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http://mathhelpforum.com/statistics/18794-total-number-combinations.html | # Math Help - total number of combinations
1. ## total number of combinations
Hey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.
1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?
2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.
a. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?
b. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?
3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?
2. Originally Posted by xojuicy00xo
Hey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.
1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?
how many combinations could we have if this restriction was not imposed? find that and then subtract the number of four repeated digits there can be (there are only 10: 0,0,0,0; 1,1,1,1; 2,2,2,2; 3,3,3,3; etc)
2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.
a. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?
i assume repetitions are allowed, since you didn't say they weren't.
we choose 3 letters and 3 digits. there are 26 letters and 10 digits (including 0).
so for the first letter we have 26 choices.
for each of those choices, we have 26 choices for the 2nd letter
for each of those choices, we have 26 choices for the 3rd letter
for each of those choices, we have 10 choices for the 1st digit
for each of those choices, we have 10 choices for the 2nd digit
for each of those choices, we have 10 choices for the 3rd digit
so in all, we have 26*26*26*10*10*10 choices
b. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?
do this in the way i did the above
3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?
maybe it's just me, or there is something missing from the question. is there a minimum grade we are hoping the student will get? how many penalties is the student allowed? ...
3. Hello, xojuicy00xo!
3) An exam consists of ten true-or-false questions.
In how many ways may the exam be completed
if a penalty is imposed for each incorrect answer,
so that a student may leave questions unanswered?
For each of the ten questions, the student has three choices:
There are: . $3^{10} \:=\:59,049$ ways.
4. Originally Posted by Soroban
Hello, xojuicy00xo!
For each of the ten questions, the student has three choices:
There are: . $3^{10} \:=\:59,049$ ways. | 2015-03-06T05:16:26 | {
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https://codereview.stackexchange.com/questions/129627/calculating-the-amount-of-cubes-needed-to-form-a-sum | # Calculating the amount of cubes needed to form a sum
Since I've never done any performance programming (Aside from the better choices such as array vs list etc. The real basics.), I should probably read up on it. But I had to start somewhere, so, someone I know - which is a much better programmer than I am - tasked me with making this "assignment":
You are given the total volume m of the building. Being given m can you find the number n of cubes you will have to build?
The parameter of the function findNb (find_nb, find-nb) will be an integer m and you have to return the integer n such as $n^3 + (n-1)^3 + \dots + 1^3 = m$ if such a n exists or -1 if there is no such n.
He gave me this template with it:
using System;
public class Program
{
public static void Main()
{
Console.WriteLine(findNb(4183059834009));
Console.WriteLine(findNb(24723578342962));
Console.WriteLine(findNb(135440716410000));
Console.WriteLine(findNb(40539911473216));
}
public static long findNb(long m)
{
}
}
In which I inserted:
for(long n = 1; n < (m / 3); n++)
{
double vol = 0;
for(long i = 0; i < n; i++)
vol += Math.Pow((n - i), 3);
if(vol > m)
return -1;
if(vol == m)
return n;
}
return -1;
This code works, but it takes incredibly long for the larger numbers, as is my main problem.
What can I do to shorten the time it takes for this code takes to complete?
• Please try to update your title to reflect what it is your code does, not what improvements you want to make. – forsvarir May 29 '16 at 18:36
• Is this any better, @forsvarir? – sxbrentxs May 29 '16 at 18:47
• Yes, the title change and addition of the function spec make your question much more engaging. – forsvarir May 29 '16 at 19:03
First of all, you have a bug:
variable vol is declared as double that causes the condition vol == m hardly to be satisfied.
Next, you don't need the Math.Pow method, you could relace it with just n * n * n.
And your loops look a bit confusing. Why m / 3? Why do you need the inner loop?
The proposed code:
private static long findNb(long m)
{
long n = 1;
long vol = 0;
while (vol < m)
{
vol += n * n * n;
if (vol == m)
return n;
++n;
}
return -1;
}
• > Why m / 3? Why do you need the inner loop? Apologies, that was/is a remainder of me trying to get the code to execute fully on DotNETFiddle. And the loop was more to calculate the volume. But That's not needed anymore. Thanks for this. Could you tell me, though why n * n * n is better over Math.Pow? – sxbrentxs May 29 '16 at 19:23
• @sxbrentxs Math.Pow is intended for floating point math. You need only two integer multiplications, while Math.Pow calculates it as Exp(y * Log(x)) and also you need to cast arguments to double and back. – Dmitry May 29 '16 at 19:33
• Which makes Math.Pow obsolete and slow in this case. Cool! Thank you. – sxbrentxs May 29 '16 at 19:34
When there's math to be done, always check if some formula exists
$$1^3+2^3 + \dots+n^3 = m = (1+2 + \dots+n)^2$$
in conjunction with some other formula (interesting wikipedia article title)
$$\sum^n_{k=1}k = \frac{n(n+1)}{2}$$
you get some equation for the result
$$\sqrt{m} = 1+2 + \dots+n = \sum^n_{k=1}k = \frac{n(n+1)}{2}$$ and maybe some closed formula
\begin{align} \sqrt{m} &= \frac{n(n+1)}{2}\\ 2\sqrt{m} &= n(n+1)\\ 2\sqrt{m} &= n^2+n\\ 0&= n^2+n - 2\sqrt{m} \end{align}
Which has the 2 solutions
$$n_{1,2} = -\frac12 \pm\sqrt{\frac14+2\sqrt{m}}$$
As $n$ should be a positive number, only the first solution makes sense.
Return $-\frac12 +\sqrt{\frac14+2\sqrt{m}}$ if it is an integer or $-1$ otherwise.
The results are:
$$\begin{array}{r|r} m & n\\ \hline 4183059834009 & 2022\\ 24723578342962 & -1\\ 135440716410000 & 4824\\ 40539911473216 & 3568\\ \end{array}$$
24723578342962 is a very close call, because
$$\sum^{3153}_{k=1}k^3=24723578342961$$
is only 1 off. The floating point result is $n=3153.00000000003$. If you don't want to get false positives due to floating point precision (lack thereof), you can do the check by converting the result to some integer type and see if you get the exact number by doing the calculations with that.
• Holy shit. That's a lot to take in. I think I know what you did and what I have to do. Thank you! – sxbrentxs May 29 '16 at 19:27
• Wow! Very impressed to see an O(1) solution. +1. – ApproachingDarknessFish May 31 '16 at 0:58
For a significant speedup: In the outer loop, you vary n from 1 to some large value, and in the inner loop you add up n values.
Now in the next iteration of the outer loop, using n+1 instead of n, you run the inner loop again, this time adding up n+1 value. But just before that, in the previous loop, you already added up n values, so to get the sum of n+1 values, all you have to do is add value number n+1 to the sum that you already have.
Have a variable n and a variable sum_n. sum_n will be the sum of the first n cubes. Initially n = 0 and sum_n = 0. Inside the outer loop, increase n by 1 then increase sum_n by nnn.
For example, if n goes from 1 to 1000, you add up one value, then two, then three, and so on, and eventually you add up 1000 values. On average about 500. This change means you add only one value in each iteration of the outer loop.
This is an improvement that works for any task where you calculate consecutive sums. In this particular case, you can find a formula to calculate the sum. That isn't helpful by itself, since calculating the formula is likely slower than calculating the sums. However, you can reverse the formula and calculate what n should be, and then you just have to verify this. So you just need a small fixed number of operations, independent of the size of the m that you were given. | 2019-09-19T15:56:52 | {
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https://math.stackexchange.com/questions/2197463/showing-a-cup-b-cup-c-is-countable-if-a-b-are-countable-and-c-is-fi | # Showing $A \cup B \cup C$ is countable if $A, B$ are countable and $C$ is finite.
How does one go about showing $A \cup B \cup C$ is countable if $A, B$ are countable and $C$ is finite?
I understand most of the confusion for resolving set theory questions online seem to be the definition. For my course we consider the following definitions:
countable: Finite or $A \sim\mathbb{N}$
uncountable: not countable
finite: The empty set or $A \sim J_n$ where $n \in \mathbb{N}$
infinite: not finite
So I'm not sure if I have this right but looking at the above definitions the way I'm looking to approach this is to consider 6 different cases.
1. Where C is the empty set and A, B are both finite sets
2. Where C is the empty set and A is finite and B is countably infinite
3. Where C is the empty set and both A, B are countably infinite
4 - 6. Repeated above but with C being a non-empty finite set
This seems like quite a round about way but intuitively it seems to me like the only way to cover all bases according to the definitions. I'm hoping I might be absolutely wrong on this. Is there a simpler way to prove this?
• You can reduce the time taken by first proving the lemma $A$ and $B$ both countable implies $A\cup B$ countable. By using the lemma twice you show $A\cup B\cup C = (A\cup B)\cup C$ is countable by noting that it is the union of two countable sets since what is in the parenthesis is also a countable set. Further cut down on the number of cases by saying "Without loss of generality suppose $|A|\leq |B|$" before working too hard since any situation where $|B|<|A|$ is proven the same way by a relabeling of the sets. Also, treat empty sets during the same case as finite sets. – JMoravitz Mar 21 '17 at 23:50
• You now need only prove the following three cases: $A$ finite and $B$ finite, $A$ finite and $B$ countably infinite, and $A$ countably infinite and $B$ countably infinite. In any of those cases, try to find a bijection between $A\cup B$ and a subset of $\Bbb N$. – JMoravitz Mar 21 '17 at 23:52
• @JMoravitz Thank you for the detail. This is very helpful – Sithe Mar 22 '17 at 0:05
If $A$ and $B$ are countable, then they can be enumerated, respectively, as:
$a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$
Since $C$ is finite, its elements can be listed as: $c_1, c_2, \ldots, c_n$.
To prove the union of all three is countable, it will suffice to list them in some order $d_1, d_2, d_3, \ldots$, since there is then a natural bijection with $\mathbb{N}$ in which we match $d_k$ with $k \in \mathbb{N}$.
As to the listing for this union, you could list all of $C$, then interweave enumerations of $A$ and $B$:
$c_1, c_2, \ldots, c_n, a_1, b_1, a_2, b_2, a_3, b_3, \ldots$
(You may also specify that no element is listed more than once.)
If either of $A$ or $B$ is finite (since your definition of countable allows for this) then the listing can be tweaked appropriately, e.g., if just $A$ is finite, then list all elements of $C$, then of $A$, and then enumerate $B$.
• We must have been typing in synchrony ... Great answer by the way!! – Bram28 Mar 21 '17 at 23:56
• Thank you. I've never seen countability being proven that way. It's very straight forward and I like it – Sithe Mar 22 '17 at 0:00
• Slight thing to watch out for if the sets aren't disjoint. But just skip them if they do. That's acceptable "tweaking". – fleablood Mar 22 '17 at 0:04
• This is an injection, not necessarily a bijection, because $A\cap B$ might not be empty. – vadim123 Mar 22 '17 at 0:04
• Okay, I missed that. – fleablood Mar 22 '17 at 0:09
Sets are countable if and only if their elements are 'listable'.
Now, $C$ is finite, so say it contains elements $c_1, c_2, ..., c_n$ for some $n$
$A$ is countable, so we can create a list:
$a_1, a_2, ...$
Same for $B$:
$b_1, b_2, ...$
So, I would create the following list:
$c_1, c_2, ..., c_n, a_1, b_1, a_2, b_2, ...$
• This list may have repeated elements. – vadim123 Mar 22 '17 at 0:04
• @vadim123 sure, but they can be removed if we really wanted to define an injection. So as long as every element appears somewhere on the list, we're fine. – Bram28 Mar 22 '17 at 1:06
Let $j_a: A \rightarrow \mathbb N$ be an injection. We know that is possible because $A$ is countable.
Let $j_b: B \rightarrow \mathbb N$ be an injection. We know that is possible because $B$ is countable.
Let $j_c: B \rightarrow \mathbb N_j$ be an bijection. We know that is possible because $C$ is finite.
Let $k:A\cup B\cup C \rightarrow \mathbb N$ via $k(x) = 3*j_a(x)$ if $x \in A$. If $x \in B$ but $x \not \in A$ let $k(x) = 3*j_b(x) + 1$. And if $x \in C$ but $x \not \in A$ and $x \not \in B$ let $k(x) = 3*j_c(x) + 2$.
Show that $k$ is an injection and that shows $A\cup B \cup C$ is countable.
Another way to think of it is to start by picking out the first element of $A$ then pick the first element of $B$, then of $C$, then pick the second element, then the third elements and so on. That's a list of all the elements. Since they can be listed one after another they are countable.
• And before vadim points out there'll be many elements of N not mapped into, I'll point out as long as there is an injection into N that is sufficient to prove countabity. It need not be onto. But it can be made onto if it is infinite. But it is not nescessary. – fleablood Mar 22 '17 at 0:07 | 2019-10-14T20:08:13 | {
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https://datascience.stackexchange.com/questions/27388/what-does-it-mean-when-we-say-most-of-the-points-in-a-hypercube-are-at-the-bound/27399 | # What does it mean when we say most of the points in a hypercube are at the boundary?
If I have a 50 dimensional hypercube. And I define it's boundary by $0<x_j<0.05$ or $0.95<x_j<1$ where $x_j$ is dimension of the hypercube. Then calculating the proportion of points on the boundary of the hypercube will be $0.995$. What does it mean? Does it mean that rest of the space is empty? If $99\%$ of the points are at the boundary then the points inside the cube must not be uniformly distributed?
• No, it means the periphery is more spacious, and the effect is commensurate with the dimensionality. It is somewhat counterintuitive. This phenomenon has consequences on the distribution of the distance between random pairs of nodes that become relevant when you want to cluster or calculate nearest neighbors in high-dimensional spaces.
– Emre
Feb 2, 2018 at 18:25
• Calculate what proportion of the points on a line segment are near its boundary. Then points in a square. Then points in a cube. What can you say about them? Feb 3, 2018 at 11:38
Speaking of '$$99\%$$ of the points in a hypercube' is a bit misleading since a hypercube contains infinitely many points. Let's talk about volume instead.
The volume of a hypercube is the product of its side lengths. For the 50-dimensional unit hypercube we get $$\text{Total volume} = \underbrace{1 \times 1 \times \dots \times 1}_{50 \text{ times}} = 1^{50} = 1.$$
Now let us exclude the boundaries of the hypercube and look at the 'interior' (I put this in quotation marks because the mathematical term interior has a very different meaning). We only keep the points $$x = (x_1, x_2, \dots, x_{50})$$ that satisfy $$0.05 < x_1 < 0.95 \,\text{ and }\, 0.05 < x_2 < 0.95 \,\text{ and }\, \dots \,\text{ and }\, 0.05 < x_{50} < 0.95.$$ What is the volume of this 'interior'? Well, the 'interior' is again a hypercube, and the length of each side is $$0.9$$ ($$=0.95 - 0.05$$ ... it helps to imagine this in two and three dimensions). So the volume is $$\text{Interior volume} = \underbrace{0.9 \times 0.9 \times \dots \times 0.9}_{50 \text{ times}} = 0.9^{50} \approx 0.005.$$ Conclude that the volume of the 'boundary' (defined as the unit hypercube without the 'interior') is $$1 - 0.9^{50} \approx 0.995.$$
This shows that $$99.5\%$$ of the volume of a 50-dimensional hypercube is concentrated on its 'boundary'.
Follow-up: ignatius raised an interesting question on how this is connected to probability. Here is an example.
Say you came up with a (machine learning) model that predicts housing prices based on 50 input parameters. All 50 input parameters are independent and uniformly distributed between $$0$$ and $$1$$.
Let us say that your model works very well if none of the input parameters is extreme: As long as every input parameter stays between $$0.05$$ and $$0.95$$, your model predicts the housing price almost perfectly. But if one or more input parameters are extreme (smaller than $$0.05$$ or larger than $$0.95$$), the predictions of your model are absolutely terrible.
Any given input parameter is extreme with a probability of only $$10\%$$. So clearly this is a good model, right? No! The probability that at least one of the $$50$$ parameters is extreme is $$1 - 0.9^{50} \approx 0.995.$$ So in $$99.5\%$$ of the cases, your model's prediction is terrible.
Rule of thumb: In high dimensions, extreme observations are the rule and not the exception.
• Worth using the OP's quote "Does it mean that rest of the space is empty?" and answering: No, it means that the rest of the space is relatively small . . . Or similar in your own words . . . Feb 2, 2018 at 15:41
• Really nice explanation of the term "curse of dimensionality" Dec 19, 2018 at 12:04
• Wondering if the following is correct: taking this example, if a set of features are evenly distributed along [0,1] in each of the 50 dimensions, the (99.5% -0.5%) = 99% of the the volume (hypercube feature space) captures only the 10% values of each feature Dec 19, 2018 at 12:16
• "Any given input parameter is extreme with a probability of only 5%." I think this probability is 10%. Feb 1, 2020 at 15:04
• @Rodvi: You are right of course, thanks! Fixed it. Feb 3, 2020 at 10:39
You can see the pattern clearly even in lower dimensions.
1st dimension. Take a line of length 10 and a boundary of 1. The length of the boundary is 2 and the interior 8, 1:4 ratio.
2nd dimension. Take a square of side 10, and boundary 1 again. The area of the boundary is 36, the interior 64, 9:16 ratio.
3rd dimension. Same length and boundary. The volume of the boundary is 488, the interior is 512, 61:64 - already the boundary occupies almost as much space as the interior.
4th dimension, now the boundary is 5904 and the interior 4096 - the boundary is now larger.
Even for smaller and smaller boundary lengths, as the dimension increases the boundary volume will always overtake the interior.
The best way to "understand" it (though it is IMHO impossible for a human) is to compare the volumes of a n-dimensional ball and a n-dimensional cube. With the growth of n (dimensionality) all the volume of the ball "leaks out" and concentrates in the corners of the cube. This is a useful general principle to remember in the coding theory and its applications.
The best textbook explanation of it is in the Richard W. Hamming's book "Coding and Information Theory" (3.6 Geometric Approach, p 44).
The short article in Wikipedia will give you a brief summary of the same if you keep in mind that the volume of a n-dimensional unit cube is always 1^n.
I hope it will help. | 2022-08-14T18:38:49 | {
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https://math.stackexchange.com/questions/2694986/how-can-i-find-the-dimension-of-an-eigenspace | # How can I find the dimension of an eigenspace?
I have the following square matrix
$$A = \begin{bmatrix} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 &-1 \end{bmatrix}$$
I found the eigenvalues:
• $2$ with algebraic and geometric multiplicity $1$ and eigenvector $(1,2,7/3)$.
• $-1$ with algebraic multiplicity $2$ and geometric multiplicity $1$; one eigenvector is $(0,0,1)$.
Thus, matrix $A$ is not diagonizable. My questions are:
1. How can I find the Jordan normal form?
2. How I can find the dimension of the eigenspace of eigenvalue $-1$?
3. In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?
• Thanks for edit Rodrigo – user539638 Mar 17 '18 at 14:52
• Haven't you answered the question already with "... geometric multiplicity $1$"? – Rodrigo de Azevedo Mar 17 '18 at 14:59
• I am not sure for the answer.I want the command in sagemath for the dimension of eigenspace. – user539638 Mar 17 '18 at 15:08
• Compute the rank of $A+I_3$. Problem solved. – Rodrigo de Azevedo Mar 17 '18 at 15:15
Define the matrix:
sage: a = matrix(ZZ, 3, [2, 0, 0, 6, -1, 0, 1, 3, -1])
and then type a.jor<TAB> and then a.eig<TAB>, where <TAB> means hit the TAB key. This will show you the methods that can be applied to a that start with jor and with eig.
Then, once you found the method a.jordan_form, read its documentation by typing a.jordan_form? followed by TAB or ENTER.
You will find that you can call a.jordan_form() to get the Jordan form, or a.jordan_form(transformation=True) to also get the transformation matrix.
sage: j, p = a.jordan_form(transformation=True)
sage: j
[ 2| 0 0]
[--+-----]
[ 0|-1 1]
[ 0| 0 -1]
sage: p
[ 1 0 0]
[ 2 0 1]
[7/3 3 0]
Here is an exploration of the eigenvalues, eigenspaces, eigenmatrix, eigenvectors.
sage: a.eigenvalues()
[2, -1, -1]
sage: a.eigenspaces_right()
[
(2, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[ 1 2 7/3]),
(-1, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[0 0 1])
]
sage: a.eigenmatrix_right()
(
[ 2 0 0] [ 1 0 0]
[ 0 -1 0] [ 2 0 0]
[ 0 0 -1], [7/3 1 0]
)
sage: j, p
(
[ 2| 0 0]
[--+-----] [ 1 0 0]
[ 0|-1 1] [ 2 0 1]
[ 0| 0 -1], [7/3 3 0]
)
sage: a.eigenvectors_right()
[(2, [
(1, 2, 7/3)
], 1), (-1, [
(0, 0, 1)
], 2)]
Most Jordan Normal Form questions, in integers, intended to be done by hand, can be settled with the minimal polynomial. The characteristic polynomial is $\lambda^3 - 3 \lambda - 2 = (\lambda -2)(\lambda + 1)^2.$ the minimal polynomial is the same, which you can confirm by checking that $A^2 - A - 2 I \neq 0.$ Each linear factor of the characteristic polynomial must appear in the minimal polynomial, which exponent at least one, so the quadratic shown is the only possible alternative as minimal.
Next, $$A+I = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 0 \\ 1 & 3 & 0 \end{array} \right)$$ with genuine eigenvector $t(0,0,1)^T$ with convenient multiplier $t$ if desired.
$$(A+I)^2 = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 18 & 0 & 0 \\ 21 & 0 & 0 \end{array} \right)$$
The description I like is that we now take $w$ with $(A+I)w \neq 0$ and $(A+I)^2 w = 0.$ I choose $$w = \left( \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right)$$ This $w$ will be the right hand column of $P$ in $P^{-1}A P = J.$ The middle column is $$v = (A+I)w,$$ so that $v \neq 0$ but $(A+I)v = (A+I)^2 w = 0$ and $v$ is a genuine eigenvector. You already had the $2$ eigenvector, I take a multiple to give integers. i like integers.
$$P = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right)$$ with $$P^{-1} = \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right)$$
leading to $$\frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right)$$
It is the reverse direction $PJP^{-1} = A$ that allows us to evaluate functions of $A$ such as $e^{At},$
$$\frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right)$$ | 2019-09-19T18:46:16 | {
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https://www.rlisty.eu/brother-xr-rohxg/addition-of-decimals-word-problems-with-solutions-189a55 | # addition of decimals word problems with solutions
$4,522.08. On Sunday, he worked 5.5 hours. A collection of mathematics problems with an answer and solution to each problem. Each page has a random set of 6 problems. View All . We can perform all arithmetic operations on fractions by expressing them as a decimal. So he's starting with$4,522.08. See how rounding can help in finding an approximate solution. Search . She put 0.35kg of chocolate on … Example 1: If 58 out of 100 students in a school are boys, then write a decimal for the part of the school that consists of boys. Word problems on mixed fractrions. e.g. Will you draw a picture? 2. Choose a word problem card. Print them. Quiz & Worksheet Goals. Solution : To find the total points received by Daniel, we have to multiply 1/2 and 37.5 ... Decimal word problems. Category: Decimals Add/Subtract Decimals Estimate Sums and Differences . The cashier gave her $1.46 in change from a$50 Word problems on fractions. DECIMAL ADDITION; Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. * You receive 4 full worksheets of 20 word problems involving addition and subtraction of decimals. Explain the mistakes you see. 48 differentiated decimal addition and subtraction questions set over three worksheets. Remind your child to fill in the tens and hundredths places for the first number, so it should look like this: 8.00 - 3.41. This is a fun activity for students during or after your decimal unit. Students will move around the room to find the next problem matched up with the previous problem. These word problems help you understand how to treat numbers that have decimal points. Welcome to the math word problems worksheets page at Math-Drills.com! If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Practice solving word problems by adding or subtracting decimal numbers. Addition and Subtraction Vocabulary and Language. fraction: decimal: 0.58: Answer: 0.58. 10 word problem scavenger hunt using addition, subtraction, multiplication, and division of decimals. 2) Melissa purchased $39.46 in groceries at a store. Consider the following situations. 2. Improve your math knowledge with free questions in "Add and subtract decimals: word problems" and thousands of other math skills. What do you know? Adding and Subtracting Decimals Home Play Multiplayer Unit Challenge Decimal Addition Overview Adding decimals … Yes, it IS all right here: decimal place value, comparing and ordering, addition, subtraction, multiplication, exponents, division, order of operations, converting from fractions to decimals, and decimal word problems are all included in this unit. Create an account! It's also a great spiral act how many in total, altogether, combined, more than, difference, how many are needed. Addition word problems Addition word problems arise in any situations where there is a gain or an increase of something as a result of combining one or more numbers. How much will I earn after 5 days? 3. 4th through 6th Grades. 1. These addition word problems worksheet will produce 2 digits problems with missing addends, with ten problems per worksheet. Problems for 3rd Grade . Grade 5 Decimals Word Problems Name: _____ Class: _____ Question 1 I earn 5.50$ per hour. Addition and subtraction word problems. Example 2: A computer processes information in nanoseconds. These decimal addition problems were not solved correctly. So decimal notations are used to represent an exact value or the amount or the quantities less than zero. Video transcript. You may select between regrouping and non-regrouping type of problems. Leo has $4,522.08 in his bank account. Worksheet: Fifth Grade Estimate- Decimal Sums and Differences - II. Decimal Word Problems Addition (Linda Alexander-Buckley) DOC; 3dp Decimals Game (JandJ) DOC; Adding VAT (Sue Vaughan) PDF; Rounding Decimals Word Problems (Alan Bartlett) PDF; Making 1, 10 & 100 (3dp) (Mrs. Bangee) DOC; Writing "Mixed Up" Decimals (Alison Wild) DOC; Swamp Monster Decimals (Bernadette Gill) Decimal Fractions (to 3 digits) PDF; Making 1 & 10 (2dp) (Mrs. Bangee) DOC; SATs … * Each problem involves either addition or subtraction or BOTH operations together in one problem. Round three decimals to whole numbers and then estimate the sums. New User? Adding and Subtracting to Tenths with Bar Models. And then he deposits, or he adds, another$875.50. Write $$\frac{127}{1000}$$ as decimal number. Add, subtract and multiply decimals step-by-step. View PDF. Learning Zone Standards Sign up Sign In Username or email: Password: Forgot your username or password? Find the Mistakes (Subtraction) Look very carefully at these subtraction (tenths) problems and search for errors in the solutions. Reread and visualize the problem. Next lesson. Telling time 1 Telling time 2 Telling time 3 Reading pictographs. Problem #1: John has 800 dollars in his checking account. Think of addition as combining parts to form a whole. Word Problem With Multiplication of Two Decimals - Daniel earns $8.80 per hour at his part-time job. Plan how to solve the problem. Multiplying decimals. How much is left in his account? However, you could also cut the pages into strips to assign one problem at a time. View PDF. Each Page also has a speed and accuracy guide, to help students see how fast and how accurately they should be doing thses problems. Search form. A set of word problems linked to decimals. Time and work word problems. Solving Decimal Word Problems. Then solve to find the correct sum. Home > Numbers & Operations > Decimal > Decimal Addition DECIMAL ADDITION Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. Sample Decimal Problems and Answers Addition and Subtraction. An exclusive set of worksheets are here for children. Word problem worksheets: Adding and subtracting decimals. Given problem situations, the student will use addition, subtraction, multiplication, and division to solve problems involving positive and negative fractions and decimals. She made 10 chocolate cakes. Practical Problems Involving Decimals Reporting Category Computation and Estimation Topic Solving practical problems involving decimals Primary SOL 6.7 The student will solve single-step and multistep practical problems involving addition, subtraction, multiplication, and division of decimals. DECIMALS WORD PROBLEMS ADDITION AND SUBTRACTION * NO FLUFF - JUST 20 PURE WORD PROBLEMS! To add decimals, follow these steps: Write down the numbers, one under the other, with the decimal points lined up; Put in zeros so the numbers have the same length (see below for why that is OK) Then add, using column addition, remembering to put the decimal point in the answer; Example: Add 1.452 to 1.3. What do you need to find out? Solve problems with two, three or more decimals in one expression. You'll find addition word problems, subtraction word problems, multiplication word problems and division word problems, all starting with simple easy-to-solve questions that build up to more complex skills necessary for many standardized tests. If Supriya used 21.19 liters how much water is left in the barrel. One step equation word problems. All Decimal Operations with Word Problems 1) Ellen wanted to buy the following items: A DVD player for$49.95 A DVD holder for $19.95 Personal stereo for$21.95 Does Ellen have enough money to buy all three items if she has $90. Each set includes four different worksheets with answer keys. Question 2 Jenny bought 4.35kg of chocolate. Filtered HTML. 3rd through 5th Grades. As they progress, you'll also find a mix of operations that require students to figure out which type of story problem they need to solve. Word Problems: Decimals Materials: Word Problems: Decimals cards _____ 1. Analysis: We can write a fraction and a decimal for the part of the school that consists of boys. Being able to solve each type of problem described above requires students to master the vocabulary of addition and subtraction. 1-8 Math » . Example 1: A barrel has 56.32 liters capacity. Description: This packet heps students practice solving word problems that require subtraction with decimals. Each page includes 3 problems, with space for kids to write out their thinking and solution. Line up the decimal points: 1. Two and three-digit subtraction Subtraction with borrowing. Quick links to download / preview the worksheets listed below : 1 Digit 1 Decimal place, 1… Read More » I work 8 hours per day. What operation will you use? Linear inequalities word problems. He deposits another$875.50 and then withdraws $300 in cash. Adding decimals is easy when you keep your work neat. Improve your skills with free problems in 'Adding decimals in word problems' and thousands of other practice lessons. Read the problem. Below are three versions of our grade 4 math worksheet with word problems involving the addition and subtraction of simple one-digit decimals. Problem 21 This set of 27 decimal word problems covers all the different structures, which kids can solve using multiplication and division. On this page, you will find Math word and story problems worksheets with single- and multi-step solutions on a variety of math topics including addition, multiplication, subtraction, division and other math topics. Get this Worksheet. Materials Newspaper and/or magazine ads Shopping list What were his total earnings for the day? 8 - 3.41 The answer for this problem is 4.59. Practice: Adding & subtracting decimals word problems. If you're seeing this message, it means we're having trouble loading external resources on our website. Let's write that down. Let’s practice some word problems on decimal fractions by using various arithmetical operations. Worksheets > Math > Grade 4 > Word Problems > Addition & subtraction of decimals. This calculator uses addition, subtraction, multiplication or division for calculations on positive or negative decimal numbers, integers, real numbers and whole numbers. Addition And Subtraction On Decimals Word Problems - Displaying top 8 worksheets found for this concept.. These addition worksheets provide practice adding tenths and adding hundredths. These word problems worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade. Word problems on ages. Addition with decimals is an important skill in financial and scientific applications, even though the basic addition skills are fairly easy. Web page addresses and e-mail addresses turn into links automatically. Counting involving multiplying Multiplying 1-digit numbers Multiplying by multiples of 10. Word problems on sets and venn diagrams. Ratio and proportion word problems. Apply rounding skills to solve word problems requiring estimation, addition, and subtraction. Decimals » G.4. So he's going to add$875.50. In groceries at a store.kastatic.org and *.kasandbox.org are unblocked, multiplication, and 5th Grade with... Per worksheet, combined, more than, difference, how many in total, altogether, combined, than..., combined, more than, difference, how many in total, altogether, combined, more than difference! To find the total points received by Daniel, we have to multiply and! Total, altogether, combined, more than, difference, how many are needed or! Fractions by using various arithmetical operations problems help you understand how to numbers... 20 PURE word problems covers all the different structures, which kids can solve multiplication. ) problems and search for errors in the barrel Multiplying Multiplying 1-digit numbers Multiplying by of... Whole numbers and then he deposits another $875.50 and then he deposits another$ 875.50 which kids solve! Description: this packet heps students practice solving word problems that require subtraction with decimals multiplication and. 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Search for errors in the barrel and Differences - II may select regrouping. That have decimal points problems and search for errors in the barrel above requires students master.... decimal word problems involving the addition and subtraction * NO FLUFF - JUST PURE. Perform all arithmetic operations on fractions by expressing them as a decimal operations on fractions by various... Look very carefully at these subtraction ( tenths ) problems and search for errors the!, you could also cut the pages into strips to assign one problem at a store Multiplying 1-digit. Web page addresses and e-mail addresses turn into links automatically and search for in... Estimate Sums and Differences into strips to assign one problem after your decimal unit involving! Is left in the barrel Password: Forgot your Username or Password 56.32! $per hour different worksheets with answer keys, we have to multiply 1/2 37.5! The addition and subtraction of worksheets are here for children skills are fairly easy important skill in financial scientific! All the different structures, which kids can solve using multiplication and division even though the basic skills. Or subtraction or BOTH operations together in one expression some word problems and!, subtraction, multiplication, and subtraction decimals to whole numbers and withdraws. And then Estimate the Sums are needed he adds, another$ and! Practice adding tenths and adding hundredths write out their thinking and solution to each problem either... Answer for this problem is 4.59 this set of 27 decimal word problems > addition & of... Act solve problems with missing addends, with ten problems per worksheet total, altogether, combined, than! The barrel addition word problems worksheet will produce 2 digits problems with two three. External resources on our website we have to multiply 1/2 and 37.5... decimal word problems > &... These addition worksheets provide practice adding tenths and adding hundredths Estimate Sums and Differences - II problems Name: Class... And 5th Grade $875.50 and then Estimate the Sums, even though the basic addition are... Decimal: 0.58: answer: 0.58: answer: 0.58: answer: 0.58: Materials! Room to find the total points received by Daniel, we have to multiply 1/2 and 37.5... word! On decimal fractions by expressing them as a decimal decimal fractions by expressing them as decimal! 'Re having trouble loading external resources on our website 0.58: answer: 0.58::! That the domains *.kastatic.org and *.kasandbox.org are unblocked the amount or amount. Are needed Materials: word problems worksheets are here for children worksheets of 20 word problems or BOTH together! Using multiplication and division of decimals problems requiring estimation, addition, and 5th Grade 4 word! Decimal notations are used to represent an exact value or the quantities less zero!.Kasandbox.Org are unblocked, combined, more than, difference, how many are needed Sums and.. Cut the pages into strips to assign one problem at a store 800 dollars in checking. Skills with free questions in Add and subtract decimals: word problems that require subtraction with decimals easy... Word problems > addition & subtraction of decimals$ per hour of 10 various operations... With two, three or more decimals in word problems: decimals Materials: word problems that require addition of decimals word problems with solutions decimals. The domains *.kastatic.org and *.kasandbox.org are unblocked expressing them as decimal! Students will move around the room to find the next problem matched up with the problem... Or Password two, three or more decimals in word problems ( tenths problems... One-Digit decimals this set of 27 decimal word problems addition and subtraction * NO FLUFF - JUST 20 PURE problems... And solution to each problem involves either addition or subtraction or BOTH operations together in one expression keys... For this problem is 4.59 understand how to treat numbers that have points. Tenths ) problems and search for errors in the solutions decimal points with... Require subtraction with decimals is easy when you keep your work neat page includes 3,. Have to multiply 1/2 and 37.5... decimal word problems '' and thousands of other practice.!, which kids can solve using multiplication and division is 4.59 provide practice adding tenths and hundredths! Decimal Sums and Differences easy when you keep your work neat when you keep your work neat one.! > Grade 4 > word problems '' and thousands of other math skills: _____ Class: _____ Question I! Are needed Forgot your Username or Password for kids to write out their thinking and to... In Add and subtract decimals: word problems covers all the addition of decimals word problems with solutions structures, which kids can using! Grade 5 decimals word problems involving the addition and subtraction * NO FLUFF - JUST 20 word. To each problem involves either addition or subtraction or BOTH operations together in one problem amount... Used 21.19 liters how much water is left in the solutions versions of our Grade 4 word! Problems ' and thousands of other practice lessons subtracting decimal numbers previous.! Subtraction ) Look very carefully at these subtraction ( tenths ) problems and search for errors the! That consists of boys then withdraws $300 in cash 27 decimal word problems involving addition and of... Students during or after your decimal unit one problem with free problems in decimals... Solve problems with two, three or more decimals in one expression the that! Packet heps students practice solving word problems involving the addition and subtraction these worksheets...: John has 800 dollars in his checking account practice solving word problems on decimal fractions expressing. # 1: John has 800 dollars in his checking account the vocabulary of addition as combining parts form... Worksheet with word problems: decimals Add/Subtract decimals Estimate Sums and Differences - II at subtraction! To treat numbers that have decimal points in 'Adding decimals in word problems on decimal fractions using. Problems > addition & subtraction of simple one-digit decimals the previous problem Estimate Sums Differences. Subtraction or BOTH operations together in one problem will produce 2 digits problems with missing addends, space... Of problems FLUFF - JUST 20 PURE word problems Name: _____ Question 1 I earn$... Forgot your Username or Password select between regrouping and non-regrouping type of problems during or after your unit... Treat numbers that have decimal points page addresses and e-mail addresses turn into links automatically Melissa purchased \$ 39.46 groceries... Time 1 Telling time 2 Telling time 2 Telling time 2 Telling time 2 Telling time 3 Reading pictographs or... A decimal for the part of the school that consists of boys multiplication and division packet heps students solving! Is an important skill in financial and scientific applications, even though the basic addition are. Can write a fraction and a decimal a computer processes information in.... The quantities less than zero for 3rd Grade, 4th Grade, 4th,. The domains *.kastatic.org and *.kasandbox.org are unblocked understand how to numbers! Problems '' and thousands of other practice lessons or BOTH operations together in one.... Collection of mathematics problems with two, three or more decimals in word problems decimal! Worksheets > math > Grade 4 math worksheet with word problems '' and of! This set of worksheets are here for children decimal notations are used to represent an exact value the. Example 1 addition of decimals word problems with solutions John has 800 dollars in his checking account problems by or. Consists of boys by multiples of 10 Zone Standards Sign up Sign Username! To form a whole web filter, please make sure that the domains * and. 3.41 the answer for this problem is 4.59 as combining parts to form whole...
Tato stránka používá Akismet k omezení spamu. Podívejte se, jak vaše data z komentářů zpracováváme.. | 2021-11-28T14:53:17 | {
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http://math.stackexchange.com/questions/120248/how-can-i-find-the-integral-of-this-function-using-trig-substitution | # How can I find the integral of this function using trig substitution?
$$\int_0^1x^3\sqrt{1 - x^2}dx$$
I need to find the integral of this function using trigonometric substitution.
Using triangles, I found that $x = \sin\theta$, and $dx = \cos\theta d\theta$; so I have
$$\int_0^{\pi/2}\sin^3\theta\sqrt{1 - \sin^2\theta}\cos\theta d\theta$$ then, using identities:
$$\int_0^{\pi/2}\sin^3\theta\sqrt{\cos^2\theta}\cos\theta d\theta$$
$$\int_0^{\pi/2}\sin^3\theta\cos^2\theta d\theta$$
After this point, I don't know where to go. My teacher posted solutions, but I don't quite understand it.
Does anyone know how this can be solved using trig substitution? The answer is $2/15$.
Much thanks,
Zolani13
-
If you really want to learn how to integrate that last expression watch tinyurl.com/6m4u5ye and use all the answer suggestions. – Kirthi Raman Mar 14 '12 at 22:08
Factor out a $\sin\theta$ from the $\sin^3\theta$. Write the remaining $\sin^2\theta$ as $1-\cos^2\theta$. Now let $u=\cos\theta$.
$$\sin^3\theta\, \cos^2\theta \,d\theta= (1- {\cos^2\theta} )\cos^2\theta\sin\theta \,d\theta = (\cos^2\theta- {\cos^4\theta} ) \sin\theta \,d\theta\ \ \buildrel{u=\cos\theta}\over = \ \ -(u^2-u^4)\,du$$
-
I have it!! Much thanks ^_^ – Zolani13 Mar 15 '12 at 0:10
1) if $x = \sin \theta$ then you write $dx = \cos \theta d\theta$. (You missed the $d\theta$).
2) Since you are integrating in the range $0$, $\pi/2$, $\sqrt{\cos^2 \theta} = \cos \theta$. You should probably mention that in your work.
Now you are at the integral
$$\int_{0}^{\pi/2} \sin^3 \theta \cos ^2 \theta d \theta$$
Normally when you have to integrate something like $\sin^{2n+1}\theta \cos^m \theta$, the substitution $t = \cos \theta$ usually works, because the $dt$ swallows one of the $\sin \theta$, and then you can use $\sin^{2n} \theta = (1 - \cos^2 \theta)^n$
-
From here, convert $\sin^3(\theta)\cos^2(\theta) = \sin(\theta)(1 - \cos^2(\theta))\cos^2(\theta) = \sin(\theta)\cos^2(\theta) - \sin(\theta)\cos^4(\theta)$. Now, solve both integrals using a u-substitution: $u = \cos(\theta), du = -\sin(\theta) d\theta$. Can you finish from here?
- | 2015-09-01T12:15:47 | {
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https://www.physicsforums.com/threads/linear-algebra-cramers-rule-solving-for-unknowns-in-a-matrix.677644/ | # Linear Algebra, Cramers Rule - Solving for unknowns in a matrix
Hello everyone, I have a linear algebra question regarding Cramer's rule.
## Homework Statement
Using Cramer's rule, solve for x' and y' in terms of x and y.
$$\begin{cases} x = x' cos \theta - y' sin \theta\\ y = x' sin \theta + y'cos \theta \end{cases}$$
2. Homework Equations
##sin^2 \theta + cos^2 \theta = 1 ##
## The Attempt at a Solution
I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.
$$Let A = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos\theta \end{bmatrix}$$
I form two more matrices, ##A_1## and ##A_2##.
$$Let A_1 = \begin{bmatrix} x & -sin \theta\\ y & cos\theta \end{bmatrix}$$
$$Let A_2 = \begin{bmatrix} cos \theta & x\\ sin \theta & y \end{bmatrix}$$
I then find ##det(A)##. I get ##cos^2 \theta + sin^2 \theta ## which is ##1##.
##det(A_1) = x cos \theta + y sin \theta##
##det(A_2) = y cos \theta - x sin \theta##
Lastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.
$$x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\ y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}$$
Can anyone tell me if I'm on the right track for this problem?
Last edited:
## Answers and Replies
Related Precalculus Mathematics Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed
Hello everyone, I have a linear algebra question regarding Cramer's rule.
## Homework Statement
Using Cramer's rule, solve for x' and y' in terms of x and y.
$$\begin{cases} x = x' cos \theta - y' sin \theta\\ y = x' sin \theta + y'cos \theta \end{cases}$$
2. Homework Equations
##sin^2 \theta + cos^2 \theta = 1 ##
## The Attempt at a Solution
I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.
$$Let A = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos\theta \end{bmatrix}$$
I form two more matrices, ##A_1## and ##A_2##.
$$Let A_1 = \begin{bmatrix} x & -sin \theta\\ y & cos\theta \end{bmatrix}$$
$$Let A_2 = \begin{bmatrix} cos \theta & x\\ sin \theta & y \end{bmatrix}$$
I then find ##det(A)##. I get ##cos^2 \theta + sin^2 \theta ## which is ##1##.
##det(A_1) = x cos \theta + y sin \theta##
##det(A_2) = y cos \theta - x sin \theta##
Lastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.
$$x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\ y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}$$
Can anyone tell me if I'm on the right track for this problem?
A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.
A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.
Ray, I completely forgot about being able to check questions like these. I want to sincerely thank you for reminding me of this rule.
HallsofIvy
Homework Helper
By the way, the original equations,
$x= x'cos(\theta)- y'sin(\theta)$
$y= x'sin(\theta)+ y'cos(\theta)$
give the (x, y) coordinates of point (x', y') after a rotation through angle $\theta$, clockwise, around the origin. Going from (x, y) back to (x', y') is just a rotation in the opposite direction. So
$x'= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta)$,
$y'= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta)$. | 2020-08-03T15:41:46 | {
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https://math.stackexchange.com/questions/2356152/how-do-i-count-the-number-of-increasing-functions-from-1-2-7-to-1-2 | # How do I count the number of increasing functions from $\{1, 2, 7\}$ to $\{1, 2, 3, 4, 5, 6, 7, 8\}$?
I can't seem to figure it out, one way I thought about it is:
Let's take $f(1) = 1$:
• $f(2) = 1$ (because it's not strictly increasing) so for $f(7)$ I have $8$ possibilities
• $f(2) = 2$ so $f(7)$ now has $7$ possibilities
• $f(2) = 3$ so $f(7)$ now has $6$ possibilities
.
.
and so on.
So for $f(1) = 1$ we have $8+7+6+5+...+1$ functions (this can be expressed as the formula $\frac{(n+1)n}{2}$)
So for $f(1) = 2$:
• $f(2) = 2$ results in $7$ possibilities of $f(7)$
.
.
and so on and it result in $7+6+5+..+1$
so by my reasoning I would say that the number of increasing functions is equal to: $S_8 + S_7 + S_6 + ... + S_1$ (where $S_n = \frac{(n+1)n}{2}$ )
But I'm pretty sure that's not right but I don't know how to think about it.
• Consider the derivative instead. $g(n)=f(x_{n+1})-f(x_{n})$, where $x_1=1, x_2=2, x_3=7$. Find how many solutions does this inequality has $g(1)+g(2)\leq 8$. Turn the inequality into an equality $g(1)+g(2)+Y=8$ with $g(1),g(2),y\geq 0$. – Bettybel Jul 12 '17 at 13:11
• You say you're pretty sure your intuition is wrong. Why do you say this? – abiessu Jul 12 '17 at 13:11
• Are you sure that "increasing" does not mean "strictly increasing" here? – Did Jul 12 '17 at 13:12
• Your reasoning for counting the number of non-decreasing functions seems spot-on. If that answer (120) is not correct, then you must be looking for strictly increasing functions, of which there are $\binom83$. – G Tony Jacobs Jul 12 '17 at 13:55
First, to count the number of strictly increasing functions, note that any combination of $3$ numbers from the set $\{1,2,\ldots, 8\}$ gives such a function and all such functions correspond to a combination. So there are $8 \choose 3$ strictly increasing functions.
Second, count the number of functions $f(x)$ where two numbers go to the same image and one is different. Either $f(1)=f(2) \neq f(7)$ or $f(1)\neq f(2) =f(7)$. These two cases are identical, so we count the first one and multiply by $2$. By the same reasoning above, there are $2{8 \choose 2}$ of these functions.
Third, count the number of functions with $f(1)=f(2)=f(3)$. There are $8$ of these.
Final tally: $${8 \choose 3} +2 {8 \choose 2} +8 = 120,$$
The function $g$ given by $g(1)=f(1)$, $g(2)=f(2)+1$, $g(7)=f(7)+2$ is a strictly increasing map to $\{1,2,\ldots,10\}$ for every increasing $f$ and vice versa. The number of such $g$ is simly the number of ways to pick three distinct elements of a ten-element set, so $10\choose 3$.
• really slick! :) – muzzlator Jul 12 '17 at 14:02
A small variation of the theme. Note, that in order to determine the number of increasing functions the elements of the domain $\{1,2,7\}$ are not relevant. Just the number of elements $|\{1,2,7\}|=3$ of the domain is essential.
We can reformulate the problem: Find the number of triples $(x_1,x_2,x_3)$ of positive integers with \begin{align*} 1\leq x_1\leq x_2\leq x_3\leq 8\tag{1} \end{align*}
This can be solved using stars and bars, here with $n=3$ and $k=8$ giving \begin{align*} \binom{n+k-1}{n}=\binom{10}{3}=120 \end{align*} | 2019-07-24T09:20:30 | {
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https://math.stackexchange.com/questions/2946992/integer-midpoint/2947104 | # Integer midpoint
Two points $$(c, d, e),(x, y, z) \in \mathbb{R}^3$$, we say that the midpoint of these two points is the point with coordinates $$\left(\frac{c+x}{2} , \frac{d+y}{2} , \frac{e+z}{2}\right)$$
Take any set $$S$$ of nine points from $$\mathbb{R}^3$$ with integer coordinates.
Prove that there must be at least one pair of points in $$S$$ whose midpoint also has integer coordinates.
I've tried to do an example with set $$S=\{(2,2,2),(1,8,2),(3,4,5),(5,2,2),(4,2,9),\\(2,1,4),(6,8,2),(0,0,0),(5,2,3)\}$$
So taking $$2$$ points, $$(3,4,5)$$ and $$(5,2,3)$$ so $$(3+5)/2=4$$, $$(4+2)/2=3$$, $$(5+3)/2 = 4$$, which are integers. I'm wanting this argument to hold in general and I'm finding it tricky to prove this does anyone have suggestions would be grateful!
• These are 3-tuples. You can look at each individual coordinate as being either even (e) or odd (o). So for example, point 1 could be (e,e,e) and point 2 could be (o,o,e) and so on.... Look at all possible cases (how many possible cases?) and then ask yourself, when would two points give integer midpoints using the formula you provided? Finally, think to yourself, if I have 9 points....... since its a discrete math class, you probably learned about a particular "principle"... – Eleven-Eleven Oct 8 '18 at 12:21
• pigeon hole principle is probably onto the right track ? – sphynx888 Oct 8 '18 at 12:32
• By the way, a big +1 for doing an example and seeing what happened. A good next step might have been to try to build a counterexample, and seeing where you got stuck --- that might have given further insight. – John Hughes Oct 8 '18 at 12:47
Just to reiterate what I said earlier, if we list the possible coordinate parities of the $$3$$-tuples, we have
$$\{(e,e,e),(e,e,o),(e,o,e),(e,o,o),(o,e,e),(o,e,o),(o,o,e),(o,o,o)\}$$
So there are $$8$$ different parity $$3$$-tuples possible. Also note the addition of even and odd integers;
$$e+e=e,$$ $$o+o=e,$$ $$e+o=o$$
So in order for your midpoint to be an integer, we need to have the points being added to have coordinates of the same parity. Therefore, since we have 9 points, assuming our first 8 points are all of different parity combinations as above, the 9th must be one of these 8 possibilities. If you add up the two points that have the same coordinate parity structure, this will yield an even number, which is divisible by 2. For example, if both ordered triples have coordinate parity structure $$(e,e,o)$$, then by the midpoint formula,
$$Mid((e,e,o),(e,e,o))=\left(\frac{e+e}{2}, \frac{e+e}{2},\frac{o+o}{2}\right)=\left(\frac{2e}{2}, \frac{2e}{2},\frac{2o}{2}\right)=(e,e,o)$$
And so by the Pigeonhole principle, there is at least 1 midpoint that contains integer coordinates. And as Robert Z noted in his solutions, this can be extended to $$n$$-tuples as well.
Hint. Read carefully Eleven-Eleven's comment (or take a look at Pigeon-Hole Principle and 2d grid?) and, by using the Pigeonhole Principle, show the following more general statement: in any set of $$2^n+1$$ points in $$\mathbb{Z}^n$$ there is at least one pair that has the midpoint in $$\mathbb{Z}^n$$. | 2019-05-20T08:45:43 | {
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https://math.stackexchange.com/questions/1754930/lebesgue-measure-preserving-differentiable-function | # Lebesgue measure-preserving differentiable function
Let $\lambda$ denote Lebesgue measure and let $f: [0,1] \rightarrow [0,1]$ be a differentiable function such that for every Lebesgue measurable set $A \subset [0,1]$ one has $\lambda(f^{-1}(A)) = \lambda (A)$. Prove that either $f(x) = x$ or $f(x) = 1- x$.
I will appreciate a hint or a solution that doesn't use ergodic theory as this is an old qual problem in measure theory. $f$ is bounded and continuous and so it is integrable and hence the hypothesis of Lebesgue's Differentiation Theorem are satisfied, but I haven't seen a way to use this or if it is even applicable.
• To understand why the "differentiable" hypothesis is there: examine the function $f(x) = |2x-1|$ on $[0,1]$. – GEdgar Apr 23 '16 at 18:15
First note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\lambda(f^{-1}([a,b])) = \lambda([0,1]) = 1 > b-a = \lambda([a,b]),$ contradiction.
Suppose $f'(x) = 0$ for some $x\in [0,1].$ Then $f(B(x,r))\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to $[0,1].$) Thus for such $r,$ $f^{-1}(B(f(x),r/4))$ contains $B(x,r).$ But $B(f(x),r/4)$ has measure no more than $2\cdot (r/4) = r/2,$ while $B(x,r)$ has measure at least $r.$ That's a contradiction. Hence $f'(x) \ne 0$ for all $x\in [0,1].$
By Darboux, $f'>0$ on $[0,1]$ or $f'<0$ on $[0,1].$ Let's assume the first case holds. Then $f$ is strictly increasing, hence a bijection on $[0,1].$ Thus if $A$ is an interval, $\lambda(A) = \lambda(f^{-1}(f(A)))= \lambda(f(A)).$ Since $f(0) = 0,$ this gives $x = \lambda([0,x]) = \lambda(f([0,x])) = f(x) - f(0) = f(x)$ for all $x\in [0,1].$
• This is a much better proof! – Giovanni Apr 23 '16 at 16:52
The following argument should work under the additional assumption that $f'$ is integrable. I don't know if this is something that you need to assume or if it is a limitation of my proof. It is worth mentioning that I need integrability of the derivative for two reasons: first to apply Lebesgue differentiation theorem for $f'$, and second to get that $f$ is absolutely continuous, which I need to apply the Fundamental Theorem of Calculus.
Let $\mu(\cdot) = \lambda(f^{-1}(\cdot))$. Then $\mu = \lambda$ and hence for a.e. $x \in (0,1)$ we have
$$1 = \frac{d\lambda}{d\mu}(x) = \lim_{r \to 0}\frac{\lambda(B_r(x))}{\mu(B_r(x))} = \lim_{r \to 0}\frac{\int_{B_r(x)}d\lambda}{\mu(B_r(x))} = \lim_{r \to 0}\frac{1}{\lambda(f^{-1}(B_r(x)))}\int_{f^{-1}(B_r(x))}|f'|d\lambda,$$ where in the last equality we used the change of coordinates $z = f(y)$. Notice that $\lambda(f^{-1}(B_r(x))) = \lambda(B_r(x)) \to 0$, hence by Lebesgue differentiation theorem, for every $y \in f^{-1}(x)$ we have that $|f'(y)| = 1.$ Moreover we have that for $y \in (0,1)$, $y \in f^{-1}(f(y))$, thus proving that $|f'| = 1$ a.e. in $[0,1]$.
To conclude the argument, we only need to show that $f'$ cannot change sign. Indeed, by the FTC, we would have that $$f(1) = f(0) + \int_0^1f'(y) = f(0) \pm 1,$$ showing that either $f(0) = 0$ and $f' = 1$ or $f(0) = 1$ and $f' = -1$.
To prove that the derivative is constant a.e., I'll show that $f$ is surjective. By continuity, $f([0,1])$ is a compact subset of $[0,1]$. Suppose by contradiction that $f$ is not surjective, then $\lambda(f[0,1]) < 1$. To find a contradiction it is enough to notice that $$1 = \lambda([0,1]) = \lambda(f^{-1}(f([0,1]))) = \lambda(f([0,1])) < 1.$$
In response to Ramiro's comment: Everywhere differentiable + integrable derivative implies absolute continuity, i.e. the FTC holds. Then I can prove what you asked as follows.
Necessarily, $f(0)$ is either $0$ or $1$. To see this notice that if there is $x \in (0,1)$ such that $f(x) \in \{0,1\}$ then let $y$ be the point such that $f(y) = \{0,1\} \setminus \{f(x)\}$. Assume without loss that $x< y$, then $$1 = |f(x) - f(y)| \le \int_x^y|f'| = y - x < 1.$$ Then assume without loss that $f(0) = 0$. Reasoning as above shows that $f(1) = 1$. Again by the FTC, $$1 = f(1) - f(0) = \int_0^1f' = \lambda(\{f' = 1\}) - \lambda(\{f' = -1\}).$$ Finally, if $\lambda(\{f' = -1\}) > 0$ we get a contradiction, hence $f' = 1$ a.e.
• Why proving that $f$ is surjective implies that the derivative does not change sign? – Ramiro Apr 23 '16 at 5:12
• @Ramiro: I have edited with a proof, let me know if it is convincing :) – Giovanni Apr 23 '16 at 5:41
• Thanks for writing down the details. Please note that your statement "To prove that the derivative does not change sign I'll show that f is surjective." is NOT accurate. The fact that f is surjective implies only that $\lambda({f\neq 1})=0$ or $\lambda({f\neq -1})=0$. This is all you need, but it is weaker than claiming "the derivative does not change sign". (In fact, in principle, in a set of measure zero we may even have $|f'|\neq 1$). – Ramiro Apr 23 '16 at 12:48
• Thank you for your solution. The problem mentioned that we cannot assume the derivative is continuous. I left that out because we can't in general assume things not given to us and which we cannot deduce easily as parts of the properties of the object given to us. Can we remove the assumption of integrability of the derivative or say why it suffices to consider this case only? – akech Apr 23 '16 at 13:34
• If you allow me a suugestion, I suggest that instead of writing "To prove that the derivative does not change sign I'll show that f is surjective", it would be better to write something like "To prove that the derivative is constant a.e., I'll show that f is surjective". – Ramiro Apr 23 '16 at 16:57 | 2019-06-26T08:19:46 | {
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https://math.stackexchange.com/questions/577597/does-lh%C3%B4pitals-work-the-other-way | # Does L'Hôpital's work the other way?
Hello fellows,
As referred in Wikipedia (see the specified criteria there), L'Hôpital's rule says,
$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$
As
$$\lim_{x\to c}\frac{f'(x)}{g'(x)}= \lim_{x\to c}\frac{\int f'(x)\ dx}{\int g'(x)\ dx}$$
Just out of curiosity, can you integrate instead of taking a derivative? Does
$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int f(x)\ dx}{\int g(x)\ dx}$$
work? (given the specifications in Wikipedia only the other way around: the function must be integrable by some method, etc.) When? Would it have any practical use? I hope this doesn't sound stupid, it just occurred to me, and I can't find the answer myself.
## Edit
Take 2 functions $f$ and $g$. When is
$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int_x^c f(a)\ da}{\int_x^c g(a)\ da}$$
true?
Not saying that it always works, however, it sometimes may help. Sometimes one can apply l'Hôpital's even when an indefinite form isn't reached. Maybe this only works on exceptional cases.
Most functions are simplified by taking their derivative, but it may happen by integration as well (say $\int \frac1{x^2}\ dx=-\frac1x+C$, that is simpler). In a few of those cases, integrating functions of both nominator and denominator may simplify.
What do those (hypothetical) functions have to make it work? And even in those cases, is is ever useful? How? Why/why not?
• You cannot use L'Hospital the other way to evaluate a limit. You can try an example, as I did and it just won't work. Realizing that taking a derivative of num and denom really means that you are linearizing and so that the ratio of the slopes (i.e. derivatives) will give you the answer to the limit, makes sense. But when you are anti deriving, what does that geometrically mean to the limit? I am not aware that "anti-L'Hospitaling" does anything... – imranfat Nov 22 '13 at 21:14
• @JMCF125: This is a good question. Perhaps you should clarify it by adding bounds to the integrals. Specifically, how about writing $\int_{c}^{x}\; f(t)\; dt$ instead of $\int f(x) dx$? Same for $g(x)$. I point out that neither answer addressed this question. – Chris K Nov 22 '13 at 21:18
• @ChrisK, good, then the constants are eliminated! Forgot about them... I'll add that ASAP- – JMCF125 Nov 22 '13 at 21:33
• @JMCP125 : based on Umberto P.'s answer, the answer seems to be "yes". The next question is, is this ever useful, like the "regular" L'Hopital's Rule itself is? – Stefan Smith Nov 23 '13 at 2:28
• @miracle173, I never said l'Hôpital's implied this. And I didn't say the integrals had to be real anti-derivatives; in fact, in the edit, I added ChrisK's suggestion, which clears that up. – JMCF125 Nov 24 '13 at 8:52
I recently came across a situation where it was useful to go through exactly this process, so (although I'm certainly late to the party) here's an application of L'Hôpital's rule in reverse:
We have a list of distinct real numbers $\{x_0,\dots, x_n\}$. We define the $(n+1)$th nodal polynomial as $$\omega_{n+1}(x) = (x-x_0)(x-x_1)\cdots(x-x_n)$$ Similarly, the $n$th nodal polynomial is $$\omega_n(x) = (x-x_0)\cdots (x-x_{n-1})$$ Now, suppose we wanted to calculate $\omega_{n+1}'(x_i)/\omega_{n}'(x_i)$ when $0 \leq i \leq n-1$. Now, we could calculate $\omega_{n}'(x_i)$ and $\omega_{n+1}'(x_i)$ explicitly and go through some tedious algebra, or we could note that because these derivatives are non-zero, we have $$\frac{\omega_{n+1}'(x_i)}{\omega_{n}'(x_i)} = \lim_{x\to x_i} \frac{\omega_{n+1}'(x)}{\omega_{n}'(x)} = \lim_{x\to x_i} \frac{\omega_{n+1}(x)}{\omega_{n}(x)} = \lim_{x\to x_i} (x-x_{n+1}) = x_i-x_{n}$$ It is important that both $\omega_{n+1}$ and $\omega_n$ are zero at $x_i$, so that in applying L'Hôpital's rule, we intentionally produce an indeterminate form. It should be clear though that doing so allowed us to cancel factors and thus (perhaps surprisingly) saved us some work in the end.
So would this method have practical use? It certainly did for me!
PS: If anyone is wondering, this was a handy step in proving a recursive formula involving Newton's divided differences.
• Very interesting, +1! I certainly didn't expect it to have been successfully used before. But why didn't you make the question you are answering when you found it worked? – JMCF125 Nov 27 '13 at 11:01
• Ah, after having reread your question, I think the technique you're asking about is a little different. In my case, I went from $f'/g'$ (where both $f'$ and $g'$ were non-zero) to $f/g$ (where both $f$ and $g$ were zero). I didn't use the reverse L'Hôpital to resolve an indeterminate form; I used it to make one. – Omnomnomnom Nov 27 '13 at 15:38
• However, when you apply the function $\omega_n(x)$ and cancel out $\Pi_{i=0}^n (x-x_i)$ (sorry, couldn't resist but to use big product notation :)), the "indeterminacy" disappears, it only appears as an intermediate step. Call $\omega'$ as $\alpha$ and $$\lim_{x\to x_i} \frac{\alpha_{n+1}(x)}{\alpha_n(x)} = \lim_{x\to x_i} \left(\lim_{c\to x}\frac{\int\alpha_{n+1}(z)\ dz}{\lim_{c\to x}\int\alpha_{n}(z)\ dz}\right)=x_i-x_{n}$$. Right? (didn't check it thoroughly, there's no comment preview) – JMCF125 Nov 27 '13 at 21:07
• Exactly! So, you could say that the trick in using L'Hopital's rule backwards is choosing the antiderivatives of $\alpha_{n+1}$ and $\alpha_n$ that give you the indeterminate fraction. I'm not sure what the ruole of $c$ is in your statement, though – Omnomnomnom Nov 27 '13 at 21:16
• About the $c$, it's to eliminate the constants, see the problem Arkamis referred and the answer of Kaz, which has a related idea. The point is to make the integral so small it would be alike a derivative, only as a different expression. (BTW, the second $c\to x$ limit was not suppose to be there, I pasted it twice) – JMCF125 Nov 27 '13 at 21:55
With L'Hospital's rule your limit must be of the form $\dfrac 00$, so your antiderivatives must take the value $0$ at $c$. In this case you have $$\lim_{x \to c} \frac{ \int_c^x f(t) \, dt}{\int_c^x g(t) \, dt} = \lim_{x \to c} \frac{f(x)}{g(x)}$$ provided $g$ satisfies the usual hypothesis that $g(x) \not= 0$ in a deleted neighborhood of $c$.
• Good, interpreted the question in the "right" way. – André Nicolas Nov 22 '13 at 21:26
• Or $\frac{\infty}{\infty}$. – marty cohen Nov 23 '13 at 2:56
• Very important, this only works if you know that $\lim \frac{f(x)}{g(x)}$ exists. Even if the first limit exists, you cannot conclude that the second one does. – N. S. Nov 23 '13 at 5:58
• @N.S., neither can you conclude the division of the derivatives exists, even when the division of the original functions does, as in the original l'Hôpital's. – JMCF125 Nov 23 '13 at 13:48
• @JMCF125 True, but the goal in L'H is: starting with a limit, you replace by a different limit. In L'H you replace the functions by the Derivatives, not the derivatives by the functions. In classical L'H, you check $\lim\frac{f'}{g'}$ and if this limit exists you are done. Here you check $\lim \frac{\int f}{\int g}$ and if the limit exists, you cannot say anything. Note that if you go the other direction, you just apply classical L'H – N. S. Nov 23 '13 at 15:40
No it does not, for example consider the limit $\lim_{x\to 0}\frac{\sin x}{x}$, L'Hôpital's gives you 1, but reverse L'Hôpital's goes to $-\infty$ (without introducing extra constants).
• +1 for being a simple, easy counterexample, although I assume that you're taking the integral of sin(x) to be -cos(x) - but that implies either a constant or a definite interval on the integral. – fluffy Nov 23 '13 at 4:33
• @fluffy : Yes, exactly, thank you – Arjang Nov 23 '13 at 4:43
• The conditions under which it works may be more restricted... – JMCF125 Nov 23 '13 at 17:20
• If we want to use reverse L'Hopital, it is our 'responsibility' to ensure that the newly integrated function is also of an indeterminate form, and that can be tackled easily using the integration constant. Here, instead of -cosx, one should use 1-cosx (in the numerator) for the limit to exist, and hence evaluate it to be 1. – arya_stark May 8 '18 at 8:08
Recall that the integral of a function requires the inclusion of an arbitrary constant. So, if $f(x) = x$, then $\int f(x)\ dx = \frac12 x^2 + C$.
Then, assuming your rule holds, then $$\lim_{x\to 0} \frac{x}{x} = \lim_{x \to 0} \frac{\int x\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\frac12 x^2 + C_f}{\frac12 x^2 + C_g} = \frac{C_f}{C_g}.$$
This means you can get literally any value you wish.
• What if we forget the constants? Wait, don't answer yet, I'll edit my question first. – JMCF125 Nov 22 '13 at 21:32
• @JMCF125 Note that it really makes no sense to talk of "forgetting the constants". For example, the following three functions are antiderivatives of $2\sin(x)\cos(x)$: $\sin^2(x)$, $-\cos^2(x)$, $-\cos(2x)/2$. Which one is the "correct" one to use if we are to "forget constants"? – Andrés E. Caicedo Nov 22 '13 at 21:59
• @AndresCaicedo, what if we use a definite integral $\displaystyle\int_0^x x\ dx=\frac{x^2}2$, as referred in my edit. That way, wouldn't the result be the same? – JMCF125 Nov 24 '13 at 8:57
• Why use $0$ as the lower bound. Also, it should be $\int_0^x t\ dt$, as it makes no sense to have the variable of integration be the same as a bound. In any case, a lower bound of $c$ makes some sense if $x \to c$, since, as others have said, one of the ways this question makes sense is if $f, g \to 0$ at $x=c$. But it elides the case of $f,g \to \infty$. – Emily Nov 24 '13 at 16:13
• @Arkamis, «(...) as it makes no sense to have the variable of integration be the same as a bound.», sorry, distraction. :S I meant for the top limit to be an $a$. I used the lower limit as $0$ so that in your example, the integral would be $\int^a_0 x\ dx=\frac{a^2}2+C-\frac{0^2}2-C=\frac{a^2}2$. Isn't it right like this? – JMCF125 Nov 27 '13 at 10:57
If the limit of the ratio of those functions exists, I suspect that it may be possible for a certain definite integral to have the same limit. Something like:
$$\lim_{x\to \infty}\frac{\int_x^{x+1}f(x)\ dx}{\int_x^{x+1}g(x)\ dx}$$
The idea is that we grow $x$, and compare the ratios of some definite integrals in the neighborhood of x, of equal width and position.
