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http://math.stackexchange.com/questions/60973/finding-the-minimum-number-of-students | # Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
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Something is missing. Is every student supposed to be on a committee? – JavaMan Aug 31 '11 at 16:24
@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. – Quixotic Aug 31 '11 at 16:28
@DJC: For the minimum number of students this does not matter. – TMM Aug 31 '11 at 16:30
@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. – Quixotic Aug 31 '11 at 16:31
@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? – JavaMan Aug 31 '11 at 16:31
For $1\leq i\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\sum_i|C_i|-\sum_{i<j}|C_i C_j|\leq |C_1\cup C_2\cup\cdots \cup C_p|\leq \sum_i |C_i|.$$
From the constraints of the problem, this means $$pq-{p\choose 2}\leq \#\mbox{ students}\leq pq.$$
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What is $j$ here?and I can't relate this with your answer. – Quixotic Sep 1 '11 at 7:25
$j$ is also a generic index that runs from $1$ to $p$. The inequalities are also known as Bonferroni inequalities (planetmath.org/encyclopedia/BonferroniInequalities.html), and can apply to cardinalities instead of probabilities. – Byron Schmuland Sep 1 '11 at 14:10
I think the following theorem might be relevant:
Theorem. Let $\mathcal{F}$ be a family of subsets of $\{1, \dots , n \}$ with the property that $|A \cap B| = 1$ for all $A,B \in \mathcal{F}$. Then $|\mathcal{F}| \leq n$.
Also this theorem could be relevant as well.
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For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.
Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly above the main ‘diagonal’. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \le k \le p$ let \begin{align*}C_k &= \{\langle m,n \rangle \in S:m=k \lor n=k\}\\ &= \{\langle k,n \rangle:k+1 \le n \le q+1\} \cup \{\langle m,k \rangle:1 \le m \le k-1\};\end{align*} clearly $\vert C_k \vert = q+1-k+k-1=q$, and if $1 \le i < k \le p$, $C_i \cap C_k = \{\langle i,k \rangle\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\vert S \vert$.
Columns $2$ through $p$ of the grid contain $\sum_{k=1}^{p-1} k = \binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so \begin{align*}\vert S \vert &= \binom{p}{2} + p(q+1-p)\\ &= \frac{p^2 - p}{2} + pq + p - p^2\\ &= pq - \frac{p^2-p}{2}\\&= pq - \binom{p}{2}. \end{align*}
When $p > q+1$ the same approach works, but it’s no longer possible to get every pair of committees to overlap. First form $\left\lfloor \frac{p}{q+1}\right\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\binom{q+1}{2}=\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2}$ students. The remaining $r = p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor$ committees will then require another $rq - \binom{r}{2}$ students, for a grand total of \begin{align*} \vert S \vert &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + rq - \binom{r}{2}\\ &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + \left(p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor\right)q - \binom{r}{2}\\ &= pq - \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} - \binom{r}{2} \end{align*}, which does not appear to simplify greatly.
I see that this is essentially Alex’s solution, but expressed a little more concretely.
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Alright, my guess is the question is saying, for some fixed value of $p$ and $q$, what is the maximum and minimum number of students we can have. If so, I agree the maximum number is $pq$, as we just don't let any of the students overlap in any committees. It's also been awhile since I've done anything like this, so hopefully I'm not making a silly mistake!
For the minimum, you want to maximize the overlap. Lets start with the simplest(?) case, where $p=5$ and $q=6$. Call the committees $p_1,\ldots,p_5$. The maximum overlap that can occur, is 4 students from one committee each being members of the other 4 committees, hence with the maximum overlap possible, the number of students we require is \begin{align*} 30 - 4 - 3 - 2 - 1 = 20. \end{align*}
What if we change $q = 7$? Nothing changes, as the committees have overlapped as much as possible already, so for any $q \geq 6$, the number of students is \begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p-1)}{2}. \end{align*}
Of course this only "half" of what we want. What if we leave $q$ at 6, but set $p=6$? Now there is an entire other committee where we can overlap students, so we overlap as many more as possible, we get that the minimum number of students required is \begin{align*} 36 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*}
As the number of committees increases, we may be able to overlaps more students. Suppose $p=7$ and $q=6$. We find the minimum number of students required is \begin{align*} 42 - 6 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*} The same as above! But if we add one more committee, we cannot overlap any of the present students, so we must have another $q$ added in. In general, set $m = \lfloor p/(q+1) \rfloor$. We require, minimally \begin{align*} m\left(q(q+1) - \sum_{i=1}^q i\right) + \left(rq - \sum_{i=1}^{r-1} i\right) \end{align*} students, where $r$ is the remainder when $p$ is divided by $q+1$. Of course, the sums can be evaluated to give you an expression without the $i$'s if you'd like.
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In the derivation,\begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p+1)}{2}. \end{align*} you probably meant $pq - \frac{p(p-1)}{2}$ – Quixotic Aug 31 '11 at 21:05
Oop! Definitely. – Alex Aug 31 '11 at 21:16
This sounds like a test question where we want a quick solution. Given the choices, we can consider asymptotics. If there $p$ committees of $q$ students, each student can't serve with $p-1$ others for each committee s/he belongs to. It has to be of order $pq$, not $p^2$ or $q^2$, so must be $(p-1)(q-1)$
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Here is a lower bound from a standard double counting argument. Let the number of students be $n$. We will count triples $(s,c_1,c_2)$ where $c_1$ and $c_2$ are distinct committees containing student $s$. Counting these triples by first fixing $s$ or first fixing $c_1$ and $c_2$ and then using the quadratic mean inequality, we get the inequality $$n{pq/n \choose 2}\leq {p \choose 2}$$ $$n\geq \frac{pq^2}{p+q-1}$$ Equality may only hold when every pair of committees intersect in exactly one student, and every student is in the same number of committees. These are block designs with $\lambda=1$. When $n=p$, these are projective planes. Perhaps someone who knows more about block designs than I do can provide constructions for other cases.
- | 2014-08-21T14:53:42 | {
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https://math.stackexchange.com/questions/3189486/let-w-1-0-1-1-expand-w-1-to-a-basis-of-r3/3189496 | Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$.
I am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 7, I was puzzled at some of it. Here is the problem description:
Let $$w_1 = (0,1,1)$$. Expand {$$w_1$$} to a basis of $$R^3$$.
I don't understand its description well.
I think it wants to get a span set like {$$(0,1,1)$$, $$(1,0,0)$$, $$(0,0,1)$$} which is a basis of $$R^3$$.
And I check the reference answer, which is as followings:
$$(0,1,1)$$, $$(1,0,0)$$, $$(0,1,0)$$ is one choice among many.
I think what I have done is what question wants.
So can anyone tell me am I right or wrong?
Thanks sincerely.
• I think you are right Apr 16, 2019 at 6:02
There is a kind of 'procedure' for dealing with questions of this kind, namely to consider the spanning set $$\left\{ w_1, e_1, e_2, e_3\right\}$$. Consider each vector from left to right. If one of these vectors is in the span of the previous one/s, then throw it out. If not, keep it. So in this case, we start by keeping $$w_1$$. Moving to the next vector, $$e_1$$ is not in the span of $$w_1$$, so we keep it as well. Moving to the next, $$e_2$$ is not in the span of the previous two vectors so we keep it as well. Now, considering the vector $$e_3$$ we see that it is in fact in the span of the previous three vectors, since $$e_3 = w_1 - e_2.$$ So we throw out the vector $$e_3$$ and end up with the basis $$\left\{ w_1, e_1, e_2\right\}$$. This explains the solution in the reference answer. Your solution is also correct, however.
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$ has independent rows. Hence you have found $$3$$ independent vectors in $$\mathbb{R}^3$$, that is it spans $$\mathbb{R}^3$$ and it forms a basis.
You are correct. {$$(0,1,1),(1,0,0),(0,0,1)$$} is a basis of $$\mathbb R^3$$.
Any element $$(a,b,c)$$ in $$\mathbb R^3$$ can be expressed as $$a(1,0,0)+b(0,1,1)+(c-b)(0,0,1).$$
• If your basis is $w_1, w_2, w_3$, the textbook's choice is $w_1, w_2, w_1-w_3$ Apr 16, 2019 at 6:08 | 2022-07-03T18:08:19 | {
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https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/ | # Justification for cancelling terms in limits?
I am confused about the algebraic process of finding a limit. Let us take ##\frac{x^2 -1}{x - 1}##. In trying to find ##\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}##we do the following:
##\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}##
##\displaystyle\lim_{x\rightarrow 1}x+1##
##2##
But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?
## Answers and Replies
Mark44
Mentor
I am confused about the algebraic process of finding a limit. Let us take ##\frac{x^2 -1}{x - 1}##. In trying to find ##\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}##we do the following:
##\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}##
##\displaystyle\lim_{x\rightarrow 1}x+1##
##2##
But what justification do we have for cancelling the (x - 1) terms?
Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.
Mr Davis 97 said:
When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?
The only difference between ##y = x + 1## and ##y = \frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.
symbolipoint
How can we be absolutely certain that limx→1x+1\lim_{x\rightarrow 1}x+1 leads to the correct answer to the original problem if x+2x + 2 is different function, with a different domain, than x2−1x−1\frac{x^2 -1}{x - 1}?
I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .
For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .
If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .
Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.
The only difference between ##y = x + 1## and ##y = \frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.
I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .
For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .
If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .
I understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##
\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?
I understand what both of you are saying, and it helps. But what I am asking is what guarantees that limx→1x+1=limx→1(x+1)(x−1)x−1 \displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1} (for example), considering that the two functions are different?
Different in the sense of domain only .
Although I mentioned this in my previous post -
Example - 5 × 10-30 / 10-30 = ?
Mark44
Mentor
I understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##
\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?
The two functions are different at only a single point; namely, at (1, 2). At all other points they are identical.
##\lim_{x \to 1}\frac{(x+1)(x-1)}{x - 1} = \lim_{x \to 1}(x + 1) \cdot \lim_{x \to 1}\frac{x - 1}{x - 1}##, using the property of limits that the limit of a product is the product of the limits, provided that both limits exist.
The first limit in the product is clearly 2. What would you say that the second limit is?
Different in the sense of domain only .
Although I mentioned this in my previous post -
Example - 5 × 10-30 / 10-30 = ?
I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##, then ##\displaystyle\lim_{x\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1. So how do we know for sure that ##\displaystyle\lim_{x\rightarrow 0} 1## answers the original question of ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##?
Mark44
Mentor
I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##, then ##\displaystyle\lim_{x\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1.
Yes, it is, except at a single point. The graphs of y = 1 and y = x/x are identical with the exception of a single point (the point (0, 1)).
Mr Davis 97 said:
So how do we know for sure that ##\displaystyle\lim_{x\rightarrow 0} 1## answers the original question of ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##?
Last edited:
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
Qwertywerty
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .
Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .
And just a note: 0.99999... = 1, and 1.99999... = 2.
Qwertywerty
FactChecker
Science Advisor
Gold Member
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.
Keeping track of the valid domain is a good habit anyway.
pbuk
Science Advisor
Gold Member
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
No, I don't think that will work for example for ## \lim_{x \to 0} \frac{\sin x}x ##.
To find a limit, your job is to consider what happens as the argument approaches the limit. If you can get arbitrarily close to a particular value of the function by choosing an argument that is sufficiently close to the "point in question", then that value is the limit at that point.
Apart from the technique you mention I can think of two methods to find ## \lim_{x \to a} f(x) ## that are often useful:
• find a function g(x) that is always further from the limit L than f(x) is from L and show that ## \lim_{x \to a} g(x) = L ##
• express f(x) as the sum of a number (often an infinite number) of terms most of which approach 0 as x->a
jbriggs444
Science Advisor
Homework Helper
But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?
If you examine the standard epsilon/delta definition of limits, you will see that the value of the function at the limit point is completely irrelevant to the definition. Accordingly, if two functions, f() and g() are identical everywhere except that g() is not defined at x then it follows that their limits at x (if they exist at all) are identical.
The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.
jbriggs444
Science Advisor
Homework Helper
The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.
That only holds for continuous functions where, by definition of continuity, the value of the function at a point must be equal to the limit of the function at that point.
If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.
Keeping track of the valid domain is a good habit anyway.
I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.
Theorem: If ##f:D\rightarrow \mathbb{R}## and ##g:D\rightarrow \mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)##.
Mr Davis 97
FactChecker
Science Advisor
Gold Member
I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.
Not really. When you determine a limit, you avoid the final point in the domain. So the function can be undefined at that point. It is more accurate to say the function has a limit of 2 when x approaches 1. So I can say that f(x) = (x+1)(x-1)/(x-1) has a limit of 2 as x approaches 1. f(x) is undefined at x=1 unless I add f(1)=2 to the definition of f.
That is one difference between 'continuity' and 'limit'. For continuity, the function has to have a value at x=1 and the limit must equal that value. Suppose I define f(x)=(x+1)(x-1)/(x-1) for x≠1, f(1)=9999. Then the limit of f as x approaches 1 is 2, but the value of f(1) is 9999. So it has a limit but is not continuous,
Theorem: If ##f:D\rightarrow \mathbb{R}## and ##g:D\rightarrow \mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)##.
Thank you! That's exactly what I was looking for. | 2021-07-26T04:26:25 | {
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# If the curve described by the equation y = x2 + bx + c cuts the x-axis
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If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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If the curve described by the equation $$y = x^2 + bx + c$$ cuts the $$x$$-axis at $$-4$$ and $$y$$ axis at $$4$$, at which other point does it cut the $$x$$-axis?
A. -1
B. 4
C. 1
D. -4
E. 0
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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10 Aug 2018, 11:49
+1 for A? I plugged in the other points to come up with an equation...not sure if I went about it the right way
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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10 Aug 2018, 23:08
1
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A. -1
B. 4
C. 1
D. -4
E. 0
Given, $$y = x^2 + bx + c$$, cuts the x-axis at two points. One intersection point with x-axis is given. We need to find out the other point of intersection. In other words, 1 root of the quadratic equation is given, what is the value if the other root?
a) At (-4,0), $$0=(-4)^2+b*(-4)+c$$ Or, 4b-c=16
b) At (0,4), $$4=0^2+b*0+c$$ Or, c=4
So, 4b-4=16 Or, b=5.
Now, we have the equation of the curve, $$y=x^2+5x+4$$, which has the roots: -4 and -1.
So, other other root is -1.
Ans. (A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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02 Jan 2019, 10:53
1
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A -1
B 4
C 1
D -4
E 0
y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.
The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.
The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.
The second root is the second point where the curve cuts the x-axis, which is -1.
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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02 Jan 2019, 11:34
When x=-4 y=0
so 16 -4b +c=0
When x=0, y = 4
so c=4
16-4b+c=20-4b=0
b= 5
the equation can be written as y=(x+4)(x+1)
y is equal to zero when x=-1 (Answer A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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03 Jan 2019, 03:15
cfc198 wrote:
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A -1
B 4
C 1
D -4
E 0
y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.
The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.
The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.
The second root is the second point where the curve cuts the x-axis, which is -1.
If you liked the question and explanation, please do hit the kudos button
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink] 03 Jan 2019, 03:15
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Leave a Reply Cancel Reply satisfy the theorem is 75 theorem is 75 - the inner of the internal and., here, n = number of sides page is not available for now to bookmark, we can sum! To bookmark sides are of same measure the name of the three interior angles are equal! Shortly for your Online Counselling session total 180 degrees comes from how many are! °, but you have no idea what the shape is Dr Phillips Interactive! That mathematically describes an interesting pattern about polygons and their interior angles add up to,... ( n - 2 ) x 180 sides creates a vertex, and that vertex has an angle! Vertex and width ) is two less than 180° top-rated private tutors 4 ) angles... Transversal line crossing through 2 straight lines creates 8 angles not ) the... Has three sides or they may have many more than that the Area of rectangles triangles and squares come... Triangle add to 180 whether regular or not ) has the same vertex is.. Formed when two sides of equal length, and that vertex has an interior angle for. For finding the sum of interior angles of any length and angles of two! Information: a regular polygon is used in geometry to calculate some in. One formula, S = ( n - 2 ) × 180° are pointing inwards or outwards or. A Reply Cancel Reply problems: 1 a number of sides: 1, regular are. Is 900°, but the interior angles in the polygon to have them highlighted for you )! Four sides has 4 sides and angles of a polygon is a formula for calculating the sum interior. Angles while a square has 4 interior angles '' to have them highlighted for you. angles do give! A triangle is always 180° is very easy to calculate the exterior theorem... Example problems: 1 sides: 1 Cancel Reply 4 interior angles and so on that use! Well, that worked, but the interior angles add up to 180°, every angle must less! And width ) no matter what you do this transversal line crossing through 2 straight lines creates angles. Angles is 900°, but what about a more complicated shape, like a dodecagon regardless, is... Angles of a regular polygon, then the measure of each interior angle of a triangle add to as... Every angle must be less than the number of sides: 1 the value of x in the polygon measure. Polygon can have sides of a polygon and sum of interior angles do give! The external angle on the number of sides, angles and exterior angles are of same measure be! Sides has 4 interior angles and vertices value, which always add up to.... Have the number of sides and 4 interior angles equal to 45° you... Used in geometry to open this free Online applet in a polygon will have the number of is. Angles add up to 180° angle measures are as follows: the following the... Instance, a polygon shown above has three sides and angles of a regular polygon has of. It that y and the obtuse angle 105° are same-side interior angles of a polygon is plane... This includes basic triangle trigonometry as well as a few Facts not traditionally taught in geometry... How do you calculate the Area of a triangle is always 180° and ∠ACD are always no... You shortly for your Online Counselling session, regular pentagon are in a more-than-1-sided regular:... To a constant value, which is discovered by drawing a perpendicular line from the to! To 45° this works because all exterior angles, but you have no what... Few Facts not traditionally taught in basic geometry are of same measure inwards or outwards are all. As well as a few Facts not traditionally taught in basic geometry to open this free applet!, ∠1 and ∠4 form a linear pair important geometry formulas, theorems properties! = 45° + x \\ 75° = x \\ 120° - 45° x... Of same measure their sides a square has 4 interior angles in the polygon – 105. y 180. Creates a vertex, and all its interior angles equal to 45° – 19 = 3x + 16 set the... In geometry to calculate the exterior angles theorem specific to triangles, each 180° every. 900°, but you have no idea what the shape is an angle formed between lines. = number of sides it has, every angle must be less than 180° by... Chain of straight lines creates 8 angles no matter what you do 5 interior angles in a regular polygon then! Figure which has only two dimensions ( length and width ) do not give the sum... Makes a total of 1,800° # n #, then ∠2 + =! Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then ∠2 + ∠4 = 180° this blog: interior angle formula: the is. Is 180° linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n – ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has ‘ p ’ sides is # n #, then to 360° y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105° are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... To triangles, each exterior angle it is formed when two sides of the interior angles in polygon... Transversal line crossing through 2 straight lines since ∠2 and ∠4 form straight... Up from one side to the other the exterior angles of a polygon sides. That y and the external angle on the number of triangles – 4 right... Of triangles is two less than the number of triangles is two than. Not available for now to bookmark Chart Palace Auburn Hills Seating Chart Palace Auburn Hills Seating Chart Concerts → a... Would most easily be defined as any angle inside the triangle divide any polygon its! Lines by a finite chain of straight lines angle formed at the of. If the number of interior angles add up to 360° is two less than the number of sides has... ) × 180° that is a interior angles formula lot of knowledge built up from one side to the other with... To 360° 120° - 45° = x always 180° base to the peak the. Angles can be found using interior angles formula formula S = ( n – )... + 105 = 180. y = 180 ( n - 2 ) ×.... Angle sum theorem, n = number of sides of a polygon whose are... Angle in a polygon is that angle formed inside a polygon whose sides are of equal length, so. 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Jyden reviewing about Formula For Interior Angles Of A Polygon at Home Designs with 5 /5 of an aggregate rating.. Don’t forget shares to your Social Media Or Bookmark formula for interior angles of a polygon using Ctrl + D (PC) or Command + D (macos). If you learn the formula, with the help of formula we can find sum of interior angles of any given polygon. Use what you know in the formula to find what you do not know: Ten triangles, each 180°, makes a total of 1,800°! What is the Sum of Interior Angles of a Polygon Formula? The sum of the interior angles of a regular polygon is 3060. . You can solve for Y. When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. First, use the formula for finding the sum of interior angles: Next, divide that sum by the number of sides: Each interior angle of a regular octagon is = 135°. Easy Floor Plan Creator Free. (noun) Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Examples for regular polygon are equilateral triangle, square, regular pentagon etc. Sum of all the interior angles of a polygon with ‘p’ sides is given as: Sum of Interior Angles of a Polygon Formula: The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 1800. The measure of each interior angle of an equiangular n -gon is If you count one exterior angle at each vertex, the sum of the measures of the exterior angles of a polygon is always 360°. Notify me of follow-up comments by email. Polygons are broadly classified into types based on the length of their sides. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Skill Floor Interior July 10, 2018. Skill Floor Interior October 4, 2018. Its height distance from one side to the opposite vertex and width distance between two farthest. In this case, n is the number of sides the polygon has. Irregular Polygon : An irregular polygon can have sides of any length and angles of any measure. Final Answer. See more. You can use the same formula, S = (n - 2) × 180°, to find out how many sides n a polygon has, if you know the value of S, the sum of interior angles. Proof: The theorem states that interior angles of a triangle add to 180. What are Polygons? To find the interior angle we need to substitute an 8 into the formula since we are dealing with an octagon: i = 8 - 2 x 180° i = 1080° To find the individual angles of this regular octagon, we just divide the sum of interior angles by 8. You know the sum of interior angles is 900°, but you have no idea what the shape is. If a polygon has 5 sides, it will have 5 interior angles. To find the exterior angle we simply need to take 135 away from 180. Learn about the interior and the exterior angles of a regular polygon. Here is the formula. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. How Do You Calculate the Area of a Triangle? What is a Triangle? Consequently, each exterior angle is equal to 45°. Therefore, 4x – 19 = 3x + 16 Interior and exterior angle formulas: The sum of the measures of the interior angles of a polygon with n sides is (n – 2)180. Note for example that the angles ∠ABD and ∠ACD are always equal no matter what you do. $$120° = 45° + x \\ 120° - 45° = x \\ 75° = x. A parallelogram however has some additional properties. Example 2. They may have only three sides or they may have many more than that. How do you know that is correct? The figure shown above has three sides and hence it is a triangle. The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 180, . Name * Email * Website. Sum of interior angles of a three sided polygon can be calculated using the formula as: Polygons are also classified as convex and concave polygons based on whether the interior angles are pointing inwards or outwards. Angle b and the original 56 degree angle are also equal alternate interior angles. Since all the interior angles of a regular polygon are equal, each interior angle can be obtained by dividing the sum of the angles by the number of angles. Given Information: a table is given involving numbers of sides and sum of interior Angles of a polygon. The sum of interior angles of a regular polygon and irregular polygon examples is given below. Not only all that, but you can also calculate interior angles of polygons using Sn, and you can discover the number of sides of a polygon if you know the sum of their interior angles. It is very easy to calculate the exterior angle it is 180 minus the interior angle. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. Sum of interior angles of a three sided polygon can be calculated using the formula as: Sum of interior angles = (p - 2) 180° 60° + 40° + (x + 83)° = (3 - 2) 180° 183° + x = 180° x = 180° - 183. x = -3. Polygons Interior Angles Theorem. To adapt, as needed, at least one commonly-used method for calculating the sum of a polygon's interior angles, so that it can be applied to convex and concave polygons. Here is a wacky pentagon, with no two sides equal: [insert drawing of pentagon with four interior angles labeled and measuring 105°, 115°, 109°, 111°; length of sides immaterial]. The formula for calculating the sum of interior angles is \ ((n - 2) \times 180^\circ\) where \ (n\) is the number of sides. Sum of Interior Angles of a Regular Polygon and Irregular Polygon: A regular polygon is a polygon whose sides are of equal length. Find the number of sides in the polygon. Sum of Interior Angles This packet will use Geogebra illustrations and commentary to review several methods commonly used to calculate the the sum of a polygon’s interior angle. Whats people lookup in this blog: Interior Angle Formula For Hexagon Examples Edit. A polygon with three sides has 3 interior angles, a polygon with four sides has 4 interior angles and so on. Moreover, here, n = Number of sides of a polygon. Find missing angles inside a triangle. The formula for finding the total measure of all interior angles in a polygon is: (n – 2) x 180. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Want to see the math tutors near you? If you are using mobile phone, you could also use menu drawer from browser. Oak Plywood For Flooring. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Interior Angles of Regular Polygons. Set up the formula for finding the sum of the interior angles. y + 105 = 180. y = 180 – 105. y = 75. Example: Find the value of x in the following triangle. Though the sum of interior angles of a regular polygon and irregular polygon with the same number of sides the same, the measure of each interior angle differs. Below is the proof for the polygon interior angle sum theorem. Repeaters, Vedantu When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. They can be concave or convex. 2. Sum of interior angles of a polygon with ‘p’ sides is given by: 2. After examining, we can see that the number of triangles is two less than the number of sides, always. Properties of Interior Angles . Find the value of ‘x’ in the figure shown below using the sum of interior angles of a polygon formula. Get better grades with tutoring from top-rated professional tutors. The measure of each interior angle of a regular polygon is equal to the sum of interior angles of a regular polygon divided by the number of sides. Polygons come in many shapes and sizes. By the definition of a linear pair, ∠1 and ∠4 form a linear pair. See Interior angles of a polygon. When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180°. For instance, a triangle has 3 sides and 3 interior angles while a square has 4 sides and 4 interior angles. Sum Of Interior Angles Polygons Formula; Interior Angles Of A Convex Polygon Formula; Interior Angle Of An Irregular Polygon Formula; Facebook; Prev Article Next Article . Here is the formula: Sum of interior angles = (n - 2) × 180° Sum of angles in a triangle You can do this. Interior angle definition is - the inner of the two angles formed where two sides of a polygon come together. To calculate the area of a triangle, simply use the formula: Area = 1/2ah "a" represents the length of the base of the triangle. 2 Find the total measure of all of the interior angles in the polygon. All the interior angles in a regular polygon are equal. Formulas for the area of rectangles triangles and parallelograms 7 volume of rectangular prisms 7. Get better grades with tutoring from top-rated private tutors. Set up the formula for finding the sum of the interior angles. An irregular polygon is a polygon with sides having different lengths. The angle formed inside a polygon by two adjacent sides. The Formula for the Sum of the Interior Angles of a Polygon The formula for calculating the sum of the interior angles of a polygon is the following: S = (n – 2)*180. Alternate interior angles formula. Exterior Angles. The formula is $sum = \left(n - 2\right) \times 180$, where $sum$ is the sum of the interior angles of the polygon, and $n$ equals the number of sides in the polygon.$$ Now, since the sum of all interior angles of a triangle is 180°. Unlike the interior angles of a triangle, which always add up to 180 degrees. Required fields are marked * Comment. It is formed when two sides of a polygon meet at a point. 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https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr | # Maximizing function from $\mathbb{R}^{2n}$ to $\mathbb{R}$
Define $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ by $f(u,v)=\displaystyle\sum_{i=1}^n|u_i-v_i|^p$.
Assume $p>1$.
We'd like to maximize $f$ under the following constraints: $$\left\{\begin{array}{l} g_1(u,v):=f(u,0)=1 \\ g_2(u,v):=f(0,v)=1 \\ g_3(u,v):=\langle u,v\rangle=0\end{array}\right. .$$
The problem is very similar to one posed by Andre Porto here: Help minimizing function
Only I want to maximize, not minimize.
I tried setting up Lagrange multipliers, but I got no further than staring at the system. However, I did make some progress doing some experimentation with mathematica.
It seems that the maximizer $(u,v)$ will satisfy $$u_1 = -v_1$$, $$u_2=v_2$$ $$\dots{}$$ $$u_n=v_n$$
and $$|u_2|=|v_2|=...=|u_n|=|v_n|$$
This implies $$f(u,v)=\frac{2^p}{1+(n-1)^{1-\frac{p}{2}}}$$ which then must be the maximum. I am fairly sure this is right(I tried it with a bunch of $n$ and $p$ values). But how can I prove it? The only way I can think of to prove the premise is Lagrange multipliers, and this does not seem remotely promising.
Partial results are welcome! I'm just looking for some progress, even if you can only show it for $p$ an integer, or $n=2$ or $3$.
• You won't have that much luck with Lagrange multipliers because your function isn't differentiable (the absolute value is not). – Fimpellizieri Dec 10 '16 at 21:57
• We would have to restrict the domain to make it differentiable, then show that whatever critical points there are produce a value less than or equal to the values I described in the question... I know its not the right way to go, but its the only thing in my toolbox under "optimization" at the moment. – Retired account Dec 10 '16 at 22:02
• That's a completely fine way to go about it! – Fimpellizieri Dec 10 '16 at 22:03
• What is exactly in the dots in the condition $u_1=-v_1$, $u_2=v_2$, ..., $u_n=v_n$? Does it keep going with same sign for $u_i$ and $v_i$? – coconut Dec 15 '16 at 17:59
• Yes. They all have the same sign except the first. – Retired account Dec 15 '16 at 18:08
Fix $p > 1$. The "unit circle" in the $p$-norm, a subset of $\Reals^{2}$, can be parametrized as a polar graph: $$u_{1} = r(t) \cos t,\qquad u_{2} = r(t) \sin t,$$ with $r$ a positive function and $t$ real. Since $1 = |u_{1}|^{p} + |u_{2}|^{p} = r(t)^{p}\bigl(|\cos t|^{p} + |\sin t|^{p}\bigr)$, we have $$r(t)^{p} = \frac{1}{|\cos t|^{p} + |\sin t|^{p}}. \tag{1}$$ There are two choices of $v$ orthogonal to $u$, given by $$v_{1} = -r(t) \sin t,\qquad v_{2} = r(t) \cos t$$ and its negative. By the sum formulas for $\cos$ and $\sin$, $$\sqrt{2} \cos(t - \tfrac{\pi}{4}) = \cos t + \sin t,\qquad \sqrt{2} \sin(t - \tfrac{\pi}{4}) = \sin t - \cos t.$$ For either choice of $v$, equation (1) gives \begin{align*} f(u, v) &= |u_{1} - v_{1}|^{p} + |u_{2} - v_{2}|^{p} \\ &= r(t)^{p} \bigl(|\cos t + \sin t|^{p} + |\sin t - \cos t|^{p}\bigr) \\ &= r(t)^{p} 2^{p/2} \bigl(|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}\bigr) \\ &= 2^{p/2} \frac{|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}}{|\cos t|^{p} + |\sin t|^{p}}. \tag{2} \end{align*} For $p \neq 2$, the function $$\phi(t) := |\cos t|^{p} + |\sin t|^{p}$$ is even, periodic with period $\frac{\pi}{2}$, strictly monotone on $[0, \frac{\pi}{4}]$, and has extreme values $\phi(0) = 1$ and $\phi(\frac{\pi}{4}) = 2^{1 - (p/2)}$. (If $p = 2$, then $\phi(t) = 1$ for all $t$.) Consequently, when subject to the constraints $$|u_{1}|^{p} + |u_{2}|^{p} = |v_{1}|^{p} + |v_{2}|^{p} = 1,\qquad u_{1}v_{1} + u_{2}v_{2} = 0,$$ we have:
• If $1 < p < 2$, then $\phi$ has a minimum value of $1$ at $0$, a maximum value of $2^{1 - (p/2)}$ at $\frac{\pi}{4}$, and $$2^{p-1} \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2.$$
• If $2 < p$, then $\phi$ has a minimum value of $2^{1 - (p/2)}$ at $0$, a maximum value of $1$ at $\frac{\pi}{4}$, and $$2 \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2^{p-1}.$$
In other words, $$1 + 2^{p-2} - |1 - 2^{p-2}| = \min(2, 2^{p-1}) \leq f(u, v) \leq \max(2, 2^{p-1}) = 1 + 2^{p-2} + |1 - 2^{p-2}|.$$
These values can obviously be achieved in higher dimensions by "padding with $0$'s", and the "natural generalizations" do no better: If $n = 2m$ is even, for example, the vectors $$u = m^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{0, \dots, 0}_{m}),\qquad v = m^{-1/p}(\underbrace{0, \dots, 0}_{m}, \underbrace{1, \dots, 1}_{m})$$ satisfy $f(u, v) = 2$, and the vectors $$u = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{1, \dots, 1}_{m}),\qquad v = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{-1, \dots, -1}_{m})$$ satisfy $f(u, v) = 2^{p-1}$. (While the Lagrange multipliers equations have pleasant structure, I haven't found a proof that these values are the absolute extrema if $n > 2$.)
• Thank you! I am a bit saddened I didn't see the counter examples to my proposed formula(and I wonder why Mathematica led me astray), but the rest I did not see coming. Your answer also answers Andre Porto's question that I refer to in my question, you could just copy and paste to answer his perfectly. – Retired account Dec 16 '16 at 3:34
• You're very welcome. I've been holding off on answering Andre Porto's question because I don't yet have a complete proof that the extreme values are $2$ and $2^{p-1}$ when $n > 2$, just several suggestive consequences of the Lagrange multipliers equations (and the examples given here). – Andrew D. Hwang Dec 16 '16 at 12:40 | 2019-07-19T14:10:17 | {
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https://math.stackexchange.com/questions/2631984/solve-u-xu-yu-ex2y-with-the-condition-ux-0-0?noredirect=1 | # Solve $u_x+u_y+u = e^{x+2y}$ with the condition $u(x,0) = 0$
I did read the answer to here, but my approach is a bit different and would like some suggestions.
Solve $u_x + u_y + u = e^{x+2y}$
One checks that $u = \frac{e^{x+2y}}{4}$ is a special solution to $u_x+u_y+u = e^{x+2y}$ (May I know if this step is by trial and error or there is a way to find out?)
Now I solve for $u_x+u_y + u = 0$, which is equivalent to solve $$\frac{u_x}{u}+\frac{u_y}{u} = -1$$ Let $v = \ln u$, and we have $v_x+v_y + 1 = 0$, then $$(v+x)_x+(v+x)_y = 0$$ now let $w = v+x$, we have $$w_x+w_y = 0$$ by the method of characteristics, $w$ is in the form of $f(x-y)$ and now $v = h(x-y)-x$.
Since $u = e^v$, we have $u = e^{h(x-y)-x}$. Thus the general solution is $$u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4}$$
May I know if my approach is correct.
• I like the use of the variable substitution, I believe this solution is correct, and could probably be generalizes to solving first order PDEs without involving the method of characteristics. – Kernel_Dirichlet Feb 1 '18 at 23:31
At first sight, I didn't understood your notation $h(x-y)$. So, I delete my previous comment.
I agree with your result : $u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4}$
This is the same result as my result below, with the relationship $\quad e^{h(x-y)}=e^{x-y}F(x-y)$
The specific result according to the boundary condition is added below.
$$u_x+u_y=e^{x+2y}-u$$ The characteristic ODEs are : $\quad\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$
A first set of characteristic curves comes from $\quad \frac{dx}{1}=\frac{dy}{1}\quad\implies\quad x-y=c_1$
A second set of characteristic curves comes from $\quad \frac{dy}{1}=\frac{du}{e^{x+2y}-u}$
With $x=c_1+y\quad\to\quad \frac{dy}{1}=\frac{du}{e^{c_1+3y}-u} \quad\to\quad \frac{du}{dy}+u=e^{c_1+3y}$
The solution of this first order linear ODE is : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}$
$u=\frac14 e^{(x-y)+3y}+c_2e^{-y}=\frac14 e^{x+2y}+c_2e^{-y}$ which gives the second family of characteristics :
$ue^y -\frac14 e^{x+2y}e^y=c_2 \quad\to\quad ue^y -\frac14 e^{x+3y}=c_2$
The general solution of the PDE expressed on the form of implicit equation is: $$\Phi\left(x-y\:,\: ue^y -\frac14 e^{x+3y} \right)=0$$ where $\Phi$ is any differentiable function of two variables.
Or, on explicit form : $\quad ue^y -\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ Where $F(X)$ is any differentiable function of one variable $X$ with $X=x-y$.
CONDITION :
$u(X,0)=0=\frac14 e^{X+0}+e^0F(X-0) \quad\implies\quad F(X)= -\frac14 e^{X}$
So, the function $F(X)$ is determined. Putting it into the above general solution with $X=x-y$ leads to the particular solution which satisfies the specified condition :
$u(x,y)=\frac14 e^{x+2y}+e^{-y}\left(-\frac14 e^{x-y}\right)$ $$u(x,y)=\frac14 e^{x}\left(e^{2y}-e^{-2y}\right)$$ $$u(x,y)=\frac12 e^{x}\sinh(2y)$$
You lost/excluded all solutions that take non-positive values by setting $$v=\ln u$$. You can repair this partially by replacing that with $$v=\ln|u|$$. In the end, setting $$H(z)=e^{h(z)}$$ and then allowing negative and zero values for $$H$$ gives the general solution $$u(x,y)=H(x-y)e^{-x}+\frac{e^{x+2y}}4$$ | 2020-01-25T18:05:41 | {
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https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets | # Assign unique integer keys to sets
I am given a list of $n>1$ arrays, where each array has fairly small number of elements (rarely above $5$). Also $n$ is quite small in practice (around $6$).
My problem is that I would like to efficiently map each element product to unique integer. Also, I would like to efficiently compute the integer, when I change the order of the arrays in the list.
I was thinking something in the line of integer factorization.
Take first $n$ primes ($p_1, p_2, ..., p_n$) and remeber the index of the array in the original list with $ind: A \rightarrow \mathbb{N}$. So the initial function would be $ind(A_i)=i$.
Also we denote with $a_{i,j}$ the $j$-th element of the $i$-th array in the list. So the function $F:A_1 \times A_2 \times \ldots \times A_n \rightarrow \mathbb{N}$ would be:
$$F(a_{1,i_1}, a_{2,i_2}, \ldots, a_{n,i_n})=\prod_{j=1}^{j=n}p_{ind(A_i)}^{i_j}$$
In more common words, we initially assign a prime number $p_i$ to array $i$ and raise $p_i$ to the index of the element in array $i$.
This guarantees us the uniqueness and also if we change the order of the arrays in the list, we just need to change the $ind$ function. But the problem here is that these products are fairly large and will overflow soon enough (even with using doubles).
Do you have any better idea how to enumerate these products without the large number constraint? I am implementing this in Java and I really would not like to use BigInteger implementation.
• Is it hashing you are after, or what is the purpose here? – Raphael Apr 5 '13 at 10:10
• Yes. Something like that. I actually need unique numbers, because these output numbers represent some kind of output (the number is actually mapped to a string value). So for example I need to get different integers for (1,2,3) and for (3,2,1). – Nejc Apr 5 '13 at 11:57
• What is the "element product", is it the product of all elements in one array? Why would this change when the order of the arrays is changed? – Juho Apr 5 '13 at 15:30
• Based on your description, I take it that an element product is taking one element from each array and multiplying (you should clarify that). – Aryabhata Apr 5 '13 at 17:26
• Sorry for not being clear! One element product of $n$ arrays is a $n$-tuple, where we take one element from each array and on the $i$-th place of the $n$-tuple is an element from $i$-th array. Clearly there are $|A_1| \cdot |A_2| \cdot \ldots \cdot |A_n|$ such products. – Nejc Apr 5 '13 at 17:45
Based on your description, I take it that an element product is taking one element from each array and multiplying?
You don't really need to multiply or use prime numbers which might overflow (but not too much for your constrains I suppose, and is a cute solution :-)).
Suppose the arrays were of fixed size $k$ (You can pad the arrays to achieve this, and based on your constraints, won't be too much of a problem).
Then you want to map all possible tuples $(x_1, x_2, \dots x_n)$ where $0 \le x_i \lt k$ to unique integers.
The mapping you need: Interpret the tuple as an integer in base-$k$.
If arrays are shuffled you can still change the $ind$ function and remap the digits according to that.
(or did you try that and didn't work for some reason?)
You want an injection between your element space and the naturals. Why not take the oldest one that happens to be a bijection, the Cantor pairing function.
Its basic form is defined to map $\mathbb{N^2}$ to $\mathbb{N}$:
$\qquad\displaystyle \pi(k_1,k_2) = \frac{1}{2}(k_1 + k_2)(k_1 + k_2 + 1)+k_2$.
It readily extends to $\mathbb{N}^k$:
$\qquad \pi^{(n)}(k_1, \ldots, k_{n-1}, k_n) = \pi ( \pi^{(n-1)}(k_1, \ldots, k_{n-1}) , k_n)$.
The Wikipedia article explains how to invert the function.
Now, in your case, $n$ is unknown. That's not a problem; we can extend above $\pi$ easily to enumerate $\mathbb{N}^+ = \bigcup_{n \geq 1} \mathbb{N}^n$ by encoding $n$ -- here the array's length -- as well:
$\qquad \pi^+(a_1,\dots,a_n) = \pi^{(n+1)}(a_1, \dots, a_n, n)$.
Since it's a bijection into $\mathbb{N}$, values are arguably as small as possible. Also, computing either direction is efficient.
When implementing it, be very carful either way. If you consider arrays over some finite set $A$, there are
$\qquad\displaystyle \sum_{i=1}^n |A|^i = \frac{|A|^{n+1} - |A|}{|A| - 1}$
many arrays of size at most $n$. $\pi^+$ has to encode them all, so the integers might overflow quickly if $|A|$ and $n$ are not both small.
• Thank you for your nice suggestion. Luckily I do not need the inverse function, so I will implement the Cantor pairing function and compare it to the suggestion of @Aryabhata. – Nejc Apr 8 '13 at 9:36 | 2019-12-12T12:04:50 | {
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https://cs.stackexchange.com/questions/58000/maximum-cost-path-in-integer-matrix | # Maximum cost path in integer matrix
In preparation for my design and algorithms exam, I encountered the following problem.
Given a $2 \times N$ integer matrix $(a[i][j] \in [-1000, 1000])$ and an unsigned integer $k$, find the maximum cost path from the top left corner $(a[1][1])$ and the bottom right corner $a[2][N]$, given the following:
$\bullet$ The path may not go through the same element more than once
$\bullet$ You can move from one cell to the other vertically or horizontally
$\bullet$ The path may not contain more than $k$ consecutive elements from the same line
$\bullet$ The cost of the path is determined by the sum of all the values stored within the cells it passes through
I've been thinking of a simple greedy approach that seems to work for the test cases I've tried, but I'm not sure if it's always optimal. Namely, while traversing the matrix, I select a the maximum cost cell on the vertical or horizontal and then go from there. I've got a counter I only increment if the current cell is on the same line with the previously examined one and reset it otherwise. If at some point the selected element happens to be one that makes the counter go over the given value of $k$, I simply go with the other option that's left.
However, I feel that I'm missing out on something terribly important here, but I just can't see what. Is there some known algorithm or approach that may be used here?
Also, the problem asks for an optimal solution (regarding temporal complexity).
This problem was basically made for dynamic programming (DP). Just review it and follow a couple of examples and solving this problem is straight-forward.
Your simple greedy approach does not always work. Consider:
[ begin ] [ 2 ]
[ 3 ] [ 10 ]
[ 1 ] [ -1000 ]
[ 1 ] [ end ]
The optimal solution would be to go right, down, left, down, down, right.
• I meant $2$ lines and $N$ columns, but I guess it doesn't make a difference in this case. – user43389 May 29 '16 at 23:28
We shall approach this using dynamic programming -- the greedy algorithm is 'myopic' in that it only considers immediate neighbors and not the path that follows those neighbors.
Let $C(i, \ j, \ m)$ be the cost of the max-cost path starting from $a[i][j]$ going towards $a[2][n]$, where $i$ is either $1$ or $2$, $\ j \in [1, n]$ and $m \in [0, k]$.
• $m$ here denotes that we have $m$ remaining consecutive steps that could be taken along row $i$.
• The additional constraint we enforce here is that when we are calculating $a[1][i]$, we have not yet visited $a[2][i]$, and similarly when we're calculating $a[2][i]$, we have not yet visited $a[1][i]$.
With this constraint in place, what recurrence can we come up with? Well,
1. $C(1, \ j, \ m) = a[1][j] + \max \{ M(1, j+1, m-1), \ a[2][j] + M(2, j+1, k)\} \$ and similarly,
2. $C(2, \ j, \ m) = a[2][j] + \max \{ M(2, j+1, m-1), \ a[1][j] + M(1, j+1, k)\} \$
Now, what's the rationale here? If we started at $a[1][j]$, we have two choices:
• continue along row $1$ : we then go to $a[1][j+1]$ with $m-1$ remaining consecutive steps that we can take along row $1$. (note here that $a[2][j+1]$ is not yet visited, maintaining the invariant).
