-value bars will all be switched to negative values. Cuemath, a student-friendly mathematics and coding platform, conducts regular Online Live Classes for academics and skill-development, and their Mental Math App, on both iOS and Android, is a one-stop solution for kids to develop multiple skills. Note that we could graph this without t-charts by plotting the vertex, flipping the parent absolute value graph, and then going over (and back) 1 and down 6 for next points down, since the “slope” is 6 (3 times 2). Complete Guide: How to multiply two numbers using Abacus? Note: Adding a positive number after the x outside the parentheses shifts the graph up, adding a negative (or subtracting) shifts the graph down. I will ask the class to recall what we learned during our last class about absolute value functions. Learn about Operations and Algebraic Thinking for grade 3. Save. We have already understood in detail about the absolute function in the blog about the absolute value function. 10 Type your answer below and then sketch a "good enough" graph of h(x) -5 -10 -5 0 5 10 Submit … This blog explains how to solve geometry proofs and also provides a list of geometry proofs. When we look at the above graph, clearly the vertex is (0, 0), Step 2: Write the given absolute value function as $$y - k = |x - h|$$. Hence, graphing absolute value functions is an important topic which we will reduce to a step by step easy process. Write the given absolute value function in the form : To get the vertex, equate (x - 1) and (y + 2) to zero. The following steps will be useful in graphing absolute value functions. This blog deals with various shapes in real life. As well as this each function has a vertex, the point where the come together. Learn about the Conversion of Units of Speed, Acceleration, and Time. This blog talks about quadratic function, inverse of a quadratic function, quadratic parent... Euclidean Geometry : History, Axioms and Postulates. Graphing absolute value functions. Home Page Describe the Transformations using the correct terminology. Learn about real-life applications of fractions. by m_13539377_87591. Using pizza to solve math? There's |x|. The constant of the function will determine whether the graph will translate up or down. Below is the graphing of absolute value equationThe graph of $$y= |x|$$ has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. Highlights common student mistakes and points out how to … The first half of the video utilizes function translations of a base function to create the sketch of the … Graphing Absolute Value Functions. … The given absolute value function is in the form : To get the vertex, equate (x - 1) and y to zero. Absolute Value Graphs & Transformations • … World cup math. To graph an absolute value function, choose several values of x and find some ordered pairs. The 3 Means: Arithmetic Mean, Geometric Mean, Harmonic Mean. This is a step by step tutorial on how to graph functions with absolute value. Example 3. This page explains how to keep track of the "V" in the absolute-value graph, and how to draw the correct shape for the quadratic equation's parabola. Graphing Absolute Value Functions Graph the absolute value functions on Desmos and list the Domain, Range, Zeros, y-intercept, End Behavior, Max or Min, Increasing and Decreasing Intervals. g (x) = f (x) + k. When k > 0, the graph of g (x) translated k units up. Graphing Absolute Value of Functions DRAFT. Practice: Graph absolute value functions. (see graph below). Absolute value graphs review. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about graphing linear functions (from Section 2.1 ) and piecewise-defined functions (from Section 1.4). This section will focus on two particular types of transformations: vertical shifts and horizontal shifts. under the reflection in the y-axis. If we are unable to determine the stretch based on the width of the graph $$f(x)=a|x-3|-2$$, \begin{align}2&=a|1-3|-2\\4&=2a\\a&=2\end{align}. From this information we can write the equation. ©K 42U0X1G2C oKsutAa A ISto8f Etvw 8a pr nee LfL CC.H p QA3lElO 2rYiNg9het Psg irpe xs DeVryvhe Id c.y k 2M 0a Wd5el 9wPiwthr kI jn cfMiHnIi qt meU yA3lwgDejb krRa Z … Complete Guide: Learn how to count numbers using Abacus now! To get the vertex, equate (x + 1) and (y + 1) to zero. 1.07 Graphing Absolute Value Functions. Introduction + Guided Notes. Understand the Cuemath Fee structure and sign up for a free trial. 10th grade. The Great Mathematician: Hypatia of Alexandria. For more apps by Peter, look under the GeoGebra user "EDC in Maine". As a result, the graph of an absolute value equation will take on the shape of the letter V . W z FM waRdCeK LwziNtphM cIXn TfUifn miktNeG NAIlGggeNb UrwaW V10. The graph of an absolute value function will intersect the vertical axis when the input is zero. Figure 8. Mathematics. In a Quadratic, it is smooth and … If c is negative, the graph is shifted down. It is intended to follow a lesson on transformations of parent functions. Learn about the different polygons, their area and perimeter with Examples. Calculus: Integral with adjustable bounds. As a distance, absolute value is always nonnegative. Let us look at the most basic absolute value function graph, $y = |x|$ Most of the absolute value function graphs will have a somewhat similar shape, a V-like structure with a vertex. by m_13539377_87591. This is the Absolute Value Function: f(x) = |x| It is also sometimes written: abs(x) This is its graph: f(x) = |x| It makes a right angle at (0,0) It is an even function. So draw the image of the graph. This means that the corner point is located at(3,4) for this transformed function. Plot one x … So, the absolute value graph of the given absolute value function is. (b) The absolute value function intersects the horizontal axis at one point. See y = |x2 - 4|.This is y = |f(x)|.Then draw, y = f(x),y = x2 - 4.At the region below the x-axis,lightly draw the graph.Quadratic Function: Vertex Form. Edit. So, the absolute value graph of the given function is. j 8 3A 1l jl r ir 5iqg ZhQtbs N ErHeHsge sr OvReMdu. Practice. Flattening the curve is a strategy to slow down the spread of COVID-19. See y = |x|.There's |x|.Absolute ValueSo draw y = xat x > 0. We now take more complex absolute function examples. Free graph paper is available. Next lesson. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points. The graph of the absolute value function for real numbers. (the right side of x = 0). This Algebra video tutorial provides a basic introduction into graphing absolute value functions. Learn about Operations and Algebraic Thinking for Grade 4. Identify the effect on the graph of replacing f (x) by f (x) + k, k f (x), f (kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Absolute Value Functions. So draw y = x. at x > 0. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. To get the vertex, equate (x + 1) and (y - 3) to zero. Next, we turn our attention to graphing absolute value functions. In mathematics, the absolute value or modulus of a real number x, denoted | x |, is the non-negative value of x without regard to its sign.Namely, | x | = x if x is positive, and | x | = −x if x is negative (in which case −x is positive), and | 0 | = 0. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. Graphing Features; Functions; Absolute Values Team Desmos December 24, 2020 16:12. Preparing For USAMO? The same trick works when graphing absolute value equations. Absolute Value Function. See (Figure). It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see (Figure)). For every point on a number line, there is another point on the opposite side of 0 … Given the graph of an absolute value function, write the function in the form g(x) = a ⎜ __ 1 b (x − h)⎟ + k. Explain 3 Modeling with Absolute Value Functions Light travels in a straight line and can be modeled by a linear function. As a distance, absolute value is always positive. To get the vertex, equate x and (y - 4) to zero. This blog deals with similar polygons including similar quadrilaterals, similar rectangles, and... Operations and Algebraic Thinking Grade 3. The abacus is usually constructed of varied sorts of hardwoods and comes in varying sizes. Learn about Vedic Math, its History and Origin. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Hence, graphing absolute value functions is an important topic which we will reduce to a step by step easy process. Learn about Parallel Lines and Perpendicular lines. This blog deals with calculus puns, calculus jokes, calculus humor, and calc puns which can be... Operations and Algebraic Thinking Grade 4. So draw the images of the graphunder the reflectionin the x-axis,in the y-axis,and in the origin. Electrical parts, such as resistors and capacitors, come with specified … No, they do not always intersect the horizontal axis. This leads to two different equations we can solve independently. The... Do you like pizza? (a) The absolute value function does not intersect the horizontal axis. We have to do the following steps to graph an absolute value function. To get the vertex, equate (x - 4) and (y + 4) to zero. Explore 1 Graphing and Analyzing the Parent Absolute Value Function Absolute value, written as ⎜x⎟,represents the distance between xand 0 on a number line. No, they do not always intersect the horizontal axis. Calculus: Fundamental Theorem of Calculus q Worksheet by Kuta Software LLC Do the graphs of absolute value functions always intersect the vertical axis? (x > 0 and y > 0). Our mission is to provide a free, world-class education to anyone, anywhere. Graph the absolute value function given below. Describe the transformation from the Absolute Value Parent Function. When you have a function in the form y = |x + h| the graph will move h units to the left. The graph of f is given by reflecting on the x axis part of the graph of y = (x - 2) 2 - 4 for which y is negative. Instead, the width is equal to 1 times the vertical distance as shown in (Figure). Learn about the different uses and applications of Conics in real life. There are |x| and |y|.So draw the images of the graphunder the reflectionin the x-axis. When light is reflected off a mirror, it travels in a straight line in a different direction. Solve an absolute value equation using the following steps: Get the absolve value expression by itself. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. Vertical Translations . Follow. Because there is a negative sign in front of the absolute sign, we have to flip the curve over. What is the equation for G of X? Absolute Value Graph. Video transcript - [Instructor] Function G can be thought of as a stretched or compressed version of F of X is equal to the absolute value of X. To graph absolute value, you can type "abs" or use pipe brackets (near the top right corner of most keyboards). Played 0 times. 1. Graph, Domain and Range of Absolute Value Functions. This Purplemath lesson provides a quick overview of graphing absolute-value and quadratic functions. Let us look at what steps are to be taken while graphing absolute value functions. Learn about the History of Eratosthenes, his Early life, his Discoveries, Character, and his Death. The graph of an absolute value function will intersect the vertical axis when the input is zero. There's |f(x)|.Then draw the image of the graphthat is below the x-axisunder the reflection in the x-axis. To get the vertex, equate (x - 2) and (y + 2) to zero. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about graphing linear functions (from Section 2.1 ) and piecewise-defined functions (from Section 1.4). ©R O2U05132 L sK GultmaU ySaotf 2tsw Aaar ie q OL8L6Ch. More References and Links to Graphing, Graphs and Absolute Value Functions Graphing Functions Graphs of Basic Functions. The graph of an absolute value function will intersect the vertical axis when the input is zero. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. Ever wondered how soccer strategy includes maths? They both have vertical symmetry, this meaning that if the graph were to fold over on itself from the Y axis each side would line up perfectly. We can express the application of horizontal shifts this way: Note: given a function $$f(x)$$, and a constant a > 0, the function $$g(x) = f(x - a)$$ represents a horizontal shift of a unit to the right from $$f(x).$$ The function $$h(x) = f(x + a)$$ represents a horizontal shift of a unit to the left. however the vertex is not the same for each. Before graphing any absolute value function, first we have to graph the absolute value parent function: \begin{align}x &= - 3 \quad\rightarrow\quad y = |-3| = 3 \quad\rightarrow\quad (-3, 3)\\x &= - 2 \quad\rightarrow\quad y = |-2| = 2 \quad\rightarrow\quad (-3, 3)\\x &= - 1 \quad\rightarrow\quad y = |-1| = 1 \quad\rightarrow\quad (-3, 3)\\x &= 0 \quad\rightarrow\quad y = |0| = 0 \quad\rightarrow\quad (0, 0)\\x &= 1 \quad\rightarrow\quad y = |1| = 1 \quad\rightarrow\quad (1, 1)\\x &= 2 \quad\rightarrow\quad y = |2| = 2 \quad\rightarrow\quad (2, 2)\\x &= 3 \quad\rightarrow\quad y = |3| = 3 \quad\rightarrow\quad (3, 3)\end{align}. Complete Guide: Construction of Abacus and its Anatomy. Graphing Absolute Value Functions. Unless you are very talented, even the most common animals can be a bit of a challenge to draw accurately (or even recognizably!). Calculating the Area and Perimeter with... Charles Babbage | Great English Mathematician. Quadratic functions and absolute value functions have many similarities. Here are four equations of absolute value functions and three coordinate pairs. So you can see F of X is equal to the absolute value of X here in blue, and then G of X, not only … The Great Mathematician: Hypatia of Alexandria, was a famous astronomer and philosopher. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. No, they do not always intersect the horizontal axis. Inside the absolute-value bars of this function, I've got a quadratic. (the upper side of y = 0). Next lesson. Writing an Equation for an Absolute Value Function Given a Graph. Let us look at the most basic absolute value function graph. Most of the absolute value function graphs will have a somewhat similar shape, a V-like structure with a vertex. Example 3; Hyperbolic half plane 4; The … Next, we turn our attention to graphing absolute value functions. The y-intercept is at y = –4. The history of Ada Lovelace that you may not know? Translating the Absolute Value Parent Function How does the graph of h(x) =- |x| compare to the graph of y = |x|? To solve an equation such as we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. ... Absolute value worksheets Solve and graph functions - Integers (1.2 MiB, 1,254 hits) There are only two more options for a simple graph with absolute value, if the variable is multiplied by a constant greater than or lesser than zero. When you have a function in the form y = |x + h| the graph will move h units to the left. 1. This blog gives an understanding of cubic function, its properties, domain and range of cubic... How is math used in soccer? The graph of the parent absolute value function is a v-shaped graph with the vertex at the origin. One trick that can help even the most "artistically challenged" to create a clearly recognizable basic sketch is in nearly all "learn to draw" courses: start with basic shapes. Robert Langlands - The man who discovered that patterns in Prime Numbers can be connected to... Access Personalised Math learning through interactive worksheets, gamified concepts and grade-wise courses, Vertical and Horizontal shift in Modulus Function graph, Function Transformations in Graphing Absolute Value Equations, Solving Absolute value equation - Math Planet. Its Domain is the Real Numbers: Its Range is the Non-Negative Real Numbers: [0, +∞) Piecewise functions. This function is kind of the opposite of the first function (above), because there is a "minus" on the absolute-value expression on the right-hand side of the equation. f (x) = - | x + 2| + 3 In general, the graph of the absolute value function f (x) = a| x - h| + k is a "V" with vertex (h, k), slope m = a on the right side of the vertex (x > h) and slope m = - a on the left side of the vertex (x < h). We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. The graph of an absolute value function will intersect the vertical axis when the input is zero. This leads to two different equations we can solve independently. To get the vertex, equate (x - 4) and y to zero. Step 3: To get the vertex of the absolute value function above, equate (x - h) and (y - k) to zero, That is. If we have negative signs in front of absolute signs, we have to flip the curve over. Sometimes you will see multiple translations in one problem. Piecewise functions. According to the vertex, we have to shift the above graph. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. 0. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points. Step 1 : Before graphing any absolute value function, first we have to graph the parent function : y = |x| Its vertex is (0,0) Let us take some random values for x. x = - 3 -----> y = |-3| = 3 -----> (-3, 3) x = - 2 -----> y = |-2| = 2 -----> (-3, 3) There's |y|.So draw the image of the graphunder the reflection in the x-axis. I can confirm (by factoring) that the x-intercepts are at x = –1 and x = 4. Graphing Absolute Value of Functions. In this example, we have the exact same shape as the graph of y = |x| only the “v” shape is upside down now.. Based on the examples we’ve seen so far, there appears to be a pattern when it comes to graphing absolute value functions.. Scaling the Graph of the Absolute Value Function. This video looks at graphing simple absolute value functions by hand. The graph of an absolute value function will intersect the vertical axis when the input is zero. m_13539377_87591. The graph of f (x) = - a| x - h| + k is an upside-down "V" with vertex (h, k), slope m = - a for x > h and slope m = a for x < h. Using these steps one will be able to reach the absolute value graph that is required to solve the absolute value equations. To translate the absolute value function f (x) = | x | vertically, you can use the function . Solo Practice. To get the vertex, equate (x + 3) and (y - 3) to zero. Graph y = –| x + 2 |. If c is positive, the graph is shifted up. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Plot one x value less than that value. Learn about the Conversion of Units of Length, Area, and Volume. The word Abacus derived from the Greek word ‘abax’, which means ‘tabular form’. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. No, they do not always intersect the horizontal axis. Vertex Form of the Function. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. Explore 1 Graphing and Analyzing the Parent Absolute Value Function Absolute value, written as ⎜x⎟,represents the distance between xand 0 on a number line. This means that the corner point is located atfor this transformed function. This quiz is incomplete! Plot one x value less than that value. This tutorial demonstrates two methods for graphing linear absolute value functions. To translate the absolute value function f (x) = … Hence, the graph of the given absolute value function is. Type in any equation to get the solution, steps and graph (the right side of x = 0). Author: Peter Tierney-Fife. Let us move on to a major aspect of solving absolute value equations which is drawing the necessary graph, looking at the intercepts and vertex. Figure 8. Properties of the graph of these functions such as domain, range, x and y intercepts are also discussed. Step 1 : Before graphing any absolute value function, first we have to graph the parent function : Graphing Absolute Value Functions Graph the absolute value functions on Desmos and list the Domain, Range, Zeros, y-intercept, End Behavior, Max or Min, Increasing and Decreasing Intervals. 0 times. It includes three examples. Donate or volunteer … Learn about Euclidean Geometry, the different Axioms, and Postulates with Exercise Questions. This point is shown at the origin in [link]. I will also ask students to use the definition of absolute value to explain the shape of the absolute value function graph. 30 … The horizontal axis? Learn about the 7 Quadrilaterals, their properties. Learn concepts, practice example... What are Quadrilaterals? cm to m, km to miles, etc... with... Why you need to learn about Percentage to Decimals? When k < 0, the graph of g (x) translated k units down. A translation is shifting a graph around the Cartesian Coordinate Plane. a) (-9, 0) \quad\quad\quad b) (9, 0) \quad\quad\quad c) (0, -9) The graph of an absolute value function will intersect the vertical axis when the input is zero. Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units. In this example, we have the exact same shape as the graph of y = |x| only the “v” shape is upside down now.. Based on the examples we’ve seen so far, there appears to be a pattern when it comes to graphing absolute value functions.. Resistance of a Resistor. 4 minutes ago. This is the currently selected item. Are you going to pay extra for it? Different Types of Bar Plots and Line Graphs. Graphing |f(x)| and f(|x|) The absolute value of x written as |x| is the "length" of x which can only be positive. 158 Chapter 3 Graphing Linear Functions CCore ore CConceptoncept Vertex Form of an Absolute Value Function An absolute value function written in the form g(x) = a ∣ x − h ∣ + k, where a ≠ 0, is in vertex form.The vertex of the graph of g is (h, k). Edit. Here are some tips you might want to know. See |y| = 2|x| - 3.There are |x| and |y|.So draw y = 2x - 3at the quadrant I. Learn to Graph Absolute Value Functions in this video by Mario's Math Tutoring. To play this quiz, … The values of h and k depend upon the translations performed on the original function. Learn Polynomial Factorization. This page explains how to keep track of the "V" in the absolute-value graph, and how to draw the correct shape for the quadratic equation's parabola. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. We have a general equation that describes the vertex as at the point (h, k). Edit. Graphing Absolute Value Funtions And Inequalities - Displaying top 8 worksheets found for this concept.. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. Yes. Horizontal Shift . Any absolute value function can be written in vertex form, and its graph is Graphing Absolute Value of Functions DRAFT. Highlights common student mistakes and points out how to avoid them. We can express the application of vertical shifts this way: Note : For any function $$f(x)$$, the function $$g(x) = f(x) + c$$ has a graph that is the same as $$f(x),$$ shifted c units vertically. Absolute value functions will graph in the shape of a V. The point of the V is called the vertex of the function. To solve an equation such as 8 = | 2 x − 6 |, 8 = | 2 x − 6 |, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. Edit. Learn about the different applications and uses of solid shapes in real life. Absolute value graphs review. Play Live Live. Write the given absolute value function in the form. A free trial • … solve an absolute value functions and three coordinate pairs upper side x. Coordinate plane: 5 Examples and their solutions and comes in varying sizes Operations... Usually constructed of varied sorts of hardwoods and comes in varying sizes straight line in quadratic! Functions with absolute value function is absolute signs, we will learn how to them... However the vertex is at ( 1.5, –6.25 ) free online cool math lessons, math. Not intersect the horizontal axis, depending on how to graph an absolute value always. One point logic ; Matrices ; Percentages ; Ratios ; Vectors ; Discover Resources and three coordinate pairs inverse a., –6.25 ) his biography, graphing absolute value functions Discoveries, Character, and... Operations and Thinking. Step by step easy process Algebraic Thinking for Grade 3 to reach the absolute value graphs visit! As at the origin function f ( x - 2 ) to zero depending on how to an... Euclidean geometry, the graph of an absolute value of a quadratic function its! … solve an absolute value function, quadratic parent... Euclidean geometry: History, Axioms and Postulates about and... Km to miles, etc ) nonprofit organization waRdCeK LwziNtphM cIXn TfUifn miktNeG UrwaW. Sketch with simple circles, ellipses, etc and quadratic functions and coordinate... Function graph Maine '' graphing linear absolute value graphs, could we ’, which means ‘ tabular form.... Know more about absolute value function travels in a different direction varying.... Including similar quadrilaterals, similar rectangles, and in the y-axis, other. Is to provide a free trial has a vertex Funtions and Inequalities - Displaying top 8 worksheets for... With... Charles Babbage | Great English Mathematician way to do the following steps to graph an absolute functions! A quadratic function, its History and origin negative sign in front of the absolute equation! The process of graphing absolute value function for real numbers here: Abacus a. Function given a graph around the Cartesian coordinate plane Matrices ; Percentages ; ;! And perimeter with... Why you need to learn about Operations and Thinking... The same trick works when graphing absolute value function will intersect the horizontal axis at zero, one, two... Way to do the following steps to graph an absolute value equation Discoveries Character. 0, the graph of an absolute value function can be written in form! Plot one x … Start studying graphing absolute value equation using the following will! Graphing absolute value function can be written in vertex form, and... Operations and Algebraic Thinking 3. Will reduce to a step by step tutorial on how the graph sheet we... Provides a quick overview of graphing absolute value equation electrical parts, such as resistors capacitors... We have to do it John Napier | the originator of Logarithms in mind in graphing absolute value.... And y > 0 function does not intersect the horizontal axis explain the shape of the the! Print ; Share ; Edit ; Delete ; Report an issue ; Start a multiplayer game flip the over! & transformations • … solve an absolute value to explain the shape of the graphunder reflection. Cartesian coordinate plane: 5 Examples and their solutions Discover Resources 3at the quadrant i ( h, )... You have a general equation that describes the vertex is at ( )... Look at what steps are to be taken while graphing absolute value functions in this video by Mario math... The form y = 2x - 3at x > 0 bars will all be switched to negative values )... Online cool math lessons, cool math lessons, cool math lessons, cool math has free online cool lessons... Input is zero located at ( 1.5, –6.25 ) + 1 ) to zero negative sign front! You can use a formula to confirm that the x-intercepts are at x 0! Of Length, Area, and Volume to negative values, a V-like structure with a vertex equate. Hyperbolic half plane 4 ; the … this tutorial demonstrates two methods for linear. Discover Resources a distance, absolute value function for real numbers Speed,,... Lesson provides a quick overview of graphing absolute-value and quadratic functions, domain and of! Shifted and reflected all his work as that user using these steps one will be useful graph. Horizontal shifts couldn ’ t observe the stretch of the letter V value functions by.... Means: Arithmetic Mean, Harmonic Mean, function graph in the Desmos keyboard in... And uses of solid shapes in real life intended to follow a lesson on transformations of parent functions,., such as domain, range, x and y > 0 basic functions Guide learn. Intended to follow a lesson on transformations of parent functions have many similarities (. As its distance from zero on how the graph ofy=\|x\| has been shifted and reflected as this function.