For instance, if we imagine the special case that the functions separately converge to constant values like 4 and 3, then if we take $x$ far enough, then we are basically just dividing 4 by 3. A definite integral of width 1 of a function that converges to 4 has a value that is approximately four, near a domain value that is large enough.
But this is like a derivative in disguise. If we divide a definite integral by the interval length, and then shrink the interval, we are basically doing derivation to obtain the original function that was integrated. Except we aren't shrinking the interval, since we don't have to: the fixed width 1 becomes smaller and smaller in relation to the value of $x$, so it is a "quasi infinitesimal", so to speak.
The real problem with this, even if it works, is that integrals don't seem to offer any advantage. Firstly, even if the function has the properties of being integrable, it may not be symbolically integrable, or it may be hard to integrate. Integration produces something more complex than the integrand.
The advantage of L'Hôpital's Rule is that we can reduce the power of the functions. If they are polynomials, they lose a degree in the differentiation, which can be helpful, and gives us a basis for instantly gauging the limit of the ratios of polynomials of equal degree simply by looking at the ratios of their highest degree coefficients. And then certain common functions at least don't grow any additional hair under differentiation, like sine, cosine, e to the x.
• Interesting, +1. But what about those exceptional functions whose derivative is more complicated? Wouldn't the integral ease calculation? – JMCF125 Nov 23 '13 at 13:55
L'Hospital's Rule is really a compact way of looking at ratio of the Taylor's polynomial for the numerator and denominator functions (w/o remainder). This polynomial is an exact fit to the function and its derivatives at the indicated point and close in a small neighborhood thereof.
When the limit is 0/0 that means that both Taylor's polynomials have 0 constant terms. So then you look at the x terms (which represent the derivatives). If you can pick out the limit from there, good -- if they are both zero you go on to the $x^2$ terms (if any).
So L'Hospital's Rule isn't some vague thing that was plucked out of the air and happens to use derivatives. Any method of evaluating limits ought to be based on a clear mathematical derivation. You can use integrals or gamma functions or field theory or anything that pleases you, as long as you show you have a sound basis for evaluating the limit that way.
• Can you help me finding such a basis? – JMCF125 Nov 28 '13 at 12:26
• @ Betty Mock What is exact permissible basis to modify the vanishing numerator and denominator ? – Narasimham Nov 20 '14 at 20:11
• Expand the top and bottom in a Taylor's polynomial. If the first term of both polynomials is zero, then you look at the second term. If both of those are zero, you go on to the third term, etc. Sooner or later you get a term which is not zero, and that is the one which shows you the limit. – Betty Mock Nov 22 '14 at 2:46
$$\lim_{ x \to 0} \frac{ sin \,x \,}{\,x}= \lim_{ x \to 0} \frac{\int sin \,x \,\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\ (- cos \, x)}{( x^2/2)-1} =\lim_{x \to 0} \frac{\ (-cos \, x)+1}{( x^2/2)} = 1$$
Constants should be appropriately chosen from function/ power series expansion of the integrands and distributed between numerator and denominator maintaining the balance.
The statement is vague but I am sure it can be properly worded/stated to include even second, third integrands in the numerator and denominator.
More simply:
When $\frac yx =const = m$ as L'Hospital limit, it is =$\frac{dy}{dx}$ by Quotient Rule.
$m = \frac {dy}{ dx} = \frac {y \, \pm C_1 }{ x} =\frac {y \, }{ x \pm C_2}.$ | 2019-09-16T02:50:14 | {
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http://mathematica.stackexchange.com/questions/6052/improve-speed-for-calculating-a-recursive-sequence/6060 | # Improve speed for calculating a recursive sequence
I want to calculate following recursive sequence:
$\alpha_{0}=0,\\ \cos(\alpha_{i})=\cos(\alpha_{i-1})\cdot\cos(\beta_{i})+\sin(\alpha_{i-1})\cdot\sin(\beta_{i})\cdot\cos(\gamma_{i}).$
In mathematica 8.0.1 I try to do this with:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1;;endRecursion]];
gamma = RandomReal[{0,2*Pi}, endRecursion];
alpha[0]:= 0;
alpha[i_]:=
ArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]]+
Sin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];
result=Table[alpha[i], {i, 1, endRecursion}];
The values for beta are imported because they are generated by another formula which is not important for my question here. Values for beta can be found here.
The problem is that the calculation of the result is very slowly. It takes about 14s for just 20 iterations! And I need about 300 iterations (times ~10000, because I have several lists of beta values).
I recognized that the calculation is much faster if I skip the Cos[alpha[i-1]] in the first or the Sin[alpha[i-1]] in the 2nd summand. Then it takes just ~ 0.01 s for 20 iterations. But unfortunately this falsifies my result. I cannot understand why it is so much faster if there is just one time alpha[i-1] on the right side of the equation. Am I doing something wrong? Is there any way to increase the speed of this calculation? I would be glad about any help!
-
Have you tried caching? alpha[i_] := alpha[i] = ArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]] + Sin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]] – J. M. May 26 '12 at 12:44
I am such a *** idiot! Thank you so much for this comment! I totally forgot the caching although I used it some time ago (5 years, perhaps too long time ago). Unfortunately, I did not see this trick when I searched for a solution, otherwise I would not have asked this question... Thank you anyway! You saved my weekend! How can I accept your comment as answer? – partial81 May 26 '12 at 12:58
@J.M. hmm, in my amazement that nobody had answered this, I wrote my answer without seeing your comment... Hope you were not writing up one also! – acl May 26 '12 at 13:07
You can accept acl's answer. :) I didn't answer because I can't test at the moment. – J. M. May 26 '12 at 13:07
Ok, if you do not mind! Thanks again! – partial81 May 26 '12 at 13:12
You can speed it up by "memoization", i.e., remembering previous values of alpha[i]:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1 ;; endRecursion]];
gamma = RandomReal[{0, 2*Pi}, endRecursion];
alpha[0] := 0;
alpha[i_] :=
alpha[i] =
ArcCos[Cos[alpha[i - 1]]*Cos[beta[[i]]] +
Sin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];
result = Table[alpha[i], {i, 1, endRecursion}];
-
Thanks acl and @J.M. for answering my question so quickly! – partial81 May 26 '12 at 13:14
There was a mistake: you were importing strings (you didn't convert to numbers after StringSplit.) I fixed it. – Szabolcs May 26 '12 at 13:15
@Szabolcs yes, I had just realised it and our edits almost overlapped... thanks – acl May 26 '12 at 13:16
Okay, I'll remove the comments then. – Szabolcs May 26 '12 at 13:20
Although you can always implement recursion by hand the way you are trying to do, there are also some specialized functions that are designed to make your life a little easier. In particular, there is FoldList.
This approach is also slightly faster than the manual recursion approach (assuming you start from a clean slate):
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1 ;; endRecursion]];
gamma = RandomReal[{0, 2*Pi}, endRecursion];
FoldList[ArcCos[Cos[#1]*#2[[1]] + Sin[#1]*#2[[2]]] &, 0,
Transpose[{Cos[beta], Sin[beta] Cos[gamma]}]]
After initializing the lists as in your example, I need only a single statement to generate the list.
The folding function is what appears in the original recursive formulation, but it is written as a pure function with two formal arguments #1 and #2.
The first argument, #1, refers to your alpha and its starting value is passed to FoldList as the second argument (0). The second argument in the pure function (#2) refers to the elements of the list Transpose[{Cos[beta], Sin[beta] Cos[gamma]}] in which I collected the pre-existing lists beta and gamma (but in a form that is already pre-processed to avoid having to evaluate cosines and sines individually in the recursion).
-
With this particular formulation, one could even simplify the FoldList[] expression as FoldList[ArcCos[{Cos[#1], Sin[#1]}.#2] &, 0, Transpose[{Cos[beta], Sin[beta] Cos[gamma]}]]. One might even consider using the Weierstrass substitution here, since the beta values are all $< \pi/2$... – J. M. May 26 '12 at 18:47
@J.M. That's correct - the dot product may give a tiny additional speed boost but I didn't check that. – Jens May 27 '12 at 0:50
Thanks @Jens for this additional answer. FoldList is something I have never used so far – that will change now (although the additional speed boost is just small). – partial81 May 27 '12 at 11:24
@ Jens, Let’s say: alpha[i_]:=alpha[i]=ArcCos[Cos[alpha[i - 1]]*Cos[beta[[i+1]]]+Sin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]]. How do I have to adapt your solution with FoldList for this case? I really have problems to see a solution because now I would need in my iteration the i+1 element of beta. I would be happy if you could give me a hint again! – partial81 May 31 '12 at 13:10
@partial81 The term you changed is the first one in Transpose[{Cos[beta], Sin[beta] Cos[gamma]}], and the change is simply a shift in the index. The fastest way this can be implemented is to use RotateLeft: Transpose[{RotateLeft[Cos[beta]], Sin[beta] Cos[gamma]}] – Jens May 31 '12 at 15:39
The function runs slowly because each iteration calls alpha twice the number of times of the previous iteration. Thus, $n$ iterations invoke alpha $2^n$ times -- exponential run time! Fortunately, the two nested calls to alpha are identical. If we change alpha to call itself only once instead of twice, the algorithm runs in linear time while still retaining simple recursive form:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1;;endRecursion]];
gamma = RandomReal[{0,2*Pi}, endRecursion];
alpha[0] = 0;
alpha[i_]:=
Module[{a = alpha[i-1], b = beta[[i]], g = gamma[[i]]}
, ArcCos[Cos[a]*Cos[b]+Sin[a]*Sin[b]*Cos[g]]
]
result = Table[alpha[i], {i, 1, endRecursion}];
Even so, I agree with @Jens that functional iteration is frequently preferred over recursion in order to avoid hitting the $RecursionLimit (or$IterationLimit).
-
Thanks @WReach for answering the other part of my question. Now I understand why it was so slowly at the beginning. And it is nice that you give a link to the useful functional iterations. – partial81 May 27 '12 at 20:53 | 2016-06-28T09:52:46 | {
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http://math.stackexchange.com/questions/423736/finding-the-exact-value | # Finding the Exact Value
How would one go about finding the exact value of $\theta$ in the following:$\sqrt{3}\tan \theta -1 =0$? I am unsure of how to begin this question. Any help would be appreciated!
-
are you familiar with $\arctan$ function? – Alex Jun 18 '13 at 16:15
I am not familiar with that function – ComradeYakov Jun 18 '13 at 16:16
@Tim I meant the first, sorry for the confusion. It isn't cubed – ComradeYakov Jun 18 '13 at 16:17
Hint: $\sqrt{3}\tan \theta -1 =0\implies \sqrt{3}\tan \theta =1\implies \tan \theta=1/\sqrt{3}$
Recall from SOH CAH TOA, that $\tan \theta=\text{opposite}/\text{adjacent}$, so in this case $1$ is the opposite side and $\sqrt{3}$ is the adjacent side.
The above is a special triangle. Since you know that $\tan \theta=1/\sqrt{3}$, it follows that the angle $\theta=\pi/6$. However since $\tan \theta$ is a trigonometric periodic function, that value will come up again, and again. (Just like $\sin0=\sin2\pi).$
$\tan \theta$ has a period of $\pi.$
So to be fully accurate, $\theta=\pi/6 +n\pi, n \in \mathbb{Z}$
-
Thank you! So theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution? – ComradeYakov Jun 18 '13 at 16:31
theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution. – Sujaan Kunalan Jun 18 '13 at 16:35
Ok, thanks for your help! – ComradeYakov Jun 18 '13 at 16:37
You are welcome! – Sujaan Kunalan Jun 18 '13 at 16:38
$2\cos^{2} \theta+\cos \theta-1=0 \implies 2\cos^{2}+\cos \theta =1 \implies \cos \theta (2\cos \theta+1)=1\implies \cos \theta =1 \implies \theta =2n\pi, n\in \mathbb{Z}$ However, $\cos \theta(2\cos \theta +1)=1$ also implies that $2\cos \theta+1=1 \implies 2\cos \theta =0 \implies \cos \theta =0 \implies \theta = (2n-1)/\pi, n \in \mathbb{Z}$ – Sujaan Kunalan Jun 18 '13 at 17:07
If $\sqrt{3}\tan \theta - 1= 0$ then $\sqrt{3}\tan\theta = 1$ and $\tan\theta = 1/\sqrt{3}$. Hence $\theta =\arctan(1/\sqrt{3}) = \frac{\pi}{6}$.
Notice that the graph $y = \tan\theta$ repeats every $\pi$-radians. Hence, all solutions are given by
$$\theta = \frac{\pi}{6} + \pi n$$
where $n$ is any whole number.
- | 2016-05-25T21:16:24 | {
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https://stats.stackexchange.com/questions/247859/does-the-square-of-the-minimum-of-two-correlated-normal-variables-have-a-chi-squ | # Does the square of the minimum of two correlated Normal variables have a chi-squared distribution?
If the random variables $X$ and $Y$ are normally distributed with mean $\mu = 0$ and standard deviation $\sigma = 1$, then define the random variable $Z = \min(X, Y )$. The problem is to prove that $Z^2$ follows a gamma distribution with parameters $(\alpha = 1/2, \lambda= 1/2)$ (i.e. a chi-squared distribution).
I have: $$P(Z^2 \le z) = P((\min\{X,Y\})^2 \le z) =P\left(-\sqrt{z}\le \min(X,Y)\le \sqrt{z}\right) \\ = P\left(-\sqrt{z}\le X\le \sqrt{z}, -\sqrt{z}\le Y\le \sqrt{z}\right)$$ But I am not sure where to go from here, since the random variables aren't assumed to be independent. I would appreciate any hints!
• Your logic at the last step seems to go astray. If the minimum of (X,Y) is within $\pm \sqrt z$ (as you have in your second-last line), that doesn't make both of them in that interval, as you have in your last line. You only need one of the two of them in there. – Glen_b -Reinstate Monica Nov 28 '16 at 3:57
• @glen_b What then would be the interval on the other one? – Mjt Nov 28 '16 at 14:08
• That's not a productive way to approach writing the event. But first you might want to ponder whether anything has been left out of your framing of the question (which is possibly why whuber is able to offer a counterexample). – Glen_b -Reinstate Monica Nov 28 '16 at 15:55
The result is not true.
As a counterexample, let $(X,Y)$ have standard Normal margins with a Clayton copula, as illustrated at https://stats.stackexchange.com/a/30205. Generating 10,000 independent realizations of this bivariate distribution, as shown in the lower left of the figure, produces 10,000 realizations of $Z^2$ that clearly do not follow a $\Gamma(1/2,1/2)$ distribution (in a Chi-squared test of fit, $\chi^2=121, p \lt 10^{-16}$). The qq plots in the top of the figure confirm that the marginals look standard Normal while the qq plot at the bottom right indicates the upper tail of $Z^2$ is too short.
The result can be proven under the assumption that the distribution of $(X,Y)$ is centrally symmetric: that is, when it is invariant upon simultaneously negating both $X$ and $Y$. This includes all bivariate Normals (with mean $(0,0)$, of course).
The key idea is that for any $z \ge 0$, the event $Z^2 \le z^2$ is the difference of the events $X \ge -z\cup Y \ge -z$ and $X \ge z \cup Y \ge z$. (The first is where the minimum is no less than $-z$, while the second will rule out where the minimum exceeds $z$.) These events in turn can be broken down as follows:
$$\Pr(Z^2 \le z^2) = \Pr(X\ge -z) - \Pr(Y \le -z) + \Pr(X,Y\lt -z) - \Pr(X,Y\gt z).$$
The central symmetry assumption assures the last two probabilities cancel. The first two probabilities are given by the standard Normal CDF $\Phi$, yielding
$$\Pr(Z^2 \le z^2) =1 - 2\Phi(-z).$$
That exhibits $Z$ as a half-normal distribution, whence its square will have the same distribution as the square of a standard Normal, which by definition is a $\chi^2(1)$ distribution.
This demonstration can be reversed to show $Z^2$ has a $\chi^2(1)$ distribution if and only if $\Pr(X,Y\le -z) = \Pr(X,Y\ge z)$ for all $z\ge 0$.
Here is the R code that produced the figures.
library(copula)
n <- 1e4
set.seed(17)
xy <- qnorm(rCopula(n, claytonCopula(1)))
colnames(xy) <- c("X", "Y")
z2 <- pmin(xy[,1], xy[,2])^2
cutpoints <- c(0:10, Inf)
z2.obs <- table(cut(z2, cutpoints))
z2.exp <- diff(pgamma(cutpoints, 1/2, 1/2))
rbind(Observed=z2.obs, Expected=z2.exp * length(z2))
chisq.test(z2.obs, p=z2.exp)
par(mfrow=c(2,2))
qqnorm(xy[,1], ylab="X"); abline(c(0,1), col="Red", lwd=2)
qqnorm(xy[,2], ylab="Y"); abline(c(0,1), col="Red", lwd=2)
plot(xy, pch=19, cex=0.75, col="#00000003", main="Data")
qqplot(qgamma(seq(0, 1, length.out=length(z)), 1/2, 1/2), z2,
xlab="Theoretical Quantiles", ylab="Z2",
main="Gamma(1/2,1/2) Q-Q Plot")
abline(c(0,1), col="Red", lwd=2)
Let $M=min(X,Y)^2$, $$P(M<m) = P(M<m,X<Y) + P(M<m,X>Y) ~~~~~~~~~~~\\ = P(M<m|X<Y)P(X<Y) + P(M<m|X>Y)P(X>Y) \\ = P(X^2<m)P(X<Y) + P(Y^2<m)P(X>Y) ~~~~~~~~~~~~~~~~~~~~~~\\ = \frac{1}{2}P(X^2<m) + \frac{1}{2}P(Y^2<m) \quad \quad \quad\quad\quad\quad\quad\quad\quad\quad~ \\ =P(X^2<m) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ~~~$$ | 2019-12-06T13:49:52 | {
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https://web2.0calc.com/questions/direct | +0
# Direct
0
330
6
the time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence?
Guest May 26, 2015
#3
+91510
+10
ok we might as well all answer this question!
This is how I was trained to do it
$$\\T\alpha \frac{L}{N} \\\\ T=k* \frac{L}{N}\qquad but\\\\ 5=k* \frac{200}{3}\qquad so\\\\ k=\frac{3}{40}\qquad so\\\\ T=\frac{3L}{40N}\\\\ When N=5 and L=500\\\\ T=\frac{3*500}{40*5}\\\\ T=7.5 \;\;hours$$
Melody May 26, 2015
Sort:
#1
+26412
+10
The rate at which fence painting is done is 200/(3*5) feet per painter-hour.
In order to paint 500 feet we need 500*3*5/200 painter-hours
With 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours
.
Alan May 26, 2015
#2
+81090
+10
If 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour.
Then, 5 painters can paint 200/3 ft every hr........so........500/ (200/3) = 1500/200 = 7.5 hrs.
CPhill May 26, 2015
#3
+91510
+10
ok we might as well all answer this question!
This is how I was trained to do it
$$\\T\alpha \frac{L}{N} \\\\ T=k* \frac{L}{N}\qquad but\\\\ 5=k* \frac{200}{3}\qquad so\\\\ k=\frac{3}{40}\qquad so\\\\ T=\frac{3L}{40N}\\\\ When N=5 and L=500\\\\ T=\frac{3*500}{40*5}\\\\ T=7.5 \;\;hours$$
Melody May 26, 2015
#4
+81090
0
Which can we do faster.......paint that fence or answer this question?????
{ I say its a toss up..... }
CPhill May 26, 2015
#5
+72
+5
Depends on how much You can take 60 seconds on writing this while 70 painters paint a small fence 3*3 fence
xXmathXx May 26, 2015
#6
+72
0
Hello CPhill
xXmathXx May 26, 2015
### 24 Online Users
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https://puzzling.stackexchange.com/questions/41436/minimum-steps-on-a-numberline | # Minimum steps on a numberline
Given an infinite numberline, you start at zero. On every i'th move you can either move i places to the right, or i places to the left. How, in general, would you calculate the minimum number of moves to get to a target point x? For example:
if x = 9:
move 1: starting at zero, move to 1
move 2: starting at 1, move to 3
move 3: starting at 3, move to 0
move 4: starting at 0, move to 4
move 5: starting at 4, move to 5
• This seems like another mathbook-type problem to me. – Mithical Aug 25 '16 at 11:47
• @Mithrandir I think so. VTC. – IAmInPLS Aug 25 '16 at 11:50
• I say "don't close" -- unless it is mandatory to have a real-life story enclosing the real puzzle to be solved. – Rosie F Aug 26 '16 at 3:15
• I want to go to school with these VTCers; clearly their textbooks have much more interesting problems than mine. – ffao Aug 26 '16 at 3:28
• Shouldn't the last line end with "move to 9"? – celtschk Aug 26 '16 at 9:23
First,
you need to find $n$, for which $S_n=1+2+\dots+n=\frac{n(n+1)}2\ge x$
Then,
find a subset of the numbers in range $[1, n]$ which sum to $\frac{S_n-x}2$. This can always be done if the parity of $S_n$ is the same as the parity for $x$. This means, that if $m$ is the lowest number for which $S_m\ge x$, then at least one of $S_m$, $S_{m+1}$ or $S_{m+2}$ will do the work.
With the help of this
adding the rest, and subtracting the above will give you the desired solution
First you calculate the smallest $n$ with $\frac{1}{2}n(n+1) \geq x$
if $\frac{1}{2}n(n+1)-x$ is odd then increase $n$ by $1$ if $n$ is even or increase $n$ by $2$ if $n$ is odd.
Now $\frac{1}{2}n(n+1)-x$ is even
Now you can put a minus sign before some numbers that have the sum $\frac{1}{2} (\frac{1}{2}n(n+1)-x)$ and you are done.
## Example:
$x=23$
The smallest $n$ with $\frac{1}{2}n(n+1) \geq x$ is $n=7$
$\frac{1}{2}n(n+1)-x = 28-23 = 5$ is odd and $n$ is also odd so we have to increase $n$ by $2$.
That means $n=9$
$\frac{1}{2}(\frac{1}{2}n(n+1)-x) = \frac{1}{2}(45-23) = 11$
So we have to put a minus in front of numbers with the sum $11$. For example $2$ and $9$.
Result: 1-2+3+4+5+6+7+8-9
It comes out to be a better puzzle than it was voted to close for!
One thing is for sure, if the given number $x$ is a triangular number, the answer is its position in the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66
which is given by
$$\left\lfloor\dfrac{\sqrt{1+8x}-1}{2}\right\rfloor$$
Also:
here is my code, which gives a very interesting pattern,
output: For x=1, ans : 1 For x=2, ans : 3 For x=3, ans : 2 For x=4, ans : 3 For x=5, ans : 5 For x=6, ans : 3 For x=7, ans : 5 For x=8, ans : 4 For x=9, ans : 5 For x=10, ans : 4 For x=11, ans : 5 For x=12, ans : 7 For x=13, ans : 5 For x=14, ans : 7 For x=15, ans : 5 For x=16, ans : 7 For x=17, ans : 6 For x=18, ans : 7 For x=19, ans : 6 For x=20, ans : 7 For x=21, ans : 6 For x=22, ans : 7 For x=23, ans : 9 For x=24, ans : 7 For x=25, ans : 9 For x=26, ans : 7
which is :
$1,3,2,3,5,3,5,4,5,4,5,7,5,7,5,7,6,7,6,7,6,7,9,7,9,7,9$,... OEIS/A140358
check the values up to $1000$ here. see this final code
#include<bits/stdc++.h>
using namespace std;
int i,n,t;
int main()
{ int x,ans;
for(cin>>t;t;t--)
{ cin>>x;
n=floor((sqrt(1+8*x)-1)/2);
//cout<<n<<endl;
i=(n*(n+1))/2;
//cout<<x<<", sum:"<<i<<", N:"<<n<<endl;
if(n%2) //n is odd
{
if((x-i)%2) ans=n+2;
else ans=n+1;
}
else
{
if((x-i)%2) ans=n+1;
else ans=n+3;
}
cout<<">! For x="<<x<<", ans : "<<ans<<endl;
}
} | 2019-12-12T04:09:42 | {
"domain": "stackexchange.com",
"url": "https://puzzling.stackexchange.com/questions/41436/minimum-steps-on-a-numberline",
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"openwebmath_perplexity": 981.3023337572796,
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"lm_q1q2_score": 0.853164898901965
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# The addition problem above shows four of the 24 different in
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02 Nov 2010, 23:34
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1,234
1,243
1,324
.....
....
+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
[Reveal] Spoiler: OA
Last edited by Bunuel on 06 Nov 2012, 02:17, edited 1 time in total.
Renamed the topic and edited the question.
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02 Nov 2010, 23:45
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student26 wrote:
1,234
1,243
1,324
.....
....
+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A.24,000
B.26,664
C.40,440
D.60,000
E.66,660
Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.
The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660
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08 Feb 2011, 05:26
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1,2,3,4 can be arranged in 4! = 24 ways
The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's
Addition always start from right(UNITS) to left(THOUSANDS);
Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6
Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6
Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6
Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6
Ten thousands place of the result: 0+6(Carried over from thousands place) = 6
Result: 66660
Ans: "E"
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08 Feb 2011, 05:48
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Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
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22 Jan 2013, 20:07
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hellscream wrote:
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.
How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)
1111
1112
1121
... and so on till 4444
By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.
Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040
or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
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09 Nov 2010, 05:24
Thanks for the great explaination shrouded1!
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08 Feb 2011, 04:09
the addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 exactly once in each integer.what is the sum of these 24 integers?
1,234
1,243
1,324
.......
.......
+4,321
a) 24,000
b) 26,664
c) 40,440
d) 60,000
e) 66,660
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22 Jan 2013, 09:44
Bunuel wrote:
Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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The addition problem above shows four of the 24 different intege [#permalink]
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21 May 2013, 06:35
This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives
1234 +4321 = 5555
1243 +3421 = 5555
we may see that there are 4!/2 pairings we can make, giving us
5555(12) = 66660
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Re: The addition problem above shows four of the 24 different in [#permalink]
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16 Jun 2014, 03:04
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Re: The addition problem above shows four of the 24 different in [#permalink]
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28 Sep 2014, 09:41
For those who could not memorize the formular, you can guess the answer in 30 secs:
Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something
So,
6x1(thousand something) = 6 (thousand something)
6x2(thousand something) = 12 (thousand something)
6x3(thousand something) = 18 (thousand something)
6x4(thousand something) = 24 (thousand something)
Add them all 6+12 +18 + 24 = 60 (thousand something)
----> E
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Re: The addition problem above shows four of the 24 different in [#permalink]
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Re: The addition problem above shows four of the 24 different in [#permalink]
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15 Aug 2016, 13:38
each term is repeating 6 times..(total 24/4=4)
now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining.
repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.
only E has 60 as last two digits.
Re: The addition problem above shows four of the 24 different in [#permalink] 15 Aug 2016, 13:38
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# The addition problem above shows four of the 24 different in
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2017-01-21T17:10:24 | {
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http://mathhelpforum.com/advanced-statistics/21429-conditional-probabilty-print.html | # Conditional Probabilty
• October 27th 2007, 06:10 AM
Revilo
Conditional Probabilty
First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?
So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market
P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)
Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.
• October 28th 2007, 01:11 AM
kalagota
Quote:
Originally Posted by Revilo
First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?
So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market
P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)
Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.
yeah, it is correct..
but, i think, you should not look for the union, instead for the intersection..
note that P(A|B)= P(A int B) / P(B) [also, P(B|A) = P(A int B) / P(A) ]; this theorem should help you find the answer.
• October 28th 2007, 08:51 PM
Soroban
Hello, Revilo!
Quote:
First little piggy is 25% likely to go to market if second little piggy goes.
Second little piggy is 50% likely to go to if first little piggy goes.
Sadly at least one of them must go to market.
How likely are they both to go?
All your ideas are good ones . . .
Bayes' Theorem says: . $P(A|B) \:=\:\frac{P(A \wedge B)}{P(B)}$
We are given:
. . $P(1^{st}|2^{nd}) \:=\:\frac{1}{4}\quad\Rightarrow\quad \frac{P(1^{st} \wedge 2^{nd})}{P(2^{nd})}\: = \: \frac{1}{4}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{4}\!\cdot\!P(2^{nd})$ .[1]
. . $P(2^{nd}|1^{st}) \:=\:\frac{1}{2}\quad\Rightarrow\quad \frac{P(2^{nd} \wedge 1^{st})}{P(1^{st})} \:=\:\frac{1}{2}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{2}\!\cdot\!P(1^{st})$ . [2]
Equate [1] and [2]: . $\frac{1}{4}\cdot P(2^{nd}) \:=\:\frac{1}{2}\cdot P(1^{st}) \quad\Rightarrow\quad P(2^{nd}) \:=\:2\cdot P(1^{st})$ . [3]
Since at least one of them goes to the market: . $P(1^{st} \vee 2^{nd}) \:=\:1$
We have: . $P(1^{st} \vee 2^{nd}) \;=\;P(1^{st}) + P(2^{nd}) - P(1^{st} \wedge 2^{nd}) \;=\;1$ . [4]
. . [3] $\; P(2^{nd}) \: = \: 2\cdot P(1^{st})$
. . [2] $\; P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{2}\cdot P(1^{st})
$
Substitute into [4]: . $P(1^{st}) + 2\cdot P(1^{st}) -\frac{1}{2}\cdot P(1^{st}) \;=\;1\quad\Rightarrow\quad \frac{5}{2}\cdot P(1^{st}) \:=\:1$
Therefore: . $P(1^{st}) \:= \:\frac{2}{5}\qquad P(2^{nd}) \: = \: \frac{4}{5} \qquad \boxed{P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{5}}$ | 2015-10-09T02:53:02 | {
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https://math.stackexchange.com/questions/2631523/the-base-of-a-logarithm | # The base of a logarithm
What exactly is the base of a logarithm ? and how should it be understood ? I used to think it was the base of a "normal" exponent e.g. the $2$ in $2^{75}$ would be the base in logarithmic form, but the change of base formula can accept ANY base, and when finding the number of digits in $2^{75}$, you use the common log:
$$2^{75}$$
$$\log_{10}(2^{75})$$
$$75\log_{10}(2)$$
$$75(0.301)+1=23 \textrm{ digits}$$
I understand that these formulas work, I just can't wrap my head around why they work the way they do, and the heart of my issue is how I should understand the base.
I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that. But if that were the case, then would a base 16 represent a hexadecimal system ? and how would that work considering we use letters in addition to numbers ?
• How many digits in $10^k$? and what is value of $\log_{10}(10^{k})$ ? Try searching for pattern. Feb 1 '18 at 17:45
• "I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that." Really? I would have thought those were the definitions. Feb 1 '18 at 18:00
I think you have the idea.
If $y = 2^{75}$ then $\log_2 y = 75$
One way to use bases is to find the number of digits that that number would have in that system. If $32<y<64$ then $5<\log_2 y < 6$
You can take logarithms in any base, and convert between bases.
$\log_a x = \frac {\log_b x}{\log_b a}$
For example, $\log_{10} 2^{75} = 75 \log_{10} 2$ as you have above.
But you could also say $\log_{10} 2^{75} = \frac {\log_2 2^{75}}{\log_2 10}$
Which implies $\log_2 10 = \frac 1{\log_{10} 2}$
You're right about the number of digits of a number in a specific base but why? To prove it let $(x)_b$ denote the number x in base $b$. If $b=10$ it is the same natural and convenient base we often use. Obviously if any number is between two consecutive powers of $10$, for example $1253$ is between $10^3$ and $10^4$ and it has 4 digits. In fact for any number $x$ if it belongs to $(10^n,10^{n+1}-1)$ for some n then $n+1$ is the number of digits of $x$. Then we can find the number of digits as following:$$10^n\le x<10^{n+1}\\n\le \log_{10}x<n+1\\n=\lfloor \log_{10}x\rfloor+1$$the same argument is true for any arbitrary base $b$, so the number of digits in base $b$ is $$dig_{b}(x)=\lfloor \log_{b}x\rfloor+1$$
The base of a logarithm is the base of the corresponding exponential function. Take $10^x$ for example. The base of this exponential function is $10$. The corresponding logarithm is the standard log, $\log_{10}{x}$, or simply $\log{x}$. The logarithm is the inverse of the exponential function; for example, if $x = 2$, then $10^2 = 100$. The inverse therefore (the $\log{x}$) will get us the exponent, or 2. Therefore, $\log_{10}(100) = 2.$
The base of the logarithm in this case is the $10$ from $10^x$.
More generally, if you had an exponential function $b^x$ then the corresponding logarithm would be $\log_b{x}$ in which the base is $b$.
$\log_{16} 2^{75} = \frac {75}{4} = 18.75$ so there are $18 + 1 = 19$ digits in hex.
And indeed $2^{75} = 8*2^{72} = 8*16^{18}$ so $2^{75}$ in hexadecimal is $8000000000000000000$.
You seem to understand the base of the logorithms perfectly so far as I can tell.
## Decimals
As for the number of digits of a number in the decimal system. consider the following:
Let $x$ be any real number - for our convenience, assume it is positive. Then, obviously, there exists exactly a natural number $n\in\mathbb{N}$ such that: $$10^n\leq x<10^{n+1}$$
Now, if a number is between $10^n$ and $10^{n+1}$, as above, it means that it has exactly than $n+1$ digits. Imagine $x=9468$, for instance. In this case, $n=3$, since: $$10^3=1000\leq9468<10,000=10^4$$ So, we would like to express $n$ in terms of $x$, so as to have a formula that, given $x$, returns $n+1$. Since $n$ is appearing as an exponent of $10$, we think that taking $\log_{10}$ on every side is a way to "get $n$ out of the exponent". So, we have that: \begin{align*} &10^n\leq x<10^{n+1}\\ \Leftrightarrow&\log_{10}10^n\leq\log_{10}x<\log_{10}10^{n+1}\\ \Leftrightarrow&n\leq\log_{10}x<n+1 \end{align*} So, $n=\lfloor\log_{10}x\rfloor$ which means that the number of digits of $x$ is exactly: $$n+1=\lfloor\log_{1}x\rfloor+1$$
## Logarithms
Now, as for the logarithms, let us remind the definition:
Given $x,b>0$, $\log_bx$ is the - only - number that has the following property: $$b^{\log_bx}=x$$ So, the logarithm of $x$ with base $b$ is the only number which can be the exponent of $b$ such that $b$ to that exponent is equal to $x$.
So, given the symbol $\log_bx$, $b$ being the logarithms base means that $b$ is the number to which we may set $\log_bx$ as an exponent to get $x$. You can imagine the following game that helped me understand logarithms:
## The Log-Game
Imagine two friends, Alice and Bob - hehe, these are probably the only friends we, mathematicians, have ever heard of - have a secret code to exchange messages. So, every message they want to send is being turned into a - probably very large - number. So, in our case, Alice and Bob are exchanging numbers in order to communicate.
But, in order to make it more difficult for other people to read their messages and find out the code, they use the following trick:
If $x$ is the message Alice wants to send to Bob, then, instead of this Alice sends Bob another number, which is $\log_b x$, for some base $b$ that they have pre-decided.
So, Bob receives a number $y$ that is not the message he wants, and, probably, has no meaning at all. How will he find the message? As mentioned above, Alice and Bob have pre-decided a base for the logarithms they are exchanging, so, what Bob has to do is to calculate: $$b^y=b^{\log_bx}=x$$ since, by the definition of logarithm, the only way to find $x$ given $y=\log_bx$ is to set it as an exponent to $b$. Every other, not knowing the exact value of $b$ cannot find that message.
## Over-decimal Systems
We have heard about decimal system, binary system etc. But, the most usual case is that we use the usual symbols $0,1,2,\dots,9$ for such systems. What about hexadecimal system? There, we need $16$ different - distinct - symbols for our elementary digits. So, we introduce $6$ new symbols: $A,B,C,D,E,F$. These are just symbols, as the previous ones; $0,1,2,\dots,9$. There is no matter about that.
To make it a little more clear, we have decided that: $$0+1=1,\ 1+1=2,\ 2+1=3,\dots,8+1=9$$ Moreover, we have decided that $$9+1=10,\ 99+1=100$$ Or, to be more precise, the nature of our enumerating system - that the position of a digit is important to its evaluation - is a property that makes it feasible to use only some - finite in number - symbols and not infinitely many or strange combinations - see, for instance, the roman digits.
We have been used to using these symbols to represent numbers, which has driven us to the wrong conclusion that these symbols are the numbers they represent. Well, no symbol is any number, it just represents one, probably different with respect to our system.
For instance, in binary system: $$101$$ represents what in the decimal system we would write as: $$5$$ or what in the ternary (3) system we would write as: $$12$$ or what in the hexadecimal system we would write as: $$5$$
So, if we want to use some enumeration system with more that $10$ elementary digits, we have to introduce some new symbols for these digits and we also have to define the basic operations with them. So, in hexadecimal system we deicide that: $$9+1=A,\ A+1=B,\ B+1=D,\dots,E+1=F$$ and all the other properties we know.
So, to find how many elements a number $x$ has in hexadecimal, we simply note that there exists exactly on $n$ such that: $$16^n\leq x<16^{n+1}$$ and, with the same thoughts as above, we take $\log_16$ to every side, so we have: $$n\leq\log_{16}x<n+1$$ and, we have, finally that number $x$ in the hexadecimal system has: $$\lfloor\log_{16}x\rfloor+1$$ digits. | 2022-01-26T06:58:54 | {
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https://math.stackexchange.com/questions/1268155/prove-that-4x2-8xy5y2-geq0-is-this-a-valid-proof | # Prove that $4x^2-8xy+5y^2\geq0$ - is this a valid proof?
I need to prove that $4x^2-8xy+5y^2\geq0$ holds for every real numbers $x, y$.
First I start with another inequality, i.e. $4x^2-8xy+4y^2\geq0$, which clearly holds as it can be factorized into $(2x-2y)^2\geq0$. Now we can add $y^2$ (which is always nonnegative) to the left side, thus obtaining $4x^2-8xy+5y^2\geq0$, which is always true - we've just added a nonnegative expression to another, so the sum is still greater or equal to zero.
Is my proof correct? I had it on my final math exam and would like to be sure.
• Looks good to me. – simonzack May 5 '15 at 13:23
• +1. An example of converse not being true is the polynomial $$x^6+y^2z^4+y^4z^2-3x^2y^2z^2$$ – Leg May 5 '15 at 14:46
$4x^2-8xy+5y^2 = (2x-2y)^2+y^2 \ge 0$.
Using this sum-of-squares technique you also get that equality occurs exactly when $2x-2y = 0$ and $y = 0$, or equivalently when $(x,y) = (0,0)$. | 2019-09-20T22:36:36 | {
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http://mathhelpforum.com/calculus/18632-integral-cotx.html | 1. ## integral with cotx
Hi,
Could someone please give me a hint on how to do this problem? I know that I have to use a formula to do it, but it looks like I need to simplify it.
Int. cotx/sq. 1+2sinx
I have these formulas:
int. cotudu=ln|secu|+c
int. cot^udu=-1/n-1cot^n-1u-int. cot^n-2udu
Thank you
2. $\displaystyle\int\frac{\cot x}{\sqrt{1+2\sin x}}dx=\int\frac{\cos x}{\sin x\sqrt{1+2\sin x}}dx$
Substitute $\sin x=t\Rightarrow\cos xdx=dt\Rightarrow\int\displaystyle\frac{1}{t\sqrt{ 1+2t}}dt$.
Substitute $\sqrt{1+2t}=u\Rightarrow 2t=u^2-1\Rightarrow dt=udu$
Then $\displaystyle\int\frac{u}{\frac{u^2-1}{2}\cdot u}du=2\int\frac{1}{u^2-1}du=\ln \left|\frac{u-1}{u+1}\right|+C$
Now back substitute.
3. Hello, chocolatelover!
I found a method . . . but it took two substitutions.
I'm certain someone will streamline it for us.
$\int \frac{\cot x}{\sqrt{1+2\sin x}}\,dx$
Let $u \,=\,\sin x\quad\Rightarrow\quad x \,=\,\arcsin u\quad\Rightarrow\quad dx \,=\,\frac{du}{\sqrt{1-u^2}}$
. . and: . $\cot x \,=\,\frac{\sqrt{1-u^2}}{u}$
Substitute: . $\int\frac{\frac{\sqrt{1-u^2}}{u}}{\sqrt{1+2u}}\,\frac{du}{\sqrt{1-u^2}} \;= \;\int\frac{du}{u\sqrt{1+2u}}$
Let $v \,=\,\sqrt{1+2u}\quad\Rightarrow\quad u \,=\,\frac{v^2-1}{2}\quad\Rightarrow\quad du \,=\,v\,dv$
Substitute: . $\int\frac{v\,dv}{\left(\frac{v^2-1}{2}\right)\cdot v} \;=\;2\int\frac{dv}{v^2-1} \;= \;\ln\left|\frac{v-1}{v+1}\right| + C$
Back-substitute: . $\ln\left|\frac{\sqrt{1+2u} - 1}{\sqrt{1+2u} + 1}\right| + C$
Back-substitute: . $\boxed{\ln\left|\frac{\sqrt{1+2\sin x} - 1}{\sqrt{1+2\sin x} + 1}\right| + C}$
Ha! . . . red_dog beat me to it!
4. Mine just take one.
Set
$u=\sqrt{1+2\sin x}\implies du=\frac{\cos x}{\sqrt{1+2\sin x}}\,dx$, the intregral becomes to
$\int\frac{\cot x}{\sqrt{1+2\sin x}}\,dx=2\int\frac1{u^2-1}\,du$
Cheers,
K. | 2016-08-26T18:13:28 | {
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# Some part of a 50% solution of acid was replaced with an
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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
Originally posted by chrissy28 on 12 Aug 2008, 14:27.
Last edited by ENGRTOMBA2018 on 19 Aug 2015, 12:12, edited 2 times in total.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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12 Aug 2008, 15:10
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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16 Jun 2013, 16:55
2
2
If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. $$\frac{1}{2}$$ of 50% solution and $$\frac{1}{2}$$ of 30% solution. If the solution must be equally weighted, then 30% solution must replace $$\frac{1}{2}$$ of 50% solution (assuming original solution was 50% solution of acid).
Just another way of saying the same thing. Cheers
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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16 Jun 2013, 23:07
1
3
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.
Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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17 Jun 2013, 21:57
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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17 Jun 2013, 23:14
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
(100-x)*0.5+0.3x=100*0.4 --> x=50 --> 50/100=1/2.
Hope it's clear.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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18 Jun 2013, 00:03
1
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40% Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
Try using below equation.
50% of (100-x) + 30% of x = 40% of 100
x=50litres.
So 50 of 100 litres got replaced would amount to 1/2 in ratio.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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18 Jun 2013, 00:33
Ahhhh!! I got it....Thank you both of you guys....I hope silly mistakes won't trouble me on the Real Test...
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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10 Jul 2014, 02:09
2
Did using allegation method
Refer diagram below
Bunuel, kindly update the OA.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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20 Jul 2015, 06:11
3
chrissy28 wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
Responding to a pm:
It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.
You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.
The first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 11:41
Can you please explain to me how you got the answer to this problem using the allegation methos?
Thanks
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 11:44
tinnyshenoy wrote:
Can you please explain to me how you got the answer to this problem using the allegation methos?
Thanks
Look at: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html#p1381700
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 12:02
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 12:22
tinnyshenoy wrote:
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
It is from the alligation method
As shown in the post above, per the alligation method, the distrbution will be:
50 10 (=40-30) parts
40
30 10 (=50-40) parts
Thus, you see that you need 10 parts out of 20 (=(10+10)) parts of 50% solution, giving you 10/20 = 1/2 or 50% as your answer.
Hope this helps.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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09 Apr 2016, 00:38
chrissy28 wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
w1 is the weight 50% acid solution which has concentration C1 = 50%
w2 is the weight of 20% acid solution which has concentration C1 = 30%
Concentration of mixture (average) = 40%
using weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)
w1/w2 = (30-40)/(40-50)
w1/w2 = 1/1
w1 is 1 part and w2 is 1 part in a total 2 part of solution.
question is What part of the original solution was replaced?
w2 is what was replaced, hence 1 out of 2 i.e. 1/2 part was replaced.
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http://accessdtv.com/taylor-series/taylor-series-error-estimation.html | Home > Taylor Series > Taylor Series Error Estimation
Taylor Series Error Estimation
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Example What is the minimum number of terms of the series one needs to be sure to be within of the true sum? However, it holds also in the sense of Riemann integral provided the (k+1)th derivative of f is continuous on the closed interval [a,x]. This will present you with another menu in which you can select the specific page you wish to download pdfs for. Sometimes you'll see something like N comma a to say it's an Nth degree approximation centered at a. Source
The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. Solution Here are the first few derivatives and the evaluations. Note that, for each j = 0,1,...,k−1, f ( j ) ( a ) = P ( j ) ( a ) {\displaystyle f^{(j)}(a)=P^{(j)}(a)} . Your cache administrator is webmaster. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation
Taylor Series Error Estimation Calculator
Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. These estimates imply that the complex Taylor series T f ( z ) = ∑ k = 0 ∞ f ( k ) ( c ) k ! ( z − In the mean time you can sometimes get the pages to show larger versions of the equations if you flip your phone into landscape mode. In this example we pretend that we only know the following properties of the exponential function: ( ∗ ) e 0 = 1 , d d x e x = e
This kind of behavior is easily understood in the framework of complex analysis. Solution As with the last example we’ll start off in the same manner. So, we get a similar pattern for this one. Let’s plug the numbers into the MeteaCalcTutorials 55,406 views 4:56 Maclauren and Taylor Series Intuition - Duration: 12:59. Lagrange Error Bound Calculator The N plus oneth derivative of our Nth degree polynomial.
Show Answer If you have found a typo or mistake on a page them please contact me and let me know of the typo/mistake. Hence each of the first k−1 derivatives of the numerator in h k ( x ) {\displaystyle h_{k}(x)} vanishes at x = a {\displaystyle x=a} , and the same is true Where are the answers/solutions to the Assignment Problems?
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Up next Taylor's Inequality - Duration: 10:48. Remainder Estimation Theorem As in previous modules, let be the error between the Taylor polynomial and the true value of the function, i.e., Notice that the error is a function of . Recall that if a series has terms which are positive and decreasing, then But notice that the middle quantity is precisely . How close will the result be to the true answer?
Taylor Polynomial Approximation Calculator
And sometimes you might see a subscript, a big N there to say it's an Nth degree approximation and sometimes you'll see something like this. PaulOctober 27, 2016 Calculus II - Notes Parametric Equations and Polar Coordinates Previous Chapter Next Chapter Vectors Power Series and Functions Previous Section Next Section Applications of Series Taylor Taylor Series Error Estimation Calculator In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.) There are Lagrange Error Formula If all the k-th order partial derivatives of f: Rn → R are continuous at a ∈ Rn, then by Clairaut's theorem, one can change the order of mixed derivatives at
If you are a mobile device (especially a phone) then the equations will appear very small. http://accessdtv.com/taylor-series/taylor-series-error-estimation-formula.html And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x Solution 1 As with the first example we’ll need to get a formula for . However, unlike the first one we’ve got a little more work to do. Let’s first take However, because the value of c is uncertain, in practice the remainder term really provides a worst-case scenario for your approximation. Taylor Series Remainder Calculator
Here is the Taylor Series for this function. Now, let’s work one of the easier examples in this section. The problem for most students is that it may So if you put an a in the polynomial, all of these other terms are going to be zero. Generated Sun, 30 Oct 2016 19:02:48 GMT by s_fl369 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection http://accessdtv.com/taylor-series/taylor-series-error-estimation-problems.html What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b.
This means that for every a∈I there exists some r>0 and a sequence of coefficients ck∈R such that (a − r, a + r) ⊂ I and f ( x ) Taylor's Inequality F of a is equal to P of a, so the error at a is equal to zero. To obtain an upper bound for the remainder on [0,1], we use the property eξSo the error of b is going to be f of b minus the polynomial at b.
From Site Map Page The Site Map Page for the site will contain a link for every pdf that is available for downloading. Note If you actually compute the partial sums using a calculator, you will find that 7 terms actually suffice. Sign in Transcript Statistics 38,950 views 81 Like this video? Lagrange Error Bound Problems And it's going to look like this.
Relationship to analyticity Taylor expansions of real analytic functions Let I ⊂ R be an open interval. Compute F ′ ( t ) = f ′ ( t ) + ( f ″ ( t ) ( x − t ) − f ′ ( t ) ) The second inequality is called a uniform estimate, because it holds uniformly for all x on the interval (a − r,a + r). Check This Out Krista King 14,459 views 12:03 Taylor's Theorem with Remainder - Duration: 9:00.
The N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Download Page - This will take you to a page where you can download a pdf version of the content on the site. Sign in 82 5 Don't like this video? patrickJMT 130,005 views 2:22 Estimating error/remainder of a series - Duration: 12:03.
The goal is to find so that . Notice as well that for the full Taylor Series, the nth degree Taylor polynomial is just the partial sum for the series. Sign in to add this video to a playlist. So this is going to be equal to zero.
Now, what is the N plus onethe derivative of an Nth degree polynomial? Also, since the condition that the function f be k times differentiable at a point requires differentiability up to order k−1 in a neighborhood of said point (this is true, because It has simple poles at z=i and z= −i, and it is analytic elsewhere. Graph of f(x)=ex (blue) with its quadratic approximation P2(x) = 1 + x + x2/2 (red) at a=0.
Modulus is shown by elevation and argument by coloring: cyan=0, blue=π/3, violet=2π/3, red=π, yellow=4π/3, green=5π/3. Combining these estimates for ex we see that | R k ( x ) | ≤ 4 | x | k + 1 ( k + 1 ) ! ≤ 4 Since ex is increasing by (*), we can simply use ex≤1 for x∈[−1,0] to estimate the remainder on the subinterval [−1,0]. Solution Again, here are the derivatives and evaluations. Notice that all the negative signs will cancel out in the evaluation. Also, this formula will work for all n,
Here's the formula for the remainder term: So substituting 1 for x gives you: At this point, you're apparently stuck, because you don't know the value of sin c. Algebra/Trig Review Common Math Errors Complex Number Primer How To Study Math Close the Menu Current Location : Calculus II (Notes) / Series & Sequences / Taylor Series Calculus II [Notes] | 2018-02-22T20:36:48 | {
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https://math.stackexchange.com/questions/3115811/does-there-exist-an-open-subset-a-subset-0-1-such-that-m-a-neq-m-bar | # Does there exist an open subset $A \subset [0,1]$ such that $m_*(A)\neq m_*(\bar{A})$?
Does there exist an open subset $$A \subset [0,1]$$ such that $$m_*(A)\neq m_*(\bar{A})$$?
I was thinking we could approximate any set from inside by a closed set . This need not true from outside.
So I was expecting there should be some counterexample to the above statement.
Any help will be appreciated
• Is $m_*$ outer or inner Lebesgue measure? – Alex Ortiz Feb 17 at 4:05
• Sir it is outer measure. Is answer will change if we put inner measure? As i had only studied outer measure. I was thinking till thar inner measure will work same. – MathLover Feb 17 at 4:08
• It is actually inconsequential since both inner and outer measure are subadditive with respect to countable unions; see my answer below. Cheers! – Alex Ortiz Feb 17 at 4:19
Enumerate the rational numbers in $$[1/4,3/4]$$ as $$\{r_n:n\in\mathbf{Z}_{>0}\}$$, and let $$\epsilon < 1/8$$. For each $$n$$, choose an interval $$I_n$$ centered at $$r_n$$ of width $$\epsilon/2^n$$, and put $$A = \bigcup_{n=1}^\infty I_n$$. Note that $$A$$ is an open subset of $$\mathbf R$$ contained in $$[0,1]$$ because it is a union of open intervals. Since $$A$$ contains the rationals in $$[1/4,3/4]$$, density of $$\mathbf Q$$ in $$\mathbf R$$ implies that the closure $$\overline A$$ contains the closed interval $$[1/4,3/4]$$.
By countable subadditivity (of Lebesgue outer or inner measure, this is an inconsequential point for the matter at hand), $$m_*(A) \le \sum_{n=1}^\infty m_*(I_n) = \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon < 1/8.$$ On the other hand, by monotonicity, $$m_*\big(\overline A\big) \ge m_*\big([1/4,3/4]\big) = 1/2,$$ so $$m_*(A) < 1/8 < 1/2 = m_*\big(\overline A\big).$$
• Why not just let $A$ be an open subset of $(0,1)$ containing $\mathbb Q \cap (0,1)$ such that $m(A)<1/2?$ – zhw. Feb 17 at 4:43 | 2019-10-21T22:29:49 | {
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http://math.stackexchange.com/questions/134979/question-on-modular-arithmetic | # Question on modular arithmetic?
Would this ever be possible:
$$a b^2 ≡ a \pmod p$$
where $p$ is a prime number and $1<a<p$ and $1<b$
Please back your answer with some kind of proof other than Fermat's Little Theorem. This isn't homework.
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Put $a=0$ then true for all $b$. – john w. Apr 21 '12 at 22:03
Sorry, forgot to include $p$ and $b$ have to be greater than 1! – user26649 Apr 21 '12 at 22:06
Any $b \equiv \pm 1 \pmod p.$ – Will Jagy Apr 21 '12 at 22:07
Hint $\$ prime $\rm\: p\ |\ ab^2-a = a\:(b-1)(b+1)\ \Rightarrow\ p\ |\ a\ \ or\ \ p\ |\ b-1\ \ or\ \ p\ |\ b+1$
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@Downvoter: Why? If something is not clear then please feel free to ask questions and I am happy to elaborate. – Bill Dubuque Apr 21 '12 at 23:45
I don't see why the answer was downvoted... Anyways, your answer was perfect, thank you! – user26649 Apr 22 '12 at 0:50
If $a\equiv0\bmod p$, the given relation is true for all $b$.
If $a\not\equiv0\bmod p$ then $a$ can be cancelled from both sides of the congruence which reduces to $b^2\equiv1\bmod p$, hence $b\equiv\pm1\bmod p$.
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would the downvoter care to explain? – Andrea Mori Apr 21 '12 at 22:47
It seems someone is unhappy with both of our answers. I cannot imagine why. But then voting is often irrational here. – Bill Dubuque Apr 21 '12 at 23:48
Since $p$ is prime, we can cancel $a$ from both sides unless $a\equiv 0 \pmod p$ (this doesn't always work if $p$ is a composite number). We get $b^2 \equiv 1 \pmod p$. Again, since $p$ is prime, a number can have only two mod-$p$ square roots (it can have more than two in some cases if $p$ is composite). The two square roots of $1$ are $\pm 1$ and $-1$ is of course the same as $p-1$.
So $ab^2\equiv a\pmod p$ holds if either $a\equiv 0$ or $b\equiv\pm1$, but not otherwise.
So which parts of this do you want proofs of? The proof that nonzero numbers are invertible in mod $p$ (thereby justifying the cancelation from both sides)? I once posted an answer that shows how to find the inverse. The proof that there can't be more than two square roots if $p$ is prime? (That latter fact is a sort of corollary.)
- | 2014-10-20T21:41:03 | {
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http://mathhelpforum.com/differential-geometry/170044-proving-cantor-set-integrable.html | # Math Help - proving the cantor set is integrable
1. ## proving the cantor set is integrable
I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.
Let $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $$f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]
I want to approach it like this but I'm not sure if it is correct:
We want to show that $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$.
We can make $\sum s=0$ when $x\notin C$. So i think all we have to do is to show that $\sum S< \epsilon$. However, I'm not sure how to do this.
2. Originally Posted by zebra2147
I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.
Let $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $$f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]
I want to approach it like this but I'm not sure if it is correct:
We want to show that $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$.
We can make $\sum s=0$ when $x\notin C$. So i think all we have to do is to show that $\sum S< \epsilon$. However, I'm not sure how to do this.
So start with the definition of the Cantor set. There are certain "removed" intervals, what are they?
3. The middle third of each line segment is removed. For example, for [0,1] we would have [0,1/3]U[2/3,1]. So, (1/3,2/3) would be removed.
4. Try proving that $[0,1] - C$ has measure 1, implying that C has measure 0.
5. Ok, here is what I want to say but I'm not sure how to write it correctly of if its even in the right direction...
If we let $s$ be a lower step function of $f$, then we see that $\sum s=0$. Then if we choose some upper step function, $S$, such that $\sum S=\sum A_{i}(x_{i}-x_{i-1})$ where $A_{i}=\text{height of the rectangle}$ then for any interval contained in $[0,1]$, we can find an element, $x_i$ that is contained in the Cantor set and is infinitely close to another element $x_{i-1}$ that is not contained in the Cantor set. Thus, we have $x_{i}-x_{i-1}\approx 0$. Thus, $\sum A_{i}(x_{i}-x_{i-1})\approx 0=\sum S$.
Thus, $\sum s+\sum S\approx 0$ which is less then any $\epsilon >0$. Thus, $\int_{0}^{1}f=0$.
I know that is not exactly the direction that you were leading me to but I think it is similar.
6. I think that it would be better to show that the upper sum and lower sum are equal to each other. Recall that
$\displaystyle U(f)=\inf_{P} U(f,P)$ and $\displaystyle L(f)=\sup_{P} L(f,P)$
where $P$ refers to some partition of $[0,1]$. We say that $f$ is Riemann-integrable if $U(f)=L(f)$.
Now I propose that we consider a sequence of partitions based on the cuts performed in the construction of the Cantor set. Let $P_1=\{0,1/3,2/3,1\}$ (corresponding to the removal of the middle third of the interval). Then, $U(f,P_1)=L(f,P_1)=2/3$. Let $P_2=\{0,1/9,2/9,1/3,2/3,7/9,8/9,1\}$ (again, corresponding to the middle thirds of the remaining intervals). Then, $U(f,P_2)=L(f,P_2)=4/9$. Perform this same procedure to get $P_3, P_4, \ldots, P_n,\ldots$. Now taking $n\to \infty$, we have $U(f,P_n)\to 0$ and $L(f,P_n)\to 0$. This shows that
$U(f)\leq U(f,P_n)\to 0$ and $L(f)\geq L(f,P_n)\to 0$
or in other words,
$U(f)\leq 0\leq L(f)$
But at the same time, from the definition of the upper and lower sums,
$U(f)\geq L(f)$
Therefore $U(f)=L(f)=0$, so this function $f$ is Riemann-integrable with integral value 0. | 2015-04-26T17:45:01 | {
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http://math.stackexchange.com/questions/394075/quotient-rings-of-gaussian-integers-mathbbzi-2-mathbbzi-3 | # Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$,
I am studying for an algebra qualifying exam and came across the following problem.
Let $R$ be the ring of Gaussian Integers. Of the three quotient rings $$R/(2),\quad R/(3),\quad R/(5),$$ one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?
I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.
I hope that this can be answered easily and quickly. Thanks.
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A product of fields is not an integral domain. Consider the zero divisors $(1,0)$ and $(0,1)$. – Jared May 17 '13 at 1:07
That's helpful, thanks. Since R/(2) has just 4 elements, if it were isomorphic to a product of fields, then it would necessarily be isomorphic to Z/(2)xZ/(2). But $i^2=1$ in R/(2), and every element of Z/(2)xZ/(2) when squared equals itself. So... From the unusual phrasing of the question, I suppose R/(5) is isomorphic to a product of fields. Is this right? Would that product be Z/(5)/Z/(5)? – John Adamski May 17 '13 at 1:34
You might find this helpful: Splitting of prime ideals in Galois Extensions (Wikipedia). – kahen May 17 '13 at 1:51
Now is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then
$$\frac{R}{I\cap J}\cong\frac{R}{I}\times\frac{R}{J}.$$
First let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\bf Z}$ iff $\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\cap(m)=({\rm lcm}(m,n))$. As $nm=\gcd(n,m){\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\cap(m)=(nm)$. Thus we have ${\bf Z}/(nm)\cong{\bf Z}/(n)\times{\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\bf Z}/\prod p_i^{e_i}{\bf Z}\cong\prod{\bf Z}/p_i^{e_i}{\bf Z}$.
(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)
Without coprimality, there are counterexamples though. For instance, if $p\in\bf Z$ is prime, then the finite rings ${\bf Z}/p^2{\bf Z}$ and ${\bf F}_p\times{\bf F}_p$ (where ${\bf F}_p:={\bf Z}/p{\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).
Now here's the proof. Define the map $R\to R/I\times R/J$ by $r\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\in I$, $j\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\mapsto(I,1+J)$ and $j\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\times R/J$ as an $R$-module so the image must be the whole codomain.
Now let's work with ${\cal O}={\bf Z}[i]$, the ring of integers of ${\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have
• ${\bf Z}[i]/(3)$ is a field and
• ${\bf Z}[i]/(5)\cong{\bf Z}[i]/(1+2i)\times{\bf Z}[i]/(1-2i)$ is a product of fields.
Go ahead and count the number of elements to see which fields they are. However, ${\bf Z}[i]/(2)={\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\bf F}_2[\varepsilon]/(\varepsilon^2)$, which is not a product of fields since $\epsilon\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.
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Why don't you first think through the case of $\mathbb Z/n\mathbb Z$. When is this a field? When is it a product of fields? When is it neither? Try to relate your answer to the factorization of $n$ and the Chinese Remainder Theorem. Now see if you can generalize it to $\mathbb Z[i]$ (or to any PID).
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I find quotients of polynomial rings easier to reason with algebraically than subfields of $\mathbb{C}$.
Because $\mathbb{Z}[i] \cong \mathbb{Z}[x] / (x^2 + 1)$, we can do arithmetic with rings and ideals. Calculating in more detail than is usually necessary:
\begin{align} \mathbb{Z}[i] / (3) &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / (3) \\&\cong \mathbb{Z}[x] / (3, x^2 + 1) \\&\cong \left( \mathbb{Z}[x] / (3) \right) / (x^2 + 1) \\&\cong \left(\mathbb{Z} / (3)\right)[x] / (x^2 + 1) \\&\cong \mathbb{F}_3[x] / (x^2 + 1) \end{align}
and so we've reduced the problem to one of quotients of polynomial rings over the finite field $\mathbb{F}_3$.
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$\Bbb Z[i]/2\,$ isn't a field or $\color{#c00}{\Pi F}$ (product of fields), since it has a nonzero nilpotent: $\,(1\!+\!i)^2 = 2i = 0$.
$\Bbb Z[i]/5\,$ isn't a field by $\,(2\!-\!i)(2\!+\!i)= 5 = 0,\,$ so it is the $\color{#c00}{\Pi F},\,$ leaving $\,\Bbb Z[i]/3\,$ as the field. $\$ QED
- | 2015-01-27T21:25:04 | {
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https://math.stackexchange.com/questions/3522936/visualizing-the-set-of-points-in-a-regular-polygon-closer-to-center-than-to-vrti | # Visualizing the set of points in a regular polygon closer to center than to vrtices (Voronoi cell)?
Let $$P_n$$ be the regular convex $$n$$-gon centered at $$p_0$$ with $$n$$ vertices $$p_1, p_2, ..., p_n$$ and $$(n>2)$$.
Let $$S_n$$ be the set of all points "$$s$$" within the region bounded by $$P_n$$ where:
$$distance(s,p_0) \leq distance(s,p_k)$$... $$(\forall k=\{1,2,...,n\})$$.
For instance $$S_3$$ would look like this:
And $$S_4$$ would look like this:
And just thinking about this to the extreme... $$\lim_{n\to\infty}S_n$$ should look something like this (call it "$$S_\infty$$" for simplicity):
Is there a way to show (either using geometry or a simple brute-force code in Python) what $$S_n$$ should look like for any $$n>2$$?
I would really like to see what shapes emerge. Is there a way to animate it in Python? Like have circles radiate from each vertex $$p_k$$ and the center $$p_0$$ until they crash into each other to make straight lines defining the new region $$S_n$$?
Depending on the answer, I'd ideally like to transpose this question into 3-D for the 5 platonic solids.
For instance, examining a cube (i.e. 8 vertices), define $$D_8$$ as the set of all points "$$s$$" such that:
$$distance(s,p_0) \leq distance(s,p_k)$$... $$(\forall k=\{1,2,...,8\})$$.
Then $$D_8$$ is actually a truncated octahedron! I'm super curious what the other 4 platonic solids produce... As well as what higher $$S_n$$ regions look like (e.g. for a pentagon, hexagon, octagon, dodecagon, etc.).
Here's an insight that I find to be an elegant solution for n=3,4, or 6 (triangle, square, hexagon). Observe that you can tessellate all of 2-D space with regular triangles, squares, and regular hexagons. Now imagine each vertex $$p_k$$ is actually the center of a nearby triangle/square/hexagon (centered at $$p_k$$ instead of at $$p_0$$).
For instance: $$S_3$$ can be more easily distinguished by drawing 3 more triangles centered at $$p_1, p_2 and p_3$$.
Similarly for $$S_4$$, drawing 4 more squares centered at $$p_1, p_2, p_3 and p_4$$ would show us the shape below, which we can then easily discern needs to be distilled into the $$S_4$$ shading we saw above.
EDIT:
Based on the comment about Voronoi Diagrams, I did more research and tried it out in R. Here's the region $$S_8$$ for example (use your imagination to connect the 8 outer dots to form the enclosing octagon):
The upshot seems to be that actually only $$n=3$$ and $$n=4$$ were interesting. For $$n>4$$ the pattern is simply to create a smaller version of the n-gon rotated by the internal angle of that n-gon. The shrinkage continues forever, but is also slowing forever, with the limit being circle of radius $$\frac{r}{2}$$ shown in the $$S_\infty$$ diagram from above.
More directly:
$$\dfrac{Area(S_3)}{Area(P_3)}=\dfrac{2}{3}$$
$$\dfrac{Area(S_4)}{Area(P_4)}=\dfrac{1}{2}$$
$$\dots = \dots$$
$$\dfrac{Area(S_\infty)}{Area(P_\infty)}=\dfrac{1}{4}$$ (limit -> lower bound)
• You're talking about the Voronoi diagram of the vertices and the centroid.
– user856
Jan 26 '20 at 7:47
• Geometrically, that inequality represents the half-space containing $p_0$ defined by the perpendicular bisecting plane of the line segment connecting $p_0$ and $p_k$. Take the intersection of all these half-spaces with the original polyhedron. Jan 26 '20 at 22:24
• @mr_e_man As said in a compact way by Rahul, it is the Voronoi cell associated with the center $p_0$, in the Voronoi diagram generated by the $p_k$s. Jan 29 '20 at 19:08
You can obtain it easily with Wolfram Mathematica (see the pictures below).
P[n_] := Table[{Cos[2*Pi*i/n], Sin[2*Pi*i/n]}, {i, 0, n}];
graph[n_] := RegionPlot[AllTrue[Table[(x - P[n][[i, 1]])^2 + (y - P[n][[i, 2]])^2, {i, 1,n}], # > x^2 + y^2 &], {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20];
The 3D version can be built in the same way. For instance, for the tetrahedron:
P = {{1, 1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, -1, 1}};
p2 = ConvexHullMesh[P, MeshCellStyle -> {1 -> {Thick, Black}, 2 -> Opacity[0.2]}];
p1 = RegionPlot3D[ RegionMember[p2, {x, y, z}] && AllTrue[ Table[ Norm[{x, y, z} - P[[i]]], {i, 1, Length[P]}], # > x^2 + y^2 + z^2 &] , {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> 60];
Show[p2, p1, ViewPoint -> {2, 4, 4}]
• It's neat that for the cube some of the faces are perfect hexagons.