• switch to row $2$ : we go to $a[2][j]$, incurring a cost of $a[2][j]$, and then have only one choice from there - to go to $a[1][j+1]$. Because we have switched rows, we can now refresh our count and take $k$ consecutive steps along row $2$. (note again that we haven't visited $a[1][j+1]$ yet).
Now if we calculate our subproblems starting from $j = n$ towards $j=1$ for every value of $m$, you shall have your answer in $C(1, \ 1, \ k)$.
I shall leave the base cases and runtime analysis to you. Note that my solution calculates the cost of the path and not the path itself; can you extend my solution to calculate the path too? | 2020-12-05T00:31:18 | {
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https://math.stackexchange.com/questions/2618387/how-to-prove-rigorously-that-the-series-1-3n-diverges | # How to prove rigorously that the series $1/(3n)$ diverges
How can I prove rigorously that the series
$$\sum_{n=1}^\infty \frac{1}{3n}$$ Diverges, assuming that I know that the harmonic series when $p = 1$ diverges,
I thought of using the property
$$\sum_{n=1}^\infty ca_n = c\sum_{n=1}^\infty a_n$$
However I think this only works when both of the series converges?
So to summarize, how can I prove that the series diverges knowing that the p-series $$\sum_{n=1}^\infty \frac{1}{n}$$ diverges?
Proof by contradiction. Let us assume that the series $$\sum_{n=1}^\infty \frac{1}{3n}$$ converges to, say, $A$. Then it would be also true that $$A = \frac{1}{3} \sum_{n=1}^\infty \frac{1}{n}$$ which implies that $$\sum_{n=1}^\infty \frac{1}{n}$$ converges to $3A$. Knowing the fact that this series diverges (we found a contradiction) completes the proof by contradiction.
• Thank you I didn't think about it just began discrete maths so I'm not used to using other forms of proof than the direct one so thank you for the tip!! Will try to consider more proof techniques in the future! – Ian Leclaire Jan 27 '18 at 23:35
You do not need convergence, just consider partial sum, you can factorize the scalar because you manipulate a finite series:
$$\sum_\limits{n=1}^N \dfrac 1{3n}=\dfrac 13\underbrace{\sum_\limits{n=1}^N \dfrac 1n}_{\to+\infty}\to+\infty$$
Thus you get that the partial sum does not have a finite limit so the series diverges.
Yes, it's true you use that property, but it's good you noticed that this equality is only guaranteed if the series converges. Actually, let's write the statement more precisely:
If $\sum_{n=1}^\infty a_n$ converges, then so does $\sum_{n=1}^\infty c\cdot a_n$ for any constant $c$.
Now think on the contrapositive of this statement:
If $\sum_{n=1}^\infty c\cdot a_n$ does not converge for some constant $c$, then $\sum_{n=1}^\infty a_n$ does not converge.
Do you see how you could use this to prove our statement?
There is a hint below.
Hint: Take $a_n = 1/(3n)$
Yeah, you can use a variation of your property. For any constant $c \neq 0$, we have $\sum_{i=1}^{\infty} ca_{n}$ converges if and only if $\sum_{i=1}^{\infty} a_{n}$ converges and likewise with divergence.
Specifically for this problem, assume your series converges. Then we have that, by definition, there exists some $L$ such that for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for $n \geq N$, we have $$\left| \sum_{i=1}^{n} \frac{1}{3n} - L \right| < \frac{1}{3}{\epsilon}.$$ But this is if and only if $$\left| \sum_{i=1}^{n} \frac{1}{n} - 3L \right| < \epsilon$$ and since $\epsilon$ was made arbitrary, this implies that $\sum_{i=1}^{\infty} \frac{1}{n}$ converges to $3L$, and you know that it doesn't. There's not much else to generalizing this concept for any series.
You can use the same proof that $\sum \frac 1n$ diverges. i.e. $1 + \frac 12 + (\frac 13 +\frac 14) + (\frac 15 + \cdots \frac 18) + (\frac 19 + \cdots + \frac 1{16})+\cdots < 1 + \frac 12 +\frac 12 + \frac 12 +\cdots$
And a divergent series multiplied by a constant (other than 0), indeed produces divergent series.
Note that by definition
$S_n(k)=\sum_{n=1}^k \frac{1}{n}$ diverges $\iff \forall M>0 \quad \exists \bar k$ such that $S_n(k)>M \quad \forall k>\bar k$
then also
$S_{3n}(k)=\frac13S_n(k)=\sum_{n=1}^k \frac{1}{3n}$ diverges since
$\forall M>0 \quad \exists \bar k$ such that $S_n(k)>3M \quad \forall k>\bar k\implies S_{3n}(k)>M$ | 2019-07-21T01:19:34 | {
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https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table | # How to properly print a table
to bootstrap my experience with Mathematica I'm trying to use it to print a table of values where the columns are different trigonometric functions and the rows contain different the values of said trig functions at some fixed angles that are between $0$ and $\frac{\pi}{2}$ .
After studying the basics I came up with this lines of code
vec = Prepend[Table[Pi/i, {i, Reverse[Range[2, 11]]}], 0];
res = Map[#, vec] & /@ {Sin, Cos, Tan, Csc, Sec, Cot};
TextGrid [res, Frame -> All]
and the output is
$\begin{array}{ccccccccccc} 0 & \sin \left(\frac{\pi }{11}\right) & \frac{1}{4} \left(\sqrt{5}-1\right) & \sin \left(\frac{\pi }{9}\right) & \sin \left(\frac{\pi }{8}\right) & \sin \left(\frac{\pi }{7}\right) & \frac{1}{2} & \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{2} & 1 \\ 1 & \cos \left(\frac{\pi }{11}\right) & \sqrt{\frac{\sqrt{5}}{8}+\frac{5}{8}} & \cos \left(\frac{\pi }{9}\right) & \cos \left(\frac{\pi }{8}\right) & \cos \left(\frac{\pi }{7}\right) & \frac{\sqrt{3}}{2} & \frac{1}{4} \left(\sqrt{5}+1\right) & \frac{1}{\sqrt{2}} & \frac{1}{2} & 0 \\ 0 & \tan \left(\frac{\pi }{11}\right) & \sqrt{1-\frac{2}{\sqrt{5}}} & \tan \left(\frac{\pi }{9}\right) & \tan \left(\frac{\pi }{8}\right) & \tan \left(\frac{\pi }{7}\right) & \frac{1}{\sqrt{3}} & \sqrt{5-2 \sqrt{5}} & 1 & \sqrt{3} & \text{ComplexInfinity} \\ \text{ComplexInfinity} & \csc \left(\frac{\pi }{11}\right) & \sqrt{5}+1 & \csc \left(\frac{\pi }{9}\right) & \csc \left(\frac{\pi }{8}\right) & \csc \left(\frac{\pi }{7}\right) & 2 & \frac{1}{\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}} & \sqrt{2} & \frac{2}{\sqrt{3}} & 1 \\ 1 & \sec \left(\frac{\pi }{11}\right) & \frac{1}{\sqrt{\frac{\sqrt{5}}{8}+\frac{5}{8}}} & \sec \left(\frac{\pi }{9}\right) & \sec \left(\frac{\pi }{8}\right) & \sec \left(\frac{\pi }{7}\right) & \frac{2}{\sqrt{3}} & \sqrt{5}-1 & \sqrt{2} & 2 & \text{ComplexInfinity} \\ \text{ComplexInfinity} & \cot \left(\frac{\pi }{11}\right) & \sqrt{2 \sqrt{5}+5} & \cot \left(\frac{\pi }{9}\right) & \cot \left(\frac{\pi }{8}\right) & \cot \left(\frac{\pi }{7}\right) & \sqrt{3} & \sqrt{1+\frac{2}{\sqrt{5}}} & 1 & \frac{1}{\sqrt{3}} & 0 \\ \end{array}$
the problems with this output are :
• some cells are in what it looks like a rational / expected form and others have a "weird" unevaluated form like $sin(\frac{\pi}{11})$ ; what is the problem with the latter ?
• I haven't found a way to print legends for the rows and the columns ( angle values and function names )
Thanks
• Not sure what you mean by "print legends" would it suffice to simply prepend your column / row values to res? Also that unevaluated form is simply the exact form for the expression. If it knows the exact ratio or whatever it'll return that, but otherwise Mathematica has a principle of remaining as exact and symbolic as possible. Mar 16, 2017 at 18:19
• @MB1965 basically any solution that could help me recognize rows and columns without making a mess with too much intricacies in the code . I would like to keep the computation separate from the the graphical layout . Regarding the 2nd part : can I force an approximation for the entire table ? Mar 16, 2017 at 18:23
• Second part, use N. That's its purpose. First part, isn't bad either. We'll do a bit of prepending. I'll knock you up a quick solution and explanation. Mar 16, 2017 at 18:25
• @MB1965 thank you Mar 16, 2017 at 18:28
So we'll just prepend your labels or whatever onto your grid. Note that you can change the dividers for Grid based constructs too, if you want a different appearance.
Here are my changes:
• Use Through to build your res in the transposed orientation
• Use N on that to get the numerical approximations you wanted
• Use Prepend to stick on the function labels, transpose the grid, then prepend on the number labels.
Looks like this in the end:
vec = Prepend[Table[Pi/i, {i, Reverse[Range[2, 11]]}], 0];
ops = {Sin, Cos, Tan, Csc, Sec, Cot};
res = Through@*ops /@ vec // N;
grid = Prepend[Transpose@Prepend[res, ops], Prepend[vec, Null]];
NumberForm[TextGrid[grid, Frame -> All], 3]
A few things to note:
I use @* (Composition) to make Through apply after the functions and NumberForm to make sure the numerical approximations only display up to 3 digits.
• @MB1965 I get the same result for res using Through[ops[#]] & /@ vec // N. I am not sure I understand the construct that you use for res. Can you enlighten me as to the differences, pros, cons? Thank you. Mar 17, 2017 at 22:18
• @JackLaVigne do you mean the Composition (@*)? What you did is another way to do it, but & is a very low precedence operator and so can often grab much more than what you intended. It just allows me to not have to use a pure function and is often clearer, in my mind. Mar 17, 2017 at 22:21
• @MB1965 Yes, the Composition. I see that it produces the desired result but can't quite figure out what is going on. Maybe you can spell it out for me? Mar 17, 2017 at 22:23
• @JackLaVigne I'd be happy to. f@*g in FullForm is Composition[f,g] and Composition[f,g][h] is f[g[h]]. Contrast that to f@g where we would get f[g][h]. It allows f to apply after g has applied instead of applying to g before hand. There is a corresponding head, RightComposition /* that is similar, but acts in reverse. Check out the docs for a more thorough treatment. Mar 17, 2017 at 22:26
• @MB1965 Got it. Map[Composition[Through, {Sin, Cos, Tan, Csc, Sec, Cot}], {x, y, z}] produces {{Sin[x], Cos[x], Tan[x], Csc[x], Sec[x], Cot[x]}, {Sin[y], Cos[y], Tan[y], Csc[y], Sec[y], Cot[y]}, {Sin[z], Cos[z], Tan[z], Csc[z], Sec[z], Cot[z]}}. Thank you for the help. Mar 17, 2017 at 22:40 | 2022-05-23T10:40:59 | {
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https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795 | # Numpy Broadcast to perform euclidean distance vectorized
I have matrices that are 2 x 4 and 3 x 4. I want to find the euclidean distance across rows, and get a 2 x 3 matrix at the end. Here is the code with one for loop that computes the euclidean distance for every row vector in a against all b row vectors. How do I do the same without using for loops?
`````` import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
dists = np.zeros((2, 3))
for i in range(2):
dists[i] = np.sqrt(np.sum(np.square(a[i] - b), axis=1))
``````
Simply use `np.newaxis` at the right place:
`````` np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
``````
• Could you please explain how `Simply using np.newaxis at the right place` works? If you could start with the fact that `a` is 2x4 and `b` is 3x4, that would be great. – stackoverflowuser2010 Jun 18 '16 at 23:45
Here are the original input variables:
``````A = np.array([[1,1,1,1],[2,2,2,2]])
B = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
A
# array([[1, 1, 1, 1],
# [2, 2, 2, 2]])
B
# array([[1, 2, 3, 4],
# [1, 1, 1, 1],
# [1, 2, 1, 9]])
``````
A is a 2x4 array. B is a 3x4 array.
We want to compute the Euclidean distance matrix operation in one entirely vectorized operation, where `dist[i,j]` contains the distance between the ith instance in A and jth instance in B. So `dist` is 2x3 in this example.
The distance
could ostensibly be written with numpy as
``````dist = np.sqrt(np.sum(np.square(A-B))) # DOES NOT WORK
# Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
# ValueError: operands could not be broadcast together with shapes (2,4) (3,4)
``````
However, as shown above, the problem is that the element-wise subtraction operation `A-B` involves incompatible array sizes, specifically the 2 and 3 in the first dimension.
``````A has dimensions 2 x 4
B has dimensions 3 x 4
``````
In order to do element-wise subtraction, we have to pad either A or B to satisfy numpy's broadcast rules. I'll choose to pad A with an extra dimension so that it becomes 2 x 1 x 4, which allows the arrays' dimensions to line up for broadcasting. For more on numpy broadcasting, see the tutorial in the scipy manual and the final example in this tutorial.
You can perform the padding with either `np.newaxis` value or with the `np.reshape` command. I show both below:
``````# First approach is to add the extra dimension to A with np.newaxis
A[:,np.newaxis,:] has dimensions 2 x 1 x 4
B has dimensions 3 x 4
# Second approach is to reshape A with np.reshape
np.reshape(A, (2,1,4)) has dimensions 2 x 1 x 4
B has dimensions 3 x 4
``````
As you can see, using either approach will allow the dimensions to line up. I'll use the first approach with `np.newaxis`. So now, this will work to create A-B, which is a 2x3x4 array:
``````diff = A[:,np.newaxis,:] - B
# Alternative approach:
# diff = np.reshape(A, (2,1,4)) - B
diff.shape
# (2, 3, 4)
``````
Now we can put that difference expression into the `dist` equation statement to get the final result:
``````dist = np.sqrt(np.sum(np.square(A[:,np.newaxis,:] - B), axis=2))
dist
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
``````
Note that the `sum` is over `axis=2`, which means take the sum over the 2x3x4 array's third axis (where the axis id starts with 0).
If your arrays are small, then the above command will work just fine. However, if you have large arrays, then you may run into memory issues. Note that in the above example, numpy internally created a 2x3x4 array to perform the broadcasting. If we generalize A to have dimensions `a x z` and B to have dimensions `b x z`, then numpy will internally create an `a x b x z` array for broadcasting.
We can avoid creating this intermediate array by doing some mathematical manipulation. Because you are computing the Euclidean distance as a sum-of-squared-differences, we can take advantage of the mathematical fact that sum-of-squared-differences can be rewritten.
Note that the middle term involves the sum over element-wise multiplication. This sum over multiplcations is better known as a dot product. Because A and B are each a matrix, then this operation is actually a matrix multiplication. We can thus rewrite the above as:
We can then write the following numpy code:
``````threeSums = np.sum(np.square(A)[:,np.newaxis,:], axis=2) - 2 * A.dot(B.T) + np.sum(np.square(B), axis=1)
dist = np.sqrt(threeSums)
dist
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
``````
Note that the answer above is exactly the same as the previous implementation. Again, the advantage here is the we do not need to create the intermediate 2x3x4 array for broadcasting.
For completeness, let's double-check that the dimensions of each summand in `threeSums` allowed broadcasting.
``````np.sum(np.square(A)[:,np.newaxis,:], axis=2) has dimensions 2 x 1
2 * A.dot(B.T) has dimensions 2 x 3
np.sum(np.square(B), axis=1) has dimensions 1 x 3
``````
So, as expected, the final `dist` array has dimensions 2x3.
This use of the dot product in lieu of sum of element-wise multiplication is also discussed in this tutorial.
• This answer is so useful, especially the part to overcome broadcasting issues. Thank you @stackoverflowuser2010 – Allen May 21 '18 at 6:22
I had the same problem recently working with deep learning(stanford cs231n,Assignment1),but when I used
`````` np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
``````
There was a error
``````MemoryError
``````
That means I ran out of memory(In fact,that produced a array of 500*5000*1024 in the middle.It's so huge!)
To prevent that error,we can use a formula to simplify:
$(a-b)^2&space;=&space;a^2&space;-&space;2ab+b^2$
code:
``````import numpy as np
aSumSquare = np.sum(np.square(a),axis=1);
bSumSquare = np.sum(np.square(b),axis=1);
mul = np.dot(a,b.T);
dists = np.sqrt(aSumSquare[:,np.newaxis]+bSumSquare-2*mul)
``````
• to add something;quoted from the official document `There are, however, cases where broadcasting is a bad idea because it leads to inefficient use of memory that slows computation.There are, however, cases where broadcasting is a bad idea because it leads to inefficient use of memory that slows computation.` – Han Qiu Mar 24 '16 at 6:40
This functionality is already included in scipy's spatial module and I recommend using it as it will be vectorized and highly optimized under the hood. But, as evident by the other answer, there are ways you can do this yourself.
``````import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
from scipy.spatial.distance import cdist
cdist(a,b)
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
``````
Using numpy.linalg.norm also works well with broadcasting. Specifying an integer value for `axis` will use a vector norm, which defaults to Euclidean norm.
``````import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
np.linalg.norm(a[:, np.newaxis] - b, axis = 2)
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
`````` | 2019-10-23T03:50:27 | {
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https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/ | # What does it mean to multiply by x / x?
1. Sep 29, 2015
### Mr Davis 97
This is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression. However, x / x also equals 1. But if we multiply it by another expression, the expression changes such that we can no longer use 0. So if the domain of an expression originally contains 0 and we multiply it by x / x, is that illegal?
2. Sep 29, 2015
### Staff: Mentor
No, you get a different expression whose value is the same.
x(x - 1) = 0 -------------- Solution set {0, 1}
If we multiply the left side by x/x, we get
$\frac{x^2(x - 1)}{x} = 0$ -------------- The numerator is zero if x = 0 or x = 1, but we can't divide by 0, so the solution set is now {1}.
By multiplying by x/x, we are tacitly assuming that $x \ne 0$, so we lose a value in the solution set.
If we graph y = x(x - 1) and $y = \frac{x^2(x - 1)}{x}$, the two graphs look exactly the same, except that the graph of the latter equation has a "hole" at (0, 0).
3. Sep 29, 2015
### Mr Davis 97
So if we divide the x's out again to get the original expression, that tacitly adds 0 back to our solution set? The only thing that confuses me is that the addition or subtraction of the domain element seems to have to be inferred by us rather than explicitly stated by the mathematics.
4. Sep 29, 2015
### Staff: Mentor
Sometimes people will expicitly write any restrictions to the domain. For example,
$y = \frac{x^2(x - 1)}{x}, x \ne 0$
$\Leftrightarrow y = x(x - 1), x \ne 0$
The two equations above are equivalent, meaning that their solution sets (and graphs) are identical.
For examples as simple as this one, some writers will omit the explicit domain restrictions, believing them to be pretty much obvious.
5. Sep 29, 2015
### Mr Davis 97
Okay, that makes sense. But what about when we are solving limits at infinity? For example, given, $\displaystyle \lim_{x\rightarrow \infty } \frac{2x - 1}{x + 1}$, the only way to solve is to multiply the fraction by $\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$. However, this changes the domain of the original function such that x = 0 is no longer defined. So why is this a rigorous way to solve limits at infinity if we're changing the domain of the function?
6. Sep 29, 2015
### Staff: Mentor
No, that's not the only way to evaluate3 this limit. You can also factor both numerator and denominator like so:
$\lim_{x\to \infty } \frac{2x - 1}{x + 1} = \lim_{x\to \infty } \frac{x(2 - 1/x)}{x(1 + 1/x)} = \lim_{x\to \infty } \frac x x \cdot \lim_{x\to \infty }\frac{x(2 - 1/x)}{x(1 + 1/x)}$.
The first limit is 1 and the second limit it 2. In the first limit, x/x is always 1 for any finite value of x (other than 0), so as x increases without bound, the value of the fraction is always 1.
Corrected the above to "such that at x= 0 the original function is no longer defined."
It doesn't matter. You're taking the limit as $x \to \infty$, so the fact that the rational expression isn't defined at x = 0 is immaterial.
7. Sep 29, 2015
### Mr Davis 97
But $\displaystyle \frac{2x - 1}{x + 1} \neq \frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}$. So how can I be sure that $\displaystyle \lim_{x\rightarrow \infty }\frac{2x - 1}{x + 1} = \lim_{x\rightarrow \infty }\frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}$?
8. Sep 29, 2015
### Staff: Mentor
Of course they are equal, except at x = 0 and x = -1. Again, the limit is as $x \to \infty$, so for any x > 0, the two expressions are identical.
I don't understand why you're so hung up on this.
9. Sep 29, 2015
### aikismos
It's easier to do polynomial division. $\frac{2x -1}{ x + 1} \rightarrow 2 + \frac{-3}{x + 1}$. Taking $\frac{\lim}{x\rightarrow \infty } \rightarrow 2$. What your worried about is the equivalence of fractions, but if you can show the steps in light of the Properties of Equations (e.g., Division Property of Equations) then you have nothing to worry about.
10. Sep 30, 2015
### HallsofIvy
I want to expand on what Mark44 said. You do NOT get "the same expression", you get another expression with the same numeric value if x is not 0.
Yes, because of your misquoting of that "rule"- you should have included "if x n is not 0" right at the start.
11. Sep 30, 2015
### DrStupid
As you can't divide by zero it is not that simple. But you can remove the singularity using L'Hôpital's rule. | 2018-07-17T02:24:48 | {
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https://ch.mathworks.com/help/symbolic/piecewise.html | # piecewise
Conditionally defined expression or function
## Syntax
``pw = piecewise(cond1,val1,cond2,val2,...)``
``pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal)``
## Description
example
````pw = piecewise(cond1,val1,cond2,val2,...)` returns the piecewise expression or function `pw` whose value is `val1` when condition `cond1` is true, is `val2` when `cond2` is true, and so on. If no condition is true, the value of `pw` is `NaN`.```
example
````pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal)` returns the piecewise expression or function `pw` that has the value `otherwiseVal` if no condition is true.```
## Examples
### Define and Evaluate Piecewise Expression
Define the following piecewise expression by using `piecewise`.
`$y=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$`
```syms x y = piecewise(x<0, -1, x>0, 1)```
```y = piecewise(x < 0, -1, 0 < x, 1)```
Evaluate `y` at `-2`, `0`, and `2` by using `subs` to substitute for `x`. Because `y` is undefined at `x = 0`, the value is `NaN`.
`subs(y, x, [-2 0 2])`
```ans = [ -1, NaN, 1]```
### Define Piecewise Function
Define the following function symbolically.
`$y\left(x\right)=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$`
```syms y(x) y(x) = piecewise(x<0, -1, x>0, 1)```
```y(x) = piecewise(x < 0, -1, 0 < x, 1)```
Because `y(x)` is a symbolic function, you can directly evaluate it for values of `x`. Evaluate `y(x)` at `-2`, `0`, and `2`. Because `y(x)` is undefined at `x = 0`, the value is `NaN`. For details, see Create Symbolic Functions.
`y([-2 0 2])`
```ans = [ -1, NaN, 1]```
### Set Value When No Conditions Is True
Set the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is `NaN`.
Define the piecewise function
`$y\left(x\right)=\left\{\begin{array}{cc}-2& x<-2\\ 0& -2`
```syms y(x) y(x) = piecewise(x<-2, -2, -2<x<0, 0, 1)```
```y(x) = piecewise(x < -2, -2, x in Dom::Interval(-2, 0), 0, 1)```
Evaluate `y(x)` between `-3` and `1` by generating values of `x` using `linspace`. At `-2` and `0`, `y(x)` evaluates to `1` because the other conditions are not true.
```xvalues = linspace(-3,1,5) yvalues = y(xvalues)```
```xvalues = -3 -2 -1 0 1 yvalues = [ -2, 1, 0, 1, 1]```
### Plot Piecewise Expression
Plot the following piecewise expression by using `fplot`.
`$y=\left\{\begin{array}{cc}-2& x<-2\\ x& -22\end{array}.$`
```syms x y = piecewise(x<-2, -2, -2<x<2, x, x>2, 2); fplot(y)```
### Assumptions and Piecewise Expressions
On creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using `simplify` on the expression.
Assume `x > 0`. Then define a piecewise expression with the same condition `x > 0`. `piecewise` automatically applies the assumption to simplify the condition.
```syms x assume(x > 0) pw = piecewise(x<0, -1, x>0, 1)```
```pw = 1```
Clear the assumption on `x` for further computations.
`assume(x,'clear')`
Create a piecewise expression `pw` with the condition `x > 0`. Then set the assumption that ```x > 0```. Apply the assumption to `pw` by using `simplify`.
```pw = piecewise(x<0, -1, x>0, 1); assume(x > 0) pw = simplify(pw)```
```pw = 1```
Clear the assumption on `x` for further computations.
`assume(x, 'clear')`
### Differentiate, Integrate, and Find Limits of Piecewise Expression
Differentiate, integrate, and find limits of a piecewise expression by using `diff`, `int`, and `limit` respectively.
Differentiate the following piecewise expression by using `diff`.
`$y=\left\{\begin{array}{cc}1/x& x<-1\\ \mathrm{sin}\left(x\right)/x& x\ge -1\end{array}$`
```syms x y = piecewise(x<-1, 1/x, x>=-1, sin(x)/x); diffy = diff(y, x)```
```diffy = piecewise(x < -1, -1/x^2, -1 < x, cos(x)/x - sin(x)/x^2)```
Integrate `y` by using `int`.
`inty = int(y, x)`
```inty = piecewise(x < -1, log(x), -1 <= x, sinint(x))```
Find the limits of `y` at `0` and `-1` by using `limit`. Because `limit` finds the double-sided limit, the piecewise expression must be defined from both sides. Alternatively, you can find the right- or left-sided limit. For details, see `limit`.
```limit(y, x, 0) limit(y, x, -1)```
```ans = 1 ans = limit(piecewise(x < -1, 1/x, -1 < x, sin(x)/x), x, -1)```
Because the two conditions meet at `-1`, the limits from both sides differ and `limit` cannot find a double-sided limit.
### Elementary Operations on Piecewise Expressions
Add, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined.
```syms x pw1 = piecewise(x<-1, -1, x>=-1, 1); pw2 = piecewise(x<0, -2, x>=0, 2); add = pw1 + pw2 sub = pw1 - pw2 mul = pw1 * pw2 div = pw1 / pw2```
```add = piecewise(x < -1, -3, x in Dom::Interval([-1], 0), -1, 0 <= x, 3) sub = piecewise(x < -1, 1, x in Dom::Interval([-1], 0), 3, 0 <= x, -1) mul = piecewise(x < -1, 2, x in Dom::Interval([-1], 0), -2, 0 <= x, 2) div = piecewise(x < -1, 1/2, x in Dom::Interval([-1], 0), -1/2, 0 <= x, 1/2)```
### Modify or Extend Piecewise Expression
Modify a piecewise expression by replacing part of the expression using `subs`. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. `piecewise` does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, `piecewise` returns the value for the first true condition.
Change the condition `x<2` in a piecewise expression to `x<0` by using `subs`.
```syms x pw = piecewise(x<2, -1, x>0, 1); pw = subs(pw, x<2, x<0)```
```pw = piecewise(x < 0, -1, 0 < x, 1)```
Add the condition `x>5` with the value `1/x` to `pw` by creating a new piecewise expression with `pw` as the otherwise value.
`pw = piecewise(x>5, 1/x, pw)`
```pw = piecewise(5 < x, 1/x, x < 0, -1, 0 < x, 1)```
## Input Arguments
collapse all
Condition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.
Example: x > 2
Value when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Value if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If `otherwiseVal` is not specified, its value is `NaN`.
## Output Arguments
collapse all
Piecewise expression or function, returned as a symbolic expression or function. The value of `pw` is the value `val` of the first condition `cond` that is true. To find the value of `pw`, use `subs` to substitute for variables in `pw`.
## Tips
• `piecewise` does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, `piecewise` mimics an if-else ladder. | 2021-02-24T21:39:07 | {
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https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs | # Considering sets in proofs
In order to show that $$\forall x\gt0, \ \exists n_0\in \mathbb N: \ n_0(n_0+1)\le x\lt (n_0+1)(n_0+2)$$, the proof in my textbook considers the set $$A=\{n\in \mathbb N :n(n+1)\le x\}$$ and then goes to show that $$A$$ has a maximal element $$n_0$$, which completes the proof.
Now, I want to know what has led to this kind of argument, because I have begun to see it very frequently and even some of my classmates often use this method. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.
I understand that such motivations usually can't be explained easily; in that case, I'm asking for other problems you know of that use a similar approach.
• The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) \leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $\left \lfloor x \right \rfloor$. We also note that by construction of $\mathbb{N}$ that $n_0 \in \mathbb{N}$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along. – Eevee Trainer Nov 6 '18 at 10:02
• It would be much easier to show that $B=\{\,n\in\Bbb N: n(n+1)>x\,\}$ has a minimal element ... – Hagen von Eitzen Nov 6 '18 at 15:30
• @HagenvonEitzen: It would be much easier to just use induction... =) – user21820 Nov 6 '18 at 16:25
## 4 Answers
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
1. I look at the statement. OK, it's saying that I can squeeze any positive real $$x$$ between two numbers. OK, let's imagine the $$x$$ as some point on the positive real line.
2. Hmm, the two numbers, $$n_0(n_0+1)$$ and $$(n_0+1)(n_0+2)$$ are both integers.
3. Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $$a_n = n(n+1)$$.
4. So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
5. So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $$x$$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $$x$$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
6. OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $$n$$ such that $$a_n < x$$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
• This has been very helpful, thank you. – FuzzyPixelz Nov 6 '18 at 10:09
For instance, let $$a,b$$ be natural numbers with $$1\leq b\leq a$$. Then consider the set $$A=\{a-qb\mid a-qb\geq 0, q\geq 0\}$$. Since the set of natural numbers is well-ordered, every nonempty subset $$A$$ of natural numbers has a least element, here $$r = a-qb$$ in $$A$$. It is then clear that $$a=qb+r$$ and $$0\leq r is the remainder.
To answer your question
Now, I want to know what has lead to this kind of argument
Look at this part of the proof
and then goes to show that $$A$$ has a maximal element $$n_0$$, which completes the proof.
The way the set is build makes it trivial to show that the set is finite thus has a maximal element. This triviality is a reason why a set is used. Other proofs requiring existence of a maximal/minimal element might follow the same approach.
Of course the thinking process is going to be as outlined by 5xum.
This is not the only way to solve the problem, and arguably uses set theory when there is no need for it. You can prove the claim directly by an easy induction (which I am sure you can do). Also note that induction is necessary, whether or not hidden. It is logically non-trivial (and requires induction) to show that the set in your question has a maximum! | 2019-07-19T13:12:07 | {
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https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1 | # Connection between universal Quantifier and implication
I remember a high school book I read a long time ago explains that every statement in the form of $\forall x\in D,P(x)$ can be turned into an implication.
For example, isn't
"For every real number $x$, $x^2$ is non-negative."
equivalent to
"If $x$ is a real number, then $x^2$ is non-negative."?
However, I came across a wikipedia article about universal quantifier that says something about bounded quantifier, which I haven't heard before. Then I thought, and confirmed, that bounded quantifier can actually be written as unbounded quantifier, and that the statement I read in the book I said above is actually a bounded version.
For example, $\forall x\in D,P(x)$ becomes $\forall x,[x\in D \implies P(x)]$ My question is: was what I read in the book correct? That every bounded universal statement can be converted into an implication? Another one is: Can or can't every (unbounded) universal statement become implication?
I ask this because I am confused. I thought every universal statement can become an implication, the wikipedia article and the fact that I cannot find (google) any other article about my thought make me want to post this question. Besides, don't we use arguably the same starting premise when we want to prove a universal statement and an implication? Both use "Let x be ..." or "Assume that ..." or "Take arbitrary x ..." (the latter especially is used most often when facing a universal statement) as starting premise, don't they?
ADDITION: Why can't $\forall x\in D,P(x)$ just become $x\in D \implies P(x)$ instead of $\forall x,[x\in D \implies P(x)]$?
• Jun 6 '18 at 13:24
• Thanks. Then, in my example, the statement "If $x$ is a real number, then $x^2$ is non-negative." is just a predicate whose truth depends on $x$. Am I reading that correctly?
– bms
Jun 6 '18 at 13:32
• Correct; but often the assertion "If $x$ is a real number, then $x^2$ is non-negative" is read as implicitly universally quantified. Jun 6 '18 at 13:36
• I see. However, does this mean we cannot simply write $\forall x,P(x)$ without knowing what or where $x$ comes from? And this means that $\forall x,P(x)$ actually means $\forall x \in X,P(x)$ for some set $X$, right?
– bms
Jun 6 '18 at 13:50
• Not clear ... the formula $\forall x P(x)$ is obviously formally correct. But it is ... a formula. How we interpret the predicate symol $P$ ? If we interpret it as "$x \text { is Even}$" in the domain or universe of natural numbers, the resulting statement will be FALSE. If instead we interpret it as "$x \text { is Mortal}$" in the domain or universe of human beings, the resulting statement will be TRUE. Jun 6 '18 at 13:54
$\forall x \in D \colon P(x)$ is simply an abbreviation for the formula $\forall x ( x \in D \implies P(x))$ -- there really isn't more to it. You will find this convention in any decent textbook that covers the basics of first order logic.
And you can't replace it by $x \in D \implies P(x)$ for the simple reason that the latter is not a sentence -- it has unbounded free variables (namely $x$). Hence it's not equivalent to the sentence $\forall x \in D \colon P(x)$.
• I see. So are you saying that every implication has to be properly quantified, or just in this case? Because I thought that the statement "If $x \in \mathbb{R}$, then $x^2$ is non-negative" is valid every day. Do you mean that it is not valid until I add "For every $x$" in the beginning?"
– bms
Jun 6 '18 at 13:16
• @bms Valid and true are different concepts: $\forall x \in \mathbb R \colon x^2 \ge 0$ is a true sentence. $x \in \mathbb R \implies x^2 \ge 0$ is not a sentence at all but it is a vaid formulae -- it's true for every assignment of $x$. Jun 6 '18 at 13:21
• Hmm.. I have more questions actually. However, if it feels like I need to go through some definitions, could you suggest me one or two college level (ungrad or grad) logic textbooks that elementary cover this topic?
– bms
Jun 6 '18 at 13:25
• @bms That's right. However most mathematicians won't be that pedantic and often implicitly assume universal quantifiers ranging over every free variable. You will find this implicit assumption spelled out at least in some textbooks by logicians -- they will talk about the 'universal closure' of these formulae. Jun 6 '18 at 13:44
• While your answer is certainly correct for typical textbooks on first-order logic, it's not in general true that the bounded quantifier is just a shorthand. If the underlying foundational system is some type theory, then unbounded quantifiers may not even be meaningful. Furthermore, in type theories that have a universal type but are based on a variant of 3-valued logic, bounded quantification can be weaker than the quantified implication. Philosophically speaking, we think in terms of bounded quantifiers. Even the symmetry in "¬∀x∈S ( P(x) ) ≡ ∃x∈S ( ¬P(x) )" is broken by the 'shorthand'. =) Jun 16 '18 at 8:34 | 2021-10-24T21:27:32 | {
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https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn | # Find solution of nonlinear ODE in terms of JacobiCN
I am trying to find a specific solution for this differential equation:
$-\frac{1}{2}\frac{d^2}{dx^2}\psi(x)-2k \; \psi(x)^3 + \frac{1}{2}k^2\; \psi(x)=0$
MMA gives me a solution in the form of a JacobiSN function that, after setting the second argument equal to one by choosing the right constant of integration, can be rewritten as a Tanh function. However, I know from a ref that also
$\psi(x)=\sqrt{\frac{k}{2}}\; \text{sech}({kx})$
solves the ode. How to make MMA show this solution? I know that this could be obtained by choosing a JacobiCN instead of a JacobiSN, but how to implement this?
The point is that I want to find the solutions of the 2D version of this equation, and MMA only finds the Tanh solution just like in 1D. How to know if there's a Sech solution (or similar) in 2D?
So I guess the general question is: is there a way to "guide" MMA when it chooses the solutions for ODEs/PDEs, for example in the above case by making it choose a JacobiCN?
EDIT
The Tanh solution I find is the following:
$\frac{\sqrt{k}}{2} \text{tanh}\left(i\frac{k|x|}{\sqrt{2}}\right)$
Here is the code:
The equation:
sol1D=DSolve[-2 k ψ[x]^3 - D[ψ[x], {x, 2}]/2 == -(1/2) k^2 ψ[x], {ψ[x]}, {x}]
The output of the above is a JacobiSN function, it's pretty big and not insightful so it makes no sense to paste it here. Now, JacobiSN[u,m] simplifies to Tanh for m=1. Important to note: JacobiSN never simplifies to Sech, but JacobiCN does. Here is what the solution becomes when you adjust the constant to get the second argument to 1 (and I choose the other constant to be zero for simplicity):
In:
Simplify[sol1D[[1]] /. C[1] -> -(k^3/8) /. C[2] -> 0, x ∈ Reals && k > 0]
Out:
ψ[x] -> 1/2 I Sqrt[k] Tan[(k Abs[x])/Sqrt[2]]}
Here below I check the Sech solution:
In:
sechSol[x_] := (Sqrt[k] Sech[k x])/Sqrt[2]
Simplify[-1/2 (D[sechSol[x], {x, 2}]) - 2 k (sechSol[x])^3] == -1/2 k^2 sechSol[x]
Out: True
As expected. Now the question is how to make this sechSol show up.
2nd EDIT
For anyone who wants to know: I found analytically that also the 2D case has a Sech solution. In two dimensions the equation reads:
$-\frac{1}{2}\frac{d^2}{dx^2}\psi(x,y)-\frac{1}{2}\frac{d^2}{dy^2}\psi(x,y)- \psi(x,y)^3 + \frac{1}{2}k^2\; \psi(x,y)=0$
The coefficient in front of the nonlinear term changes (dimensionality arguments). This equation is solved by:
$\psi(x,y)=k\;\text{sech}(\frac{k}{\sqrt{2}}(x+y))$.
• (1) You'll get more nibbles if you bait the hook: Copyable code (what you tried, for instance), would make it easy for others to cut, paste & check. (2) Traditionally your equation is called an ODE, not a PDE. You may find this this meta Q&A helpful – Michael E2 Nov 10 '15 at 15:00
• Precisely what solution did you obtain? – bbgodfrey Nov 11 '15 at 5:21
• @bbgodfrey see the edit – user50473 Nov 13 '15 at 9:45
• @MichaelE2 thanks, it should work now – user50473 Nov 13 '15 at 11:04
• @ user50473: there is a difference in your equations for 1D and 2D: the factor in front of psi^3 is 2k for D=1 and 1 for D=2. In fact, your 2D Sech expression solves the 2D equation only with the factor 1, whereas the 1D Sech expression derived from the 2D case by letting y=0 requires the factor 2k to be a soluton. So could you please clarify. – Dr. Wolfgang Hintze Nov 16 '15 at 9:42
Abstract
(Please see section 3.1 for a short answer to the original question of the OP.)
We present here the complete solution of the problem of the OP, which consists of analyzing and solving the ODE numerically and symbolically. It turns out that the essential parameter for the classification of the solutions is the energy w of the equivalent mechanical problem.
The main problem of the OP was to reconcile the basic solution given by DSolve in terms of just the function JacobiSN containing two constants of integration C[1] and C[2] with the explicitly real valued solution which, in its most compact form, is given in terms of the function JacobiCN.
We show that, as expected, the reconciliation is possible by an appropriate choice of the integration constants, but we have found no way to let Mathematica perform the necessary transformations of the Jacobi elliptic functions automatically, therefore these have been done manually.
This answer is organized as follows: we start (section 1) with a qualitative analysis of the solution types in terms of an analogue mechanical problem, then (section 2) we provide the numerical and the exact symbolical solution, and derive (section 3) the exact solution from the basic form, and accompany this with other approaches, such as the integration of the energy equation. In section 4 we provide additional information and discuss the results.
Section 1: reformulation, qualitative picture of solutions
The ODE to be solved is
eq0 = lhs == 0;
where
$$\text{lhs}=\frac{1}{2} k^2 \psi (z)-2 k \psi (z)^3-\frac{\psi ''(z)}{2}$$
First we simplify things by making the ODE dimensionless.
eq = Simplify[
0 == lhs /. {ψ[z] -> x[z k] Sqrt[k/2], ψ''[z] ->
k^2 Sqrt[k/2] x''[z k]} /. z -> t/k, k > 0]
(*
Out[54]= 2 x[t]^3 + (x^′′)[t] == x[t]
*)
It will be convenient in the following to adopt the wording of classical mechanics. The ODE eq. describes the 1D motion of a particle subject to a conservative force with the potential
U = - x^2 + x^4
The energy
w = x'[t]^2-x[t]^2+ x[t]^4
is conserved in time. Actually, the true energy is w/2, but we still keep calling w the energy.
Now let us look for the qualitative picture of the solutions. For a given energy w the motion of the particle is limited to certain intervals in x determined by the obvious requirement x'[t]^2 >= 0 as shown in the graph, in which the characteristic points are given by
solx = x /. Solve[w == -x^2 + x^4, x];
{xA, xB} = {solx[[3]], solx[[4]]};
{xC, xM, xD} = {solx[[3]], solx[[2]], solx[[4]]};
{xE, xF} = {solx[[2]], solx[[4]]};
xG = 1/Sqrt[2];
(* the cell to produce the graph has been closed to save space *)
We have to distinguish three regions of w:
Region I (w > 0)
the point oscillates between the symmetric points A and B symmetrically about x = 0
Region II (-1/4 <= w < 0)
the point oscillates between E and F (or symmetrically in the region x<0, not shown here), i.e. x is never 0. For w = -1/4 the point remains at rest at G
Region III (w = 0)
the point for x > 0 will move between D and M towards M (after having been reflexted in D, if it started out with positive velocity) in the manner of a loopswing moving slowly to its upper position at which it comes to a rest only in infinite time (this is the "soliton" solution Sech)
Furthermore we can state that the oscillations will be increasingly harmonic either for w>>1 or for w close to -1/4.
Section 2: symbolical and numerical solution for the three regions
The symbolic expressions will be derived in section 3.
Region I
xI[t_] = Sqrt[(1 + Sqrt[1 + 4 w])/2]
JacobiCN[t (1 + 4 w)^(1/4), 1/2 (1 + 1/Sqrt[1 + 4 w])]; (* w > 0 *)
With[{ww = 1/2, tt = 20},
x0 = xB /. w -> ww;
x0p = Sqrt[ww + x0^2 - x0^4];
xn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];
Plot[{xn, 0.1 + xI[t] /. w -> ww}, {t, 0, tt},
PlotRange -> {1.1 xA /. w -> ww, 1.1 xB /. w -> ww},
PlotLabel ->
"Numerical solution of ODE eq: region I\nEnergy w = " <>
ToString[ww, InputForm] <>
"\nInitial position x0 = xB\nblue curve = numerical solution\nyellow \
curve = symbolic solution (slightly shifted)", AxesLabel -> {"t", "x(t)"}]]
Region II
xII[t_] = Sqrt[(1 + Sqrt[1 + 4 w])/2]
JacobiDN[(t (1 + Sqrt[1 + 4 w])^(1/4))/2^(1/4), (1 + 4 w - Sqrt[1 + 4 w])/(
2 w)];(* -1/4 < w < 0 *)
With[{ww = -1/8, tt = 10},
x0 = xF /. w -> ww;
x0p = Sqrt[ww + x0^2 - x0^4];
xn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];
Plot[{xn, xII[t] /. w -> ww}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},
PlotLabel ->
"Solution of ODE eq: region II\nEnergy w = " <> ToString[ww, InputForm] <>
"\nInitial position x0 = xF\nblue curve = numerical solution\nyellow \
curve = symbolic solution", AxesLabel -> {"t", "x(t)"}]]
Region III
We obtain
xIII[t_] = Sech[t]; (* w = 0 *)
as a limit of region I or of region II.