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https://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex | # Uniformly distributed n-dimensional probability vectors over a simplex
What's the right way to generate a random probability vector $p={p_1,\ldots,p_n} \in {(0,1)}^n$ where $\sum_i p_i=1$, uniformly distributed over the $(n-1)$-dimensional simplex?
What I have is
Intervals = Table[{0, 1}, {i, n}]
RandomPoint := Block[{a},
a = RandomVariate[UniformDistribution[Intervals]];
a/Total[a]];
But I am unsure that this is correct. In particular, I'm unsure that it's any different from:
RandomPoint := Block[{a},
a = Table[Random[], {i, n}];
a/Total[a]];
And the latter clearly will not distribute vectors uniformly. Is the first code the right one?
• This question may be relevant. – Sjoerd C. de Vries Oct 8 '13 at 11:21
• Thanks, @SjoerdC.deVries. That question seems to suggest that my first code is also incorrect? I'm assuming that that bunch of smart guys would have stumbled upon it. – Schiphol Oct 8 '13 at 11:42
• Perhaps DirichletDistribution might help? – chuy Oct 8 '13 at 14:03
• That question involved points on a sphere. Your constraint of $\sum{p_i}=1$ is different. – Sjoerd C. de Vries Oct 8 '13 at 14:39
• Some folks here might find this StackOverflow duplicate useful (which is asking exactly the same question, but from a computer science perspective). You'll find the answer (to use the Dirichlet) is the same as well, but with a Python implementation. stackoverflow.com/questions/18659858/… – cgnorthcutt Feb 12 '18 at 18:22
#/Total[#,{2}]&@Log@RandomReal[{0,1},{m,n}] will give you a sample of m points from a uniform distribution over an n-1-dimensional regular simplex. (An equilateral triangle is a 2-dimensional regular simplex.) Here's what m = 2000, n = 3 should look like, where {x,y} = {p[[2]]-p[[1]], Sqrt@3*p[[3]]} are the barycentric coordinates of the 3-element probability vector p:
Here's what you get if you omit the Log@ and normalize Uniform(0,1) variables, which is what both of the OP's examples do:
• Thanks a lot. Could you please explain in what respects does this behave differently from RandomVariate[UniformDistribution[]]? – Schiphol Oct 9 '13 at 8:10
• See for yourself. Try it with n = 2 and make a histogram of p[[1]]. Or use n = 3 and ListPlot the barycentric coordinates: {x,y} = {p[[2]]-p[[1]],Sqrt@3*p[[3]]}. – Ray Koopman Oct 9 '13 at 18:41
• Yes, the difference is clear -- see my answer below. Actually, I meant for you to explain the difference in algorithmic terms, or perhaps provide pointers to a textbook explanation of why your method is doing what it's doing. – Schiphol Oct 11 '13 at 10:39
• I generate a Dirichlet distribution in which all the concentration parameters are 1. See the link that Jacob provided, then scroll down to this section and remember that the log of a Uniform(0,1) variable is proportional to a Gamma variable with shape parameter 1. – Ray Koopman Oct 11 '13 at 13:55
• You can also use Mathematica's built-in DirichletDistribution: points = RandomVariate[DirichletDistribution[{1, 1, 1}], 2000] /. v_?VectorQ :> {v[[2]] - v[[1]], Sqrt[3] (1 - Total[v])}; and then ListPlot[points]. – chuy Oct 11 '13 at 18:28
Old question, but I didn't see this method. Generates $n$ points uniformly randomly distributed on a simplex embedded in $d$ dimensions.
genSimplex[n_, d_] :=
Table[Differences[Sort[Flatten[{0, RandomReal[1, d – 1], 1}]]], {n}];
The algorithm generates points that are randomly distributed on an outer face of a simplex. The way to generate them is, for a d-dimensional problem…
1. Generate d-1 uniformly distributed random values in the range [0,1]
2. Add a 0 and a 1 to the list
3. Sort the list
4. Extract a list of the differences between the elements
You now have a list of random values that sum to 1 (so they are on a plane that is defined by points that sum to one) and that are otherwise independent of each other, so their dispersion is uniform.
Updating the answer with a picture of example data with 1,000 points.
This topic is well-covered here... https://stackoverflow.com/questions/3010837/sample-uniformly-at-random-from-an-n-dimensional-unit-simplex
Starting in M10.2, you can just use RandomPoint:
pts=RandomPoint[Simplex[{{0,0,1},{0,1,0},{1,0,0}}], 1000];
Graphics3D[Point[pts]]
• That's a slick capability. – MikeY Jun 7 '17 at 14:59 | 2019-12-10T07:15:59 | {
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http://b2bimpactdata.com/nhpcht7s/subset-symbol.html | # Subset symbol
Search Server Virtualization. We write B ⊆ A By definition, the empty set( { } or ∅ ) is a subset of every set Now, take a look at the following Venn diagrams. _____ Is a subset of mathematical symbol. The method you choose would depend on how you are using the degree symbol in your work. Subset. g. 228a ⊊ subset of with not equal to. This wikiHow teaches you how to place a symbol, such as the copyright symbol or the division sign, in a Microsoft Word document. Some symbols can be accessed only via function calls. 7. Our UI has a table of data, with the first row containing input fields in each column for searching based off the data in that column. 12Home > Accessing and Managing Symbols with armlink > Access symbols in another image > Outputting a subset of the global symbols6. 1; Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. Apart from the stuff "Subset of null set", let us know some other important stuff about subsets of a set. Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. Subset Definition. Otherwise, imagine someone has inquired whether set A is related to set B. How to Insert Symbols in an MS Word Document. 2289 ⊉ neither a superset of nor equal to. Download thousands of free photos on Freepik, the finder with more than 5 millions free graphic resources Use the icon on merchandise for sale (T-shirts, mugs etc. (If the window is too narrow, you see the Symbols button, from which you can choose Equation or Symbol. 27 Python representation of Symbols. , B is a super set of A. This results in a This selection of subsets is called a permutation when the order of selection is a and the number of such permutations possible is denoted by the symbol 5P2, 4 Aug 2019 If a set S is a subset of another set T, that is, S⊆T, and also: The symbol ⊊ is the usual form to use, but ⫋ is generally used on Pr∞fWiki after Some authors use the symbols ⊂ and ⊃ to indicate subset and superset respectively; that is, with the same meaning and union set union operator intersect set intersection operator minus set difference operator subset subset operator Calling Sequence Parameters Description 3 Mar 2008 So what is the minimum subset of BPMN that a process modeler flow was recorded as using the task symbol and the sequence flow symbol. So, if S is a subset of T, then S ⊆ T. We've documented and categorized hundreds of macros! Subset and Superset Subset. Flow area is used as a flow area. Symbolically, we write A ⊂ B. Population,” and “Education What's the meaning of the Proper Subset (also called a strict subset) » Proper Subset (also called a strict subset) This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Proper Subset (also called a strict subset). Ask Question as it is a symbol with fixed meaning and not a variable). \supseteq. Symbols that act on two This list is organized by symbol type and is intended to facilitate finding an unfamiliar symbol by its visual appearance. na(subset)]. Open Locate the required symbol by a simple clear-text search or perform a lookup by language or character subset. Mathematical and scientific symbols. To denote A is a subset of B the subset symbol ⊂ is used. To type a square root in Microsoft Word without using keyboard shortcuts, click the "Insert" button at the top of the screen. net dictionary. They are only interested in the subset/superset relationship. While randomly switching antenna subsets does not affect the symbol modulation for a desired receiver along the main direction, it effectively randomizes the amplitude and phase of the received symbol for an eavesdropper along a sidelobe. Info Read more in the commands section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes. Subsets: If A and B are any two sets such that every element of A is also an element of B then A is said to be subset of B and is written as A⊂B. GDB represents every variable, function and type as an entry in a symbol table. Name Unicode NOT A SUBSET OF: not a subset of : Nu: GREEK PHI SYMBOL: greek small letter phi: straightphi: Pi: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum. The symbol looks like the uppercase letters U and I placed close together in a sans-serif font , and rotated 90 degrees clockwise. A group of things or people, all of which are in a specified larger group. Some authors use the symbols ⊂ and ⊃ to indicate subset and superset respectively; that is, with the same meaning and Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number Common Symbols Used in Set Theory. The green circle is A, and the blue circle is B. Syntax --callgraph_subset=symbol[,symbol,…] where symbol is a comma-separated list of symbols. This also can be read as “A is contained in B”. List of check mark symbols, and emojis with their Unicode Hex values. My guess is that it's the symbol for "compactly contained," which would mean that the closure of the set (taken in Euclidean n-space) was a subset of U, and that Unicode Symbol Reference. The formal definition for a subset is: S ⊆ T ↔ ∀x(x∈S → x∈T) subset symbol: The subset symbol indicates a specific relationship between two sets . For example, if A =\{1,3,5\} then B=\{1,5\} is a proper subset of A. Justify your an 20 Jun 2013 The main idea in ASM is to modulate the radiation pattern at the symbol rate by driving only a subset of antennas in the array. List view. A proper subset is denoted by the subset symbol which looks like a U rotated ninety degrees to the right. Char U+2282, Encodings, HTML Entitys:⊂,⊂,⊂, UTF-8 (hex), Category: Math Symbol(Sm). If a set A is a subset of a set B and is not equal to B, then we call A a PROPER SUBSET of B, and write A ⊂ B. Figure 3: The Symbol dialog box; Options in this dialog box are explained, as marked in Figure 3, above. The symbol is sometimes read as “subset or equal to”, but in general, sets that are equal are subsets of each other. , a subset other than the set itself), this is written . It contains the zero vector. You either need to know the keyboard shortcut, or use the methods that are not very straightforward. In Word, you can insert mathematical symbols into equations or text by using the equation tools. With the release of Windows 10, the Segoe MDL2 Assets font replaced the Windows 8/8. More formally, A is a subset of B, denoted by A⊆B if, x∈A implies x∈B. Use your numeric keypad with your NUM LOCK on and you will be good to go! Symbol Description Shortcut ¶ paragraph sign ALT+0182 ± plus-or-minus sign ALT+ A subset which does not have all the elements of its superset is called a proper subset. Definition. w. what Im trying to achieve is like a commutative diagram but with with "subset" symbols instead of arrros. Font; By default, PowerPoint does not choose any particular font and any symbol you select will work with all fonts. ABOUT. This work is licensed to you under version 2 of the GNU General Public License. (Don't confuse this symbol with the letter “u. If you want to save yourselves all of these steps in the future, assign a Shortcut Key to the section symbol by clicking on the "Shortcut Key" button while the section symbol is highlighted in the list above. Subset relations form the foundation of mathematical logic, including Boolean algebra, which is important in the design of Subset - symbol description, layout, design and history from Symbols. Symbols are used to eliminate the need to write long, plain language instructions to describe calculations and other 16 Nov 2011 Hello, I've been looking all over the internet to find how to do the mathematical Subset symbol in the equation editor (this symbol: Any set with ALL the elements being a part of another set is called the subset of the latter and is The symbol – “⊂” stands for 'is a subset of' or 'is contained in. So for example, one of the ros may be A\subseteq B. \sqsubseteq. A is a subset of B if x ∈ A ⇒ x ∈ B. It is returned as the result of negating a subset expression. Then take the union of the corresponding states in the table at the left. See Example 4. I need the analogue of this but for columns. I have a vector containg about 20 unique values. Before we define subset, we need to refresh ourselves on what a set is. None. Superset is an antonym of subset. Set B is a subset of Set A if and only if ALL the elements in Set B is in Set A. Since a set is a well – defined collection of objects or elements grouped together within braces {}, it can also be disintegrated into smaller sets of its own called the subsets. That list also includes LaTeX and HTML markup, and Unicode code points for each symbol (note that this article doesn't have What is the not subset symbol [duplicate] Ask Question Asked 8 years, 1 month ago. 003F6 ϶ \backepsilon mathord amssymb wrisym GREEK REVERSED LUNATE EPSILON SYMBOL 02035 ‵ 8 \backprime mathord amssymb reverse prime, not superscripted 02102 ℂ C \mathbb{C} mathalpha mathbb = \mathds{C} (dsfont), open face C 0210C ℌ H \mathfrak{H} mathalpha eufrak /frak H, black-letter capital H The subset() function takes 3 arguments: the data frame you want subsetted, the rows corresponding to the condition by which you want it subsetted, and the columns you want returned. This is because P and C are equivalent sets (P = C). In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least Subset[x, y, ] displays as x \[Subset] y \[Subset] . A subset is a portion of a set. 0 might be better but I don't have your data to test with. NOTE: The Symbol dialog box does not close automatically when you insert a symbol. It’s just a table, which shows glyphs position to encoding system. Obviously if a technical term exists, great. Learn how to make over 100 Equal symbols of math, copy and paste text character. It is nearly impossible to meaningfully analyze all of your data at once. We want to add a dropdown that determines how that search text But it is not a proper subset. ⊂ subset of symbol, Subset Of Math Symbol Smiley Face facebook symbol, Subset Of Math Symbol Smiley Face twitter symbol, Subset Of Math Symbol Smiley Face Unicode Character U+2282, HTML Entity Hex ⊂, HTML Entity Decimal ⊂, free images, free icons, free pictures, free clipart, ALT+8834, Windows Special Characters, Windows ALT Codes, C, C++, Java "\u2282", Python u"\u2282" Figure 2: Click the Symbol button; This brings up the Symbol dialog box, that you can see in Figure 3, below. The objects or symbols are called elements of the set. 10/25/2019; 6 minutes to read; In this article. The symbol ‘ ⊂ ’ is used to denote proper subset. We use the subset() function. 2287 ⊇ superset of or equal to. That is, a subset can contain all the elements that are present in the set. I have "airports" data, data include variable "type" - small airport, large, heliports, etc. (Lists thousands of symbols and the corresponding L a T e X commands that produce them. It is notated by the symbol ⊆ which can be interpreted as "IS A PROPER SUBSET Or IS EQUAL TO". The flow reference symbol acts as a placeholder for the flow area sequence in the chart in every situation in which it is repeated. ctan. A PROPER subset is any subset of a set EXCEPT ITSELF. Info I've repaired Office and it didn't seem to make any difference but then I found out that the "from:" option in the lower right side of the Symbol panel should be "Unicode (hex)" in order to show the subset options. Share the icon nor its edited version Use the icon on merchandise for sale Check Mark Symbols. The letter Q designates the set of rational numbers. If A is not a subset of B, we write A ⊈ B. Viewed 71k times 23. Activity: Subsets . It is also always a proper subset of any set except itself. The symbols can be accessed via slots of Symbol (for example, Symbol::alpha for the symbol α) or via function calls (for example, Symbol("alpha")). A subset is a set whose elements are all members of another set. Proper-subset relations form the foundation of mathematical logic, including Boolean algebra, which is important in the design of computers and The symbol ⊆ is used to indicate a subset. Good evening, I have a strange problem. Subset definition, a set that is a part of a larger set. SYMBOL Characters and Glyphs up. MS WORD INSERT/SYMBOL/Subset by thefillanator | February 9, 2010 9:04 AM PST. ) Use the icon on mass distributed digital templates Use the icon as (part of) a logo Note: It is nice to attribute the author, but not mandatory for this license type. Symbols save time and space when writing. Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. Answer to Determine whether the following statements are true or false (in (b), the symbol means 'proper subset'). Definition of Venn Diagrams: Get the complete details on Unicode character U+2282 on FileFormat. About Segoe MDL2 Assets. The R program (as a text file) for all the code on this page. Uppercase. Subsetting is a very important component of data management and there are several ways that one can subset data in R. A "proper subset" of a set A is simply a set which contains some but not all of the objects in A. See more. In the previous section, we selected an entire column without condition. Because, { } = { } Therefore, A set which contains only one subset is called null set. The function _notsubset exists for typesetting purposes. In symbol we write x ⊆ y Proper subset definition: a subset that excludes at least one member of the containing set | Meaning, pronunciation, translations and examples To create the X bar symbol in Microsoft Word, the letter x should be used in combination with the special bar character. HTML Arrows is shared by Toptal Designers, the marketplace for hiring elite UI, UX, and Visual designers, along with top developer and finance talent. Definition Symbol-free definition. {xlx is a person living in Massachusetts}___ {yly is a person living in a New England state} Set theory - Set theory - Equivalent sets: Cantorian set theory is founded on the principles of extension and abstraction, described above. latex_name – (default: None ) string; LaTeX symbol to denote the subset; . In the course of business, you may accumulate vast amounts of varied data in Excel spreadsheets. \sqsubset. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set Set Symbols. You can put them in Facebook, Youtube or Instagram. I suggest Alt + S, since it isn't assigned to anything by Proper Subset Calculator. Commonly used mathematical symbols, such as > and < Greek Letters. This is a simple online calculator to identify the number of proper subsets can be formed with a given set of values. There are two types of subsets, proper subsets and normal subsets. org>∗ 19 January 2017 Abstract This document lists 14283 symbols and the corresponding LATEX commands that produce them. 2283 ⊃ superset of. (The notation is generally not used, since automatically means that and cannot be the same. A is said to be proper subset if A does not equal B. Please read Introduction to Sets first! This activity investigates how many subsets a set has. Example-1. The symbol font uses Adobe Symbol encoding so, for example, a lower case mu can be obtained either by the special symbol mu or by symbol("m"). This clip art is high resolution, png format and very popular on the public internet. I want to know what a mathematical symbol is? (set theory)? you know the symbol for "is a subset of" with the sideways U and the line underneath it? Well what does it mean when instead of that line there's a crossed out equals sign underneath it? List of Equal symbols with html entity, unicode number code. is a subset of (written ) iff every member of is a member of . RELATIONSHIP OF A SETS. In other words, the set A is contained inside the set B. Explore SCVMM features for VM deployment and management. A = (2,4,6,8) B = (2,4,8) Set B is a subset of Set A Do you want to add an infinity symbol on your slide? Or do you want to add one of the mathematical symbols? Or even the Yen or Rupee symbols? PowerPoint provides several ways to add such symbols, but the most straightforward option is to use the Symbol dialog box. Proper subset, superset, Venn diagrams A subset is a set of vectors. We use the symbol ⊆ to say a set is a subset of another set. A way of modifying a set by removing the elements belonging to another set. 03D2 ϒ GREEK UPSILON WITH HOOK SYMBOL. a set that is part of a larger set Not to be Subsets and Proper Subsets If every member of set A is also a member of set B, then A is a subset of B, we write A ⊆ B. We will look at the following: universal sets, subsets, equal sets and disjoint sets. The \cup and \cap sublos wont do, for they dont have a slash next to them, indicating the The conclusion would be that N is a subset of Z. Problem fixed, thanks! Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Define subset. Examples – Select the "Section" symbol by clicking on it; Click on the "Insert" button. This provides access to symbols that have no special symbol name, for example, the universal, or forall, symbol is symbol("\042"). 23. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ) Click the Symbol button see some popular or recently used symbols If you like keyboard shortcuts and use special characters, here are a few more for Microsoft Windows. My question is - whether the symbol we use to show subset is same as that for proper subset? I have seen that A is subset of A itself is shown by the same symbol as that for the subset while when we write improper subset, we usually write equal to sign with symbol of subset. External Links. List of all mathematical symbols and signs - meaning and examples. Here are the A ⊆ B, Subset: A has some (or all) elements of B, {3,4,5} ⊆ D. Even if you know the ASCII or Unicode number for the character, you can’t type it in as you can in Office for Windows. a formula: Wikimedia uses a subset Subset is an antonym of superset. Using this symbol we can express subsets as follows: A ⊆ B; which means Set A is a subset of Set B. IMPROPER SUBSET. What is a Proper Subset? Below is the complete list of Windows ALT codes for Math Symbols: Subset & Superset Relations, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. E-ER diagram –shows specialization circle (IsA relationship), and inheritance symbol (subset symbol) Specialization can also involve just one subclass – no need for circle, but show inheritance symbol The sub-entities are most likely invoking the disjointedness constraint Important APIs: Symbol enum, FontIcon class. A subset B represents the expression . This page aims to give a fairly exhaustive list of the ways in which it is possible to subset a data set in R. Definition with symbols. The elements of B are even, so I need to pick out the elements of A which are even; these will be the elements of the subset B. We can list each element (or "member") of a set inside curly brackets like this: Common Symbols Used in Set Theory proper subset symbol: The proper subset symbol indicates a specific relationship between two set s. The select argument exists only for the methods for data frames and Each circle or ellipse represents a category. You can copy & paste check mark symbols anywhere you like, or you can use their Unicode Hex values on your web page design, or computer programing. If is not a subset of , this is written . " 66 . If the text argument to one of the text-drawing functions (text, mtext, axis, legend) in R is an expression, the argument is interpreted as a mathematical expression and the output will be formatted according to TeX-like rules. From here, you can basically navigate to the checkmark symbol we used before by selecting the Wingdings 2 font and finding the checkmark symbol. But computer can understand binary code only. For example, if A Useful Mathematical Symbols Symbol What it is How it is read How it is used Sample expression + Subset symbol is a subset of Sets A B Sets Use keyboard shortcuts. Bidi The standard way to prove "A is a subset of B" is to prove "if x is in A then x is in B". 2. Q&A for Work. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wra Subset Of Above Not Equal Mathematical Symbol Comments - Mathematics Clipart is high quality free transparent clipart, which is handpicked by SeekClipart. For example; 1. The set {a, b} is a both a subset and a proper subset of {a, b, c} while the set {a, b, c} is a subset of {a, b, c} but not a proper subset of {a, b, c}. Set : It is a collection of distinct objects. In the Set section of the Symbol screen choose the Typographic Symbols and a list of symbols will display. The way that I insert a degree symbol in my version of Word (2002) is: Insert >> Symbol Font: (normal text) Subset: Latin-1 The degree symbol is the right-most symbol on the seventh line. Symbolically this is represented as A ⊆ D. ) The Comprehensive LATEX Symbol List Scott Pakin <scott+clsl@pakin. If is a proper subset of (i. . Next, click "symbol" and "more symbols. We can write it symbolically as A ⊂ B. In R the command “subset” is used to filter the data in a data frame based on the criteria you set. Subsets Recall that a set is a collection of elements. " A set, B, is not a subset set of a set, C, if one of the elements in B is not in C. Below is the complete list of Windows ALT codes for Math Symbols: Subset & Superset Relations, their corresponding HTML entity numeric character references, Subset - Tex Command - \Subset - Used to create Subset symbol. It is a factor I have a data frame with about 100000 rows. We asked a subset of the population of the town for their opinion. For example, if you enter: \sqrt. 228e ⊎ multiset The first technique selects an antenna subset randomly for every symbol. Letter-Like Symbols. A proper subset of a set A is a subset of A that is not equal to A. ”) This is a two-circle Venn diagram. are symbols for reusable sequences, such as logging in with a specific user id and password to enter the course or to initiate an on-line quiz. 5. That is, every element x of ∅ {\displaystyle \varnothing } belongs to A . 1 is in B. Therefore A is a subset of B. An "Improper Subset" is a subset which can be equal to the original set. Lowercase. Proper subsets are denoted using the symbol For example, the set {a, b} is a proper subset of the set {a, b, c}: An "improper subset" is a subset which can be equal to the original set; it is notated by the symbol. Meaning of subset. You can insert a special character or symbol in your LaTeX/Mathematics. NEW: The convenient auto-lookup feature suggests suitable symbols based on your selection! You may search for a symbol manually as well: by entering a simple letter (e. Don't forget to subscribe. Microsoft Word: . Note 3: For all coloring, the color will apply only to the text immediately following the command until the next space is encountered. The special bar character is found in the section of symbols which is accessed through the insert tab in the editing feature on Word. In our case, we take a subset of education where “Region” is equal to 2 and then we select the “State,” “Minor. \subset - Tex Command - \subset - Used to create subset symbol. How to type symbols, accents, special characters, and weird punctuation Jika A adalah sebuah subset dari B, tetapi A tidak sama dengan B (yaitu ada paling sedikit satu elemen B yang bukan elemen dari A), maka A juga merupakan suatu subset wajar (proper subset atau strict subset) dari B; ini ditulis: ⊊. 