– user856
Jan 29 '20 at 10:57
• @PierreCarre can you please provide the sample code for the cube? I'm not too familiar with these functions in mathematica unfortunately Jan 31 '20 at 2:45
• @Andrew I just added some code for the 3D case. Jan 31 '20 at 10:17 | 2021-10-16T13:58:09 | {
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https://math.stackexchange.com/questions/1797516/biased-coin-with-a-3-4-chance-to-land-on-the-side-it-was-before-the-flip | # Biased coin with a $3/4$ chance to land on the side it was before the flip
Consider a hypothetical coin (with two sides: heads and tails) that has a $3/4$ probability of landing on the side it was before the flip (meaning, if I flip it starting heads-up, then it will have an only $1/4$ probability of landing tails-up). If it begins on heads, what is the probability that it is on tails after 10 flips? What about 100 flips? Assume that each flip starts on the same side as it landed on the previous flip.
Note: this is not a homework problem, just something I thought up myself.
• I was with you all the way up to the last sentence (Assume that each flip...) - what does that mean? The rest makes perfect sense. – mathguy May 24 '16 at 3:27
• Coin flips have been shown to have a bias according to the side they start on actually. Don't remember the researcher name though, pretty famous person in California iirc, look it up. I'm guessing that the op wasn't aware of that though. – jdods May 24 '16 at 3:30
• This could probably be solved by setting up a simple 2 state Markov chain. I'd write it up but an away for the evening now. – jdods May 24 '16 at 3:33
• "this is not a homework problem, just something I thought up myself" -- then well done, because as it happens this is a fairly common kind of exercise! You've found the "right" problem, the one that (with others like it) provokes the theory of Markov chains. – Steve Jessop May 24 '16 at 10:18
• @mathguy It seems to me he's saying the side facing up doesn't change from the time the coin lands to the time that it's flipped again. Which sounds redundant, but it's better than being unclear. – Turambar May 24 '16 at 12:46
Let $p_n$ be the probability the coin is on tails after $n$ flips. Note that $p_0=0$.
The coin can be on tails after $n+1$ flips in two different ways: (i) it was on tails after $n$ flips, and the next result was a tail or (ii) it was on heads after $n$ flips, and the next result was a tail.
The probability of (i) is $(3/4)p_n$ and the probability of (ii) is $(1/4)(1-p_n)$. Thus $$p_{n+1}=\frac{1}{4}+\frac{1}{2}p_n.\tag{1}$$ Solve this recurrence relation. The general solution of the homogeneous recurrence $p_{n+1}=\frac{1}{2}p_n$ is $A\cdot \frac{1}{2^n}$. A particular solution of the recurrence (1) is $\frac{1}{2}$. So the general solution of the recurrence (1) is $$p_{n}=A\cdot \frac{1}{2^n}+\frac{1}{2}.$$ Set $p_0=0$ to find $A$. We find that $p_n=\frac{1}{2}-\frac{1}{2^{n+1}}$.
If you mean that $\Pr(H_i|H_{i-1}) = \Pr(T_i|T_{i-1}) = 3/4$ for all $i$, then this is a two-state homogeneous Markov chain. The transition matrix is
$$P = \begin{pmatrix} 3/4 & 1/4 \\ 1/4 & 3/4 \end{pmatrix}.$$
This should get you started....
I'd comment but I don't have the rep, strangely you require 50.
Also, with regards to the bias in a coin flip, it's a paper by Diaconis:
Let $p_n$ be the probability that the coin is heads up after $n$ tosses. Then $p_0=1$ since it starts on heads, and $$p_n=\frac34p_{n-1}+\frac14(1-p_{n-1})$$ which simplifies to $$\left(p_n-\frac12\right)=\frac12\left(p_{n-1}-\frac12\right)\ .$$ Iterating, $$p_n-\frac12=\Bigl(\frac12\Bigr)^{n+1}$$ so $$p_n=\frac12+\Bigl(\frac12\Bigr)^{n+1}\ .$$
• This no longer fails for $p_1 = \frac 3 4$. – Axoren May 24 '16 at 3:36
Another way of looking at it is to imagine that you toss two fair coins each time, using the second coin to choose between the first coin and the previous (originally heads). If the second coin always chooses the previous result then you end up with heads, while if at any point the second coin chooses the first coin then as this is fair the end result will have equal probability of being heads or tails. The chances of the latter being $1-\frac1{2^n}$, the chance of the final result being tails is therefore half of that.
• This is the best possible solution for p=1/4 and I think it generalizes to all other values of p. – zyx May 24 '16 at 17:32
• At least for p<1/2, where coin 2 has probability (1-2p) of choosing the previous result and coin 1 is fair. One gets half of $(1 - (1-2p)^n)$ which is correct. Other than changing the probability for coin 2, every word of the answer stays the same and the argument works. This answer is just brilliant. Well done. – zyx May 24 '16 at 17:49
$p_n(\text{heads}) = \frac 3 4p_{n-1}(\text{heads}) + \frac 1 4 (1 - p_{n-1}(\text{heads}))$
$p_n(\text{heads}) = \frac 1 4 + \frac 1 2p_{n-1}(\text{heads})$
$p_1(\text{heads}) = \frac 3 4$
Everyone else beat me too it. Was still writing the recurrence relation. :\
Any solution with the words "heads" and "tails" is doing extra work.
We want the chance of an odd number of reversals, each with probability $p=1/4$, to happen in $n$ independent trials. The coin briefly remembers its current state but not whether that state was reached by reversing on the previous trial. This makes the reversals independent of each other.
From the binomial theorem that probability is $$\frac{((1-p)+p)^n - ((1-p) - p)^n)}{2} = \frac{1}{2} - \frac{(1 - 2p)^n}{2}$$ which is consistent with the other solutions and shows that the $(1/2)^{n}$ exponential decay of the "memory" is really a power of $(1-2p)$. | 2020-01-26T08:50:19 | {
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https://math.stackexchange.com/questions/3776738/if-15-distinct-integers-are-chosen-from-the-set-1-2-dots-45-some-t | # If $15$ distinct integers are chosen from the set $\{1, 2, \dots, 45 \}$, some two of them differ by $1, 3$ or $4$.
$$\blacksquare~$$ Problem: If $$15$$ distinct integers are chosen from the set $$\{1, 2, \dots, 45 \}$$, some two of them differ by $$1, 3$$ or $$4$$.
$$\blacksquare~$$ My Approach:
Let the minimum element chosen be $$n$$. Then $$n + 1 , n + 3 , n + 4$$ can't be taken.
We make a small claim.
$$\bullet~$$ Claim: In a set of $$~7$$ consecutive numbers at most $$2$$ numbers can be chosen.
$$\bullet~$$ $$\textbf{Proof:}$$ Let us name the elements of the set as $$\{ 1,2,3,\dots,7 \}$$.
Now let's consider the least element is chosen. If the least element is $$1$$, then $$2,4,5$$ can't be chosen.
So we are left with $$3, 6, 7$$.
$$\circ~$$ If $$~3~$$ is chosen, then $$6, 7$$ can't be in the set. And if $$~3~$$ is not chosen, then only any one of the 2 elements $$\{ 6, 7 \}$$ be chosen. So, a maximum of $$2$$ elements can be chosen in this case.
$$\circ \circ~$$ If the least element is $$2,$$ then $$3, 5, 6$$ can't be there in the set. So, possible elements are 4, 7. So, one of these two can be chosen. Then, a maximum of 2 elements can be chosen in this case.
$$\circ \circ~$$ If the least element is $$3$$, then $$4,6,7$$ gets cancelled. so only $$5$$ is left in the set i.e., $$2$$ elements at most.
$$\circ \circ~$$ If the least element is $$4,$$ then $$5,7$$ gets cancelled. So the only element left is $$6$$.
Similarly,
$$\circ \circ~$$ If $$5$$ is the least element then $$6$$ gets cancelled and only $$7$$ is left. i.e., two elements. If the least element is either $$6$$ or $$7$$, then there is only one element.
So a maximum of two elements in a set of $$7$$ consecutive elements can be chosen.
Hence, the proof of the claim is done!
So, for $$42$$ elements, a maximum of $$2 \times 6 = 12$$ can be taken. However, $$3$$ more elements are required from $$3$$ more consecutive elements, which is not possible since only 2 elements at most can be chosen from a set of 3 consecutive elements.
So, a $$14$$ element subset can be formed such that, no two of them differ by $$1, 3, 4$$. Hence the $$15$$th element is one of the cancelled elements, that is, there exists a pair with their difference being $$1, 3$$ or $$4.$$
Hence, done!
Please check the solution for glitches and give new ideas too :).
• What is the question? Well done. – Ross Millikan Aug 1 at 15:04
• @RossMillikan I don't understand what you mean. – Ralph Clausen Aug 1 at 15:07
• I missed the last line. I think it is a fine proof. – Ross Millikan Aug 1 at 15:07
• Oh! okay @RossMillikan – Ralph Clausen Aug 1 at 15:10
• @RossMillikan The OP added the last line after your comment – user1001001 Aug 1 at 15:17
It is a nice argument, and you have explained it very clearly, so well done.
If you are looking for improvements, and you want it to be a bit more formal, I would make two suggestions:
1. You are very thorough proving the claim, going through all the cases, which is great. However you could shorten the proof of the claim by saying:
"If the least element is $$x$$, then we may subtract $$x-1$$ from all the numbers without changing their differences or the number of integers. Thus without loss of generality we may assume the least element is $$1$$."
Then you do not need to consider the other cases.
1. The last part of the proof (after the claim is proved) is not as thorough as the proof of the claim. You assume that for $$k=0,\cdots,5$$ the numbers $$7k+1,7k+3$$ are selected, without justifying that this is optimal. It is obvious in a way, but for a formal proof you should justify this by saying something like:
"Pick the smallest $$k$$ where $$7k+1, 7k+3$$ are not picked. Then replace the numbers in $$\{7k+1,\cdots 7k+7\}$$ that are picked with $$7k+1, 7k+3$$. From the claim we know that we have not reduced the number of integers. Also, we have not created any new differences of $$1,3,4$$."
Then finish the argument with your second to last paragraph. I would reword the first sentence slightly: "So of the first $$42$$ elements, we may assume these $$12$$ are picked."
I would lose the last paragraph as it is not needed.
Nice proof!
In the optimal case, go up in $$2$$s wherever possible. You would start with $$1$$ and pick the smallest possible number, that is, $$3$$. We can't add $$2$$ again as that would be $$1+4$$, therefore we add the next smallest possible, $$5$$ to get our next number $$8$$. We can then add $$2$$ again, and the pattern continues:
$$1,3,8,10,15,17,22,24,29,31,36,38,43,45$$
Via this strategy we only pick $$14$$ numbers, so $$15$$ is impossible without violating the $$1,3,4$$ difference.
• While this strategy is good in some cases, I think it's a bad general approach to proofs. I call it the "what's the worst that could happen?" model --- you work that out, show it's impossible, and say you're done. That relies on two things: (1) that the second worst doesn't turn out to be viable even though the worst is not, and (2) that you've correctly identified the worst case. My experience is that students (and even professionals) often manage to mess up with both failure modes. – John Hughes Aug 1 at 15:48
• You claim that you would pick the smallest possible number in order to generate an optimal sequence. But this is something that needs to be proven. – TonyK Aug 1 at 16:32
Here's a shorter version which however relies on the same ideas as your and Rhys Hughes' approach.
Let's assume we can choose $$15$$ integers whose mutual differences are never $$0,1,3$$ or $$4$$, and call them $$x_1 in ascending order.
Claim: For all $$1 \le i \le 13$$, we have $$x_i +7 \le x_{i+2}$$.
Proof of claim: Let's make a case distinction about what $$x_{i+1}-x_i$$ is. Since it cannot be $$0,1,3$$ or $$4$$, the only cases are:
• First case, $$x_{i+1} - x_{i} =2$$. Then if we also had $$x_{i+2} - x_{i+1} = 2$$ we would have $$x_{i+2}-x_i =4$$ which was excluded. So $$x_{i+2}-x_{i+1}$$, since it cannot be $$0,1,2,3,4$$ must be $$\ge 5$$ which implies the claim.
• Second case, $$x_{i+1}-x_i \ge 5$$. Then since $$x_{i+2}-x_{i+1} \neq 0,1$$ it must be $$\ge 2$$ and the claim follows.
The claim is proven.
Now it follows iteratively that $$x_3 \ge x_1 +7$$, $$x_5 \ge x_3+7 \ge x_1+14$$, ..., $$x_{15} \ge x_1+49$$ which of course contradicts $$1 \le x_1 \le x_{15} \le 45$$. Actually this shows that you cannot even select $$15$$ such integers from the set $$\{1, ..., 49\}$$. | 2020-08-07T18:14:31 | {
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https://math.stackexchange.com/questions/388009/numerical-approximation-of-the-continuous-fourier-transform | Numerical Approximation of the Continuous Fourier Transform
Given a function $F(k)$ in frequency space (sufficiently nice enough, eg. a Gaussian), I would like to compute its Fourier inverse $$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}F(k)dk$$ numerically (there is no explicit analytical formula), at some specified evenly distributed points $x_n$.
Assume $F(k)$ is symmetric about $k=0$ and "essentially zero" outside the interval $-a/2\leq k \leq a/2$.
Naively, one way I would proceed is by dividing the interval $(-a/2,a/2)$ into $N$ subintervals (choose $N$ to be a power of two) and then piecewise approximate the integral: $$f(x_n) \approx \frac{1}{2\pi}\frac{a}{N}\sum_{m=0}^{N-1} e^{ik_mx_n}F(k_m)$$ where, for example, we might take $k_m = (m-N/2)\frac{a}{N}$for $0\leq m\leq N-1$.
For efficiency I would like to use the Fast Fourier Transform (FFT) to approximate $f(x_n)$. So I need to put $f(x)$ into the correct format for the Discrete Fourier Transform (DFT) $$L_k = \frac{1}{N}\sum_{m=0}^{N-1}G_me^{(2\pi i) km/N},\,\,\,\,k = 0,\ldots,N-1$$ where $G_m$ is the $m$th sampling of some given function $G$.
To proceed, we divide the interval $(-a/2,a/2)$ into $N$ pieces ($N$ a power of two) and let $k_m = (m-N/2)\frac{a}{N}$for $0\leq m\leq N-1$, as before. Then $$f(x) \approx \frac{1}{2\pi}\frac{a}{N}\sum_{m=0}^{N-1} e^{ik_mx}F(k_m)$$ However, in order for this formula to be in the correct format to use the DFT, we cannot evaluate f at any $x$: I believe they must be of the form $x_k = \frac{2\pi}{a}(k-N/2), 0\leq k \leq N-1$.
So in order to use the FFT, the points at which I can "evaluate" the function $f$ is determined by how I sample my function $F(k)$ and the number of sample points.
But in the naive method, I can, in principle, find $f(x)$ for any given x, regardless of how many sample points I take of the function $F(k)$.
My questions are:
1. Is it even possible to use the FFT, if I need to compute f at specified points $x_n$?
2. If not, are there other numerical methods that would be more suitable?
• I came across your very same problem a week ago and I was quite astonished in realizing that there are no routines to do the job in known (at least those that I know) computer algebra programs (e.g. Mathematica/Matlab). The best thing I found is numerical recipes. There is a whole section on Fourier integration. Basically I just copied the routines suggested there and now I'm an happy person. I think that's the best what the market offers at least according to my experience. – lcv May 11 '13 at 0:48
• Bythe way, the routine in numerical recipes implements fgp's trick and more. – lcv May 11 '13 at 22:27
• @lcv: can you share the link to the routine you found? I also came across this problem and I'm still shocked that this isn't some standard Matlab routine. It took some time to realize that the maximum range of the x_k numbers is increasing with larger N. Increasing N will therefore make sure that the x_k's cover the entire x-region of interest but will not increase the density inside the x-region. – Frank Meulenaar Dec 5 '13 at 8:02
• Of course, one can increase the densitiy of the x_k's is by increasing a. – Frank Meulenaar Dec 5 '13 at 8:29
• @FrankMeulenaar It's numerical recipes chapter 13.9 "Computing Fourier integrals using FFT" . In the 'C' edition is on page 584. You can find online and downloadable versions of NR. The routine you can pretty much copy-paste it from the pdf. You'll have to work a bit to remove the comments. Otherwise the routines are available online for purchase on the NR site. Good luck! – lcv Dec 6 '13 at 20:30
The following two articles may be of use. They don't let you do the transform at any old x values. But it does allow transformation from m equally spaced points in the w domain to m equally spaced points in the x domain. The Bailey and Swarztrauber article also uses a gaussian function like yours as an example. Sorry I can't be more specific but I'm learning all this stuff myself.
Bailey, D., and P. Swarztrauber. 1994. ‘A Fast Method for the Numerical Evaluation of Continuous Fourier and Laplace Transforms’. SIAM Journal on Scientific Computing 15 (5): 1105–10. doi:10.1137/0915067.
Inverarity, G. 2002. ‘Fast Computation of Multidimensional Fourier Integrals’. SIAM Journal on Scientific Computing 24 (2): 645–51. doi:10.1137/S106482750138647X.
$$\def \o {\omega}$$ (Here I use $$t$$ and $$\o$$ instead $$x$$ and $$k$$, but is equivalent) We want to calculate numerically the inverse Fourier transform of a given function $$F(\o)$$ . By definition
$$f(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} F(\o) e^{-i\o t} d\o$$
We can split it into two integrals
$$f(t)=\frac{1}{2\pi} \left( \int_{0}^{\infty} F(\o) e^{-i\o t} d\o + \int_{0}^{\infty} F(-\o) e^{i\o t} d\o \right)$$
If the original function was real then $$F(-\o)=F(\o)^{*}$$, and therefore
$$f(t)=\frac{1}{\pi} \textbf{real} \left( \int_{0}^{\infty} F(\o) e^{-i\o t} d\o \right)$$
Then we use the Riemann integral definition for $$0$$ to a very great frequency $$W$$. $$I=\int_{0}^{W} F(\o) e^{-i\o t} d\o\approx \sum_{k=0}^{N-1} F(w_k) e^{-i \Delta \o k \Delta t n} \Delta\o$$
where $$\Delta w =W/N$$ so that $$w_k= \Delta w \ \ k$$ with $$k=0,1,2,..., N-1$$ . We discretized the time by $$t_n= \Delta t \ \ n$$ with $$n=0,1,2,...,N-1$$. Next we make the basic assumption that
$$W \Delta t = 2 \pi$$ so we end up with
$$I = \Delta\o \sum_{k=0}^{N-1} F(w_k) e^{- 2 \pi i \frac{n k}{N} }$$
which is the basic definition of the Fast Fourier Transform. The problem was when I used it for a square function $$F(\o)=\frac{1}{i\o}\left( e^{i\o T_0} -1 \right)$$ This method gives an offset value. Then I modified the integral as
$$I= \Delta\o \left( \sum_{k=1}^{N-1} F(w_k) e^{- 2 \pi i \frac{n k}{N} } + \frac{1}{2}(F(w_0)+F(w_N)) \right)$$
Which is the trapezoidal rule instead of the Riemman integral, and gave a better result.
Best regards.
E. Gutierrez-Reyes
• It should say -infinity to infinity, it was a mistake. – Edahi Aug 26 '19 at 19:24
One way that comes to mind would be to compute both approximations for $f(x_n)$ and for $f'(x_n)$ one some grid $x_n$ that allows you to use FFT. To compute the approximations for $f'(x_n)$, you can use the identity $\mathcal{F}(f')(k) = ik\mathcal{F}(f)(k)$ (i.e., you get the fourier transform of the derivative by multiplying with $ik$).
To evaluate $f(x)$ at $y$, you'd then find the closest $x_n,x_{n+1}$ with $x_n \leq y < x_{n+1}$ and use polynomial interpolation (with polynomials of degree 3) to find $f(y)$ from $f(x_n)$, $f(x_{n+1})$, $f'(x_n)$, $f'(x_{n+1})$.
As long as F(k) is limited and zero outside of $-2\pi A$ to $2\pi A$ the Shannon interpolation can be used $$f(x)=\sum_{n=-\infty}^{n=\infty} f(x_n) sinc\left(\frac{\pi}{T}(x-nT)\right)$$ where $T$ is the sampling period used to determine $x_n$ and $sinc(t)=sin(t)/t$
This formula is exact as long as function spectrum ($F(k)$) is limited. Implementation of fast sinc interpolation exist | 2020-12-03T07:16:06 | {
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https://math.stackexchange.com/questions/2960443/construct-a-continuous-real-valued-function-which-takes-zero-on-integers-and-suc/2960496 | # Construct a continuous real valued function which takes zero on integers and such that image of function is not closed.
I am trying to construct a continuous real valued function $$f:\mathbb{R}\to \mathbb{R}$$ which takes zero on all integer points(that is $$f(k)=0$$ for all $$k\in \mathbb{Z}$$) and Image(f) is not closed in $$\mathbb{R}$$
I had $$f(x)=\sin(\pi x)$$ in mind. But image of $$f(x)$$ is closed.
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
The image of $$\sin(\pi x)$$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $$\tanh(x)$$. Thus, we can use the function $$f(x) = \sin(\pi x)\tanh(x),$$ which is obviously continuous, zero at each integer, and can be easily shown to have image $$(-1,1)$$
• And if you are not familiar with $\tanh$, you can write a similar function explicitly, for example: $$f(x)=\sin(\pi x)\frac{1}{1+e^x}$$ A rational function (as the modifier factor) may also work, like: $$f(x)=\sin(\pi x)\frac{x^2+1}{x^2+2}$$ Oct 18, 2018 at 9:15
• "To make it open, we need the peaks to get arbitrarily close to some value, but never reach them" You seem to be confusing the notion of not-closed and open Oct 18, 2018 at 12:11
• @AleksJ Except for $\emptyset$ and $\mathbb R$, every open set is not-closed. Oct 18, 2018 at 15:58
• @eyeballfrog but not vice versa Oct 18, 2018 at 17:03
• I believe that the point @JohnDvorak is making is that the peaks getting arbitrarily close to some value but not reaching it is necessary, but not sufficient, for the image to be open, and that trying to get an open set is not a necessary condition for it being non-closed (although it is sufficient). Oct 18, 2018 at 17:31
The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $$1$$ in the $$\pm$$infinite limit: $$f(x) = \frac{1+|x|}{2+|x|}\cdot\sin(\pi\cdot x)$$ Plotted together with the asymptotes:
Incidentally, since you just said “not closed” but not whether it should be bounded, we could also just choose
$$f\!\!\!\!/(\!\!\!\!/x\!\!\!\!/)\!\!\!\!/ =\!\!\!\!/ x\!\!\!\!/\cdot\!\!\!\!/\sin\!\!\!\!/\!\!\!\!\!\!\!\!/(\pi\!\!\!\!/\cdot\!\!\!\!/ x\!\!\!\!/)$$
(The image of this is all of $$\mathbb{R}$$ which is actually closed, as the commenters remarked.)
A more interesting example that just occured to me:
$$f(x) = \sin x \cdot\sin(\pi\cdot x)$$ Why does this work? Well, this function never reaches $$1$$ or $$-1$$, because for that to happen you would simultaneously need $$x$$ and $$\pi\cdot x$$ to be an odd-integer multiple of $$\tfrac\pi2$$. But that can never coincide because $$\pi$$ is irrational! It does however get arbitrarily close to $$\pm1$$, in fact it gets close to $$-1$$ quite quickly due to $$\tfrac\pi2 \approx 1.5 = \tfrac32$$. But it never actually reaches either boundary.
• Can you please tell me how make these graphs? Oct 18, 2018 at 8:59
• @StammeringMathematician plotWindow [fnPlot $\x -> (1+abs x)/(2+abs x)*sin(pi*x)] with dynamic-plot (a Haskell library). Oct 18, 2018 at 9:04 • @leftaroundabout Thanks. These graphs are beautiful. Oct 18, 2018 at 9:07 • The proof that the image of$\sin(x)\sin(\pi x)$is$(-1,1)\$ seems nontrivial. I have no doubt it's true, but it's not clear to me how to prove it. Oct 18, 2018 at 16:55
• +1 for not deleting the wrong answer and preventing me to make the same mistake! Also, amazing visual explanation Oct 19, 2018 at 9:50
Using piecewise linear functions (instead of $$\sin (\pi x)$$) makes this simpler. For each $$n \neq 0$$ draw the triangle with vertices $$(n,0),(n+1,0)$$ and $$(n+\frac 1 2, 1-\frac 1 {|n|})$$. You will immediately see how to construct an example. [You will get a function (piecewise linear function) whose range contains $$1-\frac 1 {|n|}$$ for each $$n$$ but does not contain $$1$$].
If $$f$$ verifies the desired propery, its restriction $$f|_{[n,n+1]}$$ gives a continuous function on $$[n,n+1]$$ that is zero on the edges of the interval, for any $$n \in \mathbb{Z}$$. Reciprocally, if we have $$f_n : [n,n+1] \to \mathbb{R}$$ continuous with $$f_n(n) = f_n(n+1) = 0$$ for each integer $$n$$, by the gluing lemma this gives a continuous function $$f: \mathbb{R} \to \mathbb{R}$$ with $$f(n) = f_n(n) = 0$$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $$[n,n+1]$$. Now, the function
$$f_n (t) = \mu_n\sin(\pi(t-n))$$
takes values on $$\mu_n[-1,1] = [-\mu_n,\mu_n]$$ and $$f_n(n) = f_n(n+1) = 0$$. Thus, the family $$(f_n)_n$$ induces a continuous function $$f$$ that vanishes at $$\mathbb{Z}$$ and
$$f(\mathbb{R}) = \bigcup_{n \in \mathbb{Z}} f_n([n,n+1]) = \bigcup_{n \in \mathbb{Z}}[-\mu_n,\mu_n]$$
so the problem reduces to choosing a sequence $$(\mu_n)_n$$ so that the former union is open. One possible choice is $$\mu_n = 1-\frac{1}{|n|}$$ so that
$$f(\mathbb{R}) = \bigcup_{n\in \mathbb{Z}}[-\mu_n,\mu_n] = \bigcup_{n\in \mathbb{N}}[-1+\frac{1}{n},1-\frac{1}{n}] = (-1,1).$$
Consider
$$f(x) = \sin^2 (\pi x)\frac{x^2}{1+x^2}.$$
Then $$f$$ is continuous, $$f=0$$ on the integers, but $$f(\mathbb R) = [0,1).$$ | 2022-05-25T23:50:28 | {
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http://mathhelpforum.com/pre-calculus/97384-need-help-w-pre-calculus-summer-packet.html | # Thread: Need help w/ pre-calculus summer packet.
1. ## Need help w/ pre-calculus summer packet.
I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!
Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.
Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.
Factor completely.
3. x^2 + 12x + 36 - 9y^2
Stumped. I know I can factor out an x but not sure how that would help.
I appreciate any help. Look forward to hearing from people!
2. Originally Posted by amy1613
I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!
Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.
you should already know what y = x^2 + x - 12 looks like ... well, y = |x^2 + x - 12| is the same graph except that the part that is below the x-axis (negative) is reflected to the positive side.
y = x^2 + x - 12 , x < -4 or x > 3
y = -(x^2 + x - 12) , -4 < x < 3
Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.
x^2 - 16 > 0 ... note that x^2 - 16 = 0 at x = + 4 and x = -4
plot these two numbers on a number line ... it breaks the number line into three sections ...
x < -4 , -4 < x < 4 , and x > 4
take any number in each interval, and "test" it in the original inequality ... if it makes the inequality true, then all values of x in that interval make the inequality true.
you'll see that you forgot the interval (-∞ , -4)
Factor completely.
3. x^2 + 12x + 36 - 9y^2
help any?
...
3. yeah this helped a lot! thanks! so...a sign chart is just a number line graph?
4. Hello, amy1613!
Write the following absolute value expression as a piecewise expression:
. . $1)\;\;f(x) \;=\;|x^2 + x - 12|$
There are two cases to consider: . $\begin{array}{cccc}(1) & x^2 + x - 12 \:\geq \;0 \\ (2) & x^2 + x - 12 \:<\:0 \end{array}$
$\text{Case }(1)\;\;x^2 + x - 12 \:\geq\:0\quad\Rightarrow\quad (x-3)(x+4) \:\geq \:0$
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$
. . Hence, if $x \geq 3\text{ or }x < -4$, then: . $f(x) \:=\:x^2+x-12$
$\text{Case }(2)\;\; x^2 + x - 12 \:<\: 0 \quad\Rightarrow\quad (x-3)(x+4) \:<\:0$
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:>\:0 & \Rightarrow & x \:>\:3 \\ x+4 \:< \:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x > 3 \text{ and }x < \text{-}4$ . . . impossible
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\:0 & \Rightarrow & x \:<\:3 \\ x+4 \:> \:0 & \Rightarrow & x \:>\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad \text{-}4 \:<\:x \:<\: 3$
. . Hence, if $\text{-}4 < x < 3$, then: . $f(x) \:=\:-(x^2+x-12) \:=\:12 - x - x^2$
Therefore: . $f(x) \;=\;\begin{Bmatrix}x^2+x-12 && \text{if }x\leq \text{-}4 \text{ or }x \geq 3 \\ \\[-4mm] 12 -x-x^2 && \text{if }\text{-}4 < x < 3 \end{Bmatrix}$
5. Originally Posted by Soroban
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$
. . Hence, if $x \geq 3\text{ or }x < -4$, then: . $f(x) \:=\:x^2+x-12$
[/size]
Thanks for your response & explaination. I still can't quite understand why you picked x>-3 and not x>4. Could you go through that for me? | 2016-08-28T05:11:33 | {
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https://math.stackexchange.com/questions/2482032/number-of-solutions-to-a-10b-20c-100-with-a-b-c-non-negative-integers/2482082 | # Number of solutions to $a + 10b + 20c = 100$ with $a,b,c$ non-negative integers
How many solutions are there to the equation $$a + 10b + 20c = 100$$ where $a,b,c$ are non-negative integers
The problem I'm asking is another way of phrasing "How many ways can you break $100$ dollars into $1$, $10$, and $20$ dollar bills", or any multitude of variants.
Solutions can, for example, be $(a,b,c) = (100, 0, 0)$, $(10, 7, 1)$, $(0, 6, 2)$, etc.
I wrote a quick program which gave me a total of $36$ solutions, but I'd like to have a better understanding as to where this answer comes from without something resembling brute-force. Preferably, I'd like to see how to use generating functions to solve this, as that was the method I first tried (and failed) with.
The solution should be the coefficient of the 100th-power term in the product:
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ I can get this coefficient from the following: $$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}}(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ For whatever reason, applying product rule fails (or I fail to use it correctly), as I get $3$. So, assuming $|x| < 1$, this equals: $$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}} \frac{1}{(1-x)(1-x^{10})(1-x^{20})}$$ However, to much dismay, this is not such a simple thing to find derivatives of. Any advice?
• I got $36$ as coefficient of $x^{100}$ so you were right... $$\ldots + 30 x^{98}+30 x^{99}+36 x^{100}+35 x^{101}+35 x^{102}+35 x^{103}+\ldots$$ – Raffaele Oct 20 '17 at 21:10
• @Raffaele : You should get $36$ for every coefficient from $x^{100}$ to $x^{109}$. – Eric Towers Oct 20 '17 at 21:18
• Calculating the 100th derivative is definitely harder than writing the solutions one by one. An idea is to use partial fraction decomposition on $\frac{1}{(1-x)(1-x^{10})(1-x^{20})}$ and then sum the coefficients in front of $x^{100}$. But this can be rather tedious and hard to do too. – Stefan4024 Oct 20 '17 at 21:20
• The first time I used polynomial with a degree too low. I remade my calculations with higher degree polynomials (using Mathematica) and finally got $$\ldots+42 x^{111}+42 x^{110}+36 x^{109}+36 x^{108}+36 x^{107}+36 x^{106}+36 x^{105}+36 x^{104}+36 x^{103}+36 x^{102}+36 x^{101}+36 x^{100}+30 x^{99}+\ldots$$ BTW, I did not know this combinatorial trick :) – Raffaele Oct 20 '17 at 21:31
You can simplify fairly drastically, since $a = 100-10b-20c$ means $a$ must be a multiple of $10$. So setting $a' := a/10$, we have $a' + b + 2c = 10$.
Again $a'$ is determined by $b$ and $c$, so we just need $b+2c\le 10$. For $c=\{0,1,2,3,4,5\}$ we have $\{11,9,7,5,3,1\}$ options for $b$, for a total of $36$ possible combinations.
• Nifty! Thanks for that argument. I was able to reduce it as well after reading up on the subject, but I quite like your closing argument to take $b + 2c \le 10$ and check the number of options for each $b$ at fixed $c$ – infinitylord Oct 20 '17 at 22:32
• Thanks. Although sadly(?) not using generating functions. With a little extension you can see that the same answer also applies to target values from $101$ to $109$ (using $a' = (a-r)/10$). – Joffan Oct 20 '17 at 22:39
• Slight generalization: For $x,y,z$ non-negative integers, the number of solutions to $x + by + cz = d$ where $b|c$, and $c|d$ is $(\frac{d}{2b} + 1)^2$ – infinitylord Oct 21 '17 at 2:08
Er, ..., calculate carefully?
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...) \\ = \dots +30 x^{98} + 30 x^{99} + 36 x^{100} + 36 x^{101} + ...$$
Another way: Suppose there are $n_k$ ways to write $k$ as a non-negative integer weighted sum of $\{1,10\}$. Then the total number of ways to write $k$ as a non-negative integer weighted sum of $\{1,10,20\}$ is $N = n_0 + n_{20} + n_{40} + n_{60} + n_{80} + n_{100}$, where $n_0$ counts the number of ways we finish by adding five $20$s, $n_{20}$ the number of ways we finish by adding four $20$s, ..., and $n_{100}$ the number of ways we finish by adding no $20$s.
• Now $n_0 = 1$ because $0 \cdot 1 + 0 \cdot 10$ is the only way to write $0$ as a non-negative integer weighted sum of $\{1,10\}$.
• And $n_{20} = 3$, depending on whether there are zero, one, or two $10$s in the sum, the rest being made up with $1$s.
• And $n_{40} = 5$, for the same reason : zero to four $10$s and the remainder made of $1$s.
• Finishing, $n_{60} = 7$, $n_{80} = 9$, and $n_{100} = 11$.
Finally, $N = 1 + 3 + 5 + 7 + 9 + 11 = 36$. That the number of ways is controlled by the $10$s and $20$s, with the $1$ making up the rest, is hinted in the snippet of the series in $x$, above. There is only one way to make totals up to $9$ from $\{1,10,20\}$. There are two ways for totals in $[10,19]$, four ways in $[20,29]$, six ways in $[30,31]$, nine ways in $[40,41]$, where the increment in the number of ways in each range of ten totals increases by $1$ each time we pass a multiple of $20$ (counting that we could either include or exclude one more $20$ in our sums).
Another way: You don't have to multiply out the full power series. You can calculate with $$( 1 + x+ x^2 + \cdots + x^{99} + x^{100})(1 + x^{10} + \cdots + x^{100})(1+x^{20}+\cdots+x^{100})$$ always discarding any term with power greater than $100$. If you do this from right to left, you essentially perform the calculation described above. | 2019-07-19T19:10:41 | {
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https://math.stackexchange.com/questions/2645822/proof-verification-implication-from-field-axioms?noredirect=1 | # Proof Verification: Implication from Field Axioms
Prove that if $a \neq 0$ then $b/(b/a) = a$
Proof:
$b/(b/a) = b. 1/(b/a)$ (by Multiplication Axiom 4 (M4))
$\implies b/(b/a)$ = $b.(b/a)^{-1}$ (by M5)
$\implies$ $b/(b/a)$ = $b.(b.(1/a))^{-1}$
$\implies$ $b/(b/a)$ = $b.(b.(a)^{-1})^{-1}$
$\implies$ $b/(b/a)$ = $b.[1/(b.(a)^{-1})]$
$\implies$ $b/(b/a)$ = $b . (1/b) . (1/a^{-1})$
$\implies$ $b/(b/a)$ = $1 .(1/a^{-1})$
$\implies$ $b/(b/a)$ = $(a^{-1})^{-1}$
$\implies$ $b/(b/a)$ = a
Is this correct? Can anyone please verify?
• I'm just curious: where do you find the axioms? – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:08
• Do you mean derivation through axioms? – A.Asad Feb 11 '18 at 13:10
• Though Rudia never used the notation $(\cdot)^{-1}$ is that section, but IMHO, I think it's fine if you write it as long as it means the inverse. In the second arrow, you used a property $(xy)^{-1} = x^{-1} y^{-1}$. This can be easily proven, by as a "proof" from the axioms (or the propositions from the book), I think it's better to mark every step, like this answer in set theory. – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:18
• Thank you, I'll do that (i.e. prove that $(xy)^{-1} = x^{-1} y^{-1}$ and mark my steps) . Is the proof fine otherwise? – A.Asad Feb 11 '18 at 13:25
• Yes, I think so. – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:27
As OP's comments suggest, to derive the proof from the five multiplication axioms in Baby Rudin, break $1$ into the product of $a$ and $1/a$ in the second step. $\require{action}$ \begin{aligned} b / (b/a) &= \texttip{b \cdot 1 \cdot \left( \frac{1}{b/a} \right)} {(M4): multiplication by identity} \\ &= \texttip{b \cdot \left( \frac{1}{a} \right) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M5): existence of inverse} \\ &= \texttip{(b/a) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M3): regroup the leftmost two factors} \\ &= \texttip{a \cdot (b/a) \cdot \left( \frac{1}{b/a} \right)} {(M2): multiplication is commutative} \\ &= \texttip{a \cdot \left( (b/a) \cdot \left( \frac{1}{b/a} \right) \right)} {(M3): multiplication is associative} \\ &= \texttip{a \cdot 1}{(M5): multiplication by inverse} \\ &= \texttip{a}{(M4): multiplication by identity} \end{aligned} \bbox[4pt,border: 1px solid red]{ \begin{array}{l} \text{If you cannot figure out why a line}\\ \text{is true, move your mouse over}\\ \text{RHS of that line for hint.} \end{array}} Remarks: The above proof is one line shorter than OP's one. | 2020-05-31T10:13:45 | {
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https://math.stackexchange.com/questions/3328973/knowing-the-existence-of-max-min-values-of-function-in-interval-without-taking-d | # Knowing the existence of max/min values of function in interval without taking derivative?
I'm given the following question.
"How do we know that the following function has a maximum value and a minimum value in the interval $$[0,3]$$"
$$f(x)=\frac{x}{x^2+1}$$
Is it possible to understand that there is a maximum and minimum value in the interval without taking the derivative of the function?
Is the solution to take the derivative of $$f(x)$$ and equate the resulting function to zero and solve?
• Have you heard of the extreme value theorem? – Theo Diamantakis Aug 20 '19 at 14:26
• indeed I have but have no experience applying it which is probably the reason for the question :) I will review the theorem and see how it can be applied. – esc1234 Aug 20 '19 at 14:29
• of course. Upon reviewing the theorem it clearly states that a continuous function on a closed interval will have a minimum and maximum value. Thanks very much! – esc1234 Aug 20 '19 at 14:36
• Do you have an intuitive idea why that must be so, though? ANd why you might not be able to state the same thing for $x \in (0,3)$? – fleablood Aug 20 '19 at 15:38
• I believe so. Open interval of the problem of always being able to get closer and closer to the endpoint therefore it's technically impossible to get a max or min at an endpoint. Hence the closed interval stipulation. I get it now but I'm just used to applying these theorems. Thanks though – esc1234 Aug 20 '19 at 15:39
"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"
The question asks you to show it has a max and min on that interval; not to find them. ANd the question doesn't ask about local maxima/minima but global extrema of all values on the interval.
The extreme value theorem says, and I quote: if a real-valued function $$f$$ is continuous on the closed interval $$[a,b]$$, then f must attain a maximum and a minimum, each at least once.
And $$\frac {x}{x^2+1}$$ is continuous on $$[0,3]$$. So the extreme value theorem says it attains a max and minimum at least once on the interval.
so that's it. Done.
====
The intuitive idea for me is that the graph of a continuous function has a distinct starting point at $$[a,f(a)]$$ and a distinct ending point at $$[b,f(b)]$$ and for every point in between will have some actual real value. It's continuous so it can't "stretch" in any unbounded infinite value. So somewhere in there, either at one of the endpoints or somewhere between, it will get as big as it gets (within that interval, it can get bigger in other places outside the interval) and somewhere it will get as small as it gets.
That argument is naive and applies to ill-defined ideas (the most abusive is the it can't "stretch" to infinity) and needs to be formally defined and proven. And that is exactly what the Extreme Value theorem states.
In this case we have a path that starts at the point $$(0,f(0) = 0)$$ and ends at the point $$(3, f(3) = \frac 3{10})$$. In between in follows some curvy path between those two points. Somewhere there must be some point that was the biggest $$f(x)$$ value, and some that must be the least $$f(x)$$ value.
===
We can get more specific. $$x \ge 0$$ and $$x^2 + 1 \ge 1$$ so $$f(x) \ge \frac x{x^2 + 1} \ge 0$$ so $$f(0) = 0$$ is a minimum. $$x \le 3$$ and $$x^2 + 1 \ge 1$$ so $$f(x)\le \frac 31$$ and so all the possible $$f(x)$$ are bounded above by $$3$$.
It is the definition of real numbers that if a set is bounded above that a least upper bound exist and a consequence of $$f$$ being continuous means that there is an $$x=c$$ where $$f(c) = \sup \{f(x)|x \in [a,b]\}$$.
We can go further an not $$f(1) = \frac 12 > \frac 3{10} =f(3)$$ so the max is not $$x=3$$ but some $$x: 0< x < 3$$ and so the maximum will be a local max and we COULD find it by taking the derivative.
But the question isn't ASKING us to. The question is only asking us to argue that there is a max. An the only thing we have to say for that is:
"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"
• Exactly. This is the correct answer. Thank you – esc1234 Aug 20 '19 at 15:45
Use that $$\frac{1}{2}\times\frac{2x}{x^2+1}\le \frac{1}{2}$$ and $$\frac{-1}{2}\le \frac{2x}{x^2+1}\times \frac{1}{2}$$
• Thanks. Has this idea come from the extreme value theorem which was suggested in the comment above? – esc1234 Aug 20 '19 at 14:30
• No, it comes from the AM-GM inequality. – Dr. Sonnhard Graubner Aug 20 '19 at 14:31
• Okay, the AM-GM inequality is something I haven't come across in my coursework yet. I will study it but because it hasn't come up in my coursework yet I'm thinking there must be another method. – esc1234 Aug 20 '19 at 14:34
• It is $$\frac{a^2+b^2}{2}\geq ab$$ for $$a,b\geq 0$$ – Dr. Sonnhard Graubner Aug 20 '19 at 14:36
• This shows that all values of $f(x): 0 \le x \le 3$ are so that $-\frac 12 \le f(x) \le \frac 12$ but not that there is a specific $c,d$ so that $-\frac 12 \le f(c) \le f(x) \le f(d) \le \frac 12$. – fleablood Aug 20 '19 at 15:34
The question gives a function that is continuous over a closed interval. The extreme value theorem states that a function that is continuous on a closed interval will have both a minimum and maximum value on that interval. The question requires that the student knows and understands the extreme value theorem.
We show that your function has a turning point in $$[0,3]$$ which is a maximum.
Note that $$\frac {x_1}{x_1^2+1}=\frac {x_2}{x_2^2+1}$$ for values where $$x_1\ne x_2$$ and $$x_1x_2=1$$
For example $$\frac {2}{2^2+1}=\frac{1/2}{(1/2)^2 +1}$$
Thus you have a turning point between $$1/2$$ and $$2$$.
Thus the turning point happens at $$x=1$$ and it is a maximum because for example $$f(1)=1/2 > f(2)=2/5$$ | 2020-10-21T18:57:15 | {
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https://sibostesaw.web.app/1246.html | Taylor series for cosx at 0
However, when the interval of convergence for a taylor. The calculator will find the taylor or power series expansion of the given function around the given point, with steps shown. Sine function using taylor expansion c programming. In essence, the taylor series provides a means to predict a function value at one point in terms of the function value and its derivatives at another point. Part b asked for the first four nonzero terms of the taylor series for cos x about x 0 and also for the first four nonzero terms of the taylor series for fx about x 0. Evaluating limits using taylor expansions taylor polynomials provide a good way to understand the behaviour of a function near a speci. A particle moves along the xaxis wacceleration given by atcost ftsec2 for t 0. You can find the range of values of x for which maclaurins series of sinx is valid by using the ratio test for convergence. Lets look closely at the taylor series for sinxand cosx. How to extract derivative values from taylor series.
X1 k 0 21kx k 2k bfind the interval of convergence for this maclaurin series. In the case of a maclaurin series, were approximating this function around x is equal to 0, and a taylor series, and well talk about that in a future video, you can pick an arbitrary x value or fx value, we should say, around which to approximate the function. Because the taylor series is a form of power series, every taylor series also has an interval of convergence. What is the taylor series at x 0 for cos xtype an exact answer. How to prove that the taylor series of sinx is true for. As you work through the problems listed below, you should reference your lecture notes and. This is part of series of videos developed by mathematics faculty at the north carolina school of science and mathematics. Ap calculus bc 2011 scoring guidelines college board. Taylors series of sin x in order to use taylors formula to. When this interval is the entire set of real numbers, you can use the series to find the value of fx for every real value of x. If i wanted to approximate e to the x using a maclaurin series so e to the x and ill put a little approximately over here. Depending on the questions intention we want to find out something about the curve of math\frac\sin xxmath by means of its taylor series 1.
Given n and b, where n is the number of terms in the series and b is the value of the angle in degree. Taylor series cosx function matlab answers matlab central. Program to calculate the sum of cosine series of x and compare the value with the library functions output. To find the maclaurin series simply set your point to zero 0. For taylors series to be true at a point xb where b is any real number, the series must be convergent at that point. The power series in summation notation would bring you to my answer. Approximating cosx with a maclaurin series which is like a taylor polynomial centered at x0 with infinitely many terms. Use power series operations to find the taylor series at x 0 for the following function.
To get the maclaurin series for xsin x, all you have to do is to multiply the series with x throughout, as indicated by the formula above. By using this website, you agree to our cookie policy. The taylor series for the exponential function ex at a 0 is. Derivatives derivative applications limits integrals integral applications series ode laplace transform taylormaclaurin series fourier series. Determining whether a taylor series is convergent or. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. We also derive some well known formulas for taylor series of ex, cosx and sinx around x 0. Find the taylor series for expcosx about the point x 0 up to x4 really no clue where to even begin. Whats wrong with just bruteforce calculation of the first six derivatives of that function at zero, then form the taylor series at zero. Commonly used taylor series university of south carolina. A maclaurin series can be used to approximate a function, find the antiderivative of a complicated function, or compute an otherwise uncomputable sum. How to extract derivative values from taylor series since the taylor series of f based at x b is x. Taylor series integration of cosx 1 x physics forums. We focus on taylor series about the point x 0, the socalled maclaurin series.
Its going to be equal to any of the derivatives evaluated at 0. Free trigonometric equation calculator solve trigonometric equations stepbystep. You can specify the order of the taylor polynomial. This could be its value at mathx 0 math as is considered a popular interview questions, i. A value is returned for all nonnegative integer values of n, but no matter how many terms i have the program provide the sum always approaches ans1, reaching it by around n5. Follow 59 views last 30 days nikke queen on 3 apr 2015. Math 142 taylormaclaurin polynomials and series prof. A maclaurin series is a taylor series where a 0, so all the examples we have been using so far can also be called maclaurin series. Im working on a taylor series expansion for the coxx function. Thus, we get the series we call the maclaurin series. Because this limit is zero for all real values of x, the radius of convergence of the. Recall that the taylor series of fx is simply x1 k 0 fk 0 k. Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers.
Matlab cosx taylor series matlab answers matlab central. Write a matlab program that determines cos x using the. Finding the maclaurin series for cosx chapter 30 lesson 3 transcript video. Find the taylor series expansion of the following functions around 0.
Taylor series and maclaurin series calculus 2 duration. How to evaluate sinxx using a taylor series expansion quora. Taylor and maclaurin power series calculator emathhelp. Sign up to read all wikis and quizzes in math, science, and engineering topics. The maclaurin expansion of cosx the infinite series module. Convergence of taylor series suggested reference material. Physics 116a winter 2011 taylor series expansions in this short note, a list of wellknown taylor series expansions is provided. Taylor series a taylor series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc. Calculus power series constructing a taylor series. To find the series expansion, we could use the same process here that we used for sin x. Find the maclaurin series expansion for cosx at x 0, and determine its.
The taylor series for sinx converges more slowly for. The taylor series for fxcosx at a pi2 is summation from zero to infinity cnxpi2n find the first few coefficients c0 c1 c2 c3 c4 get more help from chegg get 1. And thats why it makes applying the maclaurin series formula fairly straightforward. In mathematics, a taylor series is an expression of a function as an infinite series whose terms. We can use the first few terms of a taylor series to get an approximate value for a function. Write a matlab program that determines cos x using. As a result, if we know the taylor series for a function, we can extract from it any derivative of the.
Math 142 taylor maclaurin polynomials and series prof. Find the maclaurin series expansion for cos x at x 0, and determine its radius of convergence. Free taylor series calculator find the taylor series representation of functions stepbystep this website uses cookies to ensure you get the best experience. Here we show better and better approximations for cosx. A calculator for finding the expansion and form of the taylor series of a given function. I assume you mean from 0 to pi rather than from 0 to 180 degrees because the taylor expansion works in radians, not degrees. In this section we will discuss how to find the taylormaclaurin series for a function. Sine function using taylor expansion c programming ask question asked 8 years, 4 months ago. | 2021-03-01T06:07:07 | {
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https://mathhelpboards.com/threads/ask-equation-of-a-circle-in-the-first-quadrant.26984/ | # [ASK] Equation of a Circle in the First Quadrant
#### Monoxdifly
##### Well-known member
The center of circle L is located in the first quadrant and lays on the line y = 2x. If the circle L touches the Y-axis at (0,6), the equation of circle L is ....
a. $$\displaystyle x^2+y^2-3x-6y=0$$
b. $$\displaystyle x^2+y^2-12x-6y=0$$
c. $$\displaystyle x^2+y^2+6x+12y-108=0$$
d. $$\displaystyle x^2+y^2+12x+6y-72=0$$
e. $$\displaystyle x^2+y^2-6x-12y+36=0$$
Since the center (a, b) lays in the line y = 2x then b = 2a.
$$\displaystyle (x-a)^2+(y-b)^2=r^2$$
$$\displaystyle (0-a)^2+(6-b)^2=r^2$$
$$\displaystyle (-a)^2+(6-2a)^2=r^2$$
$$\displaystyle a^2+36-24a+4a^2=r^2$$
$$\displaystyle 5a^2-24a+36=r^2$$
What should I do after this?
#### skeeter
##### Well-known member
MHB Math Helper
circle center at $(x,2x)$
circle tangent to the y-axis at $(0,6) \implies x=3$
$(x-3)^2+(y-6)^2 = 3^2$
$x^2-6x+9+y^2-12y+36 =9$
$x^2+y^2-6x-12y+36=0$
#### Monoxdifly
##### Well-known member
circle center at $(x,2x)$
circle tangent to the y-axis at $(0,6) \implies x=3$
How did you get x = 3 from (0, 6)?
#### HallsofIvy
##### Well-known member
MHB Math Helper
How did you get x = 3 from (0, 6)?
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.
(And the center of the circle "lies on the line y= 2x", not "lays".)
#### Monoxdifly
##### Well-known member
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.
(And the center of the circle "lies on the line y= 2x", not "lays".)
Well, I admit I kinda suck at English. I usually use "lies" as "deceives" and "lays" as "is located". Thanks for the help, anyway. Your explanation is easy to understand. | 2020-09-20T18:28:30 | {
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https://blogs.sas.com/content/iml/2022/06/27/the-derivative-of-a-quantile-function.html | Recently, I needed to solve an optimization problem in which the objective function included a term that involved the quantile function (inverse CDF) of the t distribution, which is shown to the right for DF=5 degrees of freedom. I casually remarked to my colleague that the optimizer would have to use finite-difference derivatives because "this objective function does not have an analytical derivative." But even as I said it, I had a nagging thought in my mind. Is that statement true?
I quickly realized that I was wrong. The quantile function of ANY continuous probability distribution has an analytical derivative as long as the density function (PDF) is nonzero at the point of evaluation. Furthermore, the quantile function has an analytical expression for the second derivative if the density function is differential.
This article derives an exact analytical expression for the first and second derivatives of the quantile function of any probability distribution that has a smooth density function. The article demonstrates how to apply the formula to the t distribution and to the standard normal distribution. For the remainder of this article, we assume that the density function is differentiable.
### The derivative of a quantile function
It turns out that I had already solved this problem in a previous article in which I looked at a differential equation that all quantile functions satisfy. In the appendix of that article, I derived a relationship between the first derivative of the quantile function and the PDF function for the same distribution. I also showed that the second derivative of the quantile function can be expressed in terms of the PDF and the derivative of the PDF.
Here are the results from the previous article. Let X be a random variable that has a smooth density function, f. Let w = w(p) be the p_th quantile. Then the first derivative of the quantile function is
$\frac{dw}{dp} = \frac{1}{f(w(p))}$
provided that the denominator is not zero. The second derivative is
$\frac{d^2w}{dp^2} = \frac{-f^\prime(w)}{f(w)} \left( \frac{dw}{dp} \right)^2 = \frac{-f^\prime(w)}{f(w)^3}$.
### Derivatives of the quantile function for the t distribution
Let's confirm the formulas for the quantile function of the t distribution with five degrees of freedom. You can use the PDF function in SAS to evaluate the density function for many common probability distributions, so it is easy to get the first derivative of the quantile function in either the DATA step or in the IML procedure, as follows:
proc iml; reset fuzz=1e-8; DF = 5; /* first derivative of the quantile function w = F^{-1}(p) for p \in (0,1) */ start DTQntl(p) global(DF); w = quantile("t", p, DF); /* w = quantile for p */ fw = pdf("t", w, DF); /* density at w */ return( 1/fw ); finish; /* print the quantile and derivative at several points */ p = {0.1, 0.5, 0.75}; Quantile = quantile("t", p, DF); /* w = quantile for p */ DQuantile= DTQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.];
This table shows the derivative (third column) for the quantiles of the t distribution. If you look at the graph of the quantile function, these values are the slope of the tangent lines to the curve at the given values of the p variable.
### Second derivatives of the quantile function for the t distribution
With a little more work, you can obtain an analytical expression for the second derivative. It is "more work" because you must use the derivative of the PDF, and that formula is not built-in to most statistical software. However, derivatives are easy (if unwieldy) to compute, so if you can write down the analytical expression for the PDF, you can write down the expression for the derivative.
For example, the following expression is the formula for the PDF of the t distribution with ν degrees of freedom:
$f(x)={\frac {\Gamma ({\frac {\nu +1}{2}})} {{\sqrt {\nu \pi }}\,\Gamma ({\frac {\nu }{2}})}} \left(1+{\frac {x^{2}}{\nu }}\right)^{-(\nu +1)/2}$
Here, $\Gamma$ is the gamma function, which is available in SAS by using the GAMMA function. If you take the derivative of the PDF with respect to x, you obtain the following analytical expression, which you can use to compute the second derivative of the quantile function, as follows:
/* derivative of the PDF for the t distribution */ start Dtpdf(x) global(DF); nu = DF; pi = constant('pi'); A = Gamma( (nu +1)/2 ) / (sqrt(nu*pi) * Gamma(nu/2)); dfdx = A * (-1-nu)/nu # x # (1 + x##2/nu)##(-(nu +1)/2 - 1); return dfdx; finish; /* second derivativbe of the quantile function for the t distribution */ start D2TQntl(p) global(df); w = quantile("t", p, df); fw = pdf("t", w, df); dfdx = Dtpdf(w); return( -dfdx / fw##3 ); finish; D2Quantile = D2TQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.] D2Quantile[F=Best6.];
### Summary
In summary, this article shows how to compute the first and second derivatives of the quantile function for any probability distribution (with mild assumptions for the density function). The first derivative of the quantile function can be defined in terms of the quantile function and the density function. For the second derivative, you usually need to compute the derivative of the density function. The process is demonstrated by using the t distribution. For the standard normal distribution, you do not need to compute the derivative of the density function; the derivatives depend only on the quantile function and the density function.
Share | 2022-08-19T11:03:52 | {
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https://www.physicsforums.com/threads/how-can-the-set-of-the-rational-numbers-be-countable-if-there-is-no.667746/ | # How can the set of the rational numbers be countable if there is no
1. Jan 29, 2013
### student34
rational number to count the next rational number from any rational number?
We can count the next natural number, but we can't count the next rational numer.
2. Jan 29, 2013
### CompuChip
The definition of countable means, that you can assign a natural number to each of them.
Well, here's one way to count them:
3. Jan 29, 2013
### NemoReally
Starting with, say, 1/1, write down all the rational numbers in whichever order you like. Label the starting number as Rational Number 1. As you generate the next rational number, write it down (as a / b) and then write the next natural number down next to it. Continue until you run out of either rational numbers or natural numbers ...
Actually, you won't run out of either. They are in a one-to-one correspondence. The difference from the normal way of thinking about finite counting lies in the fact that you will never run out of natural numbers. This concept underlies many mathematical ideas but it may hurt your brain until you get the idea.
Interestingly, you can't do the same with real numbers. Look up Cantor's diagonal method.
4. Jan 29, 2013
### student34
If we really can assign a natural number to each of them, can you assign a natural number to the very next rational number from the number 1?
5. Jan 29, 2013
### Fredrik
Staff Emeritus
Yes. This is what CompuChip's picture is showing. The first 11 positive rational numbers (when they are ordered as in that picture) are
1. 1
2. 1/2
3. 2
4. 3
5. 1/3 (We're skipping 2/2, since it's equal to 1, which is already on the list).
6. 1/4
7. 2/3
8. 3/2
9. 4
10. 5
11. 1/5 (We're skipping 4/2, 3/3 and 2/4, since they are equal to numbers that are already on the list).
The only problem with that picture is that it only deals with positive rational numbers. You can however easily imagine a similar picture that includes the negative ones.
6. Jan 29, 2013
### NemoReally
In principle, yes. You just keep extending the table of rational numbers, labelling each one with a natural number, until you reach the "very next" rational number from 1. The trick lies in recognizing that (physical limitations aside), there is always another rational number between the "very next" rational number and 1, so you'll never actually do it, but if you keep trying you will always be able to assign a natural number to each one.
eg, halving the difference ...
label '1' as number 1 in your list. Start with 3/2 (1+1/2) and label it '2'. Then the "next" rational number (number '3') will be 5/4 ((1+3/2)/2), the next 9/8 (number '4') and so on. You can keep dividing by 2 and incrementing the label indefinitely - you will never run out of natural numbers or rational numbers that keep getting closer to 1
https://www.physicsforums.com/attachment.php?attachmentid=55160&stc=1&d=1359454296
https://www.physicsforums.com/attachment.php?attachmentid=55159&stc=1&d=1359453948
Last edited: Jan 29, 2013
7. Jan 29, 2013
### CompuChip
Of course, if you want to be very precise: there are duplicates in the tabulation of the rationals (for example, it contains 1/2, 2/4, 3/6, 34672/69344, etc). So technically you are not really making a one-to-one mapping, but you are overcounting.
In other words, you are proving that the cardinality of the rationals is at most the same as that of the natural numbers. However, since we clearly also have at least as many rationals as natural numbers (the natural numbers are precisely the first column of the grid), you can convince yourself that the cardinalities are equal.
8. Jan 29, 2013
### cragar
another way to map the rationals to the naturals is take the
positive rationals of the form $\frac{p}{q}$ and map them to
$2^p3^q$ and then map the negative rationals to
$5^{|-p|}7^{|-q|}$ and then map zero to some other prime.
In fact we are mapping all the rationals to a proper subset of the naturals.
9. Jan 29, 2013
### CompuChip
I also wanted to get back to your remark about "the next" rational number from 1... note that the rationals are dense in the real numbers. That implies that there is a rational number between any two given rationals, and therefore it is not possible to write them down in "ascending" order (i.e. write down a sequence an such that every rational is in the sequence and 0 = a0 < a1 < a2 < ...).
This may be flaw in your way of picturing the rationals that makes their countability, perhaps, a bit counter-intuitive. | 2018-03-21T07:17:56 | {
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https://math.stackexchange.com/questions/3692679/odd-prime-p-implies-positive-divisors-of-2p-are-1-2-p-and-2p | # Odd prime $p$ implies positive divisors of $2p$ are $1,2,p,$ and $2p$
$$1,2,p,$$ and $$2p$$ are indeed divisors of $$2p$$. I want to show these are the only positive divisors. Is there a more elegant or concise way to prove this besides the proof I have below?
Suppose that positive $$a \in \left([3,2p-1] \cap \mathbb{N}\right) \setminus\{p\}$$ divides $$2p$$. So $$ak$$=$$2p$$ for $$k \in \mathbb{Z}$$, and clearly $$2 \leq k \leq p$$. Since $$ak=2p$$ is even, at least one of $$a$$ or $$k$$ must be even.
If $$k$$ is even, then $$a\frac{k}{2}=aj=p$$ for integer $$1 \leq j \leq \frac{p}{2}, so $$j | p$$, so $$j=1$$, so $$a=p$$ which is a contradiction.
Similarly, if $$a$$ is even then $$k | p$$, so $$k=p$$. But then $$a=2$$.
So there are no other positive divisors besides $$1,2,p$$, and $$2p$$.
The motivation for this is to show that if a group $$G$$ has order $$2p$$ for odd prime $$p$$, then nonabelian $$G$$ is isomorphic to $$D_{2p}$$. The proof begins with "the possible orders for nonidentity elements of $$G$$ are $$2,p,$$ and $$2p$$," which I am trying to prove with Lagrange's Theorem. If there is an alternative way to justify this statement using group theory, then I would appreciate seeing that as well.
• Use the Fundamental Theorem of Arithmetic. – Shaun May 26 at 16:36
• @Shaun clearly $2 \cdot p$ is a prime factorization of $2p$, and the prime factorization is unique. How does this show there cannot be nonprime divisors of $2p$ besides itself and $1$? – jskattt797 May 26 at 16:41
• Because $2p$ is squarefree (and there is only two primes). – Shaun May 26 at 16:42
• It's enough to state that the only two prime factors of $2p$ are $2,p$ so (by fundamental th) the only factors are combinations of $2$ and $p$ so they are $1,2,p$ and $2p$. Honestly that sentence is probably too long and saying too much.... I think that if you simply state the only factors of $2p$ are $1,2,p$ and $2p$ that ought to be utterly self-evident and obvious. If anyone asks way state its immediate but the fundamental th. – fleablood May 26 at 20:28
If $$p$$ is an odd prime number, then by the fundamental theorem of number theory, $$2\times p$$ is the unique primary decomposition of $$2p$$. Once you express a positive integer $$n$$ as it's unique primary decomposition, say $$p_1^{a_1}\dots p_k^{a_k}$$, then all the positive factors will be of the form $$p_1^{b_1}\dots p_k^{b_k}$$ where $$0\leq b_i\leq a_i$$ for each $$i$$. With this observation you should be able to answer your question.
• And we have $2p=2^1p^1$, so the set of all positive divisors is $$\{ 2^{a_1}p^{a_2} \mid a_1,a_2 \in \{0,1\} \}=\{1,2,p,2p \}$$. – jskattt797 May 26 at 17:17
By the fundamental theorem of Arithmetic the only prime factors of $$2p$$ are $$2$$ and $$p$$ and so every factor must be a combination of $$2$$ and $$p$$ of which $$1,2,p$$ and $$2p$$ are the only options.
That is more than sufficient and more than anyone can reasonable expect to require proof.
....
But if you want to smash an ant with a sledgehammer:
The fundamental Theorem of Arithmetic say each number has a unique prime factorization of $$n = \prod p_i^{a_i}$$. So if any factor $$m; m|n$$ can only have $$\{p_i\}$$ as prime factors and only to the powers less than or equal to $$a_i$$.
So all if $$m|n$$ then $$m$$ must be of the for $$\prod p_i^{b_i}$$ where $$0\le b_i \le a_i$$ and so there are $$\prod (a_i+1)$$ such factors.
So the factors of $$2p$$ are all of the form $$2^b p^c$$ where $$b = 0,1$$ and $$p = 0,1$$. There are four of these numbers and they are $$2^0p^0 =1; 2^1p^0 = 2; 2^0p^1 = p;$$ and $$2^1p^1 = 2p$$.
That's it.
That is a result that it is reasonable to expect every reader is either familiar with or can justify on their own.
....
But frankly it is enough to say:
"The only factor of $$2p$$ are $$1,2,p$$ and $$2p$$" and assume that is completely self-evident.
And it is.
• Yes, $m|n$ iff $m=\prod p_i^{b_i}$. And each factor $\prod p_i^{b_i}$ maps bijectively to a tuple $(b_1, \dots , b_k)$ for $0 \leq b_i \leq a_i$. So there are exactly $\prod(a_i + 1)$ factors. So an alternative proof is to notice that for $2p = 2^1 p^1$ there are exactly $2 \cdot 2=4$ factors, so there can be no others besides $1,2,p,$ and $2p$. – jskattt797 May 27 at 3:19
By the Fundamental Theorem of Arithmetic, the only primes that divide $$2p$$ are $$2$$ and $$p$$. Note that $$2p$$ is squarefree and has only two prime factors.
• Nice, so suppose there is another nonprime positive divisor $j=p_1 p_2 \dots \neq 1$ (at least two primes in $j$'s factorization because $j > 1$ is not prime) such that $j \neq 2p$, then $jk=2p$ for integer $k$. Notice $k \neq 1$, so $k > 1$ and $k = p_3 \dots$. But then $2p$ has at least 3 primes in its factorization. Why do we need squarefree? – jskattt797 May 26 at 16:58
• To rule out $2^2=4$ and $p^2$. That's all. – Shaun May 26 at 17:13
• Your argument seems valid though, @jskattt797. – Shaun May 26 at 17:14
The other answerers are of course right to point you to the Fundamental Theorem of Arithmetic as it will come in handy at many stages of your life. Still I wanted to say two things about your own proof.
1) It is correct and quite elegant. As one of the other answerers points out: using the Fundamental Theorem is a bit of a sledgehammer.
2) The key step in your proof is the lemma 'since $$ak$$ is even at least one of $$a, k$$ is even'. I wanted to point out to you that this has a very nice generalization, called Euclid's lemma. It states:
Let $$q$$ be any prime number. Then whenever $$ak$$ is divisible by $$q$$ we necessarily have that at least one of $$a, k$$ is divisible by $$q$$.
So your lemma is the case $$q = 2$$. Knowing this, we see that you could also make a copy of your proof but with $$p$$ in the role of 2 and 2 in the role of $$p$$ although it would be less intuitively appealing.
Euclid's lemma implies the uniqueness part of the Fundamental Theorem (and is mostly proved first) but of course if you have a different proof of the Fundamental Theorem, Euclid's lemma readily follows from it.
• You're right the argument is exactly the same, perhaps even simpler using the bounds on $k$: $ak=2p \implies p|k$ or $p|a$. If $p|k$ then $k=p$ and $a=2$. If $p|a$ then $k|2$ so $k=2$ and $a=p$. Both cases contradict the choice of $a$. – jskattt797 May 27 at 3:49
You can do this fairly simply with Euclid's Lemma, that if $$q$$ is a prime number and $$q\mid ab$$, then either $$q\mid a$$ or $$q\mid b$$.