Region I -> region III
xI[t] /. w -> 0
(* Out[312]= Sech[t] *)
With[{ww = 10^(-5), tt = 20},
x0 = xF /. w -> ww ;
x0p = Sqrt[ww + x0^2 - x0^4];
xn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];
Plot[{xn, xIII[t] + 0.1}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},
PlotLabel ->
"Numerical solution of ODE eq: region I close to region III\nEnergy w = " <>
ToString[ww, InputForm] <>
"\nInitial position x0 = xB\nblue curve = numerical solution\nyellow \
curve = symbolic solution (slightly shifted upwards)",
AxesLabel -> {"t", "x(t)"}]]
Region II -> region III
JacobiDN @@ Limit[List @@ (xII[t]/Sqrt[((1 + Sqrt[1 + 4 w])/2)]), w -> 0]
(* Out[313]= Sech[t] *)
With[{ww = -10^(-5), tt = 20},
x0 = xB /. w -> ww ;
x0p = Sqrt[ww + x0^2 - x0^4];
xn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];
Plot[{xn, xIII[t] + 0.1}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},
PlotLabel ->
"Numerical solution of ODE eq: region II close to region III\nEnergy w = " \
<> ToString[ww, InputForm] <>
"\nInitial position x0 = xF\nblue curve = numerical solution (= symbolic \
solution)\nyellow curve = symbolic solution (slightly shifted upwards)",
AxesLabel -> {"t", "x(t)"}]]
Section 3: Exact symbolic solution
Section 3.1: Deriving the exact solutions from the basic solution
Here we show how the solution by DSolve[eq] can be transformed, by chosing apropriate integration constants, to a simple unified form in terms of the function JacobiCN. This was the original task requested by the OP.
We repeat the ODE to be solved
eq
(*
Out[5]= 2 x[t]^3 + (x^\[Prime]\[Prime])[t] == x[t]
*)
DSolve[eq] gives two solutions which differ only in the sign. Hence it is sufficient to consider just one of them. Also, without loss of generality we rename the constants of integration.
Mathematica finds immediately
x[t] /. DSolve[eq, x[t], t][[1]] /. {C[1] -> w, C[2] -> c} // Simplify
During evaluation of In[456]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
$$- i \frac{ \text{sn}\left(\frac{\sqrt{-(c+t)^2 \left(\sqrt{4 w+1}+1\right)}}{\sqrt{2}}|\frac{1-\sqrt{4 w+1}}{\sqrt{4 w+1}+1}\right)}{\sqrt{2} \sqrt{\frac{1}{\sqrt{4 w+1}-1}}}$$
The designation of the constant C1 is not accidental; in fact it is the energy w of the mechanical problem, in other words, it is the first integral of eq.
After a manual simplification in the factors we define the basic soution as
x0[t_, w_] := -I Sqrt[(-1 + Sqrt[1 + 4 w])/2]
JacobiSN[I (c + t) Sqrt[ ((1 + Sqrt[1 + 4 w])/2)], (1 - Sqrt[1 + 4 w])/(
1 + Sqrt[1 + 4 w])]
Now it turns out that making the following choice for the integration constant c
cw = 1/(1 + 4 w)^(1/4) EllipticK[1/2 (1 + 1/Sqrt[1 + 4 w])];
repc = c -> cw;
x[t_, w_] :=
Sqrt[(1 + Sqrt[1 + 4 w])/2 ]
JacobiCN[t (1 + 4 w)^(1/4), 1/2 (1 + 1/Sqrt[1 + 4 w])]
$$x(t,w)=\sqrt{\frac{1}{2} \left(\sqrt{4 w+1}+1\right)} \text{cn}\left(t \sqrt[4]{4 w+1}|\frac{1}{2} \left(1+\frac{1}{\sqrt{4 w+1}}\right)\right)$$
It came as a surprise to me and is gratifying that this expression for the solution of the ODE is valid for all physically meaningful values of w, i.e. for w>-1/4. Hence the distinction of the expressions for the exact solutions according to regions, as was done in section 2, is not necessary.
The "unifying" idea is to place the origin of time (t=0) so that the motion starts always at the maximum attainable value of x for given w, i.e. with velocity x'[0] zero.
In the limit of w->0 we find the expected Sech
x[t, 0]
(* Out[5]= Sech[t] *)
Now we plot the solution for varying parameter w, and distingush the two regions w>0 and w<0 for better viewing:
Plot3D[x[t, w], {t, -10, 10}, {w, 0, 1}, PlotRange -> {-1.2, 1.2},
PlotLabel -> "Solution of ODE x(t,w) for w>=0",
AxesLabel -> {"t", "w", "x"}]
Plot3D[x[t, w], {t, -10, 10}, {w, -1/4, 0}, PlotRange -> {-1.2, 1.2},
PlotLabel ->
"Solution of ODE x(t,w) for -1/4 \[LessEqual] w \[LessEqual] 0",
AxesLabel -> {"t", "w", "x"}]
Unfortunately, Mathematica seems to be not familiar enough with the transformation properties of the Jacobi elliptic functions to be able to (Full)Simplify the expression x0 after the replacement of c automatically.
Therefore we shall show how to perform the steps manually indicating the relevant formulas of the "Handbook of Mathematical Functions" by Abramovich and Stegun, June 1964, called AS in what follows.
Step 0: we start out with sn(u | m) where m<0
Step 1: removal of explicitly imaginary expressions
AS 16.20
sn (I u | m) = I sc (u | n), n = 1-m > 1
Step 2: transformation to have second parameter in the "natural" region between 0 and 1
AS 16.11 and 16.3
sc = sn/cn
sn (u | n) -> n^(-1/2) sn (u n^(1/2) | 1/n)
cn (u | n) -> dn (u n^(1/2) | 1/n)
sc (u | n) -> n^(-1/2) sn (u n^(1/2) | 1/n) / dn (u n^(1/2) | 1/n)
= n^(-1/2) sd (u n^(1/2) | 1/n)
Step 3: translation of the argument by the complete elliptical integral K to have
x'(0) = 0
AS 16.8
n^(-1/2) sd (u n^(1/2) + K (1/n) n^(1/2) | 1/n)
->
n^(-1/2) (1 - 1/n)^(-1/2) cn (u n^(1/2) | 1/n)
Section 3.2: Integration of the energy equation
An alternative approach to find the exact solution is the integration of the energy equation (the first integral of the original ODE) given by
eqw = x'[t]^2 == w + x[t]^2 - x[t]^4;
Section 3.2.1 The case w = 0
We start with the most simple case of zero energy (w = 0).
As we are looking for the soliton-solution which must have x'[0]==0.
But, unfortunately, DSolve can't deal with neither the initial condition x'[0]==0 or the equivalent x[0]==1.
Hence we DSolve without initial conditions and obtain
solw0 = x[t] /. DSolve[(eqw /. w -> 0), x[t], t]
(*
Out[351]= {-I Coth[t - C[1]] Sqrt[1 - Tanh[t - C[1]]^2],
I Coth[t - C[1]] Sqrt[1 - Tanh[t - C[1]]^2], -I Coth[t + C[1]] Sqrt[
1 - Tanh[t + C[1]]^2], I Coth[t + C[1]] Sqrt[1 - Tanh[t + C[1]]^2]}
*)
It is sufficient to take just one of these and define (letting also C[1]->z)
xw0[t_] = solw0[[1]] /. C[1] -> z
(*
Out[387]= -I Coth[t - z] Sqrt[1 - Tanh[t - z]^2]
*)
This looks like a completely imaginary solution. It took me some time to get this reconciled with the expected Sech. But wait! This need not be imaginary if we allow z to be complex, as we shall see here:
Let's try to formally calculate z from the initial condition x'[0] == 0
xw0p = D[xw0[t], t] /. t -> 0 // Simplify
(*
Out[384]= I Coth[z]^2 Sqrt[Sech[z]^2]
*)
Reduce[0 == xw0p, z ]
(* Out[385]= False *)
Bad luck! OK, obviously the square root prevents Reduce[] from giving results. Hence we try the equivalent condition
Reduce[0 == xw0p^2, z, GeneratedParameters -> K ]
(*
Out[386]= K[1] ∈ Integers && z == (I π)/2 + I π K[1]
*)
Hence we have found that z must be purely imaginary.
Taking z -> I π/2 gives
xw1[t_] = Simplify[xw0[t] /. z -> I π/2, t > 0]
(*
Out[394]= Sech[t]
*)
Section 3.2.2 The energy equation for $w\neq 0$
The energy integral, which gives t + const, can be done explicitly. Here is the antidervative
FullSimplify[Integrate[1/Sqrt[w + x^2 - x^4], x], w > 0]
(*
Out[158]= -((
I Sqrt[2] EllipticF[
I ArcSinh[(Sqrt[2] x)/Sqrt[-1 + Sqrt[1 + 4 w]]], (-1 - 2 w + Sqrt[
1 + 4 w])/(2 w)])/Sqrt[1 + Sqrt[1 + 4 w]])
*)
For -1/4 < w < 0 we have the same result but now I ArcSinh is replaced by ArcSin:
(*
Out[158]= -((
I Sqrt[2] EllipticF[
ArcSin[(Sqrt[2] x)/Sqrt[-1 + Sqrt[1 + 4 w]]], (-1 - 2 w + Sqrt[
1 + 4 w])/(2 w)])/Sqrt[1 + Sqrt[1 + 4 w]])
*)
Section 4: Generalizations and discussion
Section 4.1: General potential with "confinement"
Consider a potential repelling at $x = 0$ in the simplest forU = x^2, and going to +infinity for|x| >> 1.
So the particle is "globally" confined between the "walls" of the potential on both side of the x-axis.
First, locally, the equation of motion close to $x = 0$ is
eqx0 = x''[t] == x[t];
and the solution is
x[t] /. DSolve[eqx0, x[t], t]
(* Out[104]= {E^t C[1] + E^-t C[2]} *)
Letting C[1]->0 to keep the solution finite for t -> [Infinity] we obtain
xx0[t_] = a Exp[-t]
(* Out[111]= a E^-t *)
The (positive) constant "a" is the (negative) velocity of the point at t = 0.
a /. Solve[v0 == xx0'[0], a][[1]]
(* Out[120]= -v0 *)
Hence we have the following scenario:
The point starts at x = x1 = |v0| with a negative velocity v0 towards the repelling potential at x = 0. It will approach the summit of the potential exponentially for t ->[Infinity].
This is the typical behaviour of a loop swing and of situations leading to a sech(t) solution.
Section 4.2: The soliton solution for $x''=x-x^{2 n+1}$
Now let us consider the special confinement of the form x^(2k+1)
The energy integral for w = 0 can be calculated immediately:
fi = Integrate[1/Sqrt[u^2 - u^(2 k + 2)], {u, x, 1},
Assumptions -> {k \[Element] Integers, k > 0, 0 < x < 1}]
(* Out[6]= ArcTanh[Sqrt[1 - x^(2 k)]]/k *)
Inverting gives
sol = x /. FullSimplify[Solve[t == fi, x]] // Quiet
(* Out[7]= {(-Sqrt[Sech[k t]^2])^(1/k), (Sech[k t]^2)^((1/2)/k)} *)
of which the positive solution is adequate.
Hence we have found that for a positive integer n the equation
x''[t] == x[t] - x[t]^(2n+1)
has the solution
x[t] = Sech[n t]^(1/n)
k = 1 -> Sech[t]
k = 2 -> Sech[2t]^(1/2)
...
For t -> [Infinity] we find
Exp[-t] Limit[Exp[t] Sech[k t]^(1/k), t -> \[Infinity], Assumptions -> k > 0]
(* Out[27]= 2^(1/k) E^-t *)
Which is the exponential decay expected close to x = 0.
In the limit of large n we have
Limit[Sech[n t]^(1/n), n -> \[Infinity], Assumptions -> t > 0]
(* Out[] E^-t *)
n->[Infinity] means that the potential has hard walls at x = [PlusMinus] 1. Inside the box there is just the repelling potential from x = 0.
Hence the limit of large n again leads to the expected result, the simple exponential decay.
Section 4.3: The solution of $\nabla ^2u=u-u^3$ in 2D and 3D
For higher geometrical dimensions we confine ourselves to the case of radial symmetry.
Hence for general dimension D we consider
$$\frac{(D-1) u'(r)}{r}+u''(r)=u(r)-2 u(r)^3$$
It turns out that the case D = 2 is sufficient to grasp the general picture also for D > 2.
As before we translate the problem to the mechanical analogue and write the equation of motion as
eq = x''[t] + 1/t x'[t] == x[t] - 2 x[t]^3;
The new feature is that now due to the term with the first derivative we have friction, and this dissipative process forces the "particle" to come to a rest at t = oo.
Applying the usual procedure to derive energy conservation, i.e multiplying by x'[t] we can write the equation as
$$\frac{\partial w(t)}{\partial t}=-\frac{2 x'(t)^2}{t}$$
where the conservative part of the energy is
$$w(t)=x'(t)^2+x(t)^4-x(t)^2$$
We see that the quantity w decreases from its initial value to some finite value with the passage of time. The asymptotic value is given by x'' = x' = 0 which, from the equation of motion leads to xf = 1/Sqrt2and hence wf = -1/4.
We shall now solve the equation numerically. At first sight it might seem that the factor 1/t causes problems. But this is not the case. We shall, in the numeric treatment, start with a very small time [Epsilon] instead of zero.
The solution will be displayed in two forms: the usual dependence of position as a function of time and in a non-standard form which I like to call energy trajectory. The latter has been used in the discussion of the case D=1 above. But there the trajectories were simply oscillations between turning points. Here we see the non-trivial dissipative history of the particle in the energy picture.
The normal history is that after starting with the given energy the particle may make some oscillations in the big energy container but then "decide" to take the left or the right where it comes to rest asymptotically, hereby oscillating in the trough.
However with a suitable choice of the initial conditions the particle comes to rest at x = 0. Here asymptotically the particle moves "uphill" like a loop swing (as discussd earlier). The space time diagram is then a one-sided "soliton".
The following code provides the solution in this form. We give examples of two sets of parameters here. The reader is invited to study more cases.
The oscillating case
s = 10; (* scale factor for plots *)
\[Epsilon] = 10^-2;
tmax = 100;
x0 = 1.2(*1.5959735*);
v0 = -1; (*-1*)
ww[t_] := xx'[t]^2 - xx[t]^2 + xx[t]^4
w0 = v0^2 - x0^2 + x0^4; (* initial energy *)
(* numerical solution *)
xx[t_] = x[t] /.
NDSolve[(x''[t] + x'[t]/t == x[t] - 2 x[t]^3) && x[\[Epsilon]] == x0 &&
x'[\[Epsilon]] == v0, x[t], {t, \[Epsilon], tmax}][[1]];
(* plot x(t) *)
Plot[{-1/Sqrt[2], 1/Sqrt[2], xx[t]}, {t, \[Epsilon], tmax},
AxesLabel -> {"t", "x"},
PlotLabel ->
"Solution of \!$$\*SuperscriptBox[\(x$$, $$\[Prime]\[Prime]$$,\n\
MultilineFunction->None]\)(t)+\!$$\*FractionBox[\(\*SuperscriptBox[\"x\", \"\ \[Prime]\",\nMultilineFunction->None](t)$$, $$t$$]\)\[LongEqual]x(t)-2 \
x(t\!$$\*SuperscriptBox[\()$$, $$3$$]\), x(0) = " <> ToString[x0, InputForm] <>
", v(0) = " <> ToString[v0, InputForm] <>
"\nPosition x as a function of t\n",
PlotStyle -> {{Black, Thin, Dashed}, {Black, Thin, Dashed}, Red},
PlotRange -> {-1.1, 1.1 (* x0 *)}]
(* container plot for energy trajectory *)
s = 10;
xmax = x0; xmin = -xmax;
p0 = Plot[s (-x^2 + x^4), {x, xmin, xmax},
AxesLabel -> {"x", "w (red), U (blue)"},
PlotLabel ->
"Solution of \!$$\*SuperscriptBox[\(x$$, $$\[Prime]\[Prime]$$,\n\
MultilineFunction->None]\)(t)+\!$$\*FractionBox[\(\*SuperscriptBox[\"x\", \"\ \[Prime]\",\nMultilineFunction->None](t)$$, $$t$$]\)\[LongEqual]x(t)-2 \
x(t\!$$\*SuperscriptBox[\()$$, $$3$$]\), x(0) = " <> ToString[x0, InputForm] <>
", v(0) = " <> ToString[v0, InputForm] <> ", w(0) = " <>
ToString[w0, InputForm] <>
"\nEnergy trajectory: w(t) = \!$$\*SuperscriptBox[\(v$$, \
$$2$$]\)-\!$$\*SuperscriptBox[\(x$$, $$2$$]\)+\!$$\*SuperscriptBox[\(x$$, $$4\$$]\) as a function of x(t)\n"];
(* plot energy trajectory *)
p1 = ParametricPlot[{xx[t], s ww[t]}, {t, \[Epsilon], tmax},
PlotStyle -> {Thin, Red}];
Show[{p0, p1}]
The "soliton" case
By varying the initial position x0 we arrive at the picture where the particle moves to x = 0.
s = 10; (* scale factor for plots *)
\[Epsilon] = 10^-2;
tmax = 10 (* = 30 *);
x0 = 1.5959735;
v0 = -1; (*-1*)
ww[t_] := xx'[t]^2 - xx[t]^2 + xx[t]^4
w0 = v0^2 - x0^2 + x0^4; (* initial energy *)
(* numerical solution *)
xx[t_] = x[t] /.
NDSolve[(x''[t] + x'[t]/t == x[t] - 2 x[t]^3) && x[\[Epsilon]] == x0 &&
x'[\[Epsilon]] == v0, x[t], {t, \[Epsilon], tmax}][[1]];
(* plot x(t) *)
Plot[{-1/Sqrt[2], 1/Sqrt[2], xx[t]}, {t, \[Epsilon], tmax},
AxesLabel -> {"t", "x"},
PlotLabel ->
"Solution of \!$$\*SuperscriptBox[\(x$$, $$\[Prime]\[Prime]$$,\n\
MultilineFunction->None]\)(t)+\!$$\*FractionBox[\(\*SuperscriptBox[\"x\", \"\ \[Prime]\",\nMultilineFunction->None](t)$$, $$t$$]\)\[LongEqual]x(t)-2 \
x(t\!$$\*SuperscriptBox[\()$$, $$3$$]\), x(0) = " <> ToString[x0, InputForm] <>
", v(0) = " <> ToString[v0, InputForm] <>
"\nPosition x as a function of t\n",
PlotStyle -> {{Black, Thin, Dashed}, {Black, Thin, Dashed}, Red},
PlotRange -> {-2, 1.1 x0}]
(* container plot for energy trajectory *)
s = 10;
xmax = x0; xmin = -xmax;
p0 = Plot[s (-x^2 + x^4), {x, xmin, xmax},
AxesLabel -> {"x", "w (red), U (blue)"},
PlotLabel ->
"Solution of \!$$\*SuperscriptBox[\(x$$, $$\[Prime]\[Prime]$$,\n\
MultilineFunction->None]\)(t)+\!$$\*FractionBox[\(\*SuperscriptBox[\"x\", \"\ \[Prime]\",\nMultilineFunction->None](t)$$, $$t$$]\)\[LongEqual]x(t)-2 \
x(t\!$$\*SuperscriptBox[\()$$, $$3$$]\), x(0) = " <> ToString[x0, InputForm] <>
", v(0) = " <> ToString[v0, InputForm] <> ", w(0) = " <>
ToString[w0, InputForm] <>
"\nEnergy trajectory: w(t) = \!$$\*SuperscriptBox[\(v$$, \
$$2$$]\)-\!$$\*SuperscriptBox[\(x$$, $$2$$]\)+\!$$\*SuperscriptBox[\(x$$, $$4\$$]\) as a function of x(t)\n"];
(* plot energy trajectory *)
p1 = ParametricPlot[{xx[t], s ww[t]}, {t, \[Epsilon], tmax},
PlotStyle -> {Thin, Red}];
Show[{p0, p1}]
`
But we have cheated a bit: due to numerical instability the particle finally falls into the trough and oscillates about the asymptotic position. You can see this by increasing tmax to say 30.
Section 4.4: Discussion
Some remarks concerning the soliton solution should be made.
We have seen that non oscillating solutions occur if the particle climbes the potential hill at t = 0. Actually this is the tail of the asymptotic behaviour.
We can call the finite non zero part of this structure "soliton", although it must be noted that the position of the particle does not remain finite for t = 0 except for the case D = 1.
Another point is the symmetric structure such as in the Sech[t]. This appears if we include negative times in our consideration. If we would do this for D > 1 we would also have the symmetry.
Concluding we have shown that the mechanical analogue of our ODE can be a helpful device to understand the character of the Solutions. We also have extended this approach to disspative systems by introducing the concept of an energy trajectory depending on the time as a parameter.
• There ought to be a Textbook badge. :) Sorry, can't upvote more than once. Cheers! – Michael E2 Nov 25 '15 at 21:13
• @Michael E2: thank you very much. It was great fun for me to take the opportunity to delve into Jacobi elliptic functions, elliptic functions, inverse functions, non linear differential equations, and I have still some points to make here. Don't worry, not too many ;-) By the way, I have found out (tacitly) that, despite of its name, a non linear Schrödinger equation has nothing to do with quantum mechanics. – Dr. Wolfgang Hintze Nov 25 '15 at 22:57
• @Dr.WolfgangHintze I am so waiting for section 4!!! By the way, the ref in question is this article: J. Zittartz and J. S. Langer Phys. Rev. 148, 741, 1966 where also the different coefficient of the 2D problem is clarified. I approve the Textbook badge. – user50473 Dec 1 '15 at 16:18
• @user50473: oh yes, you are right, and my bad concious is there to remind me. I ask politely for some more patience. – Dr. Wolfgang Hintze Dec 1 '15 at 20:21
• @Dr.WolfgangHintze I was reading again all this, just wondering whether you abandoned the project. Would really love to see how this ends. – user50473 Mar 1 '16 at 17:09 | 2019-12-06T15:44:37 | {
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http://math.stackexchange.com/questions/125954/bounding-n-choose-k | # Bounding ${n \choose k}$
Let $k > 0$, and $n > 2k$. Why is it necessarily true that $${n \choose k} > \frac{n^k}{2^k k!}$$ And is the condition $n > 2k$ necessary for this inequality to hold?
-
The condition is not necessary. But it makes deriving $n(n-1)(n-2)\cdots(n-k+1)>n^k/2^k$ easy. – André Nicolas Mar 29 '12 at 14:42
${n \choose k}$ increases up to $k = \lfloor n/2 \rfloor$, then goes down. So you only really need to worry about the left side up to the max. – Mitch Mar 29 '12 at 14:57
@AndréNicolas What about $\binom{6}{6} = 1 < \frac{46656}{46080} = \frac{6^6}{2^6 6!}$? – dtldarek Mar 29 '12 at 15:24
@dtldarek: Certainly the inequality does not hold unconditionally. – André Nicolas Mar 29 '12 at 15:34
@AndréNicolas Yeah, you are right, I forgot that necessary has very precise meaning in mathematics. +1 for you :-) – dtldarek Mar 29 '12 at 15:50
We have $$\binom{n}{k} = \frac{n!}{(n-k)!k!} \ge \frac{(n-k)^n}{(n-k)^{n-k}k!} \ge \frac{(n-k)^k}{k!} \ge \frac{(n/2)^k}{k!}$$ using $n > 2k \implies n-k \ge n/2$. In the more general case, as noted in the comments, you will need a more detailed analysis, e.g., using Stirling's approximation for $n!$.
-
How do you prove $\frac{n!}{(n-k)!} \geq \frac{(n-k)^n}{(n-k)^{n-k}}$? – jamaicanworm Mar 29 '12 at 15:54
Using the definition of n! and the simple inequality chain $n-k+i \ge n-k \ge n-k-i$ for $i \le k$. – Johannes Kloos Mar 29 '12 at 16:45
If $k\le\frac12n$, then $n-k+1>\frac12n$. Therefore, \begin{align} \binom{n}{k} &=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}\\ &>\frac{(n/2)^k}{k!}\\ &=\frac{n^k}{2^kk!}\tag{1} \end{align} However, the condition that $k<\frac12n$ can be extended to $k\le\frac34n$ by noting that for $0\le j\le k-1$, $(n-j)\color{red}{(n+j-k+1)}\ge n\color{red}{(n-k+1)}>n^2/4$. \begin{align} \binom{n}{k}^2 &=\frac{n\color{red}{(n-k+1)}(n-1)\color{red}{(n-k+2)}(n-2)\dots(n-k+1)\color{red}{n}}{k!^2}\\ &>\frac{(n^2/4)^k}{k!^2}\tag{2} \end{align} Taking the square root of $(2)$ yields $$\binom{n}{k}>\frac{n^k}{2^kk!}\tag{3}$$ Take It To The Limit:
Above, it is shown that the condition $k<\frac12n$ in the question can be extended to $k\le\frac34n$. One might ask how far this might be pushed. That is, what is the largest $\alpha$ so that $k<\alpha n$ implies $(3)$.
Multiplying $(3)$ by $k!$ and taking the $\log$ of both sides yields $$\sum_{j=0}^{k-1}\log(n-j)>k\log(n)-k\log(2)\tag{4}$$ Noting that $$\sum_{j=0}^{k-1}\log(n-j)\ge\int_{n-k}^n\log(x)\,\mathrm{d}x\tag{5}$$ We get that $$\int_{n-k}^n\log(x)\,\mathrm{d}x>k\log(n)-k\log(2)\tag{7}$$ implies $(3)$. Computing the integral in $(7)$ yields $$n\log(n)-n-(n-k)\log(n-k)+(n-k)>k\log(n)-k\log(2)\tag{8}$$ Subtracting $k\log(n)-k$ from both sides and dividing by $k$ yields $$\frac{n-k}{k}\log\left(\frac{n}{n-k}\right)>1-\log(2)\tag{9}$$ Substituting $\alpha=k/n$ yields $$\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)>1-\log(2)\tag{10}$$ Here is a plot of $\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)$ and $1-\log(2)$:
$\hspace{4cm}$
We can solve $\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)=\beta$ using the Lambert W function: \begin{align} \frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)&=\beta\tag{11}\\ (1-\alpha)^{1-\alpha}e^\beta&=e^{(1-\alpha)\beta}\tag{12}\\ (1-\alpha)e^{\beta/(1-\alpha)}&=e^\beta\tag{13}\\ -\beta/(1-\alpha)e^{-\beta/(1-\alpha)}&=-\beta e^{-\beta}\tag{14}\\ -\frac{\beta}{1-\alpha}&=W\left(-\beta e^{-\beta}\right)\tag{15}\\ \alpha&=1+\frac{\beta}{W\left(-\beta e^{-\beta}\right)}\tag{16} \end{align} $(12)$: multiply by $-\alpha$, exponentiate, multiply by $e^\beta$
$(13)$: raise to the $1/(1-\alpha)$ power
$(14)$: take reciprocals, multiply by $-\beta$
$(15)$: apply $W$
$(16)$: algebraically solve for $\alpha$
Plugging $\beta=1-\log(2)$ into $(16)$ gives \begin{align} \alpha &=1+\frac{1-\log(2)}{W\left(-\frac2e(1-\log(2))\right)}\\ &\approx 0.868708579475419252 \end{align} Thus, if $k<\alpha n$, $(3)$ holds.
Example:
$\left.\binom{10000}{8687}\middle/\frac{1000^{8687}}{2^{8687}8687!}\right.=3.095040785$, but $\left.\binom{10000}{8688}\middle/\frac{1000^{8688}}{2^{8688}8688!}\right.=0.8127577101$
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We have \begin{align*} \binom{n}{k} &> \frac{n^k}{2^k k!} \\ \frac{n!}{(n-k)!k!} &> \frac{n^k}{2^k k!} \\ \frac{n!}{(n-k)!} &> \frac{n^k}{2^k} \\ \prod_{i=n-k+1}^n i &> \left(\frac{n}{2}\right)^k \\ (n-k+1)^k &> \left(\frac{n}{2}\right)^k \\ n-k+1 &> \frac{n}{2} \\ n-2k+2 &> 0 \\ n+2 &> 2k \\ \end{align*}
Edit: $n > 2k$ is not a necessary condition (e.g. $n \geq 2k$ is good as well), but you need something, in general case the inequality does not hold, for example for $n \geq 6$ we have $\binom{n}{n} = 1 \leq \frac{n^n}{2^n n!}$.
- | 2015-01-31T05:32:20 | {
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https://math.stackexchange.com/questions/784508/integral-int-0-pi-2-frac-sin3-x-log-sin-x-sqrt1-sin2-xdx-frac/784682 | # Integral $\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}$
Hi I am trying to prove$$I:=\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}.$$ Thanks.
I am possibly trying to simplify this to obtain something like $2\int_0^{\pi/2} \log \sin x\, dx=-\pi \ln 2$ since this is easily integrable. However when I try to simplify the terms $$\frac{\sin^3 x}{\sqrt {1+\sin^2 x}}$$ I obtain a more complicated integrand. I am not sure how else to go about this one. I was trying to possibly write $$I(a)=\int_0^{\pi/2} \frac{\sin^3 a x\log \sin x}{\sqrt{1+\sin^2 x}}dx,\quad I'(a)=\int_0^{\pi/2} \frac{\partial}{\partial a}\left(\frac{\sin^3 ax\log \sin x}{\sqrt{1+\sin^2 x}}\right)\, dx,$$ but this didn't simplify anything for me.
I also tried the substitution $y=\sin^2 x$, but couldn't manage to get an integral because of the $\sin 2x$ from the derivative.
• Perhaps you can use IBP by taking $u=\ln\sin x$? Sorry I can't give you a complete answer for this 'claim' because I'm in the middle of the class now. :P – Tunk-Fey May 7 '14 at 5:36
The change of variables $t=\sin x$ yields $$I=\int_0^1\frac{t^3\ln t}{\sqrt{1+t^2}}\frac{dt}{\sqrt{1-t^2}} =\int_0^1\frac{t^3\ln t}{\sqrt{1-t^4}} dt$$ Then setting $t^4=u$ simplifies to \eqalign{ I&=\frac{1}{16}\int_0^1\frac{1}{\sqrt{1-u}}\ln u\, du\cr &=\left[\frac{1}{8}(1-\sqrt{1-u})\ln u\right]_0^1-\frac{1}{8}\int_0^1\frac{1-\sqrt{1-u}}{u} du\cr &=-\frac{1}{8}\int_0^1\frac{1}{1+\sqrt{1-u}} du;\qquad v\leftarrow1+\sqrt{1-u}\cr &=\frac{1}{4}\int_1^2\frac{1-v}{v} du=\frac{\ln 2-1}{4}.} and we are done. $\qquad\square$
• I was just trying to connect to wifi to post my answer and you beat me! I hate how quick you have to be on this website. – Bennett Gardiner May 7 '14 at 6:37
• Sorry. If I knew, I would have waited. – Omran Kouba May 7 '14 at 6:40
• How did the t substitution call for a 1/16 factor? Deriviating $u^4$ is $4u^3$ so shouldn't 1/4 be the balancing factor? -EDIT: Never mind I forgot the ln needed to be stripped of its 1/4 exponent. Silly me. – Nicholas Pipitone May 7 '14 at 7:03
• and the $t$ in the log!. – Omran Kouba May 7 '14 at 7:04
The integral screams for the substitution $u = \sin x$, which transforms it into $$\int_0^1\frac{u^3\log u}{\sqrt{1-u^4}} \ \mathrm{d}u,$$ another, trickier substitution, $w^2 = 1-u^4$ gives $$\frac{1}{2} \int_0^1 \log (1-w^2)^{1/4} \ \mathrm{d}w = {1\over 8} \int_0^1 \log(1+w) +\log (1-w) \ \mathrm{d}w = \frac{\ln 2 -1}{4}.$$
We can derive a more general result:
Consider the integral
\begin{align} I(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a}{\sqrt{1+\sin(x)^2}}\, dx\\ &=\int_0^{1}\, \frac{t^a}{\sqrt{1-t^4}}\, dt \tag{subst. $t=\sin(x)$}\\ &=\frac{1}{4}\int_0^{1}\, u^{(a-3)/4} (1-u)^{-1/2}\, du \tag{subst. $t^4=u$}\\ &=\frac{1}{4}\mathrm B\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{1} \end{align}
The third line represents a form of Beta function.
\begin{align} \therefore I'(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a\, \log{\sin{x}}}{\sqrt{1+\sin(x)^2}}\, dx\\ &=\frac{1}{16} \, {\left(\psi_0\left(\frac{a+1}{4} \right)-\psi_0\left(\frac{a+3}{4} \right) \right)} {\rm B}\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{$\frac{d}{da} (1)$}\\ \implies I'(3)&=\frac{\log{2}-1}{4} \end{align}
References: Beta function and Polygamma function | 2019-11-15T07:52:32 | {
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http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html | # The pigeonhole principle forms
Form,isarephrasingofthisstatement propositionphp1 (thepigeonholeprinciple,simpleversion) (the pigeonhole principle) ifn or more pigeons are distributed and placed in six pigeonholes, some pigeonhole contains two numbers by the way the. 4 the simple form of the pigeonhole principle is obtained from the strong form from math 3343 at the hong kong university of science and technology. Putnam training pigeonhole principle 2 11 prove that every convex polyhedron has at least two faces with the same number of edges. In mathematics , the pigeonhole principle states that if n items are put into m containers, with n m , then at least one container must contain more than one itemthis theorem is exemplified in real life by truisms like in any group of three gloves there must be at least two left gloves or two right gloves. The pigeonhole principle sounds trifling but its uses are deceiving astonishing thus, in our project, we intend to learn and discover more about the pigeonhole principle and illustrate its numerous interesting applications in our daily life.
The inclusion-exclusion principle and the previous parts can be used to solve this question number of arrangements with ben sitting next to alyssa or carlos is the number of arrangements with ben next to alyssa (2$$\cdot$$6) plus number of arrangements with ben next to carlos (2$$\cdot$$6) minus number of arrangements with ben next to both. Pigeonhole principle strong form – theorem: let q 1 , q 2 , , q n be positive integers if q 1 + q 2 + + q n − n + 1 objects are put into n boxes, then either the 1st box contains at least q 1 objects, or the 2nd box contains at least q 2 objects, , the nth box contains at least q n objects. The pigeons for each of these subsets, the sum of the numbers in the subset will be the pigeonhole since our numbers are all between 0 and 107, the sum of thirty of them is at most 3 108, which is less than 230 1 ˇ109thus, since there are more subsets (pigeons) than sums (holes), there. Pigeon-hole principle if nm pigeons are put into m pigeonholes, th eres a hole with more than one pigeon 2 alternative forms• if n objects are to be allocated to m containers, then at least one container must hold at least ceil(n/m) objects.
The pigeonhole principle can be phrased in terms of labels if more than n objects are to be assigned labels from a set of n labels, then there is sure to be two objects with the same label. A part of it will concentrate on the pigeonhole principle thus, i need some hard to very hard problems in the subject to solve i would be thankful if you can send me links\books\or just a lone problem. Echelon forms and the general solution to ax = b learning goals: to see that elimination is still the technique that gets our answers, but that going just by the pigeonhole principle what we will do is elimination and row swaps to clear out columns as usual if a column is missing a pivot—there. “pigeonhole live” means the pigeonhole live website and service “ pigeonhole live platform ” means the “pigeonhole live” real-time audience engagement platform and the “dashboard account.
Pigeonhole principle the pigeonhole principle (also known as the dirichlet box principle , dirichlet principle or box principle ) states that if or more pigeons are placed in holes, then one hole must contain two or more pigeons. Pigeonhole principle kin-yin li what in the world is the pigeonhole principle well this famous principle for each one, form a box containing the number and all powers of 2 times the number so the first box contains 1,2,4,8, 16, and the next box contains 3,6,12,24,48, and so on then among the 51 numbers chosen, the pigeonhole. In mathematics, the pigeonhole principle states that if items are put into containers, with , then at least one container must contain more than one item this theorem is exemplified in real life by truisms like in any group of three gloves there must be at least two left gloves or at least two right gloves.
## The pigeonhole principle forms
As @cuddlycuttlefish pointed out the pigeonhole principle with xor in it is clearly false if you replace the xor with $\vee$ and $\leftarrow \rightarrow$ with $\wedge$, however, i think your proof becomes correct. E pigeonhole principle basic geometric problems 1 five darts are thrown at a square target measuring 14 inches on a side prove that two of them must be at a distance no more than 10 inches apart. Pigeonhole (plural pigeonholes) a nook in a desk for holding papers one of an array of compartments for sorting post, messages, etc at an office, or college (for example. Pigeonhole principle to the study of efficient provability of major open problems in computational complexity, as well as some of its generaliza- tions in the form of general matching principles.
• Pigeonhole principle the following general principle was formulated by the famous german mathematician dirichlet (1805-1859): pigeonhole principle: suppose you have kpigeonholes and npigeons to be placed in them.
• Tion for the ith level of polynomial hierarchy this is because s3 2 can do the necessary minimization and paris et al [21], as presented in kraj cek [15], have shown that s3 2 proves the weak pigeonhole principle for p-time functions.
• One of the famous (although often neglected in the instructional program) problem- solving techniques is to consider the pigeonhole principle which is a powerful tool used in combinatorial mathin its simplest form, t he pigeonhole principle states that if more than n pigeons are placed into n pigeonholes, some pigeonhole must contain more than one pigeon.
By the generalized pigeonhole principle (contrapositive form), this would imply that the total number of people is at most 3 26 = 78 but this contradicts the fact that there are 85 people in all hence at least 4 people share a last initial 27. In elementary mathematics the strong form of the pigeonhole principle is most often applied in the special case when q 1 = q 2 = = q n = r in this case the principle becomes: • if n ( r - 1) + 1 objects are put into n boxes, then at least one of the boxes contains r or more of the objects. In many situations, the naive form of the pigeonhole principle can be applied directly in most problems, the objects and boxes are fairly obvious a box contains three pairs of socks colored red, blue, and green, respectively. (this story is an example of the second pigeonhole principle) 3 fundamental proof 31 first pigeonhole principle if n items are put into m pigeonholes with n m(m, n ∈ n ∗ ), then at least one pigeonhole must contain more than one item.
The pigeonhole principle forms
Rated 4/5 based on 46 review
2018. | 2018-11-19T01:49:18 | {
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http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/ | # For a bounded subset AA of R\mathbb{R}, sup(A)\sup(A) is in the closure of AA
Consider A⊂RA\subset \mathbb{R}, where AA is bounded. Since AA is bounded, sup(A)\sup(A) exists. To show that sup(A)∈Closure(A)\sup(A)\in Closure(A), we need to show that sup(A)\sup(A) is a limit point of AA. Let sup(A)=s∈R\sup(A)=s\in\mathbb{R}. If AA is a finite set then it is closed, so the closure of AA is AA, and sup(A)∈Closure(A)\sup(A)\in Closure(A). If AA is not finite, consider the sequence {s−1/n}∩A\{s-1/n\}\cap A. This sequence converges to ss and is entirely contained in AA.
But I have some doubts about my proof approach because if we consider A:=(0,1/2)∪{5}A:=(0,1/2)\cup\{5\}, the above sequence will not intersect AA. In this case, however, one can argue that {5}\{5\} is a singleton, which is closed, and so {5}\{5\} must be in the closure of AA. In all other cases, it seems, the above sequence will be contained in AA and will converge to ss. Is this true?
Another approach I could use is this. Suppose s∉Closure(A)s\not\in Closure(A). Then s∈R∖Closure(A)s\in \mathbb{R}\backslash Closure(A). Since Closure(A)Closure(A) is closed, R∖Closure(A)\mathbb{R}\backslash Closure(A) is open, thus ∃\exists an r>0r>0 such that the open ball Br(s)∩̸Closure(A)B_r(s)\not\cap Closure(A). Which is impossible if ss is not a singleton. If ss is a singleton, then s\in Closure(A)s\in Closure(A) by definition. Therefore, s\in Closure(A)s\in Closure(A) in all cases.
=================
1
The intersection \{s – 1/n \} \cap A\{s – 1/n \} \cap A can just be empty.
– xyzzyz
Oct 20 at 23:41
=================
3
=================
To show that a real number xx lies in the closure \bar{A}\bar{A} of AA, you have to show that either x\in Ax\in A or that xx is a limit point of AA. While some points in AA are limit points of AA, your example A=(0,\frac{1}{2})\cup\{5\}A=(0,\frac{1}{2})\cup\{5\} demonstrates that not every element of \bar{A}\bar{A} is a limit point of AA, as it could be an isolated point in AA.
So to show that \alpha=\sup(A)\in\bar{A}\alpha=\sup(A)\in\bar{A}, we have to show that either \alpha\in A\alpha\in A or that \alpha\alpha is a limit point of AA. If \alpha\in A\alpha\in A then we’re done, so suppose that \alpha\not\in A\alpha\not\in A.
Since \alpha-1\alpha-1 is not a upper bound for AA, it follows that we can choose some s_1\in As_1\in A such that \alpha\geq s_1>\alpha-1\alpha\geq s_1>\alpha-1. Now s_1\neq \alphas_1\neq \alpha because \alpha\not\in A\alpha\not\in A, so since neither \alpha-\frac{1}{2}\alpha-\frac{1}{2} nor s_1s_1 is an upper bound for AA, we can choose s_2\in As_2\in A such that \alpha \geq s_2>\max\{s_1,\alpha-\frac{1}{2}\}\alpha \geq s_2>\max\{s_1,\alpha-\frac{1}{2}\}.
Proceeding inductively, we can construct a sequence \{s_n\}\{s_n\} such that \alpha\geq s_n>\max\{s_{n-1},\alpha-\frac{1}{n}\}\alpha\geq s_n>\max\{s_{n-1},\alpha-\frac{1}{n}\} for all n\geq 2n\geq 2. This means that \{s_n\}\{s_n\} is a sequence of distinct elements of AA which converges to \alpha\alpha, hence \alpha\alpha is a limit point of AA.
The definition of the closure of AA is that it contains all limit points of AA. But aren’t all points of AA also its limit points (not necessarily all its limit points)?
– sequence
Oct 21 at 0:20
No, not if you require a limit point to be the limit of a sequence of distinct elements of AA (which is the definition I’m familiar with). For instance, 55 is not a limit point of (1,\frac{1}{2})\cup\{5\}(1,\frac{1}{2})\cup\{5\}.
– carmichael561
Oct 21 at 0:22
Isn’t \{5\}\{5\} a limit point of the sequence \{5,5,5,…,5\}\{5,5,5,…,5\}?
– sequence
Oct 21 at 0:28
That’s not a sequence of distinct elements.
– carmichael561
Oct 21 at 0:29
I see. Looks like the definition I was given is different.
– sequence
Oct 21 at 0:32
Because A is a bounded subset of \mathbb{R}\mathbb{R}, it has a supremum. Then see this answer. The first part of the proof, which is not shown, assumed the case where y \in\in E.
See this answer: Supremum of closed sets
By definition of sup(A), every open ball(“segment in R”) around sup(A) contains at least one element from A, thus making it by definition belonging to Closure(A). | 2018-06-24T18:20:06 | {
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https://www.physicsforums.com/threads/probability-question.876965/ | # B Probability question
1. Jun 27, 2016
### DiracPool
How do I express the following scenario mathematically?
I have access to CNN news online from two different internet sites, each of which have about an 80% reliability of actually providing the feed when I log onto the site. If I only had access to one of the sites, I'd know that I had an 80% chance that I'd get a live feed when I logged in. How does this figure change when I have two sites, both with that 80% reliability.
The temptation is to multiply the two probabilities together, i.e., .8 * .8 = .64 But that can't be right because the probability has to be greater than .8
Can someone set this scenario up for me mathematically?
2. Jun 27, 2016
### BiGyElLoWhAt
If you multiply them, you're asking for the probability of getting a news feed from one site and then the other.
I think I'm missing some detail, though. So you have to choose a site, and then log in? Or what? Do both sites pop up when you log onto whatever you're logging into?
3. Jun 27, 2016
### BiGyElLoWhAt
If that's the case, you're looking for the probability of A and/or B = A or B + A and B. At least I would think so. You only care that you get at least one, but it's also a possibility that you get two, so you need the probability of getting 1 + the probability of getting 2.
4. Jun 27, 2016
### BiGyElLoWhAt
Ok, so this is what I'm thinking, I get a value of 96% when I do this.
$A$ is getting the feed from A $A'$ is not getting the feed from A.
$P = AB' + A'B + AB$
5. Jun 27, 2016
### stevendaryl
Staff Emeritus
Let $A$ be the outcome "The first site works". Let $B$ be the outcome "The second site works". Then there are 4 possible joint outcomes: $A \wedge B, A \wedge \neg B, \neg A \wedge B, \neg A \wedge \neg B$ (where $\neg$ means "not", and $\wedge$ means "and"). So the probability that at least one of the sites works is:
$P(A \vee B) = P(A \wedge B) + P(A \wedge \neg B) + P(\neg A \wedge B)$
where $\vee$ means "or".
So do you know how to compute those three probabilities on the right side of the equals?