16 Feb 2019 List of LaTeX mathematical symbols All the predefined mathematical symbols from the TeX package are listed \subset, is proper subset of. When you add values, the initial symbol displayed is based on the default event symbol and the color ramp selected on the Color Ramp drop-down menu. The Comprehensive LATEX Symbol List Scott Pakin <pakin@uiuc. On the Symbol dialog box, select the font from which you want to select a symbol from the Font drop-down list. You can change their names, and they are sorted by use. Operators. Subsets in Math: Definition & Examples Video. Click on one of them to start using it. To prove A is NOT a subset of B is easier- you just need a counter example- find one member of A that is not in B. This is a list of your collections. " When a new window pops up, click the "subset" drop-down menu on the right and hit "number forms. a), Proper Subset. The name How to insert the “X is a subset of Y, but is not equal to Y” symbol? [duplicate] How can I insert the "X is a subset of Y, but is not equal to Y" symbol? Technically, the math symbol is NOT equivalent to and is NOT interchangeable with (Notice the equal sign at the bottom edge of the symbol is missing. SCVMM can help admins better manage their virtualization stacks by providing VM networking diagrams and templates, but there are Outlook: insert symbols of degree, trademark, emotions, and euro. MsgBox(0, Date Of Birth, 1st January 90) But I want it to be like this 1st January 90 Symbol font should not be used in Web pages. See also Symbol provides access to typesetting symbols. 228b ⊋ superset of with not equal to. HTML Arrows offers all the html symbol codes you need to simplify your site design. Population,” and “Education alef symbol is NOT the same as hebrew letter alef, U+05D0 although the same glyph could be used to depict both characters: Arrows subset of, U+2282 ISOtech A set A is a subset of another set B if all elements of the set A are elements of the set B. This page is not a demonstration of how to use Symbol font; it provides a warning of the problems that it causes, and shows how to use Unicode instead. ) Set theory was developed to explain about collections of objects, in Maths. set A is included in set B. Download thousands of free photos on Freepik, the finder with more than 4 millions free graphic resources A Proper Subset is when set A is a subset of set B but they are not equal sets. We can say A is contained in B. The symbol looks like the uppercase letter U in a sans-serif font , rotated 90 degrees clockwise. s. Note that subset will be evaluated in the data frame, so columns can be referred to (by name) as variables in the expression (see the examples). Q is equal to the set whose elements are a, b, c. And there are a lot of symbols that many people need to insert regularly, such as the degree symbol, cent symbol, delta The subset() function takes 3 arguments: the data frame you want subsetted, the rows corresponding to the condition by which you want it subsetted, and the columns you want returned. subset(x, condition) arguments: - x: data frame used to perform the subset - condition: define the conditional statement Use the icon on merchandise for sale (T-shirts, mugs etc. Information on this page was taken from these sources and may include additional informaiton not available on this page. On the far right of the Word 2016 Insert tab dwells the Symbols group. Specifying Symbol font is contrary to the published specifications, has never been a documented feature of HTML and is not reliable. You can also learn how to type them in Ms Word or Ms Excel. I'm trying to subset (in new object) ONLY large airports and Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. So the result of symbol a from {0,1} will be the set containing 0 and 1, along with 3, or {0,1,3}. Encoding takes symbol from table, and tells font what should be painted. How to use subset in a sentence. " A list of symbols will appear, one of which is the square root symbol. 2284 ⊄ not a subset of. For data frames, the subset argument works on the rows. If each element in a set A is also a member of a set B, then set A is called a subset of B. Symbols that resemble letters. e. It is like you can choose ice cream from the following flavors: {banana, chocolate, vanilla} You could choose any one flavor {banana}, {chocolate}, or {vanilla}, Subset. Common Binary Operators. BUILT-IN SYMBOL. How to insert the mathematical and other symbols into the PowerPoint slide. B C is read "B is a subset of C. This makes sure that the file can be viewed and printed as it was created by the designer. The Know the LaTeX command you want to use but can't remember how to write it? Here we present a great tips sheet produced by Dave Richeson; it's pre-loaded in Overleaf so you can see how the commands work instantly. It is possible to subset based on whether or not a certain condition was true. Table of character entity references in HTML 4 (Jukka Korpela) Windows - Alt Key Numeric Codes of All including Math, Calculus , Language,Accents ,Foreign codes [ up to mouse 50 scrolls content ] Using the Codes Windows assigns a numeric code to different accented letters, other foreign characters and special mathematical symbol Definition of subset in the Definitions. The power set must be larger than the original set and is closely related to the binomial theorem. Example. The subset relationship is denoted as A \subset B. As you can see above, a subset is a set which is entirely contained within another set. The On-Line Encyclopedia of Integer Sequences® (OEIS®) Enter a sequence, word, or sequence number: Hints Welcome Video tug. A set, B, is a subset set of a set, C, if all the elements in B are also in C. A subgroup of a group is termed proper if is not the whole of . For ordinary vectors, the result is simply x[subset & !is. The symbol for “not a subset of” is . no no. The more unusual symbols are not defined in base LATEX (NFSS) and require . \subset - Contained in , you should better use the Symbol dialog box or Alt+code The math symbol is equivalent to and is interchangeable with (notice the equal sign at the bottom edge of the symbol is crossed out, indicating the subset cannot be equal to the set). A set is a well defined group of objects or symbols. Search. Note: The empty set is a subset of every set. PDBCopy is a command-line tool that creates a stripped symbol file from a full symbol file. These characters start with character code 0300 and continue for as many as the selected font contains. a set that is part of a larger set Not to be confused with: subtext – underlying or implicit meaning, as of a literary work: What is the subtext of the Using PDBCopy. Two items are found in that group: Equation and Symbol. Usage The callgraph file: Is saved in the same directory as the generated image. Alternatively, it and the others related to it can by inserted via Insert|Symbol, choosing the MS Gothic font and, from that, the Mathematical Operators sub-set. seed When it comes to inserting symbols in Excel, things can get a bit complicated. A set is a collection of things, usually numbers. This indicates the subset cannot be equal to the set). In the dialogue box, select “Latin-1 Supplement” from Subset and select the degree symbol from all the symbols. The union of two sets is represented by ∪. Subsets synonyms, Subsets pronunciation, Subsets translation, English dictionary definition of Subsets. This is written A ⊂ B. It is defined as a subset which contains only the values which are contained in the main set, and atleast one value less than the main set. Also Z^* = Z-{0}. To create the "section" symbol (§) in Word documents that adhere to the BNC guidelines, follow these steps: The way that I insert a degree symbol in my version of Word (2002) is: Insert >> Symbol Font: (normal text) Subset: Latin-1 The degree symbol is the right-most symbol on the seventh line. Unicode is a computing standard for the consistent encoding symbols. List of Mathematical Symbols R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Symbol Symbol Name A is a subset of B. Symbol set. Subset definition is - a set each of whose elements is an element of an inclusive set. 2286 ⊆ subset of or equal to. You can jump to a group of symbols by selecting an option from the Subset drop-down list. Click on the "section" symbol (§), then click on Insert and Close. See Also. A subgroup of a group is said to be proper if it does not equal the whole group, or equivalently, if as a subset, it is a proper subset of the whole group. Instructions on how to type Section Symbol for Windows, Mac, and in HTML. All available symbols are shown below, sorted in appropriate groups. It is also always a proper subset of any set except itself. id, this guide is for you. Share the icon nor its edited version Use the icon on merchandise for sale Using this symbol, we can write the definition of subset as follows: "->" which means implies. subset synonyms, subset pronunciation, subset translation, English dictionary definition of subset. ⊂ 2282 subset, included in, proper subset ⊃ 2283 superset, includes, proper superset ⊆ 2286 subset of or equal to ⊇ 2287 superset or equal to ⊄ 2284 notin set ∩ 2229 intersection ∪ 222A union ∈ 2208 isin, isinv, Element, in, element of (large symbol) ∊ 220A element of (small symbol) ∉ 2209 notin, NotElement, notinva, not Is a subset of mathematical symbol. Subtraction of sets is indicated by either of the symbols – or \. (all the pages in this section need a unicode font installed - e. Originally called Code 128 EAN/UCC this variation is now known as 128 GS1. The set containing 1 and 3 only is a proper subset of the set of natural numbers. A set A is a subset of a set B if every element in A is also in B . Basically, the definition states it is a collection of elements. Simply click on the button above to open a version in Overleaf for editing (and to The procedure described in this section enables you to create a subset database, after which you can perform other tasks, such as editing the properties of the subset definition or exporting a subset definition. A proper subset of a set , denoted , is a subset that is strictly contained in and so necessarily excludes at least one member of . When you are answering questions in a player, you can cut, copy, and paste using the keyboard. What is a Subset? A subset is a set contained in another set. Related icons include Is a subset of mathematical symbol. This process has to terminate because there are only finitely many subsets. Active 7 years, 6 months ago. If you get as far as choosing to show Unicode character codes, then the Subset box in the upper right should allow you to select the Combining Diacritical Marks subset. See Examining the Symbol Table. com \subset: is a proper subset of The Comprehensive L a T e X Symbol List, 2017. The set could be mathematically described as: Proper Subset. The name Hi arikab, The reflexsubset (subset of or equal to) symbol can be created by typing 2286Alt-X. A set X is a subset of set Y if every element of X is also an element of Y. Venn diagrams can also demonstrate "disjoint" sets. Do this by right-clicking the symbol for each category to modify its symbol properties or to choose another symbol. The subset symbol ⊂ stands for ‘is a subset of’ or ‘is contained in’. A subset is a special, funny set. In this tutorial, I will show you these easy ways to do it (including a keyboard shortcut). Table of set theory symbols Symbol Symbol Name Meaning / definition Example { } set a collection of elements A = {3,7,9,14}, B = {9,14,28} 8834 ⊂ subset of 8835 ⊃ superset of ⊄ not a subset of ⊆ subset of or equal to 8839 ⊇ superset of or equal to 8853 ⊕ circled plus = direct sum 8855 ⊗circled times = vector product ⊥ orthogonal to = perpendicular 178 Superscript two ² 179 Superscript three ³ 0215 × 0247 ÷ 0135 ‡ 0177 ± More at: The circle is another symbol for IsA. Using Unique values, many fields to display categories I'm looking for a word to describe the relationship between two sets when set A is neither a subset nor superset of set B. Users are encouraged to submit more relevant free clipart work and manually reviewed by the SeekClipart team. Symbol-free definition. We can also use ⊂ if it is a proper subset. In mathematics we define the subset as a way of showing that all the elements of some set A are contained within some other set B. Set A is a subset of set B if all of the elements (if any) of set A are contained in set B. If every element in one set is included in another set, they are called subsets. The interface also allows you to perform inline, or at the source, masking while creating the subset definition. You can also use the Add Values button to add a unique symbol for a subset of the unique values from the selected field. For example, {1,2,3} ⊊{1,2,3,4}. Note: A subset can be equal to the set. Choose between 10569 not subset symbol icons in both vector SVG and PNG format. At each stage, you take each element of the DFA subset at the right, say it is {0,1}. Uppercase letters from the Greek alphabet. How to Subset Data in R. 1 Segoe UI Symbol icon font. See Code 128 GS1 for a discussion of Code 128 GS1 but read this page first. Lowercase letters from the Greek alphabet. In each of the following, the two images show the symbol in display mode, then in inline mode. share | improve this answer. The symbols ⊃ ⊇ are opposite - they tell us the second element is a (proper) subset of the first. Ways to type telephone symbols, it's unicode entities and more. There are two mechanisms to include fonts in a PDF: PROPER SUBSET. For example, consider a set. We can also say B ⊇ A, B is a superset of A, B includes A, or B contains A. scatter x=day y=othvar / group=group_id markerattrs=(symbol=circlefilled size=7) transparency=0. To describe some results based upon these principles, the notion of equivalence of sets will be defined. Identify the symbology you want to use to display your categories. (Segoe UI Symbol will still be available as a "legacy" resource, but we recommend updating your app to use the new Segoe MDL2 Assets. \subset. Hi, I have a problem with MS/Word 2003 running Windows Home/XP-SP3. The symbol "⊂" means "is a proper subset of". 2 Outputting a subset of the global symbols You can use a symdefs file to output a subset of the global symbols to another application. ) subset Sentence Examples. Teams. In a set theory, a subset is denoted by the symbol ⊆ and read as ‘is a subset of’. Subset a Data Frame. edu>∗ 8 October 2002 Abstract This document lists 2590 symbols and the corresponding LATEX commands that produce them. {1,3} ⊂ {1,3,5} In some examples both the subset and proper subset symbols can be used. To students of Elementary Computer Mathematics: Since writing the online textbook "Elementary Computer Mathematics" in 2002, I have made it available for free on the Internet. These elements could be numbers, alphabets, variables, etc. The symbol "⊆" means "is a subset of". Basic Math. By the definition of subset, the empty set is a subset of any set A. Selected LaTeX commands for math symbols and special typographic subset, \ subset. If A is not a subset of B we Inserting Degree Symbol in Excel. Select the desired symbol by clicking on it and then click Insert. Creates a file containing a static callgraph for one or more specified symbols. From Wikibooks, open books for an open world simplify the search for the command for a specific symbol. For example, let a set be the numbers {1, 2, 3}. But you can expand it to see other character sets: Scroll to the top and click the toggle button in the right corner: Word 2010 lets you sprinkle characters beyond the keyboard’s 26 letters of the alphabet, numbers, a smattering of symbols, and punctuation thingies. For example, A minus B can be written either A – B or A \ B. See also. When I want to insert a symbol, I seem to have You can put this solution on YOUR website! Rewrite the following using mathematical symbols: A. The set B is a subset of A, so it contains only things that are in A. A is a subset of B but B ⊇ A i. The set {x: x is a prime number greater than 10} is a proper subset of {x: x is an odd number greater than 10} The set of natural numbers is a proper subset of the set of rational numbers; likewise, the set of points in a line segment is a proper subset of the set of points in a line. This page will show you how to subset data in R. org Non-Confidential PDF version100070_0612_00_en Arm® Compiler armlink User GuideVersion 6. If you are prone to forgetting ~ is called tilde, are wondering why there are so many %s in your strings, or have a hard time googling what the (+) symbol in where users. ˆ= proper subset (not the whole thing) =subset [R] Subset by using multiple values. set. atau secara ekuivalen Then plot with different symbol size option and denser transparency setting. We will introduce simple sets, subsets, multi-dimensional sets, singleton sets feature of GAMS, and a section about Domain Defining Symbol Declarations. 1 Jul 2018 Download the royalty-free vector "Is a subset of symbol icon vector sign and symbol isolated on white background, Is a subset of symbol logo The class ManifoldSubset implements generic subsets of a topological manifold. So for instance, if you start with the set {Green Eggs, Ham, Cheese}, {Ham, Cheese} is a proper subset, but {Green Eggs, Ham, Cheese} is NOT a proper subset. Symbol: To show that set A is not a subset of set B, one must find at least one element of set A that is not an element of set B. n is the number of elements in set[]. Inserting math subset and superset symbols into office word. In the end, click on Insert and then close. If A (subset symbol) and (Subset symbol) C, what can you conclude? Why? What if A (proper subset) and (proper subset) - Answered by a verified Tutor Also missing is the important ‘subset’ pull-down list. By default, the Symbol dialog box shows emoji at first when it's opened. Mathematically, a set A is referred to as the subset of another set B, if every element of set A is also an element of set B. Proper Subset: If A and B are two sets, then A is called the proper subset of B if A ⊆ B but B ⊇ A i. What is the difference between Embedded fonts and Subset Embedded fonts? By preference any fonts that are used in a layout are also included in the PDF file itself. Subset of a set. For instance, every set in a Venn diagram is a subset of that diagram's universe. Selected LaTeX commands for math symbols and special typographic characters. 2288 ⊈ neither a subset of nor equal to. The below will show you how subset works and provides some subset examples. Universal Sets Learn about universal sets. set of integers, \mathbb{Z}, requires the amsfonts 1 Mar 2014 Remember to change the C to Normal Text (it must not appear in italic, as it is a symbol with fixed meaning and not a variable). A proper subset of a set A is a subset of A that is not equal to A. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. , A ≠ B. For a related list organized by mathematical topic, see List of mathematical symbols by subject. If Word for Mac supported the full range of characters, the subset feature would let you jump to a group of characters. Examples: The symbol "⊂" means "is a proper subset of". Set B is a subset of a set A if and only if every object of B is also an object of A. In the graphic below, A and B are disjoint: A set A is said to be a subset of a set B if every element of A is also an element of B. I can't copy paste the symbol here, but it looks like the wiki version except without a line underneath. 1. A set A is said to be a subset of a set B if every element of A is also an element of B. We've documented and categorized hundreds of macros! 2282 ⊂ subset of. Example Since all of the members of set A are members of set D, A is a subset of D. You also can use keyboard shortcuts to enter special characters and symbols. Common pronunciations (in British English - Gimson,1981) of mathematical and scientific symbols are given in the list below. The symbol ⊆ is used to indicate a subset. In example 5, you can see that G is a proper subset of C, In fact, every subset listed in example 5 is a proper subset of C, except P. Download thousands of free photos on Freepik, the finder with more than 5 millions free graphic resources. The numbers in A that are even are 2, 4, and 6, so: The subset (or powerset) of any set S is written as P(S), P(S), P(S),P(S) or 2S. Some mathematicians use the symbol to denote a subset and the symbol to denote a proper subset, with the definition for proper subsets as follows: When you are new to programming in SQL, you will come across a lot of hard-to-search-for character operators. Only one of those matches could be read as having a plural form of subset, "The largest subset are counties with large numbers of racial or ethnic populations". Then and are proper subsets, while and are not. More commonly, the empty set symbol Ø is used to show the empty set The set of integers is a subset of the set of reals. So, encoding is used number 1 or 0 to represent characters. It was created in 1991. You can also copy paste in it in other cell or even you can insert in a formula as well. Find how to type phone signs directly from your keyboard. In other words, it takes a symbol file that contains both private symbol data and a public symbol table, and creates a copy of that file that contains only the public symbol table. Q = {a, b, c} B. What does subset mean? Information and translations of subset in the most comprehensive dictionary definitions resource on the web. The symbol is sometimes read as “subset or equal to”, but in general, sets that 17 Jan 2012 This means Ω is a proper subset of T. the square root symbol appears: In this video I will explain the difference between a Subset vs a Proper Subset. The empty set is therefore a proper subset of any nonempty set. Alternatively, you may choose to receive this work under any other license that grants the right to use, copy, modify, and/or distribute the work, as long as that license imposes the restriction that derivative works have to grant the same rights and impose the same restriction. group_id(+) = group. 03D5 ϕ . In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B. "We read the above statement as "A is a subset of B if a is an element of A implies that a is also an element of B". If A is not a subset of B, we write A t B. There is a large subset of romantic cards specially dedicated to marriage proposals and wedding imagery complete with wedding scenes, Wikipedia lists the symbol for proper subset as ⊊ but my book uses a different one. empty set, \emptyset. In sets the symbol '⊂' denotes subset. In context|set theory|lang=en terms the difference between superset and subset is that superset is (set theory) (symbol: '''') with respect to another set, a set such that each of the elements of the other set is also an element of the set while subset is (set theory) with respect to another set, a set such that each of its Start studying Sets and Subsets. Code 128 Code Sets A, B, C Three different code sets are defined for Code 128 (dubbed A, B, and C) that determine how the code is interpreted by the barcode scanner. Set Subtraction. Assume a subset $V \in \Re^n$, this subset can be called a subspace if it satisfies 3 conditions: 1. \dashv lin. Symbols are a communication tool. SubsetEqual · SquareSubset · Element · Precedes · LeftTriangle · NotSubset. \subseteq. entity file. For example, Word provides foreign language letters and symbols — all sorts of fun stuff. Arial Unicode MS, Doulos SIL Unicode, Lucida Sans Unicode - see: The International Phonetic Alphabet in Unicode This site is supported by donations to The OEIS Foundation. Determine whether Subset symbol, the Proper subset symbol, both or neither can be placed in the blank to form a true statement. In both these cases, subset is singular. It is called rejectrs\$rs. In order to have the color apply to more characters, place the text you want in color in curly brackets. You can do this in Microsoft Word for both Windows and Mac. While this is the most intuitive way to insert a checkmark symbol, it does take a little bit more time to leverage, especially if it’s not one of the most recent symbols you have used. Math Symbol: Block Mathematical Operators: HTML Entity (Named) ⊇ superset of or equal to U+2287 ⊈ neither a subset of nor equal to U+2288 Unicode Data. Symbol object. Opposite The Comprehensive LATEX Symbol List Scott Pakin <pakin@uiuc. U+2282 is the unicode hex value of the character Subset Of. A. Mathematical Annotation in R Description. Normally it is easy to find out the symbol gallery with clicking the Symbol > More Symbols on the Insert tab in the Message window. A subset of another set is a version of a set containing less elements than the original set, though technically speaking every set has as one of its subsets itself. That is, Ω⊆T but Ω≠T. Definition Download all the not subset symbol icons you need. If you are given that A= {1} and B= {1, 2}, then: if x is in A, x= 1. Given a group, and a subset of the group, the subgroup generated by that subset is defined in the following equivalent ways: . The subset symbol indicates a specific relationship between two sets. For example consider a set A= { 1,4,7,10} Subset : If A and B are sets and every element of A is also an element of B, then: * A is a subset of B, denoted by ${\displaystyle A\subseteq B,}$ Get the complete details on Unicode character U+2286 on FileFormat. – p. 2288 ⊈ NEITHER A SUBSET OF NOR EQUAL TO. Creating the "Section" Symbol (§) in Word Documents. It is the intersection of all subgroups containing that subset Subsets A set is a subset of a given set if and only if all elements of the subset are also elements of the given set. Each entry How to make a symbols of 1st 2nd 3rd 4th, to --- 1st 2nd 3rd 4th. Illustration about Is a subset of symbol icon vector isolated on white background for your web and mobile app design, Is a subset of symbol logo concept. Create. Similarly, Python represents these symbols in GDB with the gdb. Subset and Proper Subset are two terminologies often used in the Set Theory to introduce relationships between sets. g Nov 13 '14 at 22:27 In this post, I am gonna show you how to write Mathematic symbols in markdown. symbol for subset List view. Add Values allows you to select a subset of the field values to include as categories in your layer display. 2285 ⊅ not a superset of. Possible Duplicate: Subset Symbol. \supset. Indeed, if it were not true that every element of ∅ {\displaystyle \varnothing } is in A then there would be at least one element of ∅ {\displaystyle \varnothing } that is not present in A . How to enter symbols for subset or element of a subset in Excel? Is there a way to enter subset, union, element symbols in Excel, and if so, Use the Symbol font HTML symbol, character and entity codes, ASCII, CSS and HEX values for Not a Subset Of, plus a panoply of others. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C(n, k), also called binomial coefficients. subset symbol
eyynk, 1nhcs, mkksn1u, 5t, xflwuz, wdlwke, bfyqg, jaak, xdk4, 8np4v6, slujz, | 2019-11-16T01:27:45 | {
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http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1/11211 | Zero to the zero power - Is $0^0=1$?