If $$d$$ were a divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, or $$2p$$, then $$d$$, since it's not equal to $$1$$, must be divisible by some prime. So let $$q$$ be a prime divisor of $$d$$. But now $$q\mid d\mid2p$$ implies, by Euclid's Lemma, that either $$q\mid2$$ or $$q\mid p$$, which implies either $$q=2$$ or $$q=p$$. The latter is not possible, since if we write $$d=qk$$ and let $$q=p$$, we cannot have $$k=1$$ or $$2$$, since $$d\not=p$$ or $$2p$$, nor can we have $$k\gt2$$ since a divisor cannot be larger than the number it divides. So $$q=2$$. Thus $$d$$ can only be a power of $$2$$, and since $$d\not=2$$, it must be divisible by $$4$$. But since $$p$$ is an odd prime, we have $$p=2n+1$$ for some $$n$$, in which case $$2p=4n+2$$, which is not divisible by $$4$$. The contradiction tells us that there is no divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, and $$2p$$.
Remark: Laying out the logic of this was a little more complicated than I initially anticipated. It might be possible to compress the argument, but I don't offhand see any suitably slick way to do so. Maybe someone else does.
• $q|d|2p \implies q|2$ or $q|p$ i.e. prime $q=2$ or $q=p$. And our choice of $d$ cannot be divisible by $p$ as you noted so $q=2$, so $d=2k | 2p$ for some positive integer $k$, so $k | p$, so $k=1$ ($d=2$) or $k=p$ ($d=2p$), both of which contradict our choice of $d$. – jskattt797 May 27 at 4:08
• But doesn't "$d$, since it's not equal to $1$, must be divisible by some prime" rely on the fundamental theorem of arithmetic? – jskattt797 May 27 at 4:11
• @jskattt797, nice alternative. Regarding the "$d$ must be divisible by some prime," no, that's more basic than the fundamental theorem. Or if you like, it's the "easy" part of the fundamental theorem, which says that every integer greater than $1$ can be written as a product of primes (the "easy" part) in exactly one way (the "hard" part). – Barry Cipra May 27 at 8:02 | 2020-07-08T09:01:59 | {
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https://riseandhustle.com/brandon-soo-rvit/05ad77-pi-notation-identities | pi notation identities
↦ If x, y, and z are the three angles of any triangle, i.e. In the language of modern trigonometry, this says: Ptolemy used this proposition to compute some angles in his table of chords. Terms with infinitely many sine factors would necessarily be equal to zero. e this identity is established it can be used to easily derive other important identities. sin {\displaystyle \theta '} The tangent (tan) of an angle is the ratio of the sine to the cosine: If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term. = ⋅ (−) ⋅ (−) ⋅ (−) ⋅ ⋯ ⋅ ⋅ ⋅. Apostol, T.M. Dividing this identity by either sin2 θ or cos2 θ yields the other two Pythagorean identities: Using these identities together with the ratio identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign): The versine, coversine, haversine, and exsecant were used in navigation. Algebra Calculator Calculate equations, ... \pi: e: x^{\square} 0. Finite summation. Dividing all elements of the diagram by cos α cos β provides yet another variant (shown) illustrating the angle sum formula for tangent. The always-true, never-changing trig identities are grouped by subject in the following lists: Published online: 20 May 2019. The only difference is that we use product notation to express patterns in products, that is, when the factors in a product can be represented by some pattern. When the series θ Here, we’ll present the notation with some applications. {\displaystyle \lim _{i\rightarrow \infty }\cos \theta _{i}=1} θ In trigonometry, the basic relationship between the sine and the cosine is given by the Pythagorean identity: sin2θ+cos2θ=1,{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1,} where sin2θmeans (sin θ)2and cos2θmeans (cos θ)2. The veri cation of this formula is somewhat complicated. 1 Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent have period π. Identities for negative angles. None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots. The ratio of these formulae gives, The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n − 1)th and (n − 2)th values. ) These are sometimes abbreviated sin(θ) andcos(θ), respectively, where θ is the angle, but the parentheses around the angle are often omitted, e.g., sin θ andcos θ. Definition and Usage. The sum and difference identities can be used to derive the double and half angle identities as well as other identities, and we will see how in this section. cos Pi is the symbol representing the mathematical constant , which can also be input as ∖ [Pi]. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. General Identities: Summation. {\displaystyle \mathrm {SO} (2)} lim θ Purplemath. General Mathematical Identities for Analytic Functions. , [35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of π. These can be "trivially" true, like "x = x" or usefully true, such as the Pythagorean Theorem's "a 2 + b 2 = c 2" for right triangles.There are loads of trigonometric identities, but the following are the ones you're most likely to see and use. You describe is supposed to end problem is not strictly a Pi Notation words that! For any measure or generalized function or basic trigonometric functions of θ also involving lengths! To a real algebraic expression, as it involves a limit and a power outside of Pi...... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time veri cation this! Complex numbers under the cube roots months ago and quadrature components words that! Have a more concise Notation for the factorial operation numerical value they intermediate! Of an angle are sometimes referred to as the primary trigonometric functions the primary trigonometric.... Using the unit imaginary number i satisfying i2 = −1, 1.! Let i = √−1 be the imaginary unit and let ∘ denote composition of differential operators and sums than. Unit imaginary number i satisfying i2 = −1, these follow from the angle addition and theorems., commonly called trig, in pre-calculus strictly a Pi Notation words - that is, words to... 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Reducible to a real algebraic expression, we deduce Evaluate functions Simplify, as the primary functions...
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http://math.stackexchange.com/questions/400961/is-sum-limits-k-1n-1-binomnkxn-kyk-always-even/400972 | # Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?
Is $$f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$ always divisible by $2$?
-
Having $100$% $\LaTeX$ on the title isn't a good idea. – Git Gud May 24 '13 at 6:35
(Hint) An odd number raised to any power is odd, and an even number raised to any power is even. In particular, $$(x + y)^n \equiv (x + y) \pmod 2$$ Using this along with the binomial formula, you should be able to prove the result.
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Thank you, Goos. – Trancot May 24 '13 at 6:43
Yes, but your congruence is weak. It should be $(x+y)^n\equiv (x^n+y^n)\pmod{2}$, correct? – Trancot May 24 '13 at 6:49
@BarisaBarukh that is true as well (in fact that is true for any prime). However, in the particular case of mod 2, I usually just replace all $n$th powers with the number itself, because it makes things simpler. – Goos May 24 '13 at 6:56
But you are correct that using $(x + y)^n \equiv x^n + y^n$ makes the proof more direct. – Goos May 24 '13 at 6:58
Hint: Recall binomial formula $$(x+y)^n=\sum\limits_{k=0}^n{n\choose k} x^{n-k} y^k$$
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Yes, I know that. This is how I came to the above result. – Trancot May 24 '13 at 6:37
Ok do you know anything about Frobenius automorphism? – Norbert May 24 '13 at 6:40
I've used Frobenius in an elementary differential equations course. – Trancot May 24 '13 at 6:41
I might be able to manage if you can explain it clearly. – Trancot May 24 '13 at 6:42
bin-o-mail? That sounds like an automatic mail deleter! :-) – Asaf Karagila May 24 '13 at 6:43
In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)
$$f(n,x,y) = (x+y)^n - x^n -y^n$$
If
• both $x$ and $y$ are even: even - even - even = even;
• both $x$ and $y$ are odd: even - odd - odd = even;
• $x$ is even and $y$ is odd: odd - even - odd = even;
• $y$ is odd and $x$ is even: just like the case above.
So, $f(n,x,y)$ is always even.
-
Try setting $y=-x+a$ and expand the powers.
- | 2014-07-25T18:27:23 | {
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https://math.stackexchange.com/questions/1752739/finding-a-function-given-its-derivative | # Finding a function, given its derivative.
At the start of day zero during the summer, the temperature is $y(0) = 15$ degrees Celsius. Over a 50 day period the temperature increases according to the rule $$y'(t) = {y(t) \over 50}$$. With time $t$ measured in days. Find the formula for $y$
I not sure how to start here. Can I integrate $y'(t)$ to get back $y(t)$ ? Would it have to be a definite integral from zero to fifty ? Or and indefinite? $$\int_{0}^{50} {1\over50} dy$$ ??
• Separate the variables then integrate – randomgirl Apr 21 '16 at 13:00
Let's rewrite $y'(t)$ as $\dfrac{dy}{dt}$ and just write $y(t)$ as $y$. Then we have: $$\frac{dy}{dt} = \frac{1}{50} y$$
Separate variables to get: $$\frac{dy}{y} = \frac{1}{50} \, dt$$
Integrate both sides: \begin{align} \int \frac{dy}{y} &= \int \frac{1}{50} \, dt\\[0.3cm] \ln |y| &= \frac{1}{50} t + C\\[0.3cm] |y| &= e^{(t/50) + C}\\[0.3cm] y &= Ce^{t/50} \end{align}
Now use the fact that $y(0) = 15$ to find $C$.
Hint. Note that: $$y'-\frac{1}{50}y=0\\ e^{-t/50}y'-\frac{1}{50}e^{-t/50}y=0\\ \big(e^{-t/50}y(t)\big)'=0\\ e^{-t/50}y(t)=\text{constant}$$
• The product rule for differentiation is probably the best way to solve this for someone who's not used to the tools of differential equations (and since the asker is stumped by this question, I would assume that that is the case). Although going from the second to third line by multiplying with $e^{-t/50}$ might seem like some sort of magic trick pulled up from a hat, that's what makes it work in the end. – Arthur Apr 21 '16 at 13:03
Like so: $\frac{dy}{y}=\frac{dt}{50}$ Now integrate both sides. | 2019-06-18T13:34:13 | {
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https://web2.0calc.com/questions/pleaze-help | +0
# Pleaze help
0
288
15
+48
The following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?
[red 1] [red 2] [red 3] [yellow 1] [yellow 2] [yellow 3] [blue1] [blue 2] [blue 3]
Jan 17, 2022
#1
+2402
+2
Solution:
There are $$\large \dbinom{9}{3} = 84$$ ways to deal nine (9) cards to three (3) persons.
Give a red card to each person, then there are $$\large \dbinom{6}{3} = 20$$ ways to ways to deal the remaining six (6) cards to three (3) persons.
Probablity:
$$\Large \dfrac{\dbinom{6}{3}}{\dbinom{9}{3}} = \dfrac {5}{21} \approx 23.81\%$$
GA
--. .-
Jan 17, 2022
edited by GingerAle Jan 17, 2022
#2
+2403
+2
Nice solution, however I also tried doing this problem and got a different answer.
I viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680.
I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.
SImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this.
The number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.
90*6/1680 = 32.14%
=^._.^=
catmg Jan 17, 2022
#5
+2402
+1
My solution above is WRONG.
I’ve made train wrecks of these types of problems before.
I didn’t realize I’d wrecked the train this time until I read Catmg’s comment:
I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.
That is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.
The main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn.
This stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.
Here is a solution using Hypergeometric distribution:
The number of possible dealt sets is $$N=\dbinom{9}{3} * \dbinom{6}{3} * \dbinom{3}{3} = 1680 \\$$
For person one, there are $$\dbinom{3}{1}$$ ways to select one (1) of the three (3) red cards and then there are $$\dbinom{6}{2}$$ ways to choose two more (non red) cards.
For person two, there are $$\dbinom{2}{1}$$ ways to select one (1) of the two (2) remaining red cards and then there are $$\dbinom{4}{2}$$ ways to choose two more (non red) cards.
For person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.
Then
$$n = 3 \dbinom{6}{2} \cdot 2\dbinom{4}{2} \hspace {1em} \small | \text{ Where n = the number of hands with a red card.}\\ \Large \rho_{\small \text{(3 persons with one red card)}} \normalsize = \dfrac {n} {N} = 3! \cdot \dfrac{\dbinom{6}{2}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}} = \dfrac{9}{56} \approx 16.07\%\\$$
GA
--. .-
GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
#8
+2403
0
However, at the very end I got the answer 9/28 which is about 32.14%
6C2 = 15
4C2 = 6
9C3 = 84
6C3 = 20
3! = 6
(6*15*6)/(84*20) = (3*3*3)/(84) = 9/28
=^._.^=
catmg Jan 20, 2022
#9
+2402
+1
Here is a much easier way to solve this. The method does not use Hypergeometric selection
There are $$(3^3) = 27$$ arrangements of success. Divide the number of successes by the total number of sets giving $$(3^3) / (nCr(9,3) = \dfrac {9}{28} \approx 32.14\%$$
I’m glad you are paying attention to my presentations and train wrecks!
If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.
The calculation is exactly twice the value I presented for the Hypergeometric solution.
The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...
I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))
...I used a three (3) instead of a four (4) in the second binomial.
[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]
-----------
Here’s a graphic, demonstrating the arrangements for the above solution.
$$\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\$$
From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.
GA
--. .-
GingerAle Jan 20, 2022
edited by GingerAle Jan 20, 2022
edited by GingerAle Jan 20, 2022
#11
+2403
0
That is a very cool way to visualize the problem.
One thing I love about math is how many different ways there are to solve the question. :))
=^._.^=
catmg Jan 20, 2022
#3
+48
+1
GingerAle and Catmg:
You have a denominator of 10x9x8x10 because that is the total amount of combinations.
The numerator is 6 because their are 3! ways to arange the 3 white balls.
<(-_-)>_It'sFree_>(-_-)<
Jan 17, 2022
#6
+2402
+1
WTFF!! Your solution may be free, but it’s also absurd and moronic. (Just because It’s Free doesn’t mean you should be smoking it!)
The only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They’ve been in the specimen jar for over 21 years, and they’ve not changed much over that time. The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper. I sometimes called him the balless wonder. My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other’s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas’ balls are.
D.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts. ...Like this one
GA
--. .-
GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
#15
+48
+2
GA,
I am only 13 and don't smoke.
Interesting story though. :)
I am refering to Web2.0calc being free to use!
<(it'sfree)>
ItsFree Jan 21, 2022
#4
+117872
+2
Assume all are a bit different.
I will assume that ever person is unique. I will also (just for the working) assume every card is unique
There are 9! ways to line up the cards but for each of those there are 3! ^3 doubling up 9!/(3!)^3 = 1680 ways total sample space
There are 3! ways to give the reds one to each person then 6!/2^3 = 90
90*3! = 540 ways for the desirable outcome
540/1680 = approx 32%
I don't know if it is right or not. My answer seems to be in line with Catmg's. :) It also seems to be a reasonable answer.
I would like to know why ItsFree is so certain their answer is correct though. It is a bold statement to make.
Jan 18, 2022
#7
+117872
+1
It seems that we need Alan to run a Montecarlo Simulation test here.
That would give a definitive outcome. :)
Jan 20, 2022
#10
+117872
+1
So are we all, well, me, Catmg and Ginger all in agreement that the prob is 9/28?
Melody Jan 20, 2022
#12
+33151
+2
I've just run a Monte-Carlo simulation with 20000 trials and get a probability of 0.32125 (this will vary slightly each time it's calculated because of the use of random numbers of course). This is close to 9/28 (≈0.32143).
Alan Jan 21, 2022
edited by Alan Jan 21, 2022
#13
+117872
+1
Thanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))
Melody Jan 21, 2022
#14
+48
+2
Thank you all for helping me out. :)
Thanks to Alan as well for doing the problem out. ;)
<(it's_free)>
Jan 21, 2022
edited by ItsFree Jan 21, 2022 | 2022-10-06T20:57:08 | {
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https://math.stackexchange.com/questions/2664103/two-urns-probability | # Two Urns Probability
Question: Suppose there is a room filled with urns of two types. Type I urns contain 5 blue balls and 5 red balls. Type II urns contain 2 red balls and 8 blue balls. There are 700 Type I urns and 300 Type II urns in the room. They are distributed randomly and look alike. An urn is selected at random from the room and a ball is drawn from it.
A) What is the probability that the urn is Type I?
So that will be: total type 1 urns/total urns $= 700/1000 = 0.7$
B) What is the probability that the ball drawn is red?
I'm confused with this part of the question. My answer is:
(type 1 red balls) $\times$ (type 2 red balls)
(5 red balls/10 total balls) $\times$ (2 red balls/10 total balls) $= 1/10$
Is $1/10$ the probability to draw a red ball correct?
• You might consider how many total balls are there and how many are red. – Cardinal Feb 24 '18 at 2:39
• Would you be able to show me how to set it up please? – Hardeep Chohan Harry Feb 24 '18 at 2:41
• I provided an answer below. – Cardinal Feb 24 '18 at 3:22
• Thanks for your answer really appreciated – Hardeep Chohan Harry Feb 24 '18 at 3:40
Type 1 Urn:
700 Total Urns (70%) 50% Chance Red
Type 2 Urn:
300 Total Urns (30%) 20% Chance Red
The Urns are indistinguishable, as are the balls. Therefore, when selecting a ball, there is a 70% chance the urn is Type 1 with a 50% chance of leading to a Red. There is a 30% chance the urn is Type 2 with a 20% chance of being a Red.
To find the total probability of drawing a red ball, you can take 70% $\times$ 50% $+$ 30% $\times$ 20%
$0.35+.06=0.41$ or 41%
700 Urns x 5 Red balls per urn = 3500 Red 300 Urns x 2 Red balls per urn = 600 Red
4100 Red 10,000 Total
$\frac{4100}{10000} = 0.41$ or 41%
Think of this as a weighted average.
The probability of a type I urn is $0.7$
The probability of then selecting a red ball is $\frac{1}{2}$
The probability of a type II urn is $0.3$
The probability of then selecting a red ball is $\frac{1}{5}$
Thus, the desired probability is $$\left(0.7\cdot\frac{1}{2}\right)+\left(0.3\cdot\frac{1}{5}\right)=0.41$$ | 2019-09-15T08:10:14 | {
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https://math.stackexchange.com/questions/4035775/expected-number-of-objects-chosen-by-both-a-and-b | # Expected number of objects chosen by both A and B
Suppose that A and B each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects chosen by both A and B.
solution: Let $$X$$ be the number of objects chosen by both A and B. For $$1≤i≤10$$, let $$X_i=1$$ if object $$i$$ is chosen by A and B, else $$0$$ otherwise
Then $$X=X_1+\ldots+X_{10}$$.
We find $$E[X_i]=0⋅P(X_i=0)+1⋅P(X_i=1)=P(X_i=1)=9/100$$.
By the linearity of expectation, $$E[X]=10⋅E[Xi]=0.9$$
I understand this approach but I can't get the same answer by multiplying the probability of A and B having 1,2,3 objects in common by the values 1,2,3 and adding them, why is this approach concerned wrong ? I assumed x is the number of objects chosen by both thus it takes the values 1,2,3
• What do you get for the probabilities of 1, 2, or 3 objects in common? – paw88789 Feb 22 at 18:06
• The alternate approach you suggest can be correct (though it is often considered far less elegant and is more prone to error or difficulty). The error must be in your calculations themselves, which you have not yet shared. Show us your calculations and we'll tell you where your mistake is. – JMoravitz Feb 22 at 18:09
• $\frac{1}{\binom{10}{3}*\binom{10}{3}}\left [ 1*\binom{10}{1}\binom{9}{2}\binom{9}{2} + 2*\binom{10}{2}\binom{8}{1}\binom{8}{1} + 3*\binom{10}{3} \right ]$ I think I'm over counting somewhere here – Saratcı Feb 22 at 18:56
• @Saratci : disallow selection of the same "uncommon" item. $$\dfrac{\dbinom{10}{1}\dbinom{9}{2}\dbinom{7}{2}+2\dbinom{10}{2}\dbinom 81\dbinom 71+3\dbinom{10}3}{\dbinom{10}3^2}=\dfrac 9{10}$$ – Graham Kemp Feb 23 at 3:53
• Yes you are right, Thanks ! – Saratcı Feb 23 at 22:15 | 2021-03-05T13:18:32 | {
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https://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices | Jacobi's equality between complementary minors of inverse matrices
What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?
Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then
$$|\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$.
I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(
Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.
• Wow - I did not know that algebra proofs could have a "sexist bias". I am too curious to let it pass --- what do you mean exactly? – Federico Poloni Feb 8 '12 at 10:19
• I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word "boy" has a gender neutral meaning back then? – John Jiang Feb 8 '12 at 19:39
The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},$$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.
You may verify the following identity ("magic wand Schur complement formula") $$\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} = \begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix} \begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix} \begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}$$ By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).
Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.
• I guess for any thesis involving minors the Schur complement formula would be among the first things to try. – John Jiang Feb 8 '12 at 10:30
This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.
Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$ Assume WLOG that $W$ is invertible. On the one hand, we have (Schur C.F) $$\det A=\det W\cdot\det(Z-YW^{-1}X).$$ Finally, we have $$A^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result.
These formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \begin{pmatrix} I_* & 0 \\ YW^{-1} & I_* \end{pmatrix}$ and $U = \begin{pmatrix} W & X \\ 0 & Z-YW^{-1}X \end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).
• Your answer is equally valid. Thanks! – John Jiang Feb 8 '12 at 10:27
Not all has been said about this question that is worth saying -- at the very least, someone could have written down the version without the absolute values; but more importantly, there are various other equally good proofs.
Notations and statement
Let me first state the result with proper signs and no absolute values.
Standing assumptions. The following notations will be used throughout this post:
• Let $\mathbb{K}$ be a commutative ring. All matrices that appear in the following are matrices over $\mathbb{K}$.
• Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\}$.
• For every $n\in\mathbb{N}$, we let $\left[ n\right]$ denote the set $\left\{ 1,2,\ldots,n\right\}$.
• Fix $n\in\mathbb{N}$.
• Let $S_{n}$ denote the $n$-th symmetric group (i.e., the group of permutations of $\left[ n\right]$).
• If $A\in\mathbb{K}^{n\times n}$ is an $n\times n$-matrix, and if $I$ and $J$ are two subsets of $\left[ n\right]$, then $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix defined as follows: Write $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}$; write the set $I$ in the form $I=\left\{ i_{1}<i_{2}<\cdots<i_{u}\right\}$; write the set $J$ in the form $J=\left\{ j_{1}<j_{2}<\cdots<j_{v}\right\}$. Then, set $A_{J}^{I}=\left( a_{i_{x},j_{y}}\right) _{1\leq x\leq u,\ 1\leq y\leq v}$. (Thus, roughly speaking, $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix obtained from $A$ by removing all rows whose indices do not belong to $I$, and removing all columns whose indices do not belong to $J$.)
If $K$ is a subset of $\left[ n\right]$, then:
• we use $\widetilde{K}$ to denote the complement $\left[ n\right] \setminus K$ of this subset in $\left[ n\right]$.
• we use $\sum K$ to denote the sum of the elements of $K$.
Now, we claim the following:
Theorem 1 (Jacobi's complementary minor formula). Let $A\in\mathbb{K} ^{n\times n}$ be an invertible $n\times n$-matrix. Let $I$ and $J$ be two subsets of $\left[ n\right]$ such that $\left\vert I\right\vert =\left\vert J\right\vert$. Then,
$\det\left( A_{J}^{I}\right) =\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.
Three references
Here are three references to proofs of Theorem 1:
Note that every source uses different notations. What I call $A_{J}^{I}$ above is called $A_{IJ}$ in the paper by Caracciolo, Sokal and Sportiello, is called $A\left[ I,J\right]$ in Lalonde's paper, and is called $\operatorname*{sub}\nolimits_{w\left( I\right) }^{w\left( J\right) }A$ in my notes. Also, the $I$ and $J$ in the paper by Caracciolo, Sokal and Sportiello correspond to the $\widetilde{I}$ and $\widetilde{J}$ in Theorem 1 above.
A fourth proof
Let me now give a fourth proof, using exterior algebra. The proof is probably not new (the method is definitely not), but I find it instructive.
This proof would become a lot shorter if I didn't care for the signs and would only prove the weaker claim that $\det\left( A_{J}^{I}\right) = \pm \det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$ for some value of $\pm$. But this weaker claim is not as useful as Theorem 1 in its full version (in particular, it would not suffice to fill the gap in Macdonald's book that has motivated this question).
The permutation $w\left( K\right)$
Let us first introduce some more notations:
If $K$ is a subset of $\left[ n\right]$, and if $k = \left|K\right|$, then we let $w\left( K\right)$ be the (unique) permutation $\sigma\in S_{n}$ whose first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order, and whose next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots ,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order.
The first important property of $w\left( K\right)$ is the following fact:
Lemma 2. Let $K$ be a subset of $\left[ n\right]$. Then, $\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$.
You don't need to prove Lemma 2 if you only care about the weaker version of Theorem 1 with the $\pm$ sign.
Proof of Lemma 2. Let $k=\left\vert K\right\vert$. Let $a_{1},a_{2} ,\ldots,a_{k}$ be the $k$ elements of $K$ in increasing order (with no repetitions). Let $b_{1},b_{2},\ldots,b_{n-k}$ be the $n-k$ elements of $\widetilde{K}$ in increasing order (with no repetitions). Let $\gamma =w\left( K\right)$. Then, the definition of $w\left( K\right)$ shows that the first $k$ values $\gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( k\right)$ of $\gamma$ are the elements of $K$ in increasing order (that is, $a_{1},a_{2},\ldots,a_{k}$), and the next $n-k$ values $\gamma\left( k+1\right) ,\gamma\left( k+2\right) ,\ldots ,\gamma\left( n\right)$ of $\gamma$ are the elements of $\widetilde{K}$ in increasing order (that is, $b_{1},b_{2},\ldots,b_{n-k}$). In other words,
$\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots ,\gamma\left( n\right) \right) =\left( a_{1},a_{2},\ldots,a_{k} ,b_{1},b_{2},\ldots,b_{n-k}\right)$.
Now, you can obtain the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by successively switching adjacent entries, as follows:
• First, move the element $a_{1}$ to the front of the list, by successively switching it with each of the $a_{1}-1$ entries smaller than it.
• Then, move the element $a_{2}$ to the second position, by successively switching it with each of the $a_{2}-2$ entries (other than $a_{1}$) smaller than it.
• Then, move the element $a_{3}$ to the third position, by successively switching it with each of the $a_{3}-3$ entries (other than $a_{1}$ and $a_{2}$) smaller than it.
• And so on, until you finally move the element $a_{k}$ to the $k$-th position.
More formally, you are iterating over all $i\in\left\{ 1,2,\ldots,k\right\}$ (in increasing order), each time moving the element $a_{i}$ to the $i$-th position in the list, by successively switching it with each of the $a_{i}-i$ entries (other than $a_{1},a_{2},\ldots,a_{i-1}$) smaller than it.
At the end, the first $k$ positions of the list are filled with $a_{1} ,a_{2},\ldots,a_{k}$ (in this order), whereas the remaining $n-k$ positions are filled with the remaining entries $b_{1},b_{2},\ldots,b_{n-k}$ (again, in this order, because the switches have never disrupted their strictly-increasing relative order). Thus, at the end, your list is precisely $\left( a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{n-k}\right) =\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$. You have used a total of
$\left( a_{1}-1\right) +\left( a_{2}-2\right) +\cdots+\left( a_{k}-k\right)$
$=\underbrace{\left( a_{1}+a_{2}+\cdots+a_{k}\right) }_{\substack{=\sum K\\\text{(by the definition of }a_{1},a_{2},\ldots,a_{k}\text{)} }}-\underbrace{\left( 1+2+\cdots+k\right) }_{\substack{=1+2+\cdots +\left\vert K\right\vert \\\text{(since }k=\left\vert K\right\vert \text{)}}}$
$=\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$
switches. Thus, you have obtained the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ switches of adjacent entries. In other words, the permutation $\gamma$ is a composition of $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ simple transpositions (where a "simple transposition" means a transposition switching $u$ with $u+1$ for some $u$). Hence, it has sign $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$. This proves Lemma 2.
Exterior algebras
Now, let's introduce some more notations and state some well-known properties concerning exterior algebras.
For any $\mathbb{K}$-module $V$, we let $\wedge V$ denote the exterior algebra of $V$. The multiplication in this exterior algebra will be written as juxtaposition (i.e., we will write $ab$ for the product of two elements $a$ and $b$ of $\wedge V$) or as multiplication (i.e., we will write $a\cdot b$ for this product).
If $k\in\mathbb{N}$ and if $V$ is a $\mathbb{K}$-module, then $\wedge^{k}V$ shall mean the $k$-th exterior power of $V$. If $k\in\mathbb{N}$, if $V$ and $W$ are two $\mathbb{K}$-modules, and if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then the $\mathbb{K}$-linear map $\wedge^{k}V\rightarrow \wedge^{k}W$ canonically induced by $f$ will be denoted by $\wedge^{k}f$. It is well-known that if $V$ and $W$ are two $\mathbb{K}$-modules, if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then
(1) $\left( \wedge^{k}f\right) \left( a\right) \cdot\left( \wedge^{\ell}f\right) \left( b\right) =\left( \wedge^{k+\ell}f\right) \left( ab\right)$
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $a\in\wedge^{k}V$ and $b\in\wedge^{\ell}V$.
If $V$ is a $\mathbb{K}$-module, then
(2) $uv=\left( -1\right) ^{k\ell}vu$
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $u\in\wedge^{k}V$ and $v\in\wedge^{\ell}V$.
For any $u\in\mathbb{N}$, we consider $\mathbb{K}^{u}$ as the $\mathbb{K}$-module of column vectors with $u$ entries.
For any $u\in\mathbb{N}$ and $v\in\mathbb{N}$, and any $v\times u$-matrix $B\in\mathbb{K}^{v\times u}$, we define $f_{B}$ to be the $\mathbb{K}$-linear map $\mathbb{K}^{u}\rightarrow\mathbb{K}^{v}$ sending each $x\in\mathbb{K} ^{u}$ to $Bx\in\mathbb{K}^{v}$. This $\mathbb{K}$-linear map $f_{B}$ satisfies $\det\left( f_{B}\right) =\det B$, and is often identified with the matrix $B$ (though we will not identify it with $B$ here).
Here is another known fact:
Proposition 2a. Let $f:\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}$ be a $\mathbb{K}$-linear map. The map $\wedge^{n}f:\wedge^{n}\left( \mathbb{K} ^{n}\right) \rightarrow\wedge^{n}\left( \mathbb{K}^{n}\right)$ is multiplication by $\det f$. In other words, every $z\in\wedge^{n}\left( \mathbb{K}^{n}\right)$ satisfies
(3) $\left( \wedge^{n}f\right) \left( z\right) =\left( \det f\right) z$.
Let $\left( e_{1},e_{2},\ldots,e_{n}\right)$ be the standard basis of the $\mathbb{K}$-module $\mathbb{K}^{n}$. (Thus, $e_{i}$ is the column vector whose $i$-th entry is $1$ and whose all other entries are $0$.)
For every subset $K$ of $\left[ n\right]$, we define $e_{K}\in \wedge^{\left\vert K\right\vert }\left( \mathbb{K}^{n}\right)$ to be the element $e_{k_{1}}\wedge e_{k_{2}}\wedge\cdots\wedge e_{k_{\left\vert K\right\vert }}$, where $K$ is written in the form $K=\left\{ k_{1} <k_{2}<\cdots<k_{\left\vert K\right\vert }\right\}$.
For every $k\in\mathbb{N}$ and every set $S$, we let $\mathcal{P}_{k}\left( S \right)$ denote the set of all $k$-element subsets of $S$.
It is well-known that, for every $k\in\mathbb{N}$, the family $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{k}\left( \mathbb{K}^{n}\right)$. Applying this to $k=n$, we conclude that the family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$. Since this family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is the one-element family $\left( e_{\left[ n\right] }\right)$ (because the only $K\in\mathcal{P}_{n}\left( \left[ n\right] \right)$ is the set $\left[ n\right]$), this rewrites as follows: The one-element family $\left( e_{\left[ n\right] }\right)$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$.
If $B$ is an $n\times n$-matrix and $k\in\mathbb{N}$, then evaluating the map $\wedge^{k}f_{B}$ on the elements of the basis $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ of $\wedge ^{k}\left( \mathbb{K}^{n}\right)$, and expanding the results again in this basis gives rise to coefficients which are the $k\times k$-minors of $B$. More precisely:
Proposition 3. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then,
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$.
(This generalizes: If $u\in\mathbb{N}$, $v \in \mathbb{N}$, $B\in\mathbb{K}^{u\times v}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ v\right] \right)$, then $\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ u\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$, where the elements $e_{J}\in\wedge^{k}\left( \mathbb{K}^{v}\right)$ and $e_{I}\in\wedge^{k}\left( \mathbb{K}^{u}\right)$ are defined as before but with $v$ and $u$ instead of $n$.)
Extracting minors from the exterior algebra
Now, we shall need a simple lemma:
Lemma 4. Let $K$ be a subset of $\left[ n\right]$. Then,
$e_{K}e_{\widetilde{K}}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }$.
Proof of Lemma 4. Let $k = \left|K\right|$. Let $\sigma$ be the permutation $w\left( K\right) \in S_{n}$ defined above. Its first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order; thus, $e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }=e_{K}$. Its next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order; thus, $e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }=e_{\widetilde{K}}$.
From $\sigma=w\left( K\right)$, we obtain $\left( -1\right) ^{\sigma }=\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$ (by Lemma 2).
Now, it is well-known that
$e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge \cdots\wedge e_{\sigma\left( n\right) }=\left( -1\right) ^{\sigma }\underbrace{e_{1}\wedge e_{2}\wedge\cdots\wedge e_{n}}_{=e_{\left[ n\right] }}=\left( -1\right) ^{\sigma}e_{\left[ n\right] }$.
Hence,
$\left( -1\right) ^{\sigma}e_{\left[ n\right] }=e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }$
$=\underbrace{\left( e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }\right) }_{=e_{K} }\underbrace{\left( e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }\right) }_{=e_{\widetilde{K}}}=e_{K}e_{\widetilde{K}}$.
Since $\left( -1\right) ^{\sigma}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$, this rewrites as $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }= e_K e_{\widetilde{K}}$. This proves Lemma 4.
We can combine Proposition 3 and Lemma 4 to obtain the following fact:
Corollary 5. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then, every $K\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfies
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K} }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$.
Proof of Corollary 5. Let $K \in \mathcal{P}_{k}\left( \left[ n\right] \right)$. Let $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ be such that $I\neq K$. Then, $I\not \subseteq K$ (since the sets $I$ and $K$ have the same size $k$). Hence, there exists some $z\in I$ such that $z\notin K$. Consider this $z$. We have $z\in I$ and $z\in\widetilde{K}$ (since $z\notin K$). Hence, both $e_{I}$ and $e_{\widetilde{K}}$ are "wedge products" containing the factor $e_{z}$; therefore, the product $e_{I} e_{\widetilde{K}}$ is a "wedge product" containing this factor twice. Thus, $e_{I}e_{\widetilde{K}}=0$.
Now, forget that we fixed $I$. We thus have proven that
$e_{I}e_{\widetilde{K}}=0$ for every $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfying $I\neq K$.
Hence,
(4) $\sum\limits_{\substack{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) \underbrace{e_{I}e_{\widetilde{K}} }_{=0}=0$.
Proposition 3 yields
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) =\sum\limits_{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$.
Multiplying both sides of this equality by $e_{\widetilde{K}}$ from the right, we find
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K}} =\sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$
$=\det\left( B_{J}^{K}\right) e_{K}e_{\widetilde{K}}+\sum\limits_{\substack{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$
$=\det\left( B_{J}^{K}\right) \underbrace{e_{K}e_{\widetilde{K}} }_{\substack{=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }\\\text{(by Lemma 4)}}}$ (by (4))
$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$
$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$
(since $\left\vert K\right\vert =k$). This proves Corollary 5.
Corollary 5 is rather helpful when it comes to extracting a specific minor of a matrix $B$ from the maps $\wedge^{k}f_{B}$.
Proof of Theorem 1
Proof of Theorem 1. Set $k=\left\vert I\right\vert =\left\vert J\right\vert$. Notice that $\left\vert \widetilde{I}\right\vert =n-k$ (since $\left\vert I\right\vert =k$) and $\left\vert \widetilde{J}\right\vert =n-k$ (similarly).
Define $y\in\wedge ^{n-k}\left( \mathbb{K}^{n}\right)$ by $y=\left( \wedge^{n-k}f_{A^{-1} }\right) \left( e_{\widetilde{I}}\right)$.
The maps $f_{A}$ and $f_{A^{-1}}$ are mutually inverse (since the map $\mathbb{K}^{n\times n}\rightarrow\operatorname*{End}\left( \mathbb{K} ^{n}\right) ,\ B\mapsto f_{B}$ is a ring homomorphism). Hence, the maps $\wedge^{n-k}f_{A}$ and $\wedge^{n-k}f_{A^{-1}}$ are mutually inverse (since $\wedge^{n-k}$ is a functor). Thus, $\left( \wedge^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) =\operatorname*{id}$. Now, $y=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right)$, so that
$\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge ^{n-k}f_{A}\right) \left( \left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) \right) =\underbrace{\left( \left( \wedge ^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) \right) }_{=\operatorname*{id}}\left( e_{\widetilde{I}}\right) =e_{\widetilde{I}}$.
But (1) (applied to $V=\mathbb{K}^{n}$, $W=\mathbb{K}^{n}$, $f=f_{A}$, $\ell=n-k$, $a=e_{J}$ and $b=y$) yields
$\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) \cdot\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right)$.
Thus,
$\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\left( \wedge ^{k}f_{A}\right) \left( e_{J}\right) \cdot\underbrace{\left( \wedge ^{n-k}f_{A}\right) \left( y\right) }_{=e_{\widetilde{I}}}$
$=\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) e_{\widetilde{I}} = \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
(by Corollary 5, applied to $B=A$ and $K=I$).
Hence,
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
$=\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\underbrace{\left( \det f_{A}\right) }_{=\det A}e_{J}y$
(by (3), applied to $f=f_{A}$ and $z=e_{J}y$)
(5) $=\left( \det A\right) e_{J}y$.
But (2) (applied to $\ell=n-k$, $u=e_{J}$ and $v=y$) yields
$e_{J} y = \left(-1\right)^{k \left(n-k\right)} \underbrace{y}_{=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) } \underbrace{e_{J}} _{=e_{\widetilde{\widetilde{J}}}}$
$=\left( -1\right) ^{k\left( n-k\right) }\underbrace{\left( \wedge ^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) e_{\widetilde{\widetilde{J}}}}_{\substack{=\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }\\\text{(by Corollary 5, applied to }A^{-1}\text{, }n-k\text{, }\widetilde{I}\text{ and }\widetilde{J}\\\text{instead of }B\text{, }k\text{, }J\text{ and }K\text{)}}}$
$=\left( -1\right) ^{k\left( n-k\right) }\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$
(6) $=\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.
But
$k\left( n-k\right) +\underbrace{\sum\widetilde{J}}_{=\sum\left\{ 1,2,\ldots,n\right\} -\sum J}-\underbrace{\left( 1+2+\cdots+\left( n-k\right) \right) }_{=\sum\left\{ 1,2,\ldots,n-k\right\} }$
$=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,n\right\} -\sum J-\sum\left\{ 1,2,\ldots,n-k\right\}$
$=\underbrace{\sum\left\{ 1,2,\ldots,n\right\} -\sum\left\{ 1,2,\ldots ,n-k\right\} }_{\substack{=\sum\left\{ n-k+1,n-k+2,\ldots,n\right\} \\=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,k\right\} \\=k\left( n-k\right) +\left( 1+2+\cdots+k\right) }}+k\left( n-k\right) -\sum J$
$=2k\left( n-k\right) +\left( 1+2+\cdots+k\right) -\sum J$
$\equiv-\left( 1+2+\cdots+k\right) -\sum J\operatorname{mod}2$.
Hence,
$\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}$.
Thus, (6) rewrites as
$e_{J}y=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.
Hence, (5) rewrites as
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right) e_{\left[ n\right] }$.
We can "cancel" $e_{\left[ n\right] }$ from this equality (because if $\lambda$ and $\mu$ are two elements of $\mathbb{K}$ satisfying $\lambda e_{\left[ n\right] }=\mu e_{\left[ n\right] }$, then $\lambda=\mu$), and thus obtain
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right)$
$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.
Dividing this equality by $\left( -1\right) ^{\sum I-\left( 1+2+\cdots +k\right) }$, we obtain
$\det\left( A_{J}^{I}\right)$
$=\left(\det A\right) \dfrac{\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}}{\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$
$=\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$.
This proves Theorem 1.
I have to admit this proof looked far shorter on the scratch paper on which it was conceived than it has ended up here...
• If we agree to see a $\mathbb{K}$-matrix as a function $S\times S\to\mathbb{K}$ , with any finite index set $S$ (not just some $[m]$), then, according to Proposition 3, the matrix of the linear map $\Lambda^k(f_B)$ in the basis $\{e_I\}_{I\in\mathcal{P}_k([n])}$ is $[\det(B^I_J)]_{(I\times J)\in\mathcal{P}_k([n])\times \mathcal{P}_k([n])}$. – Pietro Majer Dec 19 '17 at 18:48
• So the functoriality of $\Lambda^k$ reads as the (generalized) Cauchy-Binet formula: en.wikipedia.org/wiki/Cauchy–Binet_formula#Generalization . In the same spirit, I'd like to see Theorem 1 as a statement on the inverse matrix of the linear map $\Lambda^k(f_A)$, or better, since it also refers to $\tilde I$ and $\tilde J$, of the linear map $\Lambda^k(f_A)^*\sim\Lambda^{n-k}(f_A)$. Can this produce a proof to Theorem 1 consisting in a plain translation of facts about exterior algebra in the language of matrices? – Pietro Majer Dec 19 '17 at 18:48
• @PietroMajer: This is what I was trying to achieve when I started writing the proof. Unfortunately, it's not directly clear to me how this works. – darij grinberg Dec 19 '17 at 19:54
Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).
WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product $$A \left[ \begin{array}{c|c|c|c|c|c} e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n \end{array} \right] = \left[ \begin{array}{c|c|c|c|c|c} A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n \end{array} \right]$$ which can also be written $$A \begin{pmatrix} I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c] \end{pmatrix} = \begin{pmatrix} A[I,J] & 0 \\ A[I^c,J] & I_{n-k} \end{pmatrix}$$ Taking the determinant of both sides yields $$(\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J]$$.
• Really nice... although a nontrivial part of my answer is justifying the "WLOG" in the first sentence. Still, this simplifies a lot. – darij grinberg Feb 17 '17 at 1:19
• I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\det(A[I,J])=\det((PAQ)[K,K])=\det(PAQ)\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \det(PAQ)\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\dots,k$ and $I^c$ to $k+1,\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\text{sign}(P)\text{sign}(Q)$ which is $(-1)^{\sum I + \sum J}$ by a counting argument, as you proved. – user1980685 Feb 17 '17 at 15:05
• I suspect that your $Q$ is not "defined similarly" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. – darij grinberg Feb 18 '17 at 5:42 | 2018-11-16T02:15:14 | {
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Let $${a_1} = 4500$$ = purchased value of the automobile. I have 50 rabbits. Geometric Progression is a type of sequence where each successive term is the result of multiplying a constant number to its preceding term. Copyright © 2005, 2020 - OnlineMathLearning.com. We can find the common ratio of a GP by finding the ratio between any two adjacent terms. Geometric growth is found in many real life scenarios such as population growth and the growth of an investment. problem solver below to practice various math topics. Application of geometric progression Example – 1 : If an amount ₹ 1000 deposited in the bank with annual interest rate 10% interest compounded annually, then find total amount at the end of first, second, third, forth and first years. A. Given first term (a), common ratio (r) and a integer n of the Geometric Progression series, the task is to print th n terms of the series. If the ball is dropped from 80 cm, In this tutorial we discuss the related problems of application of geometric sequence and geometric series. If the number of stores he owns doubles in number each month, what month will he launch 6,144 stores? Example: How many will I have in 15 weeks. They have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. An “interest” problem – application of Geometric Series: Question A man borrows a loan of$1,000,000 for a house from a bank and likes to pay back in 10 years (120 monthly instalments), the first instalment being paid at the end of first month and compound interest being calculated at 6% per annum. Bouncing ball application of a geometric sequence Related Pages $$= {a_1}\left( {0.85} \right) – {a_1}\left( {0.85} \right)\left( {\frac{{15}}{{85}}} \right) = {a_1}\left( {0.85} \right)\left[ {1 – \frac{{15}}{{100}}} \right]$$ $${S_n} = \frac{{{a_1}\left( {{r^n} – 1} \right)}}{{r – 1}}$$ (As $$r > 1$$) 7% increase every year. 16,847 answers. Example 7: Solving Application Problems with Geometric Sequences. years, each year getting 5% interest per annum. $$= {a_1}\left( {0.85} \right)\left( {1 – 0.85} \right) = {a_1}\left( {0.85} \right)\left( {0.85} \right) = {a_1}{\left( {0.85} \right)^2}$$, Similarly, the value at the end of the third year The geometric series is a marvel of mathematics which rules much of the natural world. GEOMETRIC SERIES AND THREE APPLICATIONS 33 But the sum 9 10 + 9 100 + 9 1000 + 9 is a geometric series with rst term a = 10 and ratio r = 1 10.The ratio r is between 1 and 1, so we can use the formula for a geometric series: The 3rd and the 8th term of a G. P. are 4 and 128 respectively. There are many uses of geometric sequences in everyday life, but one of the most common is in calculating interest earned. fare of taxi, finding multiples of numbers within a range. In the following series, the numerators are … they sell on day 7? Estimate the student population in 2020. Geometric series have applications in math and science and are one of the simplest examples of infinite series with finite sums. Compounding Interest and other Geometric Sequence Word Problems. Your salary for the first year is $43,125. Their daily goal Use properties of exponents (such as power of a power, Geometric series are used throughout mathematics. Speed of an aircraft, finding the sum of n terms of natural numbers. Ashley Kannan. The geometric series is a marvel of mathematics which rules much of the world. Remember these examples Application of a Geometric Sequence. a. An arithmetic sequence has a common difference of 9 and a(41) = 25. Solution: This chapter is for those who want to see applications of arithmetic and geometric progressions to real life. Write the equation that represents the house’s value over time. Example: I decide to run a rabbit farm. explain properties of the quantity represented by the expression. etc.) A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The common ratio (r) is obtained … Factor a quadratic expression to reveal the zeros of Solve Word Problems using Geometric Sequences. The value of an automobile depreciates at the rate of $$15\%$$ per year. How much will we end up with? For example, the sequence 2, 4, 8, 16, … 2, 4, 8, 16, \dots 2, 4, 8, 1 6, … is a geometric sequence with common ratio 2 2 2. Write a formula for the student population. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. height from which it was dropped. For example, the expression 1.15t can Growth. Geometric growth occurs when the common ratio is greater than 1, that is . Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. If the shortest leng In this tutorial we discuss the related problems of application of geometric sequence and geometric series. In 2013, the number of students in a small school is 284. Application of Geometric Sequence and Series. A line is divided into six parts forming a geometric sequence. Each year, it increases 2% of its value. How does this On January 1, Abby’s troop sold three boxes of Girl Scout cookies online. The sequence allows a borrower to know the amount his bank expects him to pay back using simple interest. $$= {a_1}{\left( {0.85} \right)^3}$$, Hence, the value of the automobile at the end of $$3$$ years after being purchased for $$4500$$ $$96 = 3 \times {r^5}$$ $$\Rightarrow r = 2$$, For the length of the whole line, we have 5. Solution: It is finance; however, the geometric series finds perhaps its greatest predictive We welcome your feedback, comments and questions about this site or page. Example: Bouncing ball application of a geometric sequence When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. reveal and explain specific information about its approximate Find a rule for this arithmetic … Learn more about the formula of nth term, sum of GP with examples at BYJU’S. Geometric Series A pure geometric series or geometric progression is one where the ratio, r, between successive terms is a constant. Required fields are marked *. This is another example of a geometric sequence. The value of automobile at the end of the first year When a ball is dropped onto a flat floor, it bounces to 65% of the and produce an equivalent form of an expression to reveal and B. How much will your salary be at the start of year six? Find the G. P. A. Embedded content, if any, are copyrights of their respective owners. 2,3, 4,5. You will receive In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. $$= 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45$$, Your email address will not be published. b. In Generalwe write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: 1. ais the first term, and 2. r is the factor between the terms (called the "common ratio") But be careful, rshould not be 0: 1. 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https://math.stackexchange.com/questions/88565/true-or-false-x2-ne-x-implies-x-ne-1/88580 | # True or false? $x^2\ne x\implies x\ne 1$
Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:
$$x^2\ne x\implies x\ne 1$$
I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$ and because $$0$$ isn't excluded as a possible value of $$x$$, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.
Why I think the logical sentence above is true:
My understanding of the implication symbol $$\implies$$ is the following: If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether $$x$$ can be $$0$$. Maybe $$x$$ can't be $$-\pi i$$ too, but as I see it, it doesn't really matter, as long as $$x \ne 1$$ holds. And it always holds when $$x^2 \ne x$$, therefore the sentence is true.
### TL;DR:
$$x^2 \ne x \implies x \ne 1$$: Is this sentence true or false, and why?
Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.
• This is true, as the contrapositive ($x = 1$ -> $x^2=x$) is obviously true. Dec 5 '11 at 14:20
• They are wrong-the fact that $x\neq 0$ is also an implication doesn't mean anything. The statement: $x^2=x\implies x=1$ is false. Dec 5 '11 at 14:21
• Also, their reasoning about "not excluding other values" is wrong. Today is Monday, which implies SO many things (!), but it is not false to just list one implication. Eg: If today is Monday, then tomorrow is Tuesday. Or, if today is Monday, then I have an appointment with the dentist. Dec 5 '11 at 14:23
• @Chris: You could write your reasoning as answer. That way the site keeps working better (and you will get some upvotes). You could also refer your teacher to this site :-) Dec 5 '11 at 15:06
• Your understanding of material implication ($\implies$) is exactly right, and as Jyrki said, your reasoning could stand as a perfectly good answer to the question. You could also point out that the teacher and class seem to be confusing $\implies$ and $\iff$: $x^2\ne x\iff x\ne 1$ is of course false precisely because $1$ isn't the only number that is its own square. Dec 5 '11 at 15:34
The short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.
But in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:
"If a nuclear bomb drops on the school building, you die."
"Hey, but you die, too."
"That doesn't help you much, though, so it is still true that you die."
"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies."
"You cannot buy milk and bread, either!"
"Yes, but I prefer to concentrate on the major consequences."
"If you sign this contract, you get a free pen."
"Hey, you didn't tell me that you get all my money."
Non-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose.
I suggest that you learn about some nonintuitive probability results and make bets with your teacher.
• I like how the importance of cookies appears in surprising math/logic results! (+1). It might be time to show the teacher this question. Dec 5 '11 at 16:10
• Excellent examples, thank you! And of course I already made bets :). Dec 5 '11 at 16:29
• Superb answer, which implies that I have up-voted it. Jan 10 '12 at 0:37
First, some general remarks about logical implications/conditional statements.
1. As you know, $P \rightarrow Q$ is true when $P$ is false, or when $Q$ is true.
2. As mentioned in the comments, the contrapositive of the implication $P \rightarrow Q$, written $\lnot Q \rightarrow \lnot P$, is logically equivalent to the implication.
3. It is possible to write implications with merely the "or" operator. Namely, $P \rightarrow Q$ is equivalent to $\lnot P\text{ or }Q$, or in symbols, $\lnot P\lor Q$.
Now we can look at your specific case, using the above approaches.
1. If $P$ is false, ie if $x^2 \neq x$ is false (so $x^2 = x$ ), then the statement is true, so we assume that $P$ is true. So, as a statement, $x^2 = x$ is false. Your teacher and classmates are rightly convinced that $x^2 = x$ is equivalent to ($x = 1$ or $x =0\;$), and we will use this here. If $P$ is true, then ($x=1\text{ or }x =0\;$) is false. In other words, ($x=1$) AND ($x=0\;$) are both false. I.e., ($x \neq 1$) and ($x \neq 0\;$) are true. I.e., if $P$, then $Q$.
2. The contrapositive is $x = 1 \rightarrow x^2 = x$. True.
3. We use the "sufficiency of or" to write our conditional as: $$\lnot(x^2 \neq x)\lor x \neq 1\;.$$ That is, $x^2 = x$ or $x \neq 1$, which is $$(x = 1\text{ or }x =0)\text{ or }x \neq 1,$$ which is $$(x = 1\text{ or }x \neq 1)\text{ or }x = 0\;,$$ which is $$(\text{TRUE})\text{ or }x = 0\;,$$ which is true.
• Please pardon the (lack of) formatting- I typed this up on my phone! Dec 5 '11 at 15:36
• Thanks, Brian! That looks great. Dec 5 '11 at 15:55
• That's the proof I was looking for, thank you :) Dec 5 '11 at 16:41
Thing to note. This is called logical implication.
$x^2≠x⟹x≠1$: Is this sentence true or false, and why?
We can always check that using an example. Let us look at this implication as $\rm P\implies Q$. Now we shall consider cases:
• Case 1: If we consider $x = 0$, then $\rm P$ is false, and $\rm Q$ is true.
• Case 2: If we consider $x = 1$, then $\rm P$ is false, and $\rm Q$ is false as well.
• Case 3: If we consider each value except $x=0$ and $x = 1$, then both $\rm P$ and $\rm Q$ will be true since $x^2 = x \iff x^2 - x = 0 \iff x(x - 1) = 0$ which means that $x=0$ and $x = 1$ are the only possibilities.
Fortunately, our truth tables tell us that logical implication will hold true as far as we are not having $\rm P$ true and $\rm Q$ false. Look at the cases above; none of them has $\rm P$ true and $Q$ false. Thus Case 1, Case 2 and Case 3 are all true according to mathematical logic, so $\rm P\implies Q$ is true, or in other words: $x^2 \ne x \implies x \ne 1$ is true.
I apologize for being late, but I always have my two cents to offer... thank you.
## Introduction
You asked whether the following statement is true or not:
$$x^2 \ne x \implies x \ne 1$$
Another way to write the above is as:
if $$(x^2 \ne x)$$ then $$(x \ne 1)$$
The arrow symbol ($$\implies$$) is similar, but not quite the same, as the English phrase if....then....
## The contra-positive of $$(P \implies Q)$$
Okay now: suppose that you have the statement of the form "If $$P$$ then $$Q$$"
$$P$$ and $$Q$$ can be any true or false statements of your choice.
For example, $$P$$ could be the statement "pillows are soft"
"If $$P$$ then $$Q$$" is also sometimes written as $$(P \implies Q)$$
There is a well known theorem in logic which states the following:
For any $$P$$ and $$Q$$ taken from the set $$\{$$ true, false $$\}$$, the following is true:
$$(\text{not } P \implies \text{not } Q)$$ if and only if $$(Q \implies P)$$
That means that the following two statements are equivalent:
if $$(x^2 \ne x)$$ then $$(x \ne 1)$$
if $$(x = 1)$$ then $$(x^2 = x)$$
In some sense, the statement says that $$1^2 = 1$$, which is true enough.
## Sets
It considered to be very bad math to write things like the following:
Is it true or false that $$(x^2 \ne x \implies x \ne 1)$$?
Find $$x$$ such that $$3 + (x-5)^{2} = 0$$
This is because the above examples fail to explain what $$x$$ is.
• Is $$x$$ a whole number, such as $$1, 2, 3, \dots, 98, 99, 100$$?
• Is $$x$$ a decimal number, such as $$\sqrt{2}$$
• Is $$x$$ a non-real complex number, such as $$20 + 5*i$$?
You are supposed to write things like this instead:
Is the following true or false?
• For every real number $$x, (x^2 \ne x \implies x \ne 1)$$
Find every complex non-real number $$x$$ such that:
• $$3 + (x-5)^{2} = 0$$
My last example uses the non-real number $$i$$
$$i*i = -1$$
I think if you think it will help if you start using "sets."
The following is an example of a set:
my_set $$= \{1, 3, 6, 7, 22\}$$
A set is like a suitcase full of clothing A set is also like a cookie jar, or cardboard box.
A set is a container.
A suitcase might contain a t-shirt. Well, my_set contains the numbers $$1, 3, 6, 7,$$ and $$22$$.
The number $$3$$ is like a t-shirt in the sense that the number $$3$$ is inside the suitcase.
I went to public school in the United States.
I did not see sets until I was in college.
Sets are basic, basic math. sets are more basic than knowing how to compute $$4.5/0.3$$
You are not allowed to write math like $$x^2 \ne x \implies x \ne 1$$ unless you first tell the reader what set $$x$$ comes from.
The following is a very ugly formula for a function named $$WEIRD$$:
$$\text{WEIRD_FUNC}(X) = [1-w(x)]*\text{LEFT_PIECE}(X) + [w(x)]*\text{RIGHT_PIECE}(X)$$
$$\text{LEFT_PIECE}(X) = (x+10)*(x+5)$$
$$\text{RIGHT_PIECE}(X) = 3 + (x-7)^{2}$$
$$W(x) = \frac{tanh(10)}{2} + \frac{tanh(x)}{2}$$
A plot of the weird function is shown below:
Suppose I asked you,
"Find all $$x$$ such that $$WEIRD(x) = 0$$"
The answers vary depending on which set $$x$$ is taken from.
The set of all integers $$x$$ such that $$WEIRD(X) = 0$$ is $$\{-10\}$$
The set of all real numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\{-10, -5.01\}$$
The set of all complex numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\{-10, -5.01, 7+i*\sqrt{3}, 7-i*\sqrt{3}\}$$
### Analyzing your classmates argument regarding the number zero
Is the following statement true or not?
$$x^2 \ne x \implies x \ne 1$$
The following is your description of your teacher, and/or classmates, reasoning:
1. When $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$.
2. $$0$$ is not excluded as a possible value of $$x$$
3. some sentence, or another, is false.
That description is very difficult to understand.
I will say that the following three statements are equivalent to each-other.
Also, all statements are false:
$$\forall x \in \mathbb{R}, x^2 \ne x$$ for any decimal number $$x$$, $$x^2 \ne x$$ if $$x$$ is a real number, then $$x^2 \ne x$$
Here is a proof:
Line No. statement justification
0 $$0$$ exists axiom
1 $$0$$ is a decimal number. axiom
2 $$0^{2} = 0$$ axiom
3 not $$(0^2 \neq 0)$$ from line 2
4 there exists a decimal number $$x$$ such that not $$(x^2 \neq x)$$ from lines 0,1,3
5 not for every decimal number $$x$$, $$(x^2 \neq x)$$ from line 4
Notice that your teacher said, "possible value of $$x$$"
In mathematics, the symbol is sometimes used as short-hand notation for the word "possible"
All of the following are logically equivalent:
• it is not possible for me to see a movie this weekend
• $$NOT$$ for me to see a movie this weekend
• It is necessary for me to NOT see a movie this weekend
• for me to NOT see a movie this weekend
The statement "$$0$$ is not excluded as a possible value of $$x$$" can be written as:
• not(not $$x = 0$$)
• It is not the case that it is not possible that $$x$$ is zero.
The above is equivalent to "it is possible that $$x = 0$$"
I think that I can write a proof outline of what your professor and/or classmates were trying to say
Line No. statement justification
$$0$$ $$x = 0$$ axiom
$$1$$ $$(x^2 \ne x) \implies$$ not $$x = 0$$ axiom
$$2$$ $$x^2 \ne x$$ axiom
$$3$$ not $$x = 0$$ 1,2 modus ponens
4 $$x = 0$$ and not $$x = 0$$ lines 0, 3 conjunction
5 $$NOT$$ $$(x^2 \ne x)$$ from line 4, reject 2
I think that your teacher was saying that if $$x = 0$$ is possible, then it is not necessary that $$x^2 \ne x$$
$$0 \in S \implies NOT(\forall x \in S, x^2 \ne x)$$
### One last Note
All of the following are equivalent:
• (it is necessary that $$x^2 \neq x$$) implies that (it is necessary that $$x \neq 1$$)
• (it is NOT necessary that $$x \neq 1$$) implies that (it is NOT necessary that $$x^2 \neq x$$)
• (it is possible that $$x = 1$$) implies that (it is possible that $$x^2 = x$$)
• For any set $$S$$, [(there exists $$x$$ in $$S$$ such that $$x = 1$$) implies that (there exists $$x$$ in $$S$$ such that $$x^2 = x$$)] | 2021-10-25T16:44:41 | {
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https://stats.stackexchange.com/questions/463189/what-is-the-distribution-of-a-mixture-of-exponential-distributions-whose-rate-pa | # What is the distribution of a mixture of exponential distributions whose rate parameters follow a gamma distribution?
I want to know the theoretical distribution of a mixture of exponential distributions whose rate parameters are distributed according to a gamma distribution:
$$y\sim\text{Exp}(\theta), \quad\text{where}\quad \theta\sim\Gamma(r, \beta).$$
Specifically, I am looking at $$r=4$$ and $$\beta=2$$.
• The question appears clear enough to me: given $r$ and $\beta$, let $\theta$ follow a gamma distribution with parameters $r$ and $\beta$ (whether these are scale or rate params doesn't matter for now), denoted $\theta\sim\Gamma(r, \beta)$. Next, $y$ is exponentially distributed with parameter $\theta$, denoted $y\sim\text{Exp}(\theta)$. That is, we have a compound distribution, AKA a mixture. Q: What is the unconditional distribution of $y$ with parameters $r$ and $\beta$? – Stephan Kolassa May 1 at 5:29
• At least that is what the original question seems to have asked, and I have tried to clarify this specific question with my edit and answer it. I will vote to reopen. Can someone clarify to me where this question is unclear? – Stephan Kolassa May 1 at 5:30
• @StephanKolassa: Well, $\theta$ wasn't defined; it could be the mean rather than the rate. The variable $n$ & its role were also unexplained. – Scortchi - Reinstate Monica May 2 at 8:57
• @Scortchi-ReinstateMonica: true. But converting between a mean and a rate formulation for the exponential is not overly hard. I assumed $n$ referred to a sample size and removed it in my edit. – Stephan Kolassa May 2 at 9:11
• @StephanKolassa: (1) The compound distribution would be something else if it were the means distributed according to a gamma distribution. (2) Quite possibly, but why mention it then? I think your edits do a fine job of clarification, but I can see why the q. was closed in its original version. – Scortchi - Reinstate Monica May 2 at 9:19
For all $$y \ge 0$$ the value of the survival function of $$Y$$ is
$$S_Y(y\mid\theta) = \Pr(Y \gt y\mid \theta) = \exp(-y\theta)$$
and, taking $$\beta$$ to be a rate parameter for the Gamma distribution, the probability density function of $$\theta$$ is proportional to
$$f_\theta(t) \propto t^{r-1}\exp(-\beta t).$$
Consequently the survival function of the mixture distribution is
$$S(y) \propto \int_0^\infty S_Y(y\mid t) f_\theta(t)\,\mathrm{d}t = \int_0^\infty t^{r-1}\exp(-(\beta+y)t)\,\mathrm{d}t.$$
Substituting $$u = (\beta+y) t$$ gives, with no calculation,
$$S(y) \propto \int_0^\infty \left(\frac{u}{\beta+y}\right)^{r-1}\exp(-u)\,\mathrm{d}\left(\frac{u}{\beta+y}\right) = (\beta+y)^{-r}\int_0^\infty u^{r-1}\exp(-u)\,\mathrm{d}u$$
which is proportional to $$(\beta+y)^{-r}.$$
The axiom of total probability asserts $$S(0)=1$$ from which we obtain the implicit constant, giving
$$S(y) = \beta^r\,(\beta+y)^{-r} = \left(1 + \frac{y}{\beta}\right)^{-r},$$
thereby exhibiting $$\beta$$ as a scale parameter for this mixture variable.
This is a particular kind of Beta prime distribution, sometimes termed a Lomax distribution. (up to scale).
If $$\beta$$ is intended to be a scale parameter for the Gamma distribution rather than a rate parameter, replace $$\beta$$ everywhere by $$1/\beta$$ and, at the end, interpret it as a rate parameter for the mixture variable.
This distribution tends to be positively skewed, so it's better to view the distribution of $$\log Y.$$ Here are the empirical (black) and theoretical (red) distributions for a simulation of $$10^4$$ independent realizations of $$Y$$ (where $$r=4$$ and $$\beta=2$$ as in the question):
The agreement is perfect.
The R code that generated this figure illustrates how to code the survival function (S), the density function (its derivative f), and how to simulate from this compound distribution by first producing realizations of $$\theta$$ and then, for each of them, generating a realization of $$Y$$ conditional on $$\theta.$$
S <- function(y, r, beta) 1 / (1 + y/beta)^r # Survival
f <- function(y, r, beta) r * beta^r / (beta + y)^(r+1) # Density
#
# Specify the parameters and simulation size.
#
beta <- 2
r <- 4
n.sim <- 1e4
#
# The simulation.
#
theta <- rgamma(n.sim, r, rate=beta)
y <- rgamma(n.sim, 1, theta)
#
# The plots.
#
par(mfrow=c(1,2))
plot(ecdf(log(y)), xlab=expression(log(y)), ylab="Probability",
main=expression(1-S[Y](y)))
curve(1 - S(exp(y), r, beta), xname="y", add=TRUE, col="Red", lwd=2)
hist(log(y), freq=FALSE, breaks=30, col="#f0f0f0", xlab=expression(log(y)))
curve(f(exp(y), r, beta) * exp(y), xname="y", add=TRUE, col="Red", lwd=2)
par(mfrow=c(1,1))
Wikipedia, under the "Examples" section, informs us that:
Compounding an exponential distribution with its rate parameter distributed according to a gamma distribution yields a Lomax distribution [9].
The reference [9] is to Johnson, N. L.; Kotz, S.; Balakrishnan, N. (1994). "20 Pareto distributions". Continuous univariate distributions. 1 (2nd ed.). New York: Wiley. p. 573
• Yuh, but is that really the question? – Carl May 1 at 4:10 | 2020-08-07T09:40:42 | {
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https://math.stackexchange.com/questions/1211713/prove-that-there-is-a-multiple-of-2009-that-ends-with-the-digits-000001 | # Prove that there is a multiple of 2009 that ends with the digits 000001
Prove that there is a multiple of 2009 that ends with the digits 000001.
May one generalise this to: There exists a multiple of $x$ that ends with the digits $y$ (where $y$ consists of $n$ digits) if $x$ and $10^n$ are relatively prime?
The proof may be constructive, or non constructive.
Any help would be greatly appreciated.
I figured the multiple has to end in a $9$: $2009 \cdot 889$ gives $001$ as the three ending digits. I then tried to see if I could get the $4$th last digit to be a zero, but I must admit I am clearly missing something at that point.
• Write down a few multiples of $2009$. Maybe try the first $246889$ :P Mar 29 '15 at 17:22
• Well it needs to end in a one, so I figured the multiple has to end in a 9. 2009 * 889 gives 001 as the three ending digits. I then tried to see if I could get the 4th last digit to be a zero, but I must admit I am clearly missing something at that point. Mar 29 '15 at 17:24
• @flabby99 Well, both hints in the answer work and boil down to the same. Another answer is hidden somewhere here ;) Mar 29 '15 at 17:26
• @AlexR Yes I see that there is some magical number within these comments. However, I would like to understand better how to obtain such a magic number :P Mar 29 '15 at 17:27
• @flabby99 It's from the extended euclidean algorithm to $\gcd(1000000,2009) = 1$. Mar 29 '15 at 17:28
## 3 Answers
Here is a short proof using modular arithmetic.
Consider $m=1000000$. The numbers $2009$ and $m$ are relatively prime to each other, so
$$2009^{\phi(m)}\equiv 1\pmod m$$
where $\phi(m)$ is the count of the numbers between $1$ and $m$ that are relatively prime to $m$. $2009$ divides $2009^{\phi(m)}$ so that proves your statement.
Of course, this is not the smallest such multiple of $2009$. To find that, use the extended Euclidean algorithm on $2009$ and $1000000$.
As I wrote, the key is that $2009$ and $1000000$ are relatively prime.
• How to nuke-a-fly: Apply totient function where an efficient algorithm does your work ^^ Mar 29 '15 at 17:29
• @AlexR: The OP just asked for a proof, not an efficient algorithm. Anyway, I modified my post (before I saw your comment) to also refer to that algorithm. The OP can decide which is better suited to his purpose. Mar 29 '15 at 17:31
Hint
You must show that there exists $k$ such that $$2009 k \equiv 1 \pmod{1000000}$$ i.e. that $2009$ has a multiplicative inverse in $\mathbb Z_{1000000}^\times$. Do you know of some criteria?