6. Jun 27, 2016
### DiracPool
No I don't. But for some reason I thought intuitively the answer was what Bigyellowhat stated, 96%
I don't know where I came up with that figure, and when I tried to think about how the probability would be characterized mathematically, I drew a blank. This is especially disturbing since I just took the GRE last Summer and had these type of probability questions drilled into my head. I've already forgotten how to do them
7. Jun 27, 2016
### BiGyElLoWhAt
What's the probability of A but not B?
As for where I got that number, it's the same as what Steven Darryl is doing.
8. Jun 27, 2016
### stevendaryl
Staff Emeritus
You have to use some laws of probability:
$P(\neg X) = 1 - P(X)$
$P(X \wedge Y) = P(X) \cdot P(Y)$ (if the probabilities are independent)
9. Jun 27, 2016
### DiracPool
That looks like it adds up to .96! If there's one thing I still know how to do, it's add decimals
.64 + .16 + .16 =.96 Am I right? I think the probabilities are independent. If you have cable or satellite, you pretty much have a 100% reliability that you will get a CNN feed. But the two sites I get my CNN news feed from have essentially an 80% reliability independent of one another. I was just wondering how much closer to 100% my reliability was by having the two options.
10. Jun 27, 2016
### MrAnchovy
These answers seem somewhat over-complicated. Assuming the probabilities of NOT getting a feed from one site are independently 1 - 0.8 = 0.2, the probability of not getting a feed from both sites is 0.2 x 0.2 = 0.04. So the probability of this not happening (i.e. getting a feed from at least one site) is 1 - 0.04 = 0.96.
[edit - added this]And if you have three sites available, the probability of not getting a feed from any is 1 - 0.23 = 99.2%.
11. Jun 27, 2016
### PeroK
Yes, it's 96% if the two are independent. The simplest calculation is how often you have neither. That's 0.2 x 0.2 = 0.04.
0.8 x 0.8 = 0.64 is the probability you have both.
And 0.8 x 0.2 = 0.16 is the probability you have the first but not the second. Similarly it's the same probability you have the second but not the first. Hence:
64% both feeds available
32% only one feed available
4% neither feed available
12. Jun 27, 2016
### BiGyElLoWhAt
I suppose I overlooked that, no? lol
13. Jun 27, 2016
### DiracPool
I like how you broke that down PeroK, thanks.
14. Jun 27, 2016
### micromass
This is studied in reliability analysis. Basically, you have some subdevices (here: the two sites) which are operational which probability $p$. What is asked here is what is the probability that the entire device is operational.
If the devices are connected in series, then the probability is $p^n$ where $n$ is the number of devices.
If the devices are connected in parallel, then the probability is $1 - (1- p)^n$.
You can generalize this to $k$ out of $n$ systems where the device works if $k$ out of $n$ subdevices work. This requires the binomial distribution. Of course, you can go even more complicated than this. | 2018-01-20T14:01:01 | {
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https://math.stackexchange.com/questions/882147/find-a-conformal-map-from-semi-disc-onto-unit-disc/882199 | # Find a conformal map from semi-disc onto unit disc
This comes straight from Conway's Complex Analysis, VII.4, exercise 4.
Find an analytic function $f$ which maps $G:=$ {${z: |z| < 1, Re(z) > 0}$} onto $B(0; 1)$ in a one-one fashion.
$B(0;1)$ is the open unit disc.
My first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \subset B(0;1)$. Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.
My next thought is to modify $z^2$, something like $a(z-h)^2+k$. I've yet to work out the details, but my gut tells me this isn't the right idea.
I've been teaching myself conformal maps in preparation for a qualifying exam. So, if there's a shockingly basic, obvious solution... please patronize me.
• One thought (which I mostly pose from intuition) is seek a conformal map which maps $G$ to the right half-plane; from there it's not hard to map the right-half plane to $B(0;1)$ by a rotation. – Semiclassical Jul 30 '14 at 2:20
## 2 Answers
The following trick works for any region bounded by two circular arcs (or a circular arc and a line).
Find the points of intersection of the arc and the line. (Here, they're $i$ and $-i$.) Now pick a Mobius transformation that takes one of those points to $0$ and the other to $\infty$; here $z \mapsto \frac{z-i}{z+i}$ works. Then the arc and the line go to two rays (because a Mobius transformation sends circles in $S^2$ to circles in $S^2$, and the only circles in $S^2$ that goes through both $0$ and $\infty$ is a line in $\Bbb C$), both starting at $0$ and going off the $\infty$. Your domain maps to the region bounded by these two rays.
Let's compute the rays. It suffices to find where a single point on each arc maps; if $z_0$ is on the arc, the ray will be $\{f(z_0)t : 0 \leq t < \infty\}$. I say we pick $0$ to be our point of choice on $Re(z) = 0$ and $1$ to be the point of choice for the circular arc. These are mapped to $-1$ and $-I$ respectively; so our two arcs are the negative real axis and the negative imaginary axis. I'd like the "lower" arc to be the positive real axis, so let's multiply by $-1$ to do this.
So we have a conformal map from your half-disc to the upper-right quadrant given by $z \mapsto -\frac{z-i}{z+i}$. The upper half-plane is nicer, so let's map to that by squaring; now we have a map to the upper half plane given by $z \mapsto \frac{(z-i)^2}{(z+i)^2}$. (For other regions bounded by rays that make different angles, you get to the upper half plane by a $z \mapsto z^\beta$ for the appropriate $\beta$.)
Now there's a standard map from the upper half plane to the unit disc given by $z \mapsto \frac{z-i}{z+i}$. Composing this with our last map gives us a map from the semi-disc to the unit disc, given by $$z \mapsto -i\frac{z^2+2z-1}{z^2-2z-1}.$$
• This is more insightful. I'll delete my answer. – user138530 Jul 30 '14 at 3:06
• For fun, I added the images of your mappings as a community wiki answer. Enjoy! – Semiclassical Jul 30 '14 at 3:15
• @ChristianRemling I appreciate the compliment! – user98602 Jul 30 '14 at 3:15
• It might be worth noting that, even before computing the two rays (the negative real and imaginary axes), you knew that they had to be perpendicular, because conformal maps preserve angles. So you already know that the next step should be squaring and that it would produce a half-plane. The details of the rays and the half-plane are needed only at the very end, when you map the half-plane to the disk. – Andreas Blass Jul 30 '14 at 4:04
• @AndreasBlass That's definitely worth noting - thanks for commenting. – user98602 Jul 30 '14 at 4:31
As a supplement to the fine answers provided, here are the pictures of the conformal maps themselves:
We start with the right half of the unit disk, map it to the upper right quadrant, square this to the upper half-plane, and finally rotate onto the unit disk.
Per requests for code - the basic pattern to generate the image of a polar region $$a\leq r \leq b$$ and $$\alpha \leq \theta \leq \beta$$ under a function $$f$$ in Mathematica is:
ParametricPlot[{Re[f[r*Exp[I*theta]]], Im[f[r*Exp[I*theta]]]},
{r, a, b}, {theta, alpha, beta}, Mesh -> True]
The pictures above may be generated with the following code block:
phi1[z_] = -(z - I)/(z + I);
f[z_] = z^2;
phi2[z_] = (z - I)/(z + I);
nestedFunctions =
Composition @@@ Table[Take[{phi2, f, phi1}, -k], {k, 0, 3}];
pics = Table[
ParametricPlot[{Re[g[r*Exp[I*theta]]], Im[g[r*Exp[I*theta]]]},
{r, 0, 1}, {theta, -Pi/2, Pi/2},
Mesh -> True, PlotRange -> 2, ImageSize -> 360],
{g, nestedFunctions}];
Grid[Partition[pics, 2]]
You can find a lot more information on visualizing complex function with Mathematica in this notebook.
• It appears I can only choose one answer, but I think these images are just as essential to the solution as the explanation. With what program did you generate them? – artificial_moonlet Jul 30 '14 at 15:06
• It's from Mathematica. I'll incorporate the code for them when I get the chance. (Also, I should note that this really isn't a proper answer since I didn't remember the proper conformal maps until I saw the other answers. Hence why I did this as a community wiki answer.) – Semiclassical Jul 30 '14 at 15:08
• As a rookie mathematica user I would find it usefully to see the code for these types of mappings. – coffeebelly Oct 4 '14 at 19:10
• I wonder if you can still recall the code :) – snulty Jan 11 '17 at 8:15
• @snulty Code has been posted, if you are still interested. – Mark McClure Dec 8 '18 at 17:08 | 2021-03-07T16:22:24 | {
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http://mathhelpforum.com/algebra/148256-assistance-sketching-relations.html | # Thread: Assistance with sketching relations
1. ## Assistance with sketching relations
The question:
Sketch the set of points (x, y) which satisfy the following relations:
0 <= y <= 2x and 0 <= x <=2
I have no idea where to start. Is there a way to simultaneously equate them? Is the process supposed to be intuitive? I can't get my head around it.
Any guidance would be greatly appreciated.
2. Originally Posted by Glitch
The question:
Sketch the set of points (x, y) which satisfy the following relations:
0 <= y <= 2x and 0 <= x <=2
I have no idea where to start. Is there a way to simultaneously equate them? Is the process supposed to be intuitive? I can't get my head around it.
Any guidance would be greatly appreciated.
Hi Glitch,
If x (domain) is between 0 and 2, inclusive,
then y (range) is between 0 and 4, inclusive.
Sketch 0 to 2 on the x-axis, and 0 to 4 on the y-axis.
Complete the rectangle with corners at (0, 0), (0, 4), (2, 4), (2, 0).
All the points inside the rectangle, including the corners and edges, satisfy the given conditions.
So when determining the range, how did you know it was between 0 and 4 inclusive? Did you substitute the upper bound of x?
4. Originally Posted by Glitch
So when determining the range, how did you know it was between 0 and 4 inclusive? Did you substitute the upper bound of x?
Yes, I did.
5. Hello, Glitch!
Sketch the set of points (x, y) which satisfy: . $\begin{array}{ccccc}0 & \leq & y & \le &2x \\ 0 & \le &x & \le &2 \end{array}$
$0 \:\le\: y \:\le\:2x$ means $y$ is between the lines: . $\begin{Bmatrix}y &=& 2x \\ y &=& 0 \end{Bmatrix}$
The graph looks like this:
Code:
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| *:::::::::::::::::
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- - * - - - - - - - - - - - -
0
$0\:\le\: x \:\le\:2$ means $x$ is between the lines: . $\begin{Bmatrix}x &=&2 \\ x &=& 0\end{Bmatrix}$
Code:
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|:::::::::|
|:::::::::|
|:::::::::|
|:::::::::|
|:::::::::|
- - * - - - - + - - - - - - -
0 2
The region satisfying both inequalities is the area shaded twice.
Code:
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| *
| *:|
| *:::|
| *:::::|
| *:::::::|
- - * - - - - + - - -
0 2 | 2017-06-24T15:54:30 | {
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http://math.stackexchange.com/questions/154991/find-the-remainder-when-1214-1-is-divided-by-13 | # Find the remainder when $12!^{14!} +1$ is divided by $13$
Find the remainder when $12!^{14!} +1$ is divided by $13$
I faced this problem in one of my recent exam. It is reminiscent of Wilson's theorem. So, I was convinced that $12! \equiv -1 \pmod {13}$ after this I did some test on the exponent and it seems like $12!^{n!} +1\equiv 2\pmod {13}\forall n \in \mathbb{N}$.
After I came back home I ran some more test and I noticed that if $p$ is prime then $(p-1)!^{n!} +1\equiv 2\pmod {p}\forall n \in \mathbb{N}$.
I was wondering if this result is true, if yes how to prove it? If not what is the formal way for solving the mother problem.
-
$12!^{n!} +1\equiv 0\pmod {13}$ and $(p-1)!^{n!} +1\equiv 2\pmod {p}$ for $n=1$ (or $0$) – Henry Jun 7 '12 at 6:47
Note that the original problem can also be solved by Fermat's little theorem: $12!^{12} \equiv 1 \pmod{13}$. – Peter Taylor Jun 7 '12 at 9:18
Note that your 'for all' quantifier needs the caveat that $n\gt 1$; if $n=1$ then the sums are $0\bmod p$, not $2$. – Steven Stadnicki Jun 7 '12 at 15:07
By Wilson's Theorem, $12!\equiv -1\pmod{13}$. So for any non-negative even integer $m$, $(12!)^m+1\equiv (-1)^m+1\equiv 2\pmod{13}$. Since $0\le 2\lt 13$, it follows that $2$ is the remainder when $(12!)^m+1$ is divided by $13$.
If $m$ is odd, the same reasoning shows that $(12!)^m+1\equiv 0\pmod{13}$.
And $13$ is not particularly lucky or unlucky. The same result, with the same proof, holds if $13$ is replaced by any odd prime $p$, and $12$ is replaced by $p-1$.
The prime $2$ is slightly different. Whether $m$ is odd or even, $(1)^m +1\equiv 2\pmod{2}$, but the remainder is $0$, not $2$.
-
... and $14!$ is even ... – Robert Israel Jun 7 '12 at 6:44
Aha that helps! Thanks a lot Andre :) – VelvetThunder Jun 7 '12 at 6:50
Hint $\$ Unifying little Fermat and Wilson: $\rm C\:\!!^{\:\!C}\!\equiv 1\ mod\:\ C\!+\!1\:$ prime. Take your pick for a proof.
- | 2015-09-02T13:40:18 | {
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https://www.freemathhelp.com/forum/threads/ax-by-c-0.118450/ | # ax+by+c =0
#### apple2357
##### Full Member
So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0
The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0
Why can't this be made to work?
Tell me if i am talking nonsense!
#### pka
##### Elite Member
So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0
The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0
Why can't this be made to work?
Tell me if i am talking nonsense!
Surely you know that if a line contains $$\displaystyle (x_1,y_1)~\&~(x_2,y_2)$$ and if $$\displaystyle x_1\ne x_2$$ then its slope is $$\displaystyle m=\dfrac{y_2-y_1}{x_2-x_1}$$.
#### apple2357
##### Full Member
Yes I know all that. I just couldn’t explain why the method and approach above fails ?
#### pka
##### Elite Member
Yes I know all that. I just couldn’t explain why the method and approach above fails ?
Two linear equations in three unknowns???
#### Dr.Peterson
##### Elite Member
So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0
The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0
Why can't this be made to work?
Tell me if i am talking nonsense!
It works fine! Keep going ... (after fixing a typo).
The only problem is that the answer is not unique -- you can multiply an equation of the form ax+by+c=0 by any non-zero number and the resulting equation is equivalent, and has the same form.
So at some point in your work you will be picking an arbitrary value for either a, b, or c.
#### JeffM
##### Elite Member
So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0
The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0
Why can't this be made to work?
Tell me if i am talking nonsense!
A system of two equations with three unknowns may have an infinite number of valid solutions.
$$\displaystyle 2a + 5b + c = 0 \text { and } 3a + 10b + c = 0 \implies$$
$$\displaystyle c = -\ (2a + 5b) \implies 3a + 10b - (2a + 5b) = 0 \implies a = -\ 5b.$$
$$\displaystyle \text {Let } b = 1 \implies a = -\ 5 \implies c = -\ (-\ 10 + 5) = 5.$$
$$\displaystyle -\ 5x + y + 5 = 0 \implies y = 5x - 5.$$
$$\displaystyle \therefore x = 2 \implies 5x - 5 = 5 = y, \text { which checks, and}$$
$$\displaystyle x = 3 \implies 5x - 5 = 10 = y, \text { which also checks.}$$
#### JeffM
##### Elite Member
Two linear equations in three unknowns???
But it does work, of course. The answer simply is not unique. The uniqueness and simplicity of the resulting equation is the advantage of the traditional methods of finding the equation of a line joining two distinct points. However, simply add b = 1 as a third equation to get the traditional, simple answer.
I think you missed the thrust of the OP. It was asking why a proposed method does NOT work, but it does work. The OP incorrectly assumed that a system of two equations with three unknowns has no valid solutions. The premise of the question was therefore confusing.
Last edited:
#### apple2357
##### Full Member
Ok. So the method does completely work.
You just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?
#### JeffM
##### Elite Member
Ok. So the method does completely work.
You just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?
There is no conceptual problem, but, in terms of communication, it is very common (and therefore readily comprehensible) to state a linear equation as
$$\displaystyle y = ux + v.$$
This results from setting b = 1. This form immediately gives the slope and the y-intercept and so is mathematically useful.
Last edited:
#### HallsofIvy
##### Elite Member
The problem is that while any straight line can be written as ax+ by+ c= 0 that form is not unique! Multiplying each term by d gives a'x+ b'y+ c'= 0, where a'= ad, b'= bd, and c'= cd, a different equation for the same line. If you (arbitrarily) take a= 1 then you have x+ by+ c= 0. The line passes through (2,5) and (3, 10) so you have the two equations, 2+ 5b+ c= 0 and 3+ 10b+ c= 0, to solve for the two unknowns, b and c.
#### apple2357
##### Full Member
That makes sense!
#### Jomo
##### Elite Member
The line y=3x+4 and the line 2y=6x+8 both have the same exact points! The form y = 3x+4 or y = mx+b just insists that the coefficient of y is 1. In ax+by +c=0, b does not have to be 1!
For the record, the form is usually written as ax+by = c | 2019-10-21T12:56:59 | {
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https://web2.0calc.com/questions/11-you-deposit-200-each-month-into-an-account-earning-3-interest-compounded-monthly-a-how-much-will-you-have-in-the-account-in-30-years | +0
# 11. You deposit $200 each month into an account earning 3% interest compounded monthly. a. How much will you have in the account in 30 years 0 6713 13 11. You deposit$200 each month into an account earning 3% interest compounded monthly.
a. How much will you have in the account in 30 years?
b. How much total money will you put into the account?
c. How much total interest will you earn?
Guest Jul 31, 2015
#2
+26406
+10
.
Alan Jul 31, 2015
Sort:
#1
0
a)200*1.03^30 = $485.45 b) c) 200*1.03^30 minus$200
Guest Jul 31, 2015
#2
+26406
+10
.
Alan Jul 31, 2015
#3
+91477
+5
11. You deposit $200 each month into an account earning 3%(per annum) interest compounded monthly. I am assuming that the money goes in at the beginning of the month and the interest is paid at the end of the month. a. How much will you have in the account in 30 years? b. How much total money will you put into the account? c. How much total interest will you earn? a) You can also do this with the future value of an ordinary annuity formula C=200 n=30*12=360 i=0.03/12 = 0.0025 Amount after 30 years =$116547.38
$${\mathtt{200}}{\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{1.002\: \!5}}}^{{\mathtt{360}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.002\: \!5}}}}\right) = {\mathtt{116\,547.376\: \!919\: \!658\: \!203\: \!63}}$$
b) $${\mathtt{200}}{\mathtt{\,\times\,}}{\mathtt{360}} = {\mathtt{72\,000}}$$
You have put $72,000 into the account. c) Interest =$116547.38 - $72,000 =$44547.38
$${\mathtt{116\,547.38}}{\mathtt{\,-\,}}{\mathtt{72\,000}} = {\mathtt{44\,547.38}}$$
Melody Jul 31, 2015
#4
+26406
+5
The small difference between my result and Melody's is that I've assumed the the total includes the interest paid at the end of the last month; whereas Melody's assumes that in the last month only the $200 deposited at the beginning of the month is added in. . Alan Jul 31, 2015 #5 +1037 +5 For this type of investment, the “Annuity Due” formula returns the correct future value for standard depository account interest payments, correlating to the beginning of the month deposit and end of the month interest payment. This returns the same value as Alan’s result. $$\\ \noindent \text {Annuity Due: }{FV = D \dfrac{(1 + r)^{n} - 1} {r}(1 + r)}\hspace{20pt} |\hspace{10pt} \text {\small D=Deposit per interval}\\ \\ 200*\dfrac{(1.0025^{360}-1)}{0.0025}*(1.0025)\;=\;116838.75$$ In practice, financial institutions use the “average daily balance” to calculate the interest on deposits. The result returns a value with limits between the two formulas. $$\text {FV(annuity ordinary) \leq Total Interest \leq FV(annuity due)}$$ Nauseated Jul 31, 2015 #6 +91477 +5 Yes, sorry Alan I did not realize that our answers were different. Alan is totally corect. This is the formula for the the future value of an ordinary annuity. This is where the money is put into the account at the END of the time period instead of at the beginning. It is easy to adjust it for this question. Here we have$200 invested at the very beginning so we must ADD 200 plus the interest it will accrue for the whole 30 years that will be C(1+i)^n = 200*(1.0025^360)
but NO $200 is invested at the very end SO we must subtract C =$200 at the end
so we get
$$\\FV=\left[\frac{(1+i)^n-1}{i}\right]+C(1+i)^n-C\\\\ If you rearrange this you will find that it is identical to Nauseated's formula.\\\ FV=my original answer+C(1+i)^n-C\\\\ FV=116547.38+200(1.0025)^{360}-200\\\\$$
$${\mathtt{116\,547.38}}{\mathtt{\,\small\textbf+\,}}{\mathtt{200}}{\mathtt{\,\times\,}}{\left({\mathtt{1.002\: \!5}}\right)}^{\left({\mathtt{360}}\right)}{\mathtt{\,-\,}}{\mathtt{200}} = {\mathtt{116\,838.748\: \!442\: \!299\: \!145\: \!509\: \!1}}$$
FV = $116838.75 This is identical to Alan's answer. And Nauseated also agreed that it is correct. ------------------------------------------ Nauseated has gone one more step with this. He has said: "In practice, financial institutions use the “average daily balance” to calculate the interest on deposits." Yes this is correct Nauseated has then stated that: "The result returns a value with limits between the two formulas. " Yes I can see how this could possibly be justified.. but I would like Nauseated to justify/discuss this statement. :) Melody Aug 1, 2015 #7 0 I was in finance for years. I never heard of treating saving account like any kind of annuity. This looks like a ton of BS to me. Guest Aug 2, 2015 #8 +1037 +5 It’s not that you never heard of it, you simply forgot. That can happen when you are stoned in class. BTW, playing Monopoly is not the same as being in finance. In any case, a managed depository savings account can match the return of either type of annuity. I will demonstrate this in the next post. Nauseated Aug 2, 2015 #9 +91477 0 My goodness, that is excessively polite for you Nauseated ! Are you unwell ? Melody Aug 2, 2015 #10 +1037 +5 ### The following data sets shows first 12 months and the last 12 months of a hypothetical account of 360 interest cycles. The first column displays the$200 deposit; the second column displays the end of month interest payment on the previous balance and the $200 first day of month deposit. Column 3 displays the new balance. Column 4 is the multiplier that weights the deposit for the purpose of interest calculation. The multiplier is from 1 to 0, representing the average daily deposit of the$200, where 1 is a deposit on the first day and 0 is a deposit on the last day. (No withdraws are made from this account and only the deposit is weighted).
Column 5 is the average daily of the deposit.
Column 6 is the interest.
Column 7 is the balance.
Note that the final balance in column 3 matches the annuity due, while the final balance in column 7 matches the annuity ordinary.
$$\displaystyle \noindent \small {First data set. Months 1-12}\\ \begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 200.00 & 0.50 & 200.50 & 0.00 & 0.00 & 0.00 & 200.00 \\ 200.00 & 1.00 & 401.50 & 0.00 & 0.00 & 0.50 & 400.50 \\ 200.00 & 1.50 & 603.01 & 0.00 & 0.00 & 1.00 & 601.50 \\ 200.00 & 2.01 & 805.01 & 0.00 & 0.00 & 1.50 & 803.01 \\ 200.00 & 2.51 & 1,007.53 & 0.00 & 0.00 & 2.01 & 1,005.01 \\ 200.00 & 3.02 & 1,210.54 & 0.00 & 0.00 & 2.51 & 1,207.53 \\ 200.00 & 3.53 & 1,414.07 & 0.00 & 0.00 & 3.02 & 1,410.54 \\ 200.00 & 4.04 & 1,618.11 & 0.00 & 0.00 & 3.53 & 1,614.07 \\ 200.00 & 4.55 & 1,822.65 & 0.00 & 0.00 & 4.04 & 1,818.11 \\ 200.00 & 5.06 & 2,027.71 & 0.00 & 0.00 & 4.55 & 2,022.65 \\ 200.00 & 5.57 & 2,233.28 & 0.00 & 0.00 & 5.06 & 2,227.71 \\ 200.00 & 6.08 & 2,439.36 & 0.00 & 0.00 & 5.57 & 2,433.28 \end{tabular}$$
$$\displaystyle \noindent \small {First data set. Months 349-360}\\ \begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 200.00 & 278.06 & 111500.59 & 0.00 & 0.00 & 276.86 & 111222.53 \\ 200.00 & 279.25 & 111979.84 & 0.00 & 0.00 & 278.06 & 111700.59 \\ 200.00 & 280.45 & 112460.29 & 0.00 & 0.00 & 279.25 & 112179.84 \\ 200.00 & 281.65 & 112941.94 & 0.00 & 0.00 & 280.45 & 112660.29 \\ 200.00 & 282.85 & 113424.79 & 0.00 & 0.00 & 281.65 & 113141.94 \\ 200.00 & 284.06 & 113908.85 & 0.00 & 0.00 & 282.85 & 113624.79 \\ 200.00 & 285.27 & 114394.13 & 0.00 & 0.00 & 284.06 & 114108.85 \\ 200.00 & 286.49 & 114880.61 & 0.00 & 0.00 & 285.27 & 114594.13 \\ 200.00 & 287.70 & 115368.31 & 0.00 & 0.00 & 286.49 & 115080.61 \\ 200.00 & 288.92 & 115857.23 & 0.00 & 0.00 & 287.70 & 115568.31 \\ 200.00 & 290.14 & 116347.38 & 0.00 & 0.00 & 288.92 & 116057.23 \\ 200.00 & 291.37 & 116838.75 & 0.00 & 0.00 & 290.14 & 116547.38 \end{tabular}$$
The second data sets are the same as the first, except the multiplier is set to 0.75. This corresponds to a depositor making two deposits of $100 each on the first and 15 of each 30-day month. $$\displaystyle \noindent \small {Sceond data set. Months 1-12}\\ \begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 200.00 & 0.50 & 200.50 & 0.75 & 150.00 & 0.38 & 200.38 \\ 200.00 & 1.00 & 401.50 & 0.75 & 150.00 & 0.88 & 401.25 \\ 200.00 & 1.50 & 603.01 & 0.75 & 150.00 & 1.38 & 602.63 \\ 200.00 & 2.01 & 805.01 & 0.75 & 150.00 & 1.88 & 804.51 \\ 200.00 & 2.51 & 1007.53 & 0.75 & 150.00 & 2.39 & 1006.90 \\ 200.00 & 3.02 & 1210.54 & 0.75 & 150.00 & 2.89 & 1209.79 \\ 200.00 & 3.53 & 1414.07 & 0.75 & 150.00 & 3.40 & 1413.19 \\ 200.00 & 4.04 & 1618.11 & 0.75 & 150.00 & 3.91 & 1617.10 \\ 200.00 & 4.55 & 1822.65 & 0.75 & 150.00 & 4.42 & 1821.51 \\ 200.00 & 5.06 & 2027.71 & 0.75 & 150.00 & 4.93 & 2026.44 \\ 200.00 & 5.57 & 2233.28 & 0.75 & 150.00 & 5.44 & 2231.88 \\ 200.00 & 6.08 & 2439.36 & 0.75 & 150.00 & 5.95 & 2437.84 \end{tabular}$$ $$\displaystyle \noindent \small {Sceond data set. Months 349-360}\\ \begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 200.00 & 278.06 & 111500.59 & 0.75 & 150.00 & 277.76 & 111431.07 \\ 200.00 & 279.25 & 111979.84 & 0.75 & 150.00 & 278.95 & 111910.02 \\ 200.00 & 280.45 & 112460.29 & 0.75 & 150.00 & 280.15 & 112390.17 \\ 200.00 & 281.65 & 112941.94 & 0.75 & 150.00 & 281.35 & 112871.52 \\ 200.00 & 282.85 & 113424.79 & 0.75 & 150.00 & 282.55 & 113354.08 \\ 200.00 & 284.06 & 113908.85 & 0.75 & 150.00 & 283.76 & 113837.84 \\ 200.00 & 285.27 & 114394.13 & 0.75 & 150.00 & 284.97 & 114322.81 \\ 200.00 & 286.49 & 114880.61 & 0.75 & 150.00 & 286.18 & 114808.99 \\ 200.00 & 287.70 & 115368.31 & 0.75 & 150.00 & 287.40 & 115296.39 \\ 200.00 & 288.92 & 115857.23 & 0.75 & 150.00 & 288.62 & 115785.00 \\ 200.00 & 290.14 & 116347.38 & 0.75 & 150.00 & 289.84 & 116274.84 \\ 200.00 & 291.37 & 116838.75 & 0.75 & 150.00 & 291.06 & 116765.90 \end{tabular}$$ ... Nauseated Aug 2, 2015 #11 +1037 0 My goodness, that is excessively polite for you Nauseated ! Overt politeness is one of my faults. I try, but I am not always successful. Are you unwell ? Yes! I am unwell. I am Nauseated. It seems worse than usual. . . . Maybe it's sympathetic morning sickness. :) Nauseated Aug 2, 2015 #12 +91477 0 Is Mrs Nauseated pregnant? Congratulations ! The name you have inflicted upon her will suit her well for a while :/ Melody Aug 2, 2015 #13 +91477 +5 ok Nauseated you have a lot of figures there but I am going to be so bold as to summarize your reasoning. The question does not state WHEN the money was deposited into the account. IF it is deposited at the very beginning of the month then it is an annuitiy due question and yours and Alan figures are correct (a tiny bit too big) If it is deposited at the very end of the month then it is an ordinary annuity question and my original answer was in fact correct. (a tiny bit to small) HOWEVER If the$200 deposit is made at another time or times during the month then this will make the Future value lie between the 2 extremes.
SEE Nauseated - that was not so hard. You really did not need all those tables of values.
I am quite positive you will correct me if I have misinterpreted your logic.
-----------------------------------
The reason I have added "a tiny bit too big" and a "tiny bit too small" is because there is always going to be one day at the end, or the beginning, of the month which has the old value at the beginning and the new value at the end so effectively this day has interest paid on the new deposit of only \$100 (average of 0 and 200) [Not 0 (ordinary annuity) or 200 (annuity due) as used for the development of the formula]
Melody Aug 2, 2015
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https://math.stackexchange.com/questions/3415109/how-can-i-solve-the-following-square-root-inequality | # How can I solve the following square root inequality?
I am given the following inequality:
$$\sqrt{2-x} > x$$
And I have to solve for $$x$$.
This is what I tried:
Firstly, I applied the existence condition for the square root:
$$2-x \ge 0 \Rightarrow x \le 2 \Rightarrow x \in (- \infty, 2]$$
Then I squared both sides of the inequality:
$$2-x>x^2$$
$$x^2+x-2 < 0$$
And from this I got that:
$$x \in (-2, 1)$$
Finally, I intersected this with the condition I applied at the beginning of the exercise:
$$x \in (-2, 1) \cap (- \infty, 2]$$
$$x \in (-2, 1)$$.
The problem with this answer is that it is wrong. If I take a number like $$-10$$ and plug it into the inequality I get:
$$\sqrt{2-(-10)} > -10$$
$$\sqrt{12} > -10$$
Which is true. However $$-10$$ is not included in the interval $$(-2, 1)$$. The correct answer seems to be $$(- \infty, 1)$$, which is not what I got.
I noticed that the inequalities of before and after of the squaring are not equivalent. So:
$$\sqrt{2-x} > x$$
and
$$2-x>x^2$$
are not equivalent. If, again, I take the number $$-10$$, in the first inequality I get:
$$\sqrt{2-(-10)} > -10$$
$$\sqrt{12} > -10$$, which is true.
And in the second inequality I get:
$$2 - (-10) > (-10)^2$$
$$12 > 100$$, which is false.
So I think that's where the problem lies, but I don't know if I am correct and what should I do to get the right answer of $$(- \infty, 1)$$.
• When you're solving $$x^2 +x -2 > 0$$ you need to remember that previously you have found both $x\leq 2$ and $x>0$ (because square root cannot be negative). So in this case you're limited to $0 \leq x \leq 2$, before even looking at solving this equation. – Matti P. Oct 30 '19 at 13:29
• "Then I squared both sides of the inequality:" $x > y\not \implies x^2 > y$ unless $y \ge 0$. If $x > 0 > y$ then multiplying by $y$ flips the signs and $x>0 >y \implies xy < 0 < y^2$. And multiplying by $x$ doesn't flip and $x > 0 > y\implies x^2 > 0 >xy$ and we have $x^2 > xy$ and $xy < y^2$ and we can't conclude anything about $x^2$ compared to $y^2$. – fleablood Oct 31 '19 at 6:42
• @MattiP., the non-negativity restriction on the square root only implies $x\le2$ (as the OP found). It does not require $x\ge0$. Indeed, the inequality $\sqrt{2-x}\gt x$ is clearly true for all $x\lt0$. – Barry Cipra Oct 31 '19 at 12:45
"Then I squared both sides of the inequality:"
$$a > b\not \implies a^2 > b^2$$ unless $$b \ge 0$$.
If $$a > 0 > b$$ then multiplying by $$b$$ flips the signs and $$a>0 >b \implies ab < 0 < b^2$$. And multiplying by $$a$$ doesn't flip and $$a > 0 > b\implies a^2 > 0 >ab$$.
[Remember the reason that $$m > n > 0 \implies m^2 > n^2$$ is that because $$n > 0$$ we know $$m>n \implies mn > nn = n^2$$ and because $$m >0$$ we know $$m^2 = mm > mn$$. So we have $$m^2 >mn$$ and $$mn > n^2$$ so $$m^2 > n^2$$. That simply will not work of one of $$m,n$$ is negative. (and if both are negative you get the exact opposite result: $$0 > m>n\implies m^2 < n^2$$.]
So we have $$a^2 > ab$$ and $$ab < b^2$$ and we can't conclude anything about $$a^2$$ compared to $$b^2$$.
.... Or ....
just to be blunt.
$$3 > -987$$ by $$3^2 \not > (-987)^2$$.
....
But you can square both sides if you acknowedge you are assuming $$x \ge 0$$ as a conditional that may not be true.
so when you get $$x \in (-2,1)$$ IF $$x \ge 0$$ so $$[0,1)$$ you can intersect it with the positive values you had before. $$(-\infty, 2] \cap [0,1)=[0,1)$$ IFF $$x \ge 0$$.
But we CAN have $$x < 0$$ in which case ... squaring both sides is useless.
So either $$x < 0$$ and $$x\in (-\infty, 0)$$ OR $$x \ge 0$$ and $$x\in [0,1)$$ and so $$x \in (-\infty,0) \cup [0,1) = (-\infty, 1)$$>
The domain of the root is given by b$$x\le 2$$ Now we will consider two cases: If $$x\le 0$$ the our inequality is true. If $$x then we can square and we get $$0>x^2+x-2$$ This gives us $$-2 and $$0 and we get $$0 So the solution set is given by $$x<1$$ and $$x$$ is a real number.
$$\sqrt{2-x} > x$$ then $$\sqrt{2-x} > x-2 +2$$ , move all to the left hand side then $$2-x = (\sqrt{2-x})^2$$ $$\left(\sqrt{2-x}\right)^2 + \sqrt{2-x} - 2 > 0$$ $$( \sqrt{2-x} +2) (\sqrt{2-x} -1)>0$$ I avoided squaring both sides.
Starting from
$$\sqrt{2-x} > x$$
it is immediately clear that if $$x<0$$ then $$\sqrt{2-x} >0$$ and therefore $$x<0$$ is part of the solution set. We then need to analyze where
$$\sqrt{2-x}$$
is defined. We know that $$\sqrt{2-x}$$ will have positive real solutions if $$2-x \ge 0$$. Therefore, we need to check $$0\le x \le 2$$. In order to do this, let $$\varepsilon$$ be a small number where $$0\le\varepsilon< 1$$. Then $$\sqrt{2-\varepsilon}>\sqrt{1}> \varepsilon$$ $$\sqrt{2-1}=\sqrt{1}=1$$ $$\sqrt{2-(\varepsilon+1)}\le\sqrt{1}\le\varepsilon+1$$
so when $$0\le x < 1$$ we have $$\sqrt{2-x} > x$$ and when $$1\le x \le 2$$ we have $$\sqrt{2-x} \le x$$. Hence, the proper solution set is $$x<1$$ or $$(-\infty,1)$$.
An alternative approach is to start from the fact that we must have $$x\le 2$$ in order for $$\sqrt{2-x}$$ to exist, and write $$x=2-u^2$$ with $$u\ge0$$, which simplifies the inequality to $$u\gt2-u^2$$ (with, remember, $$u\ge0$$, which is used in removing the square root symbol to obtain $$\sqrt{u^2}=u$$). Standard algebra manipulates this to the equivalent form $$(u-1)(u+2)\gt0$$, which, given the restriction to $$u\ge0$$, implies $$u\gt1$$. Substituting back to $$x=2-u^2$$, we see we have $$x\lt2-1^2=1$$, so the solution set for the inequality is $$x\in(-\infty,1)$$.
In review, you have to do something to get rid of the square root symbol in order to find the solution set for the inequality. Squaring both sides certainly does this, but you have to contend with the fact that $$a\gt b$$ is not equivalent to $$a^2\gt b^2$$ without additional restrictions on $$a$$ and $$b$$, so that after you've found the solution set for the squared inequality you have to see how it relates to the original inquality. The approach here is somewhat simpler in that regard; in essence, it puts all its restrictive eggs into a single basket of non-negativity (i.e., $$u\ge0$$).
I think fleablood’s answer gets closest to addressing what I think OP was asking, namely why the mechanical application of squaring both sides of the equation does not yield all of the solutions. | 2020-07-03T20:45:15 | {
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https://mathhelpboards.com/threads/determining-convergence-of-series.4118/ | # Determining convergence of series
#### calcboi
##### New member
I have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test(correct me if I'm wrong) but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. I also tried Limit Comparison but the limit goes to infinity so it is also inconclusive. Can you please help?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Yes you are correct.
near infinity the term n^3 is dominant over ln (n).
#### calcboi
##### New member
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.
Right, $\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\ldots =1\neq 0$, so the series is absolutely convergent.
#### calcboi
##### New member
What test did you use to determine absolute convergence? Or was that just analyzing end behavior?
MHB Math Helper
#### Fernando Revilla
##### Well-known member
MHB Math Helper
When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?
$\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\lim_{n\to \infty}\frac{n^3}{n^3-\log n}=\lim_{n\to \infty}\frac{1}{1-(\log n/n^3)}=\frac{1}{1-0}=1$ | 2020-11-25T10:53:26 | {
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https://math.stackexchange.com/questions/2777248/when-to-use-lhospitals-rule-vs-the-limit-shortcut/2777290 | # When to use L'Hospital's rule vs the limit shortcut
Before being so quick to downvote or throw darts, please forgive my ignorance and inability to recall basic calculus atm.
Consider the limit $\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}$. It can quickly be determined that the $x$ approaches $3$ using L'Hospital's rule.
The same answer can be derived, however, by using this shortcut (I am unaware if it has a specific name):
$\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}=\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}$
Every term is divided by the highest degree of $x$ in the denominator. The terms with $x^2$ in both the numerator and denominator simplify. All other terms go to $0$. Therefore:
$\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}=\frac{3}{1}=3$
1) What are the explicit conditions in which may I use this shortcut?
2) How do I know when to use it over L'Hospital's rule, as both techniques can be used only when working with quotients?
3) What is the shortcuts name, if any?
• Your second method is the more common one --- you use L'Hopital's when it fails (provided you satisfy the right conditions for L'Hopital's). This typically works when you have a quotient of polynomials (or in cases like totoro has pointed out), whereas L'Hopital's works for other kinds of functions, too. I don't think it has a universal name. – Bill Wallis May 11 '18 at 21:35
• In class I usually recommend it when there is a quotient of sums of terms that can easily be compared in its growth to infinity. It would also be applicable, for example, to $\lim_{n\to\infty}\frac{n!+3^n-n^{50}}{2\ln(n)-e^{n^2}+2\sin(n)}$. Here 'compared' means knowing the limit of the quotient of the different terms. whether the quotient tends to $0$, to $\infty$, or if it remains bounded. – user551819 May 11 '18 at 21:37
• In my opinion L'Hopital is rarely the method of choice. Almost any other that works is more informative. This trick is good for quotients of polynomials. In general, writing down the first few terms of the power series for the various elements and looking at the leading term will often help. There are questions/answers on this site on the general theme "why not to use L'Hopital". – Ethan Bolker May 11 '18 at 21:40
• L'Hopital can also be used in not quotient indeterminate limits. – Namaste May 11 '18 at 21:42
• One can also use simplification of rational functions, e.g. $\lim_{x\to 2} \frac{x-2}{x^2-4} = \lim_{x\to 2} \frac 1{x+2} = \frac 14.$ Or $$\lim_{x\to 1}\frac{x^2+x-2}{x^3-x^2-3x+3} = \lim_{x\to 1}\frac{x+2}{x^2-3} = -\frac 32$$ – Namaste May 11 '18 at 21:51
Shorcut can be used in quotient limits anytime there exist some leading term at the numerator and denominator which becomes dominant in the limit with respect to the others terms that is, indicating with g(x) the dominant term
$$\lim \limits_{x \to \infty}\frac{f(x)}{g(x)}=0$$
To individuate the dominant term, recall that
• $\frac{x^a}{x^b} \to 0$ for $b>a$
• $\frac{x^a}{b^x} \to 0$ for $b>1$
• $\frac{\log x}{x^a} \to 0$ for $a>0$
and for sequences
• $\frac{a^n}{n!} \to 0$ for $a>1$
• $\frac{n!}{n^n} \to 0$
Recall that l'Hopital rule can be applied to limits which are expressed (or can be expressed) by quotient which are in the indeterminate form $\frac 0 0$ or $\frac{\pm \infty}{\pm \infty}$.
I don't think there is a specific name for the shorcut.
As indicated in the comment another way for rational expression can be the following
$$\frac{3x^2+14x-5}{x^2+x-12}=\frac{3x^2+3x-12+11x+7}{x^2+x-12}=3+\frac{11x+7}{x^2+x-12}$$
• @amWhy Something wrong or not clear? – gimusi May 11 '18 at 21:41
• But L'Hopital can be used in other situations too. – Namaste May 11 '18 at 21:42
• @amWhy are you referring to $0\cdot \infty$? – gimusi May 11 '18 at 21:43
On the other hand, it's not difficult to prove a general theorem about rational functions. Suppose you have $$f(x)=\frac{a_mx^m+a_{m-1}x^{m-1}+\dots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots+b_0}$$ with $a_m\ne0$ and $b_n\ne0$. Then, since it's not restrictive to assume $x>0$ when we want to compute the limit for $x\to\infty$, the numerator can be written as $$x^m\left(a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}\right)$$ and the factor in parentheses has limit $a_m$ when $x\to\infty$. Similarly for the denominator. Now $$\lim_{x\to\infty} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =\frac{a_m}{b_n}\ne0$$ Hence there are three cases.
## First case: $m>n$
$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty}x^{m-n} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} = \begin{cases} \infty & \text{if a_m/b_n>0} \\[4px] -\infty & \text{if a_m/b_n<0} \end{cases}$$ The factor $x^{m-n}$ has limit $\infty$ and the other factor is bounded.
## Second case: $m=n$
$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =\frac{a_m}{b_n}$$
## Third case: $m<n$
$$\lim_{x\to\infty}f(x)= \lim_{x\to\infty}\frac{1}{x^{n-m}} \frac{\displaystyle a_m+\frac{a_{m-1}}{x}+\dots+\frac{a_0}{x^m}} {\displaystyle b_n+\frac{b_{n-1}}{x}+\dots+\frac{b_0}{x^m}} =0$$ The factor $1/x^{n-m}$ has limit $0$ and the other factor has limit $a_m/b_n$.
This completely settles the problem and you need nothing else: your given limit is $3$ because the function is in case two. Doing umpteen times the same computations doesn't seem the best way to spend our time.
How can you remember this? The polynomial of greater degree dominates: if it is at the numerator, the limit is $\pm\infty$ (with the sign determined by $a_m/b_n$); if it is at the denominator, the limit is $0$. If the degrees are the same, the limit is $a_m/b_m$.