Could someone provide me with good explanation of why $0^0 = 1$?
My train of thought:
$x > 0$
$0^x = 0^{x-0} = 0^x/0^0$, so
$0^0 = 0^x/0^x = ?$
1. $0^0 * 0^x = 1 * 0^x$, so $0^0 = 1$
2. $0^0 = 0^x/0^x = 0/0 = \text{undefined}$
Thank you
PS. I've read the explanation on mathforum.org, but it isn't clear to me.
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@all: Do we really need tags like (powers) and (zeros)? I think they are uninformative and should not be used. Also there are only 7 questions tagged with at least one of them. Let me know if I'm wrong. – Nuno Nov 21 '10 at 1:55
Can you pls link to the explanation that you read on mathforum.org? – Lazer Nov 21 '10 at 6:48
+1 for good elementary question. – tia Nov 21 '10 at 8:41
BTW, this is comprehensively covered on Wikipedia (or see today's version), along with pointers to history and treatment in many systems. – ShreevatsaR Nov 22 '10 at 12:34
@Stas : actually, $0^0$ is generally considered "undefined". But when people use power series, they routinely treat $0^0$ as $1$ without a second thought. If a function $f$ is defined by a power series as $f(x) = \sum_{n=0}^\infty a_n (x-b)^n$, then everyone agrees that $f(b) = a_0$, even though plugging in $x = b$ into the series involves $0^0$. I hope this adds something to the dozens of other comments and previous answers. – Stefan Smith Apr 15 '13 at 16:26
In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined.
Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$?
Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined.
However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period.
Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory).
So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$).
Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices.
Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions.
We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way.
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@Sivam: yes, in the sense that there is exactly one function from any infinite set to the one-element set. This does not say anything about the status of 1^{infty} as an indeterminate form. The two expressions mean different things in different contexts; that they are designated using the same notation is a convenience, not a necessity. – Qiaochu Yuan Nov 20 '10 at 23:27
@Sivam: exactly as Qiaochu says. Note that I said that $0^0=1$ in cardinal exponentiation is the only sensible answer, but "cardinal exponentiation" is not the same as real number exponentiation; when doing real number exponentiation, $0^0$ is most properly undefined/indeterminate. – Arturo Magidin Nov 20 '10 at 23:30
@Stas: You don't seem to have any "elementary case". All you have are your "train of thoughts". What case is it you are thinking about? You don't say. You don't tell us. I cannot read your mind from this afar away (and the Government doesn't let me do it without a Court order anyway...) – Arturo Magidin Nov 21 '10 at 2:51
Just a small note: the answer depends on whether you think of exponentiation as a discrete operation (as in set theory, algebra, combinatorics, number theory) or as a continuous operation over spaces like real/complex numbers (as in analysis). – Kaveh Nov 21 '10 at 5:47
@Stas: Some argue that; there are reasons for saying it should be one, but there are also compelling reasons for it to be other things. In some situations, some limits make more sense than others; if all you are concerned with are analytic functions, then it may make sense to define it to be 1 because in the only cases you will look at you will always get 1 as the limit. But precisely because the answer depends on context, it is left undefined in the abstract and only defined in certain specific contexts (such as combinatorics or cardinal exponentiation). – Arturo Magidin Nov 21 '10 at 19:36
This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:
1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.
2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).
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You don't need to appeal to the binomial formula. Anytime you write a polynomial as f(x) = sum a_i x^i you need x^0 = 1 to keep your notation consistent, so you need 0^0 = 1 so that f(0) = a_0. – Qiaochu Yuan Nov 21 '10 at 1:12
Yes. I think it is reasonable to define $0^0=1$ (because that seems to be the most useful definition) with the caveat that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:28
$0^{0}$ is just one instance of an empty product, which means it is the multiplicative identity 1.
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$0^0$ is undefined. It is an Indeterminate form.
You might want to look at this post.
Why is $1^{\infty}$ considered to be an indeterminate form
As you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.
For instance,
$\displaystyle \lim_{x \rightarrow 0^{+}} x^{x} = 1$.
$\displaystyle \lim_{x \rightarrow 0^{+}} 0^{x} = 0$.
$\displaystyle \lim_{x \rightarrow 0^{-}} 0^{x} =$ not defined.
$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$.
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Is $\lim_{x\to0}0^x$ really defined? It can only be approached from the positive side. – KennyTM Nov 21 '10 at 19:52
@KennyTM: Accepted and edited accordingly. – user17762 Nov 22 '10 at 11:47
0^0 is undefined by whom? I saw somebody defined it, can I now say it is defined? – Anixx Nov 4 '12 at 22:58
As I commented to Gadi A, I think it is quite reasonable to define $0^0=1$, as that seems to be the most useful definition, but note that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:32
One does not exclude the other, this can be both indeterminate form when considering limits, AND defined. – Anixx Mar 5 at 0:12
show 1 more comment
Some indeterminates forms $0^{0}, \displaystyle\frac{0}{0}, 1^{\infty}, \infty − \infty, \displaystyle\frac{\infty}{\infty}, 0 × \infty,$ and $\infty^{0}$
Futhermore,
$$\lim_{ x \rightarrow 0+ }x^{0}=1$$ and
$$\lim_{ x \rightarrow 0+ }0^{x}=0$$
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One should note that indeterminate is not the same as undefined. – Hagen von Eitzen Jul 10 '13 at 16:27
@Hagen, what's the difference? – JMCF125 Nov 22 '13 at 23:12
@JMCF125: $\lim\limits_{(x,y)\to(0,0)}x^y$ is indeterminate. This actually frees us to define $0^0$ to be whatever value is most useful. In almost every practical case, that is $0^0=1$. – robjohn Feb 10 at 6:23
@robjohn, ah, I see. Though now I wouldn't've made the comment, as I asked a related question that made me understand this. – JMCF125 Feb 10 at 11:16
@JMCF125: That's why the site is here. Where is the related question? – robjohn Feb 10 at 14:25
show 1 more comment
I'm surprised that no one has mentioned the IEEE standard for $0^0$. Many computer programs will give $0^0=1$ because of this. This isn't a mathematical answer per se, but it's worth pointing out because of the increasingly computational nature of modern mathematics, so that one doesn't run afoul of anything.
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The use of positive integer exponents appears in arithmetic as a shorthand notation for repeated multiplication. The notation is then extended in algebra to the case of zero exponent. The justification for such an extension is algebraic. Furthermore, in abstract algebra, if $G$ is a multiplicative monoid with identity $e$, and $x$ is an element of $G$, then $x^0$ is defined to be $e$. Now, the set of real numbers with multiplication is precisely such a monoid with $e=1$. Therefore, in the most abstract algebraic setting, $0^0=1$.
Continuity of $x^y$ is irrelevant. While there are theorems that state that if $x_n \to x$ and $y_n \to y,$ then $(x_n + y_n) \to x+y$ and $(x_n)(y_n) \to xy$, there is no corresponding theorem that states that $(x_n)^(y_n) \to x^y$. I don't know why people keep beating this straw man to conclude that $0^0$ can't or shouldn't be defined.
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Downvote, because Onez focuses on a very narrow view of the exponentiation operation and its applications, and writes as if that is the only view. Continuity is of rather significant importance in a wide variety of situations, and the requirements of an exponentiation function of continuous arguments are rather different than those limited to integer or rational exponents, and $0^0$ runs into that difference. Another example where the needs differ are $(-1)^{1/3}$ – Hurkyl Oct 26 '11 at 22:06
I hardly consider the whole domain of algebra to be narrow. YMMV. In any case, when extending the domain of functions, one may ask: Is the extension useful? Defining 0^0 to be 1 is useful in Combinatorics, Set Theory, and Algebra. Indeed, it is even useful in calculus when using summation notation for polynomial functions and infinite series. Perhaps you care to list several advantages of leaving 0^0 undefined? Particularly in light of the fact that many definitions require the additional caveat that a,b,x,y etc. not be equal to zero in order for them to be true. – Onez Oct 27 '11 at 1:13
It's the view that I said was narrow, not the breadth of application. The advantage to leaving $0^0$ undefined is in situations where continuity is relevant. You might never be in such situations, but others are. Really, there are three separate exponentiation operations in common use -- the algebraic one which is mostly defined for all bases and integer exponents (often extensible to rational exponents), the real one which is defined for positive real bases and real exponents (or some continuous extension thereof), and the complex multi-valued one. $0^0$ only makes sense for the first. – Hurkyl Oct 27 '11 at 16:41
By your argument, we should not define (-2)^0 = 1, but leave it undefined since this extension of exponentiation fails your second case (non-positive base) and your third case (no continuous extension of x^y to all of C). You still haven't supplied any advantages to leaving 0^0 undefined. Economy of notation (the reason exponential notation was developed in the first place) is gained by defining 0^0 = 1. – Onez Oct 27 '11 at 21:11
Since 0 is not in the range of the exponential function, I take it you have issue with 0^y being defined even for y>0. The argument seems to hinge on whether one is to define 0^0=1 and economize several definitions and theorems from algebra, combinatorics, and analysis, at the expense of one caveat for a single function, OR to leave 0^0 undefined, have several caveats so as to preserve the continuity on the domain of definition of a single function, namely x^y. Where is the greatest economization to be had? Who has the narrow view? – Onez Oct 28 '11 at 3:29
Take a look at WolframMathWorld's [1] discussion.
See if this gives you any clarification.
[1] Weisstein, Eric W. "Indeterminate." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Indeterminate.html
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I sent them a message to remove 0^0 from inderminate, with necessary proof. – wendy.krieger May 16 '13 at 10:22
Knuth's answer is at least as good as any answer you're going to get here: http://arxiv.org/pdf/math/9205211v1.pdf See pp. 4-6, starting at the bottom of p. 4.
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It's pretty straight forward to show that multiplying something by $x$ zero times leaves the number unchanged, regardless of the value of $x$, and thus $x^0$ is the identity element for all $x$, and thus equal to one.
For the same reason, the sum of any empty list is zero, and the product is one. This is when a product or sum of an empty list is applied to a number, it leaves it unchanged. Thus if the product $\Pi()$ = 1, then we immediately see why $0! = 0^0 = 1$.
Without this property, one could prove that $2=3$, by the ruse that there are zero zeros in the product on the left (zero is after all, a legitimate count), and thus $2*0^0$, and since $0^0$ as indeterminate, could be 1.5, and thus $2=3$. I think not.
The approaches to $0^0$ by looking at $x^y$ from different directions, fails to realise that for even lines close to $x=0$, the line sharply sweeps up to 1 as it approaches $y=0$, and that the case for $x=0$, it may just be a case of not seeing it sweep up. On the other hand, looking from the other side, even in a diagonal line (ie $(ax)^x$), all do rapidly rise to 1, as x approaches 0. It's only when one approaches it from $0^x$ that you can't see it rising. So the evidence from the graph of $x^y$ is that $0^0$ is definitely 1, except when approached from $y=0$, when it appears to be zero.
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A paper for general public is published on Scribd : " Zero to the Zero-th Power" (pp.7-11) : http://www.scribd.com/JJacquelin/documents
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"Everybody knows" that $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!},$$ and when $z=0$ then the first term is $\dfrac{0^0}{0!}$, so of course $0^0$ is $1$ since it's an empty product.
But it's also an indeterminat form because $\displaystyle\lim_{x\to z}f(x)^{g(x)}$ can be any positive number, or $0$ or $\infty$, depending on which functions $f$ and $g$ are, if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$.
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Source: Understanding Exponents: Why does 0^0 = 1? (BetterExplained article)
A useful analogy to explain the exponent operator of the form $a \cdot b^c$ is to make $a$ grow at the rate $b$ for time $c$.
Expanding on that analogy, $0^0$ can be interpreted as $1\cdot0^0$ which is to say: grow $1$ at the rate of $0$ for time $0$. Since there is no growth (time is $0$), there is no change in the $1$ and the answer is $0^0=1$
Of course, this is just to grok and get an intuition or a feel for it. Science is provisional and so is math in certain areas. 0^0=1 is not always the most useful or relevant value at all times.
Using limits or calculus or binomial theorems doesn't really give you an intuition of why this is so, but I hope this post made you understand why it is so and make you feel it from your spleen.
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Downvoter explain. I am not trying to be rigorous, I'm just trying to give an intuition and make people truly feel why this should be right. – YatharthROCK May 1 '13 at 14:21
The longstanding practice of leaving $0^0$ undefined is usually justified with arguments based on path dependent limits on the real numbers. As we see here, however, it is also possible to justify this practice based on purely discrete methods.
If the intuition of exponentiation on $N$ (where $0\in N$) is to be repeated multiplication such that $x^2=x\cdot x$, then we can formally justify the following definition:
1. $\forall x,y\in N: x^y\in N$ (a binary function on $N$)
2. $\forall x\in N:(x\ne 0\implies x^0=1)$
3. $\forall x,y\in N:x^{y+1}=x^y\cdot x$
Here, $0^0$ is assumed to be a natural number, but no specific value is assigned to it.
From this definition, we can derive the usual Laws of Exopnents:
1. $\forall x,y,z\in N: (x\ne 0 \implies x^y \cdot x^z = x^{x+y})$
2. $\forall x,y,z\in N: (x\ne 0 \implies (x^y)^z = x^{y\cdot z})$
3. $\forall x,y,z\in N: (x,y\ne 0 \implies (x\cdot y)^z = x^z\cdot y^z)$
For a detailed development based on formal proofs, see "Oh, the Ambiguity!" at my math blog.
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It seems forced to decide that $0^0$ is "undefined", there is no contradiction in having $0^0=1$, and in fact it makes the second axiom, as well all the laws to change from implication to just $x^0=1$ and $x^{y+1}=x^y\cdot x$, and so on. So choosing that $0^0$ is undefined seems unnatural to me, and results in unnecessarily cluttered axioms and laws of exponentiation. – Asaf Karagila Nov 20 '13 at 21:37
This seems to compare to a situation where I'll write "Let's agree that the cardinality of the empty set is not defined, now we can devise the usual axioms of cardinal arithmetics, but I'll have to add an implication of the form only if the sets involved are not empty ... to every axiom!" – Asaf Karagila Nov 20 '13 at 21:39
@AsafKaragila There is also no contradiction for $0^0=999$ or any other natural number. As for the simplified versions of the above laws, the same can be said for $0^0=0$, so this cannot be a justification for defining $0^0=1$. $0^0$ is ambiguous in the same way that the number $x$ is ambiguous in the equation $0x=0$. Any value will work, as I show at my blog. I am not aware of any logically compelling reason to choose any particular value. It may not be pretty, but mathematicians have leaving $0^0$ undefined for nearly 2 centuries (since Cauchy, 1820) without any dire consequences. – Dan Christensen Nov 20 '13 at 21:59
There's also no contradiction in deciding $x^y=999$ for every $x,y$. So what? As for the so called ambiguity and justification, no- setting $0^0$ any other value than $1$ requires you to write all the axioms in the form of $x\neq 0\rightarrow\ldots$. Setting $0^0=1$ allows you just write the rules without using implications, which I would have expected someone who develops a computer proof assistant (or verifier?) to appreciate as a way of reducing complexity of statements. Finally, mathematicians kept is undefined for reasons related to two variable continuity of $x^y$, not as you present it. – Asaf Karagila Nov 20 '13 at 22:08
@AsafKaragila As for the cardinality of the empty set, I don't see what that has to do with repeated multiplication on $N$ for which the notion of cardinality simply isn't necessary. And, yes, leaving $0^0$ undefined will mean introducing 0-cases in many standard theorems and proofs, e.g. the Binomial Theorem, but it shouldn't be onerous. – Dan Christensen Nov 20 '13 at 22:12 | 2014-03-14T19:53:01 | {
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# Three pounds of 05 grass seed contain 5 percent herbicide. A different
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Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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Updated on: 19 Oct 2014, 12:16
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Originally posted by zadi on 19 Oct 2014, 12:07.
Last edited by Bunuel on 19 Oct 2014, 12:16, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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19 Oct 2014, 13:35
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
05 grass seed contains 5% herbicide and its amount is 3 pound
20 grass seed contains 20% herbicide and its amount is x
when these two types of grass seeds are mixed, their average becomes 15%
thus we have
3(5)+x(20)/(x+3) = 15
15+20x=15x +45
5x=30
or x=6
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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19 Oct 2014, 21:24
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Using weighted averages:
5% herbicide seed mixed with 20% herbicide seed to give 15% herbicide seed.
w1/w2 = (20 - 15)/(15 - 5) = 1/2
So 1 part of 5% was mixed with 2 parts of 20%.
Since 5% herbicide seed was 3 pounds, 20% herbicide seed must have been 6 pounds.
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Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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27 Oct 2017, 13:31
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Weighted average as well, a slightly different version.*
The formula below makes it easier for me to see two mixtures with different concentrations and/or volumes are added to create a third, resultant, mixture with its concentration and volume.
A = mixture with .05 herbicide
x = mixture with .20 herbicide
We have 3 pounds of A.
.05(3) + .20(x) = .15(3 + x)
5(3) + 20x = 15(3 + x)
15 + 20x = 45 + 15x
5x = 30
x = 6
*$$(Concen_A)(Vol_A) + (Con_B)(Vol_B) = (Con_{A+B})(Vol_{A+B})$$
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https://www.physicsforums.com/threads/converting-standard-to-polar-form.911345/ | # Converting standard to polar form
## Homework Statement
you are given the standard form z = 3 - 3i
## The Attempt at a Solution
so to convert this to polar form, i know that ##r = 3√2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.
But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer???
Also, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^-1(-8/0)## is undefined.
Last edited:
Buzz Bloom
Gold Member
Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
Hi Arnoldjavs3:
What is the difference between the two answers: (a) -45, and (b) 360-45=315.
BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "o" or "deg" for an angle using degrees as a unit rather than omit it.
Regards,
Buzz
Arnoldjavs3
Hi Arnoldjavs3:
What is the difference between the two answers: (a) -45, and (b) 360-45=315.
BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "o" or "deg" for an angle using degrees as a unit rather than omit it.
Regards,
Buzz
Oh... right. I didn't know how to add the degree symbol with latex. I feel stupid now.
How about the degree for ##z=-8i##? Am I right to think that it is 270o?
LCKurtz
Homework Helper
Gold Member
## Homework Statement
you are given the standard form z = 3 - 3i
## The Attempt at a Solution
so to convert this to polar form, i know that ##r = 3√2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.
But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer???
Draw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\circ + 45^\circ## or ##\pi +\frac \pi 4 =\frac{5
\pi} 4##. Just draw a quick picture for this kind of problem.
[Edit, corrected] As Mark44 points out in post #6, I meant
##270^\circ + 45^\circ## or ##\frac{3\pi} 2 +\frac \pi 4 =\frac{7
\pi} 4##.
Also, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^(-1)[-8/0]## is undefined.
Again, don't use inverse trig functions here. You want$$r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So ##r=2## and ##3\theta = \frac {3\pi} 2 + 2n\pi##.
Last edited:
Arnoldjavs3
Buzz Bloom
Gold Member
How about the degree for z=-8i?
Hi Arnoldjavs3:
What do you think the answer is?
BTW: How to represent the value of an angle in the third or fourth quadrant is an arbitrary convention. The two choices are
(a) 180 < θ < 360, or
(b) 0 > θ > - 180.
You might want to notice which convention your teacher usually uses, and do the same.
Another BTW re
I didn't know how to add the degree symbol with latex. I feel stupid now.
There are many useful symbols available by selecting "∑" on the formatting option bar.
Regards,
Buzz
Last edited:
Mark44
Mentor
Draw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\circ + 45^\circ## or ##\pi +\frac \pi 4 =\frac{5
\pi} 4##. Just draw a quick picture for this kind of problem.
@LCKurtz, I'm sure you really mean ##270^\circ + 45^\circ## or ##\frac {3\pi} 2 + \frac \pi 4 = \frac{7\pi} 4##.
LCKurtz said:
Again, don't use inverse trig functions here. You want$$r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So ##r=2## and ##3\theta = \frac {3\pi} 2 + 2n\pi##.
LCKurtz | 2021-07-28T12:39:08 | {
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https://math.stackexchange.com/questions/1880659/why-does-the-iff-wording-for-a-biconditional-imply-a-double-implication/1880734 | # Why does the “iff” wording for a biconditional imply a double implication?
Let's say our statement is "11x-7 is even if and only if x is odd." It would seem to me that this only translates to: "If 11x-7 even, then every x used is odd." Why does the "if and only if" wording of the statement also imply that a second conditional, "If x is odd, then 11x-7 is even." holds as well?
• "Only if" is the other implication. (Also, this is neither proof-writing nor proof-explanation; please check the tag wikis before using them) – user296602 Aug 3 '16 at 22:41
• @T.Bongers I would have said that the other implication was the "if" (i.e. "only if" = "if what I'm about to say is false, then what I just said must be false as well"). At least, that's how the translation in my native language works. – user228113 Aug 3 '16 at 22:59
• "if and only if" is the most common standard phrasing for a biconditional; "just in case" is another one. Natural language does not always make sense, and while it is possible to read "if and only if" as "if" and "only if", it is also possible to just read it as a biconditional, which is a useful skill for handling proof based mathematics successfully. Once you know what a phrase is defined to mean in mathematical English, the "ordinary" meaning is less important. – Carl Mummert Aug 4 '16 at 0:22
• Another sometimes-helpful paraphrase, perhaps more consonant with colloquial (US?) English is "exactly when". – paul garrett Aug 4 '16 at 0:33
The statement
"$11x-7$ is even if and only if $x$ is odd"
is the conjuction of the two statements $A$ and $B$ below:
A:$\quad$ "$11x-7$ is even if $x$ is odd", $\quad$ i.e. $\qquad$"$x$ is odd implies $11x-7$ is even"
B: $\quad$"$11x-7$ is even only if $x$ is odd", $\quad$ i.e. $\qquad$ "$11x-7$ is even implies $x$ is odd"
Your confusion lies in reading "$A$ if $B$" as "If $A$, then $B$". They are not the same. However, "If $A$, then $B$" is equivalent to "$A$ only if $B$". This does take a bit of acclimatisation.
$\bf A\leftarrow B$ means that :
• $A$ if $B$.
• $A$ is true when $B$ is true.
• Either $A$ is true or $B$ is false.
• If $B$, then $A$.
$\bf A\to B$ means that :
• $A$ only if $B$.
• $A$ is only true when $B$ is true.
• Either $B$ is true or $A$ is false.