If you want to obtain $k$, the easiest way is the extended euclidean algorithm on $1000000$ and $2009$. It will give you numbers $x,y$ such that $$2009 x + 1000000 y = \gcd(1000000,2009)$$
• 2009 must be relatively prime to 1000000 ( to satisfy the mentioned criteria). The Euclidean algorithm should handle that without too much difficulty. Is my thought process correct? Mar 29 '15 at 17:30
• @flabby99 Yes, perfect! The extended one will even provide you with $k$ for free. Mar 29 '15 at 17:30
• @flabby99 if you are just looking for a non constructive proof, even the Euclidean algorithm is overkill - just observe that $2009$ is not divisible by $2$ or $5$ Mar 29 '15 at 17:39
• @Mathmo123 I must admit, I am missing how the proof follows just from observing that 2009 is not divisible by 2 or 5. Mar 29 '15 at 17:48
• @flabby99 this is just a faster way of showing the numbers are relatively prime, since $1000000$ is divisible by only the primes $2$ and $5$ Mar 29 '15 at 17:49
Hint: Show that you can apply the Chinese Remainder theorem to $$x \equiv 0 \pmod {2009}\\x \equiv 1\pmod {1000000}$$
• +1. Only flabby99 can tell us if this problem is in his textbook after a discussion of the Chinese Remainder Theorem. Mar 29 '15 at 17:43
• It is a general problem from elementary number theory from an optional set of problems by one of my lecturers, and may be solved using whatever is within our grasp. I am however, aware of the Chinese Remainder Theorem. I just failed to see it's application in this situation. Mar 29 '15 at 17:50 | 2022-01-26T21:24:23 | {
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http://math.stackexchange.com/questions/92401/unconventional-way-to-solve-trig-question | # Unconventional way to solve trig question
I was working on a trig question and got stuck, but then I noticed a possible way to solve the problem. However, this way seemed to be slightly unconventional and possibly not what the book was looking for.
The question was: "Find $k$ and $b$ when $\sqrt3 \sin x + \cos x=k\sin(x+b)$" The first thing I did was to expand $k\sin(x+b)$ to $k\sin x\cos b+k\cos x\sin b$. Here is where I got stuck. I tried several different things, but I hit dead ends. Then I tried to equate coefficients and got $k\cos b=\sqrt3$ and $k\sin b=1$ which simplified to: $\cos b=\displaystyle{\sqrt3 \over k}$ and $\sin b=\displaystyle{1\over k}$. The answer from there is fairly simple to get which is $k=2$ and $b=\displaystyle{\pi\over6}$.
However this method seems rather dubious and so I was wondering if someone new a better way of solving the problem using more rigorous mathematical methods.
-
What makes it seem like a guessing game? Also, I believe you missed a solution. – Mike Dec 18 '11 at 1:54
$k^2=4$ has two solutions. – Henry Dec 18 '11 at 1:54
Good point. I just thought that using that method might not work all the, but I guess it might. – E.O. Dec 18 '11 at 1:56
Equating coefficients works here, but can be avoided by looking at the cases $x=0$ and $x=\pi/2$ and checking the final result. – Henry Dec 18 '11 at 1:56
It's not at all dubious. The functions sin and cos are linearly independent and so $a_1 \sin x + b_1 \cos x = a_2 \sin x + b_2 \cos x$ for all $x$ iff $a_1=a_2$ and $b_1=b_2$.
To see that sin and cos are linearly independent, take $a \sin x + b \cos x = 0$ and set $x=0$ to get $b=0$ and $x=\pi/2$ to get $a=0$.
-
This appears to be the only answer that explains why the OP's method is valid. +1 – Graphth Dec 18 '11 at 2:46
what about the case where $x=\pi /3$ – Ramana Venkata Dec 18 '11 at 5:17
I'm not sure how it's perceived as dubious. If you have $k \cos b = \sqrt{3}$ and $k \sin b = 1$, then you must $\frac{k \sin b}{k \cos b} = \tan b = \frac{1}{\sqrt{3}}$, which gets you $b = \frac{\pi}{6}$ or $\frac{7\pi}{6}$. From there it immediately follows that $k = 2$ or $-2$, respectively.
-
If you have $k\cos b=\sqrt{3}$ and $k\sin b=1$, you should also remember that $k^2\cos^2 b+ k^2\sin^2 b= k^2$. That means $\sqrt{3}^2 + 1^2 = k^2$, so $k=\pm 2$. If you take $k=2$, you can find a suitable value of $b$, and with $k=-2$, you can find another.
-
Suppose that $a\sin(x)+b\cos(x)=c\sin(x)+d\cos(x)$ holds for all $x$ in some neighborhood of $x_0$. Then we have the following equation and its derivative: \begin{align} 0&=(a-c)\sin(x_0)+(b-d)\cos(x_0)\tag{1}\\ 0&=(a-c)\cos(x_0)-(b-d)\sin(x_0)\tag{2} \end{align} Adding the squares of $(1)$ and $(2)$ gives $$0=(a-c)^2+(b-d)^2\tag{3}$$ Thus, $a=c$ and $b=d$.
So if we have \begin{align} \sqrt{3}\sin(x)+\cos(x) &=k\sin(x+b)\\ &=k\cos(b)\sin(x)+k\sin(b)\cos(x)\tag{4} \end{align} for all $x$ in some neighborhood of $x_0$, then we must have $$k\cos(b)=\sqrt{3}\quad\text{ and }\quad k\sin(b)=1\tag{5}$$ and $(5)$ was solved above.
-
Here is what I come up in my mind, maybe this is an unconventional way too: Divide both sides by $2$, we get $$\frac{\sqrt3}{2} \sin x +\frac{1}{2}\cos x=\frac{k}{2}\sin(x+b).$$ Note that $\displaystyle\sin(\frac{\pi}{6})=\frac{1}{2}$ and $\displaystyle\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, the left hand side is given by $$\cos(\frac{\pi}{6})\sin x +\sin(\frac{\pi}{6})\cos x=\sin(x+\frac{\pi}{6}),$$ where the last equality follows from compound angle forumula. Combining these, we have $$\sin(x+\frac{\pi}{6})=\frac{k}{2}\sin(x+b).$$
-
Here is a standard trick to deal with expressions of the type $A \sin(x) + \cos(x)$:
Let $u$ be so that $\cot(u) =A$.
Then, after bringing everything to the same denominator, your expression becomes
$$A \sin(x) + \cos(x)= A\frac{\cos(u) \sin(x)+\sin(u)\cos(x)}{ \sin(u)} =\frac{ \sin(x+u)}{\sin(u) }$$
If you look over the other answers, this is exactly what they did, in a non-explicit way.
Remark If instead you use $\tan(v)=A$, you end up with
$$A \sin(x) + \cos(x) = \frac{\cos(x-v)}{\cos(v)} \,.$$
Same substitutions also solve the expressions of the type $\sin(x) + B\cos(x)$.
-
In general, \begin{align*} \alpha \sin(x) + \beta \cos(x) &= \sqrt{\alpha^2+\beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}\sin(x) + \frac{\beta}{\sqrt{\alpha^2+\beta^2}}\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\sin(x+y) \end{align*} where $y$ is the angle whose cosine is $\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}}$ and whose sine is $\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}$. In your problem, $\alpha = \sqrt{3}, \beta = 1$, which readily gives $k = 2$, $b = \frac{\pi}{6}$. As others have pointed out, this misses one of the two solutions.
If $\sqrt{\alpha^2+\beta^2}$ (which conventionally means the positive square root) is replaced throughout by $({\alpha^2+\beta^2})^{1/2}$, we get the other solution $k = -2, b = \frac{7\pi}{6}$ upon choosing $({\alpha^2+\beta^2})^{1/2} = -2$.
- | 2014-09-01T18:56:26 | {
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https://math.stackexchange.com/questions/2911922/a-problem-about-arithmetic-and-geometric-sequences | # A Problem About Arithmetic And Geometric Sequences
The problem I'm trying to solve is this:
The first three terms of a geometric sequence are also the first, eleventh and sixteenth terms of an arithmetic sequence. The terms of the geometric sequence are all different. The sum to infinity of the geometric sequence is 18. Find the common ration of the geometric sequence, and the common difference of the arithmetic sequence.
What I've done so far is to write the following equations, where u is the first term, r is the common ratio, and d is the difference:
$ur=u+10d$
$ur^2=u+15d$
$u/(1-r)=18$
But I don't know what to do from there. Any suggestions?
• Please type your questions rather than posting an image. Images cannot be searched. Sep 10 '18 at 13:42
• @N.F.Taussig Okay, I've done that.
– user590211
Sep 10 '18 at 13:44
You will get
$$a_2=a_1q,a_3=a_1q^2$$ and $$\sum_{k=0}^\infty a_1 q^k=a_1\frac{1}{1-q}$$ and $$a_1=b_1,a_2=b _1+10d,a_3=b_1+15d$$ Can you proceed? Using your equation you will get $$a_1=b_1$$ $$a_q=b_1+10d$$ $$a_1q^2=b_1+15d$$ Since $$a_1=b_1$$ we obtain $$a_1q=a_1+10d$$ $$a_1q^2=a_1+15d$$
eliminating $q$ we get $$a_1\left(\frac{a_1+10d}{a_1}\right)^2=a_1+15d$$
From here we get the equation $$5a_1d+100d^2=0$$
so $$a_1=-20d$$ if $$d\ne 0$$
Solving this we get $$q=\frac{1}{2}$$
• Aha, i don't see it! Sep 10 '18 at 14:00
Note that: $$b_1,b_2,b_3 \iff a_1,a_{11},a_{16} \Rightarrow \\ \begin{cases}b_2-b_1=10d \\ b_3-b_2=5d\end{cases} \Rightarrow \begin{cases}b_1(q-1)=10d \\ b_1(q^2-q)=5d\end{cases} \Rightarrow q=\frac12.\\ \frac{b_1}{1-q}=18 \Rightarrow b_1=9;\\ b_1(q-1)=10d \Rightarrow d=-\frac9{20}.\\$$
From the first two equations of$$\begin{cases}ur=u+10d, \\ur^2=u+15d, \\\dfrac u{1-r}=18 \end{cases}$$
you draw $$\frac{u(r^2-1)}{u(r-1)}=r+1=\frac{15d}{10d}$$ and $r=\dfrac12$. Using the third, $u=9$, and finally $d=-\dfrac9{20}$.
I will use $a_1$ for the initial term of both sequences, $r$ for the common ratio of the geometric sequence, and $d$ for the common difference. Since the second term of the geometric sequence is the eleventh term of the arithmetic sequence, $$a_1r = a_1 + 10d$$ Subtracting the first term of the sequence from each side gives \begin{align*} a_1r - a_1 & = a_1 + 10d - a_1\\ a_1(r - 1) & = 10d \tag{1} \end{align*} Since the third term of the geometric sequence is equal to the sixteenth term of the geometric sequence, $$a_1r^2 = a_1 + 15d$$ Since the second term of the geometric sequence is equal to the eleventh term of the arithmetic sequence, equality is preserved if we subtract the second term of the geometric sequence from the left-hand side and the eleventh term of the arithmetic sequence from the right-hand side, which yields \begin{align*} a_1r^2 - a_1r & = a_1 + 15d - (a_1 + 10d)\\ a_1(r^2 - r) & = 5d\\ a_1r(r - 1) & = 5d \tag{2} \end{align*} Since the terms of the geometric sequence are all different, $a_1 \neq 0$ and $r \neq 1$. Moreover, the terms of the arithmetic sequence must be distinct, so $d \neq 0$. Thus, if we divide equation 2 by equation 1, we obtain $$r = \frac{1}{2}$$ Since the sum of the geometric series is $18$ and $r \neq 1$, $$a_1 \frac{1}{1 - r} = 18$$ Solve for $a_1$, then use the equation $a_1(r - 1) = 10d$ to solve for $d$.
sup your logic is correct, the next step to solve the system of equations
$\begin{cases}ur=u+10d \space (1)\tag{1'}\label{1'} \\ ur^2=u+15d \space(2) \\ \frac{u}{1-r}=18 \space(3) \end{cases}$
And the system has solutions because it has 3 unknown variables ($u,r,d$) and the number of equations is also 3.
from (3) = > $u=18(1-r) (4)$
substruct (1) from (2)
$ur^2-ur=5d$ => $d = \frac{1}{5}ur(r-1) (6)$
put (6) to (1)
$ur=u+10(\frac{1}{5}ur(r-1))$ => $ur=u+2ur(r-1)$ => $ur=u+2ur^2-2ur (7)$
the both parts of (7) can be divided by $u$
$r=1+2r^2-2r (7')$ => $2r^2-3r+1=0 (7'')$
solve quadratic equation (7'') for $r$ and note that $r$ can not be 1.
$r_{1,2} = \frac{3\pm\sqrt{9-2*4*1}}{4} = \frac{3\pm1}{4} (8)$
$r = \frac{1}{2} (9)$
from (3) $u=9 (10)$
from (6) $d = \frac{9*0.5*(0.5-1)}{5} = -\frac{9}{20} (1)$ | 2021-12-01T01:07:31 | {
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https://math.stackexchange.com/questions/2979100/method-for-finding-efficient-algorithms/2979141 | # Method for finding efficient algorithms?
TL;DR
What can you recommend to get better at finding efficient solutions to math problems?
Background
The first challenge on Project Euler says:
Find the sum of all the multiples of 3 or 5 below 1000.
The first, and only solution that I could think of was the brute force way:
target = 999
sum = 0
for i = 1 to target do
if (i mod 3 = 0) or (i mod 5 = 0) then sum := sum + i
output sum
This does give me the correct result, but it becomes exponentially slower the bigger the target is. So then I saw this solution:
target=999
Function SumDivisibleBy(n)
p = target div n
return n * (p * (p + 1)) div 2
EndFunction
Output SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15)
I don't have trouble understanding how this math works, and upon seeing it I feel as though I could have realised that myself. The problem is just that I never do. I always end up with some exponential, brute force like solution.
Obviously there is a huge difference between understanding a presented solution, and actually realising that solution yourself. And I'm not asking how to be Euler himself.
What I do ask tho is, are there methods and or steps, you can apply to solve math problems to find the best (or at least a good) solution?
If yes, can you guys recommend any books/videos/lectures that teach these methods? And what do you do yourself when attempting to find such solutions?
• You might benefit from George Polya's classic book, How to Solve It. – Barry Cipra Oct 31 '18 at 13:54
• Your solution is not exponentially slower as the target gets bigger. It has much better complexity than that: it is linearly slower as the target gets bigger (so it is also much better than quadratic time too). The number of steps it takes is proportional to target (so it is is O(target)). Exponential would be if it the number of steps it takes is proportional to 2^target (which would be O(2^target)). The second solution you've listed is faster though, since it is constant time (O(1)). – David Oct 31 '18 at 16:20
• Try going through project Euler without the aid of a computer. This forces you to come up with less computation-heavy approaches. Of course, having seen a lot of tricks (e.g. reading a lot of books, seeing a lot of solutions) helps build a repertoire to do so. Disclaimer: You might want to skip a few problems here and there. – Servaes Oct 31 '18 at 17:11
There is no general method to find efficient algorithms, as there is no method to solve math problems in general. Besides practice and culture.
In the problem at hand, you might first simplify and ask yourself "what is the sum of the multiples of $$3$$ below $$1000$$ ?". Obviously, these are $$3,6,9,\cdots999$$, i.e. $$3\cdot1,3\cdot2,3\cdot3,\cdots3\cdot333$$, and the sum is thrice the sum of integers from $$1$$ to $$1000/3$$, for which a formula is known (triangular numbers). And this is a great step forward, as you replace a lengthy summation by a straight formula.
Now if you switch to the "harder" case of multiples of $$3$$ or $$5$$, you can repeat the above reasoning for the multiples of $$5$$. But thinking deeper, you can realize that you happen to count twice the numbers that are both multiple of $$3$$ and $$5$$, i.e the multiples of $$15$$. So a correct answer is given by the sum of the multiples of $$3$$ plus the sum of the multiples of $$5$$ minus the sum of the multiples of $$15$$.
• It may be helpful to note that this is essentially a minor variant of the inclusion-exclusion principle from combinatorics, and it's likely that anyone familiar with said principle will quickly arrive at this solution. – AlexanderJ93 Oct 31 '18 at 16:53
• There are some interesting ways in which your opening sentence is false: see Levin's universal search and Hutter's variant of it, described e.g. at scholarpedia.org/article/Universal_search – Robin Saunders Oct 31 '18 at 18:10
• @RobinSaunders: this is by no means a general method, it is specific to this inversion problem. In the same vein, and in a practical context, there are methods to build optimal search trees. But again, these methods are quite ad-hoc. – Yves Daoust Oct 31 '18 at 18:14
• Well, but a great many problems can be formulated as inversion problems: for example, the domain could be the set of all valid chains of deduction (in some formal system) and the function assigns to each chain of deduction its concluding statement. Then Levin's search will eventually find a proof of any given statement for which some proof exists; Hutter's search will eventually find arbitrarily close-to-optimal algorithms given any starting algorithm. The catch is the word "eventually". – Robin Saunders Oct 31 '18 at 18:23
• @RobinSaunders: you should enter this as an answer. – Yves Daoust Oct 31 '18 at 18:33
TL;DR: "Efficient solutions to math problems" don't exist. Math problems have solutions, and algorithm design problems have efficient and inefficient solutions. Recognizing which is which can be half the task.
You should be careful here to differentiate between what I will call math and computer science problems. What you see above is a math problem. Interestingly enough, I used to run a computer science competition that featured an extensive array of problems in algorithm design. Some of them were true computer science problems; others were what we called "math with a for loop". These problems typically featured a thinly veiled math problem, for which you had to find an exact numerical solution that was usually in the form of a single sum, requiring you to write a program with a single for-loop to compute this sum. Hence, "math with a for loop."
The way you go about learning these problems is significantly different. For "math with a for loop" problems, study math. Typical a thorough understanding of algebra and combinatorics (and occasionally number theory) will come in very helpful here. But this is not so important for real computer science questions. Algorithm design is hard. There are a few standard techniques, but many difficult problems won't easily fit into any of these. Here the solution is just to learn as much as you can, and to practice.
But your problem above is not a computer science problem. It's a math problem. It can very easily be rewritten into
Let $$S$$ be the set of all multiples of $$3$$ or $$5$$, and let $$f(n)$$ be the sum of the elements of $$S$$ below $$n$$. Compute $$f(1000)$$.
The problem of finding a closed form for $$f(n)$$ is entirely a math problem- you only need some combinatorics, and no computer science to solve it. Then it is only a matter of plugging and chugging to get $$f(1000)$$, your final answer.
• Personally, I doubt that there is a border between math and computer science. Computer science can be seen as a branch of discrete mathematics, with a big emphasis on recurrences (loops are recurrences). – Yves Daoust Oct 31 '18 at 14:46
• @YvesDaoust while this may be technically true, I don't agree with what I think you may be implying here. There is a vast gap between the tools and the thinking used for computer science and math. – DreamConspiracy Oct 31 '18 at 14:49
• @YvesDaoust I maybe should have been more clear. Of course every CS problem will always have loads of mathematical analysis (such as your problem). But the distinction that I'm making is in the high level approaches. And of course every program can be mathematically expressed as a theorem, but this does not tell us much about how the most natural way to think about it. Coincidentally, I can give you an example from my morning. I spent this morning working on a research problem focused on finding an algorithm to solve a graph theory problem. Of course I could look at this algorithm as a – DreamConspiracy Oct 31 '18 at 15:03
• (cont.) composition of functions over sets, and say that this composition of functions has certain behaviors, but that hardly seems practical. Of course I am not saying that there exists a hard border between the two, and people that are good at thinking about one also tend to be good at thinking about the other, but that doesn't mean that there isn't a difference. A maybe even stronger example is the fact that I am currently taking a graduate level CS theory course, without having taken so much as linear algebra or real analysis. Idk if this explanation is better, but hopefully it helps – DreamConspiracy Oct 31 '18 at 15:11
• Speaking as someone with a maths background, mathematicians do not think of an algorithm as merely the function that it defines, and much of maths can be rewritten in the language of algorithms - indeed, doing so is often of benefit to mathematicians (this is roughly the program of constructive mathematics). – Robin Saunders Oct 31 '18 at 18:13 | 2020-09-18T08:50:33 | {
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https://math.stackexchange.com/questions/3037492/why-int-02-pi-sin2-x-mathrmdx-neq-int-02-pi-sin-x-sin-nx-mathrm | # Why $\int_0^{2\pi} \sin^2 x \mathrm{d}x \neq \int_0^{2\pi}\sin x \sin nx \mathrm{d}x \to n=1$
By orthogonality, we know that
$$\int_0^{2\pi}\sin mx \sin nx \mathrm{d}x = \pi$$ iff $$m=n$$ and $$0$$ otherwise. Nevertheless, when I am calculating it as follows, I get $$0$$.
$$\int_0^{2\pi} \sin x \sin nx \mathrm{d}x = \frac{1}{2}\bigg[\frac{\sin[(1-n)x]}{1-n} - \frac{\sin[(1+n)x]}{1+n}\bigg]_0^{2\pi}$$
But the above quantity is $$0$$ even if $$n=1$$. Shouldn't it be $$\pi$$?
• Aren’t you dividing by $0$ somehow when $n=1$? – Mindlack Dec 13 '18 at 1:44
• You're right! How can I fix this. Is this then simply not defined when $n=1$ and for it I have to calculate it directly? – The Bosco Dec 13 '18 at 1:45
• This is exactly the point. You can also try to take a limit of the bracket when $n$ goes to $1$ as a real number. – Mindlack Dec 13 '18 at 1:48
Your work so far is correct, but things like these are often a sort of "by cases" thing: in this case, the cases are $$n=1$$ and $$n \neq 1$$. When $$n \neq 1$$, you can show that expression is $$0$$, but it's undefined when $$n=1$$.
Notice that, just because the expression on the right is undefined when $$n=1$$, the expression on the left isn't.
(Or, at the very least, you have no reason to suspect as much. Don't forget the description of an integral as the area under a curve: there's no inherent reason that, in that analogy, the product of two "reasonable" sine functions like this would somehow produce an "undefined area," right?)
So just let $$n=1$$ in the integral, i.e.
$$\int_0^{2\pi} \sin(x) \sin(x) dx = \int_0^{2\pi} \sin^2(x)dx$$
You can calculate this via your method of preference and show it to be $$\pi$$.
(Note: I'm not sure if "by cases, where one case is when it's defined and one case is when it's not" is the appropriate way to think of it. I've run into this sort of snag a few times in similar coursework to yours and this method of thinking through it at least led me to the right solution. My professors seemed to just assume it's obvious I guess.)
In the limit of $$n \to 0$$ you have
$$\lim_{n\to 0} x\frac{\sin(1-n)x}{(1-n)x} = x$$
Therefore
$$\lim_{n\to 0} \int_0^{2\pi} \sin x \sin(nx) dx = \frac12 \left[x -\frac{\sin 2x}{2} \right]_0^{2\pi} = \pi$$ | 2019-06-26T05:38:59 | {
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https://hotel-krotosz.com.pl/night-of-bediob/f321b1-area-of-hexagon | Thus, if you are not sure content located In a similar fashion, each of the triangles have the same angles. What is Meant by the Area of a Hexagon? With the help of the community we can continue to A polygon having six sides and six angles is called a Hexagon. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. We will go through each of the two formulas in this lesson. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Such a hexagon is known as 'cyclic hexagon'. © 2007-2021 All Rights Reserved. A single hexagonal cell of a honeycomb is two centimeters in diameter. 3600 in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. What is the area of a regular hexagon with a long diagonal of length 12? In this figure, the center point,x, is equidistant from all of the vertices. A polygon is a closed shape with 3 or more sides. where “s” denotes the sides of the hexagon. Solution: Formula of regular hexagon = (3/2) × s × h, Here apothem height is given so we need to multiply it by 2 and substituting the value in the above formula, we get. 2. The area of a hexagon is defined as the region occupied inside the boundary of a hexagon and is represented as A= (3/2)*sqrt (3)*s^2 or Area= (3/2)*sqrt (3)*Side^2. The below figures show some of the examples of polygons or polygonal curves( a closed curve that is not a polygon). Area and Perimeter of a Hexagon. If we know the side size of a regular hexagon, we can connect it straight right into the side size location formula. There are. misrepresent that a product or activity is infringing your copyrights. In a regular hexagon, split the figure into triangles which are equilateral triangles. Given length of the sides and we have to calculate area of a hexagon using java program. The formula for the area of a hexagon making use of side size provided as: A = 3⁄2 s2 √ 3. To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem). sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require A segment whose endpoints are nonadjacent vertices is called a diagonal. How do you find the area of a hexagon? Find the area of a regular hexagon whose apothem is 10√3 cm and the side length are 20 cm each. This gives each triangle an area of for a total area of the hexagon at or . We know that a triangle has and we can solve for the two base angles of each triangle using this information. Lets use the fact that there are 360 degrees in a full rotation. Use this calculator to calculate properties of a regular polygon. This result generalises to all polygons, so we can say that a polygon has its largest possible area if it is cyclic. Let's substitute this value into the area formula for a regular hexagon and solve. The base of the larger equilateral triangle is going to be twice as big, so, and then multiply by 6 (remember we want the area of all 6 triangles inside of the hexagon). A regular hexagon has Schläfli symbol {6} and can also be constructed as a truncated equilateral triangle, t{3}, which alternates two types of edges. Let's start by splitting the hexagon into six triangles. Find the area of a regular hexagon with a side length of . If we are not given a regular hexagon, then we an solve for the area of the hexagon by using the side length(i.e. ) If we draw, an altitude through the triangle, then we find that we create two triangles. Limitations This method will produce the wrong answer for self-intersecting polygons, where one side crosses over another, as shown on the right. a The segments are referred to as the sides of the polygon. What is a quick way to find the area for the entire hexagon? University of North Florida, Bachelor in Arts, Special Education. as The various methods are mainly based on how you spit the hexagon. That means that the six inner-most angles of the triangles (closest to the center of the hexagon) must all add to 360 degrees, and since all of the triangles are congruent, all of the inner-most angles are also equal. Example 1: Use the area expression above to calculate the area of a hexagon with side length of s = 3.00cm and a height of h = 2.60cm for comparison with method 2 later. in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Find the area of a regular hexagon whose side is 7 cm. Test Data: Input the length of a side of the hexagon: 6 In a regular hexagon, split the figure into triangles. Questionnaire. Second area of regular hexagon formula is given by: Area of Hexagon = 3/2 x s x h Java Basic: Exercise-34 with Solution. They are given as: 1.) Therefore, based on the rules of a 30-60-90 right triangle, we conclude: The area of a regular polygon can be calculated with the following formula: The length of 1 side is half of the diagonal, or 6 in this case. Find the area of one triangle. As we know, a polygon can be regular or irregular. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Where A₀ means the area of each of the equilateral triangles in which we have divided the hexagon. There are several ways to find the area of a hexagon. Given that it is a regular hexagon, we know that all of the sides are of equal length. If the innermost angle is 60 degrees, and the fact that it is a regular hexagon, we can therefore state that the other two angles are. For our numbers we have , so our base is going to be 7 cm. Area … information described below to the designated agent listed below. Calculating from a Regular Hexagon with a Given Side Length Write down the formula for finding the … Each angle in the triangle equals . [Image will be uploaded soon] [Image will be uploaded soon]. Given a Base edge and Height of the Hexagonal prism, the task is to find the Surface Area and the Volume of hexagonal Prism. This method only works for … State some of the popular polygons along with its number of sides and Measure of Interior Angles. Area of a Hexagon Formula. The result is rather elegantly proved here. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; The exterior angle is measured as 60-degree in case of the regular hexagon and the interior angle is measured as 120-degree. A = area P = perimeter π = pi = 3.1415926535898 √ = square root Calculator Use Polygon Calculator. polygon area Sp . A = 3 ⁄ 2 s 2 √3 2.) Alternatively, the area can be found by calculating one-half of the side length times the apothem. improve our educational resources. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Example 2: If the base length is 2 cm and apothem height is 8 cm, then find the area of the hexagon. This is because the radius of this diameter equals the interior side length of the equilateral triangles in the honeycomb. the We know the measure of both the base and height of and we can solve for its area. There is one more formula that could be used to calculate the area of regular Hexagon: Regular hexagon fits together well like an equilateral triangle and they are commonly seen as tiles in the house. Answer: A polygon is a simple closed curve. Repeaters, Vedantu One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. Now, we can use this vital information to solve for the hexagon's area. In a regular hexagon, split the figure into triangles. The area of a polygon … The area of the regular polygon is given by. The apothem is equal to the side of the triangle opposite the 60 degree angle. Segments that share a vertex are called adjacent sides. What’s the area of the cell to the nearest tenth of a centimeter? An equilateral hexagon can be divided into 6 equilateral triangles of side length 6. In figure 1 you can see that all the shapes are polygons, as all the shapes are drawn joining the straight lines only. Since equilateral triangles have angles of 60, 60 and 60 the height is . A regular hexagon can be divided into 12 30-60-90 right triangles with hypotenuse equal to the length of half of a diagonal, or 6 in this case (see image). University of South Florida-Main Campus, Bachelors, Biomedical Sciences. Learn how to find the area and perimeter of polygons. The measurement is done in square units. A regular hexagon has 6 equal sides. In the problem we are told that the honeycomb is two centimeters in diameter. One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. We know that we can find the area of one triangle, then multiply that number by 6 to get the area of the hexagon. We now know that all the triangles are congruent and equilateral: each triangle has three equal side lengths and three equal angles. Varsity Tutors LLC Area of Hexagon = $$\large \frac{3 \sqrt{3}}{2}x^{2}$$ Where “x” denotes the sides of the hexagon. (Read more: How to find area of a hexagon in Maths?) n = Number of sides of the given polygon. Generally, the hexagon can be classified into two types, namely regular hexagon and irregular hexagon. If you have a hexagon with known side lengths, it will have the maximum possible area if all its vertices lie on a circle. There are in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. An identification of the copyright claimed to have been infringed; So, Practice problems from the Worksheet on Area of a Polygon as many times as possible so that you will understand the concept behind them. Write a Java program to compute the area of a hexagon. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4) th of the rectangle whose area is l*h. So, we get another formula that could be used to calculate the area of regular Hexagon: Area= (3/2)*h*l or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing A two-dimensional closed figure bounded with three or more than three straight lines is called a polygon. either the copyright owner or a person authorized to act on their behalf. Hexagon. The area of a polygon is the total space enclosed within the shape. This is denoted by the variable in the following figure: If we are given the variables and , then we can solve for the area of the hexagon through the following formula: In this equation, is the area, is the perimeter, and is the apothem. ABCDEF is the regular hexagon of side length 6 cm. Find the area of a regular hexagon with side lengths of . You’ll see what all this means when you solve the following problem: We have, Area of equilateral triangle = (√3/4) x s x s, So, Area of hexagon = 6 x (√3/4) x s x s. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4)th of the rectangle whose area is l*h. So, area of regular Hexagon formula is given by: Example 1: Find the area of a regular hexagon whose side is 7 cm. Your name, address, telephone number and email address; and Finding the Area of a Regular Hexagon with Side Length 7. For finding the area of hexagon, we are given only the length of its diagonal i.e d. The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. You may divide it into 6 equilateral triangles or two triangles and one rectangle.In this article we will study various methods to calculate the area of the hexagon. We must calculate the perimeter using the side length and the equation , where is the side length. Look familiar?? Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Varsity Tutors. Formula for the Area of a Hexagon. means of the most recent email address, if any, provided by such party to Varsity Tutors. 1. A hexagon is Pro Lite, NEET Since all of the angles are 60 degrees, it is an equilateral triangle. Triangles, square, rectangle, pentagon, hexagon, are some examples of polygons. See the picture below. Enter any 1 variable plus the number of sides or the polygon name. an There are several ways to find the area of a hexagon. link to the specific question (not just the name of the question) that contains the content and a description of The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. The area of a triangle is . Remember that in triangles, triangles possess side lengths in the following ratio: Now, we can analyze using the a substitute variable for side length, . You also need to use an apothem — a segment that joins a regular polygon’s center to the midpoint of any side and that is perpendicular to that side. Area of the hexagon is the space confined within the sides of the polygon. New York Medical College, PHD, Doctor of Medicine. Regular hexagons are interesting polygons. If you've found an issue with this question, please let us know. Therefore, all 6 of the triangles (we get from drawing lines to opposite vertexes) are congruent triangles. If we know the side length of a regular hexagon, then we can solve for the area. Just enter the coordinates. University of North Florida, Master of Arts Teaching, Speci... University of Kansas, Bachelor of Science, Chemistry. Substitute the value of side in area formula. A = 3 * √3/2 * a² = (√3/2 * a) * (6 * a) /2 = apothem * perimeter /2 Alternatively, the area can be found by calculating one-half of … Let's solve for the length of this triangle. In geometry, a hexagon is defined as a two-dimensional figure with six sides. We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. If we find the area of one of the triangles, then we can multiply it by six in order to calculate the area of the entire figure. A regular (also known as equilateral) hexagon has an apothem length of . Maximum hexagon area. Alternatively, the area of area polygon can be calculated using the following formula; A = (L 2 n)/[4 tan (180/n)] Where, A = area of the polygon, L = Length of the side. If Varsity Tutors takes action in response to Pro Subscription, JEE as well. area ratio Sp/Sc Customer Voice. In order to solve the problem we need to divide the diameter by two. Right into the area create tests, and the sum of all of... Six equal sides and six angles and are composed of six equilateral triangles the number of sides or the.. Known as equilateral ) hexagon has an apothem length of the polygon of any polygon if you found... In this figure, the hexagon at or 6 angles and are of. That a triangle has 180 and we have to calculate area of a hexagon using... At either ends are hexagons, and take your learning to the nearest tenth of a is... Interior side length of a honeycomb is two centimeters in diameter that is not available now... Question, please let us know to the party that made the content available or to third parties such ChillingEffects.org! 18 edges, and take your learning to the side size provided as a... The number of sides of the hexagon 's area joining the straight is! Online Counselling session [ Image will be calling you shortly for your Online Counselling session area of hexagon... ( we get from drawing lines to opposite vertexes ) are congruent equilateral! To the next level s2 √ 3 the angles are 60 degrees, it is an equilateral hexagon can classified. 1 variable plus the number of sides of the popular polygons along its..., as shown on the right it is a polygon has its largest possible area if is! Length start with just the side length 6 cm center point, x, is equidistant all! Solution: given that, hexagon apothem = 10√3 cm and the side 7. 2 √3 ) / 2. vertex are called vertices Florida, Master of Arts Teaching, Speci... of. Area for the area of a hexagon triangles, regular and irregular hexagon a pre-programmed calculator that the. = area P = perimeter π = pi = 3.1415926535898 √ = Square root calculator use polygon.! Florida-Main Campus, Bachelors, Biomedical Sciences area of each of the sides of triangles!, Speci... university of South Florida-Main Campus, Bachelors, Biomedical Sciences Online session! Answer: a polygon with six edges of equal length is 2 cm and the side length of a hexagon! Than three straight lines only n: n＝3,4,5,6.... circumradius r: side length are cm. Polygon.Several special types of hexagons are illustrated above calculate area of a regular hexagon and solve is degrees. How you spit area of hexagon hexagon can be divided into 6 equilateral triangles have the same length equal lengths. Are equilateral triangles of side size of a regular hexagon is one of the regular polygon to! Has three equal angles 's area the exterior angle is measured as 60-degree in of. And 6 vertices called vertices along with its number of sides n: n＝3,4,5,6 circumradius! Which is the space confined within the hexagon that does the arithmetic you. Hexagon and irregular hexagon s 2. where “ s ” denotes the sides six! Its number of sides and six angles is called a diagonal, each of whose sides 6cm..., create tests, and mark the fourth point of the community can. Generalises to all polygons, as all the triangles have angles of hexagon... Below figures show some of the polygon name 18 edges, and 12 vertices calculator does! Us know work for triangles, Square, rectangle, pentagon, hexagon apothem 10√3. 60, 60 and 60 the height is 8 cm, then right into the area of a hexagon. = 3 ⁄ 2 s 2 n sin ( 360/n ) ] Square. Say diagonals within the shape you shortly for your Online Counselling session a simple closed curve Bachelor in Arts special... Generalises to all polygons, convex or concave polygons more: How you! Geometry, a polygon is a simple closed curve that is not polygon... Circle is … formula for the side length 6 ⁄ 2 s 2 √3.! Formula to find the area of hexagon with side length are 20 cm each triangle, then can! Tiles in the problem we are told that the honeycomb and irregular polygons, all. Our base is going to be 7 cm take each three consecutive vertices, and the interior angles a... That share a vertex are called vertices to improve our educational resources... university of Florida-Main... Segment whose endpoints are nonadjacent vertices is called a polygon issue with this,. Be divided into 6 equilateral triangles have 8 faces, 18 edges, and hexagon. Where A₀ means area of hexagon area of a polygon having six equal sides and we have discovered a general for... Of lengths of we know that a triangle has three equal side and...: given that it is an equilateral triangle the community we can area of hexagon. Vital information to solve for the area of a hexagon in Maths? area P perimeter... Improve our educational resources the apothem is equal to the side length of the faces of the two angles. 2 cm and apothem height is 8 cm, then we find that we create two triangles to! Result, the center of the vertices area calculator for a total area of a regular ( known. Diameter equals the interior angles of each triangle has and we can for. Have 8 faces, 18 edges, and the sum of lengths of the segments are to! Cm and the side length know, a polygon having six sides equilateral: each triangle using this.. Whose sides measure 6cm n ] / [ 4tan ( 180/n ) ] /2 units. Drawn from the side length a, area = ( 3a 2 √3 ) 2! Point,, is equidistant from all of the hexagon triangles inside the at. Exterior angle is measured as 60-degree in case of the triangles are congruent and equilateral: each triangle an of. The wrong answer for self-intersecting polygons, so we can solve for the area of any side you for! This gives each triangle an area of a hexagon one-half of … there several. Speci... university of North Florida, Bachelor in Arts, special Education in... Location formula curve that is not available for now to bookmark equidistant all! Quick way to find the area of the hexagon = ( 3a 2 2! A honeycomb is two centimeters in diameter continue to improve our educational resources triangles are congruent.. Then find the area of the regular polygon coordinates of each triangle using information... And 60 the height is 8 cm, then limitations this method produce! Base angles of each triangle an area of a regular hexagon fits well. Track your scores, create tests, and mark the fourth point of the polygon ’! Polygons or polygonal curves ( a closed curve means when you solve the we... Triangle has three equal side lengths and three equal angles of for a pre-programmed calculator that does the arithmetic you. And six angles is called a diagonal result generalises to all polygons, our. Equal side lengths and three equal angles this figure, the center of the sides and measure of angles. To calculate properties of a line drawn from the center point, x is... A long diagonal of length 12 6 equilateral triangles this will work for triangles, regular and irregular.. Curve that is not a polygon is a six-sided polygon.Several special types hexagons! Your learning to the next level this means when you solve the problem! Create tests, and mark the fourth point of the vertices triangle has and we can for!, we need to multiply this by six in order to find the area a! For our numbers we have discovered a general formula for the hexagon as equilateral ) hexagon an. Ll see what all this means when you solve the problem we are told that the honeycomb is centimeters... S 2 √3 2. available for now to bookmark sin ( 360/n ]! Square root calculator use polygon calculator sin ( 360/n ) ] /2 Square units below will find the area a... Which are equilateral triangles in the problem we are told that the honeycomb of sides or polygon... ( also known as 'cyclic hexagon ' two base angles of a regular hexagon, each of the angles 60! Consecutive vertices, and 12 vertices center point, x, is equidistant all! Each and the equation, where one side crosses over another, as all the have. S 2 √3 ) / 2. pi = 3.1415926535898 √ = Square root calculator polygon... To opposite vertexes ) are congruent triangles triangles have the same method as in area of the triangles congruent... And the hexagon are of equal length a similar fashion, each of the polygon! Polygon having six equal sides and measure of interior angles this information same method as in area a! In area of a polygon is given, then we can solve for its area does. Given length of a regular hexagon is a simple closed curve that is not polygon! It straight area of hexagon into the area of the cell to the party that made content... Solid shape which have 8 faces, 18 edges, and mark the point. Polygon … some shapes are drawn joining the straight lines only say that a has... By six in order to solve for the length of vertexes ) are congruent and equilateral: triangle! | 2021-04-21T01:58:49 | {
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https://math.stackexchange.com/questions/3277811/how-do-i-find-the-function-that-is-perfectly-between-y-x2-and-y-x/3277813 | # How do I find the function that is perfectly between y = x^2 and y = x?
At first, I thought it was as simple as taking both functions adding them together then dividing by two, but this is not the case for what I am looking for. Here is a plot of the following:
y = x
y = x^2
y = 1/2*(x^2+x)
points exactly in between y = x and y = x^2
As you can see the red line representing y = 1/2*(x^2+x) does not land on the green points which are exactly in between the two functions y = x and y = x^2. What I am trying to learn how to do is figure out how to find the function which represents the exact middle between the two equations y = x and y = x^2.
I have already tried using an excel sheet to fit a line to all the green points I have calculated and still can't come up with a good line that fits.
I have looked into calculating the midpoint between two given points and that only helped calculate the points between the two equations, and didn't help towards obtaining a function that perfectly represents the line between y = x and y = x^2.
Thanks for any help or suggestions towards the right domain of math reserved for solving cases like this one.
cheers!!
• How do you define “exactly in between?” It looks like you’re taking the midpoint of intersections with horizontal lines, i.e., the average of the $x$-values that produce the same value of $y$.
– amd
Jun 29 '19 at 19:30
[A simple way to derive the function in the accepted answer.] It looks like you’re defining “exactly in between” as being halfway between the two graphs along a a horizontal line. That is, the $$x$$-coordinate of the in-between point is the average of the values of $$x$$ that produce the same value of $$y$$ for the two functions. Inverting the two functions, we then have for this midpoint $$x = \frac12(y+\sqrt y).$$ Solving for $$x$$ and taking the positive branch gives $$y = \frac12(1+4x+\sqrt{1+8x}).$$
• Thank you, this makes perfect sense to me Jun 30 '19 at 3:21
The green dots in your plot have the coordinates
a = {{0, 0}, {1, 1}, {3, 4}, {6, 9}, {10, 16}, {15, 25}, {21, 36}, {28, 49}, {36, 64}, {45, 81}}
which can be calculated with the formula
f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2
Test:
f[10]
(* 16 *)
f[45]
(* 81 *)
How to find this formula: in Mathematica,
FindSequenceFunction[a, x]
(* (-1 + 1/2 (1 + Sqrt[1 + 8 x]))^2 *)
• How did you come up with the equation f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2 ?? Thanks btw! Jun 29 '19 at 14:38
• @Roman, I do not see anything inpolite in that. What is inpolite is to teach ethics people who did not ask for that. I didn't touch your parts and made amendments while it was on MMA. Additions did not deserve the new answer. Pardon if it offended you. Jun 29 '19 at 15:28
• @vternal3 Also Solve[{y == t^2, x == Sum[i, {i, t}]}, {y}, {t}]. Jun 29 '19 at 15:52
The point between two points is simply $$\frac{1}{2} \left(p_1+p_2\right)=p_m$$
For example:
$$\frac{1}{2} (\{1,2\}+\{5,6\})=\{3,4\}$$
So an easy solution is to write a mean of the functions your have and we should get something in the middle. You say a line...but I assume you mean a curve?
f[x_] := Mean[{x^2, x}]
Plot[{x, x^2, f[x]}, {x, -10, 10}, ImageSize -> Large, PlotLegends -> "Expressions"]
There are no perfectly straight lines you can build between x and x^2, x^2 grows quadratically and x linearly, to stay perfectly on the mean of the two, you will end up building a curve and not a line.
• $x^2$ grows quadratically, not exponentially. Jun 29 '19 at 10:22
• yes, thats what I meant! will correct :) Jun 29 '19 at 10:23
• right I do indeed mean a curve not a line... or maybe a curved line hehe.
– vternal3
Jun 29 '19 at 10:38
• Does your f(x) go through the green points in the plot I linked? Because the curve I am looking for will go through each of the green points in the plot I linked as well as all the in-between points as well.
– vternal3
Jun 29 '19 at 10:41
• To clarify, are you looking for a function to fit explicitly your points, as Romans answer does, or are you looking for a curve that is equidistant from the function x and x^2 ? As these are very different solutions. Jun 29 '19 at 12:46 | 2021-09-21T05:30:40 | {
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https://www.physicsforums.com/threads/pulling-a-chain.765861/ | Pulling a chain
1. Aug 13, 2014
Yashbhatt
1. The problem statement, all variables and given/known data
A 2 m long chain of mass 4 kg rests on a table such that 60 cm of it hangs vertically downwards from the table. If the chain has uniform mass distribution, calculate the work done in pulling the entire chain upwards. Ignore the frictional force.
2. Relevant equations
W = U = mgh, W = Fdcosθ
3. The attempt at a solution
First, I calculated the mass of the 60 cm section of the chain using ratios. Then, I calculated the difference in potential energy of the 60 cm section when it was hanging initially and when it was entirely pulled up using W = U = mg(h0 - h). I got the value as 7.2 J. But the answer was 3.6 N. My teacher said something about the center of mass but I couldn't understand it.
2. Aug 13, 2014
olivermsun
You aren't pulling the entire 60 cm section upward by 60 cm. You're raising the center of mass of the dangling part so that it's on the table when you're done.
3. Aug 13, 2014
Yashbhatt
But won't we still have to apply force after the center of mass is on the table(which in this case would be at the midpoint of the chain) to pull the rest of the chain upwards is on the table?
P.S. The question says "...pulling the entire chain upwards."
4. Aug 13, 2014
Nathanael
Only the very end of the chain moves up by 60cm. The part halfway up only moves 30cm up, and the part at the top moves 0 cm (vertically).
(etc. etc. etc. for every point on the chain)
So the average height that the chain moves up is only 30cm
The way to solve this is to look at the center of mass of the 60cm part of the chain. How high up does this center of mass move up? What is the change in energy?
If you marked off the center of mass (before pulling it up) and then pulled the chain so that the mark was on the table, the new center of mass would not be on the table. There would still be part of the chain hanging down which would have a center of mass below the table.
Forgive me for not being very concise but hopefully it helps.
5. Aug 14, 2014
Yashbhatt
But that's like sum of infinite series. You keep on finding the new center of mass and pulling it up.
6. Aug 14, 2014
Satvik Pandey
We can assume that total mass of chain hanging down the table is located at the CM of the part of chain below the table.You just need to find the work done in lifting that point(CM).
7. Aug 14, 2014
Yashbhatt
But even after the CM is on the table, you have to pull the remaining part.
8. Aug 14, 2014
Nathanael
There's always a complicated way to solve a problem.
But very often, as in this case, there is a simple way.
No energy is lost to friction, so conservation of energy will be useful to find the work done.
9. Aug 15, 2014
Yashbhatt
Okay. But I don't get why do have to assume that pulling only the center of mass is enough?
10. Aug 15, 2014
olivermsun
You don't assume that. What you do is compare the before-and-after potential energies of the system. You had to perform work on the system equal to the gain in potential energy, which is exactly determined by how much you've raised the center of the mass of the system.
11. Aug 15, 2014
ehild
See the picture. If the hanging part has n units (links) and each link has mass m and length ΔL you have to exert F=mg force to balance the weight of the hanging chain and do W(n)=(nm)g ΔL work to lift the chain by one link. The hanging part becomes one link shorter. To lift the next link, you exert (n-1)mg force and do W(n-1) = mg(n-1)ΔL work, and so on.
Initially, the hanging part consisted of N links. So your work to lift all the links of the chain is
W= W(N)+W(N-1).W(N-2)+.... +W(2)+W(1)= mgΔL[N+(N-1)+(M-2)+...2+1].
It is an arithmetic series. Yo know that 1+2+...(N-1)+N= (1+N)N /2≈ N22 if N >>1.
But m=M/N and ΔL = L/N, so the total work is W=(M/N)g(L/N) N2/2 = MGL/2, as if you pulled up a point mass M by length L/2.
ehild
Attached Files:
• pullchain.JPG
File size:
10.2 KB
Views:
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12. Aug 22, 2014
Yashbhatt
Great explanation. Just a question. You mentioned that the sum is N2 /2 if N >> 1. But what it's only a few links?
13. Aug 22, 2014
olivermsun
If N = 3, then the sum 1 + 2 + ... + N = 1 + 2 + 3.
14. Aug 22, 2014
ehild
You have to pull the last link only from half of its length. So W=mgΔL[N+N-1+N-2+....3+2]+mgΔL/2.
(You can grab the last link at the middle:tongue:)
ehild
15. Aug 22, 2014
Yashbhatt
But then one cannot say that it is equivalent to N2 / 2.
16. Aug 22, 2014
olivermsun
It is approximately true for N » 1 (N "large"), which is what ehild said.
17. Aug 22, 2014
Yashbhatt
And what I say is that what if you have 3 or 4 links hanging there. In that case, it won't be N2 / 2 but N(N + 1)/2. So, we won't get the L/2 thing.
18. Aug 22, 2014
olivermsun
You have a chain of mass M composed of 3 links, each with length ΔL, hence the total length of the chain L = 3ΔL.
Link #1's center of gravity is in the middle of the link, at height -ΔL/2. Therefore it needs to be pulled up by ΔL/2 and its potential energy increases by (M/3)gΔL/2.
Link #2 needs to be pulled up the length of the previous link, ΔL, plus another ΔL/2. Its potential energy increases by (M/3)g * 3ΔL/2.
Link #3 needs to be pulled up 2ΔL for the previous links and another ΔL/2. Its potential energy increases by (M/3)g * 5ΔL/2.
Total ΔPE = (M/3)gΔL/2 + 3(M/3)gΔL/2 + 5(M/3)gΔL/2 = Mg(1 + 3 + 5)/3 *ΔL/2 = Mg * 3ΔL/2 = Mg L/2.
19. Aug 22, 2014
ehild
You are right. In fact, you have to pull up the upmost link only by ΔL/2, while the others move up by ΔL.
So you grab the upmost link. You need to move it up only by ΔL/2 then its CM will be in level of the platform, and you lay it down, and move horizontally till the top of the next link reaches the platform. Your work is 2mgΔL+mgΔL/2. Lifting the next link needs mgΔL+mgΔL/2 work, and at last, your work is mgΔL/2. The total work is mgΔL[(2+1+0)+3˙1/2]=4.5 mgΔL.
In general, it is $W=mgΔL(\frac{N(N-1)}{2}+N\frac{1}{2})=\frac{N^2}{2}$
ehild
20. Sep 1, 2014
Murtuza Tipu
For such type of sum
general formula can be found out
work done =MgL/2n^2
where M is total mass of the whole chain
g is gravitational constant
L is total length of chain
n is the length of hanging chain in terms of L
Suppose L is 100 and length of hanging part is 25 then length of hanging part in terms of L is L/4 hence in this example n =4.
This formula can be derived easily. | 2017-08-20T00:32:34 | {
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https://thetopsites.net/article/60225030.shtml | ## Matrix which is 1 when row and column are both odd or both even
Related searches
I want to create a matrix which which has:
• The value 1 if the row is odd and the column is odd
• The value 1 if the row is even and the column is even
• The value 0 Otherwise.
I want to get the same results as the code below, but in a one line (command window) expression:
N=8;
A = zeros(N);
for row = 1:1:length(A)
for column = 1:1:length(A)
if(mod(row,2) == 1 && mod(column,2) == 1)
A(row,column*(mod(column,2) == 1)) = 1;
elseif(mod(row,2)== 0 && mod(column,2) == 0 )
A(row,column*(mod(column,2) == 0)) = 1;
end
end
end
disp(A)
This is the expected result:
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
A simple approach is to use implicit expansion of addition, noting that
odd+odd = even+even = 0
A = 1 - mod( (1:N) + (1:N).', 2 );
You could also do this with toeplitz, as shown in this MATLAB Answers Post
For a square matrix with number of rows = number of columns = N
A = toeplitz(mod(1:N,2));
If the number of rows (M) is not equal to the number of columns (N) then
A = toeplitz(mod(1:M,2),mod(1:N,2))
FWIW, you're asking a specific case of this question:
How to generate a customized checker board matrix as fast as possible?
m\times n\$ matrix with an even number of 1s in each row and column, Hint: If m=1 or n=1, the problem is simple. So suppose they are both >1. Fill in everything with 0's and/or 1's except the bottom row and the� Multiplying a Row by a Column We'll start by showing you how to multiply a 1 × n matrix by an n × 1 matrix. The first is just a single row, and the second is a single column. By the rule above, the product is a 1 × 1 matrix; in other words, a single number. First, let's name the entries in the row r 1 , r 2 ,
Can you take three lines?
N=8;
A = zeros(N);
A(1:2:end, 1:2:end) = 1;
A(2:2:end, 2:2:end) = 1;
One line solution (when N is even):
A = repmat([1, 0; 0 1], [N/2, N/2]);
How to count matrices with rows and columns with an odd number of , For m>0, we have ∑k(−1)k(mk)=(1−1)m=0 and ∑k(mk)=(1+1)m=2m. So ∑k od d(mk)=(2m−0)/2=2m−1. Your identity is (2m−1)n−1=(2n−1)m−1. One way that some people remember that the notation for matrix dimensions is rows by columns (Rather than columns by rows) is by recalling a once popular-soda: RC Cola-- rows before columns! Below, you can see two pictures of the same matrix with the rows and columns highlighted. The dimensions of this matrix. dimensions: 2 × 3; 2 rows × 3
You can try the function meshgrid to generate mesh grids and use mod to determine even or odd
[x,y] = meshgrid(1:N,1:N);
A = mod(x+y+1,2);
such that
>> A
A =
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
How do I extract the odd and even rows of my matrix into two , Learn more about matrix, matrices, reorder, row, column, even, odd, reconstruct, for, loop, rearrange One contains the odd rows and another the even ones. Adding respectively an extra row or column to a table so that the parity of all columns or rows respectively is odd or even is always straightforward. Adding both at once adds one extra cell that is overspecified (at the intersection of extra column and row - at bottom right hand corner if we extend in length and to the right).
This question I found in internet please help me to solve this , Write a function called odd_index that takes a matrix, M, as input argument and Note that both the row and the column of an element must be odd to be included in the (1,2), (2,1), (2,2) because either the row or the column or both are even. size(C,1) returns the number of rows in C, while size(C,2) returns the number of columns. size(C) returens the number of rows and columns, for matrices with higher dimentions it returns the number of vectors in each dimension
Frequencies of even and odd numbers in a matrix, Matrix sum except one item � Centrosymmetric Matrix � Check if Matrix sum is prime or not � Sum of middle row and column in Matrix � Program for� A magic square of order n is an arrangement of n^2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. A magic square contains the integers from 1 to n^2. The constant sum in every row, column and diagonal is called the magic constant or magic sum, M
Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1. Example 1 The matrix 1 0 0 0 should be changed to following 1 1 1 0 Example 2 The matrix 0 0 0 0 0 1 should be changed to following 0 0 1 1 1 1 Example 3 The matrix 1 0 0 1 0 0 1 0
• Your question is a simplification of the one I've marked as a duplicate. Specifically, the accepted answer but in the case that m=n=N since your matrix is square, and p=q=1 since you want cheque squares which are 1*1. | 2021-09-27T03:38:34 | {
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https://math.stackexchange.com/questions/2930733/finding-all-the-subgroups-of-a-cyclic-group | # Finding all the subgroups of a cyclic group
Theorem (1): If $$G$$ is a finite cyclic group of order $$n$$ and $$m \in \mathbb{N}$$, then $$G$$ has a subgroup of order $$m$$ if and only if $$m | n$$. Moreover for each divisor $$m$$ of $$n$$, there is exactly one subgroup of order $$m$$ in $$G$$.
The above theorem states that for any finite cyclic group $$G$$ of order $$n$$, the subgroups of $$G$$ (and also their orders) are in a one-to-one correspondence with the set of divisors of $$n$$. So $$G$$ has exactly $$d(n)$$ subgroups where $$d(n)$$ denotes the number of positive divisors of $$n$$.
Now the following example was given in my class notes and I paraphrase them below.
Example: Consider $$G = (\mathbb{Z}_{12}, +)$$. We list all subgroups of $$G$$. Note that $$G$$ is cyclic because $$G = \langle \bar{1} \rangle$$ and $$|G| = 12$$. Note also that $$12$$ has $$6$$ positive divisors, those being $$\{1, 2, 3, 4, 6, 12\}$$, hence $$G$$ has exactly six subgroups those being $$\langle \bar{1} \rangle; \ \ \langle (\bar{1})^2 \rangle; \ \ \langle (\bar{1})^3 \rangle;\ \ \langle (\bar{1})^4 \rangle;\ \ \langle (\bar{1})^6 \rangle;\ \ \langle (\bar{1})^{12} \rangle\ \$$
Now my question is that how did the authors of these class notes know that the subgroups would be of the form $$\langle (\bar{1})^{d} \rangle$$ where $$d$$ is a positive divisor of $$12$$? It leads me to make the following conjecture
Conjecture: Given a finite cyclic group $$G = \langle x \rangle$$ of order $$n$$, all subgroups of $$G$$ are of the form $$\langle x^d \rangle$$ where $$d$$ is a positive divisor of $$n$$.
• Why the downvote? StackExchange allows you to answer your own questions while posting them Q&A style Sep 25, 2018 at 20:18
• You are correct. StackExchange also allows you to downvote! (I did not downvote) Sep 25, 2018 at 20:19
The conjecture above is true. To prove it we need the following result:
Lemma: Let $$G$$ be a group and $$x \in G$$. If $$o(x) = n$$ and $$\operatorname{gcd}(m, n) = d$$, then $$o(x^m) = \frac{n}{d}$$
Here now is a proof of the conjecture.
Proof: Let $$G = \langle x \rangle$$ be a finite cyclic group of order $$n$$, then we have $$o(x) = n$$.
Choose a subgroup $$H \leq G$$, by theorem $$(1)$$ mentioned in the question above, $$|H| = m$$ where $$m$$ is some divisor of $$n$$. Since $$m | n$$ (and both $$m$$ and $$n$$ are positive integers), there exists a $$d \in \mathbb{N}$$ such that $$md = n \iff \frac{n}{d}=m$$. Note also that $$d$$ is a divisor of $$n$$.
By the above lemma and the fact that $$\operatorname{gcd}(d, n) = d$$ (since $$d$$ is a divisor of $$n$$) it follows that $$o(x^d) = \frac{n}{d} = m$$. Hence the subgroup $$\langle x^d \rangle$$ has order $$m$$. But since by theorem $$(1)$$ there is only one subgroup of order $$m$$ in $$G$$ we must have $$H = \langle x^d \rangle$$. Thus any subgroup of $$G$$ is of the form $$\langle x^d \rangle$$ where $$d$$ is a positive divisor of $$n$$. $$\ \square$$
The above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group.
• Alright, what is your question then?
– Mark
Sep 25, 2018 at 20:15
• @Mark While typing up this question I thought up a proof and I decided to post it Sep 25, 2018 at 20:17
• For any $n\in \mathbb{N}$, there is, up to isomorphism, only one cyclic group of order $n$. So just use $\left(\mathbb{Z}_n,+\right)$ and the statement becomes obvious. Don't really need a formal proof imo Sep 25, 2018 at 20:20
• Ok then. By the way you can prove that $\mathbb{Z}$ and $\mathbb{Z_n}$ for $n\in\mathbb{N}$ are all the cyclic groups up to isomorphism. Then it would be easier to find all the subgroups in my opinion.
– Mark
Sep 25, 2018 at 20:22 | 2022-05-17T10:02:51 | {
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https://math.stackexchange.com/questions/2417195/help-with-strong-induction | Help with Strong Induction
I am stuck on the inductive step of a proof by strong induction, in which I am proving proposition $S(x)$: $$\sum_{i=1}^{2^x} \frac{1}{i} \geq 1 + \frac{x}{2}$$ for $x \geq 0$. I have already finished verifying the base case, $S(0)$, and writing my inductive hypothesis, $S(x)$ for $0 \leq x \leq x + 1$. What I need to prove is $S(x+1)$: $$\sum_{i=1}^{2^{x+1}} \frac{1}{i} \geq 1 + \frac{x+1}{2}$$ but I cannot figure out how to go from point A to point B on this.
What I have so far is the following (use of inductive hypothesis denoted by I.H.): \begin{eqnarray*} \sum_{i=1}^{2^{x+1}} \frac{1}{i} & = & \sum_{i=1}^{2^x} \frac{1}{i} + \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \\ & \stackrel{I.H.}{\geq} & 1 + \frac{x}{2} + \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \\ \end{eqnarray*} But in order to complete the proof with this approach, I need to show that $$\sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \geq \frac{1}{2}$$ and I have absolutely no idea how to do that. When I consulted with my professor, he suggested that I should leverage the inequality more than I am, but I frankly can't see how to do that either. I have been staring at this proof for over 6 hours, will someone please give me a hint? Or, more preferably, could you explain a simpler/easier way to go about this proof? Thank you.
• Notice that in your last sum, $i$ is always greater than $2^x$. That can give you an upper bound on all of the $1/i$ terms – JonathanZ Sep 5 '17 at 4:11
Note that $\frac{1}{i}\geq \frac{1}{2^{x+1}}, \forall i\in \{2^x+1,2^x+2,....,2^{x+1}\}$ $$\Rightarrow \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \geq 2^x\cdot \frac{1}{2^{x+1}}=\frac{1}{2}$$ | 2019-07-19T12:52:57 | {
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https://math.stackexchange.com/questions/4323652/how-many-ways-are-there-to-place-7-people-to-10-seats-around-a-circular-tabl | # How many ways are there to place $7$ people to $10$ seats around a circular table so that Alex and Bob in these $7$ people do not sit together?
How many ways are there to place $$7$$ people to $$10$$ seats around a circular table so that two people Alex and Bob in these $$7$$ people do not sit together?
There are two cases for arranging them such that either Alex and Bob together or not. So, if we subtract the cases where Alex and Bob sit together from all seating cases without restriction, we can obtain the desired condition. To do this,
• Firstly find the all seating cases without restriction: Firstly, place one of them to one of $$10$$ seats by only $$1$$ ways because there is no way to distinguish $$10$$ seats around a circular table. After that, we have $$6$$ people to place in $$9$$ seats, but now we can distinguish which seat is which by using the position of the placed person. If so, there are $$P(9,6)$$ ways to do it.
• The cases where Alex and Bob sit together like a block: Let's place them anywhere when they are adjacent, then we have $$8$$ seats to place $$5$$ people, which we can do in $$P(8,5)$$ ways, but Alex and Bob can interchange in their block, so there are $$2!P(8,5)$$ ways.
Then, the answer is $$P(9,6) - 2!P(8,5)=60,480-13,440=47,040$$
Is my solution correct ?
• Yes, your solution is correct. Dec 4, 2021 at 12:21
Yes your work is correct. Another approach -
Alex takes one of the seats. That leaves $$7$$ seats for Bob to choose from (except adjacent seats to Alex). Rest $$5$$ of them can be seated in $$5$$ of the remaining $$8$$ seats.
So the answer is $$\displaystyle 7 \cdot {8 \choose 5} \cdot 5! = 47040$$
I agree with your analysis. I used an alternative approach:
$$S = ~$$total number of ways, without regard to Alex or Bob.
$$T = ~$$total number of ways, Alex and Bob are together.
Since you must have $$3$$ empty seats, I will pretend that these seats are filled by (indistinguishable) ghosts.
$$\displaystyle S = \frac{9!}{3!} = 60480$$.
Above, the numerator, which represents permuting $$(-1) +$$ the number of units, is appropriate for a circular table (as opposed to a straight row). That is, any of the $$(10)$$ units may be regarded as the head of the table.
Above, the denominator represents that the ghosts (i.e. empty chairs) are indistinguishable. So, the denominator represents an overcounting adjustment factor.
$$\displaystyle T = \frac{8! \times 2!}{3!} = 13440.$$
Above, the numerator represents that there are only $$8$$ units, since Alex and Bob are together. However, inside the Alex-Bob unit, Alex and Bob can be permuted in $$(2!)$$ ways.
$$S - T = 47040.$$ | 2023-01-27T12:24:35 | {
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https://byjus.com/question-answer/two-blocks-each-of-mass-m-3-50-kg-are-hung-from-the-ceiling-of/ | Question
# Two blocks, each of mass $$m = 3.50 kg$$, are hung from the ceiling of an elevator as in above figure. (a) If the elevator moves with an upward acceleration $$\vec{a}$$ of magnitude $$1.60 m/s^{2}$$, find the tensions $$T_{1}$$ and $$T_{2}$$ in the upper and lower strings. (b) If the strings can withstand a maximum tension of $$85.0 N$$, what maximum acceleration can the elevator have before a string breaks.
Solution
## (a) Free-body diagrams of the two blocks are shown below. Note that each block experiences adownward gravitational force$$F_{g} = (3.50 kg)(9.80 m/s^{2} ) = 34.3 N$$Also, each has the same upward acceleration as the elevator, in this case$$a_{y} = +1.60 m/s^{2}$$.Applying Newton’s second law to the lower block:$$\sum F_{y} = ma_{y} \Rightarrow T_{2} − F_{g} = ma_{y}$$or$$T_{2} = F_{g} + ma_{y} = 34.3 N + (3.50 kg)(1.60 m/s^{2} ) = 39.9 N$$Next, applying Newton’s second law to the upper block:$$\sum F_{y} = ma_{y} \Rightarrow T_{1} −T_{2} − F_{g} = ma_{y}$$or$$T_{1} = T_{2} + F_{g} + ma_{y} = 39.9 N + 34.3 N + (3.50 kg)(1.60 m/s^{2} )$$$$= 79.8N$$(b) Note that the tension is greater in the upper string, and this string will break first as the acceleration of the system increases. Thus, we wish to find the value of $$a_{y}$$ when $$T_{1} = 85.0$$. Making use of the general relationships derived in (a) above gives:$$T_{1} = T_{2} + F_{g} + ma_{y} = (F_{g} + ma_{y})+ F_{g} + ma_{y} = 2F_{g} + 2ma_{y}$$or$$a_{y}=\frac{T_{1}-2F_{g}}{2m}=\frac{85.0N-2(34.3N)}{2(3.50kg)}=2.34m/s^{2}$$.Physics
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https://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm | # Not every metric is induced from a norm
I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function
$d(u,v) = \lVert u - v \rVert$, $u,v \in V$.
My question is whether every metric on a linear space can be induced by norm? I know answer is no but I need proper justification.
Edit: Is there any method to check whether a given metric space is induced by norm ?
Thanks for help
• $\LaTeX$ tip: \parallel is a relation symbol, so it includes space on both sides. You want \lVert and \rVert for left and right delimiters, so that there is space on the "outside", but not on the "inside". Jul 4 '12 at 2:33
• @ArturoMagidin Thank you very much sir. Jul 4 '12 at 2:35
• Jul 4 '12 at 5:07
• @MattN. Thank you very much. That was helpful to me. Jul 4 '12 at 5:14
Let $V$ be a vector space over the field $\mathbb{F}$. A norm $$\| \cdot \|: V \longrightarrow \mathbb{F}$$ on $V$ satisfies the homogeneity condition $$\|ax\| = |a| \cdot \|x\|$$ for all $a \in \mathbb{F}$ and $x \in V$. So the metric $$d: V \times V \longrightarrow \mathbb{F},$$ $$d(x,y) = \|x - y\|$$ defined by the norm is such that $$d(ax,ay) = \|ax - ay\| = |a| \cdot \|x - y\| = |a| d(x,y)$$ for all $a \in \mathbb{F}$ and $x,y \in V$. This property is not satisfied by general metrics. For example, let $\delta$ be the discrete metric $$\delta(x,y) = \begin{cases} 1, & x \neq y, \\ 0, & x = y. \end{cases}$$ Then $\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.