• This is essentially the mnemonic “Bobo Bots eats DC”—bigger on bottom: $0$; bigger on top: slant; exponents are the same: divide coefficients. – gen-z ready to perish May 11 '18 at 22:23
• @ChaseRyanTaylor I don't like such kind of mnemonics, TBH. But this is quite a simple case to handle and a general rule is surely handy. There's a similar case for integrals of the form $\int x^ne^x\,dx$; rather than doing umpteen integrations by parts, it's much easier to remember the antiderivative is of the form $P(x)e^x$ with $P$ of degree $n$ and do the derivative. – egreg May 11 '18 at 22:32
In general, suppose you have an expression of the form
$$f_1(x)+f_2(x)\over g_1(x)+g_2(x)$$
for which
$$\lim_{x\to c}{f_2(x)\over f_1(x)}=\lim_{x\to c}{g_2(x)\over g_1(x)}=0$$ Then
$$\lim_{x\to c}{f_1(x)+f_2(x)\over g_1(x)+g_2(x)}=\lim_{x\to c}{f_1(x)\over g_1(x)}$$
The proof amounts to inserting the intermediate expression
$$\lim_{x\to c}{f_1(x)\over g_1(x)}\cdot{1+{f_2(x)\over f_1(x)}\over1+{g_2(x)\over g_1(x)}}$$
A nice example, in which L'Hopital gets you nowhere, is
$$\lim_{x\to\infty}{5e^{3x}+2e^x\over2e^{3x}+7e^{2x}}=\lim_{x\to\infty}{5e^{3x}\over2e^{3x}}=\lim_{x\to\infty}{5\over2}={5\over2}$$
The trick, in general, is to recognize a dominant term in the numerator and/or denominator. In essence, the shortcut says you can ignore all the other stuff. But you have to make sure that you pick out terms that really do dominate the other stuff; sometimes you can eyeball it, and sometimes you make mistakes (at least I do).
• Have you got an explicit description of what a dominant term is? – rainier May 11 '18 at 22:44
• @rainier, it's what the second display says: the quotient of the "other stuff" to the "dominant" term tends to $0$ in the limit. In the example, we have $(2e^x)/(5e^{3x})={2\over5}e^{-2x}\to0$ and $(7e^{2x})/(2e^{3x})={7\over2}e^{-x}\to0$ as $x\to\infty$. – Barry Cipra May 11 '18 at 22:49
• Got it, thanks! – rainier May 11 '18 at 22:53 | 2019-08-18T19:20:41 | {
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https://math.stackexchange.com/questions/1727586/number-of-ways-in-which-they-can-be-seated-if-the-2-girls-are-together-and-the | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2\times 6!=3600$$ suitable arrangements.
Now, sticking to these pairs but varying the order gets us $$4\times 3600=14,400$$
And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3\times 14,400= 43,200$$
Method II: (closer to what you were trying) Arrange the kids as $$-\;AB-CD\;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $$\{0,5,0\},\;\{1,4,0\},\;\{0,4,1\},\;\{2,3,0\},\;\{0,3,2\},\;\{1,3,1\},\;\{3,2,0\},\;\{0,2,3\},\;\{2,2,1\},\;\{1,2,2\},\;\{3,1,1\},\;\{1,1,3\},\{2,1,2\},\;\{4,1,0\},\;\{0,1,4\}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15\times 4!\times 5!=43,200$$
First we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them. $$\text{b}\qquad\text{b}\qquad\text{b}\qquad\text{b}\qquad\text{b}$$ These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $\binom{6}{2}$ ways.
We have found the number of "patterns." Now permute the girls, permute the boys. The number of arrangements is $\binom{6}{2}4!5!$. | 2019-05-21T09:30:51 | {
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https://zingale.github.io/phy504/floating-point.html | # Floating Point¶
## Storage overview¶
We can think of a floating point number as having the form:
$\mbox{significand} \times 10^\mbox{exponent}$
Most computers follow the IEEE 754 standard for floating point, and we commonly work with 32-bit and 64-bit floating point numbers (single and double precision). These bits are split between the signifcand and exponent as well as a single bit for the sign:
Since the number is stored in binary, we can think about expanding a number in powers of 2:
$0.1 \sim (1 + 1 \cdot 2^{-1} + 0 \cdot 2^{-2} + 0 \cdot 2^{-3} + 1 \cdot 2^{-4} + 1 \cdot 2^{-5} + \ldots) \times 2^{-4}$
In fact, 0.1 cannot be exactly represented in floating point:
Listing 52 simple_roundoff.cpp
#include <iostream>
#include <iomanip>
int main() {
double a = 0.1;
std::cout << std::setprecision(19) << a << std::endl;
}
### Precision¶
With 52 bits for the significand, the smallest number compared to 1 we can represent is
$2^{-52} \approx 2.22\times 10^{-16}$
but the IEEE 754 format always expresses the significant such that the first bit is 1 (by adjusting the exponent) and then doesn’t need to store that 1, giving us an extra bit of precision, so the machine epsilon is
$2^{-53} \approx 1.11\times 10^{-16}$
We already saw how to access the limits of the data type via std::numeric_limits. When we looked at machine epsilon, we saw that for a double it was about $$1.1\times 10^{-16}$$.
Note that this is a relative error, so for a number like 1000 we could only add 1.1e-13 to it before it became indistinguishable from 1000.
$\mbox{relative roundoff error} = \frac{|\mbox{true number} - \mbox{computer representation} |} {|\mbox{true number}|} \le \epsilon$
Note that there are varying definitions of machine epsilon which differ by a factor of 2.
### Range¶
Now consider the exponent, we use 11 bits to store it in double precision. Two are reserved for special numbers, so out of the 2048 possible exponent values, one is 0, and 2 are reserved, leaving 2045 to split between positive and negative exponents. These are set as:
$2^{-1022} \mbox{ to } 2^{1023}$
converting to base 10, this is
$\sim 10^{-308} \mbox{ to } \sim 10^{308}$
### Reporting values¶
We can use std::numeric_limits<double> to query these floating point properties:
Listing 53 limits.cpp
#include <iostream>
#include <limits>
int main() {
std::cout << "maximum double = " << std::numeric_limits<double>::max() << std::endl;
std::cout << "maximum double base-10 exponent = " << std::numeric_limits<double>::max_exponent10 << std::endl;
std::cout << "smallest (abs) double = " << std::numeric_limits<double>::min() << std::endl;
std::cout << "minimim double base-10 exponent = " << std::numeric_limits<double>::min_exponent10 << std::endl;
std::cout << "machine epsilon (double) = " << std::numeric_limits<double>::epsilon() << std::endl;
}
## Roundoff vs. truncation error¶
Consider the Taylor expansion of $$f(x)$$ about some point $$x_0$$:
$f(x) = f(x_0 + \Delta x) = f(x_0) + \left . \frac{df}{dx} \right |_{x_0} \Delta x + \mathcal{O}(\Delta x^2)$
where $$\Delta x = x - x_0$$
We can solve for the derivative to find an approximation for the first derivative:
$\left . \frac{df}{dx} \right |_{x_0} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} + \mathcal{O}(\Delta x)$
This shows that this approximation for the derivative is first-order accurate in $$\Delta x$$ – that is the truncation error of the approximation.
We can see the relative size of roundoff and truncation error by using this approximation to compute a derivative for different values of $$\Delta x$$:
Listing 54 truncation_vs_roundoff.cpp
#include <iostream>
#include <iomanip>
#include <cmath>
#include <limits>
#include <vector>
double f(double x) {
return std::sin(x);
}
double dfdx_true(double x) {
return std::cos(x);
}
struct point {
double dx;
double err;
};
int main() {
double dx = 0.1;
double x0 = 1.0;
std::vector<point> data;
while (dx > std::numeric_limits<double>::epsilon()) {
point p;
p.dx = dx;
double dfdx_approx = (f(x0 + dx) - f(x0)) / dx;
double err = std::abs(dfdx_approx - dfdx_true(x0));
p.err = err;
data.push_back(p);
dx /= 2.0;
}
std::cout << std::setprecision(8) << std::scientific;
for (auto p : data) {
std::cout << std::setw(10) << p.dx << std::setw(15) << p.err << std::endl;
}
}
It is easier to see the behavior if we make a plot of the output:
Let’s discuss the trends:
• Starting with the largest value of $$\Delta x$$, as we make $$\Delta x$$ smaller, we see that the error decreases. This is following the expected behavior of the truncation error derived above.
• Once our $$\Delta x$$ becomes really small, roundoff error starts to dominate. In effect, we are seeing that:
$(x_0 + \Delta x) - x_0 \ne 0$
because of roundoff error.
• The minimum error here is around $$\sqrt{\epsilon}$$, where $$\epsilon$$ is machine epsilon.
## Testing for equality¶
Because of roundoff error, we should never exactly compare two floating point numbers, but instead ask they they agree within some tolerance, e.g., test equality as:
$| x - y | < \epsilon$
For example:
Listing 55 comparing.cpp
#include <iostream>
int main() {
double h{0.01};
double sum{0.0};
for (int n = 0; n < 100; ++n) {
sum += h;
}
std::cout << "sum == 1: " << (sum == 1.0) << std::endl;
std::cout << "|sum - 1| < tol: " << (std::abs(sum - 1.0) < 1.e-10) << std::endl;
}
## Minimizing roundoff¶
Consider subtracting the square of two numbers – taking the difference of two very close-in-value numbers is a prime place where roundoff can come into play.
$x^2 - y^2$
$(x - y)(x + y)$
by factoring this, we are subtracting more reasonably sized numbers, reducing the roundoff.
We can see this directly by doing this with single precision (float) and comparing to an answer computed via double precious (double)
Here’s an example:
Listing 56 subtraction.cpp
#include <iostream>
int main() {
float x{1.000001e15};
float y{1.0000e15};
std::cout << "x^2 - y^2 : " << x*x - y*y << std::endl;
std::cout << "(x - y)(x + y) : " << (x - y) * (x + y) << std::endl;
double m{1.000001e15};
double n{1.0000e15};
std::cout << "double precision value: " << (m - n) * (m + n) << std::endl;
}
As another example, consider computing 1:
$\sqrt{x + 1} - \sqrt{x}$
We can alternately rewrite this to avoid the subtraction of two close numbers:
$\sqrt{x + 1} - \sqrt{x} = (\sqrt{x + 1} - \sqrt{x}) \left ( \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}} \right ) = \frac{1}{\sqrt{x+1} + \sqrt{x}}$
Again we’ll compare a single-precision calculation using each of these methods to a double precision “correct answer”. To ensure that we use the single-precision version of the std::sqrt() function, we will use single precision literal suffix, e.g., 1.0f tells the compiler that this is a single-precision constant.
Listing 57 squareroots.cpp
#include <iostream>
#include <iomanip>
#include <cmath>
int main() {
float x{201010.0f};
std::cout << std::setprecision(10);
float r1 = std::sqrt(x + 1.0f) - std::sqrt(x);
float r2 = 1.0f / (std::sqrt(x + 1.0f) + std::sqrt(x));
std::cout << "r1 = " << r1 << " r2 = " << r2 << std::endl;
double xd{201010.0};
double d = std::sqrt(xd + 1.0) - std::sqrt(xd);
std::cout << "d = " << d << std::endl;
}
Notice that we get several more significant digits correct when we compute it with the second formulation compared to the original form.
### Summation algorithms¶
Summing a sequence of numbers is a common place where roundoff error comes into play, especially if the numbers all vary in magnitude and you do not attempt to add them in a sorted fashion. There are a number of different summation algorithms that keep track of the loss due to roundoff and compensate the sum, for example the Kahan summation algorithm.
## Special numbers¶
IEEE 754 defines a few special quantities:
• NaN (not a number) is the result of 0.0/0.0 or std::sqrt(-1.0)
• Inf (infinity) is the result of 1.0/0.0
• -0 is a valid number and the standard says that -0 is equivalent to 0
## Trapping floating point exceptions¶
What happens when we do something bad? Consider this example:
Listing 58 undefined.cpp
#include <iostream>
#include <cmath>
double trouble(double x) {
return std::sqrt(x);
}
int main() {
double x{-1};
double y = trouble(x);
for (int i = 0; i < 10; ++i) {
y += std::pow(x, i);
}
std::cout << y << std::endl;
}
Here, we pass -1 to trouble() which then takes the square root of it – this results in a NaN. But if we run the code, it goes merrily about its way, using that result in the later computations.
Unix uses signals to indicate that a problem has happened during the code execution. If a program created a signal handler then that signal can be trapped and any desired action can be taken.
Note
This example was only tested on a Linux machine with GCC. Other OSes or compilers might have slightly different headers or functionality.
There are a few parts to trapping a floating point exception (FPE). First we need to enable exception trapping via:
feenableexcept(FE_INVALID|FE_DIVBYZERO|FE_OVERFLOW);
That catches 3 different types of floating point exceptions – invalid, divide-by-zero, and overflows.
Next we need to add a handler to deal with the exception:
signal(SIGFPE, fpe_handler);
Here, SIGFPE is the standard name for a floating point exception, and fpe_handler is the name of a function that will be called when we detect a SIGFPE.
In our handler, we use the Linux backtrace() function to access the stack of our program execution. This is really a C-function, so we need to use C-style arrays here.
Here’s the new version of our code:
Listing 59 undefined_trap.cpp
#include <iostream>
#include <cmath>
#include <csignal>
#include <cfenv>
#include <execinfo.h>
void fpe_handler(int s) {
std::cout << "floating point exception, signal " << s << std::endl;
const int nbuf = 64;
void *bt_buffer[nbuf];
int nentries = backtrace(bt_buffer, nbuf);
char **strings = backtrace_symbols(bt_buffer, nentries);
for (int i = 0; i < nentries; ++i) {
std::cout << i << ": " << strings[i] << std::endl;
}
abort();
}
double trouble(double x) {
return std::sqrt(x);
}
int main() {
feenableexcept(FE_INVALID|FE_DIVBYZERO|FE_OVERFLOW);
signal(SIGFPE, fpe_handler);
double x{-1};
double y = trouble(x);
for (int i = 0; i < 10; ++i) {
y += std::pow(x, i);
}
std::cout << y << std::endl;
}
When we compile the code, we want to add the -g option to store the symbols in the code – this allows us to understand where problems arise:
span.prompt1:before {
content: "\$ ";
}
g++ -g -o undefined_trap undefined_trap.cpp
Now when we run this, the program aborts and we see:
floating point exception, signal 8
0: ./undefined_trap() [0x401261]
1: /lib64/libc.so.6(+0x42750) [0x7f3dc35dc750]
2: /lib64/libm.so.6(+0x1435c) [0x7f3dc37d335c]
3: ./undefined_trap() [0x4012ff]
4: ./undefined_trap() [0x401347]
5: /lib64/libc.so.6(+0x2d560) [0x7f3dc35c7560]
6: /lib64/libc.so.6(__libc_start_main+0x7c) [0x7f3dc35c760c]
7: ./undefined_trap() [0x401145]
Aborted (core dumped)
This is the call stack for our program. In the brackets are the address in the program where the execution was when the FPE occurred. These are ordered such that the calling function is below the function where the execution is. So it usually is best to look at the addresses near the top.
We can turn those into line numbers using addr2line:
addr2line -e undefined_trap 0x4012ff
gives:
/home/zingale/classes/phy504/examples/floating_point/undefined_trap.cpp:23
and that line is precisely where the sqrt() is!
We can get slightly nicer output (including the function name) by doing:
addr2line -C -f -i -p -e undefined_trap 0x4012ff
which gives:
trouble(double) at /home/zingale/classes/phy504/examples/floating_point/undefined_trap.cpp:23
Note
On the MathLab machines, the stack trace seems to include an offset, like:
floating point exception, signal 8
0: ./undefined_trap(+0xc03) [0x561d8799dc03]
1: /lib/x86_64-linux-gnu/libc.so.6(+0x3ef10) [0x7f5461e3df10]
2: /lib/x86_64-linux-gnu/libm.so.6(+0x11397) [0x7f5462201397]
3: ./undefined_trap(+0xccf) [0x561d8799dccf]
4: ./undefined_trap(+0xd21) [0x561d8799dd21]
5: /lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xe7) [0x7f5461e20c87]
6: ./undefined_trap(+0xaaa) [0x561d8799daaa]
Aborted (core dumped)
and we need to use that offset instead with addr2line, like:
addr2line -a -f -e ./undefined_trap +0xcd5
1
this example is based on Yakowitz & Szidarovszky | 2022-06-29T03:44:06 | {
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https://math.stackexchange.com/questions/3887771/when-does-improper-riemann-integration-fail-for-computing-expected-values/3887777 | When does improper Riemann integration fail for computing expected values?
Riemann integration requires bounded intervals of integration. However, some continuous random variables (like those from the normal or gamma distribution) have unbounded support sets.
For these 2 distributions, would improper Riemann integrals be sufficient to get their expected values? If not, why not?
Usually, improper Riemann integration will correctly provide the expectation of a continuous random variable with an unbounded support. Where you will usually encounter problems with improper Riemann integrals is when the integral of interest does not converge absolutely. In this case the improper integral may have some value, but this value has limited probabilistic meaning; for example the law of large numbers doesn't necessarily hold.
As an example of this to get a feel for it, you can look at the evaluation of the expected value of $$X \sin(X)$$ where $$X$$ has PDF $$\frac{1}{x^2}$$ on $$[1,\infty)$$ and $$0$$ otherwise. This expectation is formally $$\int_1^\infty \frac{\sin(x)}{x} dx$$ which has a value in improper Riemann integration. But actually this random variable will not satisfy the law of large numbers, as you can observe numerically, so it doesn't make a whole lot of sense to say that it has an expected value.
Indeed, if you have Matlab or Octave, try running the following:
x=1./rand(1000,1);
plot(cumsum(x.*sin(x))./(1:1000)');
This shows sample means from this distribution for progressively larger samples, and you see that they don't converge. (Off-topic: the weird 1/rand trick being used there is called the probability integral transformation, which is extremely useful for numerical work in probability.)
Note that this will never be a problem for a random variable of one sign such as your example of the Gamma distribution. It is also not a problem for the normal distribution.
The exception to the second "usually" above is when the improper Riemann integral doesn't exist but the Lebesgue integral does. This is a pretty uncommon situation in practice.
• @Iterator516 No, the integrals in both of those cases converge absolutely. – Ian Oct 30 at 15:30
• Awesome! Thank you! – Iterator516 Oct 30 at 15:30
• I mistakenly deleted my initial question to Ian. I asked if the integrals for the expectations of the normal and the gamma distributions do converge. – Iterator516 Oct 30 at 16:29 | 2020-12-03T16:05:27 | {
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https://math.stackexchange.com/questions/2433888/greatest-common-divisor-of-prime-numbers | # Greatest Common Divisor of prime numbers
What is the greatest common divisor of $2^4\cdot3^4\cdot25\cdot7$ and $2\cdot12^2\cdot15$?
I know how to find the GCD of a problem like this, but I didn't know what to do since $12$ is not a prime number. Should I say $12^2 = 2^2\cdot2^2\cdot3^2$?
• Hint: $12^2 = (2^2*3)^2 = 2^4*3^2$ – AnalyticHarmony Sep 18 '17 at 2:22
• that's a start. $12^2 =2^2*2^2*3^2=2^4*3^2$. And $15=3*5$ so $12^2*15=2^4*3^2*3*5 = 2^4*3^3*5$. Basically the first step is to rewrite everything as its unique prime factorization. $2^4*3^4*25*7 = 2^4*3^4*5^2*7$ and $2*12^2*15 = 2*2^4*3^2*3*5=2^4*3^3*5$. Then... you know what to do. – fleablood Sep 18 '17 at 3:49
The first one equal to $2^4*3^4*5^2*7$
The second one equals to $2^5*3^3*5$
Therefore the GCD is $2^4*3^3*5=2160$
Yes, that is precisely what you should be doing.
In short, when you want to find the gcd of two numbers directly, you must first represent each one as a product of prime powers, before matching powers and primes.
So in our case, we can write: $$2^4 \times 3^4 \times \color{blue}{25} \times 7 = 2^4 \times 3^4 \times \color{blue}{5^2} \times 7 \\ 2 \times \color{red}{12^2} \times \color{green}{15} = 2 \times \color{red}{2^2 \times 2^2 \times 3^2} \times \color{green}{3 \times 5} = 2^5 \times 3^3 \times 5$$
where, I highlight by color the terms that are expanded on both the LHS and RHS.
Now, we are permitted to compare powers directly. This gives us the answer as $2^4 \times 3^3 \times 5$.
• Nicely explained and nice use of colors. The OP should note when you express a number and a product of prime powers, there will always be exactly one unique way to do it. So doing this will always be a failsafe that will work. – fleablood Sep 18 '17 at 3:54 | 2019-09-15T08:27:31 | {
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https://math.stackexchange.com/questions/3569668/eigenvalues-of-block-toeplitz-matrix-with-toeplitz-blocks | # Eigenvalues of block Toeplitz matrix with Toeplitz blocks
Consider integers $$m,n$$ and a $$m \times m$$-block Toeplitz matrix $$A$$ consisting of two different types of blocks as follows
\begin{align} A_{mn \times mn} &= \begin{bmatrix} B & C & C & \cdots & \cdots & C \\ C & B & C & C & \cdots & C \\ C & C & B & C & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & C \\ C & \cdots & \cdots & C & B & C \\ C & \cdots & \cdots & \cdots & C & B \end{bmatrix} _{mn \times mn} , \end{align}
where $$B$$'s are diagonal blocks with $$B=\frac{1}{m}I_n$$ and $$C$$'s are multiples of the all-ones matrix $$J_n$$, specifically $$C=\frac{1}{mn}J_n$$.
I want to compute the eigenvalues of $$A$$ (I am mainly interested in the value of the 2nd largest eigenvalue since it has a special meaning in graph expansion applications).
Note that in my problem the following conditions also hold for $$m,n$$:
• $$m$$ is odd.
• $$n$$ is prime.
• $$m.
I have experimented with such matrices on the computer and I have observed a trend for the spectrum of $$A$$ which consists of the following eigenvalues:
• $$\lambda_1=0$$ with algebraic multiplicity $$m-1$$.
• $$\lambda_2=1/m$$ with algebraic multiplicity $$m(n-1)$$.
• $$\lambda_3=1$$ with algebraic multiplicity $$1$$.
I do not claim that this is necessarily the answer but at least it was consistent for the pairs of $$m,n$$ I tried.
Can you suggest how one can go and prove the above claim (if correct) or pinpoint other known results?
EDIT
After Omnomnomnom's note that $$$$A = \frac 1{mn}\underbrace{\pmatrix{ 0&1&\cdots & 1\\ 1&0&\ddots&\vdots\\ \vdots&\ddots&\ddots&1\\ 1&\cdots&1&0}}_{= C_{m \times m}} \otimes J_n + \frac 1m I_{mn}$$$$
I did some computation of the spectrums of the individual matrices. First, the characteristic polynomial of the all-ones $$J_n$$ is $$(\lambda-n)\lambda^{n-1}$$ and hence its spectrum (with the multiplicities) is $$$$\sigma(J_n)=\{(n,1),(0,n-1)\}.$$$$ For $$C$$, assume that $$\lambda_1,\dots,\lambda_m$$ are its eigenvalues. By the facts that $$\mathrm{det}(C-(-1)I_m)=det(J_m)=0$$, $$C\mathbf{1}_m=(m-1)\mathbf{1}_m$$ and $$\mathrm{trace}(C)=\sum_i\lambda_i=0$$ it turns out that $$$$\sigma(C)=\{(m-1,1),(-1,m-1)\}.$$$$ Suppose that $$\mu_1,\dots,\mu_n$$ are the eigenvalues of $$J_n$$ then by the Kronecker product's properties the spectrum of $$CJ_n$$ consists of the pairwise products $$\lambda_i\mu_j, \forall i,j$$.
Your observations are correct and hold for arbitrary $$m,n$$. It suffices to note that $$A = \frac 1{mn}\pmatrix{ 0&1&\cdots & 1\\ 1&0&\ddots&\vdots\\ \vdots&\ddots&\ddots&1\\ 1&\cdots&1&0} \otimes J_n + \frac 1m I_{mn}$$ and use the properties of the Kronecker product.
In more detail: $$C_{m \times m}$$ is a rank 1 update of a scalar matrix, so we find that its eigenvalues are $$-1$$ with multiplicity $$m-1$$ and $$m-1$$ with multiplicity $$1$$. On the other hand, $$J_n$$ has eigenvalues $$0$$ with multiplicty $$n-1$$ and $$n$$ with multiplicity $$1$$.
It follows that $$C \otimes J$$ has eigenvalues $$0$$ with multiplicity $$m(n-1)$$, $$-n$$ with multiplicity $$m-1$$, and $$n(m-1)$$ with multiplicity $$1$$.
From there, it suffices to note that $$\lambda$$ is an eigenvalue of $$A$$ if and only if $$c \lambda + d$$ is an eigenvalue of $$c A + dI$$.
• I have made some edits based on your response (please see edit). However, the resulting sum of the Kronecker product and the diagonal matrix confuses me. How can I proceed?
– mgus
Mar 5 '20 at 3:35
• @mgus I'll add in what I had in mind when I have the chance Mar 5 '20 at 10:08
• See my latest edit Mar 5 '20 at 14:21
• Thanks for your response. Now it is clear to me except one thing: if the spectrums of $C$ and $J_n$ are indeed $\sigma(C)=\{(m-1,1),(-1,m-1)\}$ and $\sigma(J_n)=\{(n,1),(0,n-1)\}$, respectively then where is the eigenvalue $-n(m-1)$ of $C \otimes J$ is coming from? I thought it would be $n(m-1)$ with multiplicity $1$. After doing the remaining of the calculations for the eigenvalues of $A$ it seems that this should be true but how? Am I missing some plus or minus somewhere?
– mgus
Mar 5 '20 at 17:37
• @mgus You're right about $n(m-1)$; I just had an extra minus sign there. Mar 5 '20 at 17:39 | 2021-12-08T16:46:57 | {
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http://math.stackexchange.com/questions/261100/optimization-of-the-area-of-a-cross-inscribed-in-a-circle | # Optimization of the area of a cross inscribed in a circle
I've really been scratching my head over this optimization problem. "Consider a symmetric cross inscribed in a circle of radius $r$." The length from the center of the cross to the middle of one of its arms is $x$. Also, the angle between two line segments drawn from the cross's center to the vertices of one of its arms has a measure of $\theta$. Here's a diagram:
There are three parts to the problem: "(a) Write the area $A$ of the cross as a function of $x$ and find the value of $x$ that maximizes the area. (b) Write the area $A$ of the cross as a function of $\theta$ and find the value of $\theta$ that maximizes the area. (c) Show that the critical numbers of parts (a) and (b) yield the same maximum area. What is that area?"
So, let me show you what I've done so far. For part (a), I decided to break the cross into two middle rectangles and two side rectangles. I saw that a middle rectangle (from the center to the top) would have an area of
$$x \cdot 2 \sqrt{r^2 - x^2}$$
using the Pythagorean theorem. I worked out that a side rectangle (the remaining area on the right, adjacent to the middle rectangles) would have an area of
$$2 \sqrt{r^2-x^2} \cdot \left( x - \sqrt{r^2 - x^2} \right) .$$
So, the area of the cross is
$$A = 2 \bigg( x \cdot 2 \sqrt{r^2 - x^2} + 2 \sqrt{r^2 - x^2} \cdot \Big( x - \sqrt{r^2 - x^2} \Big) \bigg) = 8x \sqrt{r^2 - x^2} - 4r^2 + 4x^2 .$$
If my math is right there (fingers crossed), then I'll take the first derivative to locate a maximum.
$$A^\prime = 8 \sqrt{r^2 - x^2} + 8x \left( 1 \over 2 \right) \left( r^2 - x^2 \right)^{- {1 \over 2}} \left( -2x \right) + 8x.$$
I was a little unsure about what to do at this point. I plugged the $A^\prime$ equation into my graphing calculator, substituting $1^2$ for $r^2$ (for a radius of $1$). The graph crosses the $x$-axis at $x \approx 0.85$. Substituting $2^2$ for $r^2$ (for a radius of $2$) gives me $x \approx 1.70$. From this, I concluded that
$$A^\prime = 0 \; \mathbf{at} \; x \approx 0.85r.$$
Analysis of graphs of $A$ for various values of $r$ concludes that, indeed, maxima do appear at $x \approx 0.85r$. So, I have the function $A$ in terms of $x$, but I'm curious: What should my final answer be for the second part of (a)? All I have is $x \approx 0.85r$. Is that a sufficient answer?
As for part (b), I really have no idea how to write $A$ in terms of $\theta$. I know that $\text{area} = {1 \over 2} b \cdot c \cdot \sin A$ for triangles, but I really need help writing the area of this cross in terms of $\theta$.
Part (c) should be easy enough once I finish (b).
If you got to the end of this, I sincerely thank you for reading, and I would really appreciate an answer (and any corrections to my math). Thanks!
-
You can simplify the formula for $A'$ if you set it to zero and multiply with $\sqrt{1-x^2}$ (with $r=1$). In this way you get $0=1-2x^2+8x\sqrt{1-x^2}$. This is still tricky, but you might want to try a tailor expansion until $x^2$ to simplify further. – Konstantin Schubert Dec 18 '12 at 1:07
@Konstantin I computed $x\sqrt{1-x^2}=2x^2-1$. Square both sides and you have a quadratic in $x^2$. The solutions are $x^2={1\over2}\pm{\sqrt5\over10}$. – David Mitra Dec 18 '12 at 1:22
@DavidMitra aww thanks – Konstantin Schubert Dec 18 '12 at 1:28
Without loss of generality we may assume that the radius is $1$: we can scale area by $r^2$ later.
By the formula you quoted, the area of the triangle enclosed by the two lines that form the angle $\theta$ is $\frac{1}{2}\sin\theta$. The area covered by one of the two full arms of the cross is therefore $4$ times this, which is $2\sin\theta$.
Double this to get the sum $4\sin\theta$ of the areas covered by the two full arms. Unfortunately, this sum counts the area of the middle square twice. So we will need to subtract the area of that square.
Note that by trigonometry, the horizontal segment at the top of the cross has length $2\sin(\theta/2)$. Thus the middle square has area $4\sin^2(\theta/2)$. Using the trigonometric identity $\cos 2\phi=1-2\sin^2\phi$, we find that the middle square has area $2-2\cos \theta$.
So the area of the cross is $4\sin\theta +2\cos\theta-2$. Maximizing should be straightforward.
Remarks: $1.$ We don't really need calculus. Look at the equivalent problem of maximizing $4\sin\2\theta+2\cos\theta$. Rewrite this as $$2\sqrt{5}\left(\frac{2}{\sqrt{5}}\sin \theta+\frac{1}{\sqrt{5}}\cos\theta\right),$$ and let $\psi$ be the angle whose cosine is $2/\sqrt{5}$ and whose sine is $1/\sqrt{5}$. Then our expression becomes $2\sqrt{5}\sin(\theta+\psi)$. The maximum possible value of the sine function is $1$. So the maximum area is $2\sqrt{5}-2$.
$2.$ We can use the above calculation to answer your question about $x$. Alternately, set the derivative equal to $0$, as you did. Manipulation will yield an explicit expression for the root.
$3.$ At a certain stage you were maximizing $8x\sqrt{r^2-x^2}-4r^2+4x^2$. Let $x=r\sin t$. We want to maximize $8r^2\sin t\cos t-4r^2+4r^2\sin^2 t$. Forget about the $r^2$ part, it is a constant multiplier. Now simplify to $8\sin t\cos t-4\cos^2 t$, differentiate. (Using double angle identities to simplify first is a good idea.) We can think of this as just a technical device to ensure we end up with a simple equation.
-
Could you show me how $\sqrt{2 - 2\cos\theta} = 2\sin(\theta/2)$? – Jackson Dec 18 '12 at 1:44
Recall the trig identity $\cos 2t=2\cos^2 t-1=1-2\sin^2 t$. Rewrite as $2\sin^2 t=1-\cos 2t$. Double. Get $4\sin^2 t=2-2\cos 2t$. Finally, let $t=\theta/2$. – André Nicolas Dec 18 '12 at 1:47
Thank you so much! It took me a while, but I got everything figured out. I cannot thank you enough :) – Jackson Dec 18 '12 at 2:07
I will add a last remark to my post in a few minutes. – André Nicolas Dec 18 '12 at 2:15
I'm trying to solve the same problem, and got to the same concusions. The thing is, if you graph the function you get this :
as you can see, the graph takes negative values when x gets smaller, but if the function represents the area of the cross, how come that area is negative ??
I even made a little program to test this, here it is: http://www.khanacademy.org/cs/calculus-maxarea-circle-v00x/1512261081
does anybody have an idea of what's going on ?
- | 2016-07-27T17:31:28 | {
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https://www.physicsforums.com/threads/easiest-way-to-learn-exact-values-for-trig-functions.840460/ | # Easiest way to learn exact values for trig functions?
1. Oct 30, 2015
### cmkluza
I'm realizing now how much I need to know the exact values of various trigonometric functions, as shown in various trig tables. Memorizing is pretty arduous, and I'd prefer to understand it, so how can I learn all of these?
2. Oct 30, 2015
### Staff: Mentor
A long long time ago, my math teacher made us build our own trigonometric circles. Having a geometrical representation really helped me, and the values of the functions have stuck with me ever since.
3. Oct 30, 2015
### cmkluza
Thanks! I guess I'd forgotten that these circles existed. Does this mean that if I memorize the important angles/values from the first quadrant I can use the identities of -sin(θ) = sin(-θ) and cos(-θ) = cos(θ) to figure out the values in each other quadrant?
4. Oct 30, 2015
### Staff: Mentor
It's even more than that. Consider for instance π/6 (30°): you get halfway up on the y axis, so obviously sin π/6 = 1/2. Likewise, for π/4, you can see it as building a square of side 1/2 (along x and y), so the diagonal is (according to Pythagoras) $\sqrt{(1/2)^2 + (1/2)^2} = 1/\sqrt{2} = \sqrt{2}/{2}$. All of these geometric equivalents have really helped me.
5. Oct 30, 2015
### symbolipoint
That graphical circle representation is typically called, the Unit Circle, and is a very important part of Trigonometry instruction. Is is used frequently. The picture can often help because of how the circle is symmetric. Full ray rotation, 2*pi radians. Same function values as for zero degrees rotation.
6. Oct 30, 2015
### cmkluza
What would you guys say are the most important angles to learn? I've seen some charts that go from 0, adding π/12 each segment, and that seems like an awful lot to go through in order to find these angles, are the three, π/6, π/4, and π/3 and their counterparts typically sufficient?
7. Oct 30, 2015
### symbolipoint
THE MOST COMMON you must know are 0, 2*pi, pi/2, pi/4, pi/3, pi/6.
8. Oct 30, 2015
### Staff: Mentor
I realize that this might not be as clear as it can be. Rather, take the radius as the hypotenuse of a right triangle with two equal sides of length $a$ (along the x and y axes), so by Pythagoras $2a^2 = 1 \Rightarrow a = 1/\sqrt{2}$, thus $\cos \pi/4 = \sin \pi/4 = \sqrt{2}/2$.
9. Oct 30, 2015
### Staff: Mentor
πTo expand on what symbolipoint said, the important angles in the first quadrant, and their sines and cosines are:
$\sin(0) = \frac {\sqrt{0}} 2 =\cos(\pi/2)$
$\sin(\pi/6) = \frac {\sqrt{1}} 2 =\cos(\pi/3)$
$\sin(\pi/4) = \frac {\sqrt{2}} 2 =\cos(\pi/4)$
$\sin(\pi/3) = \frac {\sqrt{3}} 2 =\cos(\pi/6)$
$\sin(\pi/2) = \frac {\sqrt{4}} 2 =\cos(0)$
These values should be memorized, but you can use symmetry to figure out the sines, cosines, tangents, etc. of the counterparts of these angles in the other three quadrants.
10. Oct 30, 2015
Staff Emeritus
There are 7 special angles worth memorizing:0, 15, 30, 45, 60, 75 and 90. Sines going up are cosines coming down. Tangent=sine/cosine may be a bit slow for calculating in your head for 15 and 75, and arguably 30 and 60, but the other three are trivial. So you have at most 10 - possibly 8 - facts to remember. That shouldn't be too arduous.
11. Oct 30, 2015
### Student100
Another method that may help you should you forget is to draw a picture and use geometry/trig:
From my poorly drawn circle you can see you want to find the point, (x, y) on the unit circle that intersects with a line of the radial length drawn 30 degrees from the origin. The right triangle formed by hypotenuse (which is our radius of 1) forms a special 30, 60, 90 triangle. A property of such a triangle is that the hypotenuse is twice the length of the shorter leg, and the longer leg is $\sqrt{3}$ times the length of the shorter leg. Since the hypotenuse is one, the length of the shorter leg, $y=\frac{1}{2}$ and the longer leg, x is $\sqrt{3}$ times more so, $x=\frac{\sqrt{3}}{2}$. So the $sin(30)=\frac{1}{2}$ while the $cosine(30)=\frac{\sqrt{3}}{2}$
You can repeat the same logic for 60 degrees, or just realize that the x and y's are swapped. For a 45-45-90 triangle, the legs are congruent, so the hypotenuse is $\sqrt{2}$ the length of either leg. So lets say the hypotenuse has length $x\sqrt{2}$ in this case, since we know the length to be one, we can simply say $x\sqrt{2}=1$ or $x=\frac{1}{\sqrt{2}}$. The legs are congruent so $y=\frac{1}{\sqrt{2}}$.
For 15 and 75 degrees you can use $sin(45-30)$ and solve using trig rules. For 0 and 90 you can again draw a line from the origin to the corresponding point on the circle and see either, x=1 in the case of an angle of 0, or x=0 in the case of an angle of 90 or vice versus.
Not sure if this will help or not, hopefully it'll allow you to re-derive these values should you forget them on a test.
12. Oct 30, 2015
### symbolipoint
The common reference angles that are learned on the Unit Circle come from a couple of Special Triangles.
13. Nov 4, 2015
### rs1n
If you look at the picture here: http://etc.usf.edu/clipart/43200/43216/unit-circle8_43216_lg.gif you will see that only the first quadrant is needed (and even then, only the angles between 0 and 45 degrees are needed (everything else can be derived by symmetry of the circle. As for the coordinates, I find it easier to remember them as:
0: $\left( \sqrt{\frac{4}{4}}, \sqrt{\frac{0}{4}}\right)$
pi/6: $\left( \sqrt{\frac{3}{4}}, \sqrt{\frac{1}{4}}\right)$
pi/4: $\left( \sqrt{\frac{2}{4}}, \sqrt{\frac{2}{4}}\right)$
pi/3: $\left( \sqrt{\frac{1}{4}}, \sqrt{\frac{3}{4}}\right)$
pi/2: $\left( \sqrt{\frac{0}{4}}, \sqrt{\frac{4}{4}}\right)$
14. Nov 5, 2015
### Erland
Easiest way is to use to draw a half square and a half equiliteral triangle, and use the Pythagorean theorem to obtaim sin, cos and tan for the angles in tjose triangles, i.e. 45, 30 and 60 degrees.
15. Nov 5, 2015
### Fredrik
Staff Emeritus
And once you have those, you can use them to find the values for some other angles. For example,
\begin{align*}
&\sin 15=\sin(45-30)=\sin 45\cos 30-\cos 45\sin 30\\
&\sin 120=\sin(2\cdot 60)=2\sin 60\cos 60
\end{align*} The only thing worth committing to memory in my opinion, is (what Erland said) that you start with half a square and half an equilateral triangle.
16. Nov 5, 2015
### symbolipoint
The unit circle and its typically shown values can be easily memorized.
THIS here is essential: | 2017-11-21T14:17:32 | {
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https://math.stackexchange.com/questions/2822356/binomial-coefficient-real-life-example | Binomial coefficient real life example.
"At a university, $15$ juniors and $20$ seniors volunteer to serve as a special committee that requires $8$ members. A lottery is used to select the committee from among the volunteers. Suppose the chosen students consists of six juniors and two seniors.
(a) For a test of homogeneity, what are the expected counts?
This question I understand.
(b) If the selection had been random, what is the probability of the committee having exactly two seniors?
My answer was that the probability is binomial, with Binom$(k=2, n=8, p=0.57)$, but this is apparently wrong. Instead the correct answer is: $$\frac{\binom{20}{2}\binom{15}{6}}{\binom{35}{8}}$$.
Can anyone explain the difference between this and standard binomial distribution?
• I suppose that you made a type and that there are $15$ junior volunteers, not $13$. Am I right? – José Carlos Santos Jun 17 '18 at 7:12
• It's not a straight binomial since the trials are not independent. Knowing that the first choice was a senior changes the probability that the second was a senior. – lulu Jun 17 '18 at 7:12
• Ahh. Thanks! I understand now :-) – Mathe Jun 17 '18 at 7:20
• This is the hypergeometric distribution – Lord Shark the Unknown Jun 17 '18 at 7:21
• To be sure of useful answers, you should state null and alternative hypotheses for your 'test of homogeneity'. You don't have quite large enough expected counts for the chi-squared goodness-of-fit statistic to truly have a chi-squared distribution. // Using the hypergeometric distribution, it seems you are aiming at "Fisher's exact test' which you can google. – BruceET Jun 17 '18 at 8:04
The issue is that the trials are not independent from each other. Having chosen a junior first, for example, changes the probability of now choosing a senior.
You must look at how many ways are there to choose two seniors (which is exactly $\binom{20}{2}$) and how many ways are there to choose six juniors (which is exactly $\binom{15}{6}$) and multiply them. This gives you the overall number of valid arrangements.
To get the probability, simply divide by the number of overall arrangements possible (which is $\binom{35}{8}$).
• To get a p-value you also need to include the probabilities of more extreme cases. – BruceET Jun 17 '18 at 8:47
In case it is helpful, here is Minitab output for testing two proportions. It attempts a normal test (with a warning about sample sizes being too small) and does Fisher's exact test.
Test and CI for Two Proportions
Sample X N Sample p
1 6 15 0.400000
2 2 20 0.100000
Difference = p (1) - p (2)
Estimate for difference: 0.3
95% CI for difference: (0.0193759, 0.580624)
Test for difference = 0 (vs ≠ 0): Z = 2.10 P-Value = 0.036
* NOTE * The normal approximation may be inaccurate for small samples.
Fisher’s exact test: P-Value = 0.051 | 2019-09-23T21:09:32 | {
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https://stats.stackexchange.com/questions/263931/calculation-of-transition-probabilities-of-markov-chain-problem | # Calculation of transition probabilities of Markov Chain problem
In the step of learning Markov chains, I came across few questions from assignments on UTDallas website. Finding the transition probabilities seems a bit hard for me in this one particular question only. The question is
(Sec 7.3, page 355, #1) Consider a system with two components. We observe the state of the system every hour. A given component operating at time $n$ has probability $p$ of failing before the next observation at time $n + 1$. A component that was in a failed condition at time $n$ has a probability $r$ of being repaired by time $n + 1$, independent of how long the component has been in a failed state. The component failures and repairs are mutually independent events. Let $X_n$ be the number of components in operation at time $n$. The process $\{X_n, n = 0, 1, . . .\}$ is a discrete time homogeneous Markov chain with state space $I = \{0, 1, 2\}$.
a) Determine its transition probability matrix, and draw the state diagram.
b) Obtain the steady state probability vector, if it exists
Although the answers are given, but I cannot understand that on what basis the transition probabilities are calculated. Can someone help me in this...
I had the following guesses my ownself (which mostly proved to be wrong)
\begin{bmatrix} 1-r-r^2 & r & r^2\\ ??& ?? & r(1-p) \\ p^2 & ?? & ?? \end{bmatrix}
The actual solution that the website pose is following...
\begin{bmatrix} (1-r)^2 & 2r(1-r) & r^2\\ p(1-r)& pr + (1-p)(1-r) & r(1-p) \\ p^2 & 2p(1-p) & (1-p)^2 \end{bmatrix}
• Thanks Yes it is self-study question. What I am asking is not the solution since I am not taking this exam, I am trying to learn the probability, in general, and markov chains, in specific. Appreciate your comment @Tavrock. – Kashan Feb 25 '17 at 4:11
• I am looking if someone can help me to learn HERE how transition probabilities are calculated...Since to my understanding, transitional probabilities are different than normal probabilities... – Kashan Feb 25 '17 at 10:59
• Can you explain the reasoning behind your guesses? (Also, based on the last comment, do you know what a transition probability is?) – Juho Kokkala Feb 25 '17 at 14:27
Note that, there are two components and each of them individually could be either in working or failure state. Then the states of the Markov chain be designated as $ff$ both components failed, $wf$ one component working and one component failed, and $ww$ both components working. Given that, the components function independently of each other.
• If the chain is in state $ff$ at time $n$, at the next time point it could be in the state
• $ff$ with probability $(1-r)(1-r)=(1-r)^2$, as the probability of not being repaired by next time point is $(1-r)$;
• $wf$ with probability $2r(1-r)$, as it may be the first component
that was repaired or the the second component that was repaired, so
that only one of the two components will be working;
• $ww$ with probability $r^2$, if both the components got repaired.