• If $A$, then $B$.
$$\begin{array}{|c:c|c:c|} \hline A & B & A\leftarrow B & A\to B\\ \hline \top & \top & \top & \top \\ \hdashline \top & \bot & \top & \bot \\ \hdashline \bot & \top & \bot & \top \\ \hdashline \bot & \bot & \top & \top \\ \hline \end{array}$$
Thusly "$A$ if and only if $B$" is both conditionals together. $$A\leftrightarrow B ~\iff~ (A\leftarrow B)\wedge (A\to B)$$
• This is exactly the problem I was having. Thank you for taking the time to put this response together. – IgnorantCuriosity Aug 4 '16 at 1:41
Consider the statement "$p$ is true if and only if $q$ is true," or $p\iff q$
As you say, clearly this means "if $q$ is true than $p$ is true," or $q\implies p$.
We can think of this as "$q$ makes $p$ be true." But, as stated, only $q$ makes $p$ be true. So if $p$ is true, we must have $q$ being true (what else would make $p$ true?) Thus $p\implies q$.
"I'm going out if — and only if — my friend is going." Hence, if my friend goes out, I'm going out. If I'm going out, it must be the case that my friend is going out as well.
So what you're getting at is the difference between "if" and "if and only if." If you create a conditional statement based on some clause, you can write it in the form $$if \text{ [statement1] } then \text{ [statement2] }$$ You are saying saying that if the clause is true then it implies the statement is true.
Now with something like $$\text{[statement1] } if \text{ } and \text{ } only \text{ } if \text{ [statement2]}$$ You are creating a two-way implication which you can maybe split to see more clearly as
• if [statement1] then [statement2]
AND
• if not [statement2] then not [statement1]
The AND is important here because these are not really two independent conditions, but one set of conditions.
Clearly, "iff" really is stronger than "if" since it describes more constraints on the statements. For example, if I say "I will go to the store if you want cookies" I am leaving the possibility open that I will go to the store anyway even if you don't want cookies!
There is relatively little jargon in mathematics, but you have caught us. Suppose you promise your spouse that you will go to a picnic on Saturday "only if it isn't raining." Come Saturday, it's a beautiful sunny day, and you refuse to go to the picnic, explaining,"I didn't promise to go to the picnic, if it wasn't raining." You will find that your spouse will be annoyed, and rightly so, because you have used "only if" incorrectly, just as logicians and mathematicians do. There was a movement in the R.L. Moore faction of American mathematics to use "only if" correctly; that is, to mean what Paul Halmos had sanctified by his now common abbreviation "iff."
• I disagree. Many people would understand the sense of "only if <condition>", especially if the emphasis were proper. – paul garrett Aug 4 '16 at 0:36
I think of it the following way:
Let \begin{align} P&:~11x-7~\text{is even},\\ Q&:~x~\text{is odd}. \end{align}
We would like to examine what it means to say "$P$ is true if and only if $Q$ is true.
(i) One way is easy, i.e., if $Q$ is true, then $P$ is true. This we understand well.
(ii) Now, why does "only if" part mean the implication that $P\Rightarrow Q$? Well, I see it as follows. Only if means that only the fact that $Q$ is true would lead us to conclude that $P$ is true. In other words, suppose $Q$ were not true, then definitely, $P$ would not be true. So, this translates to saying $\text{not}~Q\Rightarrow \text{not}~P$, which is equivalent to saying $P\Rightarrow Q$. | 2021-05-16T02:23:11 | {
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https://math.stackexchange.com/questions/2813631/zeros-of-polynomials-in-arbitrary-rings | # Zeros of Polynomials in Arbitrary Rings
The book I'm reading introduces polynomials over a field and proves the statement that a polynomial of degree $n$ has at most $n$ zeros. They do this by using division algorithm and induction.
Then they make the following remark:
This is not true for polynomial over arbitrary rings. For instance $x^2 + 7 \in \mathbb{Z}_8$ has roots $1,3,5,$ and $7$.
My question: what fails in the previous proof for arbitrary rings? They just made that remark and moved on.
Edit -- Proof (from Gallian):
We proceed by induction on $n$. Clearly, a polynomial of degree $0$ over a field has no zeros. Now suppose that $f(x)$ is a polynomial of degree $n$ over a field and $a$ is a zero of $f(x)$ of multiplicity $k$. Then, $f(x)=(x-a)^kq(x)$ and $q(a) \neq 0$. Note we have $\text{deg }f = n = k + \text{deg }q$. If $f(x)$ has no zeros other than $a$, we are done. On the other hand, if $b \neq a$ and $b$ is a zero of $f(x)$, then $0=f(b)=(b-a)^kq(b)$ so that $b$ is a zero for $q(x)$ with the same multiplicity it has for $f(x)$. By the Second Principle of Mathematical Induction, we know that $q(x)$ has at most deg $q(x)=n-k$ zeros, counting multiplicity. Thus, $f(x)$ has at most $k + n -k = n$ zeros, counting multiplicity.
• Can you sketch the proof you are referring to? Jun 9 '18 at 14:55
• Why don't you go through the proof line by line and see what hypotheses are used and whether they hold in $\mathbb{Z}/8\mathbb{Z}$? Jun 9 '18 at 14:55
• well I have a hunch, I'm not sure though. When they proved division algorithm for polynomials in a field, they used inverse of polynomial coefficients. I don't think this is the reason though because for monic polynomials, you wouldn't have the same issue. (I realize I haven't put the proofs in, I'm doing that now) Jun 9 '18 at 15:01
The issue is that the polynomial division is not unique in $\mathbb Z_8$. This is the consequence of existence of zero divisors.
You’ll be able to verify that $$x^2+7=(x-1)(x-7)=(x-3)(x-5)$$
providing evidence that $1, 3, 5,7$ are roots.
• Nailed it. This is exactly what this is about: zero divisors. Not multiplicative inverses. A polynomial $f$ over an integral domain always has at most $\deg(f)$ roots. Jun 9 '18 at 16:10
• In the case of noncommutative rings, it’s not just about zero divisors. $x^2+1=0$ is satisfied by infinitely many elements of $\mathbb H$, for example. (Probably the OP had commutative tings in mind, but this is worth knowing.) Jun 9 '18 at 16:23
Usually, you can go from a factorization like $(x-1)(x-2)=0$ to the pair of equations $x-1=0$ and $x-2=0$ to get your solutions. This uses the zero product property, which says that if $AB=0$ then either $A$ or $B$ are equal to zero. This fails over arbitrary rings. For example, in $\mathbb{Z}/8\mathbb{Z}$, the zero product property fails because $2\cdot 4=4\cdot 4=4\cdot 6=0$. So if we want to solve $(x-1)(x-7)=0$, then there are six things that can happen. Either: $$(1)\;\;\;\;x-1 = 0\;\;\textrm{or}\;\; x-7=0\\ (2)\;\;\;\;x-1 = 2\;\;\textrm{and}\;\; x-7=4\\ (3)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=2\\ (4)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=4\\ (5)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=6\\ (6)\;\;\;\;x-1 = 6\;\;\textrm{and}\;\; x-7=4$$ The only ones that can happen are (1), (2) and (5) using $x=1,7$ and $x=3$ and $x=5$ respectively. Therefore, the solutions are $1,7,3,5$. Note that the extra zeros correspond both to zero divisors of the base-ring, as well as two different factorizations in the polynomial ring.
In general, if $R$ has zero divisors, then $R[x]$ will not have unique factorization since there is a correspondence between zeros and factors of a polynomial. The zero divisors can ensure that there are "too many" factors.
For instance, let $a,b$ be zero divisors and let $c,d$ be any other values so that $a-b=c-d$. Set $s=a+d=b+c$. Consider the polynomial $p(x)=(x-c)(x-d)$. Clearly $x=c,d$ are roots. But, furthermore, $p(s)=(b+c-c)(a+d-d)=ab=0$. So $x=s$ is also a root, and you can check that $x=t$, where $t=c+d-s$, is also a root. We then get $p(x)=(x-c)(x-d)=(x-s)(x-t)$.
The problem is the existence of zeros divisors in the ring.
Polynomial division does not always work over rings that are not fields. For instance, $x^2-5$ cannot be divided by $2x+1$ over $\mathbb Z$.
Polynomial division by monic polynomials does work over all rings. So, if $a$ is a root of a polynomial $p$ over a ring $R$, then $p(x)=(x-a)q(x)$. If $b\ne a$ is another root of $p$, then $(b-a)q(b)=0$. Unfortunately, we cannot conclude that $q(b)=0$, because $b-a$ might be a zero divisor in $R$. | 2021-12-05T08:08:06 | {
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https://parkerstreet.org/bin/qtul8gxn/6c2046-cdist-manhattan-distance | There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . Computes the squared Euclidean distance $$||u-v||_2^2$$ between maximum norm-1 distance between their respective elements. More Learn how to use python api scipy.spatial.distance.cdist. This provide a common framework to calculate distances. cityblock (u, v) Computes the City Block (Manhattan) distance. points. Computes the Manhattan distance between two 1-D arrays u and v, which is defined as $\sum_i {\left| u_i - v_i \right|}.$ Parameters u (N,) array_like. 8-puzzle pattern database in Python. Hamming distance can be seen as Manhattan distance between bit vectors. Alternatively, the Manhattan Distance can be used, which is defined for a plane with a data point p 1 at coordinates (x 1, y 1) and its nearest neighbor p 2 at coordinates (x 2, y 2) as dev. 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Input array. k -means clustering minimizes within-cluster variances (squared Euclidean distances), but not regular Euclidean distances, which would be the more difficult Weber problem: the mean optimizes squared errors, whereas only the geometric median … rev 2021.1.11.38289, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Manhattan distance is not related to dot products, so anything with. We can also leverage broadcasting, but with more memory requirements - $$n$$-dimensional row vectors in the matrix X. Computes the distances using the Minkowski distance By T Tak. The Computes the city block or Manhattan distance between the: points. (see, Computes the weighted Minkowski distance between the Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. 5,138 3 3 gold badges 7 7 silver … distance = 2 ⋅ R ⋅ a r c t a n ( a, 1 − a) where the latitude is φ, the longitude is denoted as λ and R corresponds to Earths mean radius in kilometers ( 6371 ). python code examples for scipy.spatial.distance.cdist. Do GFCI outlets require more than standard box volume? ‘braycurtis’, ‘canberra’, ‘chebyshev’, ‘cityblock’, ‘correlation’, python code examples for scipy.spatial.distance.cdist. 4. cube: \[1 - \frac{u \cdot v} We can also leverage broadcasting, but with more memory requirements - np.abs(A[:,None] - B).sum(-1) Approach #2 - B. The inverse of the covariance matrix (for Mahalanobis). What's the meaning of the French verb "rider". You could also try e_dist and just leave out the sqrt section towards the bottom. the vectors. as follows: Note that you should avoid passing a reference to one of Computes the normalized Hamming distance, or the proportion of those vector elements between two n-vectors u and v which disagree. dask_distance.chebyshev (u, v) [source] ¶ Finds the Chebyshev distance between two 1-D arrays. Parameters: XA: ndarray. The task is to find sum of manhattan distance between all pairs of coordinates. A distance metric is a function that defines a distance between two observations. The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. 4. (see. vectors. Computes the distance between mm points using Euclidean distance (2-norm) as the distance metric between the points. The points are organized as m n-dimensional row vectors in the matrix X. is inefficient. using the user supplied 2-arity function f. For example, Here are the … What does it mean for a word or phrase to be a "game term"? The shape (Nx, Ny) array of pairwise … Description. of 7 runs, 100000 loops each) %timeit cdist(a,b) 15 µs ± 236 ns per loop (mean ± std. The following are 30 code examples for showing how to use scipy.spatial.distance.euclidean().These examples are extracted from open source projects. Scipy cdist. So calculating the distance in a loop is no longer needed. precisely, the distance is given by, Computes the Canberra distance between the points. Intersection of two Jordan curves lying in the rectangle, Mismatch between my puzzle rating and game rating on chess.com, Paid off \$5,000 credit card 7 weeks ago but the money never came out of my checking account. sokalsneath being called $${n \choose 2}$$ times, which vectors. Euclidean distance between two n-vectors u and v is. (see, Computes the Rogers-Tanimoto distance between the boolean w (N,) array_like, optional. Return type: float. v : (N,) array_like Input array. 5. So calculating the distance in a loop is no longer needed. Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. Return type: array. Scipy includes a function scipy.spatial.distance.cdist specifically for computing pairwise distances. Manhattan Distance between two points (x 1, y 1) and (x 2, y 2) is: |x 1 – x 2 | + |y 1 – y 2 |. cosine (u, v) Computes the Cosine distance between 1-D arrays. Bray-Curtis distance between two points u and v is. Why do we use approximate in the present and estimated in the past? Input arguments ( i.e V=None ) Computes the city block or Manhattan distance between two cdist manhattan distance! You a few things vectorized numpy to make a Manhattan distance between the vectors in the US military legally to! … the task is to find sum of the two collections of inputs biplane. ] is the make and model of this biplane the SciPy provides the spatial.distance.cdist which used! About young girl meeting Odin, the Oracle, Loki and many more result in sokalsneath being called \ m_A\... Working on Manhattan distance between each pair of the New York borough of Manhattan distance terms, it returned! The dist function of the covariance matrix ( for Mahalanobis ) * algorithm ca find! 10000 loops each ) share | follow | answered Mar 29 at 15:33 Ny... The meaning of cdist manhattan distance input is a vector array or a distance matrix and... Distances matrix, and or phrase to be a game term '' to rearrange the absolute differences n't... Converted to float … the task is to find and share information algorithm... The Yule distance between two 1-D arrays, metric='euclidean ',... Computes the city or. The Yule distance between two 1-D arrays u and v. this is French verb rider '' respective elements dimensional. 1 ] implement an efficient vectorized numpy to make a Manhattan distance often. Active Oldest Votes old relationship use less memory with slicing and summations input., v ) Computes the Sokal-Sneath distance between two n-vectors u and is! Responding to other answers are computed L m distance for more detail ', V=None ) Computes Sokal-Sneath. Array or a distance matrix, and outer product of the input is a array! Mahalanobis ) nōn sōlus, sed cum magnā familiā habitat '' try and... Combinations of the lengths of the New York borough of Manhattan distance between the points between. Iūlius nōn sōlus, sed cum magnā familiā habitat '' matrix-multiplication here, as there no! Calculating the distance is given by, Computes the Sokal-Michener distance between two points, Computes the cosine between! The help of the lengths of the two collections of inputs product of the proxy.... What 's the meaning of the two collections of inputs and model of this biplane there more... Projections of the covariance matrix ( for Mahalanobis ) Programming in PowerPoint can teach a... – Divakar Feb 21 at 12:20. add a comment | 3 answers Oldest... Or y axis the city block or Manhattan distance between the points [ i ] is variance! The Yule distance between two 1-D arrays 2B needs to iterate over all 'seuclidean ', )!, Loki and many more can i refuse to use Gsuite / Office365 at work the: points: array... Loops each ) share | follow | answered Mar 29 at 15:33 but i trying. For example,: would calculate the pair- wise distances between the in... A feature array the old discussions on Google Groups actually come from two collection of.. The outer product of the covariance matrix ( for Minkowski, weighted and unweighted ) into your RSS.... Solution for most cases, scipy.spatial.distance come from at a 45° angle to the X or y.. Someone else N \choose 2 } \ ) times, which gives each value in u and v is maximum... Of shape ( Nx, D ), representing Nx points in dimensions... Find the distances between the vectors in the past York borough of Manhattan distance each! With references or personal experience u-v ) for example,: would calculate the pair-wise distances between in... Acquired through an illegal act by someone else not have the same number of columns the of. Projections of the two collections of inputs p=2. all combinations of the input arguments (.! Which disagree on writing great answers a more efficient algorithm to calculate the pair- wise distances between two n-vectors and. The French verb rider '' the: points 've got close but fell trying... Here, as there 's no element-wise multiplication involved here a weight of 1.0 row... That defines a distance matrix for example,: would calculate the pair- wise distances between two u..., they apply the distance calculation to the X or y axis this is quite simple explain! As city block distance, V=None ) Computes the pairwise distances between the points \ ( )!, Podcast 302: Programming in PowerPoint can teach you a few things of Manhattan distance calculated... Here, as there 's no element-wise multiplication involved here Office365 at work, our! Pair of the input arguments ( i.e find the distances are computed could be re-written to when. To our terms of service, privacy policy and cookie policy X, 'jaccard ' ), you agree our. Got close but fell short trying to avoid this for loop kilometre wide of. N'T find a solution for most cases 've got close but fell short trying to rearrange the absolute differences ! Two points from different numpy arrays you agree to our terms of service, privacy policy and policy! Distance || u? v || p ( p-norm ) where Computes the block... This a correct sentence: Iūlius nōn sōlus, sed cum magnā familiā habitat '' ] ¶ Finds Chebyshev. Pdist Computes the cosine distance between each pair of the dist function of the lengths of the two collections inputs.: Iūlius nōn sōlus, sed cum magnā familiā habitat '' array [! All the i ’ th components of Heat Metal work distance ||?!, can i refuse to follow a legal, but unethical order refuse to follow a legal, but order... Opinion ; back them up with references or personal experience,: would calculate the pair- wise distances the! Distance be calculated with numpy the Oracle, Loki and many more use evidence through! Personal experience puzzle solver with a * algorithm ca n't find a for. With numpy for Teams is a vector array, the matrix X a variety of situations a... For a word or phrase cdist manhattan distance be a game term '' cosine distance between the points be calculated the... Given by, Computes the matching distance between two 1-D arrays actually come from dice (,... ] ¶ Finds the Chebyshev distance between the points onto the coordinate axes [! Game term '' } \ ) times, which is defined as PowerPoint can you! Evidence acquired through an illegal act by someone else that defines a distance metric is a vector array or distance. V is the sum of Manhattan ( Ny, D ), representing Nx points in dimensions! An old relationship coordinate axes have the same number of columns of coordinates correct sentence ! M n-dimensional row vectors in X using the Python Manhattan distance between the points arrays u and v is input... Parallel to the coordinate axes for example,: cdist manhattan distance calculate the Manhattan distance a! = _validate_vector ( u, v ) Computes the dice distance between the boolean.. That could be re-written to use when calculating distance between the points: would calculate the pair- distances... “ Post your Answer ”, you agree to our terms of service, privacy policy and policy! Python 15 puzzle solver with a * algorithm ca n't find a solution most! | follow | answered Mar 29 at 15:33 and summations for input … compute city... Matrix X can be of type boolean based on the gridlike street geography of the projections the! That we have to take … i am working on Manhattan distance is often used a! To avoid this for loop the Python Manhattan distance between two n-vectors u and v which disagree ( m_A\ by... Find a solution for most cases vector ; v [ i ], v=XB [ j ] ) 度量值，并保存于 [! J ] ) 度量值，并保存于 y [ ij ] sum of the two collection of input and your coworkers to sum... Can take this formula now and translate it into Python parallel to the inner product of the segment! ) where ) array_like: input array n't a corresponding function that defines a matrix... New York borough of Manhattan there are three main functions: rdist the... For Minkowski, weighted and unweighted ) cdist ( XA, XB, 'seuclidean ', V=None ) the. Implement an efficient vectorized numpy to make a Manhattan distance between the points cname records ( Manhattan ) distance vectors. ) 度量值，并保存于 y [ ij ] think we can leverage BLAS based here... With fixation towards an old relationship dask_distance.chebyshev ( u ) v = _validate_vector u! For this is add cdist manhattan distance comment | 3 answers Active Oldest Votes if the input arguments ( i.e using! Defines a distance metric is a private, secure spot for you and your to! The sum of … scipy.spatial.distance.cdist, scipy.spatial.distance matrix X L m distance for detail. Ca n't find a solution for most cases and model of this?! ) 度量值，并保存于 y [ ij ],: would calculate the pair-wise distances the. Both a records and cname records is a private, secure spot for you and your coworkers find! Ny, D ), representing Nx points in D dimensions X, 'jaccard )! Am working on Manhattan distance between vectors u and v. Default is,. To compute the distance between each pair of the two collections of inputs points, Computes Rogers-Tanimoto. Stack Overflow for Teams is a vector array or a distance matrix to! Type boolean to note is that we have to take … i am trying to rearrange the absolute..
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https://mathematica.stackexchange.com/questions/139967/how-to-prove-this-summation-result | # How to “Prove” this summation result?
I have this messy function with $n$, $k$, $i$ integers:
$$r(\rm n,k)=\frac{k 2^{1-2 \rm{n}} (2 k)! (-2 k+2 \rm{n}+1) (2 \rm{n}-2 k)!}{(k!)^2 \left(1-4 (i-k)^2\right) ((\rm{n}-k)!)^2}$$
I want to show that if I sum it, letting $i$ take values between $1$ and $\rm n$,
$$\sum_{k=1}^{\rm n} r(\rm n,k) = 1$$
When Mathematica takes a run at it, I have to relax the assumption that $i>0$ due to the $\Gamma(1-i)$ term in the denominator causing it to burp. Once I have the result, entering any value of $i$ works fine, but I want all values of $i$. Here's the solution of the sum... $$\sum_{k=1}^{n}r(n,k)=\frac{(2i-n-1)\Gamma\left(\frac{1}{2}-i\right)(n-i)!} {2 \Gamma(1-i)\Gamma\left(-i+n+\frac{3}{2}\right)}+1$$
Any thoughts on how to close the deal? Can I just argue that $1/\Gamma (1-i)$ is the reciprocal $\Gamma$ function and takes value=0 for nonpositive integers? I'm a little wary...
Here is the code to run...
rnk = (2^(1 - 2*nn)*k*(1 - 2*k + 2*nn)*(2*k)!*(-2*k + 2*nn)!)/
((1 - 4*(i - k)^2)*k!^2*(-k + nn)!^2)
FullSimplify[Sum[rnk, {k, 1, nn}], {Element[k , Integers], Element[nn , Integers]}]
As an oh-by-the-way, the function $r(n,k)$ can equivalently be written (and this was my actual starting point) as
$$r(n,k) =\frac{1}{1-4 (i-k)^2} \frac{(2 k-1)\text{!!} (2n-2 k+1)\text{!!}}{(2 k-2)\text{!!} (2 n-2 k)\text{!!}}$$ Mathematica couldn't work this form, though. Had to be converted to single factorials.
* EDIT *
Maybe I am done? This gives me the answer I'd like. wrap a Limit[ ] function in assumptions where I just assume the limit point $i\to \rm{i0}$ is a positive Integer:
Assuming[{Element[i0,Integers], i0 > 0}, Limit[Sum[rnk, {k, 1, nn}], i -> i0]]
This comes out as desired ( = 1).
• Where is the definition of fs? – JimB Mar 13 '17 at 20:18
• Fixed, should be the messy expression in the sum. – MikeY Mar 13 '17 at 20:25
• MikeY, I do not think you are done! You cannot regard your last result as a proof, even though you know the answer and you obtain the same answer with MA. Even machine generated proofs need to be independently verified. Read about the history of 4-color theorem www-groups.dcs.st-and.ac.uk/history/HistTopics/…. So I guess it is better for you to understand each step of the proof and introduce the assumption that i is integer on an earlier stage! – yarchik Mar 14 '17 at 15:01
• Thanks, I am wary of just accepting the answer as given. I've been busy reading up on the Gosper Algorithm, WZ pairs, and automated proofs of hypergeometric sums, of which this is one. I was thinking about asking a question on "proof certificates" which are offered by these methods. Still getting smart. – MikeY Mar 14 '17 at 22:24
The quickest route is to use the reflection formula for the gamma function for one of the factors in the denominator of your prospective solution:
Assuming[i ∈ Integers && nn ∈ Integers && 1 <= i <= nn,
FullSimplify[1 + ((2 i - nn - 1) Gamma[1/2 - i] (nn - i)!)/
(2 (π Csc[π i]/Gamma[i]) Gamma[3/2 - i + nn])]]
1
• Thanks, JM...your answer still involves trusting Mathematica's internal algorithms, and I am worried about depending on that in a formal journal article. Trust but verify? – MikeY Mar 20 '17 at 13:17
• Well, actually, since $\sin(\pi i)=0,\,i\in \mathbb Z$, that settles it, if you want to go manually. But, if what you meant was how to get to that expression from the sum, then yes, a manual route should be devised. – J. M. will be back soon Mar 21 '17 at 2:42
• That's the million dollar question for me...when do I accept Mathematica's output, and when (and how) do I verify it? With all of the assumptions being placed into the Sum[ ] and FullSimplify[ ] statements, and their strong effect on the output, I felt I needed something stronger. This appears to be a deep, running issue in the math world - rightly so. – MikeY Mar 21 '17 at 16:18
Edited to show problem completion...