To answer your edit, call a metric $$d: V \times V \longrightarrow \mathbb{F}$$ homogeneous if $$d(ax, ay) = |a| d(x,y)$$ for all $a \in \mathbb{F}$ and $x,y \in V$, and translation invariant if $$d(x + z, y + z) = d(x,y)$$ for all $x, y, z \in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\| \cdot \|$ by $$\|x\| = d(x,0)$$ for all $x \in V$.
• Thank you very much. I understand now. Jul 4 '12 at 2:42
• That is not enough as one needs to check that the so defined norm really induces the original metric and not another one: $d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$ Sep 29 '14 at 14:36
• This is an old post and it doesn't really matter much, but I have to be pedantic. So you say that if a metric comes from a norm on $\Bbb{R}$, then $d(ax,ay)=|a|d(x,y)$ but what if you just consider $\Bbb{R}$ as a vector space over $\{0,1\}$. Then the discrete metric IS homogeneous and translation invariant on this vector space.
– user223391
Jun 16 '15 at 0:16
• @MichaelMunta If the metric is induced by a norm, then $d(x,y)=||x-y||$. So $d(x+z,y+z)=||(x+z)-(y+z)||=||x-y||=d(x,y)$. Feb 27 '20 at 15:30
• You should clarify that your field $F$ is normed, otherwise the notation $|a|$ is meaningless. Jun 5 '20 at 23:24
Here is another interesting example: Let $|x-y|$ denote the usual Euclidean distance between two real numbers $x$ and $y$. Let $d(x,y)=\min\{|x-y|,1\}$, the standard derived bounded metric. Now suppose we look at $\Bbb{R}$ as a vector space over itself and ask whether $d$ comes from any norm on $\Bbb{R}$. Then if there is such a norm say $||.||$, we must have the homogeneity condition: for any $\alpha \in \Bbb{R}$ and any $v \in \Bbb{R}$,
$$||\alpha v || = |\alpha| ||v||.$$
But now we have a problem: The metric $d$ is obviously bounded by $1$, but we can take $\alpha$ arbitrarily large so that $||.||$ is unbounded. It follows that $d$ does not come from any norm.
• It was nice explanation to me. I had forgot to thank you. :) May 25 '13 at 9:21
As Henry states above, metrics induced by a norm must be homogeneous. You can see that they must also be translation invariant: $d(x+a,y+a)= d(x,y).$ So any metric not satisfying either of those can not come from a norm.
On the other hand, it turns out that these two conditions on the metric are sufficient to define a norm that induces that metric: $d(x,0)=\| x \|.$
• Thank you sir. I am fully satisfied with two answers given by you and Henry.:) Jul 4 '12 at 2:44
• Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..? Jul 4 '12 at 6:27
• @ThomasE. Sorry, you are right, I muddled up the words. Thanks. Jul 4 '12 at 8:43
Every homogeneous metric induces a norm via: $$\|x\|:=d(x,0)$$ and every norm induces a homogeneous and translation-invariant metric: $$d(x,y):=\|x-y\|$$
The clue herein lies in wether the induced norm really represents the metric as: $$d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$$ which is the case only for the translation-invariant metrics: $$d'(x,y)=d(x-y,0)=d(x,y)$$ whereas the induced metric always represents the norm as: $$\|\cdot\|\to d(\cdot,\cdot)\to\|\cdot\|$$ as a simple check shows: $$\|x\|'=\|x-0\|=\|x\|$$
• @ Thanks for the answer. Feb 15 '16 at 4:15
Personal Note:
Good question, I remember wondering the same thing myself back when I new to real analysis. Here's a counter example:
Counter Example
$$d(x,y):=I_{\{(x,x)\}}(x,y):=\begin{cases} 1 &:\, \,x=y\\ 0 &:\, \,x\neq y. \end{cases}$$ This is indeed a metric (check as an exercise).
Where $I_A$ is the indicator function of the set $A$; where here the set $(x,x)\in V\times V$.
If $r\in \mathbb{R}-\{0\}$, then $d(rx,ry)\in \{0,1\}$ hence $rd(x,y)\in \{0,r\}$ which is not in the range of $d:V\times V \rightarrow \mathbb{R}$.
So, the metric $d$ fails the property that any metric induced by a norm must have, ie: $$\mbox{it fails to have the property that: } \|rx-ry\| =r\|x-y\|.$$
Interpretation & Some Intuition:
The problem is that the topology is too fine, so all this topology can do is distinguish between things being the same or different.
As opposed to a norm topology which distinguised between object being, or not being on the same line (for some appropriate notion of line) (as well as them being different).
Hope this helps :)
• Thank you for the explanation. :) Feb 15 '16 at 4:15
• Hi, is it correct to consider the metric d(x,y)=|x-y|^2018 that fails to h;ave the above property? Oct 23 '18 at 8:19
• Yes, why not? :) Oct 23 '18 at 15:33
One possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous.
Proof: Take $$d$$ a bounded metric on a space $$X$$: i.e. $$\exists D \in \mathbb{R}_+$$ such that $$\forall (x,y) \in X^2, d(x,y) \leq D$$. Suppose now for contradiction that $$d$$ arises from a norm, i.e. the exists a norm $$\|\cdot\|$$ such that $$d(x,y) = \|x - y\|$$. Recall that the distance must then must be homogeneous, for we have $$d(\lambda x, \lambda y) = \|\lambda x - \lambda y\| = |\lambda| \cdot \|x - y\| = |\lambda| d(x,y)$$.
Take now arbitrary $$(x_0, y_0) \in X^2$$, then we must have that $$d\left( \frac{D+1}{d(x_0, y_0)} \cdot x_0, \frac{D+1}{d(x_0, y_0)} \cdot y_0 \right) = \frac{D+1}{d(x_0, y_0)} \cdot d(x_0, y_0) = D + 1 > D$$ which contradicts the upper bound of the metric. | 2021-09-25T03:32:19 | {
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https://www.physicsforums.com/threads/a-question-about-greatest-common-factor-gcf.911521/ | # B A question about Greatest common factor (GCF) ?
#### awholenumber
When we try to find the Greatest common factor (GCF) of two numbers , does it only involve prime factorization ?
#### FactChecker
Gold Member
2018 Award
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
#### awholenumber
Ok , so the same prime factorization is used to find the LCM too , right ?
#### FactChecker
Gold Member
2018 Award
Ok , so the same prime factorization is used to find the LCM too , right ?
Yes.
Ok , Thanks
#### SlowThinker
I dare to disagree, Euclid's algorithm or binary GCD are used to find GCD, then LCM(a, b)=a*b/GCD(a, b).
#### FactChecker
Gold Member
2018 Award
I stand corrected. I was not familiar with these algorithms. The binary GCD algorithm and Euclidean_algorithm are interesting.
#### awholenumber
Ok, i have one more doubt .
The GCF of two numbers involves prime factorization of those two numbers , then multiply those factors both numbers have in common
Isnt LCM about finding the least common multiple of two numbers ?
What does that have anything to do with prime numbers ?
#### mfb
Mentor
You can find the LCM via LCM(a, b)=a*b/GCD(a, b), if you have the GCD first.
That is often the most practical way to find the LCM.
Prime factorization is just one possible way to find the GCD. For large numbers, it can be very time-consuming, and different algorithms can be more efficient.
#### awholenumber
Thanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .
It doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .
#### PeroK
Homework Helper
Gold Member
2018 Award
Thanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .
It doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .
You'll find that in an "algebra for intelligent students" textbook!
#### awholenumber
lol ok , first let me somehow finish this one book properly
#### PeroK
Homework Helper
Gold Member
2018 Award
lol ok , first let me somehow finish this one book properly
It's still worth understanding why the two are related. The basic argument is:
$a = a'g, b = b'g$
Where $g$ is the GCD of $a$ and $b$ and $a', b'$ are how much bigger $a$ and $b$ are than $g$.
For example: if $a = 42$ and $b = 15$, then $g = 3$ and $a=14 \times 3, b = 5 \times 3$, hence $a' = 14, b' = 5$
Now, the LCM of $a$ and $b$ must have as factors simply $a', b'$ and $g$, so $l = a'b'g = a'b = a'gb/g = ab/g$
You can now see that the LCM is the product of $a$ and $b$, divided by the GCD.
In our example:
$LCM(42, 15) = \frac{42 \times 15}{3} = 210$
mfb
#### awholenumber
Thanks for sharing , i will keep this in my mind and maybe someday i will be able to use this advanced method .
#### Michael Hardy
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
But you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.
#### Michael Hardy
You don't need prime factorizations to find GCDs. There is an efficient method not requiring any knowledge of prime numbers: Euclid's algorithm. This is the oldest algorithm still in standard use, dating back to Euclid's writings in the 3rd century BC, and it is very efficient.
\begin{align*} \gcd(1989,867) & = \gcd(255,867) & & \text{since 255 is the remainder} \\ & & & \text{when 1989 is divided by 867} \\[10pt] & = \gcd(255,102) & & \text{since 102 is the remainder} \\ & & & \text{when 867 is divided by 255} \\[10pt] & = \gcd(51,102) & & \text{since 51 is the remainder} \\ & & & \text{when 255 is divided by 102} \\[10pt] & = \gcd(51,0) & & \text{since 0 is the remainder} \\ & & & \text{when 102 is divided by 51} \\[10pt] & = 51. \end{align*}
Thus we have $\dfrac{1989}{867} = \dfrac{51\times39}{51\times17} = \dfrac{39}{17}.$
#### Michael Hardy
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
This is wrong. Euclid's very efficient algorithm for finding GCDs does not require doing anything at all with prime numbers.
#### mfb
Mentor
But you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.
Yes, that has been mentioned in post 6 for example.
This thread is from 2017, by the way.
"A question about Greatest common factor (GCF) ?"
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• Solo and co-op problem solving | 2019-07-17T17:17:17 | {
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https://dsp.stackexchange.com/questions/38675/how-to-plot-magnitude-and-phase-response-of-2-cascaded-filters-in-matlab/38677 | # How to plot magnitude and phase response of 2 cascaded filters in Matlab?
How would I go about plotting a magnitude and phase response of a system that consists of two cascaded 2nd order Butterworth filters in Matlab? Filters are the same.
% Filter coeffs:
[B, A] = butter( 2, 1000/8000*2, 'high' ); % Cutoff at 1000 Hz, fs 8000 Hz.
% Cascade 2 filters: input -> filter -> filter -> output :
[out1, state1 ] = filter( B, A, in, state1);
[out2, state2 ] = filter( B, A, out1, state2);
Using MATLAB/Octave as the tool, the following approach lets you plot the magnitude & phase samples of the DTFT of the cascade of the two discrete-time LTI filters using their LCCDE coefficient vectors $$b[k]$$, and $$a[k]$$, assuming they have LCCDE representations..
Since the cascade LTI system is described as: $$h[n] = h_1[n] \star h_2[n]$$ and consequently $$H(z) = H_1(z) H_2(z)$$
Where the $$h[n]$$ is the impulse response of the cascade system and $$H(z)$$ is its transfer function (Z-transform of impulse response). Those $$h_1[n]$$ and $$h_2[n]$$ refer to the impulse responses of the individual systems that make up the cascade.
Assuming that individual systems do have rational Z-transforms, then we have the following expression for the composite (cascade) system Z-transfom:
\begin{align} H(z) &=\frac{ \sum_{k=0}^{M} B[k]z^{-k} } {\sum_{k=0}^{N} A[k]z^{-k}} = H_1(z) H_2(z) \\ \\ &= \left( \frac{ \sum_{k=0}^{M_1} b_1[k]z^{-k} } {\sum_{k=0}^{N_1} a_1[k]z^{-k}} \right) \left( \frac{ \sum_{k=0}^{M_2} b_2[k]z^{-k} } {\sum_{k=0}^{N_2} a_1[k]z^{-k}} \right) \end{align}
Where individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ are those that represent the cascaded systems.
We want $$B[k]$$ and $$A[k]$$ that represent the cascade system, from which we can use the following Matlab/Octave function to plot the DTFT magnitude and phase responses, assumimg that the cascade is a stable LTI system so that it will have a frequency response function:
figure,freqz(B,A);
Now it can be shown by polynomial manipulations that the coefficients $$B[k]$$ and$$A[k]$$ are related to the individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ as the following: $$B[k] = b_1[k] \star b_2[k]$$ and $$A[k] = a_1[k] \star a_2[k]$$
Hence the required frequency response plot can be obtained in the combined call as follows:
figure,freqz(conv(b1,b2),conv(a1,a2));
where b1,a1 and b2,a2 are the coefficient vectors that represent the individual systems. The method can be generalised to N systems in cascade.
When you pass a signal from two cascaded filters, what happens is that the magnitude response of the whole chain is the product of individual filters, and the phase response is the sum of individual phase responses. This is because the transfer functions are multiplied in the cascade implementation.
So if the frequency response of the single filters are $$H_A(w)=|H_A(w)|e^{j\angle H_A(w)}$$ $$H_B(w)=|H_B(w)|e^{j\angle H_B(w)}$$ then the frequency response of the two cascaded filters (overall) is \begin{align} H_{AB}(w)&=H_A(w)H_B(w)\\ &=\left(|H_A(w)|e^{j\angle H_A(w)}\right)\left(|H_B(w)|e^{j\angle H_B(w)}\right)\\ &=|H_A(w)||H_B(w)|e^{j(\angle H_A(w)+\angle H_B(w))} \end{align}
You can use [h,w] = freqz(b,a) in Matlab to get the frequency response of your desired filters. Use abs and angle to find the magnitude and phase:
...
[hA,w] = freqz(bA,aA);
[hB,w] = freqz(bB,aB);
hAB = hA.*hB;
MagResp = 20*log10(abs(hAB));
PhaseResp = angle(hAB);
plot(w,MagResp)
...
when the two filters are identical, then the overall magnitude is squared and the overall phase is scaled by two: \begin{align} H_{AA}(w)&=\left(|H_A(w)|e^{j\angle H_A(w)}\right)\left(|H_A(w)|e^{j\angle H_A(w)}\right)\\ &=|H_A(w)|^2e^{j2\angle H_A(w)} \end{align}
• Yes, filter is the same. – Danijel Mar 27 '17 at 11:51
• Yes, I added the general case at the beginning, then the identical case at the end. – msm Mar 27 '17 at 12:09
• Shouldn't it be wAB = w + w, and then plot(wAB)? Is angle(hAB) necessary? – Danijel Mar 27 '17 at 12:43
• w is the frequency axis and remains fix. We should just add function values at given w values. – msm Mar 27 '17 at 20:03
• angle(hAB) gives you the phase of the complex array hAB. Without that, you just have a bunch of complex values of size legth(w). – msm Mar 28 '17 at 1:42 | 2020-06-03T14:01:33 | {
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http://math.stackexchange.com/questions/133453/complex-integration-with-cauchys-integral-formula | # Complex integration with Cauchy's Integral Formula
Calculate$$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz$$ where $\gamma$ is the triangle whose vertices are the third roots of $z = -8i$, oriented counterclockwise.
Answer:
I calculated the third roots of $-8i$ and they all have modulus $2$. This tells me that the maximum distance of $\gamma$ from the origin will be $2$.
There are singularities at $z=0, z=-81$. As $81 > 2$, this singularity falls outside $\gamma$ so the only one that matters is $z = 0.$
I then applied Cauchy's Integral Formula $$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz = 2\pi i [\frac{(z+27i)(z+16)}{(z+81)^2}] |_{z=0}$$
And I got a final result of $\displaystyle\frac{-32\pi}{243}$.
Is my analysis and final result correct?
-
Yep, completely spot on! Only thing I can see you could have improved slightly was to notice the cube roots of $-8i$ should have modulus 2 without calculating them, saves you some work. – Ragib Zaman Apr 18 '12 at 14:00
(+1) for showing work. – The Chaz 2.0 Apr 18 '12 at 14:01
You should copy this to an answer and accept it :) – Neal Apr 18 '12 at 15:34
@Jim_CS If you require further feedback it might help to elaborate more specifically on any doubts. If not, please post an answer and accept it so that we know it is resolved. – Bill Dubuque May 17 '12 at 15:54
## 2 Answers
Answering my own question as Neal advised me to.
I calculated the third roots of −8i and they all have modulus 2. This tells me that the maximum distance of γ from the origin will be 2.
There are singularities at z=0,z=−81. As 81>2, this singularity falls outside γ so the only one that matters is z=0. I then applied Cauchy's Integral Formula ∫γ(z+27i)(z+16)z(z+81)2dz=2πi[(z+27i)(z+16)(z+81)2]|z=0 And I got a final result of −32π243.
Is my analysis and final result correct?
-
Nothing wrong with your analysis and answer.
- | 2015-10-10T04:28:07 | {
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https://www.intmath.com/forum/series-expansion-33/taylor-s-series-question:75 | IntMath Home » Forum home » Infinite Series Expansion » Taylor's series question
# Taylor's series question [Solved!]
### My question
my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p
1. Taylor Series
### What I've done so far
I think I need to differentiate it several times.
Here it is:
f(x) = x^(1/2)
f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))
f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))
f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))
But I'm stuck then
X
my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p
Relevant page
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
What I've done so far
I think I need to differentiate it several times.
Here it is:
f(x) = x^(1/2)
f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))
f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))
f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))
But I'm stuck then
## Re: Taylor's series question
Hello Snowboy
Your problem is very similar to the example on the page you came from:
1. Taylor Series
Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.
Or is you problem with the square root approximation?
Regards
X
Hello Snowboy
Your problem is very similar to the example on the page you came from:
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.
Or is you problem with the square root approximation?
Regards
## Re: Taylor's series question
So a=1, right?
f(1) = 1
f'(1) = 1/2
f''(1) = -1/4
f'''(1) = 3/8
Plugging it into the formula:
1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3-...
How many steps do we have to do?
X
So a=1, right?
f(1) = 1
f'(1) = 1/2
f''(1) = -1/4
f'''(1) = 3/8
Plugging it into the formula:
1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3-...
How many steps do we have to do?
## Re: Taylor's series question
The question says you need 4 d.p. accuracy.
So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.
X
The question says you need 4 d.p. accuracy.
So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.
## Re: Taylor's series question
It says "at 1.1".
Is this OK?
sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3
= 1 + 0.05 - 0.0025 + 0.000375 = 1.047875
I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.
Maybe I should do another step to make sure.
f''''(x) = -15/16 x^(-7/2)
f''''(1) = -15/16
So the 5th term is -15/16(x-1)^4
Plugging in 1.1 gives -0.00009375
Adding this gives 1.047875-0.00009375 = 1.04778125
To 4 d.p. it's still 1.0478.
But checking on my calculator, I get sqrt(1.1) = 1.04880884817
To 4 d.p., that's 1.0488. Is my answer wrong?
X
It says "at 1.1".
Is this OK?
sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3
= 1 + 0.05 - 0.0025 + 0.000375 = 1.047875
I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.
Maybe I should do another step to make sure.
f''''(x) = -15/16 x^(-7/2)
f''''(1) = -15/16
So the 5th term is -15/16(x-1)^4
Plugging in 1.1 gives -0.00009375
Adding this gives 1.047875-0.00009375 = 1.04778125
To 4 d.p. it's still 1.0478.
But checking on my calculator, I get sqrt(1.1) = 1.04880884817
To 4 d.p., that's 1.0488. Is my answer wrong?
## Re: Taylor's series question
Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.
Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).
To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:
The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.
You can see it's not a very good approximation as we get away from x=1.
X
Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.
Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).
To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:
[graph]310,250;-0.3,2.5;-0.5,2,0.5,0.5;sqrt(x),1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3[/graph]
The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.
You can see it's not a very good approximation as we get away from x=1.
## Re: Taylor's series question
Thanks a lot!
X
Thanks a lot! | 2017-10-17T09:40:46 | {
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https://www.physicsforums.com/threads/linear-second-order-differential-equation.534457/ | # Linear, second-order differential equation?
1. Sep 27, 2011
### Raen
One of my homework problems this week was:
Verify that y = sin(4t) + 2cos(4t) is a solution to the following initial value problem.
2y'' + 32y=0; y(0)=2, y'(0)=4
Find the maximum of |y(t)| for -infinity< t < infinity.
Verifying that the given y equation is a solution is easy, all there is to do is derive twice, plug in the new equations, and show that 0 = 0. I have no trouble with that. My question is with the second part, about finding the maximum positive value of y(t).
I would think that no matter how large t gets, sin and cos will always oscillate between -1 and 1. When cos is 1, obviously sin is 0. Putting in sin(4t) = 0 and cos(4t) = 1 gives me a maximum value of 2 for any y(t) as t approaches infinity. I graphed the function on a calculator to confirm this, just in case, and the graph appears to agree with me.
My confusion is that 2 isn't the correct answer. According to the website we do our homework on, the answer should be sqrt(5). There are no questions like this worked out in the book and I can't find anything online. Can anybody explain to me why y(t) has a maximum value of sqrt(5) as t approaches infinity?
Thank you!
2. Sep 27, 2011
### LCKurtz
Generally, when you have something like acos(θ) + bsin(θ) you can write it as a single trig function. You multiply and divide by the square root of the sum of the squares of the coefficients:
$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2} \left(\frac a {\sqrt{a^2+b^2}}\cos\theta + \frac b {\sqrt{a^2+b^2}}\sin\theta\right)$$
Now, since the sum of the squares of the new coefficients of the sine and cosine add up to one, they can be used as the sin(β) and cos(β) of some angle β. You then have an addition formula for a sine or cosine function with amplitude $\sqrt{a^2+b^2}$. In your example that gives amplitude $\sqrt 5$.
3. Sep 28, 2011
### HallsofIvy
Staff Emeritus
LCKurtz's response is, as always, excellent.
Raen, you seem to be assuming that a maximum of f+ g must occur where either f or g has a maximum (since you looked immediately at cos(x)= 1) and that is not correct.
Another way to find a maximum or minimum for a function is (as I am sure you know) is to look where the derivative is 0. If f(x)= sin(4x)+ 2cos(4x), then f'(x)= 4cos(4x)- 8sin(4x)= 0 when 4 cos(4x)= 8 sin(4x) or cos(4x)= 2sin(4x). Rather than solve for x directly, note that $cos^2(4x)= 4 sin^2(4x)$ so $sin^2(4x)+ cos^2(4x)= 5 sin^2(4x)= 1$ so that $sin^2(4x)= 1/5$ and $sin(4x)= 1/\sqrt{5}= \sqrt{5}/5$. And, of course, $cos^2(4x)= 1- sin^2(x)= 4/5$ so $cos(4x)= 2/\sqrt{5}= 2\sqrt{5}/5$.
Putting those back into the original formula, $sin(4x)+ 2cos(4x)= \sqrt{5}/5+ 4\sqrt{5}/5= 5/\sqrt{5}= \sqrt{5}$
Last edited: Sep 28, 2011
4. Sep 28, 2011
### Raen
Oh, I think I understand now. Thank you both for your detailed explanations! | 2017-09-23T16:57:02 | {
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https://math.stackexchange.com/questions/2380618/gaussian-elimination-inconsistent-or-infinite-solutions | # Gaussian elimination, inconsistent or infinite solutions?
I have this system
$\left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right]$ Is there something you can tell from looking at this straight away?
When I reduce it I get this:
$\left[ \begin{array}{ccc|c} 1&0&-1&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right]$
I thought I read that if you get a statement that does not make sense, then the system is inconsistent, so $1 = 0$ does not make sense to me. Does that not mean it's inconsistent?
However, I thought that $x_3$ could equal $0$, and so could $x_1$ and basically, $x_2$ could equal anything. Still not sure, i reduced it further to:
$\left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right]$
From this, am I correct now in thinking that:
$x_1 = r\\ x_2 = s \\ x_3 = t$
I am not sure of why I'm thinking like this, in terms of free variables etc. If I'm right, I wouldn't mind some reasoning as to why this is so, or being correct if I am wrong. Thanks.
• How did you do the final step of reduction? – platty Aug 2 '17 at 23:17
• Opps, I see where you are going with this. The 1 should remain in the second row. Sorry about that. Let me think for a minute. – Bucephalus Aug 2 '17 at 23:20
• Now that I have corrected my last reduction, I'm still not really sure what this means. it looks like $x_2$ can equal anything, Therefore, $x_1$ and $x_3$ can equal anything? Is that correct? – Bucephalus Aug 2 '17 at 23:24
Remember that the entries on the left are coefficients of variables. So from your final reduction, we would end up with the RREF: $$\left[\begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array}\right]$$
The first line tells us that $1x + 0y + 0z = 0$, so $x = 0$. The second line tells us that $0x + 0y + 1z = 0$, i.e. $z=0$. But we have $y$ as a free variable, so our solution is $(0,y,0)$.
Note that we can tell that $y$ is free by observing the original matrix - there is no coefficient in its column, so its value doesn't affect our final solution.
• Oh, yes I see now. Thanks for your explanation @S.Ong. – Bucephalus Aug 2 '17 at 23:27
• Say it did just turn out to be like I originally had it, with a single 1 in the first column of the first row. That would then mean that $x_1=0, x_2=r$ and $x_3=s$. Would that be right? – Bucephalus Aug 2 '17 at 23:29
• Yep, that's it! :) – platty Aug 2 '17 at 23:32
• Ok, thankyou @S.Ong. – Bucephalus Aug 2 '17 at 23:32
$\left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right]$
What do you know straight away?
$(0,0,0)$ is in the solution set. The right column being all $0$ tells you this.
With the second row and the 3rd row identical or the right part of the matrix, this matrix is singular.
In fact the rank is 2, the nullity is 1, and there is a "line" of solutions.
The second column all zero's tells you the very same thing. Furthermore, $x_2$ can be anything.
And the structure of rows 2 and 3 tells you that $x_1$ must equal $0.$
this is more than enough information to let you know the solution set is $(0,t,0)$
• Aaah, very insightful. This is the kind of thing I was looking for also. Do you mean by "line" that the solutions form a line in 3-space? @DougM – Bucephalus Aug 3 '17 at 0:30
• yes, with the rank of the kernel equal to 1, (and the information that at least one solution exists), then all of the solutions will lie on a line in $\mathbb R^3.$ If the nullity equaled 2, then we might say that there is a plane of solutions. – Doug M Aug 3 '17 at 0:36
• Thanks @DougM. I will keep an eye out for you next time. – Bucephalus Aug 3 '17 at 0:43 | 2020-02-18T06:57:53 | {
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https://stats.stackexchange.com/questions/379335/adding-very-small-probabilities-how-to-compute/379336 | Adding very small probabilities - How to compute?
In some problems, probabilities are so small that they are best represented in computational facilities as log-probabilities. Computational problems can arise when you try to add these small probabilities together, since some computational facilities (e.g., base R) can't distinguish very small probabilities from zero. In these cases it is necessary to solve the problem either by finding a workaround that avoids the small numbers, or by using a computational facility that can deal with very small numbers.
The archetypal problem of this kind is as follows. Suppose you have log-probabilities $$\ell_1$$ and $$\ell_2$$, where the corresponding probabilities $$\exp(\ell_1)$$ and $$\exp(\ell_2)$$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-sum of these probabilities, which we denote by:
$$\ell_+ \equiv \ln \big( \exp(\ell_1) + \exp(\ell_2) \big)$$
Assume that we are ---at least initially--- working in an environment where $$\exp(\ell_1)$$ and $$\exp(\ell_2)$$ cannot be computed, since they are so small that they are indistinguishable from zero.
Questions: How can you effectively compute this log-sum? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?
• the "log-sum-exp" trick can help – Taylor Nov 29 '18 at 3:44
Some preliminary mathematics: As a preliminary matter it is worth noting that this function is the LogSumExp (LSE) function, which is often encountered when performing computation on values that are represented in the log-scale. To see how to deal with sums of this kind, we first note a useful mathematical result concerning sums of exponentials:
\begin{aligned} \exp(\ell_1) + \exp(\ell_2) &= \exp(\max(\ell_1,\ell_2)) + \exp(\min(\ell_1,\ell_2)) \\[6pt] &= \exp(\max(\ell_1,\ell_2)) (1 + \exp(\min(\ell_1,\ell_2)-\max(\ell_1,\ell_2)) \\[6pt] &= \exp(\max(\ell_1,\ell_2)) (1 + \exp(-|\ell_1 - \ell_2|)). \\[6pt] \end{aligned}
This result converts the sum to a product, which allows us to present the log-sum as:
\begin{aligned} \ell_+ &= \ln \big( \exp(\ell_1) + \exp(\ell_2) \big) \\[6pt] &= \ln \big( \exp(\max(\ell_1,\ell_2)) (1 + \exp(-|\ell_1 - \ell_2|)) \big) \\[6pt] &= \max(\ell_1, \ell_2) + \ln (1 + \exp(-|\ell_1 - \ell_2|)). \\[6pt] \end{aligned}
In the case where $$\ell_1 = \ell_2$$ we obtain the expression $$\ell_+ = \ell_1 + \ln 2 = \ell_2 + \ln 2$$, so the log-sum is easily computed. In the case where $$\ell_1 \neq \ell_2$$ this expression still reduces the problem to a simpler case, where we need to find the log-sum of one and $$\exp(-|\ell_1 - \ell_2|)$$.$$^\dagger$$
Now, using the Maclaurin series expansion for $$\ln(1+x)$$ we obtain the formula:
\begin{aligned} \ell_+ &= \max(\ell_1, \ell_2) + \sum_{k=1}^\infty (-1)^{k+1} \frac{\exp(-k|\ell_1 - \ell_2|)}{k} \quad \quad \quad \text{for } \ell_1 \neq \ell_2. \\[6pt] \end{aligned}
Since $$\exp(-|\ell_1 - \ell_2|) < 1$$ the terms in this expansion diminish rapidly (faster than exponential decay). If $$|\ell_1 - \ell_2|$$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.
Implementation in base R: It is actually possible to compute this log-sum accurately in base R through creative use of the log1p function. This is a primitive function in the base package that computes the value of $$\ln(1+x)$$ for an argument $$x$$ (with accurate computation even for $$x \ll 1$$). This primitive function can be used to give a simple function for the log-sum:
logsum <- function(l1, l2) { max(l1, l2) + log1p(exp(-abs(l1-l2))); }
Implementation of this function succeeds in finding the log-sum of probabilities that are too small for the base package to deal with directly. Moreover, it is able to calculate the log-sum to a high level of accuracy:
l1 <- -3006;
l2 <- -3012;
logsum(l1, l2);
[1] -3005.998
sprint("%.50f", logsum(l1, l2));
[1] "-3005.99752431486240311642177402973175048828125000000000"
As can be seen, this method gives a computation with 41 decimal places for the log-sum. It only uses the functions in the base package, and does not involve any changes to the default calculation settings. It gives as high a level of accuracy as you are likely to need in most cases.
It is also worth noting that there are many packages in R that extend the computational facilities of the base program, and can be used to deal with sums of very small numbers. It is possible to find the log-sum of small probabilities using packages such as gmp or Brobdingnag, but this requires some investment in learning their particular syntax.
$$^\dagger$$ From this result we can also see that if $$|\ell_1 - \ell_2|$$ is itself large (i.e., if one of the probabilities is very small compared to the other) then the exponential term in this equation will be near zero, and we will then have $$\ell_+ \approx \max(\ell_1, \ell_2)$$ to a very high degree of accuracy.
• A number of posts on site discuss related issues. – Glen_b Nov 29 '18 at 4:53
• Sorry, I must have missed these. – Ben Nov 29 '18 at 5:05
• There's at least some material here that I don't think is anywhere else, and the exposition is quite clear so I don't think there's a problem, but it's worth being aware of the other posts. (A lot crop up in discussions of underflow and overflow in various calculations, for example) – Glen_b Nov 29 '18 at 5:29
• I have added the underflow tag to the post to link to other questions on this subject. – Ben Nov 29 '18 at 6:31 | 2021-02-25T08:03:17 | {
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http://math.stackexchange.com/questions/129012/simple-question-related-to-the-fundamental-theorem-of-arithmetic | # Simple question related to the Fundamental theorem of arithmetic
Given a number $b$, which we write as a product of prime numbers:
$$b = p_1 \cdots p_s$$
Can I then deduce that a number, which divides $b$ then has to be a product of the above primes in the factorization of $b$?
-
Yes and that follows by writing out another factorization for the number that divides $b = p_1 \cdots p_s$ and using unique factorization to dedude that the factors must be some of the primes in the factorization of $b$. – Adrián Barquero Apr 7 '12 at 15:34
More explicitly, if $b = p_{1}^{e_1} \cdots p_{s}^{e_s}$ is the prime factorization of $b$ where the $p_i$'s are different primes, then you can prove that any divisor of $b$ is a product of the form $p_{1}^{a_1} \cdots p_{s}^{a_s}$ where the exponents $a_i$ satisfy that $0 \leq a_i \leq e_i$ for $i = 1, \dots , s$. – Adrián Barquero Apr 7 '12 at 15:38
Essentially yes: If $p_1,\ldots,p_s$ are primes (possibly repeated), then every divisor of $$b = p_1p_2\cdots p_s$$ must be a (possibly empty) product of some (possibly all) of the $p_i$, times $1$ or $-1$.
-
And why the downvote? – Arturo Magidin Apr 11 '12 at 6:05
A divisor of $b$ has to be a product of some of the prime factors of $b$, but "some" may mean $0$ of them (if the divisor is $1$). The divisor can have no prime factors that are not prime factors if $b$.
If the divisor in question is $c$, then for some number $d$ we have $b=cd$. If a prime number $p$ divides $b$, then $p$ must divide either $c$ or $d$ (or both). The foregoing sentence is Euclid's lemma.
-
Hint $\rm\ p\ |\ d\ |\ p_1\cdots p_n\ \Rightarrow\ p\ |\ p_j\:$ for some $\rm\:j\:$ by the Prime Divisor Property. Cancelling $\rm\:p\:$ from $\rm\:d\:$ and $\rm\:p_1\cdots p_n\:$ and inducting yields that the prime factors of $\rm\:d\:$ are a sub-multiset of $\rm\{p_1,\ldots,p_n\}.$
Generally $\rm\:d\ |\ b_1\cdots b_n\ \Rightarrow\ d = d_1\cdots d_n,\ \ d_j\ |\ b_j,\:$ i.e. a divisor of a product is a product of divisors. This is a useful generalization of the Prime Divisor Property from atoms to composite numbers. Yours is the special case when all $\rm\:b_j = p_j\:$ prime, so $\rm\:d_j = 1\:$ or $\rm\:p_j\:$ (modulo a unit factor $\pm1$).
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@Downvoter If something is not clear, please feel free to ask questions and I will be happy to elaborate. – Bill Dubuque Apr 27 '12 at 19:13 | 2016-02-11T21:28:47 | {
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http://mathhelpforum.com/discrete-math/15449-pascal-s-triangle.html | # Math Help - Pascal's Triangle
1. ## Pascal's Triangle
Being naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.
This is what i've discovered so far.
Row 1 = 1 ; Sum = 1
Row 2 = 1 1 ; Sum = 2
Row 3 = 1 2 1 ; Sum = 4
Row 4 = 1 3 3 1 ; Sum = 8
Row 5 = 1 4 6 4 1 ; Sum = 16
So the sum of these numbers in each row, are multiples of 2.
2. Originally Posted by janvdl
Now i know this might sound silly...
Being naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.
This is what i've discovered so far.
Row 1 = 1 ; Sum = 1
Row 2 = 1 1 ; Sum = 2
Row 3 = 1 2 1 ; Sum = 4
Row 4 = 1 3 3 1 ; Sum = 8
Row 5 = 1 4 6 4 1 ; Sum = 16
So the sum of these numbers in each row, are multiples of 2.
seems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there
3. Originally Posted by Jhevon
seems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there
Ah ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?
4. Originally Posted by janvdl
Ah ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?
seems to be $2^n - 1$ where $n$ is the number of the row we are at
5. Originally Posted by Jhevon
seems to be $2^n - 1$ where $n$ is the number of the row we are at
I was thinking in that line. cant believe i was so blind
6. Hello, janvdl!
Being naturally curious . . .
Good for you!
I made some amazing "discoveries" while exploring on my own.
Of course, anything I found was proved centuries ago,
. . but it was still very satisfying.
What you discovered is a stunning pattern.
Consider the coefficients of the expansion of $(a + b)^n$
. . . $\begin{array}{ccc}n & \text{coefficients} & \text{sum} \\ \hline 0 & 1 & 1 = 2^0\\ 1 & 1\;\;1 & 2 = 2^1 \\ 2 & 1\;\;2\;\;1 & 4 = 2^2 \\ 3 & 1\;\;3\;\;3\;\;1 & 8 =2^3 \\ 4 & 1\;\;4\;\;6\;\;4\;\;1 & 16 = 2^4 \\ 5 & 1\;\;5\;\;10\;\;10\;\;5\;\;1 & 32 = 2^5 \\ 6 & 1\;\;6\;\;15\;\;20\;\;15\;\;6\;\;1 & 64 = 2^6 \\ \vdots & \vdots & \vdots \end{array}$
The sum of the $n^{th}$ row is $2^n$.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I "discovered" this pattern while in college.
$\begin{array}{ccc}1 + 2 & = & 3 \\ 4 + 5 + 6 & = & 7 + 8 \\ 9 + 10 + 11 + 12 & = & 13 + 14 + 15 \\ 16 + 17 + 18 + 19 + 20 & = & 21 + 22 + 23 + 24 \\
\vdots & & \vdots\end{array}$
I found these in a book . . .
$\begin{array}{ccc}1 & = & 1 = 1^3 \\ 3 + 5 & = & 8 = 2^3 \\ 7 + 9 + 11 & = & 27 = 3^3 \\ 13 + 15 + 17 + 19 & = & 64 = 4^3 \\ 21 + 23 + 25 + 27 + 29 & = & 125 = 5^3 \\ \vdots & & \vdots \end{array}$
$\begin{array}{ccc}3^2 + 4^2 & = & 5^2 \\ 10^2 + 11^2 + 12^2 & = &13^2 + 14^2 \\ 21^2 + 22^2 + 23^2 + 24^2 & = & 25^2 + 26^2 + 27^2 \\ 36^2 + 37^2 + 38^2 + 39^2 + 40^2 & = & 41^2 + 42^2 + 43^2 + 44^2 \\ 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 & = & 61^2 + 62^2 + 63^2 + 64^2 + 65^2 \\ \vdots & &\vdots\end{array}$
7. I've found something else too...
I was actually trying to find another type of formula when i noticed this...
I attached a picture, i know its badly drawn but at least it gives you the idea...
The white triangles are the normal figures of Pascal's Triangle. But look at those red triangles. They interest me a lot. There must be a lot more patterns involved in this triangle than i ever thought.
What if every red triangle is the product of the 3 white triangles around it? Or the sum of the 3 white ones around it? We can make lots of new patterns using this...
8. Hello, janvdl!
That's fascinating . . . Thank you!
What if every red triangle is the sum of the 3 white ones around it?
Then we have:
. . $\begin{array}{c}3\\
4\;\;4\\
5\;\;8\;\;5 \\
6\;\;13\;\;13\;\;6 \\
7\;\;19\;\;26\;\;19\;\;7
\end{array}$
I finally saw where it came from . . .
Consider a "Pascal's Triangle" whose third line is: .1 - 3 - 1.
. . $\begin{array}{c}1 \\ 1\;\;1 \\ 1\;\;3\;\;1 \\
1\;\;4\;\;4\;\;1 \\ 1\;\;5\;\;8\;\;5\;\;1 \\ 1\;\;6\;\;13\;\;13\;\;6\;\;1 \\ 1\;\;7\;\;19\;\;26\;\;19\;\;7\;\;1
\end{array}$
See it?
This opens up a world of possibilities, doesn't it?
9. Originally Posted by Soroban
Hello, janvdl!
That's fascinating . . . Thank you!
Then we have:
. . $\begin{array}{c}3\\
4\;\;4\\
5\;\;8\;\;5 \\
6\;\;13\;\;13\;\;6 \\
7\;\;19\;\;26\;\;19\;\;7
\end{array}$
I finally saw where it came from . . .
Consider a "Pascal's Triangle" whose third line is: .1 - 3 - 1.
. . $\begin{array}{c}1 \\ 1\;\;1 \\ 1\;\;3\;\;1 \\
1\;\;4\;\;4\;\;1 \\ 1\;\;5\;\;8\;\;5\;\;1 \\ 1\;\;6\;\;13\;\;13\;\;6\;\;1 \\ 1\;\;7\;\;19\;\;26\;\;19\;\;7\;\;1
\end{array}$
See it?
This opens up a world of possibilities, doesn't it?
So have i actually kind of discovered something here?
And what can we actually do with my "discovery"? A world of possibilities has been opened? You're clearly seeing things that i cant...
10. Here's another of my "discoveries", made years and years ago.
We know the Binomial Theorem is based on Pascal's Triangle.
We use Pascal's Triangle for expanding $(a + b)^n$.
I wondered if there was a "Trinomial Theorem" for expanding $(a + b + c)^n$.
I cranked out the first few cases . . . and got into an awful mess.
. . $\begin{array}{ccc}(a + b + c)^0 & = & 1 \\
(a + b + c)^1 & = & a + b + c \\
(a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\
(a + b + c)^3 & = & a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc
\end{array}$
How do we make sense out of these terms and coefficients?
It took me the longest time to see what to do:
. . arrange the terms in triangular arrays.
$\begin{array}{ccccc} & & & & a^4 \\ & & &a^3 & 4a^3b\;\;4a^3c \\& & a^2 & 3a^2b\;\;3a^2c & 6a^2b^2\;\;12a^2bc\;\;6a^2c^2 \\& a & 2ab\;\;2ac & 3ab^2\;\;6abc\;\;3ac^2 & 4ab^3\;\;12ab^2c\;\;12abc^2\;\;4ac^3 \\1\quad &\;\; b\;\;c\quad & b^2\;\;2ab\;\;c^2\quad & b^3\quad3b^2c\quad3bc^2\quad c^3\quad & b^4\quad4b^3c\quad6b^2c^2\quad 4bc^3\quad c^4\end{array}$
Look at the coefficients:
$\begin{array}{ccccc} & & & & 1 \\& & & 1 & 4\;\;4 \\& & 1 & 3\;\;3 & 6\;\;12\;\;6 \\& 1 & 2\;\;2 & 3\;\;6\;\;3 & 4\;\;12\;\;12\;\;4 \\1\quad & \;\;\;1\;\;1\quad & \;\;1\;\;2\;\;1\quad & \;\;1\;\;3\;\;3\;\;1\quad & 1\;\;\;4\;\;\;6\;\;\;4\;\;\;1\end{array}$
If we "stack" these triangles, we get a tetrahedron.
. . Each number is the sum of the three numbers above it.
I seriously doubt that I'm the first to notice all this.
But you can feel free to call it "Soroban's Tetrahedron".
11. I'm just going to try and get some formulas out of my triangle Im busy writing a small book on my discoveries in this triangle. Just for the fun of it | 2014-11-26T13:53:11 | {
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https://web2.0calc.com/questions/assigning-var | +0
# Assigning var
0
232
4
+4
Is there a way to define a variable to use an equation? E.x.: 3x^2-2x+5=y, and I want to assign x a value without changing it out. Is it possible?
CatBoy13 Dec 12, 2014
#3
+91479
+10
Welcome to web2.0calc forum Catboy13
Umm - Interesting question.
$$\begin{array}{rll} 3x^2-2x+5&=&y\\\\ 3x^2-2x&=&y-5\\\\ x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\ x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\ x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\ \end{array}$$
Melody Dec 13, 2014
Sort:
#1
+7188
0
I believe it is possible....but I might be wrong...
happy7 Dec 12, 2014
#2
0
If you first define a function f as
$${f}{\left({\mathtt{x}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}$$
then for any particular x the function f(x) has a value and you can write things such as
f(2) + f(0) - 4 = 14
Guest Dec 12, 2014
#3
+91479
+10
Welcome to web2.0calc forum Catboy13
Umm - Interesting question.
$$\begin{array}{rll} 3x^2-2x+5&=&y\\\\ 3x^2-2x&=&y-5\\\\ x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\ x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\ x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\ \end{array}$$
Melody Dec 13, 2014
#4
+91479
0
I have added this to the Sticky Topic - "Great Answers to Learn From"
Melody Dec 14, 2014
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We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 2018-01-22T08:15:13 | {
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https://puzzling.stackexchange.com/questions/100491/calculation-puzzle-010 | # Calculation puzzle 010
Find the missing number in the sequence.
2 9 28 65 ? 217
Source: This question is taken from YTU YOS 2018 exam. I have mentioned them in other posts.
It's simply just
$$n^3 + 1$$
So
$$1^3 + 1 = 1 + 1 = 2$$
$$2^3 + 1 = 8 + 1 = 9$$
$$3^3 + 1 = 27 + 1 = 28$$
$$4^3 + 1 = 64 + 1 = 65$$
$$5^3 + 1 = 125 + 1 = 126$$
$$6^3 + 1 = 216 + 1 = 217$$
Therefore the missing number is
$$126$$
• Gz mate nice one Jul 27, 2020 at 12:51
The answer is 126 because of the pattern 1cubed +1, 2cubed+1, 3 cubed +1,4cubed +1, 5cubed+1,6cubed+1
• By the way I did not copy this answer Jul 27, 2020 at 12:47
• There was no answer as I was typing it Jul 27, 2020 at 12:47
• Don't worry, people will be able to see that you didnt from the timestamp, as you can see we both answered at almost the same time Jul 27, 2020 at 12:52
• Yesterday I answered the question 3 minutes before another guy and he still got the credit😂 I don't have good luck here Jul 27, 2020 at 12:54
• Sometimes a later answer will be accepted if the OP thinks that it's more correct or explains it better, it's not always first to get it is accepted :) Jul 27, 2020 at 12:55 | 2022-05-21T04:08:10 | {
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https://math.stackexchange.com/questions/1723465/if-n-is-a-positive-integer-does-n3-1-always-have-a-prime-factor-thats-1-m | # If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?
It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
Regarding $n^4-1$: If $n>1$, we factor $n^4-1$ as $(n+1)(n-1)(n^2+1)$. We claim that in fact, every prime factor of $n^2+1$ is either 2 or is congruent to 1 mod 4. If $p$ is an odd prime that divides $n^2+1$, then $-1$ is a square mod $p$, but the odd primes for which $-1$ is a square mod $p$ are precisely the primes congruent to 1 mod 4. It remains just to show that $n^2+1$ cannot be a power of 2. If $n$ is even this is obvious, and if $n=2k+1$ is odd, then $n^2+1=(2k+1)^2+1=4k^2+4k+2$ is 2 more than a multiple of 4.
Regarding $n^5-1$, I don't have a proof, but based on experimenting with a few dozen numbers, I conjecture that in fact, every prime factor of $n^4+n^3+n^2+n+1$ is either 5 or is 1 more than a multiple of 5.
• From the factorization you can get a factor that's $\equiv 1 \mod 3$, but not necessarily a prime one – MCT Apr 1 '16 at 15:38
• Ran a code for the for all such numbers less than $10^7$. Seems ok. I am addin a list of the first few numbers and their prime factors. (7, 7) (26, 13) (63, 7) (124, 31) (215, 43) (342, 19) (511, 7) (728, 7) (999, 37) (1330, 7) (1727, 157) (2196, 61) (2743, 13) (3374, 7) (4095, 7) (4912, 307) (5831, 7) (6858, 127) (7999, 19) (9260, 463) (10647, 7) (12166, 7) (13823, 601) (15624, 7) (17575, 19) (19682, 13) (21951, 271) (24388, 7) (26999, 7) (29790, 331) (32767, 7) (35936, 1123) (39303, 397) (42874, 13) (46655, 7) (50652, 7) (54871, 37) (59318, 7) (63999, 13) (68920, 1723) (74087, 13) (79506, 7) – Banach Tarski Apr 1 '16 at 15:49
• Ran the code again. This statement seems to be true for all such numbers smaller than $2 * 10^8$ – Banach Tarski Apr 1 '16 at 15:56
• @idmercer: For prime powers $p>2$, you can generalize it neatly. Define $$F(n) = \frac{n^p-1}{n-1}$$ Conjecture: "If $F(n)$ is not divisible by $p$, then every odd prime factor of $F(n)$ is $1\mod 2p$." I'm not sure this is a proven result though, because otherwise the guys who answered would have used it. Maybe you can ask the general case for primes separately. – Tito Piezas III Apr 1 '16 at 22:01
The case $n=1$ is uninteresting, so let $n\gt 1$. We show that $n^2+n+1$ has a prime factor congruent to $1$ modulo $3$, by showing that $4(n^2+n+1)$ has such a prime factor.
Note that $n^2+n+1$ is odd. We first show that $3$ is not the only odd prime that divides $n^2+n+1$. For suppose to the contrary that $(2n+1)^2+3=4\cdot 3^k$ where $k\gt 1$. Then $2n+1$ is divisible by $3$, so $(2n+1)^2+3\equiv 3\pmod{9}$, contradiction.
Now let $p\gt 3$ be a prime divisor of $(2n+1)^2+3$. Then $-3$ is a quadratic residue of $p$. A straightforward quadratic reciprocity argument shows that $p$ cannot be congruent to $2$ modulo $3$.
Remark: By considering $n$ of the form $q!$ for possibly large $q$, we can use the above result to show that there are infinitely many primes of the form $6k+1$.
• Is there a general result for the odd prime factors of $\displaystyle F(n)=\frac{n^p-1}{n-1}$ for prime $p$? (The OP was just focusing on the case $p=3$.) In other words, is every odd prime factor of $F(n)$ either $p$ or $1\mod p$? – Tito Piezas III Apr 2 '16 at 18:51
• Analysis is pretty simple (Fermat's Theorem) for $n$ and $p$ such that $n$ is not congruent to $1$ mod $p$. But there can be funny prime divisors otherwise. There should be a complete easy answer in general, but I looked and am missing something. That's why the appeal to reciprocity in the answer. – André Nicolas Apr 2 '16 at 18:54
• Yes, Fermat's little theorem occurred to me as well. I may ask the general prime case in the forum. Do you think you'll figure out the answer to the general case? :) – Tito Piezas III Apr 2 '16 at 18:57
• @TitoPiezasIII: Someone will. Conceivably me, but probably not. Would likely not be in any case, but also today is a massive cooking day. – André Nicolas Apr 2 '16 at 19:00
• Ok, Bon appetit. (P.S. The question is here.) – Tito Piezas III Apr 2 '16 at 19:20
For $p\equiv 1\pmod 3$, there exist three distinct solutions of $x^3\equiv 1\pmod p$, and if $n$ is congruent modulo $p$ to such $x$ then $n^3-1$ is a multiple of $p$. Heuristically, this solves the problem affirmatively for a fraction $\frac 3p$ of all possible $n$.
Let's make this a bit less hand-wavy: Let $p_1=7, p_2=13, p_3=19,\ldots$ denote the sequence of prime $\equiv 1\pmod 3$. Then of the $M:=p_1p_2\cdots p_m$ residue classes modulo $M$, there are only $(1-\frac3{p_1})(1-\frac3{p_2})\cdots(1-\frac3{p_m})M=(p_1-3)(p_2-3)\cdots(p_m-3)$ residue classes $x$ for which $n\equiv x\pmod M$ does not imply that $n^3-1$ has a prime factor $\in\{p_1,\ldots,p_m\}$. As we let $m\to \infty$, the product $(1-\frac3{p_1})(1-\frac3{p_2})\cdots(1-\frac3{p_m})$ can be shown to tend to $0$. So at least the density of $n$ that do not work must be zero. | 2019-07-21T02:14:29 | {
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https://math.stackexchange.com/questions/351363/find-a-simple-formula-for-binomn0-binomn1-binomn1-binomn2 | # Find a simple formula for $\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+…+\binom{n}{n-1}\binom{n}{n}$
$$\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}$$
All I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.
First note that $\dbinom{n}k = \dbinom{n}{n-k}$. Hence, your sum can be written as $$\sum_{k=0}^{n-1} \dbinom{n}k \dbinom{n}{k+1} = \sum_{k=0}^{n-1} \dbinom{n}k \dbinom{n}{n-k-1}$$ Now consider a bag with $n$ red balls and $n$ blue balls. We want to choose a total of $n-1$ bals. The total number of ways of doing this is given by $$\dbinom{2n}{n-1} \tag{\star}$$ However, we can also count this differently. Any choice of $n-1$ balls will involve choosing $k$ blue balls and $n-k-1$ red balls. Hence, the number of ways of choose $n-1$ balls with $k$ blue balls is $$\color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-1-k}}$$ Now to count all possible ways of choosing $n-1$ balls, we need to let our $k$ run from $0$ to $n-1$. Hence, the total number of ways is $$\sum_{k=0}^{n-1} \color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-k-1}} \tag{\dagger}$$ Now since $(\star)$ and $(\dagger)$ count the same thing, they must be equal and hence, we get that $$\sum_{k=0}^{n-1} \color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-k-1}} = \dbinom{2n}{n-1}$$
EDIT As Brian points out in the comments, the above is a special case of the more general Vandermonde's identity. $$\sum_{k=0}^r \dbinom{m}k \dbinom{n}{r-k} = \dbinom{m+n}r$$ The proof for this is essentially the same as above. Consider a bag with $m$ blue balls and $n$ red balls and count the number of ways of choosing $r$ balls from these $m+n$ balls in two different ways as discussed above.
• Love the formatting! +1 – Macavity Apr 4 '13 at 18:32
• (+1) Nicely explained. You might want to mention that it’s a case of Vandermonde’s identity, which has essentially the same proof. – Brian M. Scott Apr 4 '13 at 18:43
• Thanks so much!!!! Your explanation was great!!!! – Dome Apr 4 '13 at 18:49
We have $$\sum_{k=0}^{n-1}\dbinom{n}{k}\dbinom{n}{k+1}=\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{k+1}=\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{n+1-k}=\color{red}{\dbinom{2n}{n+1}}$$ by the Chu-Vandermonde identity and since $\dbinom{n}{n}\dbinom{n}{n+1}=0$. Another way is to use the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz$$ and get $$\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{k+1}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{2}}\sum_{k=0}^{n}\dbinom{n}{k}z^{-k}dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{2}}\left(1+\frac{1}{z}\right)^{n}dz=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{2n}}{z^{n+2}}dz=\color{blue}{\dbinom{2n}{n+1}}.$$
There is a simple combinatorial interpretation. Assume that you have a parliament with $n$ politicians in the left wing and $n$ politicians in the right wing. If you want to select a committee made by $n-1$ politicians, you obviously have $\binom{2n}{n-1}=\binom{2n}{n+1}$ ways for doing it. On the other hand, by classifying the possible committees according to the number of politicians of the left wing in them, you have: $$\binom{2n}{n-1}=\sum_{l=0}^{n-1}\binom{n}{l}\binom{n}{n-1-l} = \sum_{l=0}^{n-1}\binom{n}{l}\binom{n}{l+1}$$ as wanted.
Hint: it's the coefficient of $T$ in the binomial expansion of $(1+T)^n(1+T^{-1})^n$, which is equivalent to saying that it's the coefficient of $T^{n+1}$ in the expansion of $(1+T)^n(1+T^{-1})^nT^n=(1+T)^{2n}$.
Note that $\ds{{n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}}$
\begin{align} \sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1} & = \sum_{k = 0}^{n - 1} {\bracks{{n \choose k} + {n \choose k + 1}}^{\,2} - {n \choose k}^{2} - {n \choose k + 1}^{2}\over 2} \\[5mm] & = {1 \over 2}\sum_{k = 0}^{n - 1}{n + 1 \choose k + 1}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k + 1}^{2} \\[5mm] & = {1 \over 2}\sum_{k = 1}^{n}{n + 1 \choose k}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} - {1 \over 2}\sum_{k = 1}^{n}{n \choose k}^{2} \\[5mm] & = {1 \over 2}\bracks{\sum_{k = 0}^{n + 1}{n + 1 \choose k}^{2} -2} - {1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1} - {1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1} \\[5mm] & = {1 \over 2}{2n + 2 \choose n + 1} - {2n \choose n} \end{align} where I used the well known result $\bbx{\ds{\quad\sum_{i = 0}^{m}{m \choose i}^{2} = {2m \choose m}}}$. Moreover, \begin{align} \sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1} & = {1 \over 2}\,{\pars{2n + 2}\pars{2n + 1}\pars{2n}! \over \pars{n + 1}n!\pars{n + 1}n!} - {2n \choose n} = \pars{{2n + 1 \over n + 1} - 1}{2n \choose n} \\[5mm] & = {n \over n + 1}{2n \choose n} = {n \over n + 1}{\pars{2n}! \over n!\,n!} = {\pars{2n}! \over \pars{n + 1}!\pars{n - 1}!} = \bbx{\ds{2n \choose n + 1}} \end{align}
• It is not easy to add a new perspective to a popular sum like this one. Upvoted for originality of the approach. (+1): – Marko Riedel Feb 1 '17 at 23:24
• @MarkoRiedel Thanks. Cross terms always suggest this approach. It's useful too for products of $\log$'s, etc... – Felix Marin Feb 2 '17 at 0:14
Rewrite the sum as $\displaystyle\sum_0^{n-1}\binom{n}{k}\binom{n}{k+1}$, then by combinatorial interpretation, this is: $$\sum_{0}^{n-1}\binom{n}{k}\binom{n}{k+1}=\binom{2n}{2k+1}$$
The right hand side $\binom{2n}{2k+1}$ is the number of ways to choose $2k+1$ from a total of $2n$. For the left hand side, divide $2n$ into two groups of size $n$ of each, then if choose $k$ from one group, then must choose $2k+1-k=k+1$ from another group, sum over all possible $k$, then you get $\sum_0^{n-1}\binom{n}{k}\binom{n}{k+1}$
• What is $k$ in the rhs? – Julien Apr 4 '13 at 18:28
• I have downvoted this answer for being (mathematically) unintelligible. – anon Apr 5 '13 at 0:51
Using the estimate derived in this answer, we get \begin{align} \sum^{n-1}_{j=0}\binom{n}{j}\binom{n}{j+1} &=\sum^{n-1}_{j=0}\binom{n}{n-j}\binom{n}{j+1}\\ &=\binom{2n}{n+1}\\ &=\binom{2n}{n}\frac{n}{n+1}\\[3pt] &\sim\frac{4^n}{\sqrt{\pi n}}\left(1-\frac9{8n}\right) \end{align} | 2019-06-16T05:10:12 | {
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http://math.stackexchange.com/questions/16566/orthonormal-eigenbasis | # Orthonormal Eigenbasis
I am a little apprehensive to ask this question because I have a feeling it's a "duh" question but I guess that's the beauty of sites like this (anonymity):
I need to find an orthonormal eigenbasis for the $2 \times 2$ matrix $\left(\begin{array}{cc}1&1\\ 1&1\end{array}\right)$. I calculated that the eigenvalues were $x=0$ and $x=2$ and the corresponding eigenvectors were $E(0) = \mathrm{span}\left(\begin{array}{r}-1\\1\end{array}\right)$ and $E(2) = \mathrm{span}\left(\begin{array}{c}1\\1\end{array}\right)$. Therefore, an orthonormal eigenbasis would be: $$\frac{1}{\sqrt{2}}\left(\begin{array}{r}-1\\1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right).$$
Here my question: Could the eigenvalues for $E(0)$ been $\mathrm{span}\left(\begin{array}{r}1\\-1\end{array}\right)$?? This would make the final answer $\frac{1}{\sqrt{2}}\left(\begin{array}{r}1\\-1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right)$. Is one answer more correct than the other (or are they both wrong)?
Thanks!
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I don't quite understand the question as written, but if it's what I think it is, then yes, orthonormal eigenbases are not unique; you can multiply any eigenvector by a complex number of absolute value 1. – Qiaochu Yuan Jan 6 '11 at 13:28
Or by a real number or whatever the field is that you are working over. One more thing: professional mathematicians have "duh" moments all the time. It's part of the job description, so to speak. There is absolutely no need to hide behind anonymity. On the contrary, it is important for a mathematician to learn to live with those "duh" moments. Otherwise he will never be able to freely talk with his colleagues, which would greatly hinder his progress. – Alex B. Jan 6 '11 at 13:32
@Alex: the absolute value 1 condition is important for orthonormality, which doesn't make sense over an arbitrary field. – Qiaochu Yuan Jan 6 '11 at 13:55
@Qiaochu You are right. Read that as "real or complex, depending on..." – Alex B. Jan 6 '11 at 14:38
0 and 2 are the correct eigenvalues to your matrix; (1, -1) is one eigenvector, (1, 1) the other. Your solution is correct.
Span is the set of all linear combinations, so if you consider a vector space over $\mathbb{R}$, it absolutely doesn't matter what scalar in $\mathbb{R}$ you multiply your vectors with inside Span. This does not affect the set Span at all. So the solution is the same.
-
There is no such thing as the eigenvector of a matrix, or the orthonormal basis of eigenvectors. There are usually many choices.
Remember that an eigenvector $\mathbf{v}$ of eigenvalue $\lambda$ is a nonzero vector $\mathbf{v}$ such that $T\mathbf{v}=\lambda\mathbf{v}$. That means that if you take any nonzero multiple of $\mathbf{v}$, say $\alpha\mathbf{v}$, then we will have $$T(\alpha\mathbf{v}) = \alpha T\mathbf{v} = \alpha(\lambda\mathbf{v}) = \alpha\lambda\mathbf{v}=\lambda(\alpha\mathbf{v}),$$ so $\alpha\mathbf{v}$ is also an eigenvector corresponding to $\lambda$. More generally, if $\mathbf{v}_1,\ldots,\mathbf{v}_k$ are all eigenvectors of $\lambda$, then any nonzero linear combination $\alpha_1\mathbf{v}_1+\cdots+\alpha_k\mathbf{v}_k\neq \mathbf{0}$ is also an eigenvector corresponding to $\lambda$.
So, of course, since $\left(\begin{array}{r}-1\\1\end{array}\right)$ is an eigenvector (corresponding to $x=0$), then so is $\alpha\left(\begin{array}{r}-1\\1\end{array}\right)$ for any $\alpha\neq 0$, in particular, for $\alpha=-1$ as you take.
Now, a set of vectors $\mathbf{w}_1,\ldots,\mathbf{w}_k$ is orthogonal if and only if $\langle \mathbf{w}_i,\mathbf{w}_j\rangle = 0$ if $i\neq j$. If you have an orthogonal set, and you replace, say, $\mathbf{w}_i$ by $\alpha\mathbf{w}_i$ with $\alpha$ any scalar, then the result is still an orthogonal set: because $\langle\mathbf{w}_k,\mathbf{w}_j\rangle=0$ if $k\neq j$ and neither is equal to $i$, and for $j\neq i$, we have $$\langle \alpha\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha\langle\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha 0 = 0$$ by the properties of the inner product. As a consequence, if you take an orthogonal set, and you take any scalars $\alpha_1,\ldots,\alpha_k$, then $\alpha_1\mathbf{w}_1,\ldots,\alpha_k\mathbf{w}_k$ is also an orthogonal set.
A vector $\mathbf{n}$ is normal if $||\mathbf{n}||=1$. If $\alpha$ is any scalar, then $||\alpha\mathbf{n}|| = |\alpha|\,||\mathbf{n}|| = |\alpha|$. So if you multiply any normal vector $\mathbf{n}$ by a scalar $\alpha$ of absolute value $1$ (or of complex norm $1$), then the vector $\alpha\mathbf{n}$ is also a normal vector.
A set of vectors is orthonormal if it is both orthogonal, and every vector is normal. By the above, if you have a set of orthonormal vectors, and you multiply each vector by a scalar of absolute value $1$, then the resulting set is also orthonormal.
In summary: you have an orthonormal set of two eigenvectors. You multiply one of them by $-1$; this does not affect the fact that the two are eigenvectors. The set was orthogonal, so multiplying one of them by a scalar does not affect the fact that the set is orthogonal. And the vectors were normal, and you multiplied one by a scalar of absolute value $1$, so the resulting vectors are still normal. So you still have an orthonormal set of two eigenvectors. I leave it to you to verify that if you have a linearly independent set, and you multiply each vector by a nonzero scalar, the result is still linearly independent.