• If the chain is in the state $wf$ at time $n$, then at the next time point it could be in the state
• $ff$ with probability $p(1-r)$, as the working component failed and the component requiring repair was not yet repaired;
• $wf$ with probability $pr+(1-p)(1-r)$, as the working component was
not failed and the component requiring repair is not repaired, so
that only one component is working, which has a probability
$(1-p)(1-r)$; or the component working at the previous time has
failed and the component requiring repair has got repaired, which has a probability $pr$; mutually exclusiveness of the events results in the required probability;
• $ww$ with probability $r(1-p)$, as the working component does not
require repair and the failed component got repaired.
• If the chain is in the state $ww$ at time $n$, then at the next time point it could be in the state
• $ff$ with probability $p^2$, as both components have failed;
• $wf$ with probability $2p(1-p)$, which is the sum of probabilities
of two mutually exclusive events, viz., the first component failed
and the second working or the first component working and the
second component failed;
• $ww$ with probability $(1-p)^2$, as both components are still
working.
Hence, the transition matrix:
\begin{equation*} P=\begin{array}{c|ccc} &ff & wf & ww\\ \hline ff & (1-r)^2& 2r(1-r) & r^2\\ wf & p(1-r) & pr+(1-p)(1-r) & r(1-p) \\ ww & p^2 & 2p(1-p) & (1-p)^2 \end{array} \end{equation*}
• Thanks.... Helped me a lot in developing understanding and brushing off some dust over my mind – Kashan Feb 26 '17 at 2:20 | 2019-12-10T07:49:53 | {
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https://www.physicsforums.com/threads/reduction-of-order.778044/ | # Homework Help: Reduction of Order
1. Oct 24, 2014
### QuantumCurt
1. The problem statement, all variables and given/known data
Given - $$y_1(x)=sin(2x)$$, find a second linearly independent solution to $$y''+4y=0$$
2. Relevant equations
I'm using reduction of order to write a second solution as a multiple of the first solution. I think I've gotten to the right answer, my main question is in how I should write the general solution of the equation.
3. The attempt at a solution
Using the fact that $y_1(x)=sin(2x)$, and that $y_2(x)=vy_1(x)$ I'm writing a second solution as $$y_2(x)=vsin(2x)$$
Now taking the first and second derivatives I get $$y_2'(x)=v'sin(2x)+2vcos(2x)$$
and $$y_2''(x)=v''sin(2x)+4v'cos(2x)-4vsin(2x)$$
Now I substitute back into the original equation and get, $$v''sin(2x)+4v'cos(2x)=0$$
Now to reduce order; $z=v' and z'=v''$
Then substituting - $$z'sin(2x)+4zcos(2x)=0$$
Which leads to $$\frac{dz}{z}=-4cot(2x)dx$$
Integrating this to logarithms and then exponentiating etc. leads me to - $$z=csc^2(2x)$$
Since $z=v'$ I can substitute again
$$v=\int csc^2(2x)dx$$
$$v=-\frac{1}{2}cot(2x)$$
Now since $y_2(x)=vy_1(x)$, I can say that,
$$y_2(x)=[-\frac{1}{2}cot(2x)][sin(2x)]$$
which simplifies to
$$y_2(x)=-\frac{1}{2}cos(2x)$$
The formula for the general solution is $y(x)=c_1y_1(x)+c_2y_2(x)$. Given this fact, my general solution is written as - $$y(x)=c_1sin(2x)-\frac{1}{2}c_2cos(2x)$$
Now...my question -
Since $-\frac{1}{2}c_1$ is really just an arbitrary constant, can I write my general solution in terms of just a constant rather than a fraction times a constant? This would give me the solution of -
$$y(x)=c_1sin(2x)+c_2cos(2x)$$
Is this essentially the same solution, or would it be better to write it in terms of the negative fractional constant?
Any help would be much appreciated. :)
Last edited: Oct 24, 2014
2. Oct 24, 2014
### vela
Staff Emeritus
Just absorb the -1/2 into c2.
3. Oct 24, 2014
### QuantumCurt
Thank you. I figured that was the case, but I wanted to be sure. | 2018-05-25T21:02:18 | {
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https://mathematica.stackexchange.com/questions/166711/how-to-create-a-particular-5-x-5-square-colour-map-of-the-mean-value-of-data-poi | # How to create a particular 5 x 5 square colour map of the mean value of data points
I am trying to create a specific type of colour map for some data. The data is in the form of a list with dimensions 300000 x 3, that is 300,000 sets of {x, y, n} where x and y are basically the $xy$ cartesian coordinates and n is a value either 0 or 1. Doing a list plot of the 300,000 points gives the following figure:
What I am trying to make is a colour map of this figure, that is made up of 5 x 5 evenly sized squares (so 25 all up), in which the colour-key denotes the mean of the n values of all the points within that square. Is this possible?
Please let me know if any additional information is needed,
## 1 Answer
SeedRandom[1]
data = Join[RandomReal[1, {300000, 2}], RandomChoice[{0, 1}, {300000, 1}], 2];
Dimensions[data]
{300000, 3}
nbins = 5;
binlims = Through[{Floor[Min@#, .01] &, Ceiling[Max@#, .01] &}@#] & /@ Transpose[data];
{xbins, ybins} = {##, -Subtract[##]/nbins} & @@@ Most[binlims];
binlists = BinLists[data, xbins, ybins, {0, 2, 2}];
binmeans = Flatten /@ Map[Mean, binlists[[All, All, All, All, -1]], {-2}];
cft = ChartingFindTicks[{0, nbins}, {0, 1}];
MatrixPlot[binmeans, DataReversed -> True, ColorFunction -> "Rainbow",
FrameTicks -> {{cft, cft}, {cft, cft}}]
With nbins = 25 we get
• I take it that if I would like to increase the number of squares, I just alter that 5 in the second line? Also, I tried changing it to 25 and it looks good, but there are a lot of data ticks. Is it possible to maybe only show every second, or even third data tick? – Lagiacrus Feb 27 '18 at 18:38
• @Lagiacrus, right; changing 5 changes the number of bins. Re ticks, please see the updated version. – kglr Feb 27 '18 at 20:17
• So far the code appears to have been working well, but I just tried it for a similar situation in which the third element of the list can be either 0, 1, 2 or 3 (instead of the original 0 or 1) and in this case the binmeans never ends up with a value higher than 1. Is there a limit on the value that binmeans can take? – Lagiacrus Mar 3 '18 at 12:57
• @Lagiacrus, you can change {0, 2, 2} in the definition of binlists to {0,, 4, 4} if the third list can take values in {0,1,2,3}`. – kglr Mar 3 '18 at 14:00 | 2020-01-28T22:03:35 | {
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https://math.stackexchange.com/questions/3926426/probability-of-n-same-type-pairs-from-a-heterogeneous-group | # probability of N same type pairs from a heterogeneous group
There are 16 people: 8 males and 8 females. The people are randomly divided to pairs, each pair get the exact same mission. Let $$X$$ be the number of male-male pairs.
1. calculate $$P\{X=3\}$$
2. calculate $$P\{X=i\}$$ where $$i$$ is each value $$X$$ can take
3. find $$E[X]$$
for (1) there are $$\binom{8}{2,2,2,1,1}$$ ways to divide the males to 3 all male pairs and $$\binom{8}{2,2,2,1,1}$$ ways to divide the females to 3 all female pairs and there are 4 ways to pair the remaining males and females so I expected the probability to be $$P\{X=3\}=\frac{\binom{8}{2,2,2,1,1}\cdot \binom{8}{2,2,2,1,1} \cdot 4}{\binom{16}{2,2,2,2,2,2,2,2}} = \frac{\frac{8!}{8}\cdot \frac{8!}{8}\cdot 4}{\frac{16!}{2^8}}=\frac{8!^2\cdot 4\cdot 64 \cdot 4}{16! \cdot 64}=\frac{8!^2}{15!} \approx 0.0012$$ but it is incorrect so instead I tried, I choose 6 males and pair them in $$\binom{8}{6}\binom{6}{2,2,2}$$ and the 2 remaining males I have $$\binom{2}{1}\binom{8}{1}$$ and $$\binom{7}{1}$$ ways to pair with a female, and the remaining 6 females I have $$\binom{6}{2,2,2}$$ ways to pair so $$\frac{\binom{8}{6}\binom{6}{2,2,2}\binom{2}{1}\binom{8}{1}\binom{7}{1}\binom{6}{2,2,2}}{\binom{16}{2,2,2,2,2,2,2,2}}=\frac{\frac{8!}{16}\cdot 16 \cdot 7\cdot\frac{6!}{8}}{\frac{16!}{2^8}}=\frac{7!\cdot 7! \cdot 2^8}{16!}\approx 0.0003$$ which is also incorrect. I'm not sure how I can calculate $$P\{X=3\}$$ or any other $$P\{X=i\}$$
after reading @saulspatz's answer I found a possible solution: all possible pairing is given by $$\binom{16}{2,2,2,2,2,2,2,2}\cdot \frac{1}{8!}$$ because all the pairs have the same role but the same works for each of the 3 male-male or female-female pairs, hence $$\binom{6}{2,2,2}\cdot \frac{1}{3!}$$ and because the 2 male-female pairs also have the same role I get $$P\{X=3\}=\frac{\frac{\binom{8}{6}\binom{6}{2,2,2}}{3!}\frac{\binom{2}{1}\binom{8}{1}\binom{7}{1}}{2!}\frac{\binom{6}{2,2,2}}{3!}}{\frac{\binom{16}{2,2,2,2,2,2,2,2}}{8!}}=\frac{\frac{8!}{16\cdot 3!}\cdot \frac{16\cdot 7}{2!}\cdot\frac{6!}{8\cdot 3!}}{\frac{16!}{2^8 \cdot 8!}}=\frac{8!\cdot 7! \cdot 7!\cdot 2^8}{16!\cdot 3!\cdot 3!\cdot 2!}\approx 0.1740$$ and from here calculating $$P\{X=i\}$$ is the same and and there is no problem finding $$E[X]$$
• What is a mission and what role does it play in your question (if any)? Also, you say some results you calculated are incorrect. How do you know they are incorrect? (I am not saying they are correct, I am just pointing out that you made unsupported claims in your post.) – mathguy Nov 28 '20 at 16:34
• @mathguy I think the mission part is so that the order of the pairs doesn't matter. I have 4 possible answers and I need to choose one – CforLinux Nov 28 '20 at 16:48
The number of ways to divide the $$16$$ people is not pairs is not $$\binom{16}{2,2,2,2,2,2,2,2}$$ but $$\frac1{8!}\binom{16}{2,2,2,2,2,2,2,2}$$ To divide the people into pairs, arrange them into a line, and pair the first and second, the third and fourth, and so on. The order of the tow people in the pairs doesn't matter, so we have $$\frac{16!}{(2!)^8}$$ but also the order of the pairs themselves doesn't matter, so we must divide by $$8!$$. | 2021-06-13T09:20:14 | {
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https://math.stackexchange.com/questions/1809061/describe-the-structure-operatornamegal-mathbbq-zeta-4-mathbbq | # Describe the structure $\operatorname{Gal}(\mathbb{Q}(\zeta_4)/\mathbb{Q})$
I know that if $n$ is prime then $G=\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq \Bbb Z_{n-1}$
But I am unsure what $G$ is when $n$ is not prime. For example when $n=4$:
$\operatorname{Gal}(\mathbb{Q}(\zeta_4)/\mathbb{Q}) \simeq \Bbb Z_4^*$
What is the structure of the group $\Bbb Z_4^*$? I am guessing it is cyclic, but I do not know how to explicitly find its generators.
Also, what is the basis of $\mathbb{Q}(\zeta_4)$ over $\mathbb{Q}$? Is it $\{\zeta, ..., \zeta^4\}$?
For general $n$, the units of the ring $\Bbb Z/n\Bbb Z$, which I’ll write $(\Bbb Z/n\Bbb Z)^*$, do not form a cyclic group. For, when $n=p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$, then $(\Bbb Z/n\Bbb Z)^*\cong(\Bbb Z/p_1^{e_1}\Bbb Z)^*\oplus(\Bbb Z/p_2^{e_2}\Bbb Z)^*\oplus\cdots\oplus(\Bbb Z/p_m^{e_m}\Bbb Z)^*$, the $p_i$ all being prime.
For odd primes $p$, $(\Bbb Z/p^e\Bbb Z)^*$ is always cyclic, but the structure of $(\Bbb Z/2^e\Bbb Z)^*$ is $C_2\oplus C_{2^{e-2}}$, where by $C_m$ I mean an abstract cyclic group of order $m$. (The explanation of why these groups have this structure is a story for another night.)
• Thank you. So if I we take $n=18=3^2 \times 2 \implies Gal(\mathbb{Q}(\zeta_{18}/\mathbb{Q}) \simeq \mathbb{Z}/9\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ – amiz9 Jun 2 '16 at 13:17
• Take $n=8=2^3 \implies Gal(\mathbb{Q}(\zeta_{8}/\mathbb{Q}) \simeq \mathbb{C_2} \oplus \mathbb{C_2}$? – amiz9 Jun 2 '16 at 13:19
• Are these correct? This is a nice classification - can you point me to somewhere where I can read more about the generators of these groups and their structure? Do all cyclotomic extension fields have a similar structure? – amiz9 Jun 2 '16 at 13:21
• Don’t forget that (in your notation) $\Bbb Z_2^*$ is trivial, not order two. In fact, $\Bbb Q(\zeta_{18})=\Bbb Q(\zeta_9)$. And yes, $\Bbb Z_8^*\cong C_2\oplus C_2$, generators $3$ and $7$. – Lubin Jun 2 '16 at 13:32
• Structure of $\Bbb Q(\zeta_n)$ should be in most books on algebraic number theory. – Lubin Jun 2 '16 at 13:35
I mean, $\zeta_4 = i$ so the extension has degree $2$, so the Galois group is a finite group of order $2$ and there's only one of those, $\Bbb Z/2\Bbb Z$. Like all quadratic extensions, the basis is $\{1, i\}= \{1,\sqrt{-1}\}$. If you want the exact structure, it's mercifully simple: $i\mapsto \overline{i}$ is the automorphism, i.e. complex conjugation is the only one.
• OK thanks, that all makes sense, except I am unsure why $\zeta_4=i \implies [\mathbb{Q}(\zeta_4) : \mathbb{Q}]=2$.. could you help me understand this please? And is the $i \rightarrow i$ the automorphism because $i \notin \mathbb{Q}$? Thanks – amiz9 Jun 2 '16 at 0:16
• Is it because the we must take $i$ to the power of $2$ to return back to our field $\mathbb{Q}$? – amiz9 Jun 2 '16 at 0:17
• @amiz9 that's one way to think of it, yeah. $\Bbb Q(i) \cong \Bbb Q[x]/(x^2+1)$. And look at the little bar over the $i$, it means complex conjugate. You can also think of it just as $i\mapsto -i$. – Adam Hughes Jun 2 '16 at 0:18
• ok thanks. So in general to find $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, we can calculate the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$. The degree of this polynomial will be the order of our group. Is that right? – amiz9 Jun 2 '16 at 0:23
• it will always be $\varphi(n)$ so you don't really need the polynomial. Generally speaking all automorphisms are $\zeta_n\mapsto \zeta_n^k$ for $k\in\Bbb Z/n\Bbb Z^*$. In this case $i\mapsto i^3 = i\mapsto (i)^2\cdot i = -i$ is how it works. – Adam Hughes Jun 2 '16 at 0:25
If $\zeta_n$ is a primitive $n^{th}$ root of unity, then $$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}_n)^*.$$
What the elements of the group look like:
If $\sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, then there is an $a \in \{1,2,\dots,n\}$ with $a$ relatively prime to $n$ such that $$\sigma(\zeta_n) = (\zeta_n)^a$$
For the structure of $(\mathbb{Z}_n)^*$, see Lubin's answer.
• Thank you. What can we say about the generators that generate these groups? – amiz9 Jun 2 '16 at 13:21 | 2019-12-11T03:17:47 | {
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https://www.physicsforums.com/threads/moment-of-inertia-of-a-disk-by-integration.349466/ | # Moment of inertia of a disk by integration
1. ### soopo
226
1. The problem statement, all variables and given/known data
Show that the moment of inertia of a disk is $0.5 mr^2$.
3. The attempt at a solution
$$I = \int R^2 dm$$
Using $dm = \lambda dr$ such that $m = \lambda r$:
$$= \int_{-r}^{r} R^2 \lambda dr$$
$$= \frac { \lambda } {3} ( 2r^3 )$$
$$= \frac {2} {3} (\lambda r ) (r^2)$$
$$= \frac {2} {3} M R^2$$
which should be the moment of inertia for a ring.
Integrating this from 0 to 2pii relative to the angle gives me $\frac {4} {9} m r^3 [/tex], which is wrong. How can you calculate the moment of inertia for a disk? 2. ### jdwood983 383 If you have a disc of radius $$r$$, how can you integrate from $$-r$$ to $$r$$? 3. ### jdwood983 383 The infinitesimal change in mass is given by $$dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr$$ If you use that in your integral and integrate from $$0$$ to $$R$$, you should get the desired result. 4. ### soopo 226 Your result gives me a wrong result: $$I = \int R^2 dm$$ $$= \int R^2 \frac { 2m } {R^2} r dr$$ $$= \int_{0}^{r} 2mr dr$$ $$= [ m r^2 ]^{r}_{0}$$ $$= mr^2$$ The result shoud be $$I = .5 mr^2$$ 5. ### soopo 226 I set the null point to the center of the circle such that I am integrating from -r to r. I am not sure why I cannot do that. 6. ### jdwood983 383 If you set the origin to the center of the circle (which you should always try to do), the smallest value that $$r$$ can be is 0. So it is physically impossible to integrate from $$-r$$ to $$r$$, that is why you can't do it. 7. ### soopo 226 I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards $$r$$, while going to left means towards $$-r$$. Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative $$-r$$. I feel that it is possible to integrate from $$-r$$ to $$r$$ in a cartesian coordinate system. 8. ### jdwood983 383 If you want Cartesian coordinates, then you'll need two integrals: one over $$x$$ and one over $$y$$. While technically you have two integrals in polar, $$r\, \mathrm{and}\, \theta$$, one is already done for you and reduces the integration to just one term: $$r$$. This problem is by far easier in polar coordinates: $$\begin{array}{ll}I&=\int_0^R r^2\frac{2m}{R^2}rdr \\ &=\frac{2m}{R^2}\int_0^Rr^3dr \\ &=\frac{2m}{R^2}\left(\frac{R^4}{4}-0\right) \\ &=\frac{2m}{R^2}\cdpt\frac{R^4}{4} \\ &=\frac{1}{2}mR^2$$ 9. ### soopo 226 If you use symmetry, it is enough to consider only the first quadrant that is where x > 0 and y > 0 such that four of these quadrants form the area of the disk. You do not get the x- and y -coordinates easily from the definition of the moment of inertia. You would get $$I = \int (x^2 + y^2) dm \\ &= \int (x^2 + y^2) m \frac { 2r } {R^2} dr$$ The calculations seem to get challenging, since we need to use Pythogoras such that $$r = \sqrt{ x^2 + y^2 }$$ which implies $$dr = \frac { 1 } { \sqrt {x^2 + y^2} } * 2x$$ We can get similarly the relation relative to $$y$$. The next step is not fun at all: $$I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx$$, where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4$, since it is the maximum radius.
This way the two 2s cancel out.
I do not even know how to integrate this!
Polar coordinate system really seems to be better in this case.
Last edited: Oct 27, 2009
10. ### jdwood983
383
Not quite. The integral you need is given by
$$I=\frac{m}{A}\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\left(x^2+y^2\right)dydx$$
$$I=\frac{m}{\pi R^2}\int_{-R}^R \frac{2\sqrt{R^2-x^2}\left(R^2+2x^2\right)}{3}dx$$
$$I=\frac{m}{\pi R^2}\cdot\frac{\pi R^4}{2}$$
$$I=\frac{mR^2}{2}$$
For most moment of inertia problems, spherical or cylindrical coordinates are the best.
11. ### soopo
226
How did you solve this part?
It has taken my some effort in trying to solve it by hand.
12. ### jdwood983
383
There's a few extra steps between the two lines, like making a change of variables. But to be honest I used Mathematica and just wrote the lines because I forget what changes needed to be made. While it may be good to know the form of the equation, I'm not sure you would need the solution since it is far easier to do it in polar coordinates than in Cartesian coordinates. | 2015-02-28T13:57:35 | {
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http://3429851863.srv040042.webreus.net/ryefz16/symmetric-relation-graph-12be00 | So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. This means drawing a point (or small blob) for each element of X and joining two of these if the corresponding elements are related. Explore anything with the first computational knowledge engine. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. The symmetric relations on nodes are isomorphic Thus, symmetric relations and undirected … directed graph of R. EXAMPLE: Let A = {1,2,3} and R = {(1,3), (2,1), (2,3), (3,2)} be represented by the. Terminology: Vocabulary for graphs often different from that for relations. PROOF. Terminology: Vocabulary for graphs often different from that for relations. Symmetric and antisymmetric (where the only way a can be related to b and b be related to a is if a = b) are actually independent of each other, as these examples show. Symmetric Division Deg Energy of a Graph K. N. Prakash a 1 , P. Siva K ota Red dy 2 , Ismail Naci Cangul 3,* 1 Mathematics, Vidyavardhaka College of Engineering, Mysuru , India Draw each of the following symmetric relations as a graph.' However, it is still challenging for many existing methods to model diverse relational patterns, es-pecially symmetric and antisymmetric relations. This module exposes the implementation of symmetric binary relation data type. Discrete Mathematics Questions and Answers – Relations. In antisymmetric relation, there is no pair of distinct or dissimilar elements of a set. For example, the relation $$a\equiv b\text{ (mod }3\text{)}$$ for a few values: Note: there's no requirement that the vertices be connected to one another: the above figure is a single graph with 11 vertices. 1, April 2004, pp. Conversely, if R is a symmetric relation over a set X, one can interpret it as describing an undirected graph with the elements of X as the vertices and the pairs in R as the edges. Suppose f: R !R is de ned by f(x) = bx=2c. Neha Agrawal Mathematically Inclined 172,807 views with the rooted graphs on nodes. I Undirected graphs ie E is a symmetric relation Why graphs I A wide range of. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. 5 shows the SLGS operator’s operation. Suppose we also have some equivalence relation on these objects. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) ∈ R for every a ∈ ASymmetricRelation is symmetric,If (a, b) ∈ R, then (b, a) ∈ RTransitiveRelation is transitive,If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ RIf relation is reflexive, symmetric and transitive,it is anequivalence relation A symmetric relation can be represented using an undirected graph. Converting a relation to a graph might result in an overly complex graph (or vice-versa). And similarly with the other closure notions. directed graph. Example # 2. may or may not have a property , such as reflexivity, symmetry, or transitivity. Symmetric with respect to x-axis Algebraically Because 2 x 2 + 3 (− y) 2 = 16 is equivalent to 2 x 2 + 3 y 2 = 16, the graph is symmetric with respect to x-axis. Learn its definition with examples and also compare it with symmetric and asymmetric relation … This definition of a symmetric graph boils down to the definition of an unoriented graph, but it is nevertheless used in the math literature. Robb T. Koether (Hampden-Sydney College) Reflexivity, Symmetry, and Transitivity Mon, Apr 1, 2013 12 / 23 The Graph of the Symmetric … consists of two real number lines that intersect at a right angle. 2. definition, no element of. Symmetry, along with reflexivity and transitivity, are the three defining properties of an equivalence relation. One way to conceptualize a symmetric relation in graph theory is that a symmetric relation is an edge, with the edge's two vertices being the two entities so related. For undirected graph, the matrix is symmetric since an edge { u , v } can be taken in either direction. Notice the previous example illustrates that any function has a relation that is associated with it. Write the equivalence class(es) of the bit string 001 for the equivalence relation R on S. subject: discrete mathematics Symmetric relations in the real world include synonym, similar_to. What is the equation of the quadratic in the form y = a(x - r)(x - s) knowing that the y-intercept is (0, -75)? Why study binary relations and graphs separately? I Undirected graphs, i.e., E is a symmetric relation. A symmetric relation is a type of binary relation. Its graph is depicted below: Note that the arrow from 1 to 2 corresponds to the tuple , whereas the reverse arrow from to corresponds to the tuple . Let 0be a non-edge-transitive graph. DIRECTED GRAPH OF AN IRREFLEXIVE RELATION: Let R be an irreflexive relation on a set A. Any relation R in a set A is said to be symmetric if (a, b) ∈ R. This implies that $(b, a) ∈ R$ In other words, a relation R in a set A is said to be in a symmetric relationship only if every value of a,b ∈ A, (a, b) ∈ R then it should be (b, a) ∈ R. Symmetric Division Deg Energy of a Graph K. N. Prakash a 1 , P. Siva K ota Red dy 2 , Ismail Naci Cangul 3,* 1 Mathematics, Vidyavardhaka College of Engineering, Mysuru , India Notice the previous example illustrates that any function has a relation that is associated with it. , v n , this is an n × n array whose ( i , j )th entry is a ij = ( 1 if there is an edge from v i to v j 0 otherwise . This is distinct from the symmetric closure of the transitive closure. This phenomenon causes subsequent tasks, e.g. A relation R is irreflexive if the matrix diagonal elements are 0. Formally, a binary relation R over a set X is symmetric if: If RT represents the converse of R, then R is symmetric if and only if R = RT. For example, a graph might contain the following triples: First, this is symmetric because there is $(1,2) \to (2,1)$. For a relation R in set A Reflexive Relation is reflexive If (a, a) ∈ R for every a ∈ A Symmetric Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R Transitive Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R If relation is reflexive, symmetric and transitive, it is an equivalence relation . Walk through homework problems step-by-step from beginning to end. It's also the definition that appears on French wiktionnary. You can use information about symmetry to draw the graph of a relation. 12-15. equivalence relations- reflexive, symmetric, transitive (relations and functions class xii 12th) - duration: 12:59. Thus, symmetric relations and undirected graphs are combinatorially equivalent objects. Edges that start and end at the same vertex are called loops. However, there is a general phenomenon in most of KGEs, as the training progresses, the symmetric relations tend to zero vector, if the symmetric triples ratio is high enough in the dataset. Let’s understand whether this is a symmetry relation or not. From MathWorld--A Wolfram Web Resource. A relation R is irreflexive if there is no loop at any node of directed graphs. 6 4 2-2-4-6-5 5 Figure 1-x1-y1 y1 x1 y = k x; k > 0 P Q. Fig. The rectangular coordinate system A system with two number lines at right angles specifying points in a plane using ordered pairs (x, y). In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, i.e., g(y) = x if and only if f(x) = y. Let 0have n vertices, and let 00be the hull of 0. In §5, using the analytic approach, we identify the Cheeger constant of a symmetric graph with that of the quotient graph, Theorem 1.3. 1. Symmetry can be useful in graphing an equation since it says that if we know one portion of the graph then we will also know the remaining (and symmetric) portion of the graph as well. Why graphs? You should use the non-internal module Algebra.Graph.Relation.Symmetric instead. The symmetric relations on nodes are isomorphic with the rooted graphs on nodes. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. 2-congruence (n,r)-congruence. Knowledge-based programming for everyone. I undirected graphs ie e is a symmetric relation why. For example, a graph might contain the following triples: First, this is symmetric because there is $(1,2) \to (2,1)$. Section focuses on relations '' in Discrete Mathematics symmetric and off-diagonal Figure 1-x1-y1 y1 x1 =! Included in relation or not is using a directed graph of a in the let! 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https://primer-computational-mathematics.github.io/book/c_mathematics/numerical_methods/14_ill_conditioning_errors.html | # Ill-conditioning and roundoff errors#
Numerical Methods
## Ill-conditioned matrices#
The conditioning (or lack of, i.e. the ill-conditioning) of matrices we are trying to invert is incredibly important for the success of any algorithm.
As long as the matrix is non-singular, i.e. $$\det(A)\ne 0$$, then an inverse exists, and a linear system with that $$A$$ has a unique solution. What happens when we consider a matrix that is nearly singular, i.e. $$\det(A)$$ is very small?
Well smallness is a relative term and so we need to ask the question of how large or small $$\det(A)$$ is compared to something. That something is the norm of the matrix.
Matrices come in all shape and sizes, and their determinants come in all kinds of values. We know that a ill conditioned matrix has a determinant that is small in absolute terms, but the size of determinants is a relative thing, and we need some kind of comparison to determine what is “small” and what is “large”. Thus, we can create such a reference calculating the norms of the matrix. In this notebook, we will explore how to find the norm and how does the norm relate to the ill conditioning of the matrix.
## Vector norms#
Just as for vectors $$\pmb{v}$$ (assumed as a $$n\times 1$$ column vector) where we have multiple possible norms to help us decide quantify the magnitude of a vector:
$\begin{split} ||\pmb{v}||_2 = \sqrt{v_1^2 + v_2^2 + \ldots + v_n^2} = \left(\sum_{i=1}^n v_i^2 \right)^{1/2}, \quad{\textrm{the two-norm or Euclidean norm}}\\\\\\ ||\pmb{v}||_1 = |v_1| + |v_2| + \ldots + |v_n| = \sum_{i=1}^n |v_i|, \quad{\textrm{the one-norm or taxi-cab norm}}\\\\\\ ||\pmb{v}||_{\infty} = \max\{|v_1|,|v_2|, \ldots, |v_n| = \max_{i=1}^n |v_i|,\quad{\textrm{the max-norm or infinity norm}} \end{split}$
## Matrix norms#
We can define measures of the size of matrices, e.g. for $$A$$ which for complete generality we will assume is of shape $$m\times n$$:
$\begin{split} ||A||_F = \left(\sum_{i=1}^m \sum_{j=1}^n A_{ij}^2 \right)^{1/2}, \quad{\textrm{the matrix Euclidean or Frobenius norm}}\\\\\\ ||A||_{\infty} = \max_{i=1}^m \sum_{j=1}^n|A_{i,j}|, \quad{\textrm{the maximum absolute row-sum norm}}\\\\\\ \end{split}$
Note that while these norms give different results (in both the vector and matrix cases), they are consistent or equivalent in that they are always within a constant factor of one another (a result that is true for finite-dimensional or discrete problems as here). This means we don’t really need to worry too much about which norm we’re using.
Let’s evaluate some examples.
import numpy as np
import scipy.linalg as sl
A = np.array([[10., 2., 1.],
[6., 5., 4.],
[1., 4., 7.]])
print("A =", A)
# The Frobenius norm (default)
# equivalent to sl.norm(A)
print("SciPy norm = ", sl.norm(A, 'fro'))
# The maximum absolute row-sum
print("Maximum absolute row-sum = ", sl.norm(A,np.inf))
# The maximum absolute column-sum
print("Maximum absolute column-sum", sl.norm(A,1))
# The two-norm - note not the same as the Frobenius norm
# also termed the spectral norm
print("SciPy spectral norm =", sl.norm(A,2))
# Spectral norm definition
print("Spectral norm by hand =", np.sqrt(np.real((np.max(sl.eigvals( A.T @ A))))))
A = [[10. 2. 1.]
[ 6. 5. 4.]
[ 1. 4. 7.]]
SciPy norm = 15.748015748023622
Maximum absolute row-sum = 15.0
Maximum absolute column-sum 17.0
SciPy spectral norm = 13.793091098640064
Spectral norm by hand = 13.793091098640065
## Norm implementation#
We will write some code to explicitly compute the two matrix norms defined mathematically above (i.e. the Frobenius and the maximum absolute row-sum norms) and compare against the values found above using in-built scipy functions.
def frob(A):
m, n = A.shape
squsum = 0.
for i in range(m):
for j in range(n):
squsum += A[i,j]**2
return np.sqrt(squsum)
def mars(A):
m, n = A.shape
maxarsum = 0.
for i in range(m):
arsum = np.sum(np.abs(A[i]))
maxarsum = arsum if arsum > maxarsum else maxarsum
return maxarsum
A = np.array([[10., 2., 1.],
[6., 5., 4.],
[1., 4., 7.]])
print("A =", A)
print("Are our norms the same as SciPy?",
frob(A) == sl.norm(A,'fro') and mars(A) == sl.norm(A,np.inf))
A = [[10. 2. 1.]
[ 6. 5. 4.]
[ 1. 4. 7.]]
Are our norms the same as SciPy? True
## Matrix conditioning#
The (ill-)conditioning of a matrix is measured with the matrix condition number:
$\textrm{cond}(A) = |A||A^{-1}|.$
If this is close to one then $$A$$ is termed well-conditioned; the value increases with the degree of ill-conditioning, reaching infinity for a singular matrix.
Let’s evaluate the condition number for the matrix above.
A = np.array([[10., 2., 1.],[6., 5., 4.],[1., 4., 7.]])
print("A =", A)
print("SciPy cond(A) =", np.linalg.cond(A))
print("Default condition number uses matrix two-norm =", sl.norm(A,2)*sl.norm(sl.inv(A),2))
print("sl.norm(A,2)*sl.norm(sl.inv(A),2) =", sl.norm(A,2)*sl.norm(sl.inv(A),2))
print("SciPy Frobenius cond(A) = ", np.linalg.cond(A,'fro'))
print("sl.norm(A,'fro')*sl.norm(sl.inv(A),'fro') =", sl.norm(A,'fro')*sl.norm(sl.inv(A),'fro'))
A = [[10. 2. 1.]
[ 6. 5. 4.]
[ 1. 4. 7.]]
SciPy cond(A) = 10.71337188134679
Default condition number uses matrix two-norm = 10.71337188134679
sl.norm(A,2)*sl.norm(sl.inv(A),2) = 10.71337188134679
SciPy Frobenius cond(A) = 12.463616561943587
sl.norm(A,'fro')*sl.norm(sl.inv(A),'fro') = 12.463616561943585
The condition number is expensive to compute, and so in practice the relative size of the determinant of the matrix can be gauged based on the magnitude of the entries of the matrix.
### Example#
We know that a singular matrix does not result in a unique solution to its corresponding linear matrix system. But what are the consequences of near-singularity (ill-conditioning)?
Consider the following example
$\left( \begin{array}{cc} 2 & 1 \\ 2 & 1 + \epsilon \\ \end{array} \right)\left( \begin{array}{c} x \\ y \\ \end{array} \right) = \left( \begin{array}{c} 3 \\ 0 \\ \end{array} \right)$
When $$\epsilon=0$$ the two columns/rows are not linear independent, and hence the determinant of this matrix is zero, the condition number is infinite, and the linear system does not have a solution (as the two equations would be telling us the contradictory information that $$2x+y$$ is equal to 3 and is also equal to 0).
Let’s consider a range of values $$\epsilon$$ and calculate matrix deteterminant and condition number:
A = np.array([[2.,1.],
[2.,1.]])
b = np.array([3.,0.])
print("Matrix is singular, det(A) = ", sl.det(A))
for i in range(3):
A[1,1] += 0.001
epsilon = A[1,1]-1.0
print("Epsilon = %g, det(A) = %g, cond(A) = %g." % (epsilon, sl.det(A), np.linalg.cond(A)),
"inv(A)*b =", sl.inv(A) @ b)
Matrix is singular, det(A) = 0.0
Epsilon = 0.001, det(A) = 0.002, cond(A) = 5001. inv(A)*b = [ 1501.5 -3000. ]
Epsilon = 0.002, det(A) = 0.004, cond(A) = 2501. inv(A)*b = [ 751.5 -1500. ]
Epsilon = 0.003, det(A) = 0.006, cond(A) = 1667.67. inv(A)*b = [ 501.5 -1000. ]
We find for $$\epsilon=0.001$$ that $$\det(A)=0.002$$ (i.e. quite a lot smaller than the other coefficients in the matrix) and $$\textrm{cond}(A)\approx 5000$$.
Change to $$\epsilon=0.002$$ causes 100% change in both components of the solution. This is the consequence of the matrix being ill-conditioned - we should not trust the numerical solution to ill-conditioned problems.
A way to see this is to recognise that computers do not perform arithmetic exactly - they necessarily have to truncate numbers at a certain number of significant figures, performing multiple operations with these truncated numbers can lead to an erosion of accuracy. Often this is not a problem, but these so-called roundoff errors in algorithms generating $$A$$, or operating on $$A$$ as in Gaussian elimination, will lead to small inaccuracies in the coefficients of the matrix. Hence, in the case of ill-conditioned problems, will fall foul of the issue seen above where a very small error in an input to the algorithm led to a far larger error in an output.
## Roundoff errors#
As an example, consider the mathematical formula
$f(x)=(1-x)^{10}.$
We can relatively easily expand this out by hand
$f(x)=1- 10x + 45x^2 - 120x^3 + 210x^4 - 252x^5 + 210x^6 - 120x^7 + 45x^8 - 10x^9 + x^{10}.$
Mathematically these two expressions for $$f(x)$$ are identical; when evaluated by a computer different operations will be performed, which should give the same answer. For numbers $$x$$ away from $$1$$ these two expressions do return (pretty much) the same answer.
However, for $$x$$ close to 1 the second expression adds and subtracts individual terms of increasing size which should largely cancel out, but they don’t to sufficient accuracy due to round off errors; these errors accumulate with more and more operations, leading a loss of significance.
import matplotlib.pyplot as plt
def f1(x):
return (1. - x)**10
def f2(x):
return (1. - 10.*x + 45.*x**2 - 120.*x**3 +
210.*x**4 - 252.*x**5 + 210.*x**6 -
120.*x**7 + 45.*x**8 - 10.*x**9 + x**10)
xi = np.linspace(0, 2, 1000)
fig, axes = plt.subplots(1, 3, figsize=(14, 3))
ax1 = axes[0]
ax2 = axes[1]
ax3 = axes[2]
ax1.plot(xi, f1(xi), label = "unexpanded")
ax1.plot(xi, f2(xi), label = "expanded")
ax1.legend(loc="best")
ax1.set_ylabel("$f(x)$", fontsize=14)
ax2.plot(xi, 1.-f1(xi)/f2(xi) * 100, label="Relative\ndifference\nin %")
ax2.legend(loc="best")
ax2.set_xlabel("x", fontsize=14)
ax2.set_ylabel(r"$1-\frac{unexpanded}{expanded}$", fontsize=14)
ax3.set_xlim(0.75, 1.25)
ax3.plot(xi, 1.-f1(xi)/f2(xi) * 100, label="Relative\ndifference\nin %")
ax3.legend(loc="best")
ax3.set_ylabel(r"$1-\frac{unexpanded}{expanded}$", fontsize=14)
plt.suptitle("Comparison of $(1-x)^{10}$ expansion", fontsize=14)
As we can see on the graph, for most of the domain, i.e. far away from 1.0, the expansion is almost the same as the unexpanded version. Near $$x=1$$, the expansion creates huge errors in terms of relative difference. | 2023-02-04T15:00:49 | {
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https://math.stackexchange.com/questions/1373760/logic-quantifier-position-nuances | # Logic quantifier position nuances
I've been learning about translating English to logic and vice versa and I have to say that I find this quite difficult.
On a paper that I'm examining , I've across the following logic formula:
$\forall x (circle(x) \Rightarrow \exists y (circle(y) \land above(x,y)))$
Each circle has a circle somewhere below it
With predicates
$circle(x)$ - $x$ is a circle.
$above(x, y)$ - $x$ is above $y$.
I started to wonder what the nuance is if the exists quantifier was moved outside of the entire bracket like this:
$\forall x \exists y (circle(x) \Rightarrow circle(y) \land above(x,y))$
But my feeling is that this formula I says the same thing as the former sentence but I also think it doesn't as I'm unsure of the meaning in natural English.
Former formula: For all $x$ , if $x$ is a circle then there exists a $y$ such that $y$ is a circle and it is above $x$.
Latter formula: For all $x$ , there exists a $y$ such that if $x$ is a circle then $y$ is a circle and is above $x$.
My question is , what the difference between these two formulas in logic and natural English?
Translating from logic to English is quite hard I think and even harder from English to logic.
Are there any tips for translating from logic to English and vice versa? How do you work out the nuance if two formulas seem alike when you read them?
I've been learning about translating English to logic and vice versa and I have to say that I find this quite difficult.
Indeed. It is difficult. Logical expressions in native languages are often very imprecise, and formal mathematics is all about precision. When converting to and from mathematical expressions you will encounter many trips and traps.
Every student has trouble with this. Don't be discouraged; just be careful. It takes practice to master, that's all.
In this case your intuition is quite correct. The two expressions are correct. Generally speaking, as long as $y$ does not occur free in $P(x)$, then we have the following equivalence: $$\forall x\Big( P(x) \to \exists y \big(Q(x,y)\big)\big) \iff \forall x\exists y \Big(P(x)\to Q(x,y)\Big)$$
It is important that you don't change the order of nesting the quantifiers when shifting to and from PreNex form of the expression. Be careful about this.
Note also, however, the following equivalence.
$$\forall x \Big(\exists y\big(Q(x,y)\big) \to P(x)\Big) \iff \forall x \color{red}{\forall} y\Big( Q(x,y) \to P(x)\Big) \\ \Updownarrow \\ \forall x \Big(\neg \exists y\big(Q(x,y)\big) \vee P(x)\Big) \iff \forall x \color{red}{\forall} y\Big( \neg Q(x,y) \vee P(x)\Big)$$
As you see, the placement of the quantifier in the implication matters. Be careful.
Consider the first formula.
$\forall X (circle(X) \implies \exists Y(circle(Y) \land above(X, Y))$
Let $x$ be an instantiation of $X$.
$circle(x) \implies \exists Y(circle(Y) \land above(x, Y))$
Let $y$ be an instantiation of $Y$.
$circle(x) \implies (circle(y) \land above(x,y))$
So clearly we have a $y$ such that
$circle(x) \implies (circle(y) \land above(x,y))$
So we may existentially generalize this so that
$\exists Y (circle(x) \implies circle(Y) \land above(x, Y))$
And since no conditions were placed on $x$, we may universally generalize so that
$\forall X \exists Y (circle(X) \implies circle(Y) \land above(X, Y))$
To answer your question, in general yes $\forall x \exists y(A(x) \implies B(x,y)) \equiv \forall x (A(x) \implies \exists y B(x, y))$ so long as A does not depend on y; however, it is also VERY important that you do not changer the order in which the quantifiers occur.
Any formula in logic is equivalent to a formula in Pre-Nex normal form, which begins with a string of quantifiers, followed by a quantifier-free expression. The two formulas you give are equivalent. I think there must be a published algorithm .Even professional mathematicians can find parsing a formula to be less than easy.
• So is it safe to say that since they are equivalent they have the same meaning in natural English? How did you figure out that they are equivalent? By using equivalences? $\forall x (A \Rightarrow \exists y B)$ and $\forall x \exists y (A \Rightarrow B)$ ? – Nubcake Jul 25 '15 at 19:33 | 2019-09-20T21:10:59 | {
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https://math.stackexchange.com/questions/2793350/mean-distance-between-2-points-in-adjacent-voxels-cubes/2796405 | # Mean distance between 2 points in adjacent voxels (cubes)
Let's imagine our central unit voxel $$V_1=[0,1]^3$$.
We allow the second one to be in $$V_2=[-1,2]^3 \setminus [0,1]^3~$$, the surrounding voxel (center removed).
My question:
What is the mean distance between a point in $$V_1$$ and a point in $$V_2$$?
Of course, the direct approach would be to write an integral:
$$\frac{1}{1^3}\frac{1}{3^3-1^3} \times$$ $$\iiint_{(x_1,y_1,z_1)\in V_1} \iiint_{(x_2, y_2,z_2) \in V_2} \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\mathrm{d}x_{1,2} \mathrm{d}y_{1,2}\mathrm{d}z_{1,2}$$
but the one-variable function does not seem to admit a nice integral.
Has anyone gone through this calculation and would you be able to suggest any helpful manipulation before I dive into a numeric simulation?
Thanks.
• Small update : succeeding a numeric simulation, I got $\approx 1.8288$. If this rings a bell to you or if it roughly seems odd, please tell me so that I can correct. I expected a smaller number but the distribution's boundaries I obtained seemed consistent with the problem. – Wyllich May 24 '18 at 15:21
• An analytic solution is indeed available (using known results), but before I post an answer I need to make sure: if I understand your setup correctly, the volume of $V_2$ is $3^3 - 1$ so shouldn't the normalization constant be $\frac1{26}$? – Lee David Chung Lin May 24 '18 at 18:46
• Yes, the volume is equal to that fraction. I will correct that right away. – Wyllich May 24 '18 at 19:37
• Can you get access (by whatever means you feel appropriate) to the paper by Robbins and Bolis cited in the wiki page for Robbbins' constant? If not, in addition to me having to type up the equation (for box of general side lengths), you'll have to trust me that I quote their result correctly. – Lee David Chung Lin May 24 '18 at 21:31
• Thank you very much for mentioning the name of Robbin's constant. I now possess a starting point from which I may derive the solution. Thank you again. I'll try to find the source material if possible, count on me. – Wyllich May 24 '18 at 21:41
## Notations
Denote $$E(a,b,c)$$ as the expectation seen in the middle of p.278 (right before Eq.(2)) of the short paper (communication) by Robbins and Bolis. It is the average distance for a box of side lengths that are double of the inputs: $$2a, 2b$$, and $$2c$$.