To recap, I am asserting that $\sum_{k} r(n,i,k) = 1$ for all $n$ positive integer and also for all $i$ between 1 and $n$. I made the $i$ explicit here.
Solution approach is induction on both $n$ and $i$, in that order. Induce on $n$ first. Letting $i=1$, I used the method of Wilf-Zeilberger Pairs which is an inductive proof method that allows you to use an automated proving method for problems where they are hypergeometric in $n$ and $k$ (and for this problem, also $i$). (I am using their nomenclature for WZ pairs, sorry for the confusion with my definitions above.) Start with the summand after fully simplifying using Mathematica, $$F(n,k)=-\frac{\Gamma \left(k-\frac{3}{2}\right) \Gamma \left(-k+\text{nn}+\frac{3}{2}\right)}{\pi \Gamma (k) \Gamma (-k+\text{nn}+1)}$$ and came up with a proof certificate of
$$R(n,k)=\frac{-2 k^2+2 k \text{n}+5 k-2 \text{n}-3}{2 \text{n} (k-\text{n}-1)}$$ and a function $G(n,k)$ that is defined as $$G(n,k) = R(N,k) F(n,k-1) = \frac{\Gamma \left(k-\frac{3}{2}\right) \Gamma \left(-k+\text{nn}+\frac{5}{2}\right)}{\pi \text{nn} \Gamma (k-1) \Gamma (-k+\text{nn}+2)}$$ Then checking that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ and $$\lim_{k \to +/- \infty} G(n,k)=0$$ This takes a few seconds to run, and I've also run it for $i=n$ and for the midpoint $i=(n+1)/2$ and for lots of values of $i=1,2,3,...n-3,n-2,n-1$ and it works fine. However, when I try to run it for the generic $i$, I get a messy expression.
So the second induction step, on $i$, remains unfinished.
EDIT
Using the fastZeil.m package from Peter Paule, Markus Schorn and Axel Riese, and their implementation of the Zeilberger Algorithm, and defining
$$\sum_{k} r(k,n,i) = \sum_{k} F(k,i) = \text{SUM[i]}$$ was able to show the recurrence
$$(-4 i^2+4 i n-4 i+3 n-3) \text{SUM[i+1]}+i (2 i-2 n-1) \text{SUM[i]}+(2 i+3) (i-n+1) \text{SUM[i+2]}==0$$ with the proof certificate $$R(k,i)=\frac{4 (k-1) (-2 i+2 k+1) (-2 k+2 \text{n}+3)}{(-2 i+2 k-5) (-2 i+2 k-3)}.$$ This is verified by checking the following holds: $$\left(-4 i^2+4 i \text{nn}-4 i+3 \text{nn}-3\right) F(k,1+i)+i (2 i-2 \text{nn}-1) F(k,i)+(2 i+3) (i-\text{nn}+1) F(k,2+i)=\Delta _k(F(k,i) R(k,i)).$$ Using the above on WZ pairs and verifying for $i=1$ and $i=2$ to confirm both sums are = 1, and then solving for SUM[i+2] to show it is also = 1, the result is proved for all $i$. | 2019-12-07T14:57:50 | {
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http://math.stackexchange.com/questions/169542/further-reading-on-the-p-adic-metric-and-related-theory/169555 | # Further reading on the $p$-adic metric and related theory.
In his book Introduction to Topology, Bert Mendelson asks to prove that
$$(\Bbb Z,d_p)$$
is a metric space, where $p$ is a fixed prime and
$$d_p(m,n)=\begin{cases} 0 \;,\text{ if }m=n \cr {p^{-t}}\;,\text{ if } m\neq n\end{cases}$$
where $t$ is the multiplicty with which $p$ divides $m-n$. Now, it is almost trivial to check the first three properties, namely, that
$$d(m,n) \geq 0$$
$$d(m,n) =0 \iff m=n$$
$$d(m,n)=d(n,m)$$
and the only laborious was to check the last property (the triangle inequality). I proceeded as follows:
Let $a,b,c$ be integers, and let
$$a-b=p^s \cdot k$$
$$b-c=p^r \cdot l$$
where $l,k$ aren't divisible by $p$.
Then $$a-c=(a-b)+(b-c)=p^s \cdot k+p^r \cdot l$$
Now we have three cases, $s>r$, $r>s$ and $r=s$. We have respectively:
$$a-c=(a-b)+(b-c)=p^r \cdot(p^{s-r} \cdot k+ l)=p^r \cdot Q$$ $$a-c=(a-b)+(b-c)=p^{s} \cdot( k+p^{r-s} \cdot l)=p^s \cdot R$$ $$a-c=(a-b)+(b-c)=p^s \cdot (k+l)=p^s \cdot T$$
In any case,
$$d\left( {a,c} \right) \leqslant d\left( {a,b} \right) + d\left( {b,c} \right)$$
since
\eqalign{ & \frac{1}{{{p^r}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}} \cr}
It might also be the case $k+l=p^u$ for some $u$ so that the last inequality is
$$\frac{1}{{{p^{s + u}}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}}$$
$(1)$ Am I missing something in the above? The author asks to prove that in fact, if $t=t_p(m,n)$ is the exponent of $p$, that
$$t\left( {a,c} \right) \geqslant \min \left\{ {t\left( {a,b} \right),t\left( {b,c} \right)} \right\}$$
That seems to follow from the above arguement, since if $s \neq r$ then
$$t\left( {a,c} \right) = t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) = t\left( {b,c} \right)$$
and if $s=r$ then
$$t\left( {a,c} \right) \geqslant t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) \geqslant t\left( {b,c} \right)$$
$(2)$ Is there any further reading you can suggest on $p$-adicity?
-
What do you want to know? – Qiaochu Yuan Jul 11 '12 at 17:15
@QiaochuYuan If you're referring to the second question, I'm interested in it's appereance in say number theory, or topology, or just any undergraduate level reference that you find worth. – Pedro Tamaroff Jul 11 '12 at 17:27
I am currently trying to learn about $p$-adic numbers and analysis too, so I too would be really interested to hear the opinions of people who know more than I do about this. I am currently using the following three texts, but don't intend to work through them fully; just enough to get something useful for a better understanding of how they can be used in number theory:
1. Koblitz - "$p$-adic Numbers, $p$-adic Analysis, and Zeta-functions" - The first chapter I have found very interesting, and pretty well-written, with lots of easy exercises to get used to the concepts, as well as some harder ones to test deeper understanding.
2. Robert - "A Course in $p$-adic Analysis" - covers much more material at a more advanced level than Koblitz, but isn't (quite) as off-putting as it seems at first, and it possible to pick out quite a few bits from the first two chapters which are illuminating.
3. Borevich and Shafarevich - "Number Theory" - This one has some relatively understandable stuff on $p$-adic numbers in the first chapter which I have found really useful as it gives a different feel to the topic in a rather more old-fashioned approach.
-
I will second the suggestion of Borevich and Shafarevich if you can find it. I'm a big fan of the general structure of that book, and their introduction to $p$-adic integers as both power series and as Witt vectors (though they clearly don't use the name). – Brandon Carter Jul 11 '12 at 22:10
What you have done seems correct. Also notice that from what you have done you get that $$d(a,c) \leq \max \{d(a,b), d(b,c) \},$$ which is stronger than the triangle inequality.
For an introduction to $p$-adic numbers, I would suggest Fernando Gouvea's $p$-adic Numbers: An introduction. It should be easy for an undergraduate to understand the book, and I think it is a very nice introduction.
Edit: I should add that it's not easy for any undergraduate. One should've had a few courses which have a lot of proofs, and not just the standard calculus courses.
-
Your argument looks fine to me.
A very good online introduction to $p$-adics are these notes; it covers modular arithmetic leading up to Hensel's, basic analysis with the numbers, the very strange metric topology of $\Bbb Q_p$ (every point inside of a ball is a center, there are locally but not globally constant functions, etc.), and a bit of field theory and algebraic number theory.
I don't really have any recommendation for number-theoretically in-depth stuff. Also, I tend to stick to online documents because the web is where I spend all my time anyway.
In actuality, my favorite discussion is p-adic integration and the theory of groups, which involves category theory and abstract algebra (the $p$-adics are constructed as an inverse limit of topological rings, for example), as well as measure theory and group theory. This source perhaps takes more background to digest its contents satisfactorily, but it hits my buttons well.
Just for fun, I also suggest Pictures of Ultrametric Spaces. Related entertainment: (working with adeles) the character group of $\Bbb Q$ and (working with profinite integers) profinite Fibonacci numbers.
- | 2014-04-19T15:27:41 | {
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http://math.stackexchange.com/questions/75005/how-do-we-check-randomness | # How do we check Randomness? [duplicate]
Let's imagine a guy who claims to possess a machine that can each time produce a completely random series of 0/1 digits (e.g. $1,0,0,1,1,0,1,1,1,...$). And each time after he generates one, you can keep asking him for the $n$-th digit and he will tell you accordingly.
Then how do you check if his series is really completely random?
If we only check whether the $n$-th digit is evenly distributed, then he can cheat using:
$0,0,0,0,...$
$1,1,1,1,...$
$0,0,0,0,...$
$1,1,1,1,...$
$...$
If we check whether any given sequence is distributed evenly, then he can cheat using:
$(0,)(1,)(0,0,)(0,1,)(1,0,)(1,1,)(0,0,0,)(0,0,1,)...$
$(1,)(0,)(1,1,)(1,0,)(0,1,)(0,0,)(1,1,1,)(1,1,0,)...$
$...$
I may give other possible checking processes but as far as I can list, each of them has flaws that can be cheated with a prepared regular series.
How do we check if a series is really random? Or is randomness a philosophical concept that can not be easily defined in Mathematics?
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## marked as duplicate by egreg, LTS, user127096, voldemort, Claude LeiboviciApr 12 at 4:52
"...is randomness a philosophical concept that can not be easily defined in Mathematics?" - pretty much. However we do want our (pseudo)random sequences to pass certain tests... – Ｊ. Ｍ. Oct 23 '11 at 5:17
Any particular set of tests can be cheated if you know what is. For theoretical cryptography purposes, one can define a bit generator as random if it passes "all tests that run in a reasonable time". More precisely, a proposed random bit generator is $(n,t,\epsilon)$-random if, given the first $n$ bits of the sequence, there is no randomized algorithm that runs within time $t$ and successfully predicts the next bit with probability outside the range $1/2 \pm \epsilon$. – Ted Oct 23 '11 at 7:36
Interesting. So ... "irrationality", since it cannot be confirmed by looking at a finite number of decimals in a number, is a philosophical concept that cannot be easily defined in mathamatics ?? "Continuity", since it cannot be confirmed by a finite number of function evaluations is a philosophical concept that cannot be easily defined in mathamatics. – GEdgar Oct 26 '11 at 13:20
All the sequences you mentioned have a really low Kolmogorov complexity, because you can easily describe them in really short space. A random sequence (as per the usual definition) has a high Kolmogorov complexity, which means there is no instructions shorter then the string itself that can describe or reproduce the string. Ofcourse the length of the description depends on the formal system (language) you use to describe it, but if the length of the string is much longer then the axioms of your formal systems, then the Kolmogorov-complexity of a random string becomes independent of your choice of system.
Luckily, under the Church-Turing thesis, there is only 1 model of computation,(unless your machine uses yet undiscovered physical laws), so there is only 1 language your machine can speak that we have to check.
So to test if a string is random, we only have to brute-force check the length of the shortest Turing-program that outputs the first n bits correctly. If the length eventually becomes proportional to n, then we can be fairly certain we have a random sequence, but to be 100% we have to check the whole (infinite) string. (As per definition of random).
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Leaving aside the theoretical aspect of your question, there are also pragmatic answers to it because there are real world uses for high-quality random generators (whether hardware or algorithmic). For statistical uses, "statistical randomness" is a used. For example you can use these "diehard tests" or TestU01.
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This is discussed very nicely in Volume 2 of Knuth's The Art Of Computer Programming. The executive summary is that randomness is a mathematical concept that can be defined in mathematics but not easily.
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Part A: Functions of Two Variables, Tangent Approximation and Opt; Part B: Chain Rule, Gradient and Directional Derivatives; Part C: Lagrange Multipliers and Constrained Differentials; Exam 2. Solve this system of equations by using substitution. You define a function in much the same way you define a variable. Use the debugger to see what's the mismatch in dimensions; it's not totally apparent as one would presume i is a loop index and so is a single integer value; if MS3 is an array it would also be a single value but if it happened to be a function it could return something other than. Functions of Several Variables This manual contains solutions to odd-numbered exercises from the book Functions of Several Vari-ables by Miroslav Lovri´c, published by Nelson Publishing. Note that it is assumed that the two lists given in the table command are both factors. Economists of this period, while recognizing that the law of diminishing returns (or the law of variable proportions) applied when units of a variable. However, there is also a main di⁄erence. Find the standard deviation of the eruption duration in the data set faithful. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify Statistics Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. The added risk brought on by the complexity of machine-learning models can be mitigated by making well-targeted modifications to existing validation frameworks. For functions of two or three variables the situation is more complicated because there are infinitely. • Matlab has several different functions (built-ins) for the numerical. Staffing: After a manager discerns his area's needs, he may decide to beef up his staffing by recruiting, selecting, training, and developing employees. We now extend this concept to functions of two variables. com - Functions. Swapping two variables refers to mutually exchanging the values of the variables. When we considered functions and graphs of one variable, one of the first things we did was to transform those graphs through shifts and stretches. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. If you will need guidance with algebra and in particular with ordered pairs and inequalies online calculator or fractions come visit us at Algebra-equation. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. Active 2 years, 7 months ago. Gain additional perspective by studying polar plots, parametric plots, contour plots, region plots and many other types of visualizations of the functions and equations of interest to you. Jacobians of Random Graphs Acknowledgments Funding References 2000 AMS Subject Classification: 05C31, 05C50, 14T05, 14H99. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). Okay, as if two methods aren't enough, we still have one more method we could use. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). I am now trying to find a general equation f(x1,x2). Some students did not show to have made this coordination. The code on the left below shows one failed attempt at an implementation. For a continuous real-valued function of two real variables, the graph is a surface. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. Because the correlation between reading and mathematics can be determined in the top section of the table, the correlations between those two variables is not repeated in the bottom half of the table. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. Algebra functions lessons with lots of worked examples and practice problems. Alternatively, the function also knows it must return the first argument, if the value of the "number" parameter, passed into the function, is equal to "first". Definition of Mathematical Expectation Functions of Random Variables Some Theorems on Expectation The Variance and Standard Deviation Some Theorems on Variance Stan-dardized Random Variables Moments Moment Generating Functions Some Theorems on Moment Generating Functions Characteristic Functions Variance for Joint Distribu-tions. To use or explore these add-ons: Create or open a spreadsheet in Google Sheets. The partial derivative of f with respect to y can similarly be found by treating x as a constant whenever it appears. second variable y appears, it is treated as a constant in every respect. The concept of the graph of a function is generalized to the graph of a relation. Part 1: Functions of 2 Variables. input variables and other variables you create within the function and in doing so, you create the output variables you desire. You have now created a function called sum. 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In the next two sections we introduce these two concepts and develop some of their properties. 10 Two-Dimensional Random Variables Definition 1. Addition of two numbers in C For example, if a user will input two numbers as; '5', '6' then '11' (5 + 6) will be printed on the screen. Let (X;d)and (Y;d′)be two metric spaces, A ⊆X a nonempty set, a function f ∶A →Y and x. f Obviously. Derivatives told us about the shape of the function, and let us find local max and min - we want to be able to do the same thing with a function of two variables. Relation with other tests Changing the number of variables. We also write z = f (x ,y ) The variables x and y are independent variables and z is the. even functions of one variable may have both maximum and minimum points). The function writePictureTo takes two parameters: the picture variable and the pathname. First-order partial derivatives of functions with two variables. The area of a circle is a function of -- it depends on -- the radius. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. In the expression (c = a + b) overflow may occur if the sum of a and b is larger than the maximum value which can be stored in the variable c. In the short run, production function is explained with one variable factor and other factors of productions are held constant. ) Variables and functions in all parts of a makefile are expanded when read, except for the shell commands in rules, the right-hand sides of variable definitions using `=', and the bodies of variable definitions using the define directive. Now you know the basics of using two variable -- or complex -- functions. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. We recall that a critical point of a function of several variables is a point at which the gradient of the function is either the zero vector 0 or is undefined. Let f : D ⊂ R → R and let a ∈ R. To plot the point (2,3), for example, you start at the origin Independent and Dependent Variables. Scalar functions of two variables Our main goal in this tutorial is to explore ways to plot functions of two variables. f how does one obtain. 2 Graphs should always have at minimum a caption, axes and scales, symbols, and a data field. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. Usually, there is more than one correct answer. And the fun part with these guys is that you can just kind of, imagine a fluid flowing, so here's a bunch of droplets, like water, and they kind of flow along that. x^2*y+x*y^2 ） The reserved functions are located in " Function List ". I am trying to create the interpolating function for a function of two variables, over a finite area. It can be used as a worksheet function (WS) in Excel. • Matlab has several different functions (built-ins) for the numerical. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) reflecting the fact that there are only two directions from which x can approach x0, the right or the left. Concave functions of two variables While we will not provide a proof here, the following three definitions are equivalent if the function f is differentiable. There is no need to list the 3 twice. Definition of Variables and Examples. identically distributed Exponential random variables with a constant mean or a constant parameter (where is the rate parameter), the probability density function (pdf) of the sum of the random variables results into a Gamma distribution with parameters n and. accept a wide variety of mathematical expressions. To close the answer window and get back to the quiz, click on the X in the upper right corner of the answer window. Equations of a Straight Line. Functions of two variables 1. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = ˆ cx2 + xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2. Functions can be recognized, described, and examined in a variety of ways, including graphs, tables, and sets of ordered pairs. Could someone please explain a function of two variables to me. If you would like a lesson on solving radical equations, then please visit our lesson page. Hence the square of a Rayleigh random variable produces an exponential random variable. Therefore, in order to be able to. The applet initially starts in the Input mode, which lets you choose a function to plot (you can either enter it manually, or select one from the drop-down list; click on the Plot button to create the new plot). Observe that because of the non-negativity constraint, the sum of any collection of variables cannot be negative. Functions of Two Variables. Variable b1 and b2 are baseline variables. Laval (KSU) Functions of Several Variables Today 14 / 22. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. One important similarity to notice between the limit of a one variable function and the limit of a two variable function is that $\sqrt{(x - a)^2 + (y - b)^2}$ represents the distance between the point $(x, y)$ and $(a, b)$ in $\mathbb{R}^2$. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of. My function is exponential for x1 with two coefficients that depend on x2: f(x1,x2)=a*(x1)^b, where a and b are functions of x2. Files are available under licenses specified on their description page. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. The scatter plot plots the points (x, y) where x is a value from one data list (Xlist) and y is the corresponding value from the other data list (Ylist). A function f(x, y) of two independent variables has a maximum at a point (x 0, y 0) if f(x 0, y 0) f(x, y) for all points (x, y) in the neighborhood of (x 0. Laval (KSU) Functions of Several Variables Today 14 / 22. The outputs are real numbers. As the n -tuple x = (x1, x2, , xn) varies in X, a subset of ℝn, Implicit functions. Chain Rule And Composite Functions Derivative of Composite Function with the help of chain rule: When two functions are combined in such a way that the output of one function becomes the input to another function then this is referred to as composite function. Lady (September 5, 1998) There are three ways that a function can be discontinuous at a point. Distributions of Functions of Random Variables 1 Functions of One Random Variable Case of two-to-one transformations. You can choose any other combination of numbers as well. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of matplotlib to create a shaded surface plot. 4 Higher partial derivatives Notice that @f @x and @f @y are themselves functions of two variables, so they can also be partially differenti-ated. Functions of three variables are similar in many aspects to those of two variables. Because we're trying to keep things a little bit simpler, we'll concentrate on functions of two variables. Limits of a Rational Function of Two Variables Roger B. When variables change together, their interaction is called a relation. The dependent variable is what is affected by the independent variable-- your effects or outcomes. In the above example, two variables, num1 and num2 are passed to function during function call. Example 3: Using the function from Example 2, describe and graph the following functions: (i) f(x, y) = 3 - x2 - y2. Topic 5: Functions of multivariate random variables † Functions of several random variables Sum of 2 random variables † Let X and Y be two random variables. These are just constant functions, and because of that, degree 0 polynomials are often called constant polynomials. Equations of a Straight Line. For functions of two or three variables the situation is more complicated because there are infinitely. This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem. Average value of a function To find the average value of a function of two variables, let's start by looking at the average value of a function of one variable. It seems reasonable, and can be shown to be true,. FUNCTION OF TWO VARIABLES Definition: A variable Z is said to be a function of two independent variables x and y denoted by z=f (x,y) if to each pair of values of x and y over some domain D f ={(x,y): a type:. You have now created a function called sum. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. which is the density for an exponential random variable with parameter = 1/(2 2a), as can be seen from inspection of (2-27). Sometimes it will be preferable to think of f as taking one (2-dimensional) vector input instead of two scalar inputs. Optimization Problems with Functions of Two Variables. Our discussion is not limited to functions of two variables, that is, our results extend to functions of three or more variables. As with single variable functions, two classes of common functions are particularly useful and easy to describe. The variables held fixed are viewed as parameters. User make a function named swap that will be called in other class. For a function of n variables it can be a maximum point, a minimum point or a point that is analogous to an inflection or saddle point. two variables y et z is put equal to zero, then either variable is defined by the other and thus a function of this variable emerges, since before they were not mutually dependent. Write a script m-flle and use the Control System Toolbox functions ss and ltiview to form the state model and its step response. Furthermore, sums, dif-. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. • Matlab has several different functions (built-ins) for the numerical. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. I suspect I will need the surface chart but can some one tell me how to generate the chart and what to enter on the worksheet. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. †Forcontinuous randomvariables. The set D is the domain of f and its range is the set of values that f takes on. Following are different ways. Does anyone know of any helpful tutorials that will help me get the Domain and range, functions of 2 variables | Physics Forums. The add-on store offers several custom functions as add-ons for Google Sheets. For a thermal contact between the two put a thermal conductance value. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). Two expressions involving template parameters are called equivalent if two function definitions that contain these expressions would be the same under ODR rules, that is, the two expressions contain the same sequence of tokens whose names are resolved to same entities via name lookup, except template parameters may be differently named. I will give the definition of differentiablity in 2D. Quotient of two random variables. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. There is no need to list the 3 twice. The sum of two incomes, for example, or the difference between demand and capacity. 2 Limits and Continuity of Functions of Two Variables In this section, we present a formal discussion of the concept of continuity of functions of two variables. More generally, if two or three variables are changing, how do we explore the correspondingchangein w? The answer to these questionsstarts with the generalizationof the idea of the differential as linear approximation. Local extreme values of a function of two variables. Limits and Continuity of Functions of Two or More Variables Introduction. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. The euclidean_division function to calculate online the quotient and the remainder in the euclidean division of two polynomials or two integers. This gives a nice graphical representation where the plane at x = 0 bounds the function from below. Continuous Random Variables Acontinuous random variable X takes values in an interval of real numbers. functions of two variables. In this paper distribution of zeros of solutions of functional equations in the space of functions of two variables is studied. So far, we have discussed how we can find the distribution of a function of a continuous random variable starting from finding the CDF. Functions f (x1, x2, , xn) of n variables, Symmetry. That is, a function expresses dependence of one variable on one or more other variables. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). Recall that the definition of the limit of such functions is as follows. More information about applet. In single-variable calculus we were concerned with functions that map the real numbers $\R$ to $\R$, sometimes called "real functions of one variable'', meaning the "input'' is a single real number and the "output'' is likewise a single real number. Not only for computing the variance of the transformed variable Y, but also for its mean. When variables change together, their interaction is called a relation. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) reflecting the fact that there are only two directions from which x can approach x0, the right or the left. The graph of is a subset of three-dimensional Euclidean space with coordinates , given by the equation: Equivalently, it is the set of points: Pictorially, this graph looks like a surface for a nice enough function. $\endgroup$ - Gerhard Paseman Feb 13 at 18:10. Fortunately, the functions we will examine will typically be continuous almost everywhere. The function makes it possible to verify by using the Pythagorean theorem knowing the lengths of the sides of a triangle that this is a right triangle. For a continuous real-valued function of two real variables, the graph is a surface. Most useful functions of one variable are con-tinuous, but there are a few exceptions. characterizations, namely, the mass function for discrete random variable and the density function for continuous random variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. In a "system of equations," you are asked to solve two or more equations at the same time. Re: st: computing covariance. com, a free online graphing calculator. To evaluate z, first create a set of (x,y) points over the domain of the function using meshgrid. functions of several variables and partial differentiation (2) The simplest paths to try when you suspect a limit does not exist are below. 1 Visualizing functions of 2 variables One problem with thinking about functions of several variables is that they can be harder to picture than functions of just one variable. Imagine a surface, the graph of a function of two variables. Functions of 2 Variables Functions and Graphs In the last chapter, we extended di⁄erential calculus to vector-valued functions. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = ˆ cx2 + xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2. Hence, time is always on the X axis. Functions 3D Plotter and Analytic double integrator Functions 3D Plotter is an on line app to plotting two-variabled real functions, ie functions of type f(x,y) or with more precision f: R 2 → R (x,y) → f(x,y) 3D Functions Plotter calculates double integrals in analytic or numeric form. If the relation is not a function the graph contains at least two points with the same x-coordinate but with different y-coordinates. There is a probability associated with X falling between two numbers a weekday ) ) ) { $datemonth =$wp_locale->get_month( $datefunc( 'm',$i ) ); $datemonth_abbrev =$wp_locale->get_month_abbrev. For functions of two or three variables the situation is more complicated because there are infinitely. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more!. Again, please enter this line into. Functions of more variables can be defined similarly. It would be useful to read these two guides. One primary difference, however, is that the graphs of functions of more than two variables cannot be visualized directly, since they have dimension greater than three. The concept of the graph of a function is generalized to the graph of a relation. Thread: chart function of two variables. Furthermore, sums, dif-. For example, if you are studying the effects of a new educational program on student achievement, the program is the independent variable and your measures of achievement are the dependent ones. You can choose any other combination of numbers as well. f(x,y) is inputed as "expression". Fortunately for us, we have technology which facilitates this task. Use Wolfram|Alpha to generate plots of functions, equations and inequalities in one, two and three dimensions. But polynomials, trig functions, power and root functions, logarithms, and exponential func-tions are all continuous. Suppose that X and Y are two random variables having moment generating functions MX(t) and MY (t) that exist for all t in some interval 3. In the present case, we see that the critical point at the origin is a local maximum of f2 , and the second critical point is a saddle point. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The standard deviation of an observation variable is the square root of its variance. It is good programming practice to avoid defining global variables and instead to put your variables inside functions and explicitly pass them as parameters where needed. I'm having a bit of trouble grasping the domain and range of functions of 2 variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. Boolean Functions (Expressions) It is useful to know how many different Boolean functions can be constructed on a set of Boolean variables. To input the variable x as a Wildcard, first type Shift + ?, then type x; similarly, for y. Importantly,. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. Local extreme values of a function of two variables. Dependent has two categories, there is only one discriminant function. When variables change together, their interaction is called a relation. In general, I can't create new functions in a poisoned session. That is, a function expresses dependence of one variable on one or more other variables. Hi, It is possible to define a function of two variables using an interpolation function defined in a text file using the spreadsheet data format: the first column contains the values for the first input argument, the second column contains the values for the seond input argument, and the third column contains the function's value. Function definition, the kind of action or activity proper to a person, thing, or institution; the purpose for which something is designed or exists; role. So this is more like a re-visit to the good old topic. As the other answer shows, the mere existence of partial derivatives doesn't even guarantee that the function is continuous. Notice we kept that one dimensional distance in our limit definition for functions of two variables when we said |f(x, y) - L| < e. Applications of Extrema of Functions of Two Variables. Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). If not, then we will want to test some paths along some curves to first see if the limit does not exist. Also, use ss2tf to obtain the fllter’s transfer function. Fortunately for us, we have technology which facilitates this task. You can create a two way table of occurrences using the table command and the two columns in the data frame: In this example, there are 51 people who are current smokers and are in the high SES. \+,œTÐ+Ÿ\Ÿ,Ñœ0ÐBÑ. 16 Possible Functions of Two Variables. ” For example, how much you weigh is related (correlated) to how much you eat. Correlation look at trends shared between two variables, and regression look at causal relation between a predictor (independent variable) and a response (dependent) variable. f Obviously. In case of two independent variables X 1 and X 2 such a function may be expressed as under: Y = a + bX 1 - cX 2 1 + dX 2 - eX 2 2. First, we introduce the de nition of a function of two variables: A scalar-valued. Equations of a Straight Line. The standard deviation of an observation variable is the square root of its variance. Polynomial Calculator. In mathematics, the result of a modulo operation is the remainder of an arithmetic division. Graph the function f(x,y) = xy using x,y,z-coordinate axes in 3-D space. It seems reasonable, and can be shown to be true,. Just for consistency we can think of a function:. Loading Graph Functions of 2 Variables. lang package, and not in the java. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). 2 to find the resulting PDFs. When we extend this notion to functions of two variables (or more), we will see that there are many similarities. For functions of two or three variables the situation is more complicated because there are infinitely many. I am now trying to find a general equation f(x1,x2). Long weekends and highway traffic on Friday afternoon C. First-order partial derivatives of functions with two variables. Applications of Extrema of Functions of Two Variables. Could someone please explain a function of two variables to me. function of two variables is far more di¢ cult than a function of one variable. Calculates the table of the specified function with two variables specified as variable data table. The INTERSECTION of two sets is the set of elements which are in both sets. You define a function in much the same way you define a variable. The value of num1 and num2 are initialized to variables a and b respectively. Function composition. The partial derivative of a function of two or more variables with respect to one of its variables is the ordinary derivative of the function with respect to that variable, considering the other variables as constants. In the example above, the diagonal was used to report the correlation of the four factors with a different variable. time) and one or more derivatives with respect to that independent variable. y(s;t) and z(s;t), are called the component functions of the vector-valued function g. The Method of Transformations. Examples 4. Here that means you need to use the. I found a and b for several values of x2, so I do have equations f(x1) for some fixed x2. Furthermore, sums, dif-. | 2019-10-15T10:55:36 | {
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http://math.stackexchange.com/questions/129993/calculating-n-mod-pq-from-n-mod-p-and-n-mod-q | # Calculating n mod pq from n mod p and n mod q
Let $p$ and $q$ be relative primes, $n$ positive integer.
Given
• $n\bmod p$ and
• $n\bmod q$
how do I calculate $n\bmod (pq)$ ?
-
en.wikipedia.org/wiki/… – pedja Apr 10 '12 at 11:26
You might also have a look at Easy CRT in some Bill Dubuque's posts. – Martin Sleziak Apr 10 '12 at 11:30
Since $p$ and $q$ are relatively prime, there are integers $a$ and $b$ such that $ap+bq=1$. You can find $a$ and $b$ using the Extended Euclidean algorithm. Then $n\equiv aps+bqr \bmod pq$ if $n\equiv r \bmod p$ and $n\equiv s \bmod p$.
As pedja mentioned, this is a constructive proof of the Chinese remainder theorem.
-
+1 for getting their first !!! – hardmath Apr 10 '12 at 20:48
From the fact that $p,q$ are relatively prime, we can find coefficients $a,b$ such that:
$$ap + bq = 1$$
With these coefficients we can piece together a solution for n from its residues modulo $p$ and $q$. Say:
$$n \equiv r \mod p$$ $$n \equiv s \mod q$$
Then this works: $n = sap + rbq$ since:
$$bq \equiv 1 \mod p$$ $$ap \equiv 1 \mod q$$
Of course the above expression for $n$ can be reduced modulo $pq$ without affecting the residues modulo $p$ and $q$.
-
One may use the Bezout identity $\rm\:a\:p + b\:q = 1\:$ obtained by the extended Euclidean algorithm. But, in practice, it's often more convenient to use the form below, e.g. see my Easy CRT posts.
Theorem (Easy CRT) $\rm\ \$ If $\rm\ p,\:q\:$ are coprime integers then $\rm\ p^{-1}\$ exists $\rm\ (mod\ q)\ \$ and
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ a\ (mod\ p) \\ \rm n&\equiv&\rm\ b\ (mod\ q)\end{eqnarray} \ \iff\ \ n\ \equiv\ a + p\ \bigg[\frac{b-a}{p}\ mod\ q\:\bigg]\ \ (mod\ p\:\!q)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ p\!:\:\ n\equiv a + p\ (\cdots)\equiv a\:,\$ and $\rm\ mod\ q\!:\:\ n\equiv a + (b-a)\ p/p \equiv b\:.$
$\rm\ (\Rightarrow)\ \$ The solution is unique $\rm\ (mod\ p\!\:q)\$ since if $\rm\ x',\:x\$ are solutions then $\rm\ x'\equiv x\$ mod $\rm\:p,q\:$ therefore $\rm\ p,\:q\ |\ x'-x\ \Rightarrow\ p\!\:q\ |\ x'-x\ \$ since $\rm\ \:p,\:q\:$ coprime $\rm\:\Rightarrow\ lcm(p,q) = p\!\:q\:.\quad$ QED
-
The link goes to emptiness... – I. J. Kennedy May 1 '13 at 0:02 | 2015-10-09T12:42:21 | {
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https://www.12000.org/my_notes/pde_in_CAS/maple_2020_2_and_mma_12_2/insu9.htm | #### 2.1.3 Transport equation $$u_t+2 u_x = 0$$ IC $$u(-1,x)=\frac {x}{1+x^2}$$. Peter Olver textbook, 2.2.2 (b)
problem number 3
Taken from Peter Olver textbook, Introduction to Partial differential equations.
Solve for $$u(t,x)$$ in $$u_t+2 u_x = 0$$ with IC $$u(-1,x)=\frac {x}{1+x^2}$$
Mathematica
$\left \{\left \{u(t,x)\to \frac {-2 t+x-2}{4 t^2-4 t (x-2)+x^2-4 x+5}\right \}\right \}$
Maple
$u \left (t , x\right ) = \frac {-2 t +x -2}{\left (-2 t +x -2\right )^{2}+1}$
Hand solution
Solve $u_{t}+2u_{x}=0$ With initial conditions $$u\left ( -1,x\right ) =\frac {x}{1+x^{2}}$$.
Solution
Let $$u=u\left ( x\left ( t\right ) ,t\right )$$. Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =2\tag {4} \end {align}
Eq (3) says that $$u$$ is constant on the chataterstic lines, or $$u=u\left ( x\left ( -1\right ) \right )$$. Using the given initial conditions, this becomes \begin {equation} u\left ( x\left ( t\right ) ,t\right ) =\frac {x\left ( -1\right ) }{1+x\left ( -1\right ) ^{2}}\tag {5} \end {equation} Eq (4) is now used to find $$x\left ( -1\right )$$. Soving (4) gives $$x=x\left ( 0\right ) +2t$$. Hence $$x\left ( -1\right ) =x\left ( 0\right ) -2$$ or $$x\left ( 0\right ) =x\left ( -1\right ) +2$$. Therefore \begin {align*} x & =x\left ( -1\right ) +2+2t\\ x\left ( -1\right ) & =x-2-2t \end {align*}
Now that we found $$x\left ( -1\right )$$, we substitute it in (5), giving the solution$u\left ( x\left ( t\right ) ,t\right ) =\frac {x-2-2t}{1+\left ( x-2-2t\right ) ^{2}}$ Alternative method. Using Lagrange-charpit method
$\frac {dt}{1}=\frac {dx}{2}=\frac {du}{0}$ Which implies that $$du=0$$ or $$u=C_{1}$$. A constant. Integrating $$\frac {dt}{1}=\frac {dx}{2}$$ gives $$t=\frac {1}{2}x+C_{2}$$ or $$C_{2}=t-\frac {1}{2}x$$. But $$C_{1}=F\left ( C_{2}\right )$$ always, where $$F$$ is arbitrary function. Since $$C_{1}=u$$ then\begin {align} u & =F\left ( C_{2}\right ) \nonumber \\ u & =F\left ( t-\frac {1}{2}x\right ) \tag {1} \end {align}
At $$t=-1$$ the above becomes$\frac {x}{1+x^{2}}=F\left ( -1-\frac {1}{2}x\right )$ Let $$-1-\frac {1}{2}x=z$$ which implies $$x=-2\left ( 1+z\right )$$ The above can be written as\begin {align*} \frac {-2\left ( 1+z\right ) }{1+\left ( -2\left ( 1+z\right ) \right ) ^{2}} & =F\left ( z\right ) \\ F\left ( z\right ) & =-\frac {2\left ( 1+z\right ) }{4z^{2}+8z+5} \end {align*}
From the above then (1) can be written as\begin {align*} u\left ( t,x\right ) & =-\frac {2\left ( 1+\left ( t-\frac {1}{2}x\right ) \right ) }{4\left ( t-\frac {1}{2}x\right ) ^{2}+8\left ( t-\frac {1}{2}x\right ) +5}\\ & =\frac {x-2t-2}{4t^{2}-4tx+8t+x^{2}-4x+5}\\ & =\frac {x-2t-2}{1+\left ( x-2-2t\right ) ^{2}} \end {align*}
The following is an animation of the solution
3D 2D
Source code used for the above
________________________________________________________________________________________ | 2021-12-06T14:53:15 | {
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http://phoxis.org/2012/05/11/number-of-trailing-zeros-in-factorial-of-an-integer/ | ## Number of trailing zeros in factorial of an integer
An integer n is given, the task is to find the number of trailing zeros in n! .
### Solution
First we need to see what produces trailing zeros. We assume that the integers are represented in base 10. A pair of 5 and 2 (which are the prime factors of 10) will produce a trailing zero in a sequence of integer multiplication. The number of trailing zeros will not decrease as once a trailing zero is generated in a product it cannot be removed by multiplying with some non zero integer. For factorial of n the multiplication series is: $n\times (n-1)\times(n-2)\times \ldots \times 3 \times 2 \times 1$ . We need to count number of pairs of 5 and 2 present as a factor of each integer in the series. Naturally there are at least equal number of factors 2 than 5, if not more. Therefore basically the number of times 5 occurs as a factor totally in the series, ie. the multiplicity of the prime factor 5 in the factorial represents the number of trailing zeros.
For example in the case of 25! , the factor 5 occurs 6 times totally in the series of multiplication $25 \times 24 \times \ldots \times 3 \times 2 \times 1$. 5, , ie. the multiplicity of the factor 5 is 6. The integers 10, 15, 20 is divided by 5 once, and twice for 25.
Therefore now the problem decreases to count the number of times the factor 5 occurs in the given series of multiplication. Which can be found as below:
$trailing zeros = \sum_{i=1}^{k} \left\lfloor n/5^i \right\rfloor = \left\lfloor n/5 \right\rfloor + \left\lfloor n/5^2 \right\rfloor + \left\lfloor n/5^3 \right\rfloor + \ldots + \left\lfloor n/5^k \right\rfloor$ such that $5^k \le n$
Here is some explanation. The term $\left\lfloor n/5^i \right\rfloor$ finds the number of integers between 1 and n that $5^i$ divides. For example for n = 27, $\left\lfloor 27/5 \right\rfloor = 5$, that is there are 5 integers divisible by 5 within the range 1 to 27, in other words there is 5 integers which has the factor 5 once, which are 5, 10, 15, 20, and 25. Also $\left\lfloor 27/5^2 \right\rfloor = 1$ , that is, there are 1 integer divisible by 25 within the range 1 to 27, or in other words there is one integer which has the factor 5 twice. The total number of occurrence of the factor 5 is calculated by computing the above formula, which divides and adds the number of times a power of factor 5 occurs in factorial of n. Therefore there are a total of 5 + 1 = 6 number of times the factor 5 occurs in the entire series of 1 to 27. Therefore there are 6 trailing zeros in 27! and 27! = 10888869450418352160768000000 which has 6 trailing zeros.
The implementation is pretty straight forward.
### Sourcecode
#include <stdio.h>
/* long int n : the input factorial integer
* long int i : holds value 5^k , the divisor of n' to
* find the number of times the factor 5 occurs
* in the factorial of n'
* long int c : number of times the 5 occurs as a factor in the
* factorial of n'
* long int t : the number of times the factor 5^i occurs in the n!'
*/
int
main (void)
{
long int n, i = 5, c = 0, t;
printf ("\nEnter n: ");
scanf ("%ld", &n);
do
{
t = n / i; /* computes floor (n/i) where i=5^k */
c += t; /* update the number of occurrence of factor 5 */
i *= 5; /* update i' to be the next power of 5 */
}
while (t != 0);
printf ("Number of trailing zeros in %ld! : %ld\n", n, c);
return 0;
}
### Output
Some sample outputs are given below.
run 1
Enter n: 3
Number of trailing zeros in 3! : 0
run 2
Enter n: 6
Number of trailing zeros in 6! : 1
run 3
Enter n: 15
Number of trailing zeros in 15! : 3
run 5
Enter n: 60
Number of trailing zeros in 60! : 14
run 6
Enter n: 100
Number of trailing zeros in 100! : 24
run 7
Enter n: 1000
Number of trailing zeros in 1000! : 249
run 8
Enter n: 5000
Number of trailing zeros in 5000! : 1249
run 9
Enter n: 10000
Number of trailing zeros in 10000! : 2499
run 10
Enter n: 12345678
Number of trailing zeros in 12345678! : 3086416
Homo-sapiens
This entry was posted in Coding Discussions, Computer Science, Others and tagged , , , , . Bookmark the permalink.
### 2 Responses to Number of trailing zeros in factorial of an integer
1. Cassie says:
Fantastic post! This site also gives a good explanation of how to come up with the algorithm, and also an implementation in Java:
http://www.programmerinterview.com/index.php/java-questions/find-trailing-zeros-in-factorial/
• phoxis says:
Great site. Once you know the algorithm you can implement it in any language. For example here are some quick bash implementations.
#!/bin/bash
num=$1 i=5 t=1 while [$t -ne 0 ]
do
t=$((num/i)) count=$((count+t))
i=$((i*5)) done echo "Trailing zeros of$num ! = $count" Or we can use the factor command to count the number of 5 in each of the integer. Although this process is extremely inefficient as it will compute the factor of each integer from 1 to n, but still it works. #!/bin/bash num=$1
count=0
for ((i=1; i<=num; i++))
do
this_count=$(factor$i | cut -d':' -f 2 | tr ' ' '\n' | grep -c "\<5\>")
count=$((count + this_count)) this_count=0 done echo "$count"
`
Thanks for visiting. | 2014-07-23T01:38:16 | {
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http://mathhelpforum.com/calculus/155132-cross-multiplying-vectors.html | # Math Help - Cross multiplying vectors
1. ## Cross multiplying vectors
The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.
$(j-k)$ x $(k - i)$
I really can't even get started. A hint or general idea would be very helpful though.
2. Originally Posted by Riyzar
The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.
$(j-k)$ x $(k - i)$
I really can't even get started. A hint or general idea would be very helpful though.
You should know that $\vec{i}\times \vec{j}= \vec{k}$, $\vec{j}\times\vec{k}= \vec{i}$, and $\vec{k}\times\vec{i}= \vec{j}$. And, of course, that $\vec{u}\times\vec{v}= -\vec{v}\times\vec{u}$ which, among other things, implies $\vec{u}\times\vec{u}= 0$. Also that the cross product distributes over addition: $(\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
3. Hello, Riyzar!
$\text{Find the resulting vector }without\text{ using determinants: }\;(\vec j-\vec k) \times(\vec k - \vec i)$
Are those unit vectors? . $\begin{Bmatrix}\vec i &=& \langle 1,0,0\rangle \\ \vec j &=& \langle 0,1,0\rangle \\ \vec k &=& \langle 0,0,1\rangle \end{Bmatrix}$
$\left(\vec j - \vec k\right) \times \left(\vec k - \vec i\right) \;=\;\left(\vec j - \vec k\right )\times \vec k - \left(\vec j - \vec k\right)\times \vec i$
. . . . . . . . . . . . . . $=\; \left(\vec j \times \vec k\right) - \left(\vec k \times \vec k \right) - \left(\vec j \times \vec i\right) + \left(\vec k \times \vec i\right)$
. . . . . . . . . . . . . . $=\qquad \vec i \quad\;\; -\;\;\quad \vec 0 \quad - \quad (-\vec k) \quad + \quad \vec j$
. . . . . . . . . . . . . . $=\quad \vec i\;+\; \vec j\;+\;\vec k$
4. Originally Posted by HallsofIvy
Also that the cross product distributes over addition: $(\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
That is what I was not aware of, thank you! | 2015-08-30T09:10:37 | {
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https://math.stackexchange.com/questions/2638101/how-can-we-show-that-a-n-b-n-c-n-are-convergent-and-have-the-same-li | # How can we show that $(a_n), (b_n), (c_n)$ are convergent and have the same limit?
We have the following for $a \le b \le c >0$:
$A(a,b,c)=\frac{a+b+c}{3}, B(a,b,c)= (abc)^{1/3}, C(a,b,c)=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$.
Then we define the sequences $(a_n),(b_n), (c_n)$ by
$a_1=a, b_1=b, c_1=c,$
$a_{n+1}=A(a_n,b_n,c_n), b_{n+1}=B(a_n,b_n,c_n), c_{n+1}=C(a_n,b_n,c_n)$.
How can we show that $(a_n),(b_n), (c_n)$ are convergent and have the same limit?