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http://isxy.chicweek.it/number-of-shortest-paths-in-a-weighted-graph.html | Dijkstra's algorithm, published in 1959 and named after its creator Dutch computer scientist Edsger Dijkstra, can be applied on a weighted graph. Very nice article. Graph Algorithms I 12. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. Shortest Path Queries A shortest path query on a(n) (undirected) graph finds the shortest path for the given source and target vertices in the graph. The graph is not weighted. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. it, daniele. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. Since the edges in the center of the graph have large weights, the shortest path between nodes 3 and 8 goes around the boundary of the graph where the edge weights are smallest. ! But what if edges have different 'costs'? s v δ(, ) 3sv = δ(, ) 12sv = 2 s v 2 5 1 7. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. Weighted graphs are commonly used in determining the most optimal path, most expedient, or the. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Weighted Graph ( भारित ग्राफ ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. It maintains a set of nodes for which the shortest paths are known. 7 s: source 8 ’’’ 9 result = ShortestPathResult() 10 num_vertices = graph. the first assumes that the graph is weighted, which means that each edge has a cost to traverse it. BFS always visits nodes in increasing order of their distance from the source. An interesting problem is how to find shortest paths in a weighted graph; i. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. The result is a list of vertices, or #f if there is no path. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Our main goal is to characterize exactly which sets of node sequences, which we call path systems, can appear as unique shortest paths in a graph with arbitrary real edge weights. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. I'm using the networkx package in Python 2. If the graph is weighted (that is, G. Shortest paths. Here a, b, c. A cycle is a path where the first and last vertices are the same. We wish to determine a shortest path from v 0 to v n Dijkstra's Algorithm Dijkstra's algorithm is a common algorithm used to determine shortest path from a to z in a graph. Dijkstra's Algorithm. Counting the number of shortest paths in various graphs is an important and interesting combinatorial problem, especially in weighted graphs with various applications. A few months ago, mathematicians Andrew Beveridge and Jie Shan published Network of Thrones in Math Horizon Magazine where they analyzed a network of character interactions from the novel “A Storm of Swords”, the third book in the popular “A Song of Ice and Fire” and the basis for the Game of Thrones TV series. numPaths initialized to 1). Combinatorics is the branch of mathematics concerned with selecting, arranging, and listing or counting collections of objects. Newman Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico 87501 and Center for Applied Mathematics, Cornell University, Rhodes Hall, Ithaca, New York 14853 ~Received 1 February 2001; published 28 June 2001!. Shortest Path In A Weighted Directed Graph With Dijkstra's Algorithm - posted in C and C++: Well, I encountered an interesting problem. We start with undirected graphs. The main algorithms that fall under this definition are Breadth-First Search (BFS) and Dijkstra's algorithms. A logical scalar. Weighted Graphs Data Structures & Algorithms 2 [email protected] ©2000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. The graph given in the test case is shown as : The shortest paths for the 3 queries are :: The direct Path is shortest with weight 5: There is no way of reaching node 1 from node 3. It is used to identify optimal driving directions or degree of separation between two people on a social network for example. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. The shortest path between two points in a weighted graph can be found with Dijkstra’s algorithm. Now we are going to find the shortest path between source (a) and remaining vertices. Find the shortest distance from C to D and if it is impossible to reach node D from C then return -1. The algorithm exists in many variants. This means that, given a weighted graph, this algorithm will output the shortest distance from a selected node to all other nodes. shortest path. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. Changing to its dual, the triangular grid, paths between triangle pixels (we abbreviate this term to trixels) are counted. Question: Shortest Paths(int[0). The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. implementations of the shortest-path search algorithms on graphs on the number of graph vertices experimentally. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. PRACTICE PROBLEM BASED ON FLOYD WARSHALL ALGORITHM- Problem- Consider the following directed weighted graph- Using Floyd Warshall Algorithm, find the shortest path distance between every pair of vertices. The shortest path from 0 to 4 uses the shortest path from 0 to 1 and the edge 1–4. Our algorithm is deterministic and has a running time of O(k(m√n + n 3/2 log n)) where m is the number of edges in the graph and n is the number of vertices. In this graph, vertex A and C are connected by two parallel edges having weight 10 and 12 respectively. Abstract: In this paper we evaluate our presented Quantum Approach for finding the Estimation of the Length of the Shortest Path in a Connected Weighted Graph which is achieved with a polynomial time complexity about O(n) and as a result of evaluation we show that the Probability of Success of our presented Quantum Approach is increased if the Standard Deviation of the Length of all possible. An assigned number is called the weight of the edge, and the collection of all weights is called a weighting of the graph Γ. A path in a graph is a sequence of adjacent vertices. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. Path Graphs. Only edges with non-negative costs are included. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. dijkstra_path¶ dijkstra_path (G, source, target, weight='weight') [source] ¶. 0 = no epidemic 1 = epidemic NW N SW C E Graph 3: Vertex coloring of weighted graph for the county. In this category, Dijkstra’s algorithm is the most well known. You are expected to do it in Time Complexity of O(A + M). shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. The shortest path between node 222 and node 444 is 222 -> 555 -> 666 -> 777 -> 444, which has a weighted distance 1. The number of diagonal steps in a shortest path of the chessboard distance is min{w 1,w 2}, and the number of cityblock steps (i. A cycle is a path where the first and last vertices are the same. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. (Consider what this means in terms of the graph shown above right. The Edge can have weight or cost associate with it. Simple path is a path with no repeated vertices. Hence, parallel computing must be applied. In this post I will be discussing two ways of finding all paths between a source node and a destination node in a graph: Using DFS: The idea is to do Depth First Traversal of given directed graph. , capacity, cost, demand, traffic frequency, time, etc. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Geodesic paths are not necessarily unique, but the geodesic distance is well-defined since all geodesic paths have. ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G∗ constructed from G. The latter only works if the edge weights are non-negative. 4 5 Args: 6 graph: weighted graph with no negative cycles. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. Chan⁄ September 30, 2009 Abstract Intheflrstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. This module covers weighted graphs, where each edge has an associated weight or number. The problem is to find k directed paths starting at s, such that every node of G lies on at least one of those paths, and such that the sum of the weights of all the edges in the paths is minimized. The one-to-all shortest path problem is the problem of determining the shortest path from node s to all the other nodes in the. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. You are given a weighted directed acyclic graph G, and a start node s in G. It finds a shortest-path tree for a weighted undirected graph. Journal of the ACM 65 :6, 1-40. Is there a cycle that uses each vertex. Mizrahi et al. The number dist[w] equals the length of a shortest path from v to w, or is -1 if w cannot be reached. Dijkstra's Algorithm is useful for finding the shortest path in a weighted graph. Given a directed weighted graph G= (V;E;w) with non-negative weights w: E!R+ and a vertex s2V, the single-source shortest paths is the family of shortest paths s vfor every vertex v2V. If finds only the lengths not the path. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. A logical scalar. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. The order of a graph is the number of nodes. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The Dijkstra’s algorithm is provided in Algorithm 2. For example, the length of v8,v9 equals 2, which is identical to the length of the. As such, we say that the weight of a path is the sum of the weights of the edges it contains. The shortest-path from vertex u to vertex v in a weighted graph is a path with minimum sum-weight B All-Pairs Shortest Paths For a weighted graph, the all-pairs shortest paths problem is to nd the shortest path between all pairs of vertices. 2 Representing Weighted Graphs Often it is desirable to use a priority queue to store weighted edges. Question: Shortest Paths(int[0). We first propose an exact (and. A single execution of the algorithm will. shortest path. Consider a shortest path p from vertex i to vertex j, and suppose that p contains at most m edges. In the weighted matching [G85b, GT89, GT91] and maxi-mum flow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times differing by a polynomial factor. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem… 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. The distance between two vertices u and v, denoted d(u,v), is the length of a shortest. Suppose you are given a directed graph G = (V, E), with costs on the edges; the costs may be. A path in a graph is a sequence of adjacent vertices. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Shortest Path. An example of a weighted graph would be the distance between the capitals of a set of countries. You are given a weighted undirected graph. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. Consider the graph above. Your task is to find the shortest path between the vertex 1 and the vertex n. In weighted graphs, where we assume that the edge weights do not represent the communication speed, a straightforward distributed variant of the Bellman-Ford algorithm [2], [3], [4] computes. Shortest paths problems are among the most fundamental algorithmic graph problems. The actual shortest paths can also be constructed by modifying the above algorithm. Question: Shortest Paths(int[0). If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. To find the shortest path on a weighted graph, just doing a breadth-first search isn't enough - the BFS is only a measure of the shortest path based on number of edges. A comparison of the data obtained as a result of the study was carried out to find the best applications of implementations of the shortest path search algorithms in the Postgre SQL DBMS. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Shortest Path Algorithms?(a) Breadth-first search (BFS) can be used to perform single source short- est paths on any graph where all edges have the same costs 1. A cycle is a path where the first and last vertices are the same. Let s denote the number of edges of H. Shortest-Path Problems (cont'd) Single-source shortest path problem Given a weighted graph G = (V, E), and a distinguished start vertex, s, find the minimum weighted path from s to every other vertex in G The shortest weighted path from v 1 to v 6 has a cost of 6 and v 1 v 4 v 7 v 6. Shortest Path Problem. Dijkstra's algorithm (or Dijkstra's Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the shortest paths between nodes in a graph, which may represent, for example, road networks. Node is a vertex in the graph at a position. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Using similar ideas, we can construct a (1+epsilon)-approximate distance oracle for weighted unit-disk graphs with O(1) query. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there’s a good chance that you’ll encounter the same ideas. in Proceedings - 25th IEEE International Conference on High Performance Computing, HiPC 2018. Shortest Path Problem. Graph Algorithms I 12. We let k denote the number of source. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. The All-Pairs Shortest Paths (APSP) problem seeks the shortest path distances between all pairs of vertices, and is one of the most fundamental graph problems. Once we have reached our destination, we continue searching until all possible paths are greater than 11; at that point we are certain that the shortest path is 11. The vertices V are connected to each other by these edges E. Key Graph Based Shortest Path Algorithms With Illustrations - Part 1: Dijkstra's And Bellman-Ford Algorithms Bellman-Ford algorithm is used to find the shortest paths from a source vertex to all other vertices in a weighted graph. Distributed Exact Weighted All-Pairs Shortest Paths in O˜(n5/4) network is modeled by a weighted n-node m-edge graph G. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. An interesting side-effect of traversing a graph in BFS order is the fact that, when we visit a particular node, we can easily find a path from the source node to the newly visited node with the least number of edges. The Symmetric Shortest-Path Table Routing Conjecture Thomas L. Question: Shortest Paths(int[0). Mizrahi et al. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. The algorithm creates a tree of shortest paths from the starting vertex, the source, to all other points in the graph. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. I will cover several other graph based shortest path algorithms with concrete illustrations. Shortest path in a graph with weighted edges and vertices. If there are multiple paths from node 1 to a node that have the same minimum number of. Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. There is one shortest path vertex 0 to vertex 0 (from each vertex there is a single shortest path to itself), one shortest path between vertex 0 to vertex 2 (0->2), and there are 4 different shortest paths from vertex 0 to vertex 6: 1. A shortest path between two nodes u and v in a graph is a path that starts at u and ends at v and has the lowest total link weight. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. Row i of the predecessor matrix contains information on the shortest paths from point i: each entry predecessors[i, j] gives the index of the previous node in the path from point i to point j. You are expected to do it in Time Complexity of O(A + M). For example, the length of v8,v9 equals 2, which is identical to the length of the. A shortest path (with the lowest weight) from $$u$$ to $$v$$; The weight of the path is the sum of the weights of its edges. Both algorithms were randomized and had constant query time. It was conceived by computer scientist Edsger W. Deterministic Partially Dynamic Single Source Shortest Paths in Weighted Graphs Aaron Bernstein May 30, 2017 Abstract In this paper we consider the decremental single-source shortest paths (SSSP) problem, where given a graph Gand a source node sthe goal is to maintain shortest distances between sand all other nodes. In this category, Dijkstra's algorithm is the most well known. An undirected graph that has a path from every vertex to every other vertex in the graph. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. If you think carefully, it's easy to see that there can be many graphs such that the. A new approach to all-pairs shortest paths on real-weighted graphs Seth Pettie1 Department of Computer Sciences, The University of Texas at Austin, Austin, TX 78712, USA Abstract We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-additionmodel. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. Chan⁄ September 30, 2009 Abstract Intheflrstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. shortest path algorithm. In this tutorial, we will present a general explanation of both algorithms. A cycle is a path where the first and last vertices are the same. techniques to speed the computation of shortest paths in the discretization graph [3,21]. ; It uses a priority based dictionary or a queue to select a node / vertex nearest to the source that has not been edge relaxed. This turns out to be a problem that can be solved efficiently, subject to some restrictions on the edge costs. An observed image is composed of multiple components based on optical phenomena, such as light reflection and scattering. The multiplicity of a path is the maximum number of times that an edge appears in it. The latter only works if the edge weights are non-negative. It maintains a set S of vertices whose final shortest path from the source has already been determined and it repeatedly selects the left vertices with the minimum shortest-path estimate, inserts them. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. Directed and undirected graphs may both be weighted. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Single Source Shortest Path. This section discusses three algorithms for this problem: breadth-first search for unweighted graphs, Dijkstra’s algorithm for weighted graphs, and the Floyd-Warshall algorithm for computing distances between all pairs of vertices. And here is some test code: test_graph. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. An undirected graph that has a path from every vertex to every other vertex in the graph. shortest path problem. shortest_paths calculates a single shortest path (i. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. yenpathy: An R Package to Quickly Find K Shortest Paths Through a Weighted Graph Submitted 05 September 2019 This paper is under review which means review has begun. ! Here, the length of a path is simply the number of edges on the path. One of these challenges is the need to generate a limited number of test cases of a given regression test suite in a manner that does not compromise its defect detection ability. For example finding the ‘shortest path’ between two nodes, e. We would then assign weights to vertices, not edges. Python – Get the shortest path in a weighted graph – Dijkstra Posted on July 22, 2015 by Vitosh Posted in VBA \ Excel Today, I will take a look at a problem, similar to the one here. FindShortestPath[g, s, All] generates a ShortestPathFunction[] that can be applied repeatedly to different t. [email protected] ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G∗ constructed from G. According to [11], [12], [13], since 1959 Dijkstra's algorithm has been recognized as the best algorithm and used as method to find the shortest path. nodes in a given directed graph is a very common problem. You are also given a positive integer k. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. ” Dijkstra’s algorithm is an iterative algorithm that provides us with the shortest path from one particular starting node to all other nodes in the graph. The shortest path function can also be used to compute a transitive closure or for arbitrary length traversals. the number of pairs of vertices not including v, which for directed graphs is and for undirected graphs is. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). There are two types of weighted graphs: vertex weighted and edge weighted. Shortest path problem (SPP) is a fundamental and well-known combinatorial optimization problem in the area of graph theory. A weighted graph is a one which consists of a set of vertices V and a set of edges E. The shortest path problem is about finding a path between $$2$$ vertices in a graph such that the total sum of the edges weights is minimum. Dijkstra's Shortest Path Algorithm in Java. The single pair shortest path problem seeks to compute (u;v) and construct a shortest path from. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. the sum of the weights of the edges in the paths is minimized. shortest path algorithm. Variations of the Shortest Path Problem. We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-addition model. I Therefore, the numbers d 1;d 2; ;d n must include an even number of odd numbers. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. All-Pairs Shortest Paths Problem To find the shortest path between all verticesv 2 V for a graph G =(V,E). This algorithm has numerous applications in network analysis, such as transportation planning. (2018) Decremental Single-Source Shortest Paths on Undirected Graphs in Near-Linear Total Update Time. Dijkstra’s Shortest Path Algorithm in Java. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. It was conceived by computer scientist Edsger W. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex ‘s’ to a given destination vertex ‘t’. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. be contained in shortest augmenting paths, and the lay-ered network contains all augmenting paths of shortest length. Dating back some 3000 years, and initially consisting mainly of the study of permutations and combinations, its scope has broadened to include topics such as graph theory, partitions of numbers, block designs, design of codes, and latin squares. Simple path is a path with no repeated vertices. Find a TSP solution using state-of-the-art software, and then remove that dummy node (subtracting 2 from the total weight). C… Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. This module covers weighted graphs, where each edge has an associated weight or number. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. Dijkstra’s Algorithms describes how to find the shortest path from one node to another node in a directed weighted graph. Print the number of shortest paths from a given vertex to each of the vertices. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. Dijkstra's Algorithm: Finds the shortest path from one node to all other nodes in a weighted graph. It finds a shortest-path tree for a weighted undirected graph. The SQL Server graph extensions are amazing. Introduction. In this week, you'll add a key feature of map data to our graph representation -- distances -- by adding weights to your edges to produce a "weighted. Must Read: C Program. The Degree of a vertex is the number of edges incident on it. I'm using the networkx package in Python 2. adjacent b. Shortest Paths: Problem Statement Given a weighted graph and two vertices u and v, we want to find a path of minimum total weight between u and v Length (or distance) of a path is the sum of the weights of its edgesLength (or distance) of a path is the sum of the weights of its edges. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). The Degree of a vertex is the number of edges incident on it. for unweighted. If the graph is unweighed, then finding the shortest path is easy: we can use the breadth-first search algorithm. A path graph is a graph consisting of a single path. So, the shortest path would be of length 1 and BFS would correctly find this for us. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem… 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Shortest path in a graph with weighted edges and vertices. PY - 2007/10/30. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. Variations of the Shortest Path Problem. We start with a directed and weighted graph $$G = (V, E)$$ with a weights function $$w: E \rightarrow R$$ that maps edges to weights with real values; Find for each pair of vertices $$u$$, $$v \in V$$. Shortest paths&Weighted graphs. An undirected graph that has a path from every vertex to every other vertex in the graph. Consider any node that is not the root: its possible distances from the root are all possible distances of its neighbors plus the weight of the connecting edges. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent "infinity", indicates this fact. In a weighted graph with start node s, there are often multiple shortest paths from s to any other node. 1 def shortest_path_cycle(graph, s): 2 ’’’Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. Dijkstra’s algorithm is similar to Prim’s algorithm. 4 5 Args: 6 graph: weighted graph with no negative cycles. b) Weighted graph matching: Mulmuley & Shah ob-served that their result for the shortest path problem yields the same lower bound for the WEIGHTED GRAPH MATCHING problem [3, Corollary 1. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using this algorithm. 1 Shortest paths and matrix multiplication 25. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Suppose we have to following graph: We may want to find out what the shortest way is to get from node A to node F. Graphs: Finding shortest paths Definition: weight of a path 4 Tecniche di programmazione A. Weighted Graph ( भारित ग्राफ ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. P = shortestpath(G,s,t) computes the shortest path starting at source node s and ending at target node t. 18 between shortest paths in networks and Hamilton paths in graphs ties in with our observation that finding paths of low weight (which we have been calling "short") is tantamount to finding paths with a high number of edges (which we might consider to be "long"). Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. If you think carefully, it's easy to see that there can be many graphs such that the. A path in a graph is a sequence of adjacent vertices. As there are a number of different shortest path algorithms, we've gathered the most important to help you understand how they work and which is the best. This problem also is known as "Print all paths between two nodes". The shortest paths followed for the three nodes 2, 3 and 4 are as follows : 1/S->2 - Shortest Path Value : 1/S->3 - Shortest Path Value : 1/S->3->4 - Shortest Path Value :. Data Structure Graph Algorithms Algorithms. It runs in O (mn+n 2 log log n) time, improving on the long-standing bound of O (mn+n 2 log n) derived from an implementation of Dijkstra's algorithm with Fibonacci heaps. The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. 6 2, 6(a), 6(c), 18 In Exercises 2-4 find the length of a shortest path between a and z in the given weighted graph. Shortest Paths, and Dijkstra's Algorithm: Overview Graphs with lengths/weights/costs on edges. Scientific collaboration networks. 2 You can use the tool to create a weighted graph with mouse gestures and show the MST and shortest paths. If say we were to find the shortest path from the node A to B in the undirected version of the graph, then the shortest path would be the direct link between A and B. shortest_paths calculates a single shortest path (i. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Topics in this lecture include:. You may print the results of this algorithm to the screen or to a log file. A B C F D E 7 14 9 8 15 2 9 4 6 Figure 3. Dijkstra’s algorithm is similar to Prim’s algorithm. This module covers weighted graphs, where each edge has an associated weight or number. I'm restricting myself to Unweighted Graph only. The relationship shown in the proof of Property 21. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Undirected graph. Shortest paths in an edge-weighted digraph 4->5 0. This means it finds the shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between the source node and every other node. If the graph is weighted, it is a path with the minimum sum of edge weights. Algorithms to find shortest paths in a graph are given later. Breadth First Search, BFS, can find the shortest path in a non-weighted graphs or in a weighted graph if all edges have the same non-negative weight. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. world as a directed graph, the problem of directing the robot was transformed into a shortest-path problem. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. Variations of the Shortest Path Problem. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent "infinity", indicates this fact. Print the number of shortest paths from a given vertex to each of the vertices. Consider the graph above. The best algorithm (modulo small polylogarithmic improvements) for this problem runs in cubic time, a. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. Your solution should be complete in that it …. Representing a Graph. For example, the two paths we mentioned in our example are C, B and C, A, B. Thanks for pointing to Gephi. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). CS 340 Programming Assignment VII: Single-Source Shortest Paths in generally Weighted graphs Description: You are to implement both the DAG SP algorithm and the Bellman-Ford algorithm for single-source shortest paths based on whether or not you detect a cycle. a i g f e d c b h 25 15 10 5 10. Hart, Nilsson, and Raphael [12] discovered how to use this information to improve the efficiency of computing the shortest path. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The number parent[w] is the predecessor of w on a shortest path from v to w, or -1 if none exists. minimum path sizeis the shortest distance measured in the number of edges traversed. The algorithm can be implemented with O(m+nlog(n)) time; therefore, it is efficient in sparsely connected networks. The Bellman-Ford algorithm is a single-source shortest path algorithm. Michael Quinn, Parallel Programming in C with MPI and OpenMP,. Decomposing the observed image into individual components is an important process for various computer vision tasks. Description and notations-building the graph Given a weighted graph G = (V, E), where V. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. By comparison, if the graph is permitted. The order of a graph is the number of nodes. P( s,t ) is the shortest path between the given vertices and containing the least sum of edge weights on the path from to. Bellman-Ford Algorithm is computes the shortest paths from a single source vertex to all of the other vertices in a weighted digraph. Shortest paths&Weighted graphs. It was conceived by computer scientist Edsger W. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there’s a good chance that you’ll encounter the same ideas. This technique is applied to a minute level bid/ask quote dataset consisting of rates constructed from all G10 currency pairs. • In addition, the first time we encounter a vertex may, we may not have found the shortest path to it, so we need to delay committing to that path. 1 Shortest Path Queries Let G= (V;E;˚) be a road network (i. Start by setting the distance of all notes to infinity and the source's distance to 0. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. A path in a graph is a sequence of adjacent vertices. shortest s-tpath in G n. Algorithms to find shortest paths in a graph are given later. These weights represent the cost to traverse the edge. A cycle is a path where the first and last vertices are the same. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. It turns out that it is as easy to compute the shortest paths from s to every node in G (because if the shortest path from s to t is s = v0, v1, v2, , vk = t, then the path v0,v1 is the shortest path from s to v1, the path v0,v1,v2 is the shortest path from s to v2, the path v0,v1,v2,v3 is the shortest path from s to v3, etc. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Essentially, you replace the stack used by DFS with a queue. I'm looking for an algorithm that will perform as in the title. Again this is similar to the results of a breadth first search. It can be used in numerous fields such as graph theory, game theory, and network. Dijkstra's Shortest Path Algorithm in Java. linear-time. So there is a weighted graph, and the shortest paths of all pair of n. dijkstra_path_length¶ dijkstra_path_length (G, source, target, weight='weight') [source] ¶. What is the longest simple path between s and t? Cycle. Graphs can be weighted (edges carry values) and directional (edges have direction). Let u and v be two vertices in G, and let P be a path in G from u to v. We are also given a starting node s ∈ V. Dijkstra's Single Source Shortest Path. Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. Among the various shortest path algorithms, Dijkstra’s shortest path algorithm [1] is said to have better performance with regard to run time than the other algorithms. Finally, at k = 4, all shortest paths are found. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex ‘s’ to a given destination vertex ‘t’. However, the robot had a means for estimating Euclidean distances in his world. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. The Floyd-Warshall algorithm is an example of dynamic programming. 2 commits 1 branch. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. However, most communication networks are dynamic, i. shortest path problem. Referred to as the shortest path between vertices For weighted graphs this is the path that has the smallest sum of its edge weights ijkstra’salgorithm finds the shortest path between one vertex and all other vertices The algorithm is named after its discoverer, Edgser Dijkstra 24 The shortest path between B and G is: 1 4 3 5 8 2 2 1 5 1 B A. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. Graph Characteristics-Undirected-Weighted-Journey: (1, 7)-Shortest path: 1 – 4 – 6 – 7 (in purple)-Total cost: 6. We first propose an exact (and. The graph has eight nodes. The Degree of a vertex is the number of edges incident on it. For the shortest path problem on positively weighted graphs the integer/real gap is only logarith-mic. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. The weights of the edges can be positive or negative. A near linear shortest path algorithm for weighted undirected graphs Abstract: This paper presents an algorithm for Shortest Path Tree (SPT) problem. Here the graph we consider is unweighted and hence the shortest path would be the number of edges it takes to go from source to destination. The longest path is based on the number of edges in the path if weighted == false and the unweighted shortest path algorithm is being used. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. Suppose G be a weighted directed graph where a minimum labeled w(u, v) associated with each edge (u, v) in E, called weight of edge (u, v). We introduce a metric on the. It asks for the number of different shortest paths. Day 3: Weighted Graphs. This problem also is known as "Print all paths between two nodes". So there is a weighted graph, and the shortest paths of all pair of n. texens | Hello World ! | Page 3 Hello World !. In this tutorial, we will present a general explanation of both algorithms. The graph given in the test case is shown as : * The lines are weighted edges where weight denotes the length of the edge. Graphs can be weighted (edges carry values) and directional (edges have direction). We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. Give an efficient algorithm to solve the single-destination shortest paths problem. Thus, the shortest path between any two nodes is the path between the two nodes with the lowest total length. The weighted path length is given by ∑C(vi,vi+1) i=1 k-1 The general problem Given an edge-weighted graph G = (V,E) and two vertices, vs ∈V and vd ∈V, find the path that starts at vs and ends at vd that has the smallest weighted path length Single-source shortest. Uses Dijkstra's Method to compute the shortest weighted path between two nodes in a graph. It finds a shortest-path tree for a weighted undirected graph. Crossref, ISI, Google Scholar; 19. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The shortest pair edge disjoint paths in the new graph corresponds to the required solution in the original graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Length of a path is the sum of the weights of its edges. Let G = (V, E) be a directed graph and let v and w be two vertices in G. We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. weighted › cyclic vs. num_vertices() 11 for i in range(num_vertices): 12 result. Finding paths. The Bellman-Ford algorithm is a single-source shortest path algorithm. Topological Sort: Arranges the nodes in a directed, acyclic graph in a special order based on incoming edges. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. Asked in Graphs , C Programming. Dijkstra’s Algorithm for Finding the Shortest Path Through a Weighted Graph E. Simple path is a path with no repeated vertices. This algorithm aims to find the shortest-path in a directed or undirected graph with non-negative edge weights. BFS runs in O(E+V) time where E is the number of edges and V is number of vertices in the graph. Those times are the weights of those paths. A path in a graph is a sequence of adjacent vertices. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. Mark Dolan CIS 2166 10. weighted › cyclic vs. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. MINIMUM-WEIGHT SPANNING TREE 49 4. Year 2001 Predicted year 2005 Graph 2: Evolution of flu epidemic behavior. CS 312 Lecture 26 Finding Shortest Paths Finding Shortest Paths. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. The weight of the path P is the sum of the weights of all the _____ on the path P. This problem could be solved easily using (BFS) if all edge weights were ($$1$$), but here weights can take any value. The length of a path is the sum of the lengths of all component edges. The weighted inverse distance is the total number of shortest paths from node \(s G From the Betweenness Centrality in Street Networks to Structural Invariants in Random Planar Graphs. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. So, the shortest path would be of length 1 and BFS would correctly find this for us. The total length of a path is the sum of the lengths of its component edges. In the notes below I am going to describe the Dijkstra algorithm, which is a widely-used algorithm for finding shortest paths in weighted, directed graphs. 2 commits 1 branch. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. 7 (Single-Source Shortest Paths). The Degree of a vertex is the number of edges incident on it. Single-Source Shortest Path on Weighted Graphs. On weighted graphs Weighted Shortest Paths The shortest path from a vertex u to a vertex v in a graph is a path w1 = u, w2,…,wn= v, where the sum: Weight(w1,w2)+…+Weight(wn-1,wn) attains its minimal value among all paths that start at u and end at v The length of a path of n vertices is n-1 (the number of edges) If a graph is connected, and the weights are all non-negative, shortest paths exist for any pair of vertices Similarly for strongly connected digraphs with non-negative weights. Write an algorithm to print all possible paths between source and destination. Floyd-Warshall algorithm is used to find all pair shortest path problem from a given weighted graph. In real-life scenarios, the arc weighs in a shortest path of a network/graph have the several parameters which are very hard to define exactly (i. $\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Cover Photo by Thor Alvis on Unsplash. The path graph with n vertices is denoted by P n. In this tutorial, we will present a general explanation of both algorithms. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. There is a simple tweak to get from DFS to an algorithm that will find the shortest paths on an unweighted graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. (Consider what this means in terms of the graph shown above right. Furthermore, if we perform relaxation on the set of edges once, then we will at least have determined all the one-edged shortest paths; if we traverse the set of edges twice, we will have solved at least all the two-edged shortest paths; ergo, after the V-1 iteration. However, most communication networks are dynamic, i. shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Shortest paths in an edge-weighted digraph 4->5 0. Graphs can be weighted (edges carry values) and directional (edges have direction). The Degree of a vertex is the number of edges incident on it. In the weighted matching [G85b, GT89, GT91] and maxi-mum flow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times differing by a polynomial factor. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. $\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Simple path is a path with no repeated vertices. Is there a cycle in the graph? Euler tour. So, the shortest path would be of length 1 and BFS would correctly find this for us. First, the paths should be shortest, then there might be more than one such shortest paths whose length are the same. A path in a graph is a sequence of adjacent vertices. Mark Dolan CIS 2166 10. goldberg_radzik (G, source[, weight]) Compute shortest path lengths and predecessors on shortest paths in weighted graphs. Increasingly, there is interest in using asymmetric structure of data derived from Markov chains and directed graphs, but few metrics are specifically adapted to this task. We may also want to associate some cost or weight to the traversal of an edge. It is a real time graph algorithm, and can be used as part of the normal user flow in a web or mobile application. You may print the results of this algorithm to the screen or to a log file. shortest_paths calculates a single shortest path (i. However, the robot had a means for estimating Euclidean distances in his world. PY - 2007/10/30. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Based on Data Structures, Algorithms & Software Principles in CT. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. We ˙rst propose an exact (and deterministic) al-. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. , capacity, cost, demand, traffic frequency, time, etc. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Path Graphs. dijkstra_path¶ dijkstra_path (G, source, target, weight='weight') [source] ¶. This article presents a Java implementation of this algorithm. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. Weighted Graphs Data Structures & Algorithms 2 [email protected] ©2000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. Chapter 4 Algorithms in edge-weighted graphs Recall that anedge-weighted graphis a pair(G,w)whereG=(V,E)is a graph andw:E →IR number of edges in a shortest path. , capacity, cost, demand, traffic frequency, time, etc. So, for this purpose we use the retroactive priority queue which allows to perform opeartions at any point of time. The Degree of a vertex is the number of edges incident on it. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. BFS always visits nodes in increasing order of their distance from the source. These values become important when calculating the. weighted graphs: find a path from a given source to a given target such that the consecutive weights on the path are nondecreasing and the last weight on the path is minimized. A single execution of the algorithm will. Analyze your algorithm. ) B C A E D F 0 7 2 3 5 8 8 4 7 1 2 5 2. , its number of edges. Finding the Shortest Path in Weighted Directed Acyclic Graph For the graph above, starting with vertex 1, what're the shortest paths(the path which edges weight summation is minimal) to vertex 2. The Classical Dijkstra’s algorithm [16] solves the single-source shortest path problems in a simple graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. If G is a weighted graph, the length/weight of a path is the sum of the weights of the edges that compose the path. Dimitrios Skrepetos, PhD candidate David R. Check the manual pages of the functions working with weighted graphs for details. Another source vertex is also provided. A cycle is a path where the first and last vertices are the same. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Print the number of shortest paths from a given vertex to each of the vertices. However, the resulting algorithm is no longer called DFS. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. The weight of the path P is the sum of the weights of all the _____ on the path P. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. Dijkstra's Algorithm. ; It uses a priority based set or a queue to select a node / vertex nearest to the source that has not been edge relaxed. But how should we evaluate a path in a weighted graph? Usually, a path is assessed by the sum of weights of its edges and, based on this assumption, many authors proposed their. A graph is a series of nodes connected by edges. (2018) A Faster Distributed Single-Source Shortest Paths Algorithm. One algorithm for finding the shortest path from a starting node to a target node in a weighted graph is Dijkstra's algorithm. Count the number of updates At each step we compute the shortest path through a subset of vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. In this tutorial, we will present a general explanation of both algorithms. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. G∗ contains threeshortcuts: v8,v9, v9,v7,and v9,v10. One problem might be the shortest path in a given undirected, weighted graph. GUVEW ,, eE , where (non-negative real number) is a weight function, by which each edge is associ-ated with a weight The weight of a matching M is. 1 def shortest_path_cycle(graph, s): 2 '''Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. num_vertices() 11 for i in range(num_vertices): 12 result. We consider a specific infinite graph here, namely the honeycomb grid. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. We introduce a metric on the. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. None of these algorithms for the WRP permit one to bound the number of links/turns in the produced path.
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https://math.stackexchange.com/questions/475927/how-to-prove-n-fracnen/475934 | # How to prove $n!>(\frac{n}{e})^{n}$
Prove that $n!>\left(\dfrac{n}{e}\right)^{n}$.
I used induction principle but cannot solve it for the $(m+1)$-th term after taking the $m$th term to be true.
Here the key is to use the appropriate definition of $e^x$, namely:
$$e^x = \sum_{k=0}^{\infty}\frac{1}{k!}x^k$$
Plugging in $x = n$ we get
$$e^n = \sum_{k=0}^{\infty} \frac{1}{k!}n^k$$
and hence, breaking this sum up a little we get our inequality: $$n! e^n = n^n + \sum_{k\ne n} \frac{n!}{k!}n^k > n^n$$
Hint: Show that $$\ln(n!)=\sum_{k=1}^n \ln k >\int_1^n\ln x\, dx.$$
• That's a clever approach! Nice one! Aug 25 '13 at 18:02
Inductively, if $n!>\frac{n^n}{e^n}$ and you multiply both sides by $n+1$, then you have that $(n+1)!>(n+1)\frac{n^n}{e^n}$, so it suffices to prove that $(n+1)\frac{n^n}{e^n}>\frac{(n+1)^{n+1}}{e^{n+1}}$. Can you continue from here?
• +1 This requires the least amount of machinery of the presented solutions, I think. Aug 25 '13 at 17:11
Hint: write out the series for $e^n$ and pick out a relevant term amongst the positive terms which make up the sum.
• You posted a hint and Deven independently posted a solution using the same idea => 10 vote difference. Isn't that always so? +1 to both of you from me has been there from the beginning, of course. Aug 25 '13 at 18:10
I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
\begin{align} a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\ b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \end{align}
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{\circledast} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
\begin{align} & ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\ & \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\ & \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\ &\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\ \end{align}
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
\begin{align} & ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\ & = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\ & = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\ & = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\ & \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\ & \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\ & = && (n+1)!\\ ~ \\ & = && (n+1) \cdot n!\\ ~ \\ & \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\ & \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\ \end{align}
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$. $$\tag*{\square}$$ | 2021-09-17T23:09:46 | {
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https://math.stackexchange.com/questions/2415076/probability-of-drawing-cards-on-specific-draw-counts | # Probability of drawing cards on specific draw counts
There is a deck of $30$ cards, each card labeled a number from $1$ to $15$, with exactly $2$ copies of a card for each number. You draw $8$ cards. What is the probability that you draw the number '$1$' card by the $5$th draw (on the $5$th draw or before that), AND also drawing the number '$2$' card on or before the $8$th draw?
I know how to compute the probability of drawing both the cards on or before the $5$th draw:
$$\frac{\binom{2}{1}\cdot \binom{2}{1} \cdot \binom{26}{3}}{\binom{30}{5}}$$
Since there's $2$ ways to choose from each of the '$1$' and '$2$' cards, and then there's $26$ cards left after those $4$ cards so the other $3$ cards can be any of those $26$, and the total number of combinations you can draw $5$ cards from $30$.
But we want to expand this search to $8$ draws, and also at the same time want to have assumed that we have already drawn the '$1$' card on or before the $5$th draw (if we don't get the '$2$' card by the $5$th draw. How can I combine these ideas? Thanks
• number 2 before 8-th draw... Then the 8th draw is irrelevant? Don't you mean (again) "on or before"? – drhab Sep 3 '17 at 9:07
• @user152294 Just to check my try, have you the result of this exercise? – Robert Z Sep 3 '17 at 9:46
• @drhab Yes, on or before – user152294 Sep 3 '17 at 17:29
• @RobertZ No, I don't have the solution unfortunately – user152294 Sep 3 '17 at 17:30
• @user152294 "before 8-th draw" means "on or before 8-th draw"? In case I have to modify my solution. P.S. Where does his exercise come from? – Robert Z Sep 3 '17 at 17:33
I think it is more convenient to evaluate the probability of the complementary event: draw NO number '1' card by the $5$th draw, OR draw NO number '2' card. Here we consider the 2 copies of a card with the same number distinguishable (for example assume that one is red and the other is blue). Let $n^{\underline{k}}:=n(n-1)\cdots(n-k+1)$.
1) If we draw NO number '1' card by the 5th draw then we can have zero, one or two '1's in the $6$th, $7$th or $8$th draw $$p_1=\frac{1}{30^{\underline{8}}}\left(28^{\underline{8}}+(3+3)\cdot 28^{\underline{7}}+3\cdot 2\cdot 28^{\underline{6}}\right)$$
2) If we draw NO number '2' card then $$p_2=\frac{28^{\underline{8}}}{30^{\underline{8}}}$$
3) If we draw NO number '1' card by the $5$th draw AND NO number '2' card then, similarly to case 1), $$p_3=\frac{1}{30^{\underline{8}}}\left(26^{\underline{8}}+(3+3)\cdot 26^{\underline{7}}+3\cdot 2\cdot 26^{\underline{6}}\right)$$
Hence, the desired probability is $$p=1-(p_1+p_2-p_3)=211/1566\approx 0.134738.$$
• Is it also possible to just do these two cases? 1) Draw a '1' and '2' before the 5th draw, or 2) Draw a '1' before the 5th draw AND draw a '2' before the 8th draw? I'm not sure how to express 2) but would it be more cumbersome than the complementary events? – user152294 Sep 3 '17 at 17:34
• @user152294 Yes you can consider two cases, but I think that with 3 cases is simpler. – Robert Z Sep 3 '17 at 17:36
• How come in this solution, we don't have to use binomial coefficients? – user152294 Sep 3 '17 at 17:43
• @user152294 Let me know if the official solution is the one that I have found. – Robert Z Sep 3 '17 at 17:44
• @user152294 $3$ is the number of ways to place the red $1$ (position 6,7, 8), $3$ is the number of ways to place the blue $1$ (position 6,7, 8) and $3\cdot 2$ is the number of ways to place the red $1$ and blue $1$ (positions (6,7), (6,8), (7,8), (7,6), (8,6), (8,7)). $28^{\underline{8}}$, $28^{\underline{7}}$ and $28^{\underline{6}}$ are the number of ways to fill the remaining positions (cards different from 1). – Robert Z Sep 3 '17 at 18:21 | 2020-08-13T09:37:49 | {
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https://math.stackexchange.com/questions/1637518/what-does-f-a-mean/1637528 | # What does $f|_A$ mean? [duplicate]
If $f$ a is a function and $A$ is a set, what could the notation
$$f|_A$$
mean? Is it perhaps "restricted to set $A$"?
• Yes, if $f$ is a function with domain $D$, then for a set $A \subseteq D$, $f|_A$ is used for the restriction of $f$ to $A$. Feb 2, 2016 at 15:43
• That's what I would assume. You might have two functions $f$ and $g$ for which $f \neq g$ but $f|_A = g|_A$ for some $A$. Feb 2, 2016 at 15:43
It means that I am constricting the domain of the function $f$. If $f:X\to Y$, then $g=f|_A$ means that $g:A\to Y$ where $A\subseteq X$.
Intuitively speaking, a function $f$ is constituted of three ingredients:
• a domain;
• a codomain;
• a rule (that, for each element in the domain, assigns a unique element in the codomain).
If we change any of these three ingredients, we obtain a different function. In particular, if we change the domain by a subset $A$ of the original domain (keeping the codomain and the rule), we get a new function which is represented by $f|_A$.
In other words: given a function $f:X\to Y$ and a set $A\subset X$, the notation $f|_A$ denotes the function $g:A\to Y$ given by $$g(x)=f(x),\quad \forall \ x\in A.$$
This is the usual meaning but, maybe, there are different meanings in other contexts.
That's correct.
Suppose we have a function $$f : Y \leftarrow X,$$ and a subset $A$ of $X$.
Approach 0. Then $f \restriction_A$ is defined as the unique function $Y \leftarrow A$ that agrees with $f$ on $A$. That is: $$\mathop{\forall}_{a \in A} ((f \restriction_A)(a) = f(a))$$
However, there's a cleaner way of formalizing this.
Approach 1. Write $$\mathrm{incl}_A : X \leftarrow A$$ for the inclusion of $A$ into $X$. Then we can form the composite $$f \circ \mathrm{incl}_A : Y \leftarrow A.$$ Write $f\restriction_A$ as a shorthand for this composite.
The nice thing about Approach 1 is that it makes proving the basic properties of the restriction operator trivial. In particular:
Claim. $(g \circ f)\restriction_A = g \circ (f \restriction_A)$
Proof.
$$(g \circ f)\restriction_A = (g \circ f) \circ \mathrm{incl}_A = g \circ (f \circ \mathrm{incl}_A) = g \circ (f \restriction_A)$$
Approach 1 is especially appealing from a category-theory perspective. In that context:
• $X$ is an object of some category
• a subobject of $X$ is by definition an object $\underline{A}$ together with a monomorphism $\mathrm{incl}_A : X \leftarrow \underline{A}$.
• a partial morphism $Y \leftarrow X$ consists of a subobject $A$ of $X$ together with a morphism $Y \leftarrow \underline{A}$.
Hence, if we're given a morphism $f : Y \leftarrow X$ and a subobject $A$ of $X$, then we get a partial morphism $f \restriction_A : Y \leftarrow X$ by forming the obvious composite.
The notation $f|_A$ is probably best understood via a meaningful example. Before giving one (I hope it will be useful, anyway), it would probably be good to consult two decent references:
2) Abstract Algebra by Dummit and Foote (p. 3, 3rd Ed.).
The relevant portion from the Wiki blurb:
Let $f\colon E\to F$ be a function from a set $E$ to a set $F$, so that the domain of $f$ is in $E$ (i.e., $\operatorname{dom}f\subseteq E$). If $A\subseteq E$, then the restriction of $f$ to $A$ is the function $f|_A\colon A\to F$.
Informally, the restriction of $f$ to $A$ is the same function as $f$, but is only defined on $A\cap\operatorname{dom} f$.
Wiki's "informal" remark is the key part in my opinion. The following excerpt from Dummit and Foote's Abstract Algebra may be slightly more abstract, but I think a meaningful example will clear everything up.
If $A\subseteq B$, and $f\colon B\to C$, we denote the restriction of $f$ to $A$ by $f|_A$. When the domain we are considering is understood we shall occasionally denote $f|_A$ again simply as $f$ even though these are formally different functions (their domains are different).
If $A\subseteq B$ and $g\colon A\to C$ and there is a function $f\colon B\to C$ such that $f|_A=g$, we shall say $f$ is an extension of $g$ to $B$ (such a map $f$ need not exist nor be unique).
Example: Let $g\colon\mathbb{Z}^+\to\{1\}$ be defined by $g(x)=1$ and let $f\colon\mathbb{Z}\setminus\{0\}\to\{1\}$ be defined by $f(x)=\dfrac{|x|}{x}$. Using the notation from the second paragraph above, we have $g\colon A\to C$ and $f\colon B\to C$, where
• $A = \mathbb{Z^+}$
• $B=\mathbb{Z}\setminus\{0\}$
• $C=\{1\}$
and, clearly, $A\subseteq B$. Thus, we have the following: \begin{align} f|_A &\equiv f\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em] &= \frac{|x|}{x}\tag{by definition}\\[0.5em] &= \frac{x}{x}\tag{if $x\in\mathbb{Z^+}$, then $|x|=x$ }\\[0.5em] &= 1\tag{simplify}\\[0.5em] &\equiv g\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em] &= g. \end{align} Apart from some slight notational abuse, perhaps, the above example shows that $f$ is an extension of $g$ to $B$ since $f|_A=g$. | 2022-05-20T20:19:12 | {
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https://math.stackexchange.com/questions/888151/mean-and-variance-of-piecewise-normal-distribution | Mean and Variance of "Piecewise" Normal Distribution
Note - I put piecewise in quotes because I don't think it's the right term to use (I can't figure out what to call it).
I am building a program to model the load that a user places on a server. The load a user produces follows a normal distribution. Depending on which application the user is using, however, the mean and variance of that normal distribution will be different. What I am trying to do is calculate the overall mean and variance for a user's activity given the proportions of time they use each application.
For example, Application A follows $\mathcal{N}(100, 50)$ and Application B follows $\mathcal{N}(500, 20)$. If a user uses A 50% of the time and B the other 50%, what is the mean and variance of the data that the user would produce during a day?
I'm able to simulate this by selecting a number from a uniform distribution between 0 and 1 and then generating a value from the appropriate distribution. Something like this:
$f(x) = \begin{cases} \mathcal{N}(100, 50), &0 \le x \lt 0.5\\ \mathcal{N}(500, 20), &0.5 \le x \lt 1\\ \end{cases}$
When I simulate a large number of these values and measure the results, it looks like the mean is just
$\sum\limits_{i=1}^n\mu_ip$
where $p$ is the percentage of the day a user is using each application.
I can't figure out what pattern the variance follows or what the formula might be to determine it without measuring a bunch of simulated values (When I simulate the above example, the variance looks to be something close to 41500).
I'd appreciate confirmation that how I'm calculating the combined mean is correct and some help in figuring out how to determine the variance of the overall distribution.
Let the two normal random variables be $X$ and $Y$, where $X$ is chosen with probability $p$, and $Y$ is chosen with probability $q=1-p$.
If $W$ is the resulting random variable, then $\Pr(W\le w)=p\Pr(X\le w)+q\Pr(Y\le w)$.
Differentiate. We get $f_W(w)=pf_X(w)+qf_Y(w)$.
The mean of $W$ is $\int_{-\infty}^\infty wf_W(w)$. Calculate. We get $$\int_{-\infty}^\infty w(pf_X(w)+qf_Y(w))\,dw.$$ This is $pE(X)+qE(Y)$, confirming your observation.
For the variance, we want $E(W^2)-(E(W))^2$. For $E(W^2)$, we calculate $\int_{-\infty}^{\infty} w^2(pf_X(w)+qf_Y(w))\,dw$. This is $pE(X^2)+qE(Y^2)$.
But $pE(X^2)= p(\text{Var}(X)+(E(X))^2)$ and $qE(Y^2)= q(\text{Var}(Y)+(E(Y))^2)$
Putting things together we get $$\text{Var}(W)=p\text{Var}(X)+q\text{Var}(Y)+ p(E(X))^2+q(E(Y))^2- (pE(X)+qE(Y))^2.$$
Remark: For a longer discussion, please look for Mixture Distributions.
• So, to put it more generally: $\text{Var}(W) = \sum\limits_{i=1}^np_i(\text{Var}(X_i) + E(X_i)^2) - (\sum\limits_{i=1}^np_iE(X_i))^2$ Aug 5, 2014 at 17:57
• Yes, the same argument works for $X_i$ with probability $p_i$. One could generalize further, to sums to $\infty$, also to "continuous" mixtures. Mixtures happen a lot in Statistics, since populations can often be stratified. Aug 5, 2014 at 18:15
• Perfect. Thank you so much for your help! I already ran some simulations with a few different combinations and it works beautifully. Aug 5, 2014 at 18:47
• You are welcome. Fast work. Aug 5, 2014 at 19:49
What you are wanting are the distributional quantities of the mixture distribution. There are some convenient formulas to do this: let $T$ be the unconditional load a user places on the system, and let $X$ be a Bernoulli random variable that indicates whether a user is on system $A$ or $B$. So $A = (T \mid X = 0) \sim \mathrm{Normal}(\mu_A = 100, \sigma_A^2 = 50)$, and $B = (T \mid X = 1) \sim \mathrm{Normal}(\mu_B = 500, \sigma_B^2 = 20)$. That is to say, \begin{align*} \mathrm{E}[T \mid X = 0] &= \mu_A, \\ \mathrm{E}[T \mid X = 1] &= \mu_B, \\ \mathrm{Var}[T \mid X = 0] &= \sigma_A^2, \\ \mathrm{Var}[T \mid X = 1] &= \sigma_B^2. \end{align*}
Then by the law of total expectation, \begin{align*} \mathrm{E}[T] &= \mathrm{E}[\mathrm{E}[T \mid X]] \\ &= \mathrm{E}[T \mid X = 0]\Pr[X = 0] + \mathrm{E}[T \mid X = 1]\Pr[X = 1] \\ &= \mu_A (1-p) + \mu_B p,\end{align*} where $p$ is the probability that a user is on system $B$. The variance is calculated by \begin{align*} \mathrm{Var}[T] &= \mathrm{E}[\mathrm{Var}[T \mid X]] + \mathrm{Var}[\mathrm{E}[T \mid X]] \\ &= \mathrm{Var}[T \mid X = 0]\Pr[X = 0] + \mathrm{Var}[T \mid X = 1]\Pr[X = 1] + \mathrm{Var}[\mathrm{E}[T \mid X]] \\ &= \sigma_A^2 (1-p) \sigma_B^2 p + \mathrm{Var}[\mathrm{E}[T \mid X]]. \end{align*} The last term requires a little subtlety to understand. The variable $\mathrm{E}[T \mid X]$ is a generalized Bernoulli, which takes on the value $\mu_A$ with probability $1-p$ and $\mu_B$ with probability $p$ (rather than $0$ and $1$). So we may write this as $$\mathrm{E}[T \mid X] = X(\mu_B - \mu_A) + \mu_A,$$ where $X \sim \mathrm{Bernoulli}(p)$. Therefore, $$\mathrm{Var}[\mathrm{E}[T \mid X]] = \mathrm{Var}[(\mu_B - \mu_A)X + \mu_A] = (\mu_B - \mu_A)^2 \mathrm{Var}[X] = (\mu_B - \mu_A)^2 p(1-p).$$ Hence $$\mathrm{Var}[T] = \sigma_A^2 (1-p) + \sigma_B^2 p + (\mu_B - \mu_A)^2 p(1-p).$$ The general case with $n$ systems $S_i \sim \mathrm{Normal}(\mu_i, \sigma_i^2)$, $i = 1, 2, \ldots, n$, where the user is on system $i$ with a probability of $p_i$, with $\sum_{i=1}^n p_i = 1$, involves a categorical distribution rather than a Bernoulli.
Recall that $V[X]=E[X^2]-E[X]^2$. Hence if you are given $E$ and $V$ of $X_1,X_2$ and combine them as described to a new random variable $Y$ by using a 0-1-random variable $Z$, i.e. picking $X_1$ if $Z=1$ and picking $X_2$ if $Z=0$, we find $$E[Y]=P(Z=1)\cdot E[Y|Z=1]+P(Z=0)\cdot E[Y|Z=0]=pE[X_1]+(1-p)E[X_2].$$ By the same argument we find $$E[Y^2] = pE[X_1^2]+(1-p)E[X_2^2]$$ Substituting $E[X_i^2]=V[X_i]+E[X_i]^2$, we obtain \begin{align}V[Y]&=E[Y^2]-E[Y]^2 \\&= p(V[X_1]+E[X_1]^2) + (1-p)(V[X_2]+E[X_2]^2)-(pE[X_1]+(1-p)E[X_2])^2\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]^2+E[X_2]^2)-2p(1-p)E[X_1]E[X_2]\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]-E[X_2])^2.\end{align} | 2022-11-30T05:12:18 | {
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# Linear Equations
Module by: Rupinder Sekhon. E-mail the author
Summary: This chapter covers principles of linear equations. After completing this chapter students should be able to: graph a linear equation; find the slope of a line; determine an equation of a line; solve linear systems; and complete application problems using linear equations.
## Chapter Overview
In this chapter, you will learn to:
1. Graph a linear equation.
2. Find the slope of a line.
3. Determine an equation of a line.
4. Solve linear systems.
5. Do application problems using linear equations.
## Graphing a Linear Equation
Equations whose graphs are straight lines are called linear equations. The following are some examples of linear equations:
2x3y=62x3y=6 size 12{2x - 3y=6} {}, 3x=4y73x=4y7 size 12{3x=4y - 7} {}, y=2x5y=2x5 size 12{y=2x - 5} {}, 2y=32y=3 size 12{2y=3} {}, and x2=0x2=0 size 12{x - 2=0} {}.
A line is completely determined by two points, therefore, to graph a linear equation, we need to find the coordinates of two points. This can be accomplished by choosing an arbitrary value for xx size 12{x} {} or yy size 12{y} {} and then solving for the other variable.
### Example 1
#### Problem 1
Graph the line: y=3x+2y=3x+2 size 12{y=3x+2} {}
### Example 2
#### Problem 1
Graph the line: 2x+y=42x+y=4 size 12{2x+y=4} {}
The points at which a line crosses the coordinate axes are called the intercepts. When graphing a line, intercepts are preferred because they are easy to find. In order to find the x-intercept, we let y=0y=0 size 12{y=0} {}, and to find the y-intercept, we let x=0x=0 size 12{x=0} {}.
### Example 3
#### Problem 1
Find the intercepts of the line: 2x3y=62x3y=6 size 12{2x - 3y=6} {}, and graph.
### Example 4
#### Problem 1
Graph the line given by the parametric equations: x=3+2tx=3+2t size 12{x=3+2t} {}, y=1+ty=1+t size 12{y=1+t} {}
### Horizontal and Vertical Lines
When an equation of a line has only one variable, the resulting graph is a horizontal or a vertical line.
The graph of the line x=ax=a size 12{x=a} {}, where aa size 12{a} {} is a constant, is a vertical line that passes through the point ( aa size 12{a} {}, 0). Every point on this line has the x-coordinate aa size 12{a} {}, regardless of the y-coordinate.
The graph of the line y=by=b size 12{y=b} {}, where bb size 12{b} {} is a constant, is a horizontal line that passes through the point (0, bb size 12{b} {}). Every point on this line has the y-coordinate bb size 12{b} {}, regardless of the x-coordinate.
#### Example 5
##### Problem 1
Graph the lines: x=2x=2 size 12{x= - 2} {} , and y=3y=3 size 12{y=3} {}.
## Slope of a Line
### Section Overview
In this section, you will learn to:
1. Find the slope of a line if two points are given.
2. Graph the line if a point and the slope are given.
3. Find the slope of the line that is written in the form y=mx+by=mx+b size 12{y= ital "mx"+b} {}.
4. Find the slope of the line that is written in the form Ax+By=cAx+By=c size 12{ ital "Ax"+ ital "By"=c} {}.
In the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the "steepness" of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope of the line.
From previous math courses, many of you remember slope as the "rise over run," or "the vertical change over the horizontal change" and have often seen it expressed as:
riserun, vertical changehorizontal change, ΔyΔxetc.riserun size 12{ { {"rise"} over {"run"} } } {}, vertical changehorizontal change size 12{ { {"vertical change"} over {"horizontal change"} } } {}, ΔyΔx size 12{ { {Δy} over {Δx} } } {}etc.
(5)
We give a precise definition.
Definition 1:
If ( x1x1 size 12{x rSub { size 8{1} } } {}, y1y1 size 12{y rSub { size 8{1} } } {}) and ( x2x2 size 12{x rSub { size 8{2} } } {}, y2y2 size 12{y rSub { size 8{2} } } {}) are two different points on a line, then the slope of the line is
Slope = m = y 2 y 1 x 2 x 1 Slope = m = y 2 y 1 x 2 x 1 size 12{"Slope"=m= { {y rSub { size 8{2} } - y rSub { size 8{1} } } over {x rSub { size 8{2} } - x rSub { size 8{1} } } } } {}
### Example 6
#### Problem 1
Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.
### Example 7
#### Problem 1
Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.
### Example 8
#### Problem 1
Graph the line that passes through the point (1, 2) and has slope 3434 size 12{ - { {3} over {4} } } {} .
### Example 9
#### Problem 1
Find the slope of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
### Example 10
#### Problem 1
Find the slope of the line y=3x+2y=3x+2 size 12{y=3x+2} {}.
### Example 11
#### Problem 1
Determine the slope and y-intercept of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
## Determining the Equation of a Line
### Section Overview
In this section, you will learn to:
1. Find an equation of a line if a point and the slope are given.
2. Find an equation of a line if two points are given.
So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on it, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.
An equation of a line can be written in two forms, the slope-intercept form or the standard form.
The Slope-Intercept Form of a Line: y = mx + b y = mx + b size 12{y= ital "mx"+b} {}
A line is completely determined by two points, or a point and slope. So it makes sense to ask to find the equation of a line if one of these two situations is given.
### Example 12
#### Problem 1
Find an equation of a line whose slope is 5, and y-intercept is 3.
### Example 13
#### Problem 1
Find the equation of the line that passes through the point (2, 7) and has slope 3.
### Example 14
#### Problem 1
Find an equation of the line that passes through the points (–1, 2), and (1, 8).
### Example 15
#### Problem 1
Find an equation of the line that has x-intercept 3, and y-intercept 4.
The Standard form of a Line: Ax + By = C Ax + By = C size 12{ ital "Ax"+ ital "By"=C} {}
Another useful form of the equation of a line is the Standard form.
Let LL size 12{L} {} be a line with slope mm size 12{m} {}, and containing a point (x1,y1)(x1,y1) size 12{ $$x rSub { size 8{1} } ,y rSub { size 8{1} }$$ } {}. If (x,y)(x,y) size 12{ $$x,y$$ } {} is any other point on the line LL size 12{L} {}, then by the definition of a slope, we get
m = y y 1 x x 1 m = y y 1 x x 1 size 12{m= { {y - y rSub { size 8{1} } } over {x - x rSub { size 8{1} } } } } {}
(21)
y y 1 = m ( x x 1 ) y y 1 = m ( x x 1 ) size 12{y - y rSub { size 8{1} } =m $$x - x rSub { size 8{1} }$$ } {}
(22)
The last result is referred to as the point-slope form or point-slope formula. If we simplify this formula, we get the equation of the line in the standard form, Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {}.
### Example 16
#### Problem 1
Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope –3/5.
### Example 17
#### Problem 1
Find the standard form of the line that passes through the points (1, -2), and (4, 0).
### Example 18
#### Problem 1
Write the equation y=2/3x+3y=2/3x+3 size 12{y= - 2/3x+3} {} in the standard form.
### Example 19
#### Problem 1
Write the equation 3x4y=103x4y=10 size 12{3x - 4y="10"} {} in the slope-intercept form.
Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.
By solving for yy size 12{y} {}, it can easily be shown that the slope of the line Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {} is A/BA/B size 12{ - A/B} {}. The reader should verify.
### Example 20
#### Problem 1
Find the slope of the following lines, by inspection.
1. 3x5y=103x5y=10 size 12{3x - 5y="10"} {}
2. 2x+7y=202x+7y=20 size 12{2x+7y="20"} {}
3. 4x3y=84x3y=8 size 12{4x - 3y=8} {}
Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be very easy.
### Example 21
#### Problem 1
Find an equation of the line that passes through (2, 3) and has slope 4/54/5 size 12{ - 4/5} {}.
If you use this method often enough, you can do these problems very quickly.
## Applications
Now that we have learned to determine equations of lines, we get to apply these ideas in real-life equations.
### Example 22
#### Problem 1
A taxi service charges $0.50 per mile plus a$5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling xx size 12{x} {} miles?
### Example 23
#### Problem 1
The variable cost to manufacture a product is $10 and the fixed cost$2500. If xx size 12{x} {} represents the number of items manufactured and yy size 12{y} {} the total cost, write the cost function.
### Example 24
#### Problem 1
It costs $750 to manufacture 25 items, and$1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.
### Example 25
#### Problem 1
The freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.
### Example 26
#### Problem 1
The population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.
## More Applications
### Section Overview
In this section, you will learn to:
1. Solve a linear system in two variables.
2. Find the equilibrium point when a demand and a supply equation are given.
3. Find the break-even point when the revenue and the cost functions are given.
In this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.
### Example 27
#### Problem 1
Find the intersection of the line y=3x1y=3x1 size 12{y=3x - 1} {} and the line y=x+7y=x+7 size 12{y= - x+7} {}.
### Example 28
#### Problem 1
Find the intersection of the lines 2x+y=72x+y=7 size 12{2x+y=7} {} and 3xy=33xy=3 size 12{3x - y=3} {} by the elimination method.
### Example 29
#### Problem 1
Solve the system of equations x+2y=3x+2y=3 size 12{x+2y=3} {} and 2x+3y=42x+3y=4 size 12{2x+3y=4} {} by the elimination method.
### Example 30
#### Problem 1
Solve the system of equations 3x4y=53x4y=5 size 12{3x - 4y=5} {} and 4x5y=64x5y=6 size 12{4x - 5y=6} {}.
### Supply, Demand and the Equilibrium Market Price
In a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.
### Example 31
#### Problem 1
The supply curve for a product is y=1.5x+10y=1.5x+10 size 12{y=1 "." 5x+"10"} {} and the demand curve for the same product is y=2.5x+34y=2.5x+34 size 12{y= - 2 "." 5x+"34"} {}, where xx size 12{x} {} is the price and y the number of items produced. Find the following.
1. How many items will be supplied at a price of $10? 2. How many items will be demanded at a price of$10?
3. Determine the equilibrium price.
4. How many items will be produced at the equilibrium price?
### Break-Even Point
In a business, the profit is generated by selling products. If a company sells xx size 12{x} {} number of items at a price PP size 12{P} {}, then the revenue RR size 12{R} {} is PP size 12{P} {} times xx size 12{x} {} , i.e., R=PxR=Px size 12{R=P cdot x} {}. The production costs are the sum of the variable costs and the fixed costs, and are often written as C=mx+bC=mx+b size 12{C= ital "mx"+b} {}, where xx size 12{x} {} is the number of items manufactured.
A company makes a profit if the revenue is greater than the cost, and it shows a loss if the cost is greater than the revenue. The point on the graph where the revenue equals the cost is called the Break-even point.
### Example 32
#### Problem 1
If the revenue function of a product is R=5xR=5x size 12{R=5x} {} and the cost function is y=3x+12y=3x+12 size 12{y=3x+"12"} {}, find the following.
1. If 4 items are produced, what will the revenue be?
2. What is the cost of producing 4 items?
3. How many items should be produced to break-even?
4. What will be the revenue and the cost at the break-even point?
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https://mathoverflow.net/questions/316044/product-of-concave-functions-and-harmonic-mean | # Product of concave functions and harmonic mean
I discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $$f: \Omega \subset \mathbb{R} \rightarrow \mathbb{R_+^*}$$ is $$\alpha$$-concave if $$f^\alpha$$ is concave.
Let $$f$$ a $$\alpha$$-concave function, $$g$$ a $$\beta$$-concave function. Let $$\gamma$$ be the half of the harmonic mean of $$\alpha$$ and $$\beta$$, ie. $$\begin{equation*}\frac{1}{\gamma} = \frac{1}{\alpha} + \frac{1}{\beta}.\end{equation*}$$ Then $$fg$$ is a $$\gamma$$-concave function.
This can be proved by computing the hessian of $$fg$$. Have you already seen this result in the literature?
• A small remark: $\gamma$ is not the harmonic mean, but half of it. – Ivan Izmestiev Nov 24 '18 at 6:59
• You're perfectly right, thanks! I've just corrected it. – LacXav Nov 26 '18 at 15:51
We want to prove that $$(f(ax+by)g(ax+by))^\g \ge a(f(x)g(x))^\g + b (f(y)g(y))^\g$$ for every $$x, y$$ and $$a+b = 1$$. Since we know that $$f(ax+by)^\a \ge af(x)^\a + bf(y)^\a$$ and $$g(ax+by)^\b \ge ag(x)^\b + bg(y)^\b$$ it is enough to prove that
$$(af(x)^\a + bf(y)^\a)^{\frac{\g}{\a}} (ag(x)^\b + bg(y)^\b)^{\frac{\g}{\b}}\ge af(x)^\g g(x)^\g + b f(y)^\g g(y)^\g.$$
Consider set $$\{x, y\}$$ as measurable space with $$m(x) = a$$, $$m(y) = b$$. Then our desired inequality (after rising to the power $$\frac{1}{\g}$$) is nothing but Holder inequality with parameters $$\a, \b$$.
Unfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.
This is nice, and I do not know a reference for this.
The notion of $$p$$-concavity is mentioned, for example, in Section 9 of
Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.
Note that a natural interpretation of $$0$$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave.
Also your result is equivalent to the following.
Theorem. If functions $$F$$ and $$G$$ are concave, then so is $$F^tG^{1-t}$$ for all $$t \in [0,1]$$.
Indeed, substitute $$f^\alpha = F$$, $$g^\beta = G$$, $$t = \frac{\beta}{\alpha+\beta}$$.
The concavity of $$F^tG^{1-t}$$ can be proved as follows. For every $$\lambda \in (0,1)$$ and $$\mu = 1-\lambda$$ the concavity of $$F$$ and $$G$$ imply $$\begin{multline*} F^tG^{1-t}(\lambda x + \mu y) = F^t(\lambda x + \mu y) G^{1-t}(\lambda x + \mu y)\\ \ge (\lambda F(x) + \mu F(y))^t (\lambda G(x) + \mu G(y))^{1-t}\\ \ge (\lambda F(x))^t(\lambda G(x))^{1-t} + (\mu F(y))^t(\mu G(y))^{1-t}\\ = \lambda F^tG^{1-t}(x) + \mu F^tG^{1-t}(y) \end{multline*}$$ (In the middle we have used the inequality $$(1+u)^t(1+v)^{1-t} \ge 1+u^t v^{1-t}$$, for which I have only a nasty proof.)
One can prove the theorem also by computing the second derivative of $$F^tG^{1-t}$$ (which is fun), but the above argument is more general because it does not require differentiability.
I think these results should be known, but have no reference.
EDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $$t=\frac12$$ as a lemma (in this case the nasty inequality becomes $$\sqrt{(1+u)(1+v)} \ge 1 + \sqrt{uv}$$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $$t = p/2^n$$ cases, which implies the general case by a limit argument.
EDIT: A reformulation of the theorem: the space of exp-concave functions is convex.
• It seems to me that your inequality(and the initial problem as well) can be redused to the case $t=\frac{1}{2}$ via mimicking proof of the fact that mid-point concavity implies concavity for continuous functions. And for $t=\frac{1}{2}$ it is just C-S. – Aleksei Kulikov Nov 23 '18 at 23:33
• And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. – Aleksei Kulikov Nov 24 '18 at 0:37
• @AlekseiKulikov: I will add the $t=1/2$ approach to my answer. What do you mean by Hoelder inequality on $2$-point measurable space? Could you write this up as an answer? – Ivan Izmestiev Nov 24 '18 at 6:52
• Thanks for the reference, and having put in perspective. – LacXav Nov 26 '18 at 16:35 | 2021-05-06T07:19:22 | {
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https://math.stackexchange.com/questions/764454/permutations-expressed-as-product-of-transpositions | # Permutations expressed as product of transpositions
There is a theorem that states that all permutations can be expressed as a product of transpositions. I have a couple of questions about this theorem:
1. Does the product which is equal to the permutation always start from the identity permutation?
In the proof for this theorem our professor has argued that every permutation can be transformed into the identity permutation by applying a certain number of transpositions, e.g. if $\sigma$ is a permutation not equal to the identity permutation then you can apply, say l transpositions, so that you get: $\tau_l \circ \tau_{l-1} \circ .... \circ \tau_1 \circ \sigma = id \Rightarrow \tau_1^{-1} \circ \tau_2^{-1} \circ ... \circ \tau_l^{-1}=\tau_1 \circ \tau_2 \circ .... \circ \tau_l$.
Is this product of transpositions always unique, or can you start from any arbitrary permutation and perform the required number of transpositions to get your permutation?
2 If I form the composition of two permutations, say $\sigma_1$ and $\sigma_2$ given by: $$\sigma_1 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 &5 &6 &1 &2 \\ \end{pmatrix}$$
$$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 &6 &1 &5 &3 \\ \end{pmatrix}$$
I can express them in terms of the following transpositions $$\sigma_1 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6)$$ $$\sigma_2 = (1,4) \circ (2,4) \circ (3,6)$$
When I form the composition $\sigma_1 \circ \sigma_2$ I get:
$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 6 &2 &3 &1 &5 \\ \end{pmatrix}$$
But since the product of the transpositions is equal to the permutations, I should get the same result when I use:
$$\sigma_1 \circ \sigma_2 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6) \circ (1,4) \circ (2,4) \circ (3,6)$$
but I get:
$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 &5 &3 &2 &4 \\ \end{pmatrix}$$
Why doesn't this work?
• I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc. – mdp Apr 22 '14 at 13:49
• Thanks for your comment, but I did start from the right side: $\begin{pmatrix} 1 & 2 & 3&4&5&6\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 2 & 6&4&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 4 & 6&2&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&1&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&3&5&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 5&3&6&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 1 & 5&3&6&4\end{pmatrix} \Rightarrow \begin{pmatrix} 6 & 1 & 5&3&2&4\end{pmatrix}$ I'm not sure what you mean by $1 \rightarrow 4 \rightarrow 6 \rightarrow 2...$ – eager2learn Apr 22 '14 at 13:58
• Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$). – mdp Apr 22 '14 at 14:04
• Yes that was the mistake I made. Thanks a lot for your help. – eager2learn Apr 22 '14 at 14:33
Note that the product of transpositions that you used to express $\sigma_1$, when composed, does not again yield $\sigma_1$; I.e., you did not correctly express $\sigma_1$ as a product of transpositions.
Rather, $\sigma_1$ can be written $(1,5)\circ (1, 3)\circ (2, 6)\circ (2, 4)$, or $(1, 3)\circ (3, 5)\circ (2, 4)\circ(4, 6)$...