Note that the inverse hyperbolic sine has a logarithmic form: $$\sinh^{-1} x = \ln (x + \sqrt{x^2 + 1})$$. That's how one ends up seeing those logs when $$E(a,b,c)$$ is evaluated numerically.
Denote $$\mathcal{I}$$ as the integral for your desired average but without the $$1/26$$ normalization.
Similarly, we will have the integrals $$I_{f}$$ for face-to-face, $$I_{e}$$ for edge-to-edge, and $$I_{v}$$ for vertex-to-vertex. All these are just integrals without normalization (thus not averages themselves). Details continue in the following sections.
Denote $$R_b$$ as the Robbins' constant, namely, $$R_b = E(\frac12, \frac12, \frac12) \approx 0.6617$$. Note that $$R_b$$ is the average, while it equals in value to its corresponding integral, which shall be denoted $$I_0$$. That is, numerically $$R_b = I_0$$ because the volume is $$1$$.
## Strategy Outline
As you have noticed, there are obstacles in directly setting up the density. The symmetry and nice numbers (integers) strongly suggest that one should solve this via decomposing the space into blocks of unit cubes.
Any cube has $$6$$ faces, $$12$$ edges (sides), and $$8$$ vertices, therefore
$$\mathcal{I} = 6 I_f + 12 I_e + 8 I_v \tag{1} \label{Eq_26-decomp}$$
Your $$V_1$$ is the unit cube at the "center", surrounded by $$26$$ unit cubes that are disjoint and which adds up exactly to your $$V_2$$. For a different configuration this approach needs modification since the space doesn't decompose nicely in general.
The $$26$$ blocks consist of three groups: six of them are face-to-face with $$V_1$$, twelve of them are edge-touching-edge-only with $$V_1$$, and the remaining eight are vertex-touching-vertex-only with $$V_1$$.
The value of each $$I_f$$, $$I_e$$, and $$I_v$$ will be obtained using $$E(a,b,c)$$, successively building up starting from $$I_f$$.
It will involve a $$2$$-$$1$$-$$1$$ long box, a $$2$$-$$2$$-$$1$$ flat box, then finally the $$2$$-$$2$$-$$2$$ "double cube".
## Calculation Details
Formally, the integrals can be defined as (with more shorthands in additional to yours) D_{1,2} \equiv \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2} \\ \begin{align*} I_f = \iiint\limits_{ V_1} \iiint\limits_{(x_2, y_2,z_2) \in V_{\mathbf{F}}} D_{1,2} \,\mathrm{d}x_{1,2}\,\mathrm{d}y_{1,2}\, \mathrm{d}z_{1,2} \\ I_e = \iiint\limits_{ V_1} \iiint\limits_{(x_2, y_2,z_2) \in V_{\mathbf{E}}} D_{1,2} \,\mathrm{d}x_{1,2}\,\mathrm{d}y_{1,2}\, \mathrm{d}z_{1,2} \\ I_v = \iiint\limits_{ V_1} \iiint\limits_{(x_2, y_2,z_2) \in V_{\mathbf{V}}} D_{1,2} \,\mathrm{d}x_{1,2}\,\mathrm{d}y_{1,2}\, \mathrm{d}z_{1,2} \end{align*} The volume $$V_F$$ can be any unit cube adjacent and face-to-face with $$V_1$$. For example, $$V_F \equiv \bigl\{(x,y,z) \big|~~ x \in [1,2],\, y \in [0,1],\, z \in [0,1]\bigr\}$$
Similarly, the edge-to-edge volume $$V_E$$ can be, for example, $$\bigl\{x \in [1,2],\, y \in [0,1],\, z \in [1,2]\bigr\}$$, and the vertex-to-vertex volume $$V_V$$ can be $$\bigl\{x \in [1,2],\, y \in [1,2],\, z \in [1,2]\bigr\}$$.
The formulation of $$I_f$$ (or $$I_e$$, or $$I_v$$) is not unique (up to $$6$$, $$12$$, $$8$$ multiplicity for any $$V_1$$, which itself can resides anywhere), but it doesn't matter. The value of each integral is unique and can be obtained using $$E(a,b,c)$$.
### Successive Build-up (1): $$~$$ starting from $$I_f$$
Consider a $$2$$-by-$$1$$-by-$$1$$ long box, which is equivalent to two $$V_1$$ joined face-to-face. Denote the box as $$V_{_{2.1.1}}$$ . We know that its volume is $$|V_{_{2.1.1}}| = 2$$ .
For the average distance, decompose the integral (space) into two within-cube and two across-cubes.
$$|V_{_{2.1.1}}|^2 \cdot E( 1,\tfrac12, \tfrac12) = 2 I_0 + 2 I_f \tag{2} \label{Eq_long-box_decomp}$$
There are two $$I_0$$ because each half of $$V_{_{2.1.1}}$$ are physically distinct so that ordering matters. For the same reason, there are two $$I_f$$.
Note that $$I_0$$ here emphasizes that it is the integral and not the average $$R_b$$, although numerically they are the same.
Now is the time we must make explicit use of $$E(a,b,c)$$ by Robbins and Bolis for the general box. $$\begin{multline*} E(1, \tfrac12, \tfrac12) = \frac1{1260} \Bigl[520 + 17\sqrt{2} - 20\sqrt{5} - 81 \sqrt{6} + 21 \log\left(\sqrt{2}+1\right) \\ + 168 \log \left(\sqrt{2}+\sqrt{3}\right) + 735 \log\bigl( \frac{ \sqrt{6} + 1 }{ \sqrt{5} }\bigr) + 1344 \log \left(\frac{1}{2} \left(\sqrt{5}+1\right)\right) \\ + 42 \log \left(\sqrt{5}+2\right) - 1344 \arcsin\bigl( \frac15 \bigr) - 168 \arcsin\bigl( \sqrt{\frac25 } \bigr) \Bigr] \end{multline*}$$ I coded $$E(a,b,c)$$ with Mathematica and would never consider doing this by hand. The $$E(1,\frac12,\frac12)$$ here as well as later $$E(a,b,c)$$ evaluations are done by the computer algebraic system.
The above display is just to demonstrate what we are dealing with here. To save space, the exact expression for future $$E(a,b,c)$$ will be omitted.
Moving on, $$E(1,\frac12,\frac12) \approx 0.91455248459664313930734$$ and we have \begin{align*} I_f = \frac{ |V_{_{2.1.1}}|^2 }2 E( 1,\tfrac12, \tfrac12) - I_0 &= 2 E( 1,\tfrac12, \tfrac12) - Rb \tag{3.a} \label{Eq_Int-face_in_Rb} \\ &\approx 1.16739778692611 \tag{3.b} \label{Eq_Int-face_value} \end{align*}
### Successive Build-up (2): $$~$$ from $$I_f$$ to $$I_e$$
Consider a $$2$$-by-$$\mathbf{2}$$-by-$$1$$ flat box, which is equivalent to four $$V_1$$ joined together with pairwise face-to-face. Denote this box as $$V_{_{2.2.1}}$$. We know that its volume is $$|V_{_{2.2.1}}| = 4$$.
The decomposition of the integral (space) goes like this: for each of the four unit cubes, there is one within-cube integral, two face-to-face integral, and one edge-to-edge integral. $$|V_{_{2.2.1}}|^2 \cdot E( 1, 1, \tfrac12) = 4 \left( I_0 + 2 I_f + I_e \right) \tag{4} \label{Eq_flat-box_decomp}$$ Again, the cubes are physically distinct and order matters. Note that the decomposition counts the spaces as $$4^2 = 4\cdot (1 + 2 + 1)$$.
Plugin the $$E(a,b,c)$$ formula to get $$E(1, 1,\frac12) \approx 1.1320874696118224667$$ and then replacing $$I_f$$ with Eq\eqref{Eq_Int-face_in_Rb}, we have: \begin{align*} I_e = \frac{ |V_{_{2.2.1}}|^2 }4 E( 1, 1, \tfrac12) - I_0 - 2I_f &= 4E( 1,1, \tfrac12) - R_b - 2 \bigl( 2 E( 1,\tfrac12, \tfrac12) - R_b \bigr) \\ &= 4 \bigl( E( 1, 1, \tfrac12) - E( 1,\tfrac12, \tfrac12) \bigr) + R_b \tag{5.a} \label{Eq_Int-edge_in_Rb} \\ &\approx 1.5318471223279 \tag{5.b} \label{Eq_Int-edge_value} \end{align*}
### Successive Build-up (3): $$~$$ get $$I_v$$ from $$I_f$$ and $$I_e$$
Consider a $$2$$-by-$$2$$-by-$$2$$ "double cube", which is equivalent to eight $$V_1$$ joined together with pairwise face-to-face. It is also just $$V_1$$ scaled up by a factor of two. Denote this box as $$V_{_{2.2.2}}$$.
We know that its volume is $$|V_{_{2.2.2}}| = 2^3 = 8$$, and this time we also know that $$E(1,1,1) = 2 E( \frac12, \frac12, \frac12) = 2R_b$$ simply due to the linear scalability of the geometry.
The integrals (spaces) decompose similarly. For each of the eight unit cubes, there is one within-cube integral, three face-to-face integral, three edge-to-edge integral, and one vertex-to-vertex integral. $$|V_{_{2.2.2}}|^2 \cdot E( 1, 1, 1) = 8 \left( I_0 + 3 I_f + 3I_e + I_v \right) \tag{6} \label{Eq_double-cube_decomp}$$ As before, the order matters and the decomposition says $$8^2 = 8\cdot (1 + 3 + 3 + 1)$$.
Take $$E( 1, 1, 1) = 2R_b\,,\,$$ take $$I_f$$ as Eq\eqref{Eq_Int-face_in_Rb}, and take $$I_e$$ as Eq\eqref{Eq_Int-edge_in_Rb}, we end up with
\begin{align*} I_v &= \frac{ |V_{_{2.2.2}}|^2 }8 E( 1, 1, 1) - I_0 - 3I_f - 3 I_e \\ &= 8 \cdot 2R_b - R_b - 3 \Bigl[ 2 E( 1,\tfrac12, \tfrac12) - R_b + 4 \bigl( E( 1, 1, \tfrac12) - E( 1,\tfrac12, \tfrac12) \bigr) + R_b \Bigr] \\ &= 15 R_b - 6\bigl[ 2E( 1, 1, \tfrac12) - E( 1,\tfrac12, \tfrac12) \bigr] \tag{6.a} \label{Eq_Int-vertex_in_Rb} \\ &\approx 1.82787300624563276 \tag{6.b} \label{Eq_Int-vertex_value} \end{align*}
Finally, putting things together into Eq\eqref{Eq_26-decomp} we get:
\begin{align*} \mathcal{I} &= 6I_f + 12I_e + 8 I_v \\ &= 2 \Bigg\{ 3\bigl( 2 E(1,\tfrac12,\tfrac12 ) - R_b\bigr) + 6 \Bigl[ R_b + 4 \bigl( E(1, 1,\tfrac12 ) - E(1,\tfrac12,\tfrac12 )\bigr) \Bigr] + 4 \Bigl[ 15R_b - 6 \bigl( 2E(1, 1,\tfrac12 ) - E(1,\tfrac12,\tfrac12 )\bigr) \Bigr]\Bigg\} \\ &= 2 \bigl( 63R_b + 6 E(1,\tfrac12,\tfrac12 ) - 24 E(1,1,\tfrac12 ) \bigr) \tag{7.a} \label{Eq_Int-desired_in_Rb} \\ &\approx 40.0095362394564449 \tag{7.b} \label{Eq_Int-desired_value} \\ \end{align*}
The desired average is $$\frac1{26} \mathcal{I}\approx 1.53882831690217095767637627$$
The respective sub-equations Eq\eqref{Eq_Int-face_in_Rb}, Eq\eqref{Eq_Int-edge_in_Rb}, Eq\eqref{Eq_Int-vertex_in_Rb}, and Eq\eqref{Eq_Int-desired_in_Rb} showcase the expression in minimal terms of $$E(a,b,c)$$ where terms are converted to $$R_b$$ whenever possible.
These are the reminders that the exact solution is obtainable, that the functional form still exhibits some recognizable geometry, and that one can calculate any intermediate or final results to as many digits as one like.
## Concluding Remarks
It should be clear that one cannot arrive at the desired integral solely by using $$R_b$$. After carving out the contribution of $$V_1$$ at the center from a triple-cube, one still has to deal with the $$V_2$$-to-$$V_2$$ cross terms. In the end, one will have to do something similar to the derivation above.
When the space is in a nice configuration (smooth boundary with no zig-zags, no holes within ,etc), the exact integral in general is manageable, yet already requiring some attention to details. It is very telling that for the derivation of $$E(a,b,c)$$, Robbins and Bolis wrote that "By tedious yet routine successive integration one finds that ..."
For the sake of completeness, for related in-site posts one can start with rectangle in $$2$$-dim and more generally for hypercubes.
It is totally feasible to derive the density from scratch for this particular $$\{ V_1, \, V_2 \}$$ pair. The transformation of random variables involved is pretty basic. It is just that due to the "hole" in $$V_2$$, inevitably one will be forced to decompose the space.
"Decomposing the space into disjoint spaces" is a standard technique in dealing with probability inquiries, and the decomposition is not necessarily into integer valued partitions. There are indeed limitations to this approach, but the concept is rather simple.
In terms of numerical calculation, once you have coded the $$E(a,b,c)$$ in whatever system you're using, this procedure is pretty straightforward.
I don't know your original motivation for doing this calculation, but I'd say this approach is practically useful across disciplines, from theoretical to applications.
• What a ride! Thank you for posting such a detailed and comprehensive answer. I will definitely remember the idea of dividing the integral space as you masterfully did to obtain intermediate integrals through multiple iterations. I guess I owe you an explanation about the whys. In LiDAR image processing, I asked myself the possibility of predicting the resolution (here understand mean distance between a point and its closest neighbour) of a downsampled cloud using a voxel grid. In spite of my hypothesis based on the distance you calculated being disproved by experience, I still believe it (1) – Wyllich May 28 '18 at 12:46
• ... was a good experience to have. Thank you very much again. (2) – Wyllich May 28 '18 at 12:47
• Very glad I could help. I know only a little bit about image process or data compression in general, but what you described sounds no lesser than some innovative ideas that eventually became successful. – Lee David Chung Lin May 28 '18 at 13:12
• Now that I think about it, your method gave me new ideas to precise my hypothesis which is based on isotropy (hence V2 containing V1). However, I may add some coefficients in front of each term of equations (2), (4), (6) to take into account anisotropy and capture an idea about the cloud's global structure. I must repeat myself : your dedication gave birth to many new ideas I may now explore. – Wyllich May 28 '18 at 13:25
• Thank you for saying that. It's been a real pleasure. – Lee David Chung Lin May 28 '18 at 13:46 | 2020-04-04T13:13:45 | {
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https://math.stackexchange.com/questions/1117948/how-do-i-solve-this-olympiad-question-with-floor-functions | # How do I solve this Olympiad question with floor functions?
Emmy is playing with a calculator. She enters an integer, and takes its square root. Then she repeats the process with the integer part of the answer. After the third repetition, the integer part equals 1 for the first time. What is the difference between the largest and the smallest number Emmy could have started with?
$$\text{(A)} \; 229 \qquad \text{(B)} \; 231 \qquad \text{(C)} \; 239 \qquad \text{(D)} \; 241 \qquad \text{(E)} \; 254$$
This was Problem 19 from the first round of the Norwegian Mathematical Olympiad, 2014–15.
I think the floor function applies well here.
The integer part obviously is the floor part of it.
Let $x_1$ be the initial.
$x_2 = \left \lfloor{{\sqrt{x_1}}}\right \rfloor$
$x_3 = \left \lfloor{{\sqrt{x_2}}}\right \rfloor$
$x_4 = \left \lfloor{{\sqrt{x_3}}}\right \rfloor = 1$ <-- Last one.
So we need to find $b \le x_1 \le c$ such that $b$ is the smallest number to begin with and $c$ is the largest number to begin with.
$\sqrt{b} \le \sqrt{x_1} \le \sqrt{c}$
Clearly we must have $x_3<4$, $x_2<16$, and $x_1<256$, so the largest possible value of $x_1$ is $255$. Since $x_3>1$, we know that $x_3\ge 2$ and hence that $x_2\ge 4$. This implies that $x_1\ge 16$. Thus, the range is from $16$ through $255$, the difference being $239$.
• @BrianScott, Thanks. Clear up, why must we have those conditions? Why $x_1 < 256$? – Robart Jan 24 '15 at 18:05
• We know that $x_4=1$, so $1\le\sqrt{x_3}<2$, and therefore $x^3<2^2=4$. Now repeat the reasoning back up the line: $x_3<4$, so $\sqrt{x_2}<4$, and $x_2<16$. And so on. – Brian M. Scott Jan 24 '15 at 18:06
• @Robart: Sorry: I forgot to ping you with my answer. See the comment immediately above this one. – Brian M. Scott Jan 24 '15 at 18:13
• why cant she start with $x_1 = 1$ so the difference is $255 - 1 = 254$? – Robart Jan 24 '15 at 18:43
• @Robart: Because we’re told that $x_4$ is the first of these integers to be equal to $1$. – Brian M. Scott Jan 24 '15 at 18:45
We can work backwards more easily than we can work forwards. Firstly, what does a number $x$ need to satisfy to have $\lfloor x \rfloor = 1$? Easy. We need: $$1\leq x < 2.$$ Well, suppose that $\sqrt{y}=x$ or, equivalently, $y=x^2$. Well, obviously we just square the above equation (as all of its terms are positive): $$1\leq x^2 < 2^2.$$ Suppose $\sqrt{z}=y$ or $z=y^2=x^4$. Square the above again! $$1\leq x^4 < 2^4.$$ And finally let $\sqrt{w}=z$ or $w=z^2=y^4=x^8$. Square the above $$1\leq x^8 < 2^8.$$
So, if she chose $z$ to be any integer in $[1,2^8)$, she will get $1$ as the final result. However, if $z$ were in the interval $[1,2^4)$, she would have gotten $1$ as the second result as well. So, we restrict ourselves to the integers in $[2^4,2^8)$, as these will satisfy the desired condition.
• And how do you pick the smallesT? Its supposed to be $16$? – Robart Jan 24 '15 at 18:33
• Yes, the condition that the third time is the first time that the number is $<2$ means that $x^2\geq 2$. – Thomas Andrews Jan 24 '15 at 18:35
• Why does it have to be $x^2 \ge 2$? then $x ge \sqrt{2}$ – Robart Jan 24 '15 at 18:41
• It should be 254 I believe. The smallest is $1$ since $\sqrt{1} = 1$ for $n$ square roots. – Robart Jan 24 '15 at 18:43
• @Robart The question says that the third time is the first for which the result is $1$ - the greatest possible number is $255$, as higher numbers don't reach $1$ fast enough, and the least is $16$, as lower numbers reach $1$ too fast. – Milo Brandt Jan 24 '15 at 18:46 | 2020-02-22T02:12:35 | {
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http://math.stackexchange.com/questions/414776/what-is-the-use-of-the-dot-product-of-two-vectors/414785 | # What is the use of the Dot Product of two vectors?
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.
My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).
-
You can use it to find the angle between any two vectors. $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ where $\theta$ is the angle between the two vectors. This is a better approach than using the cross product as the cross product can only be defined in a few dimensions (normally only 3 dimensions). Obviously this is just one simple use of the dot product which is a special case of a more general phenomenon known as an inner product en.wikipedia.org/wiki/Inner_product_space . – Daniel Rust Jun 8 '13 at 15:35
You should take a look at how it is derived as it can be very enlightening. Take a look at Lang's Linear Algebra (2E) pg.19-20. – 12F8031 Jun 8 '13 at 20:18
I always found that the dot product is a good way to measure how "parallel" two vectors are – Dan Jun 9 '13 at 7:15
Re: "[the dot product] seems almost useless to me compared with the cross product of two vectors ".
Please see the Wikipedia entry for Dot Product to learn more about the significance of the dot-product, and for graphic displays which help visualize what the dot product signifies (particularly the geometric interpretation). Also, you'll learn more there about how it's used. E.g., Scroll down to "Physics" (in the linked entry) to read some of its uses:
Mechanical work is the dot product of force and displacement vectors.
Magnetic flux is the dot product of the magnetic field and the area vectors.
You've shared the algebraic definition of the dot product: how to compute it as the sum of the product of corresponding entries in two vectors: essentially, computing $\;\mathbf A \cdot \mathbf B = {\mathbf A}{\mathbf B}^T.\;$ But the dot product also has an equivalent geometric definition:
In Euclidean space, a Euclidean vector is a geometrical object that possesses both a magnitude and a direction. A vector can be pictured as an arrow. Its magnitude is its length, and its direction is the direction the arrow points. The magnitude of a vector A is denoted by $\|\mathbf{A}\|.$ The dot product of two Euclidean vectors A and B is defined by
$$\mathbf A\cdot\mathbf B = \|\mathbf A\|\,\|\mathbf B\|\cos\theta,\quad\text{where \theta is the angle between A and B.} \tag{1}$$
With $(1)$, e.g., we see that we can compute (determine) the angle between two vectors, given their coordinates: $$\cos \theta = \frac{\mathbf A\cdot\mathbf B}{\|\mathbf A\|\,\|\mathbf B\|}$$
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I am not stating what the dot product signifies, in fact that is the essence of this question, I did not know that the dot product has an equivalent geometric definition, or that it could be used to calculate the angle between two vectors that is extremely useful. – Jonn_Underwood Jun 8 '13 at 15:45
I didn't intend to criticize. I'm sorry if that's how the post comes across! Indeed, I upvoted the question because it's a fine question, and you put thought into writing it. – amWhy Jun 8 '13 at 15:47
The dot product is also the product $\bf A B^T$ of vector A with the transpose of vector B. – amWhy Jun 8 '13 at 15:49
@amWhy: Thank you, I had an upvote mind - trigger finger! :-) Fixed! – Amzoti Jun 9 '13 at 1:37
@amWhy thank you for your suburb answer , and also thank you for adding that formula at the bottom for finding the cosine of theta this alone is extremely useful to me due to my pursuit of physics! – Jonn_Underwood Jun 9 '13 at 21:24
show 1 more comment
The original motivation is a geometric one: The dot product can be used for computing the angle $\alpha$ between two vectors $a$ and $b$:
$a\cdot b=|a|\cdot|b|\cdot \cos(\alpha)$.
Note the sign of this expression depends only on the angle's cosine, therefore the dot product is
• $<0$ if the angle is obtuse,
• $>0$ if the angle is acute,
• $=0$ if the $a$ and $b$ are orthogonal.
Another important special case appears when $a=b$: The root of the scalar product of a vector with itself is the length of a vector:
$a\cdot a=|a|\cdot|a|\cdot1=|a|^2$.
There's another interesting application of the dot product, in combination with the cross product: If you have three vectors $a$, $b$ and $c$, they define a parallelepiped, and you can compute its (signed) volume $V$ as follows using the so-called scalar triple product:
$V=(a\times b)\cdot c$
(Note that this is a generalization of $|a\times b|$ being the area of the parallelogram defined by $a$ and $b$.)
-
There's nothing special about 2d/3d here; this means of finding the angle between vectors applies to an arbitrary number of dimensions. – Muphrid Jun 8 '13 at 15:50
@Muphrid: Of course you can define something like an "angle" in arbitrary dimensions. I've changed "geometrical" to "visual", I think that makes more sense. – zero-divisor Jun 9 '13 at 6:12
Perhaps. But two vectors define only a plane, so even in a 4d space or higher, the geometry basically isn't changing: you have two vectors in some common plane, and the dot product tells us how alike they are. I think you're making it out to be more complicated than it really is. – Muphrid Jun 9 '13 at 6:17
@Muprid: Good point. I've deleted the sentence now. – zero-divisor Jun 9 '13 at 6:19
The dot product is an essential ingredient in matrix product. The product of the two matrices $A$ and $B$ (of compatible sizes, that is, the number of columns of $A$ equals the number of rows of $B$) is a matrix whose $(i, j)$ component is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
Among the many applications, consider this simple one.
You have students $s_{1}, \dots, s_{n}$ taking courses $c_{1}, \dots, c_{m}$. Consider the matrix $A$ which has $0$ everywhere, except that the $(i,j)$ coefficient is $1$ if the student $s_{j}$ takes the course $c_{i}$. Now note that the dot product of the $i_{1}$-th row of $A$ with the $i_{2}$-th row gives the number of students that take both $c_{i_{1}}$ and $c_{i_{2}}$. In other words, the matrix $A A^{t}$ has in its $(i, j)$ position the number of students taking both $c_{i}$ and $c_{j}$.
This matrix is of course useful in building a course timetable.
-
Before addressing your question, i want to say that this is a very good question and you are right to expect that the dot product has use/significance.
First, it is important that you think about vectors separate from their coordinates. While it is true that we often represent vectors as a series of coordinates along well-defined axes, this is merely for computational reasons. A vector as an idea "exists" in a space without any predefined coordinate system. I say this because there are two definitions of the dot product, one is coordinate free (i.e. $\mathbf a\cdot\mathbf b = \|\mathbf a\|\,\|\mathbf b\|\cos\theta$) and the other is based on coordinates (i.e. $\mathbf a\cdot\mathbf b = \sum_i{a_i b_i}$). Of these two, it is best to think of the dot product in terms of the former as it does not depend on a coordinate system. (It is relatively easy to show that the latter may be derived from the former, but in that derivation is an implicit assumption that the coordinate system being used to represent the dot product is orthogonal.)
Second, given the coordinate-free definition, the fundamental idea of the dot product is that of projection. By this it gives a single number which indicates the component of a vector in the direction of another vector. Your observation of the dissimilarity between the dot and cross product is correct, however, the dot product is used to produce a vector as well, it just does it component-by-component. Let's suppose that we have a vector $\mathbf v$ represented by its components in a given coordinate system. Let's further suppose that we have an orthonormal basis defined in that same coordinate system as the set of column vectors $\{\mathbf u_1, \mathbf u_2, \ldots, \mathbf u_n\}$. Finally, suppose that we want to represent $\mathbf v$ in this basis as $\mathbf w$. The question is how do we do that? We use the dot product of course! So the first component of $\mathbf w$ would then be $w_1 = \mathbf u_1\cdot \mathbf v$, and the second component would be $w_2 = \mathbf u_2\cdot \mathbf v$ and so on. (Note that because $\|\mathbf u_i\| = 1$, we have $\mathbf u_1\cdot \mathbf v= \|\mathbf v\|\cos\theta_i$.) If we then think of the vector $\mathbf w$ defined as such we have
$$\mathbf w = \left[ \begin{array}{c} w_1 \\ w_2 \\ \vdots \\ w_n \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1\cdot \mathbf v \\ \mathbf u_2\cdot \mathbf v \\ \vdots \\ \mathbf u_n\cdot \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \mathbf v \\ \mathbf u_2^T \mathbf v \\ \vdots \\ \mathbf u_n^T \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \\ \mathbf u_2^T \\ \vdots \\ \mathbf u_n^T \end{array} \right]\mathbf v = \left[ \begin{array}{cccc} \mathbf u_1 & \mathbf u_2 & \cdots &\mathbf u_n \end{array} \right]^T\mathbf v$$
Finally, we conclude that the dot product plays a key role in the transformation of a vector from one basis to another and that the dot product is hidden in the definition of matrix multiplication in that one view of a matrix-vector product is that each element in the product represents a dot product between a row of the left and a column of the right.
I hope this helps.
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The geometric idea of the dot product has been touched upon, but there is a vast generalization of this product in geometric algebra, the algebra of not only oriented lines (vectors) but planes, volumes, and more (called blades).
In geometric algebra, we have a generalized dot product of a vector $a$ and another blade $B$ denoted $a \cdot B$. This product has a simple geometric interpretation as the part of $B$ orthogonal to the projection of $a$, with magnitude $|a||B|| \cos \theta|$, where $\theta$ is the angle $a$ forms with its projection in $B$.
(Note: as a point of fact, $a \cdot B$ is orthogonal to $a$ also, not just its projection into $B$, but thinking about this leads to some difficulties, while thinking about what's perpendicular to the projection does not.)
If $B$ is a 2-blade (also called a bivector), then you should be able to imagine this directly: if $a$ lies entirely in $B$, then $a \cdot B$ is just the vector perpendicular to $a$ in $B$. If $a$ does not lie entirely in $B$, then it can be decomposed into a tangential part and a normal part. We throw away the normal part, and the previous logic applies for the tangential part.
If $B$ is a 3-blade (a trivector), then in 3d space $a$ must lie in $B$ (for there is no 3d volume that a vector does not help span), and the product $a \cdot B$ is the "Hodge dual", or the plane perpendicular to $a$.
In this light, the dot product of vectors may actually be the most non-intuitive part of this reasoning. When you take the dot product, there's only a scalar left--there's no vector or other higher dimensional object left to be orthogonal to $a$. Again, this is why I emphasize that $a \cdot B$ is the part of $B$ orthogonal to the projection of $a$ onto $B$. When $B$ is a vector, it's clear there is no other vector or anything else that can be orthogonal to the projection of $a$, for $B$ and that projection are parallel, so the result is necessarily just a scalar.
- | 2014-03-10T21:38:34 | {
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https://math.stackexchange.com/questions/3506020/every-natural-number-is-covered-by-consecutive-numbers-that-sum-to-a-prime-power | # Every natural number is covered by consecutive numbers that sum to a prime power.
Conjecture. For every natural number $$n \in \Bbb{N}$$, there exists a finite set of consecutive numbers $$C\subset \Bbb{N}$$ containing $$n$$ such that $$\sum\limits_{c\in C} c$$ is a prime power.
A list of the first few numbers in $$\Bbb{N}$$ has several different covers by such consecutive number sets.
One such is:
3 7 5 13 8 19 11 25 29 16 37 41 49 53
___ ___ _ ___ _ ____ __ _____ _____ __ __ _____ _____ _____ _____ __
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
___ ___ _____ _____ __ ________
11 17 31 43 81
59 71 3^5
___ __ _____ _________________
30 31 32 33 34 35 36 37 38 39 40 41 42 43 .....
_____ _____ _____
61 67 73
• Apparently, primes are allowed as well. Not sure, whether it has been proven, but this is very likely to be the case. More interesting is to exclude the primes. – Peter Jan 12 '20 at 10:34
• @Peter: The sum can only be a prime if it has at most two terms. (Since $\sum_{k=m}^nk=(n-m+1)(n+m)/2$, which factorizes unless $n-m+1\le2$.) The density of one- and two-term sums that are primes goes to zero, so the brunt of the conjecture has to be borne by the proper prime powers, anyway. – joriki Jan 12 '20 at 13:02
• @Peter a prime power by definition includes the exponent $1$. Therefore, for sake of simplicity, we include the primes. – StudySmarterNotHarder Jan 12 '20 at 17:32
For any odd prime $$p$$, there are $$p$$ consecutive integers centred on $$p$$ that sum to $$p^2$$.
$$2+3+4=3^2$$
$$3+4+5+6+7=5^2$$
$$4+5+6+7+8+9+10=7^2$$
etc.
Let $$p_n$$ be the $$n$$-th prime. Then, using Bertrand's postulate in the form $$p_{n+1}<2p_n$$we know that the above sums for consecutive primes overlap.
Finally, we note that $$1+2=3$$ to complete the proof.
I don't know if this has been shown before, but the proof seems straightforward.
While nickgard's answer shows how to solve the problem using sums being squares of increasing primes, this answer shows how to do it using the sums being just odd powers of $$3$$.
As suggested in joriki's question comment, for any integers $$1 \le j \le k$$, you have
\begin{aligned} \sum_{i=j}^{k}i & = \sum_{i=1}^{k}i - \sum_{i=1}^{j-1}i \\ & = \frac{k(k+1)}{2} - \frac{(j-1)(j)}{2} \\ & = \frac{k^2 + k - j^2 + j}{2} \\ & = \frac{(k-j)(k+j) + k + j}{2} \\ & = \frac{(k+j)(k-j+1)}{2} \end{aligned}\tag{1}\label{eq1A}
Consider the ranges $$\left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$$ for $$m = 0, 1, 2, \ldots$$. The union of these disjoint subsets cover all positive integers. Thus, for any $$n \ge 1$$, there's a unique $$m$$ where $$n \in \left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$$. For that $$m$$, since $$\frac{5\left(3^{m}\right)-1}{2} \gt \frac{3^{m+1} - 1}{2}$$, you can have $$j = \frac{3^m + 1}{2}$$ and $$k = \frac{5\left(3^{m}\right)-1}{2}$$ with $$n \in [j,k]$$. Using this in \eqref{eq1A} gives
\begin{aligned} \sum_{i=j}^{k}i & = \frac{(k+j)(k-j+1)}{2} \\ & = \frac{\left(\frac{5\left(3^{m}\right)-1}{2}+\frac{3^m + 1}{2}\right)\left(\frac{5\left(3^{m}\right)-1}{2}-\frac{3^m + 1}{2}+1\right)}{2} \\ & = \frac{\left(\frac{6\left(3^{m}\right)}{2}\right)\left(\frac{4\left(3^{m}\right)}{2}-\frac{2}{2}+1\right)}{2} \\ & = \frac{\left(3\left(3^{m}\right)\right)\left(2\left(3^{m}\right)\right)}{2} \\ & = \left(3^{m+1}\right)\left(3^{m}\right) \\ & = 3^{2m+1} \end{aligned}\tag{2}\label{eq2A}
The first few examples for $$m = 0, 1$$ and $$2$$ are
$$1 + 2 = 3 = 3^{1} \tag{3}\label{eq3A}$$
$$2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^{3} \tag{4}\label{eq4A}$$
$$5 + 6 + \ldots + 21 + 22 = 243 = 3^{5} \tag{5}\label{eq5A}$$ | 2021-02-27T19:23:02 | {
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https://leanprover-community.github.io/archive/stream/217875-Is-there-code-for-X%3F/topic/Classification.20of.20finite-dimensional.20vector.20spaces.html | ## Stream: Is there code for X?
### Topic: Classification of finite-dimensional vector spaces
#### Heather Macbeth (Jan 01 2021 at 16:02):
Do we have, there exists a linear equivalence between two finite_dimensional vector spaces of the same findim?
#### Johan Commelin (Jan 01 2021 at 16:03):
yes, I think Markus Himmel did that
#### Johan Commelin (Jan 01 2021 at 16:04):
I forgot the name of the result...
#### Heather Macbeth (Jan 01 2021 at 16:05):
I tried searching for linear_equiv in linear_algebra.finite_dimensional and couldn't see it.
#### Adam Topaz (Jan 01 2021 at 16:06):
This is true in general, not just for finite dimensional. Maybe that helps in finding it?
#### Johan Commelin (Jan 01 2021 at 16:06):
Is this what you want: src/linear_algebra/invariant_basis_number.lean ?
#### Johan Commelin (Jan 01 2021 at 16:07):
or is that in the wrong direction?
#### Adam Topaz (Jan 01 2021 at 16:08):
(Heather are you formalizing projective spaces? :smile: )
#### Heather Macbeth (Jan 01 2021 at 16:08):
I think it might be in the wrong direction, yes.
#### Heather Macbeth (Jan 01 2021 at 16:09):
(Heather are you formalizing projective spaces? :smile: )
#### Adam Topaz (Jan 01 2021 at 16:10):
Oh nice! I figured it was projective spaces because of the span notation we were discussing a few day ago.
#### Adam Topaz (Jan 01 2021 at 16:12):
anyway, the closest I can find so far to the problem at hand is this docs#equiv_of_is_basis
#### Heather Macbeth (Jan 01 2021 at 16:14):
That looks good, thanks! I don't understand why it's not named linear_equiv_of_is_basis, though :)
Yeah I agree.
#### Adam Topaz (Jan 01 2021 at 16:15):
docs#cardinal.eq gets you essentially all the way to what you want.
#### Adam Topaz (Jan 01 2021 at 16:16):
And this docs#is_basis.mk_eq_dim''
#### Adam Topaz (Jan 01 2021 at 16:27):
import linear_algebra
import linear_algebra.dimension
open_locale classical
universe u
variables (K : Type*) [field K] (V W : Type u)
example (cond : vector_space.dim K V = vector_space.dim K W) : nonempty (V ≃ₗ[K] W) :=
begin
obtain ⟨B,hB⟩ := exists_is_basis K V,
obtain ⟨C,hC⟩ := exists_is_basis K W,
have h1 := is_basis.mk_eq_dim'' hB,
have h2 := is_basis.mk_eq_dim'' hC,
have h3 : cardinal.mk B = cardinal.mk C, by cc,
rw cardinal.eq at h3,
cases h3,
refine ⟨equiv_of_is_basis hB hC h3⟩,
end
#### Heather Macbeth (Jan 01 2021 at 16:28):
Wonderful! Can you PR it?
sure.
#### Adam Topaz (Jan 01 2021 at 16:29):
Which file should it go in?
#### Adam Topaz (Jan 01 2021 at 16:30):
Oh, there is one annoying thing -- to even state that the two vector spaces have the same dimension, they need to come from the same universe.
#### Heather Macbeth (Jan 01 2021 at 16:30):
I think linear_algebra.dimension
#### Heather Macbeth (Jan 01 2021 at 16:31):
Oh, there is one annoying thing -- to even state that the two vector spaces have the same dimension, they need to come from the same universe.
We just need a version for finite_dimensional and findim, that avoids all those issues.
#### Adam Topaz (Jan 01 2021 at 16:31):
You can still have finite dimensional vector spaces in different universes.
#### Adam Topaz (Jan 01 2021 at 16:32):
Oh I see what you mean.
#### Adam Topaz (Jan 01 2021 at 16:32):
The dimension would be in nat, not in cardinal
#### Adam Topaz (Jan 01 2021 at 16:48):
Okay, I pushed to this branch branch#lin_equiv_of_basis_equiv
#### Adam Topaz (Jan 01 2021 at 16:49):
Is there a finite_dimensional analogue of docs#linear_equiv.dim_eq with findim?
#### Heather Macbeth (Jan 01 2021 at 17:22):
Thanks!
Is there a finite_dimensional analogue of docs#linear_equiv.dim_eq with findim?
#### Adam Topaz (Jan 01 2021 at 17:24):
I have the finite-dimensional case essentially done too (with no universe limitations). I'll open a PR soon.
#5559
#### Patrick Massot (Jan 01 2021 at 18:39):
I'm skeptical that you really want the conclusion of the lemma to be nonempty (V ≃ₗ[K] V₂). The first thing you'll do when using the lemma will be to extract an isomorphism. Why not making it a def?
#### Patrick Massot (Jan 01 2021 at 18:42):
I'm also surprised those lemmas are no already there somehow. I would wait for @Anne Baanen to comment on this.
#### Adam Topaz (Jan 01 2021 at 18:46):
I mostly used nonempty to make it a Prop, which is useful for the if and only if statement. I can make a def to get the linear equiv as well, but I agree that we should wait for @Anne Baanen :)
#### Anne Baanen (Jan 02 2021 at 06:05):
Patrick Massot said:
I'm also surprised those lemmas are no already there somehow. I would wait for Anne Baanen to comment on this.
I don't recall seeing something like linear_equiv.of_findim_eq, only the equivalences assuming a basis, so I think this is indeed a hole in the library.
#### Anne Baanen (Jan 02 2021 at 06:18):
I mostly used nonempty to make it a Prop, which is useful for the if and only if statement. I can make a def to get the linear equiv as well, but I agree that we should wait for Anne Baanen :)
The nonempty situation as it is in the PR right now seems optimal to me: cardinal equality is expressed in nonempty, so the first result also uses nonempty and the next ones apply classical.choice :+1: | 2021-05-19T02:24:24 | {
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https://mathhelpboards.com/threads/pr-of-exactly-one-dice-showing-2.24196/ | # Pr of exactly one dice showing 2
#### navi
##### New member
So I ran into this problem:
You roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?
I thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.
#### Opalg
##### MHB Oldtimer
Staff member
So I ran into this problem:
You roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?
I thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.
You have given two answers, but the question wants a single answer.
Let $A$ stand for "the sum being at least 8", and let $B$ stand for "exactly one of the dice showing 2". You have correctly found the probabilities $\def\Prob{\operatorname{Prob}\,}\Prob(A) = \frac{15}{36}$ and $\Prob(B) = \frac{10}{36}.$ But the question asks for $\Prob(A\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\Prob(A\text{ or }B) = \Prob(A) + \Prob(B) - \Prob(A\text{ and }B)$.
#### navi
##### New member
You have given two answers, but the question wants a single answer.
Let $A$ stand for "the sum being at least 8", and let $B$ stand for "exactly one of the dice showing 2". You have correctly found the probabilities $\def\Prob{\operatorname{Prob}\,}\Prob(A) = \frac{15}{36}$ and $\Prob(B) = \frac{10}{36}.$ But the question asks for $\Prob(A\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\Prob(A\text{ or }B) = \Prob(A) + \Prob(B) - \Prob(A\text{ and }B)$.
Thank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?
#### Opalg
##### MHB Oldtimer
Staff member
Thank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?
It's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.
#### navi
##### New member
It's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.
Okay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.
#### Opalg
##### MHB Oldtimer
Staff member
Okay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.
It should indeed be 2/36.
#### navi
##### New member
It should indeed be 2/36.
It is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36
#### Opalg
##### MHB Oldtimer
Staff member
It is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36
Here is a possible explanation for that. You entitled this thread "Pr of exactly one dice showing 2", but the statent of the problem only refers to "one of the dice showing 2". I suspect that this may be intended to cover the case when both dice show 2. In that case, the probability of that event goes up from $\frac{10}{36}$ to $\frac{11}{36}$, and the probability of "$A$ or $B$" becomes $\frac{11}{36} + \frac{15}{36} - \frac{2}{36}$, which agrees with the answer accepted by WebWork.
#### Country Boy
##### Well-known member
MHB Math Helper
By the way, your title, "Pr of one dice showing 2", should be "Pr of one die showing 2" and "you roll two fair die" should be "you roll two fair dice". "Dice" is the plural of the singular "die".
#### Wilmer
##### In Memoriam
1st roll: must be 2 or 6: probability = 2/6 = 1/3
2nd roll: must be 2 if 1st was 6, or 6 if 1st was 2: probability = 1/6
1/3 * 1/6 = 1/18 | 2020-07-14T09:26:44 | {
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https://math.stackexchange.com/questions/2838491/trick-or-indefinite-integration | # Trick, or Indefinite Integration?
Evaluate $$\int_{0}^{\pi/2}\dfrac{\sin^6 x}{\sin x + \cos x}\text{ d}x$$
Is there a nice, elegant way of solving the above integral?
Here's what I did - Replaced $f(x)$ by $f(\pi/2-x)$, added and got rid of some common factors in numerator and denominator. This was followed by indefinite integration, and ultimately substituting the limits in the final expression.
The above method is kind of long, though - and I'm wondering if there's a quicker, shorter and rather elegant way of getting around this integral. Could someone post a solution, and share ideas?
Thanks a lot.
• Please use MathJax to format your equations so that they are searchable, rather than images. – Clarinetist Jul 2 '18 at 13:30
• If your integral is denoted by $I$, then $$2I=\int_0^{\pi/2} \frac{\sin^6 x + \cos^6 x}{\sin x + \cos x} \mathrm{d}x$$ – Crostul Jul 2 '18 at 13:43
• Did that, what's next? – arya_stark Jul 2 '18 at 13:44
• Is it true that $$8I=\int_0^{2\pi} \frac{\sin^6 x + \cos^6 x}{\sin x + \cos x} \mathrm{d}x$$ ?. I think it won't be a bad idea to use integration in the complex circle (see math.stackexchange.com/questions/308693/…) then, if it's possible. – Gonzalo Benavides Jul 2 '18 at 17:01
## 4 Answers
I have solved it first as a indefinite integral and then as a definite integral. I am not sure how different it is with yours. Good luck. Let me know if you have any questions.
A standard way for dealing with trigonometric integrals is to exploit the substitution $x=2\arctan\frac{t}{2}$.