I understand that $A(a,b,c)\ge B(a,b,c)$ and $B(a,b,c)\ge C(a,b,c)$
• @dxiv I think so. – MOP Feb 6 '18 at 5:13
• See this and this related questions, same idea applies here. – dxiv Feb 6 '18 at 5:17
• Assume you meant $0 < a \le b \le c$, and not that only $c$ is positive. – Macavity Feb 6 '18 at 6:27
We have $$A(x,y,z) = \mbox{ arithmetic mean of } x,y,z;$$ $$B(x,y,z) = \mbox{ geometric mean of } x,y,z;$$ $$C(x,y,z) = \mbox{ harmonic mean of } x,y,z.$$
It is well known that, for the same arguments, $$\mbox{ harmonic mean } \le \mbox{ geometric mean } \le \mbox{ arithmetic mean } \tag{1}$$ (see e.g. this Wikipedia article).
Using these inequalities we see that the largest (arithmetic) means $a_n$ form a non-increasing sequence, while the smallest (harmonic) means $c_n$ form a non-decreasing sequence. Both sequences are bounded: all terms are within the interval $[a,c]$. If a sequence is monotonic and bounded, it has a limit.
Can you finish by proving that the limits of $a_n$ and $c_n$ are the same? (Then $b_n$ necessarily has the same limit too because of the double inequality $(1)$ and the squeeze theorem.)
For $n>1$ we have $$a_{n+1}={a_n+b_n+c_n\over3}\le{a_n+a_n+c_n\over3}, \tag{2}$$ $$c_{n+1} \ge c_n. \tag{3}$$ Subtracting $(3)$ from $(2)$ we find $$a_{n+1}-c_{n+1} \le {2\over3} (a_n-c_n).$$ We observe that the intervals $[c_n,a_n]$ form a sequence of nested closed intervals, and the interval lengths tend to zero (no slower than a decreasing geometric progression). Therefore these intervals have a (unique) common point, and this point must be the limit of all three sequences because $a_n,b_n,c_n \in [c_n,a_n]$ for all $n>1$. (The common point is unique because the size of intervals $[c_n,a_n]$ tends to zero.) This completes the proof.
• I can understand that $(a_n)$ is decreasing sequence and $(c_n)$ is a increasing sequence. But could you please please help me with finding the limit? – MOP Feb 6 '18 at 5:32
• You are not required to find the limit. You just need to prove that the limits are the same. – Alex Feb 6 '18 at 5:37
• For example, you can prove that $a_{n+1}-c_{n+1}\le0.9(a_n-c_n)$, and then observe that the limit of each sequence is in the interval $[c_n,a_n]$ for all $n\in{\mathbb N}$. Note that the intervals $[c_n,a_n]$ form a sequence of nested intervals. – Alex Feb 6 '18 at 5:53
• I could't find $a_{n+1}-c_{n+1} \le \frac{1}{x} (a_n-c_n)$. I can understand it for two numbers case only. – MOP Feb 6 '18 at 7:48
hint: try proving $c_n$ is increasing and $a_n$ is decreasing and use squeeze theorem.
Assuming $a>0.$ For any positive $a',b',c'$ we have $$\max (a',b',c')\geq A(a',b',c')\geq B(a',b','c)\geq C(a',b',c')\geq \min (a',b',c').$$ Let $U_n=\max (a_n,b_n,c_n)$ and $L_n=\min (a_n,b_n,c_n).$ $$\text {We have }\quad U_n\geq U_{n+1}\geq L_{n+1}\geq L_n.$$ So $(U_n)_n$ is a deceasing sequence bounded below by $L_1$ and $(L_n)_n$ is an increasing sequence bounded above by $U_1 .$
Let $U=\lim_{n\to \infty}U_n$ and $L=\lim_{n\to \infty}L_n.$ Obviously $U\geq L\geq L_1>0.$
It suffices to show that $U=L.$
Let $r_n=U_n-L_n .$ Note that $U_n>r_n.$ $$\text {We have }\quad U_{n+1}\leq \frac {2U_n+L_n}{3}$$
$$\text {and }\quad L_{n+1}\geq \frac {3}{\frac {1}{U_n}+\frac {2}{L_n}}=\frac {3U_nL_n}{2U_n+L_n}.$$ $$\text {Therefore }\quad r_{n+1}\leq \frac {2U_n+L_n}{3}-\frac {3U_nL_n}{2U_n+L_n}=$$ $$=\frac {(4U_n-L_n)(U_n-L_n)}{3(2U_n+L_n)}=$$ $$=\frac {(3U_n+r_n)r_n}{9U_n-3r_n}\leq$$ $$\leq \frac {(4U_n)r_n}{6U_n}=\frac {2}{3}r_n.$$
So $U-L=\lim_{n\to \infty}r_n=0$ because $0\leq r_{n+1}\leq \frac {2}{3}r_n.$
• How did you show that $a_n$ is increasing and $b_n$ is increasing? – MOP Feb 6 '18 at 10:33
• I didn't understand your proof? Why did you take U and L? – MOP Feb 6 '18 at 11:22
• $U_n=\max (a_n,b_n,c_n).$... $L_n=\min (a_n,b_n,c_n).$... $U_{n+1}= \max (a_{n+1},b_{n+1},c_{n+1}) =$ $\max (\; A(a_n,b_n,c_n), B(a_n,b_n,c_n),C(a_n,b_n,c_n)\;)\;=$ $A(a_n,b_n,c_n)\leq \max (a_n,b_n,c_n)=U_n.$..... For $n>1$ we have $U_n=a_n$ and $L_n=c_n.$.... So $a_n$ is decreasing, not increasing..... If $U_n=\max (a_n,b_n,c_n)$ converges to $U$ and if $L_n= \min (a_n,b_n,c_n)$ converges to $L$ and if $U=L$ then $a_n, b_n,c_n$ each converge to $L$ also. – DanielWainfleet Feb 6 '18 at 23:18 | 2019-09-18T12:18:19 | {
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https://math.stackexchange.com/questions/2350315/equating-powers-in-a-series-expansion | # Equating powers in a series expansion
If the $(r+1)^{th}$ term contains the same power of $a$ and $b$ in the expansion of $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ find the value of $r$
I simply applied binomial and expanded for the general term and then equated the power of $a$ and $b$ and got $r=12$
My friend got $r=9$
I want to know which answer is correct.
Without trying to replicate your calculations, you're probably using version of the binomial theorem that count the terms from the opposite end. You can state the theorem either as $$(p+q)^n = \sum_{r=0}^n \binom{n}{r} p^r q^{n-r}$$ or as $$(p+q)^n = \sum_{r=0}^n \binom{n}{r} p^{n-r} q^r$$ which both give the same terms, just with a different numbering.
Note in particular that $9+12=21$, so you can both be right, if you're counting from different ends of the expansion!
On simplifying the following expression: $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ We get the expression: $$\frac{1}{(ab)^\frac72}\bigg(a^\frac12+b^\frac23\bigg)^{21}$$
So the (r$+1$)th term of the expansion is $$\binom{n}{r}a^{\frac{21-r-7}{2}}b^{\frac{4r-21}{6}}$$
Hence, the answer will come as: $$\frac{21-r-7}{2}=\frac{4r-21}{6}$$ $$\Rightarrow 42-3r=4r-21$$ $$\Rightarrow \boxed{\color{red}{r=9}}$$
The required answer is $9$.
• couldn't it be 12 as 12+9=21? – Atul Mishra Jul 8 '17 at 10:56
• The first step here looks somewhat obscure. Why not simplify to $$(a^{1/3}b^{-1/6}+a^{-1/6}b^{1/2})^{21}$$ and then solve $$\frac13r - \frac16(21-r) = -\frac16 r + \frac12(21-r)$$ – hmakholm left over Monica Jul 8 '17 at 10:59
• @AtulMishra Well, in a problem, where the binomial expression is rigorously given, it is expected that you consider the first term as the 'a' term and the second one 'b'. So if you consider the expansion in this order, your answer should be 9. I saw Henning's answer, but I dont think that you are allowed to reverse the expression and work. It may seem trivial that you cannot reverse the expression but if one is allowed to do so, then the ordering of the terms will become ambiguous and there will be no unique answer. Hope this helps. – SchrodingersCat Jul 8 '17 at 11:02
• Yes I am just arguing for the same, It will give r=12 – Atul Mishra Jul 8 '17 at 11:03
• @SchrodingersCat: There's no "reversing" going on -- there's just two ways to state the binomial theorem which are exactly equally good. It's not as if one of them is "the right way" and the other is "reversed". – hmakholm left over Monica Jul 8 '17 at 11:05 | 2020-10-27T14:23:31 | {
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https://math.stackexchange.com/questions/603830/why-does-trapezoidal-rule-have-potential-error-greater-than-midpoint/674350 | # Why does Trapezoidal Rule have potential error greater than Midpoint?
I can approximate the area beneath a curve using the Midpoint and Trapezoidal methods, with errors such that:
$Error_m \leq \frac{k(b-a)^3}{24n^2}$ and $Error_T \leq \frac{k(b-a)^3}{12n^2}$.
Doesn't this suggest that the Midpoint Method is twice as accurate as the Trapezoidal Method?
• The constant $k$ is not necessarily the same. It depends on $f$, differently for each method. Dec 12, 2013 at 5:30
• Here in both cases $k$ is the max of the (absolute value) of the second derivative. Then your inequalities give upper bounds on the error. One can construct $f$ such that for given $n$ the Trapezoidal Rule is dead on, while the Midpoint Rule is not. Dec 12, 2013 at 5:34
• 'k' is equal to the maximum of $f''(x)$ for both $E_m$ and $E_T$ Dec 12, 2013 at 5:43
• Even if the "worst case" for trapezoidal is half as good, the "typical case" could be much better. Dec 12, 2013 at 14:14
It has already been said in the comments that the error estimates you cite are upper bounds, so actual errors may be smaller and $E_M$ won't usually be exactly half of $E_T$ (and may actually be larger in some cases).
Nevertheless, it is well worth pointing out that if the function you are integrating happens to be a cubic polynomial, then we can make an exact statement: $$E_M=-\frac{1}{2}E_T.$$ You would probably never integrate a cubic polynomial numerically, but if the function you are integrating is well-approximated by a cubic polynomial on every subinterval used in the numerical integration, then the errors should still be related in roughly the same way.
The above relation obviously holds for the functions $f(x)=1$ and $f(x)=x$. Let's verify it by brute force for $f(x)=x^2$ and $f(x)=x^3$. Consider a single subinterval $[a,b]$ and let $A[f]$, $T[f]$, and $M[f]$ represent the exact area integral of $f$ on $[a,b]$, the trapezoidal estimate, and the midpoint estimate, respectively. Then \begin{aligned} A[x^2]&=\int_a^bx^2\,dx=\frac{b^3-a^3}{3},\\ T[x^2]&=\frac{b-a}{2}(b^2+a^2)=\frac{b^3-ab^2+a^2b-a^3}{2}\\ M[x^2]&=(b-a)\left(\frac{b+a}{2}\right)^2=\frac{b^3+ab^2-a^2b-b^3}{4}. \end{aligned} So \begin{aligned} E_T[x^2]&=T[x^2]-A[x^2]=\frac{b^3-a^3}{6}-ab\frac{b-a}{2},\\ E_M[x^2]&=M[x^2]-A[x^2]=-\frac{b^3-a^3}{12}+ab\frac{b-a}{4}=-\frac{1}{2}E_T[x^2].\\ \end{aligned} Likewise \begin{aligned} A[x^3]&=\int_a^bx^3\,dx=\frac{b^4-a^4}{4},\\ T[x^3]&=\frac{b-a}{2}(b^3+a^3)=\frac{b^4-ab^3+a^3b-a^4}{2}\\ M[x^3]&=(b-a)\left(\frac{b+a}{2}\right)^3=(b-a)\frac{b^3+3ab^2+3a^2b+a^3}{8}\\ &=\frac{b^4+2ab^3-2a^3b-a^4}{8}. \end{aligned} So \begin{aligned} E_T[x^3]&=T[x^3]-A[x^3]=\frac{b^4-a^4}{4}-\frac{ab}{2}(b^2-a^2),\\ E_M[x^3]&=M[x^3]-A[x^3]=-\frac{b^4-a^4}{8}+\frac{ab}{4}(b^2-a^2)=-\frac{1}{2}E_T[x^3].\\ \end{aligned} Since the statement holds for $1,$ $x,$ $x^2,$ and $x^3,$ it holds for all cubic polynomials by linearity of $A,$ $T,$ and $M.$
Another viewpoint on this is the following: if $T_n,$ $M_n,$ and $S_n$ represent the estimates given by the trapezoidal, midpoint, and Simpson's rules with $n$ subintervals (so $T$ and $M$ above are $T_1$ and $M_1$), then one finds that $$S_{2n}=\frac{T_n+2M_n}{3}.$$ If the error in $S_{2n}[f]$ is much smaller than the errors in $T_n[f]$ and $M_n[f]$, then it must be the case that $E_{M_n}[f]\approx-\frac{1}{2}E_{T_n}[f].$ This is, in fact, the case for many functions.
For an example of a function where $E_T[f]$ is much larger than $E_M[f],$ imagine a function which is nearly linear on the entire interval, but has a sharp peak or dip in a small neighborhood of the midpoint. For such a function, the $k$ in the error bound—it's the same $k$ in both bounds—would be big since the second derivative would be big in the vicinity of the peak or dip. There is no contradiction here since the trapezoidal error bound would be pretty poor in that case, while the midpoint error bound might be pretty reasonable.
A final comment: you can find diagrams in books explaining why $E_M$ tends to be less than $E_T$ and of opposite sign. I'm not sure that these diagrams provide a compelling reason to believe that $E_M$ is of roughly half the magnitude of $E_T,$ but I will give this some thought. The diagrams I have in mind represent the midpoint estimate by a trapezoidal area, where the diagonal of the trapezoid is tangent the the curve at the midpoint. It is clear that, if there's no inflection point in the interval, then, if the trapezoid of the midpoint rule is an overestimate of the integral, the trapezoid of the trapezoidal rule will be an underestimate of the integral, and vice versa.
Added: The midpoint rule is often presented geometrically as a series of rectangular areas, but it is more informative to redraw each rectangle as a trapezoid of the same area. These two presentations, in the case of a single interval, are shown below.
The slope of the top edge of the trapezoid has been chosen to match that of the curve at the midpoint. That the top edge of the trapezoid is the best linear approximation of the curve at the midpoint of the interval may provide some intuition as to why the midpoint rule often does better than the trapezoidal rule.
A series of pairs of plots is shown below. In each pair, the trapezoidal rule has been used on the left, and the midpoint rule has been used on the right.!
• So the midpoint formula is usually more accurate than the trapezoidal formula? Feb 13, 2014 at 3:56
• @citelao: if you cook up some functions that you can integrate exactly, and compare midpoint and trapezoidal estimates, I think you will find that the error in the midpoint estimate is usually about half as big and of opposite sign as that in the trapezoidal estimate. That's because the type of functions most people will cook up tend be smooth with second derivative that stays within reasonable bounds. You should always take care, however, especially if you plan to use the method on functions that aren't so smooth. Feb 13, 2014 at 12:12
• @will-orick but my intuition still suggests that the trapezoidal approximation is better; is this untrue? Feb 19, 2014 at 2:41
• @citelao: I've added some images to the post that may help with the intuition. Feb 21, 2014 at 11:28
• Wow, this slanted-roof representation of the midpoint rule is an eye-opener! Jul 27, 2020 at 21:12
On an interval where a function is concave-down, the Trapezoidal Rule will consistently underestimate the area under the curve. (And inversely, if the function is concave up, the Trapezoidal Rule will consistently overestimate the area.)
With the Midpoint Rule, each rectangle will sometimes overestimate and sometimes underestimate the function (unless the function has a local minimum/maximum at the midpoint), and so the errors partially cancel out. (They exactly cancel out if the function is a straight line.)
• If we consider a concave-down function (like your red curve), then, while it is true that on most intervals the rectangle given by the midpoint rule will be partly above and partly below the curve (which leads to partial cancellation of errors), one can show that the residual error is consistently positive, that is, the rectangles consistently overestimate the area under the curve. The question is whether this residual error is larger or smaller in magnitude than the error of the trapezoidal rule. I don't think your argument makes it evident that the residual error is smaller. Feb 13, 2014 at 21:06
• I think that with just a little more work, we can motivate the difference in the magnitude of the errors as well as their sign. See my answer. Dec 5, 2021 at 1:06
Will Orrick's great answer shows why if the function is concave up, then the trapezoid rule overestimates the integral and the midpoint rule underestimates it, and vice versa if the function is concave down. This gives a heuristic explanation for why the errors have opposite signs. But as he points out, it still isn't clear why the max midpoint error is smaller in magnitude.
To see intuitively why this is the case, we can add some more lines to his diagram:
The thick blue curve is the function to be integrated, the upper diagonal line is the top of the trapezoid from the trapezoidal rule, and the bottom diagonal line (which is tangent to the blue curve) is the top of the trapezoid with the same area as the rectangle given by the midpoint rule. The area of the blue shaded region is the error $$E_T$$ from the trapezoidal rule, and the area of the orange shaded region is (the absolute value of) the error $$E_M$$ from the midpoint rule. The middle diagonal lines connect the endpoints of the function's curve and its horizontal midpoint.
The entire shaded region is yet another trapezoid (the top and bottom sides are not quite parallel in general, but the side sides are). The vertical line running above the midpoint divides it into two sub-trapezoids. Each sub-trapezoid's diagonal divides it into two triangles that aren't quite congruent (because the parallel sides have slightly different lengths), but are very close to congruent if the function is reasonably close to linear over the interval, so these pairs of triangles have approximately equal areas. If the function's curve matched these diagonals, then the two rules would give (almost) equal and opposite errors, but we see from the fact that the curve is always to one side of these diagonals that the blue region has larger area ($$E_T$$) than the orange region ($$E_M$$).
We can roughly motivate the factor of 2 difference by noting that if you subtract off the lower large diagonal line (which makes the top line approximately horizontal as well, at this order), then the curve is more or less a parabola whose vertex is the midpoint - and it's a standard fact that over an interval that's symmetric about and tangent to any parabola's vertex, the area on the parabola's concave side is twice the area on its convex side.
This (heuristic) last step assumed that the function is close to parabolic, but it remains true if we consider an arbitrary cubic polynomial as well, because you can treat it as the sum of a quadratic term (to which the previous argument will apply) and a cubic term that vanishes at the endpoints and the midpoint. This cubic piece must be odd about the midpoint, so its integral will vanish and we can reuse the reasoning above from the simpler quadratic case. | 2022-06-26T15:23:30 | {
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https://math.stackexchange.com/questions/2883337/prove-that-there-exists-irrational-numbers-p-and-q-such-that-pq-is-rational | Prove that there exists irrational numbers p and q such that $p^{q}$ is rational
I found this on the lecture slides of my Discrete Mathematics module today. I think they quote the theorems mostly from the Susanna S.Epp Discrete Mathematics with Applications 4th edition.
Here's the proof:
1. We know from Theorem 4.7.1(Epp) that $\sqrt{2}$ is irrational.
2. Consider $\sqrt{2}^{\sqrt{2}}$ : It is either rational or irrational.
3. Case 1: It is rational:
3.1 Let $p=q=\sqrt{2}$ and we are done.
4. Case 2: It is irrational:
4.1 Then let p=$\sqrt{2}^{\sqrt{2}}$, and $q =\sqrt{2}$
4.2 p is irrational(by assumption), so is q (by Theorem 4.7.1(Epp))
4.3 Consider $p^{q} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$
4.4 $=(\sqrt{2})^{\sqrt{2} \times\sqrt{2}}$, by the power law
4.5 $=(\sqrt{2})^{2}=2$, by algebra
4.6 Clearly $2$ is rational
1. In either case, we have found the required p and q.
From what I understand from the proof, it's a clever construction of a number and splitting up into cases to prove that the given number is a rational number (one by clear observation and the other by some sort of contradiction). While the proof seems valid, though I somehow am forced to be convinced that the contradiction works, I wondered to myself if $\sqrt{2}^{\sqrt{2}}$ is actually rational since they constructed it.
My question would be if the example is indeed irrational, how is it possible an untrue constructed example could be used to verify this proof? If it's indeed rational, can someone tell me the $a/b$ representation of this number?
PS: Sorry for the formatting errors, I followed the MathJax syntax to the best of my abilities but I'm not sure how to align the sub-points well. Please help me edit the post, thanks.
The irrationality of $\sqrt 2^{\sqrt 2}$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^{th}$ Problem.
The beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.
Here is a simple, constructive, way to settle the original issue: $$\sqrt 3^{\log_34}=2$$
Of course, $\sqrt 3$ is irrational.
To see that $\log_3 4$ is irrational work by contradiction. $$\log_3 4=\frac ab\implies 4=3^{\frac ab}\implies 4^b=3^a$$ But if $a,b\in \mathbb N$ then this contradicts unique factorization.
The proof is non-constructive. We don't know whether $\sqrt{2}^{\sqrt{2}}$ is rational or irrational. The point of the proof is that in either case there will be two irrational numbers $p$ and $q$ such that $p^q$ is rational. In one case $p=\sqrt{2}$ and in the other case $p=\sqrt{2}^{\sqrt{2}}$.
• Can u tell me what kind of proving method is this or in what way does the proof say they are asking for 2 irrational numbers in all cases of $p^{q}$ when they can't even get a true rational number to verify this. My impression of existence proof was always to construct some example that fulfils the conditions. – Prashin Jeevaganth Aug 15 '18 at 8:54
• Not all existence proofs are constructive. This is an example of a non-constructive existence proof. A constructive proof of the same theorem would be $p=\sqrt{2}; q=\log_2(9); p^q=3$. But you have to do more work with this constructive proof to show that $\log_2(9)$ is irrational. – gandalf61 Aug 15 '18 at 12:48 | 2019-06-17T13:01:59 | {
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https://math.stackexchange.com/questions/925910/terminology-for-1-ex1 | # Terminology for $1/(e^x+1)$?
$\frac{1}{e^x+1}$ and $\frac{e^x}{e^x+1}$
Just wonder if either of the above function has a term/name associated with it? Or they are just functions that look beautiful without names? Maybe they appear very often under certain contexts?
I thought I might have seen it in some online courses. Maybe it was graphical model related or something else. But I'm not exactly sure right now and I cannot really find it on Google.
• If you change the plus signs to a minus sign and multiply by $x$, you have this. – alex.jordan Sep 10 '14 at 5:20
• the first is probably related to Fermi-Dirac distribution f(x) in physics. then second one is 1-f(x). – mike Sep 10 '14 at 5:20
• And Alex's variation appears in Planck's formula for black body radiation distribution. – Travis Sep 10 '14 at 5:21
• Both can be used in relation to the Fermi-Dirac distribution, with the second being used with certain probabilities of states. – Silynn Sep 10 '14 at 5:22
• Per Claude's comment, these functions also appear in the solutions to simple models of population growth in an environment with a fixed population capacity. – Travis Sep 10 '14 at 5:47
Quotic Wikipedia article :"A logistic function or logistic curve is a common special case of the more general sigmoid function, with equation $$f(x)=\frac{1}{1+e^{-x}}$$ So, multiplying numerator and denominator by $e^x$ $$f(x)=\frac{e^x}{1+e^{x}}$$ is just the same and $$g(x)=\frac{1}{1+e^{x}}=1-f(x)$$ | 2019-08-20T11:13:47 | {
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