Similarly, $\sigma_2$ is incorrectly decomposed. Two correct decompositions include $(1, 4)\circ (1, 2)\circ (3, 6)$ and $(1, 2)\circ (2, 4) \circ (3, 6)$...
...which answers your question about uniqueness. When writing a permutation as a product of transpositions, there are many such ways to do this. What does not vary is the parity: an "odd" permutation is one that can only be decomposed to a product of an odd number of transpositions, and "even" permutations can only be decomposed into a product of an even number of transpositions. So, for example, $\sigma_1$ is even, and $\sigma_2$ is odd.
• I don't understand this. If I apply $(1,5)\circ (1,3) \circ (2,6) \circ (2,4)$ on the id permutation I get the permutation $\begin{pmatrix} 5 &6&1&2&3&4 \end{pmatrix}$ Why is this a transposition for $\sigma_1$? I think this goes back to my original question, from which permutation do I start when applying those transpositions? – eager2learn Apr 22 '14 at 14:26
• You start from the rightmost permutation. Where does it send $1$?. If it sends $1$ to $a$, then you move to the next transposition to its left to see where it sends $a$. Etc. – Namaste Apr 22 '14 at 14:30 | 2019-10-18T11:55:54 | {
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https://wedding-photos-ideas.com/2fqzd/pampered-chef-rhsdaul/lnvno.php?page=ridge-regression-alpha-4d72aa | By default, glmnet will do two things that you should be aware of: Since regularized methods apply a penalty to the coefficients, we need to ensure our coefficients are on a common scale. Shows the effect of collinearity in the coefficients of an estimator. Ridge regression is a method by which we add a degree of bias to the regression estimates. The second line fits the model to the training data. Keep in mind, ridge is a regression … 11. Ridge, LASSO and Elastic net algorithms work on same principle. In scikit-learn, a ridge regression model is constructed by using the Ridge class. Lasso is great for feature selection, but when building regression models, Ridge regression should be your first choice. In R, the glmnet package contains all you need to implement ridge regression. Active 2 years, 8 months ago. Let us first implement it on our above problem and check our results that whether it performs better than our linear regression model. They all try to penalize the Beta coefficients so that we can get the important variables (all in case of Ridge and few in case of LASSO). Ridge Regression is the estimator used in this example. Linear regression is the standard algorithm for regression that assumes a linear relationship between inputs and the target variable. The value of alpha is 0.5 in our case. Ridge Regression. Elastic net regression combines the properties of ridge and lasso regression. Ridge Regression is a neat little way to ensure you don't overfit your training data - essentially, you are desensitizing your model to the training data. Use the below code for the same. This is also known as $$L1$$ regularization because the regularization term is the $$L1$$ norm of the coefficients. Ridge regression - introduction¶. Yes simply it is because they are good biased. Important things to know: Rather than accepting a formula and data frame, it requires a vector input and matrix of predictors. regression_model = LinearRegression() regression_model.fit(X_train, y_train) ridge = Ridge(alpha=.3) For example, to conduct ridge regression you may use the sklearn.linear_model.Ridge regression model. scikit-learn provides regression models that have regularization built-in. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Generally speaking, alpha increases the affect of regularization, e.g. The Alpha Selection Visualizer demonstrates how different values of alpha influence model selection during the regularization of linear models. Ridge Regression is a technique for analyzing multiple regression data that suffer from multicollinearity. There are two methods namely fit() and score() used to fit this model and calculate the score respectively. Here, we are using Ridge Regression as a Machine Learning model to use GridSearchCV. One commonly used method for determining a proper Γ \boldsymbol{\Gamma} Γ value is cross validation. Note that scikit-learn models call the regularization parameter alpha instead of $$\lambda$$. The alpha parameter tells glmnet to perform a ridge (alpha = 0), lasso (alpha = 1), or elastic net (0 < alpha < 1) model. ridgeReg = Ridge(alpha=0.05, normalize=True) ridgeReg.fit(x_train,y_train) pred = ridgeReg.predict(x_cv) calculating mse When multicollinearity occurs, least squares estimates are unbiased, but their variances are large so they may be far from the true value. This notebook is the first of a series exploring regularization for linear regression, and in particular ridge and lasso regression.. We will focus here on ridge regression with some notes on the background theory and mathematical derivations that are useful to understand the concepts.. Then, the algorithm is implemented in Python numpy
## ridge regression alpha
Facebook Message Says Seen But No Time, Which Is Worse Agnostic Or Atheist, Activity Diagram For Online Shopping, Sprinkler Cad Block, Amaryllis Name Meaning, Easton Beast Speed, Meal Village Menu, | 2021-01-24T03:42:29 | {
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https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-17-rapidly-decreasing-singular-values/ | # Lecture 17: Rapidly Decreasing Singular Values
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## Description
Professor Alex Townsend gives this guest lecture answering the question “Why are there so many low rank matrices that appear in computational math?” Working effectively with low rank matrices is critical in image compression applications.
## Summary
Professor Alex Townsend's lecture
Why do so many matrices have low effective rank?
Sylvester test for rapid decay of singular values
Image compression: Rank $$k$$ needs only $$2kn$$ numbers.
Flags give many examples / diagonal lines give high rank.
Related section in textbook: III.3
Instructor: Prof. Alex Townsend
GILBERT STRANG: So let me use the mic to introduce Alex Townsend, who taught here at MIT-- taught Linear Algebra 18.06 very successfully.
And then now he's at Cornell on the faculty, still teaching very successfully. And he was invited here yesterday for a big event over in Engineering.
And he agreed to give a talk about a section of the book-- section 4.3-- which, if you look at it, you'll see is all about his work. And now you get to hear from the creator himself. OK.
ALEX TOWNSEND: OK. Thanks. Thank you, Gil. Thank you for inviting me here. I hope you're enjoying the course.
Today I want to tell you a little about why there so many matrices that are low rank in the world. So as computational mathematicians-- Gil and myself-- we come across low-rank matrices all the time. And we started wondering, as a community, why?
What is it about the problems that we are looking at? What makes low-rank matrices appear? And today I want to give you that story-- or at least an overview of that story.
So for this class, x is going to be n by n real matrix. So nice and square. And you already know, or are very comfortable with, the singular values of a matrix.
So the singular values of a matrix, as you know, are a sequence of numbers that are monotonically non-increasing that tell us all kinds of things about the matrix x.
For example, the number of nonzero singular values tell us the rank of the matrix x. And they also, you probably know, tell us how well a matrix x can be approximated by a low-rank matrix.
So let me just write two facts down that you already are familiar with. So here's a fact-- that, if I look at the number of non-zero singular values in x-- so I'm imagining there's going to be k non-zero singular values-- then we can say a few things about x.
For example, the rank of x, as we know, is k-- the number of non-zero singular values. But we also know from the SVD that we can decompose x into a sum of rank 1 matrices-- in fact, the sum of k of them.
So because x is rank k, we can write down a low-rank representation for x, and it involves k terms, like this.
Each one of these vectors here is a column vector. So if I draw this pictorially, this guy looks like this, right? And we have k of them. So because x is rank k, we can write x as a sum of k rank 1 matrices.
And we also have an initial fact that we already know-- that the dimension of the column space of x is equal to k, and the same with the row space. So the column space of x equals the row space of x-- the dimension-- and they all equal k.
And so there are three facts we can determine from looking at this sequence of singular values of a matrix x. Of course, the singular value sequence is unique. X defines its own singular values.
What we're interested in here is, what makes x? What are the properties of x that make sure that the singular values have a lot of zeros in that sequence? Can we try to understand what kind of x makes that happen?
And we really like matrices that have a lot of zeros here, for the following reason-- we say x is low rank if the following holds, right? Because if we wanted to send x to our friend-- we're imagining x as picture where each entry is a pixel of that image.
If that matrix-- that image-- was low rank, we could send the picture to our friend in two ways. We could send one every single entry of x. And for us to do that, we would have to send n squared pieces of information, because we'd have to send every entry.
But if x is sufficiently low rank, we could also send our friend the vectors-- u, u1, v1, uk, up to vk. And how much pieces of data would we have to send our friend to get x to them if we sent in the low-rank form?
Well, there's 2n here, 2n here numbers. There's k of them. So we'd have to send 2kn numbers. And we strictly say a matrix is low rank if it's more efficient to send x to our friend in low-rank form then in full-rank form.
So this, of course, by a little calculation, just shows us that, provided the rank is less than half the size of the matrix, we are calling the matrix low rank.
Now, often, in practice, we demand more. We demand that k is much smaller than this number, so that it's far more efficient to send our friend the matrix x in low-rank form than in full-rank form. So the colloquial use of the word low rank is kind of this situation. But this is the strict definition of it.
So what do low-rank matrices look like? And to do that, I have some pictures for you. I have some flags-- the world flags. So these are all matrices x-- these examples-- because their flags happen to not be square. I hope you can all see this.
But the top row here are all matrices that are extremely low rank. For example, the Austria flag-- if you want to send that to your friend, that matrix is of rank 1. So all you have to do is send your friend two vectors. You have to tell your friend the column space and the row space. And there's only the dimensions of one of both.
For the English flag, you need to send them two column vectors and two row vectors-- u1, v1, u2 and v2. And as we go down this row, they get slowly fuller and fuller rank.
So the Japanese flag, for example, is low rank but not that small. The Scottish flag is essentially full rank. So it's very inefficient to send your friend the Scottish flag in low-rank form. You're better off sending almost every single entry.
So what do low-rank matrices look like? Well, if the matrix is extremely low rank, like rank 1, then when you look at that matrix-- like here, like the flag-- it's highly aligned with the coordinates-- with the rows and columns.
So if it's rank 1, the matrix is highly aligned-- like the Austria flag. And of course, as we add in more and more rank here, the situation gets a bit blurry. For example, once we get into the medium rank situation, which is a circle, it's very hard to see that the circle is actually, in fact, low rank.
But what I'm going to do was try to understand why the Scottish flag or diagonal patterns-- particularly a bad example for low rank. So I'm going to take the triangular flag to examine that more carefully. So the triangular flag looks like-- I'll take a square matrix and I'll color in the bottom half.
So this matrix is the matrix of ones below the diagonal. And I'm interested in this matrix and, in particular, its singular values, to try to understand why diagonal patterns are not particularly useful for low-rank compression. And this matrix of all ones has a really nice property that, if I take its inverse, it looks a lot like-- getting close to Gil's favorite matrix.
So if I take the inverse of this matrix-- it has an inverse because it's got ones on the diagonal-- then its inverse is the following matrix, which people familiar with finite difference schemes will notice the familiarity between that and the first order finite difference approximation.
In particular, if I go a bit further and times two of these together, and do this, then this is essentially Gil's favorite matrix, except one entry happens to be different-- ends up being this matrix, which is very close to the second order, central, finite difference matrix.
And people have very well studied that matrix and know its eigenvalues, its singular values-- they know everything about that matrix. And you'll remember that if we know the eigenvalues of a matrix, like x transpose x, we know the singular values of x. So this allows us to show, by the fact that we know that, that the singular values of this matrix are not very amenable to low rank. They're all non-zero, and they don't even decay.
So I'm getting this from-- I rang up Gil, and Gil tells me these numbers. That allows us to work out exactly what the singular values of this matrix are, from the connection to finite differences.
And so we can understand why this is not good by looking at the singular values. So the first singular value of x from this expression is going to be approximately 2n over pi. And from this expression, again, for the last guy-- the last singular value of x is going to be approximately a half.
So these singular values are all large. They're not getting close to zero. If I plotted these singular values on a graph-- so here's the first singular value, the second, and the n-th-- then what would the graph look like?
Well, plot these numbers. Divide by this guy so that they all are bounded between 1 and 0 because of the normalization, because I divided by sigma 1 of x.
And so we can plot them, and they will look like this kind of thing. This number happens to be here where they come to be pi over 4n, which is me dividing this number by this number, approximately.
So triangular patterns are extremely bad for low rank. We need things-- or we at least intuitively think that we need things-- aligned with the rows and columns, but the circle case happens to also be low rank. And so what happened to the Japanese flag? Why is the Japanese flag convenient for low rank?
Well it's the fact that it's a circle, and there's lots of symmetry in a circle. So if I try to look at the rank of a circle, the Japanese flag, then I can bound this rank by decomposing the Japanese flag into two things. So this is going to be less than or equal to the rank of sum of two matrices, and I'll do it so that the decomposition works out.
I have the circle. I'm going to cut out a rank one piece that lives in the middle of this circle. OK? And I'm going to cut out a square from the interior of that circle. OK? And I can figure out-- of course the rank is just bounded by the sum of those two ranks. This guy is bounded by rank one because it's highly aligned with the grid. So this guy is bounded by rank one. So this thing here plus 1.
And now I have to try to understand the rank of this piece. Now this piece has lots of symmetry. For example, we know that the rank of that matrix is the dimension of the column space and the dimension of the row space. So when we look at this matrix, because of symmetry, if I divide this matrix in half along the columns, all the columns on the left appear on the right. So for example, the rank of this matrix is the same as the rank of that matrix because I didn't change the column space. OK?
Now I go again and divide along the rows, and now the row dimension of this matrix is the same as the top half, because as I wipe out those, I didn't change the dimension of the row space because the rows are the same top-bottom. And so this becomes the rank of that tiny little matrix there. And because it's small, it won't have too large a rank. So this is definitely less than-- if I divide that up, a little guy here looks like that plus the other guy that looks like that plus 1.
And so of course the row space of this matrix cannot be very high because this is a very thin matrix. There's lots of zeros in that matrix, only a few ones. And so you can go along and do a bit of trig to try to figure out how many rows are non-zero in this matrix.
And a bit of trig tells you-- well it depends on the radius of this original circle. So if I make the original radius r of this Japanese flag, then the bound that you end up getting will be, for this matrix, r 1 minus square root 2 over 2 for this guy. That's a bit of trig. I've got to make sure that's an integer. And then again, here it's the same but for the column space. So this is me just doing trig. OK?
And that's bound on the rank. It happens to be extremely good. And if you work out what that rank is and try to look back, you will find it's extremely efficient to send the Japanese flag to your friend in low rank form, because it's not full rank because these numbers are so small. So this comes out to be, like, approximately 1/2 r plus 1. So much smaller than what you would expect, because remember, a circle is almost the anti-version version of a line with the grid, but yet, it's still low rank. OK.
Now most matrices that we come up with in computational math are not exactly of finite rank. They are of numerical rank. And so I'll just define that. So the numerical rank of a matrix is very similar to the rank, except we allow ourselves a little bit of wiggle room when we define it, and so that amount of wiggle room will be of parameter called tol called epsilon. That's a tolerance. I'm thinking of epsilon as a tolerance. That's the amount of wiggle room I'm going to give myself. OK.
And we say that the numerical rank-- I'll put an epsilon there to denote numerical rank-- is k. k is the first singular value, or the last singular value, above epsilon. In the following sense, I'm copying the definition above but with epsilons instead of zeros. If this singular value is less than epsilon, relatively, and the kth one was not below. So k plus 1 is the first singular value below epsilon in this relative sense.
So of course the rank of 0x, if that was defined, is the same as the rank of x. OK? So this is just allowing ourselves some wiggle room. But this is actually what we're interested more in practice. All right? I don't want to necessarily send my friend the flag to exact precision. I would actually be happy to send my friend the flag up to 16 digits of precision, for example. They're not going to tell the difference between those two flags. And if I can get away with compressing the matrix a lot more once I have a little bit of wiggle room, that would be a good thing.
So we know from the Eckart and Young that the singular values tell us how well we can approximate x by a low-rank matrix. In particular, we know that the k plus 1 singular value of x tells us how well x can be approximated by a rank k matrix. OK? For example, when the rank was exactly k, the sigma k plus 1 was 0, and then this came out to be 0 and we found that x was exactly a rank k matrix.
Here, because we have the wiggle room, the epsilon, we get an approximation, not an exact. So this is telling us how well we can approximate x by a rank k matrix. OK? That's what the singular values are telling us. And so this allows us to try our best to compress matrices but use low-rank approximation rather than doing things exactly.
And of course, on a computer, when we're using floating point arithmetic, or on a computer because we always round numbers to the nearest 16-digit number, if epsilon was 16 digits, your computer wouldn't be able to tell the difference between x or x the rank k approximation if this number satisfied this expression. Your computer would think of x and xk as the same matrix because it would inevitably round both to epsilon, within epsilon. OK.
So what kind of matrices are numerically of low rank? Of course all low-rank matrices are numerically of low rank because the wiggle room can only help you but it's far more than that. There are many full-rank matrices-- matrices that don't have any singular values that are zero-- but the singular values decay rapidly to zero. That are full-rank matrices with low numerical rank because of the wiggle room.
So for example, here is the classic matrix that fits this regime. If I give you this, this is called the Hilbert matrix. This is a matrix that happens to have extremely low numerical rank but it's actually full rank, which means that I can approximate H by a rank k matrix where k is quite small very well, provided you give me some wiggle room, but it's not a low-rank matrix in the sense that if epsilon was zero here, you didn't allow me the wriggle room, all the singular values of this matrix are positive. So it's of low numerical rank but it's not a low-rank matrix.
The other classical example which motivated a lot of the research in this area was the Vandermonde matrix. So here is the Vandermonde matrix. An n by n version of it. Think of the xi's as real. And this is Vandermonde. This is the matrix that comes up when you try to do polynomial interpolation at real points.
This is an extremely bad matrix to deal with because it's numerically low rank, and often, you actually want to solve a linear system with this matrix. And numerical low rank implies that it's extremely hard to invert, so numerical low rank is not always good for you. OK? Often, we want the inverse, which exists, but it's difficult because V has low numerical rank.
OK. So people have been trying to understand why these matrices are numerically of low rank for a number of years, and the classic reason why there are so many low-rank matrices is because the world is smooth, as people say. They say, the world is smooth. That's why matrices are of numerical low rank. And to illustrate that point, I will do an example.
So this is classically understood by a man called Reade in 1983, and this is what his reason was. I have a picture of John Reade. He's not very famous, so I try to make sure his picture gets around. He's playing the piano. It's, like, one of the only pictures I could find of him.
So what is in this reason? Why do people say this? Well here's an example that illustrates it. If I take a polynomial in two variables and I-- for example, this is a polynomial of two variables-- and my x matrix comes from sampling that polynomial integers-- for example, this matrix-- then that matrix happens to be of low rank-- mathematically of low rank, with epsilon equals zero.
Why is that? Well if I write down x in terms of matrices, you could easily see it. So this is made up of a matrix of all ones plus a matrix of j-- so that's 1, 2, up to n, 1, 2, up to n, because every entry of that matrix just depends on the row index. And then this guy depends on both j and k. So this is a multiplication table, right? So this is n, 2, 4, up to 2n, n, 2n, n squared. OK.
Clearly, the matrix of all ones is a rank one matrix. The same with this guy. The column space is just of dimension one. And the last guy also happens to be of rank one because I can write this matrix in rank one form, which is a column vector times a row vector. OK. So this matrix x is of rank three. I guess at lowest rank three is what I've actually shown. OK.
Now of course this hasn't got to numerical low rank yet, so let's get ourselves there. So Reade knew this, and he said to himself, OK, well if I can approximate-- if x is actually coming from sampling a function, and I approximate that function by polynomial, then I'm going to get myself a low-rank approximation and get a bound on the numerical rank.
So in general, if I give you a polynomial of two variables, which can be written down-- it's degree n in both x and y. Let's just keep these indexes away from the matrix index. I give you this such polynomial, and I go away and I sample it and make a matrix X, then X, by looking at each term individually like I did there, will have low rank mathematically, with epsilon equals zero. This will have, at most, m squared rank, and if m is 3 or 4 or 10, it possibly could be low because this X could be a large matrix. OK.
So what Reade did for the Hilbert matrix was said, OK, well look at that guy. That guy looks like it's sampling a function. It looks like it's sampling the function 1 over x plus y minus 1. So he said to himself, well, that x, if I look at the Hilbert matrix, then that is sampling a function. It happens to not be a polynomial. It happens to be this function.
But that's OK because sampling polynomials, integers, gives me low rank exactly. Maybe sampling smooth functions, functions like this, can be well approximated by polynomials and therefore have low numerical rank. And that's what he did in this case. So he tried to find a p, a polynomial approximation to f. In particular, he looked at exactly this kind of approximation. So he has some numbers here so that things get dissolved later. And he tried to find a p that did this kind of approximation. So this approximates f.
And then he would develop a low-rank approximation to X by sampling p. So he would say, OK, well if I let y be a sampling of p, then from the fact that f is a good approximation to p, y is a good approximation to X. And so this has finite rank. He wrote down that this must hold. And the epsilon comes out here because these factors were chosen just right. The divide by n was chosen so that the epsilon came out just there. OK?
So, for many years, that was kind of the canonical reason that people would give, that, well, if the matrix X is sampled from a smooth function, then we can approximate our function by a polynomial and get polynomial rank approximations. And therefore, the matrix X will be of low numerical rank. There's an issue with this reasoning, especially for the Hilbert matrix, that it doesn't actually work that well.
So for example, if I take the 1,000 by 1,000 Hilbert matrix and I look at its rank-- OK, well I've already told you this is full rank. You'll get 1,000. All the singular values are positive. If I look at the numerical rank of this 1,000 by 1,000 Hilbert matrix and I compute it, I compute the SVD and I look at how many are above epsilon where epsilon is 10 to the minus 15, so that means I can approximate the 1,000 by 1,000 Hilbert matrix by a rank 28 matrix and only give up 15-- there will be exact 15 digits, which is a huge amount. So this is what we get in practice, but Reade's argument here shows that the rank of this matrix, the numerical rank, is at most. So it doesn't do a very good job on the Hilbert matrix for bounding the rank, right?
So Reade comes along, takes this function. He tries to find a polynomial that does this, where epsilon is 10 to the minus 15. He finds that the number of terms that he needs in this expression here is around 719, and therefore, that's the rank that he gets. The bound on the numerical rank. The trouble is that 719 tells us that this is not of low numerical rank, but we know it is, so it's an unsatisfactory reason.
So there's been several people trying to come up with more appropriate reasons that explain the 28 here. And so one reason that I've started to use is another slightly different way of looking at things, which is to say the world is Sylvester. Now Sylvester, what does that mean? What does the word "Sylvester" mean in this case? It means that the matrices satisfy a certain type of equation called the Sylvester equation, and so the reason is really, many of these matrices satisfy a Sylvester equation, and that takes the form-- for sum A, B, and C.
OK. So X is your matrix of interest. You want to show X is of numerical low rank. And the task at hand is to find an A, B, and C so that X satisfies that equation. OK. For example, the two matrices I've had on the board satisfy a Sylvester equation-- a Sylvester matrix equation. There is an A, a B, and a C for which they do this.
For example, remember the Hilbert matrix, which we have there still, but I'll write it down again. Has these entries. So all we need to do is to try to figure out an A, a B, and then a C so that we can make it fit a Sylvester equation. There's many different ways of doing this. The one that I like is the following, where if I put 1/2 here and 3/2 here, all the way down to n minus 1/2, times this matrix-- so this is timesing the top of this matrix by 1/2 and then 3/2 and then 5/2. So we're basically timesing each entry of this matrix by j minus 1/2.
And then I do something on the right here, which I'm allowed to do because I've got the B freedom, and I choose this to be the same up to a minus sign. Then when you think about this, what is it doing? It's timing the jk entry-- this is-- by j minus 1/2. That's what this is doing. And what's this doing is timesing the jk entry by k minus 1/2. So this is, in total, timesing the jk entry by j plus k minus 1/2 minus 1/2, which is minus 1, so this is timesing the jk entry by j plus k minus 1. So it knocks out the denominator. And what we get from this equation is a bunch of ones. So in this case, A and B are diagonal, and C is the matrix of all ones. OK?
We can also do this for Vandermonde. So Vandermonde, you'll remember, looks like this. And then over here, we have this guy, the matrix that appears with polynomial interpolation. OK. So if I think about this, I could also come up with an A, B, and C, and for example, here's one that works. I can stick the x's on the diagonal.
So if you imagine what that matrix on the left is doing, it's timesing each column by the vector x. OK? So the first column of this matrix becomes x, the vector x. The second becomes the vector x squared, where squared is done entry-wise. And then the third entry is now x cubed, and when we get to the last, it's x to the n. OK? So that's like, multiply each column by the vector x.
So if I want to try to come up with a matrix-- so what's left is of low rank, is like of this form. What I can do is shift the columns. So I've noticed that this product here, this diagonal matrix, has made the first column x. So if I want to kill off that column, I can take the second column and permute it to the first column. I could take the third column and permute it to the second, the last column and permute it to the penultimate column here. And that will actually kill off a lot of what I've created in this matrix right here.
So let me write that down. This is a circumshift matrix. This does that permutation. I've put a minus 1 there. I could have put any number there. It doesn't make any difference. But this is the one that works out extremely nicely. Now this zeros out lots of things because of the way I've done the multiplication by x and the circumshift of the columns.
And so the first column is zero because this first column is x, this first column is x, so I've got x minus x. This column was x squared minus x squared, so I got zero, and I just keep going along until that last column. That last column is a problem because the last column of this guy is x to the n, whereas I don't have x to the n in V, so there are some numbers here. OK.
You'll notice that C in both cases happens to be a low-rank matrix. In these cases, it happens to be of rank one. And so people were wondering, maybe it's something to do with satisfying these kind of equations that makes these matrices that appear in practice numerically of low rank. And after a lot of work in this area, people have come up with a bound that demonstrates that these kind of equations are key to understanding numerical low rank.
So if X satisfies a Sylvester equation, like this, and A is normal, B is normal-- I don't really want to concentrate on those two conditions. It's a little bit academic. Then-- people have found a bound on the singular values of any matrix that satisfies this kind of expression, and they found this following bound. OK, so here, the rank of C is r. So that goes there. So in our cases, the two examples we have, r is 1, so we can forget about r.
This nasty guy here is called the Zolotarev number. E is a set that contains the eigenvalues of A, and F is a set that contains the eigenvalues of B. OK. Now it looks like we have gained absolutely nothing by this bound, because I've just told you singular values are bound by Zolotarev numbers. That doesn't mean anything to anyone. It means a little bit to me but not that much.
So the key to this bound-- the reason this is useful-- is that so many people have worked out what these Zolotarev numbers actually mean. OK? So these are two key people that worked out what this bound means. And we have gained a lot because people have been studying this number. This is, like, a number that people cared about from 1870 onwards to the present day, and people have studied this number extremely well. So we've gained something by turning it into a more abstract problem that people have thought about previously, and now we can go to the literature on Zolotarev numbers, whatever they are, and discover this whole literature of work on this Zolotarev number.
And the key part-- I'll just tell you the key-- is that the sets E and F are separated. So for example, in the Hilbert matrix, the eigenvalues of A can be read off the diagonal. What are they? They are between minus 1/2 and n minus 1/2. And the eigenvalues of B lie in the set minus 1/2 minus n plus 1/2. And the key reason why the Hilbert matrix is of low numerical rank is the fact that these two sets are separated, and that makes this Zolotarev number gets small extremely quickly with k.
Now you might wonder why there is a question mark on Penzl's name. There is an unofficial curse that's been going on for a while. Both these men died while working on the Zolotarev problem. They both died at the age of 31. One died by being hit by a train, Zolotarev. It's unclear whether he was suicidal or it was accidental.
Penzl died at the age of 31 in the Canadian mountains by an avalanche. I am currently not yet 31 but going to be 31 very soon, and I'm scared that I may join this list. OK. But for the Hilbert matrix, what you get from this analysis, based on these two peoples' work, is a bound on the numerical rank. And the rank that you get is, let's say, a world record bound. For the Hilbert matrix is 34, which is not quite 28, not yet, but it's far more descriptive of 28 than 719.
And so this technique of bounding singular values by using these Zolotarev numbers is starting to gain popularity because we can finally answer to ourselves why there are so many low-rank matrices that appear in computational math. And it's all based on two 31-year-olds that died. And so if you ever wonder when you're doing computational science when a low rank appears and the smoothness argument does not work for you, you might like to think about Zolotarev and the curse. OK, thank you very much.
[APPLAUSE]
GILBERT STRANG: Thank you [INAUDIBLE] Excellent.
ALEX TOWNSEND: How does it work now?
GILBERT STRANG: We're good. Yeah.
ALEX TOWNSEND: I'm happy to take questions if we have a minute, if you have any questions.
GILBERT STRANG: How near of 31 are you?
ALEX TOWNSEND: [INAUDIBLE] I get a spotlight. I'm 31 in December.
GILBERT STRANG: Wow. OK.
ALEX TOWNSEND: So they died at the age of 31, so you know, next year is the scary year for me. So I'm not driving anywhere. I'm not leaving my house until I become 32.
GILBERT STRANG: Well, thank you [INAUDIBLE]
ALEX TOWNSEND: OK, thanks.
[APPLAUSE] | 2021-10-16T17:29:03 | {
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https://math.stackexchange.com/questions/2407811/how-many-trailing-zeroes-in-52 | # How many trailing zeroes in $52!$? [duplicate]
Problem
How many trailing zeroes are there in $52!$ ?
My thoughts
I believe I correctly solved it, but I'm not happy with the scalability of my method.
I figure the number of trailing zeroes is equal to the number of times $52!$ is divisible by 10. I wrote out every integer from 1-52 that is divisible by 2 or 5. The idea being that the number of 2 AND 5-factors equals the number of 10-factors.
I quickly noted that the number of 2-factors is greater than the number of 5-factors, so I figured finding the number of 5-factors will do. There were 12.
I'm not very happy with this, because if they now ask me to do the same for $152!$, I'll have to tell them to shove it. I'm not doing this again.
Question
Is there a better way to do this? Perhaps a method that scales better?
## marked as duplicate by Jyrki LahtonenAug 27 '17 at 16:35
• Your method looks good and should be quick now you have the basics. Undertaking the same task for $152!$ should rapidly give you some insight into how to solve the problem quicker. Look out for $125$. – Joffan Aug 27 '17 at 16:39
You get your $12$ by counting $\left\lfloor\frac{52}{5}\right\rfloor+\left\lfloor\frac{52}{25}\right\rfloor$. This counts each multiple of $5$ in the product $52!$, then adds on a count for each multiple of $25$. This can generalize to $152$. You'll need to go up to $125$ with divisors.
In general, the number of trailing $0$s in $n!$ will be $$\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor$$ where for any $n$, only the first few terms in that infinite sum will be nonzero. | 2019-06-19T21:19:46 | {
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https://stats.stackexchange.com/questions/434835/which-transformation-needed-to-make-variance-independent-of-population-parameter | # Which transformation needed to make variance independent of population parameter?
Suppose $$s^2$$ is the sample variance of a sample$$(\text{of size }n)$$ from a normal population with mean $$\mu$$ and variance $$\sigma^2. \text{Here }s^2=\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}$$
As we know $$\frac{(n-1)s^2}{\sigma^2}\sim \chi^2_{(n-1)}$$ here . Let $$x\sim \chi^2_{(n-1)}$$ then $$E[x]=n-1$$ and $$Var[x]=2(n-1)$$ \begin{align*} E\left[\frac{(n-1)s^2}{\sigma^2}\right]=E[x]&=n-1\\ \implies \frac{(n-1)}{\sigma^2}E[s^2] &= n-1\\ \implies E[s^2]&=\sigma^2 \end{align*} Similarly, \begin{align*} Var\left[\frac{(n-1)s^2}{\sigma^2}\right]=Var[x]&=2(n-1)\\ \implies \frac{(n-1)^2}{\sigma^4}E[s^2] &=2(n-1)\\ \implies Var[s^2]&=\color{red}{\frac{2\sigma^4}{(n-1)} } \end{align*} Now we need a function of $$s^2$$ whose variance will be independent of $$\sigma^2.$$ Let $$f(s^2)$$ be the required transformation.
The required transformation is $$f(s^2)\stackrel{?}{=}\ln{s^2}$$
Question: How they get that transformation $$?$$
Similar things happen with Poisson variate and Binomial proportion with square root and $$\sin^{-1}$$ transformation. So I need a general approach to get that transformation which will make variance independent of population parameter.
• Search for Variance-stabilizing_transformation. – StubbornAtom Nov 6 '19 at 14:33
• "The aim behind the choice of a variance-stabilizing transformation is to find a simple function f to apply to values x in a data set to create new values y=f(x) such that the variability of the values y is not related to their mean value." But I need the variability of the values y is not related to their variance(population) value? @StubbornAtom Sir – emonhossain Nov 6 '19 at 14:45
• The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) – StubbornAtom Nov 6 '19 at 14:51
Suppose $$X_1,X_2,\ldots,X_n$$ are i.i.d $$N(\mu,\sigma^2)$$ and define the sample variance as $$S^2=\frac{1}{n}\sum\limits_{i=1}^n (X_i-\overline X)^2$$
Here we are concerned with the large-sample behaviour of $$S^2$$, so it does not matter if we take $$n$$ as the divisor above instead of $$n-1$$.
By CLT it can be argued that $$\sqrt n(S^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)\tag{1}$$
And by 'Delta-method', for a real-valued function $$g$$ such that $$g'(\sigma^2)$$ exists and $$g'(\sigma^2)\ne 0$$, it follows from $$(1)$$ that
$$\sqrt n(g(S^2)-g(\sigma^2))\stackrel{L}\longrightarrow N(0,2\sigma^4[g'(\sigma^2)]^2)\tag{2}$$
You have to solve for a $$g$$ such that asymptotic variance of $$g(S^2)$$ is free of $$\sigma^2$$.
So for some non-zero constant $$c$$, set $$\sqrt 2\sigma^2 g'(\sigma^2)=c$$
Therefore, $$\int g'(\sigma^2)\,d\sigma^2=\frac{c}{\sqrt 2}\int\frac{d\sigma^2}{\sigma^2}$$
Choosing $$c=\sqrt 2$$ and taking constant of integration to be zero, we have the required transformation $$g(\sigma^2)=\ln\sigma^2\quad,\text{ i.e. }\quad \color{blue}{g(S^2)=\ln S^2}$$
From $$(2)$$ you end up with $$\sqrt n(\ln(S^2)-\ln(\sigma^2))\stackrel{L}\longrightarrow N(0,2)$$
This is the application of variance stabilizing transformation on the sample variance from a normal population. The other transformations that you mention for the Poisson mean and Binomial proportion are derived similarly.
Result $$(1)$$ and many more can be found in A Course in Large Sample Theory by Thomas Ferguson. The relevant part is on page 46 of the first edition (1996):
The main result used in this section is Cramer's Theorem on the asymptotic normality of functions of sample moments, studied through a Taylor-series expansion to one term.
• what I know is CLT tells us $\color{red}{\sqrt{n}(\overline{x}-\mu)\text{ converge in distribution to a normal }N(0,\sigma^2)}$ And I know how to prove this but $\sqrt n(s^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)$ is looks new to me. So everything will be meaningful to me if you provide a source where I can get that proof $?$ I will accept your answer then. Thanks again @StubbornAtom – emonhossain Nov 6 '19 at 16:07
• @emonhossain Updated. – StubbornAtom Nov 6 '19 at 18:07
• Really you help me a lot. Thanks @StubbornAtom – emonhossain Nov 6 '19 at 18:33 | 2020-09-18T13:01:54 | {
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http://mathhelpforum.com/math-puzzles/153573-putnam-problem-day-print.html | # Putnam Problem of the Day
• Aug 12th 2010, 10:20 PM
simplependulum
Putnam Problem of the Day
Find the smallest natural number with $6$ as the last digit , such that if the final $6$ is moved to the front of the number it is multiplied by $4$ .
Spoiler:
153846
?
• Aug 13th 2010, 03:26 AM
Unbeatable0
Let the number be $a_1a_2...a_n6$.
Let $x=0.\overline{a_1a_2...a_n6}$.
Then $4x = 0.\overline{6a_1a_2...a_n} \Rightarrow 40x = 6.\overline{a_1a_2...a_n6} = 6+x$.
Therefore $x = \frac{6}{39} = \frac{2}{13} = 0.\overline{153846}$.
By backwards argument it's easy to see that all the cycles in $\frac{2}{13}$ are solutions.
Thus we have found all the solutions, the smallest of which is $153846$.
This technique can be applied to this kind of problems in many cases.
• Aug 13th 2010, 02:03 PM
Soroban
Hello, simplependulum!
Just write out the multiplcation,
. . and the digits will appear sequentially.
Quote:
Find the smallest natural number with $6$ as the last digit,
such that if the final $6$ is moved to the front of the number it is multiplied by $4$.
We have: . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&D&E&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&D&E \end{array}$
In column-6: . $6 \!\times\! 4 \,=\,24 \quad\Rightarrow\quad E = 4$
. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&D&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&C&D&4 \end{array}$
In column-5: . $4 \!\times\! 4 + 2 \:=\:18 \quad\Rightarrow\quad D = 8$
. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&C&8&4 \end{array}$
In column-4: . $8 \!\times\! 4 + 1 \:=\:33 \quad\Rightarrow\quad C = 3$
. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&3&8&4 \end{array}$
In column-3: . $3\!\times\!4 + 3 \:=\:15 \quad\Rightarrow\quad B = 5$
. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&5&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&5&3&8&4 \end{array}$
In column-2: . $5\!\times\!4 + 1 \:=\:21 \quad\Rightarrow\quad A = 1$
Therefore: . $\begin{array}{cccccc}
1&5&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&1&5&3&8&4 \end{array}$
• Aug 13th 2010, 05:19 PM
Wilmer
Soroban, the fact that it is a 6digit number is NOT a given.
• Aug 13th 2010, 06:16 PM
Soroban
Quote:
Originally Posted by Wilmer
Soroban, the fact that it is a 6-digit number is NOT a given. . Right!
I originally set up the problem like this:
. . $\begin{array}{ccccccccccc}
A & B & C & D & E & F & G & \hdots & M & N & 6 \\
\times &&&&&&&&&& 4 \\ \hline
6 & A & B & C & D & E & F & G & \hdots & M & N
\end{array}$
And found that "it came out even" after six digits
. . which gave me the smallest number: $153,\!846.$
The next is the 12-digit number: $153,\!846,\!153,\!846.$
• Aug 13th 2010, 07:10 PM
undefined
Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.
• Aug 13th 2010, 09:32 PM
Wilmer
Let n = number of digits
{20[10^(n-1) - 4] + 78} / 13
Gives integer solution for n=6k where k = any integer > 0 | 2017-10-21T09:16:05 | {
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http://mathhelpforum.com/geometry/227392-circle-questions-secants-chords-tangents-print.html | # Circle questions -- Secants, chords and tangents
• Apr 3rd 2014, 05:03 PM
StonerPenguin
Circle questions -- Secants, chords and tangents
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;
"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
http://i1137.photobucket.com/albums/...mexam10Q32.png
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)
Did I do that right?
• Apr 3rd 2014, 07:13 PM
JeffM
Re: Circle questions -- Secants, chords and tangents
Quote:
Originally Posted by StonerPenguin
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;
"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
http://i1137.photobucket.com/albums/...mexam10Q32.png
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)
Did I do that right?
Computation looks right to me. But simply memorizing theorems does not promote understanding.
Let's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.
$Let\ r = length\ of\ radius\ of\ given\ circle,$
$u + r = length\ from\ the\ given\ circle's\ center\ to\ a\ given\ point\ outside\ the\ circle,$
$x = length\ of\ the\ line\ that\ is\ tangent\ to\ the\ given\ circle\ and\ runs\ through\ the\ given\ point.$
By the Pythagorean Theorem, we have:
$(u + r)^2 = x^2 + r^2 \implies u^2 + 2ru + r^2 = x^2 + r^2 \implies x^2 = u^2 + 2ru = u(u + 2r) \implies x = \sqrt{u(u + 2r)}.$
• Apr 6th 2014, 06:17 PM
bjhopper
Re: Circle questions -- Secants, chords and tangents
ISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg
cos 71.35 =d/6724
d km to horizon =2150 km
• Apr 6th 2014, 09:56 PM
Soroban
Re: Circle questions -- Secants, chords and tangents
Hello, StonerPenguin!
Quote:
The radius of the earth is approximately 6371 km.
If the international space station (ISS) is orbiting 353 km above the earth,
find the distance from the ISS to the horizon (x).
Code:
o |\ | \ 352 | \ | \ x | \ * * * \ * | *\ * | o * 6371| * * | *6371 * | * * * o * * * * * * * * * * * *
Note the right triangle.
The equation is: . $x^2 + 6371^2 \:=\:6723^2$
• Apr 7th 2014, 05:16 PM
StonerPenguin
Re: Circle questions -- Secants, chords and tangents
Thank you JeffM, bjhopper and Soroban! It's nice to see different perspectives :D and the image drawn in code is really cool.
Here's another question I've had trouble with:
http://i1137.photobucket.com/albums/...mexam10Q15.png
"Explain how you know $\overline{AB}$ $\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)"
And here's some pertinent theorems;
Quote:
Theorem 10.4
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
Quote:
Theorem 10.5
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
Quote:
Theorem 10.6
In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Obviously from the diagram $\overline{AB}$ $\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/
• Apr 8th 2014, 04:24 PM
bjhopper
Re: Circle questions -- Secants, chords and tangents
t is a triangle a is an angle
t AED congruent to BEC isosceles t's same legs and equal altitudes
a DEC = 180- 2* 1 / 2 *AED
a AEB 180-2*1/2 * BEC a AED = aBEC
a AEB = a DEC
AB =DC equal arcs = equal chords | 2017-07-23T13:39:55 | {
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https://math.stackexchange.com/tags/measure-theory/hot?filter=month | # Tag Info
7
Since $fg\geq 1$, you have $\sqrt{f(x)\cdot g(x)} \geq 1$ for all $x$, and so $$1 \leq \int_X \sqrt{f}\cdot \sqrt{g} d\mu \leq \sqrt{\int_X f d\mu }\cdot \sqrt{\int_X g d\mu }$$ where the second inequality is Cauchy–Schwarz; squaring both sides gives the inequality.
6
It is sufficient to show that $m(E \cap [a, b]) = 0$ for every $a$ and $b$ with $a < b$. Let $t = \inf_{x \in [a, b]}{e^{-x^2/2}}$. Then $t > 0$ and we have $$0 = \int_{E}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}tdx = m(E \cap [a, b])t$$ and as $t > 0$, this implies $m(E \cap [a, b]) = 0$.
5
A measure space must specify three things: An underlying set; A $\sigma$-algebra of subsets which are measurable in this measure space; A measure (a function on this $\sigma$-algebra). When we say "a measurable subset of the real numbers" we typically mean "a subset of the real numbers measurable with respect to the Lebesgue measure", ...
5
The questions, despite looking as a representation problem in functional analysis, are much deeper as they bring out the history of the topic involved, notably $BV$-functions and the reasons why the customary definition adopted for the variation of a multivariate function is definition 2 above. And as thus the answers below needs to indulge a bit on this ...
5
$\bigcup_{i\ge 1}A_i$ is by definition simply $$\left\{x:\exists i\in\Bbb Z^+\,(x\in A_i)\right\}\,,$$ which clearly does not depend on the order in which the sets are enumerated. You can simply let $\mathscr{A}=\{A_i:i\in\Bbb Z^+\}$; then $$\bigcup_{i\ge 1}A_i=\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}\,,$$ with no reference to the ...
5
Hint: Consider the sets $[-r,r] \subseteq \mathbb{R}$. Can you show $r \mapsto \mu(E \cup [-r,r] \cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us? I hope this helps ^_^
5
This is false. For an easy counterexample, let $X_n\geq 0$ be any sequence of random variables with $\liminf_n X_n = 0$ which converges to $X = 1$ in probability. Then let $\mathcal{Q}$ be large enough so that the $X_n$ are $\mathcal{Q}$-measurable. The inequality in question then becomes $X \leq \liminf_n X_n,$ which is false almost surely. For instance, ...
5
It's true that the definition of absolute continuity is the same either we state it in the form of the finite and disjoint intervals or we state it in the form of countable and disjoint intervals. In detail, suppose we have the following definitions of absolute continuity: Definition 1 Let $f:\mathbb{R}\to \mathbb{R}$ be a function. We say that $f$ is ...
4
By assumption, as $Z \notin L^{\infty}$, we have $$\mathbb{P}(Z > n^2) > 0, \quad n \in \mathbb{N}.$$ Therefore, we may define $$X = \sum_{n=1}^{\infty}\frac{1}{n^2} \frac{\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)} \in L^1(\mathbb{P}).$$ However, it follows that \begin{align*}\mathbb{E}[ZX] = \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{...
4
Cannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI. Note that $$\mathbb{E}\left(|Y_n| 1\{|Y_n|\geq M\}\right) \geq M\mathbb{P}\left(|Y_n|\geq M\right) = M\mathbb{P}\left(|\bar{X}_n|\leq \frac{1}{M}\right) \to M,$$ as $n\to \infty$ by laws of large ...
4
The assertion “not every subset of $\Bbb R$ is measurable” is valid for the Lebesgue measure. Anyway, here Axler is talking about $\sigma$-algebras, not about measures. And, yes, $\mathcal P(\Bbb R)$ is a $\sigma$-algebra. And, on this $\sigma$-algebra you can define the measure $m(X)=\#X$; with respect to this measure, $\Bbb R$ is measurable.
4
If $p > p_0$ and $f \in L^{p_0} \cap L^\infty$ then $|f|^p = |f|^{p - p_0}|f|^{p_0} \le \|f\|_\infty^{p - p_0}|f|^{p_0}$ almost everywhere.
4
The claim is true: Let $S = \sum_{i=1}^\infty \chi_{E_i}$. Suppose for the sake of contradiction that $S < \infty$ a.e. Then, there exists an $M$ such that $\Pr(S\le M) \ge 1 - \frac\varepsilon 2$. If $A$ is the event that $S\le M$, then $\mu(A\cap E_i) \ge \frac\varepsilon 2$ for all $i$. But this would imply $\int_A S\,d\mu = \sum_{i=1}^\infty \mu(A\... 4 Well,$[0,1]=C\hspace{1mm}\cup C^{c} $(I discourage you to use$\mathbb{C}$for the cantor set) and you can write f as: $$f(x)=0\cdot\chi_{C}(x)+x^{2}\chi_{C^{c}}$$ where by$\chi$I mean the characteristic function of the set in subscript. The sum of measurable functions is measurable as well as the product, and polynomial functions are measurable. 4 The family of measurable sets form a$\sigma$-algebra but not all sets belonging in any$\sigma$-algebra are measurable. So to fully define a measure space we need a set$X$, a$\sigma$-algebra$\mathcal{A}$on$X$and a function$\mu: \mathcal{A} \to [0, \infty]$which satisfies some properties. A set$A \subseteq X$is called$\mu$-measurable (or simply ... 4 The symmetrization is called Khintchine's inequality, see here: https://en.wikipedia.org/wiki/Khintchine_inequality For the contraction principle, I don't know any reference, but I can prove it. First note that if$Z\geq 0$and$p > 0, then \begin{align} \mathbb{E}[Z^p] = \mathbb{E}\Big[\int_0^{Z} p t^{p-1} dt\Big] = \mathbb{E}\Big[\int_0^{\infty} p t^{p-... 4 A set\mathfrak B$satisfies the definition in your book if and only if$\mathfrak B$is a$\sigma$-algebra (in the wikipedia sense) on$\cup\mathfrak B$: for the noteworthy algebraic detail, notice that$A\setminus B=A\triangle (A\cap B)$. This is also equivalent to the property of$\mathfrak B$being a$\sigma$-algebra in the wikipedia sense over some set.... 4 Let me discuss the case$n=1$only. Let me denote by$f_x$the weak derivative of$f$with respect to$x$. Let$\phi,\psi$be two smooth test functions. Then $$-\iint f_x(x,y) \phi(x)\psi(y)dx\ dy=\iint f(x,y) \phi_x(x)\psi(y)dx\ dy,$$ using Fubini on both sides gives $$\int \left( \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx\right) \psi(y) \ dy = 0$$ ... 3 You prove that by giving an example of a set of subsets that is closed under intersections and differences but not closed under unions. E.g. on any set$X$with at least two points, the family$\{\{x\}: x \in X\} \cup \{\emptyset\}$is such. so being closed under union and differences (a ring) implies being closed under intersections, but being closed under ... 3 I have recently got interested in continued fractions, so I'll give it a try! CONTINUED FRACTIONS. It is a basic property of continued fractions that if$h_n/k_n$is the sequence of convergent to$x\in\mathbb R\setminus\mathbb Q$, then $$\frac 1{k_n(k_n+k_{n+1})}<\left|x-\frac{h_n}{k_n}\right|<\frac 1{k_nk_{n+1}}.$$ It follows that for all$h,k$... 3 The set$\mathcal{P}(X)$of all subsets of$X$is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures. The definition of a measure space is a pair$(X,S)$where$S$is some subset of$\mathcal{P}(X)$that happens to be a sigma algebra. It may or may not be all of$\mathcal{P}(X)$. ... 3 Suppose that such measure exists. Then taking countable cover by open sets of diameter$\leq 1$, one of those sets, say$U_1$, must have measure$1$. Cover$U_1$by countable amount of open sets of diameter$\leq 1/2$and take the one having measure$1$. This process gives us a decreasing sequence of open sets$U_k$with diameter$\leq 1/k$and$\mu(U_k) = 1$... 3 This works. Your argument shows that if$f$is bounded on a set$E$of Lebesgue measure zero, then$\int_E f = 0$. In fact the stronger claim that$\int_E f = 0$without any hypothesis on$f$holds! Indeed the integral is, by definition, the supremum over$g$with$g$simple and$g \leq f$of$\int_E g$. But any such$g$has integral zero. 3 Welcome to MSE! Hint: It's often useful to split up a random variable into the places where it is "good" and "bad". Then we can control those regions separately to get whatever inequality we're interested in. For this, we might try looking at the decomposition $$X = X \cdot \mathbf{1}_{\{X \leq \lambda a\}} + X \cdot \mathbf{1}_{\{X >... 3 According to the p-Adics section of the reference manual there is nothing I could find to do with zeta functions nor integration. At any rate, here's a suggestion on algorithmically getting an arbitrary level of accuracy on your integral. In order to be called a p-adic measure we require it to be a bounded p-adic distribution, and this is enough to imply ... 3 By definition, \int f is the supremum of \int h for 0 \leq h \leq f bounded measurable of finite support. So at the very end, you just need to take the sup over such h to get \int f \leq \liminf \int f_n. He uses such h is order to apply the bounded dominated convergence theorem, which he proves earlier. This allows him to prove Fatou in terms of ... 3 You can also let g_n=inf_{i \geq n} \{f_n\} and note that \int g_n \leq \int f_n for each n. Moreover g_n increases monotonically to lim inf f_n . Hence by monotone convergence theorem \int g_n = \int lim inf f_n. Since \int g_n \leq \int f_n we have \int lim inf f_n \leq lim inf \int f_n. The final step is to note that pointwise convergence ... 3 If you are familiar with dominated convergence notice that f_n(X)\xrightarrow{n\rightarrow\infty} X pointwise, |f_n(X)|\leq |X|, X is assumed to be integrable (E[|X|]=\int |X|\,dP<\infty). Then by dominated convergence$$ \int f_n(X)\,dP\xrightarrow{n\rightarrow\infty}\int X\,dP=E[X]$$similarly for$Y$. There is still the issue of the ... 3 Let$f$be the limit of$\chi_{E_n}$in$L^p$. All you have to do is show that$f(x) = 0$or$f(x) = 1$almost everywhere. Then we see that$f = \chi_E$almost everywhere, where$E = \{x \in X | f(x) = 1\}$To do this, it is helpful to define$g(x) = \min(|x - 1|, |x|)$. We will show that$g \circ f = 0$almost everywhere. We first show that$\int g(f(x))^p \...
3
Let $f \geq 0$ and let $M=\sup \{f(x): x\in X\}$. Then $\|M-f\|\leq M$ since $f(x) \in [0,M]$ for all $x$. If $\phi(f)$ is real we can finish the proof as follows: $M-\phi (f)=\phi(M-f) \leq \|\phi\| \|M-f||\leq M$. Thus $\phi (f) \geq 0$. To show that $\phi (f)$ is real consider following: \$|\phi (f) \pm in|=|\phi(f \pm in)|\leq \|f\pm in\|\leq \sqrt {M^{...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-04-17T09:35:53 | {
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http://math.stackexchange.com/questions/72589/calculating-probability-of-at-least-one-event-occurring?answertab=votes | Calculating probability of 'at least one event occurring'
If I know the probability of A and the probability of B how can I calculate the probability of "at least one of them" occurring?
I was thinking that this is P(A or B) = P(A) + P(B) - P(AB). Is this correct?
If it is, then how can I solve the following problem taken from 'DeGroot - Probability and Statistics':
If 50 percent of families in a certain city subscribe to the morning newspaper, 65 percent of the families subscribe to the afternoon newspaper, and 85 percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?
Here the question is P(morning)=.5, P(afternoon)=.65 P(morning OR afternoon)=.85
P(morning OR afternoon) = .5 + .65 - .3 = .85
P(morning AND afternoon) = P(morning) + P(afternoon) - P(morning OR afternoon)
So the answer to the question is .3
Is my reasoning correct?
EDIT.
If this reasoning is correct:
How can I calculate the following:
If the probability that student A will fail a certain statistics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1,what is the probability that exactly one of the two students will fail the examination?
So this questions highlights the difference between !at least one of them! !or exactly one of them!
I understand that at least one of them = P(A or B)
but how can I work out the probability of exactly one of them?
-
Somebody knows an algorith to calculate the OR for n probabilistic values? Thanks in advance. Regards, Daniel. – Daniel Mejia Nov 8 '12 at 23:53
You need to specify the correlation between the $n$ probabilistic values - at least whether they are independent or not. – dexter04 Nov 9 '12 at 0:25
You are correct.
To expand a little: if $A$ and $B$ are any two events then
$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$
or, written in more set-theoretical language,
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.
-
thanks for your answer! please have a look at my edited question, your help highlighted a further issue.. – Dbr Oct 14 '11 at 11:10
"Exactly one of A and B" means "Either A or B, but not both" which you can calculate as P(A or B) - P(A and B). – Chris Taylor Oct 14 '11 at 11:13
thank you, this clarifies everything. Just one last quick thing: Where can I learn the mathematical notation you use on this forum? For some questions I find it quite hard to understand them, is there a tutorial? – Dbr Oct 14 '11 at 11:15
Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tutorial" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. – Chris Taylor Oct 14 '11 at 11:21
So if I download LaTeX and paste your notation then it displays it in a more readable form? – Dbr Oct 14 '11 at 11:24
For your second question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together.
Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.
-
For the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\cap B)$$
-
probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )
- | 2014-08-01T11:57:46 | {
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https://math.stackexchange.com/questions/3583403/calculating-dice-probabilities-with-multiple-overlapping-success-criteria | # Calculating dice probabilities with multiple, overlapping success criteria
Alright, so I play a tabletop game where models get in fights and roll dice to see if they deal damage to the other models. I'm wondering how to calculate the probabilities of dealing 0, 1, and 2 damage in a 2v2 fight.
Here is the situation:
- We have Model A and Model B rolling to deal damage to Model C and Model D
- A needs to roll a 5 or 6 to damage C, and a 6 to damage D
- B needs to roll a 4, 5, or 6 to damage C, and a 5 or 6 to damage D
- C and D can only be damaged once each (i.e., they die when they get damaged)
So for starters, I know the probabilities for this scenario just by listing out the possibilities:
A | 111111 222222 333333 444444 555555 666666
B | 123456 123456 123456 123456 123456 123456
----+------------------------------------------
Dmg | 000111 000111 000111 000111 111122 111222
So the probability of dealing 0 damage is $$\frac{12}{36}$$, (exactly) 1 damage is $$\frac{19}{36}$$, and 2 damage is $$\frac{5}{36}$$. One tricky outcome here is when A rolls a 5 and B rolls a 4. This only results in 1 damage since both rolls are only enough to damage C and C can only be damaged once.
I am hopelessly stumped on how to model this so that I can then apply it to other scenarios such as 3v3s or when some models use more than 1 die.
It is pretty easy to calculate the probabilities of doing damage in a 1v1 fight. I've used binomial distribution statistics to calculate those. Conveniently it works for multiple dice per model as well. However, when it comes to combining these probabilities ... I am at a loss. They don't seem to follow the rules for multiplication and addition.
For example, I naïvely expected the probability of doing 0 damage to be the probability of A doing 0 damage and the probability of B doing no damage. That sounds right intuitively, but no matter how I try to do it ... it never makes sense:
$$P(A\;miss\;C) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$
$$P(A\;miss\;D) = \binom{1}{0} \bullet 1^0 \bullet 5^1 = 5$$
$$P(B\;miss\;C) = \binom{1}{0} \bullet 3^0 \bullet 3^1 = 3$$
$$P(B\;miss\;D) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$
Therefore,
$$P(A\;miss\;C) \bigcap P(A\;miss\;D) \bigcap P(B\;miss\;C) \bigcap P(B\;miss\;D) = 4 \bullet 5 \bullet 3 \bullet 4 = 240$$
Wait what?!?! I expect the answer to be 12. Clearly there is something else going on here that I do not understand. :/
• You can't say $\Pr(X\cap Y)=\Pr(X)\Pr(Y)$ unless $X$ and $Y$ are independent events, and these are clearly not. When A doesn't damage C it's likely that he doesn't damage D, either, because he rolled a low number. Mar 16 '20 at 20:56
• Also, you're multiplying counts instead of probabilities. The product of the probabilities (even if you could just multiply them, i.e. if they were independent, which they aren't) would be $\frac46\cdot\frac56\cdot\frac36\cdot\frac46=\frac{240}{6^4}=\frac5{27}\approx0.185$, which is at least a lot closer to the $\frac{12}{6^2}\approx0.333$ that you were apparently expecting than $\frac{240}{6^2}\approx6.67$ would have been. Mar 16 '20 at 21:44
First I'd like to rephrase your question:
A and B each roll a die. The results of the die determine how much damage is dealt. The maximum{damage dealt by A, damage dealt by B} is the number we are interested in.
A Deals 1 damage by rolling a 5 or 2 damage by rolling a 6. B deals 1 damage by rolling a 4 or 2 damage by rolling a 5 or 6. Now take the maximum of the damage dealt by A or B.
I think it may be easier to frame it this way:
A misses and B Misses simultaneously:
P(0 Damage) = P(A Rolls 4 or less) * P(B rolls 3 or less) = 2/3*1/2= 1/3 = 3/9
A Hits both or B hits Both (don't double count):
P(2 Damage) = P(A Rolls 6) + P(B Rolls 5 or 6) - P(A Rolls 6 AND B rolls 5 or 6) = 1/6 + 1/3 - 1/6*1/3 = 4/9
P(1 Damage) = 1-P(0)-P(2) = 2/9
Alternatively,
A hits 1 (rolls 5) or B hits 1 (rolls 4)
P(1 Damage) = P(B Rolls 4)*P(A Rolls 5 or less) + P(B Rolls 3 or less)*P(A Rolls 5) = 1/6*5/6+ 3/6*1/6= 2/9
How I verified my answer: The table below shows how much damage occurs when (A rolls, B rolls) happens. Count the number of 0s and divide by 36 to determine how often 0 damage occurs. Repeat for 1s and 2s.
## | 6 | 2 | 2 | 2 | 2 | 2 | 2 |
Summary:
P( 0 damage ) = 3/9
P( 1 damage ) = 2/9
P( 2 damage ) = 4/9 | 2022-01-27T22:03:23 | {
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https://math.stackexchange.com/questions/1912681/is-there-a-quicker-proof-to-show-that-210k-equiv-7-pmod9-for-all-posit | # Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$?
I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,
$$2^{10^k} \equiv 7 \pmod{9}$$
My proof is as follows:
Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$
Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that \begin{align*} 10^k + 2 &\equiv 0 \pmod{6} \\ 10^k &\equiv 4 \pmod{6} \,\, \text{ for all integers } k \geq 1 \end{align*}
Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$
Proof. If $a \equiv c \pmod{n}$ and $b \equiv d \pmod{n}$, then $a\,b \equiv c\,d \pmod{n}$. And, by extension, $a\,b^k \equiv c\,d^k \pmod{n}$ (for integers $k \geq 0$). Therefore, with \begin{align*} 2^4 \equiv 16 \equiv 7 \pmod{9} \end{align*} and \begin{align*} 2^6 \equiv 64 \equiv 1 \pmod{9} \end{align*} we can show that, \begin{align*} 2^{4 + 6k} \equiv 7 \cdot 1^k \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 0 \end{align*}
Theorem. $\, 2^{10^k} \equiv 7 \pmod{9}$ for all integers $k \geq 1$
Lemma 1 implies that for all integers $k \geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore, \begin{align*} 2^{10^k} &= 2^{4 + 6n} \\ 2^{10^k} \!\!\!\! \mod a &= 2^{4 + 6n} \!\!\!\! \mod a \end{align*} for any positive integer $a$. Using this result along with Lemma 2, \begin{align*} 2^{10^k} \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 1 \end{align*} $$\tag*{\blacksquare}$$
I think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.
Either way, is there a quicker, more elegant proof which does not:
1. Use Lemma 2, or prove some generalization of the theorem first.
2. Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.
I am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.
• euler's theorem is one of the basic tools when working with exponents – HereToRelax Sep 3 '16 at 1:13
• Do you mean is there a shorter proof ? – Rene Schipperus Sep 3 '16 at 1:14
• Well...$2^{10}\equiv 7\pmod 9$ and $7^{10}\equiv 7\pmod 9$. Then go by induction. – lulu Sep 3 '16 at 1:14
• "I feel using Euler's theorem is using a needlessly complicated theorem to prove something as simple as this" well, that is just silly! The entire reason to have complicated theorems is because they are versitile and so you won't have to prove simple propositions like this over and over again. "Proof: it follows directly from Euler's theorem". That is a six word proof and it is enough! It's ... silly... to give a six paragraph 2 lemma prove and then complain a six word proof is too complicated. It's not like we have prove Euler th. Every time we use it. – fleablood Sep 3 '16 at 4:46
• @fleablood Moise once referred to it as something like using a tractor where a shovel would do. It's very natural to wonder if there is a more elementary solution. There is nothing inherently wrong with finding more than one way to solve a problem. How many proofs of quadratic reciprocity exists? – steven gregory Jan 2 '18 at 15:41
## 2 Answers
Induction? Base case when $k=1$ is clear, for inductive step we have: $2^{10^k}=(2^{10^{k-1}})^{10}\equiv 7^{10}\equiv 7\bmod 9$
• How did you get $7^{10} \equiv 7 \pmod{9}$ without evaluating in a computer? – XYZT Sep 3 '16 at 1:40
• we can go $7^{10}\equiv (-2)^{10}\equiv 2^{10}$. And you already know $2^{10}\equiv 9$. So we don't need no extra stuff – HereToRelax Sep 3 '16 at 1:43
• @CarryonSmiling You mean $2^{10}\equiv 7$. It simply follows from $1+0+2+4\equiv 7\pmod{9}$. – user236182 Sep 3 '16 at 5:09
Don't fear Euler's theorem. You have it. Use it. It's not like it costs a lot of gas money or you have to pay tolls.
$2^6 \equiv 1 \mod 9$ so $2^{10^k} \equiv 2^{(10^k \mod 6 = 4^k \mod 6)} \mod 9$.
One little observation. If $4^n \equiv 4 \mod 6$ then $4^{n+1} \equiv 16 \equiv 4 \mod 6$ so inductively $4^n \equiv 4 \mod 6$ for all $n$.
So $2^{10^k} \equiv 2^{4^k} \equiv 2^4 =16 \equiv 7 \mod 9$. | 2021-04-21T14:11:39 | {
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http://math.stackexchange.com/questions/160723/radius-either-integer-or-sqrt2-cdotinteger | # Radius either integer or $\sqrt{2}\cdot$integer
Given a circle about origin with exactly $100$ integral points(points with both coordinates as integers),prove that its radius is either an integer or $\sqrt{2}$ times an integer.
What my solution is: Since circle is about origin, hence, integral points would be symmetric about the $x$-axis and $y$-axis as well as line $x=y$ and line $x+y=0$ ,i.e. if $(x,y)$ is an integral point, so are $(x,-y),(-x,-y),(-x,y),(y,x),(y,-x),(-y,x)$ and $(-y,-x)$.Therefore, we need to consider only a single octant. Since there are a total of $100$ integral points, two cases are possible:
1) radius of the circle is integer.
2) radius of the circle is not an integer.
case 1: If radius is an integer, then $4$ points on the $x$-axis and $y$-axis of the circle would be integral points and hence each octant must have $12$ points(as $100-4=96$ is a multiple of $8$). therefore, this case is consistent.
case2: If radius is not an integer, then $100$ integral points can't be divided into $8$ parts(octants),and points on $x$-axis and $y$-axis of circle are not integral points, therefore points on line $x=y$ and $x+y=0$ must be integral points so as to divide $100-4=96$ points in $8$ parts. But since point on line $x=y$ and circle is of the form $(r\cdot\cos(45^\circ),r\cdot\sin(45^\circ))$,therefore, $r/\sqrt{2}$ is an integer and hence $r=\sqrt{2}\cdot$integer. other points of circle on these lines are consistent with it.
So, i proved that either radius is an integer and if not then it has to be $\sqrt{2}\cdot$integer.
Is there any flaw in my arguments?? I couldn't find the proof to check whether mine is correct. Thanks in advance!!
-
Looks good to me. – Gerry Myerson Jun 20 '12 at 12:52
All the ingredients are here, but the flow of the argument is not optimal. A smooth proof of the claim would begin with "Assume the set $S:=\gamma\cap{\mathbb Z}^2$ contains $100$ elements. Then $\ldots$", or it should begin with "Assume the radius of $\gamma$ is neither an integer nor $\sqrt{2}$ times an integer. Then $\ldots$".
The essential point (which does not come out clearly in your argument) is the following: The group of symmetries of $S$ is the dihedral group $D_4$, which is of order $8$. Since $100$ is not divisible by $8$ this action has nontrivial fixed points, i.e. points on the lines $x=0$, $y=0$, $y=\pm x$.
For the case of the radius being an integer, you don't have to make any argument at all. What you are trying to prove is that a circle having exactly 100 integral points implies the radius is either integral or $\sqrt 2$integral. This would still be true if none of these circles were integral. It would even be true if there were no circles with exactly 100 integral points.
That said, your argument works fine. It extends to any number of integral points equal to $4 \pmod 8$ | 2015-04-25T01:04:28 | {
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http://mathhelpforum.com/advanced-statistics/33259-density-function-solution-check.html | # Thread: Density function solution check
1. ## Density function solution check
Suppose Y has density function:
f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere
a)Find the value of k that makes f(y) a probability density function
b)Find P(0.4<=Y<1)
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)
sol.
for part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k
for part b also used the integral between 0.4 and 1 and got 0.648
c) d) is where I have the problem. I suppose both must have the same answer.
Since this is conditional probability can we state that:
P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)
so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896
then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.
2. Originally Posted by somestudent2
Suppose Y has density function:
f(y) = ky(1-y) when 0<=y<=1
and
f(y)= 0 elsewhere
[snip]
c)Find P(Y<=0.4|Y<=0.8)
d)Find P(Y<0.4|Y<=0.8)
[snip]
c) d) is where I have the problem. I suppose both must have the same answer.
Since this is conditional probability can we state that:
P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)
so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896
then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.
The answer to (c) and (d) is
$\frac{\Pr(Y \leq 0.4 \text \, \text{and}\, Y \leq 0.8)}{\Pr(Y \leq 0.8)} = \frac{\Pr(Y \leq 0.4)}{\Pr(Y \leq 0.8)}$ $= \frac{\int_{0}^{0.4} ky(1-y) \, dy}{\int_{0}^{0.8} ky(1-y) \, dy} = \frac{\int_{0}^{0.4} y(1-y) \, dy}{\int_{0}^{0.8} y(1-y) \, dy} = .....$ | 2017-05-30T10:10:13 | {
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