Here it leads to $$\int_{0}^{\pi/2}\frac{\sin^6 x}{\sin x+\cos x}\,dx = 128\int_{0}^{1}\frac{t^6}{(2t+1-t^2)(t^2+1)^6}\,dt$$ and the last integral can be computed by partial fraction decomposition. It equals $$\frac{1}{8}\left(2+\sqrt{2}\,\text{arctanh}\frac{1}{\sqrt{2}}\right)\approx\frac{28}{69}.$$ This also follows from $\sin x+\cos x=\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$ and $$\int_{-\pi/4}^{\pi/4}\frac{\sin^6\left(x+\frac{\pi}{4}\right)}{\sqrt{2}\cos x}\,dx=\frac{1}{8\sqrt{2}}\sum_{k=0}^{6}\binom{6}{k}\int_{-\pi/4}^{\pi/4}\sin^k(x)\cos^{5-k}(x)\,dx$$ where the contribution provided by odd $k$s is zero and $$2\int_{0}^{\pi/4}\cos(x)^5\,dx =\frac{43}{30\sqrt{2}},\qquad 2\int_{0}^{\pi/4}\sin(x)^2\cos(x)^3\,dx =\frac{7}{30\sqrt{2}}$$ $$2\int_{0}^{\pi/4}\sin(x)^4\cos(x)^1\,dx =\frac{3}{30\sqrt{2}}$$ $$2\int_{0}^{\pi/4}\sin(x)^6\cos(x)^{-1}\,dx =-\frac{73}{30\sqrt{2}}+4\,\text{arctanh}\tan\frac{\pi}{8}.$$
Hint:
$$2I=\int_0^{\pi/2}\dfrac{\sin^6x+\cos^6x}{\sin x+\cos x}dx$$
Now $\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$
and $\sin^2x\cos^2x=\dfrac{\sin^22x}4=\dfrac{1-\cos4x}8$
For $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos4x}{\sin x+\cos x}dx=\int_0^{\pi/2}\dfrac{\cos4x}{\sqrt2\cos\left(x-\dfrac\pi4\right)}dx,$
set $x-\dfrac\pi4=y,\cos4x=\cos4\left(y+\dfrac\pi4\right)=\cos(4y+\pi)=-\cos4y$
$\implies-\sqrt2J=\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{\cos4ydy}{\cos y}=\int_{-\pi/4}^{\pi/4}\dfrac{1-8\cos^2y+8\cos^4y}{\cos y}dy$
Now use $\cos3y=4\cos^3y-3\cos y$
Let $J$ be the integral to be calculated, we want to get the computer result, here using sage:
sage: var('x');
sage: J = integral( sin(x)^6/(sin(x)+cos(x)), x, 0, pi/2 )
sage: J
1/16*sqrt(2)*log((2*sqrt(2)) + 3) + 1/4
For this, we use the simple idea of getting a $\sin(x+\pi/4)$ (up to a multiplicative constant) in the denominator, then do trivial computations...: \begin{aligned} J &=\int _0^{\pi/2} \frac{\sin^6 x}{\sin x + \cos x}\;dx \\ &=\int _0^{\pi/2}\frac 1{\sqrt 2} \frac{\sin^6 x}{\frac 1{\sqrt 2}\left(\sin x + \cos x\right)}\;dx \\ &=\frac 1{\sqrt 2} \int _0^{\pi/2} \frac {\sin^6 x} {\sin \left(x + \frac\pi4\right)}\;dx \\ &\qquad\text{Substitution: }y = x+\frac \pi 4\ , \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac {\sin^6 \left(y- \frac\pi4\right)} {\sin y}\;dy \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac 18\cdot\frac 1{\sin y} \Big(\ \sin y-\cos y\ \Big)^6\;dy \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac 18\cdot\frac 1{\sin y} \Big(\ \sin^6 y -6\sin^5y\cos y +15\sin^4y\cos^2 y \\&\qquad\qquad -20\sin^3y\cos^3 y +15\sin^2y\cos^4 y -6\sin y\cos^5 y +\cos^6 y \ \Big)\;dy \\ &=\frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4} \Big(\ \sin^5 y -6\sin^4y\cos y +15\sin^3y\cos^2 y \\&\qquad\qquad -20\sin^2y\cos^3 y +15\sin y\cos^4 y -6\cos^5 y +\frac{\cos^6 y}{\sin y} \ \Big)\;dy \\ &= \frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4}\left[\ \begin{aligned} -(1-\cos^2y)^2 &\,\cdot\,\cos 'y \\ - 6\sin^4y &\,\cdot\,\sin'y \\ - 15(1-\cos^2y)\cos^2y &\,\cdot\,\cos'y \\ - 20\sin^2y(1-\sin^2 y) &\,\cdot\,\sin'y \\ - 15\cos^4y &\,\cdot\,\cos'y \\ - 6(1-\sin^2 y)^2 &\,\cdot\,\sin'y \\ -\frac{\cos ^6y}{1-\cos^2 y}&\,\cdot\,\cos'y \end{aligned} \ \right]\; dy \\ &= \frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4}\left[\ \begin{aligned} -(1-\cos^2y)^2 & \\ - 15(1-\cos^2y)\cos^2y & \\ - 15\cos^4y & \\ -\frac{\cos ^6y}{1-\cos^2 y}& \end{aligned} \ \right]\,\cdot\,\cos 'y\; dy \\ &\qquad\text{because for the \sin'y-terms we have \sin(\pi/4)=\sin(3\pi/4)...} \\ &\qquad\text{Substitution: }u=\cos t\ ,\ du=\cos't\, dt\ , \\ &= \frac 1{8\sqrt 2} \int_{-1/\sqrt2}^{1/\sqrt2}\left[\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 +\frac {u^6-1+1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \int_0^{1/\sqrt2}\left[\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 -(1+u^2+u^4) +\frac {1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \int_0^{1/\sqrt2}\left[\ 12u^2 +\frac {1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \left[\ 4\left(\frac 1{\sqrt2}\right)^3 +\frac 12\log \frac{1+\frac 1{\sqrt 2}}{1-\frac 1{\sqrt 2}} \ \right] \\ &= \frac 14 + \frac 1{4\sqrt 2}\log(\sqrt2+1)) \ . \end{aligned} | 2019-05-20T03:18:08 | {
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https://math.stackexchange.com/questions/2794259/what-is-lim-x-to-infty-sin-x | # What is $\lim_{x\to\infty} \sin x$?
What is
$$\lim_{x\to\infty} \sin x$$
?
I always thought it was undefined; however, Wolfram|Alpha says that
$$\lim_{x\to\infty} \sin x = -1 \text{ to } 1$$
Now, what is the correct answer? What does it even mean if the limit is not a single, distinct value, but rather an interval?
Other than that, am I right in assuming that
$$\liminf_{x\to\infty} \sin x = -1$$
and
$$\limsup_{x\to\infty} \sin x = 1$$ ?
• They mean that the values of $\sin(x)$ for $x\to+\infty$ accumulate to every value in $[-1,1]$. In particular, that implies the $\limsup$, $\liminf$ equations that you wrote, and that in the context of an introduction to calculus, the limit doesn't exist. – user561348 May 24 '18 at 12:37
• @arugula so is it in fact undefined? – user500664 May 24 '18 at 12:37
• Who knows what Alpha means, but I guess you could interpret it as "for every point $p$ between -1 and 1, there's a sequence tending to infinity, such that $\sin$ of that sequence is constantly $p$". And yes you're right about the sup and inf limits. – Steve D May 24 '18 at 12:38
• To find what Wolfram Alpha means may be difficult? But presumably this is the same answer as Mathematica itself, which does have extensive documentation you can look at. – GEdgar May 24 '18 at 12:45
Take note that the sine function takes values between $-1$ and $1$. When you are talking about a limit to $\infty$, it's an undetermined number, which is infinitely large. Now, taking into account the periodicity of the sine function, there is no possible way to determine a specific value, as it entirely depends on the nature of the "infinite" number.
More specifically, $-1 \leq \sin(x) \leq 1, \; \; \forall x \in \mathbb R$. This means that for any given $x$ over the real numbers, the sine function is bounded. Thus, all you can say about an undetermined infinite limit (it does not exist talking strictly mathematics), is :
$$-1 \leq \lim_{x \to \infty} \sin(x) \leq 1$$
What you mentioned though is indeed true :
$$\liminf_{x\to\infty} \sin x = -1$$
$$\limsup_{x\to\infty} \sin x = 1$$
• So it can be understood as "the limit doesn't exist, but if it existed it would be between -1 and 1", right? – user500664 May 24 '18 at 12:42
• Essentially, yes. The limit does not exist, but it surely is bounded. – Rebellos May 24 '18 at 12:43
• alright, thank you. I will accept your answer as soon as it will let me. – user500664 May 24 '18 at 12:43
The limit doesn't exist and it can be proved formally by 2 subsequences with different limits, that is
• $x_n=2n\pi+\frac{\pi}2 \to \infty \implies \sin(x_n)=1$
• $x_n=2n\pi+\frac{3\pi}2 \to \infty \implies \sin(x_n)=-1$
Yes what is true is that $\liminf=-1$ and $\limsup=1$ indeed
$$-1\le\sin x \le 1$$
and we have found 2 subsequences which tends to those limits.
• Thank you for the proof as well. Your answer is very good, but I promised Rebellos to accept his answer earlier, so I am sorry. Thank you very much though. – user500664 May 24 '18 at 12:55
• @ThomasFlinkow You are welcome! The choice of the answer to be accepted is absolutely up to you! Bye – user May 24 '18 at 13:05 | 2019-10-17T10:37:15 | {
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https://math.stackexchange.com/questions/2058605/calculating-a-cumulative-distribution-function-cdf | # Calculating a Cumulative Distribution Function (CDF)
Let $Z$ be a continuous random variable with probability density function:
$$f_Z(z) = \begin{cases} \gamma(1 + z^2) & \mbox{ if } -2 < z < 1, \\ 0 & \mbox{ otherwise} \end{cases}$$
a) For what $γ$ is this possible?
b) Find the cumulative distribution function of $Z$.
• For question a do you know what a probability distribution is and the requirements? for question b do you know what a cumulative distribution is and the integral/summation involved? – Chinny84 Dec 14 '16 at 15:47
• I know that to solve part b I need to solve the integral from negative infinity to z of fZ(z), but I don't know how to solve part a, which is why I am stuck. – Monil Dec 14 '16 at 15:53
• Hint: which value should the integral of the PDF have over all domain? – Jesús Ros Dec 14 '16 at 15:55
• I'm sorry, I still don't understand how to do it. – Monil Dec 14 '16 at 16:02
• If $f_Z(z)$ is a PDF, which is the value of $P[-\infty<z<\infty]=\int_{-\infty}^{\infty}f_Z(z)dz$? Or in other words, which is the probability that $z$ lies between $(-\infty,\infty)$? – Jesús Ros Dec 14 '16 at 16:04
$a)$ $$\int_{-∞}^∞\ f_Z(z)\ dz\ = 1$$ $$\int_{-2}^1\ γ(1 + z^2)\ dz= 1$$ $$γ(z + \frac{z^3}3)\ |_{-2}^1\ = 1$$ $$γ\ = \frac16$$
$b)$ $$F_Z(z)\ =\ \int_{-2}^z\ f_Z(z)\ dz$$ $$F_Z(z) = \int_{-2}^z\ \frac16(1 + z^2)\ dz$$ $$F_Z(z) = \frac16(z + \frac{z^3}3) |_{-2}^z$$ $$F_Z(z) = \frac16[(z + 2) + (\frac{z^3 + 8}3)]$$
• Notice that $F_Z(-2) = 0$ and $F_Z(1) = 1.$ That's as it must be. Do you you understand why I made sure of that? Guessing that you do. (+1) – BruceET Dec 15 '16 at 6:44 | 2019-06-18T02:59:31 | {
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A(t) = a(1 + r)t Sometimes you have to use given values to find r! 15. 23)* Initial Value Initial Value GROWTH or DECAY Rate GROWTH or DECAY Rate 3. Exponential growth and decay by a factor. Exponential Functions Part 1 - Graphing - Duration: 14:34. Logistic Growth And Decay By Phillip Neman Logistic Decay Example Problem ( b ) What is the percentage of remaining wood products after 10 years? History of the Logistic Function The logistic function is sometimes called the Verhulst model because it was first published by Pierre. Description of Changes. the x/r is the percent it grew or decreased divided by the rate you're doing at, so one year would be r=1, one month would be r=12, one day would be r=365. Page 5 of 25 Additional Note: Just like a growth factor, you need to be able to identify a decay factor. 98 Which is an exponential decay function?. 3) Given, The initial number for blog, P = 20, Rate per week, r = 20% = 0. Shows growth/decay at consecutive intervals Looks like a normal or backwards "L" or "r" graphically Key words: increasing, decreasing, lost, depreciate Linear Functions: y = mx + b Shows a constant rate of change Looks like a straight line graphically Key words: each 6. What is the rate of growth or rate of decay? 3. InCA Assignment for 3/13. 02) to the 𝘵 power, 𝘺 = (0. h OR one value of the function at a future time. Human population also grows exponentially. Exponential models that use e as the base are called continuous growth or decay models. Does this function represent exponential growth or exponential decay? B. No obligation, cancel anytime. the x/r is the percent it grew or decreased divided by the rate you're doing at, so one year would be r=1, one month would be r=12, one day would be r=365. Hello All, I am trying to fit a curve by exponential growth. Start studying Exponential Growth & Decay. 24 #13, 14, 16, 19, 20a. The Exponential function in Excel is often used with the Log function, for example, in case, if we want to find the rate of growth or decay, in that case, we will use the EXP and the LOG function together. An algebra equation involves a variable representing an un-known number, often denoted by x; and to solve the equation means to nd the nu-merical values of x which make the equation true. "Kavod" = "Respect" btw, YAY MATH!. 2)(t/10), and classify them as representing exponential growth and decay. For example, identify percent rate of change in functions such as y = (1. Determine whether each function is an example of exponential growth or decay. It can be used as a worksheet function (WS) in Excel. Viral growth. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. The questions in this maze are of "Parent Exponential Equations". If you graph this function, you will see it decays really fast, but it actually does not have exponential decay. 2 Graph the following Exponential Functions 8) Function: y=3X 9) Function: Decide if the following tables are exponential or Linear and then write the equation that corresponds with the table. The Exponential Growth Calculator. MEMORY METER. 1%; exponential decay. Exponential functions are commonly used in the biological sciences to model the amount of a particular quantity being modeled, such as population size, over time. function; exponential growth/decay; asymptotes; and end behavior. Exponential Functions, exponential growth, exponential decay, decay factor, growth rates, exponents, rules of exponents, scientific notation. The number C gives the initial value of the function (when t = 0) and the number a is the growth (or decay) factor. Tell whether the function y = 2( 5 ) shows growth or decay. Exponential growth / decay is a specific way that a quantity may increase / decrease over time. Harold’s Exponential Growth and Decay Cheat Sheet 16 May 2016 Discrete Continuous 𝐴=𝑃(1+ N J) 𝑛𝑡 𝐴=𝑃 𝑟𝑡 Simple Interest: 𝐴=𝑃+ =𝑃+𝑃=𝑃(1+ N P) A = Amount after time t P = Original amount, such as principle e = The natural number (~2. 5, is between greater than 1 than 0. The population in the town of Deersburgh is presently 42,500. The first step is to enter the initial value (x0). The next step is to find the trend by noting that we are left with a certain percentage of the substance. Is the pictured graph growth, decay, or linear or none? answer. $\endgroup$ – Matthew Leingang Jan 15 '15 at 17:01 $\begingroup$ Although your biggest problem is important, you will also need to figure out how to handle the “compounded quarterly” part of the problem. Property #1) rate of growth starts slow and increases (Read on, to learn more about this property, which is the primary focus of this web page) Of note: Exponential decay is not the inverse of exponential growth. Since there is a negative exponent, it reverses what the base should be. Under Construction. Modeling Exponential Growth and Decay NOTES In the exponential growth and decay formulas, y = final amount, a = original amount, r = rate of growth or decay, and t = time. com - View the original, and get the already-completed solution here! Solve for solutions: The exponential models describe the population of the indicated country, A, in millions, t years after 2003. -1-1 I a 34 3 4-2 Solve each of the following equations for x. In today’s post, I’ll discuss radioactive decay and the half-life formula. Xtra Gr 11 Maths: In this lesson on Financial Maths we consider the following: Different compounding periods, nominal and annual effective rates, depreciation, linear depreciation as well as reducing balance depreciation. What is the value of the car when it is 12 years old?. For a decay function, the range includes all numbers between the starting value and O. Substituting into a table of values gives us: We plot these points to give:. Differential Equations In Section 6. Example #1 - £5000 is invested in a savings bond, which pays 7% p. That's the difference between a positive and negative growth rate. You will also decide whether or not investing large amounts of money into a home or automobile are wise financial decisions. Exponential growth and decay by a factor. The main purpose of this task is to introduce the decay graph and highlight its differences and similarities to a growth graph. This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. Exponential Growth B. Define exponential decay. 98 Which is an exponential decay function?. Concepts from the last few videos, particularly exponential growth and decay functions, are covered in this video about word problems. notebook January 30, 2015 Writing Exponential Growth and Decay Functions in Real Life!!! Remember this from yesterday? A(t) = a(1 + r)t Sometimes you have to use given values to find r! 15. Lesson #42: Exponential Growth and Decay Functions. How do you know if L : ; represents an exponential growth or exponential decay function? Exponential _____ if Exponential _____ if. Left 2 units c. 99)^x\)? Determine whether the function represents exponential growth or exponential decay. Exponential growth and decay can be determined with the following equation: N = (NI)(e^kt). The base b determines the rate of growth or decay: If 0 b 1 , the function decays as x increases. t is the time in discrete intervals and selected time units. r = the growth rate. − If k < 0, the function will always be negative (it is flipped upside down). Chapter 4 4. Choose a topic that can be related to exponential growth or decay. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. Human population also grows exponentially. Primary objective of this lecture is to present on Exponential Growth and Decay. Property #1) rate of growth starts slow and increases (Read on, to learn more about this property, which is the primary focus of this web page) Of note: Exponential decay is not the inverse of exponential growth. What values of b make it an exponential growth function? C b. Find the value of the car after 10 years. In this section, we will study some of the applications of exponential and logarithmic functions. Series of questions lead through this fascinating topic so that formula becomes clear and connected to the physical world. When it's a rate of decrease, you have an exponential decay function! Check out these kinds of exponential functions in this tutorial!. Sketch an exponential decay function. Explain 2 Modeling Exponential Decay Recall that a function of the form y = a b x represents exponential decay when a > 0 and 0 < b < 1. A di erential equation (DE) involves. f(x) = 2 x 62/87,21 Make a table of values. Another convention is to always write the exponential functions in terms of the natural constant e. An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. And I just came back from this teaching conference, where one of the sessions included a few handouts of the types of problems that this one charter school uses. You must use your knowledge of exponential growth and decay functions to make some important financial decisions. Quantitative Reasoning Linear Models, Scatterplots, Exponential Growth, Exponential Decay Questions 1-3. a = value at the start. 8% per year and the current population is 1543, what will the population be 5. x(t) = x 0 × (1 + r) t. Exponential decay models decrease very rapidly, and then level off to become asymptotic towards the x-axis. 25)^x growth or decay y-intercept: asked by Aaron on April 9, 2017; Algebra. That is, the rate of change is always the same, and each value is determined from the previous value by adding the same amount. Tell whether the equation represents exponential growth, exponential decay, or neither. Exponential decay and be used to model radioactive decay and depreciation. Geometric growth or decay is the same as exponential growth or decay except the function is only evaluated at discrete values. Write!an!exponential!function!to!model!each!situation. ' Notice that exponential growth word problems often involve compound interest, and exponential decay word problems often involve depreciation. Exponential decay occurs when 0 < b< 1. Ma 112 Precalculus Section 5. Choose the word or phrase from the list that best completes each sentence. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. In the Exponential Growth and Decay Gizmo™, you can explore the effects of C and r in the function y = C(1 + r)t. p(t) = 50(0. Exponential Decay. " "We just looked at an exponential growth function. Determine if the equation y = 2. Differentiate between growth and decay by examining functions % Progress. Exponential growth and decay by percentage. Exponential Growth and Decay Formulas The formula that represents exponential growth or decay is y = a bx, where a = original (initial) amount b x = growth factor (1 + r) final amount unit of measure (usually time) PART I: Determining Exponential Growth and Decay. Graph the function. The odds of you using them on your next trip to the supermarket are fairly slim (more on using odds later), but some professions simply would not be able to do the things they do without them. Growth Function in Excel. The instructions are “The function below represents a growth or decay What is the initial quantity?. amount after. Exponential Decay functions model many real world scenarios. The GROWTH Function is categorized under Excel Statistical functions. Exponential decay functions also cross the y-axis at (0, 1), but they go up to the left forever, and crawl along the x-axis to the right. P(t)= Poe^(-kt). - when 0 b < 1, the function represents a decay function Writing an Equation Using y — abl There are many applications of exponential fimctions in real. 2 Exponential Decay: Worked Examples. Microsoft Word - Exponential growth and decay problems 1 Author: kstewart Created Date: 4/7/2020 7:02:47 AM. Exponential growth refers to an amount of substance increasing exponentially. It has grown from 30 grams to 50 grams in 48 hours. Online exponential growth/decay calculator. If A is decaying continuously at rate r, then A may be written as follows A = A0e−rt. 32 MB] A Powerpoint presentation on Exponential Growth and Decay. A) 3) 4) y SECTION 3: Determine which functions are exponential growth and which are exponential decay. Exponential Growth and Decay Make Up 1. Write a function V that will predict the value of a home after x years. , a function in which the time value is the exponent. Exponential Growth and Decay Functions An exponential function has the form y = abx, where a ≠ 0 and the base b is a positive real number other than 1. Then find the painting's value in 15 years. Worksheet 9 Memorandum: Finance, Growth and Decay. 4 Graph an exponential function of the form f(x) = ab^x. growth or exponential decay. A town had a population of 53700 in 1996 and a population of 58100 in 2000. These exponential functions describe both the rise and the fall of particular systems, especially systems that can be described through mathematical concepts, such as population or radioactive decay. Does this function represent exponential growth or exponential decay? B. In this case, each number, starting at 5 is multiplied by 2 to an exponent power such as 2. 4% compounded. Exponential Growth and Decay Word Problems Write an equation for each situation and answer the question. , 5% becomes 1. 1 Examples of exponential growth or decay. Step 3: Substitute for t. Probably the most well known example of exponential decay in the real world involves the half-life of radioactive substances. These included the growth of populations and the decay of radioactive substances. Typically what you need to do is: 1. Determine whether the function represents exponential growth or decay. Formula to calculate exponential growth and decay is given by:. Exponential Growth and Decay. In the same way, whenever the base is less than 1 and greater than 0, the exponential function shows decay. Chapter 4 4. It gets rapidly smaller as x increases, as illustrated by its graph. 5% every year. It is a decreasing function,because, for The Curve has one Asymptote which is, y=0. The simplest type of exponential growth function has the form y = Core Concept Core Concept parent. Write!an!exponential!function!to!model!each!situation. If you're behind a web filter, please make sure that the domains *. December 14, 2018 December 14, 2018 mshallmaa Leave a comment. Exponential functions tell the stories of explosive change. Exponential Functions - Day 1 Homework Growth & Decay Name Date Block 1. The formulas below are used in calculations involving the exponential decay of, for example, radioactive materials Let’s code a Matlab function to calculate the final amount of a substance, given the elapsed time, half-life and initial amount. Parameter a (called the amplitude) affects the height of the wave,. A model for decay of a quantity for which the rate of decay is directly proportional to the amount present. Exponential Decay Formula: Make a substitution for A and t since it is known that the half-life is 1690 years and : Solve for the decay rate k: Start by dividing both sides by the coefficient to isolate the exponential factor. amount after. The function g(x)=(1 2)x is an example of exponential decay. The stuffed animal was originally worth $6, so. growth function. Try it risk-free. Described as a function, a quantity undergoing exponential growth is an exponential function of time, that is, the variable representing time is the exponent (in contrast. 2 1536 12288 98304 786432 0. Scribd is the world's largest social reading and publishing site. In 2000, there were 236,000 people in the city. When I graph points relating to the function, as. As, x→∞, So, when , x→∞,y→0. Exponential growth: y = a (1 + r)t Exponential decay: y = a (1 − r)t The population of a city is increasing at a rate of 4% each year. Find the value ot the base. ) Comparing the rates of growth (or decay) will only yield the same results when both rates are converted to the same compounding period. Graph the function. Exponential growth (including exponential decay) occurs when the growth rate of a mathematical function is proportional to the function's current value. It is a decreasing function,because, for The Curve has one Asymptote which is, y=0. 71 Graphing Exponential Functions. Exponential growth and decay by percentage. An exponential function where a > 0 and 0 < b < 1 represents an exponential decay and the graph of an exponential decay function falls from left to right. b) c) 2) A car was purchased for$20,000. • Fill in the table, showing the increase in area of the bacteria over eight weeks. Understanding how to graph exponential growth and decay functions is a very important tool for future courses. 43%) At close: 4:00PM EST. 4 Modeling Exponentials; 4. Exponential Growth: a function or model which begins increasing at a very slow rate, then keeps growing faster and faster; functions in the form of y = P(1 + r) t. Continuing with the unit on exponential functions, today we spent some time working on creating functions that model other real world scenarios. This function is useful for describing many very different observations in science. A 5 step guide for sketching exponential functions is discussed to help students remember. Exponential models that use e as the base are called continuous growth or decay models. 1 as x gets bigger so does 5^x and so does the expression : Growth. 3c Use the properties of exponents to transform expressions for exponential functions. In the Exponential Growth and Decay Gizmo™, you can explore the effects of C and r in the function y = C(1 + r)t. To solve real-life problems, such as modeling the height of a sunflower in Example 5. One Bernard Baruch Way (55 Lexington Ave. Damping of oscillating system. The mathematical model of exponential growth is used to describe real-world situations in population biology, finance and other fields. 825 13) Identify the RATE as a percent for the. How Do You Solve a Word Problem with Exponential Growth? If something increases at a constant rate, you may have exponential growth on your hands. Exponential Growth and Decay. You must use your knowledge of exponential growth and decay functions to make some important financial decisions. unit 5 worksheet 6 graphing exponential growth and decay functions UNIT 5 WORKSHEET 6 GRAPHING EXPONENTIAL GROWTH_DECAY. What type of function is y = 7(5/4)x? Exponential Growth and Decay DRAFT. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. Differential Equations In Section 6. Identifying Exponential Growth and Decay. Company Outlook. The most important feature of an exponential function is that the independent variable is in the exponent of some base, usually an agreed-upon (but constant) base, like 2, 10 or the special number "e". The table shows the world population of the lynx in 2003 and 2004. what multiplicative rate of change should Hal use in his function? 0. ) Write an. In 2013, a research company found that smartphone shipments (units sold) were up 32. Exponential growth and decay is a concept that comes up over and over in introductory geoscience: Radioactive decay, population growth, CO 2 increase, etc. It is generally used to express a graph in many applications like Compound interest, radioactive decay, or growth of population etc. 7% worldwide from 2012, with an expectation for the trend to continue. Identify the asymptote of each graph. amount after. V(x) = 165,000 ( 1. Exponential growth and exponential decay questions Basic steps Most basic exponential growth or exponential decay questions in HSC Maths (2 Unit) exams involve a series of substitutions to get the answers. Exponential growth is a specific way that a quantity may increase over time. 9th - University grade. Exponential decay and be used to model radioactive decay and depreciation. Exponential Decay of Quantum Wave Functions I’ve no doubt that for ODEs, the questions and techniques for exponential decay of solutions go back a long way, maybe even to the nineteenth century. Exponential Growth and Decay Name_____ Date_____ Period____ Solve each exponential growth/decay problem. A 5 step guide for sketching exponential functions is discussed to help students remember. Growth is also known as 'appreciation' when it is said that something has appreciated in value, meaning that its value has increased. Exponential Growth and Decay DRAFT. The Microsoft Excel GROWTH function returns the predicted exponential growth based on existing values provided. Classify the model as. Worksheets are Exponential growth and decay, Graphing exponential, Exponential growth and decay work, Concept 17 write exponential equations, Exponential growth and decay word problems, Exponential functions work 1, Exponential population growth, Honors pre calculus d1 work. Finance: Compound interest. Shows growth/decay at consecutive intervals Looks like a normal or backwards "L" or "r" graphically Key words: increasing, decreasing, lost, depreciate Linear Functions: y = mx + b Shows a constant rate of change Looks like a straight line graphically Key words: each 6. Exponential functions have many scientific applications, such as population growth and radioactive decay. Formula to calculate exponential growth and decay is given by:. a) Write an equation that gives the value of the car, y, after x years. a logarithmic functions. Exponential Functions: Graphs. Exponential Decay Formula: Make a substitution for A and t since it is known that the half-life is 1690 years and : Solve for the decay rate k: Start by dividing both sides by the coefficient to isolate the exponential factor. ANSWER KEY 1. And I just came back from this teaching conference, where one of the sessions included a few handouts of the types of problems that this one charter school uses. Research your topic and learn how it is related to exponential functions. Exponential Growth and Decay Word Problems Worksheet Answers – Begin customizing it and you could also to open it on your document window when you find a template that you would like to use! You will discover others call for a premium account and that a number of the templates are free to use. If the rate of increase is 8% annually, how many stores does the restaurant operate. Then, solve the function and get the answer!. Student Exploration: Exponential Growth and Decay. Growth or Decay? Write a function that represents this situation Answer: 1. Exponential Functions Part 1 - Graphing - Duration: 14:34. The parent exponential function is,f(x) = bx, where the base, b, is a constant and the. 5 Exponential Functions Identify whether the equation represents a growth or decay function. In the Exponential Growth and Decay Gizmo™, you can explore the effects of C and r in the function y = C(1 + r)t. The basic parent function of any exponential function is f ( x) = bx, where b is the base. It appreciates 6% per year. 1 Exponential Models Every model is, at the very end, an approximation of real phenomena. While function with exponential decay DO decay really fast, not all functions that decay really fast have exponential decay. Students will cut out the pieces and match a word problem to the type of exponential function (growth or decay) and the answer. We see these models in finance, computer science, and most of the sciences, such as physics. In 2000, there were 236,000 people in the city. 7) Worksheet 6-6. where C is the initial amount (the amount before any decay occurs), r is the. That is, the rate of change is always the same, and each value is determined from the previous value by adding the same amount. In fact, we can use the Exponential Growth and Decay Formula to find snow depth levels, the magnitude of a star, how temperature affects a body, or how a fast-food chain expands its business as Khan. His biomedical. – The graph decreases at a decreasing rate. First, they evaluate for a variable given the second variables value. Bank accounts that accrue interest represent another example of exponential growth. 825 13) Identify the RATE as a percent for the. Lesson #3: Applications of Exponential Growth or Decay Review An exponential function is a function whose equation is of the form y —abl - when b > 1, the function represents a function. Exponential growth is a specific way that a quantity may increase over time. Also, the a value can tell us if the exponential curve is concave up (opening upwards) or concave down (opening downwards). Yes, any teacher can have a random Britney Spears moment. Explaining the shape of exponential functions, and why exponential functions illustrate growth and decay. to have exponential growth or exponential decay. Exponential Growth Formula Find the Exponential Growth Function That Models Given Data Example 4. 2 b 1, so the function shows exponential growth. Base Function: y = ½x a. Find a function that models the population after t years. In talking about problems like population growth, we needed to learn about the exponential function. the set of all real numbers). Is this exponential growth or decay? Preview this quiz on Quizizz. In this section, we will study some of the applications of exponential and logarithmic functions. The exponential curve is especially important in mathematics. But as time continues, the quadratic function changes direction and grows to infinity, while the exponential decay function approaches an asymptote. 2x 10) Write and solve an exponential function to model the situation and find the value after the given time, Round to the nearest whole number. In a straight line, the “rate of change” is the same across the graph. If 0 b 1, the function shows decay. No matter how many are in the population at some point in time, the percent that leave the population in the next period. What type of function is y = 7(5/4)x? Exponential Growth and Decay DRAFT. 8 Logistic Growth Functions 517 Evaluate and graph logistic growth functions. Determine the exponential functions that model the populations of both cities. The function helps calculate the predicted exponential growth by using the existing data. Exponential Decay Function - Half Life This video explains how to determine an exponential decay function from given information. function; exponential growth/decay; asymptotes; and end behavior. Remember that there are two ways to tell whether a graph is an example of exponential decay or exponential growth. at 24th St) New York, NY 10010 646-312-1000. Then graph. – The graph decreases at a decreasing rate. 1: Exponential Growth and Decay For each problem use a function of the form f(t)=a⋅bt orf(t)=a⋅bkt, where k is a constant that describes the situation and t is time. We discuss exponential growth, exponential decay, domain, and range as well. Shortcut for exponential shrinkage. what multiplicative rate of change should Hal use in his function? 0. Exponential Growth And Decay Functions. is known as an exponential function. a) let P represent the population “t” years after 2000. 4% compounded. MAT 111 Exponential Growth and Decay Problems. If you graph this function, you will see it decays really fast, but it actually does not have exponential decay. For example, in algebra 2 the students will be learning about logarithms and exponentials, and will have to graph both of them and know the difference between. Exponential Growth and Decay. They asked us graph the following exponential function. Exponential growth and decay can be determined with the following equation: N = (NI)(e^kt). P 0 = initial amount at time t = 0. M&M Lab (Exponential Growth and Decay) Part I: Modeling Exponential Growth M&M Activity The purpose of this lab is to provide a simple model to illustrate exponential growth of cancerous cells. 95% interest yearly, will. An exponent is part of the equation as well, so for instance, an equation could be y = 5*2x. Welcome: Exponential Growth and Decay Description: The student will learn exponential growth and decay and solve related application problems involving compound interest, half-life, doubling, and carbon dating. Part 1: How are exponential growth and decay present in the real world? Give at least 2 examples for exponential growth and 2 examples of exponential decay. 8b Use the properties of exponents to interpret expressions for exponential functions. Exponential Growth and Decay. 65)𝑥, where x is the number of years since he made his investment. The equation for the line of an asymptote for a function in the form of f(x) = abx is always y = _____. 3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. Milton has his money invested in a stock portfolio. Find the value of the car after 10 years. On the other hand, exponential functions where 0 < b < 1 represent exponential decay. Growth formula in Excel helps in financial and statistical analysis, it helps to predict revenue targets, sales. Graphing Exponential Functions What is an Exponential Function? Exponential functions are one of the most important functions in mathematics. This is due to the exponential curve, never reaching a number. Make sure you have memorized this equation, along with. Exponential function definition is - a mathematical function in which an independent variable appears in one of the exponents —called also exponential. An algebra equation involves a variable representing an un-known number, often denoted by x; and to solve the equation means to nd the nu-merical values of x which make the equation true. ←Identifying Exponential Growth and Decay. Exponential+Growthand+DecayWord+Problems+!!! 4. This is an exponential growth function. Sinusoidal functions are useful for describing anything that has a wave shape with respect to position or time. The number C gives the initial value of the function (when t = 0) and the number a is the growth (or decay) factor. When it's a rate of decrease, you have an exponential decay function! Check out these kinds of exponential functions in this tutorial!. r = the decay rate. As, x→∞, So, when , x→∞,y→0. What is your initial value? C. Exponential Growth and Decay Functions An exponential function has the form y = abr, where a O and the base b is a positive real number other than I. In other words, $$y′=ky$$. Here are the basics that you should know if you want to get a perfect SAT score: A general exponential function has the form f(t) = a(1 + r) ct, where a = f(0) is the initial amount and r is the growth rate. Exponential Decay B. P (t) is an increasing function if b > 0 and a decreasing function if b < 0. For example, you will have to decide where you will bank and invest your money. The base (5 in this case) determine the growth or decay of the function. 01)(12t), y = (1. 06 the rate of decay of a chemical compound that has a half-life of 5730 years the rate of growth of a bacteria strain. function; exponential growth/decay; asymptotes; and end behavior. Exponential Decay 2. Typical problems involve population, radioactive decay, and Newton's Law of Cooling. From the U. Of note: Exponential Growth is not the inverse of exponential decay. The equation can be written in the form $f(x)=a(1+r)^x$ or $f(x)=ab^x$ where $$b=1+r$$. Graph the function. a) y = 2(4)^x growth or decay y-intercept: b) y = 3(. Decay function : if 0 < b < 1. Figure a, for instance, shows the graph of …. U4D8_T Applications - Exponential Growth and Decay Problems: Duo-tang Worksheet for U4D8 Applications Exponentials (Exponential Growth & Decay) Do: Growth Worksheet # 1 - 4 & Decay Worksheet # 1 - 4. Worksheet 9 Memorandum: Finance, Growth and Decay. In many cases "a" represents a starting or initial value, "b" represents the multiplier or. This calculator has three text fields and two active controls that perform independent functions of the calculator. Exponential Growth and Decay Make Up 1. IE7e EXAMPLE - Graphing f(x) = If for b > 1 Graph f(x) = 2*. Use the properties of exponents to interpret expressions for exponential functions. To vary the values of C and r, drag the sliders. edu: 8-1 Linear Growth and Decay. Exponential Growth and Decay Calculator Use our online exponential growth and decay calculator by entering the initial value (x 0), decay rate (r) and time (t) in the below calculator and click calculate button to find the answer. 3 4 4 4-4. Exponential Growth Rate of Growth Function, Inhibited Growth, development of the Quantity Formula, Logistic Growth, … Download [1. The value of the property in thousands of dollars, t years after 2001 is given by the exponential growth model. Exponential growth and exponential decay are two of the most common applications of exponential functions. $\endgroup$ – Matthew Leingang Jan 15 '15 at 17:01 $\begingroup$ Although your biggest problem is important, you will also need to figure out how to handle the “compounded quarterly” part of the problem. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. The table shows the world population of the lynx in 2003 and 2004. The instructions are “The function below represents a growth or decay What is the initial quantity?. In the previous posts in this series, I considered financial applications. Modeling Exponential Growth and Decay. Exponential Decay 5. Is the function. The growth "rate" (r) is determined as b = 1 + r. asked • 04/06/16 Find the initial value a, growth/decay factor b, and growth/decay rate r for the following exponential function:Q(t)=1950(1. The equation for "continual" growth (or decay) is A = Pe rt, where "A", is the ending amount, "P" is the beginning amount (principal, in the case of money), "r" is the growth or decay rate (expressed as a decimal), and "t" is the time (in whatever unit was used on the growth/decay rate). For example, a \$2000 deposit, earning. Write an exponential decay function to model this situation. One Bernard Baruch Way (55 Lexington Ave. 4 Exponential growth and decay THE MAIN QUESTIONS 1 of 4 - Duration: 7:25. However, for each unit increase in t, 2 units are added to the value of L (t), whereas the value of E (t) is multiplied by 2. Name: Period: Question Exponential. Sketch the graph of. Also, if you substitute any x-value from the table into the function rule, it produces the correct y-value. Logarithmic growth is the inverse of exponential growth and is very slow. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. Example: Graphing Exponential Growth. Example, recall a geometric series that has a start value of 5 and a common ratio of 2, i. t = time (number of periods) Exponential Growth Calculator. As the value of the variable changes, the value of the function increases (or decreases) in proportion to its current value. growth or exponential decay. 2 Graph Exponential Decay Functions Georgia Performance Standard(s) MM3A2e, MM3A2f, MM3A2g Your Notes Goal p Graph and use exponential decay functions. The unrestricted growth of bacteria is an example of exponential population growth. Exponential Decay is an decrease in a quantity P. Write an exponential decay function to model this situation. The odds of you using them on your next trip to the supermarket are fairly slim (more on using odds later), but some professions simply would not be able to do the things they do without them. So, the number of blocks after x weeks, Hence, the number of blocks. bacteria/disease, money, population, etc. Download here: Worksheet 9: Finance, Growth and Decay. Before showing how these models are set up, it is good to recall some basic background ideas from algebra and calculus. If we start with only one bacteria which can double every hour, how many bacteria will we have by the end of one day?. Write an exponential growth function to model this situation. Unit 4 lesson 8 (Growth and Decay Textbook. Get access risk-free for 30 days, just create an account. Write!an!exponential!function!to!model!each!situation. 9th - University grade. Exponential decay models decrease very rapidly, and then level off to become asymptotic towards the x-axis. Exponential growth and decay often involve very large or very small numbers. Get access risk-free for 30 days, just create an account. Solve the di erential equation y0= ky. Exponential growth occurs when the growth rate of the value of a mathematical function is proportional to the function's current value, resulting in its growth with time being an exponential function, i. 00 KB] If you found these worksheets useful, please check out Finding the Area between Curves, Hyperbolic Functions Worksheet. Someone help me please! 1. All models are wrong, some models are more wrong than others. We start with the basic exponential growth and decay models. Professor Elvis Zap 16,078. This is a PPT I put together for my Year 11 top set to cover off the new GCSE topic of exponential growth and decay. t = time (number of periods) Exponential Growth Calculator. It can be used as a worksheet function (WS) in Excel. The graphs of such functions are like exponential growth functions in reverse. If 0 b 1, the function shows decay. Given a function P(t), where P is a function of the time t, the rate of change. Exponential Growth B. Let's consider exponential decay and exponential growth by inspecting their respective general shapes of their graphical representations. determine if it represents exponential growth or. Finance, growth and decay tests simple and compound appreciation and depreciation, as well as gives story sums and timeline questions. Does the function f (x) = 5(0. It provides the formulas and equations / functions that you need to solve it. It depreciates at a rate of 13% a year. (The results from compounding monthly are increasing more rapidly over time. Follow these steps to write an exponential equation if you know the rate at which the function is growing or decaying, and the initial value of the group. y represents the size of the population or amount at time x. When b is less than 1, you have an exponential decay function. Exponential Decay functions model many real world scenarios. When does f' (x) = f (x)? Quiz: Writing Exponential Growth & Decay Functions. Popular Tutorials in Exponential Growth and Decay. Chapter 4 4. Is the graph below exponential growth or decay? One might think this was exponential decay due to the severe plunge of the y values. t is the time of growth/decay. Use logistic growth functions to model real-life quantities, such as a yeast population in Exs. 31 KB (Last Modified on January 18, 2019) Comments (-1). Exponential growth and decay functions are written as: F(t) = A₀b^(kt) We know that A₀ is the initial amount (at the time equal to zero) b is the growth factor, k is the growth rate, and t is the time (the independent variable) Note that b is a real and positive number, while A₀ and k can be also negative. Identify the growth or decay factor, the growth or decay rate, and the initial value. You will also decide whether or not investing large amounts of money into a home or automobile are wise financial decisions. Exponential growth vs. In fact, any change in the number of objects which depends proportionally on the number of objects present will be described by an exponential. Grade Level: 6-8 Curriculum: Math Keywords: exponents, exponential, functions, rapid, growth, decay Author(s): Dominador Guillermo. In this tutorial, learn how to turn a word problem into an exponential growth function. 8b Use the properties of exponents to interpret expressions for exponential functions.
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http://techiemathteacher.com/2014/07/28/phi-function/ | The Phi(ϕ) Function
Phi Function is define as the number of positive integers less than or equal to and coprime to the number. The other name is Euler’s Totient Function.
The application of this function is used to finding the remainders of large numbers, last digits of a large number, and more to be found.
Coprime numbers – these are pairs or set of numbers that don’t have a common divisor. Example are {3,5}, {2,3,7}, {12,7}. Pairs that is not coprime are {3,6}, {12,9}, {20,15}.
Formula:
Given $n=a^xb^yc^x\ldots$, then
$Phi(\varphi) n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$n,a,b,c,x,y,z$ are all positive intergers.
To test the formula, let’s try the obvious one.
Worked Problem 1:
How many positive integers less than or equal to and co-prime to 12?
Solution:
Listing all numbers that satisfy this condition we have 1,5,7,11. There are four positive integers less than or equal to and co-prime to 12.
Using the ϕ function we have,
$12=2^2\cdot 3$
$\varphi n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$\varphi 12=12(1-\displaystyle\frac{1}{2})((1-\displaystyle\frac{1}{3})$
$\varphi 12=12(\displaystyle\frac{1}{2})(\displaystyle\frac{2}{3})$
$\varphi 12=4$
The formula satisfies the answer. This means that we can rely on this to calculate same problem with large numbers without exhausting our time to count them individually.
Worked Problem 2:
Find the number of positive integers less than or equal to and shares no common divisor with 2016.
Solution:
This problem is just exactly the same as the definition of ϕ function.
$2016=2^5\cdot 3^2\cdot 7$
$Phi(\varphi) n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$\varphi 2016=2016(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{7})$
$Phi(\varphi) 2016=576$
Worked Problem 3:
Find the number positive integers less than or equal to and co-prime to 390.
Solution:
$390=2\cdot 3\cdot 5\cdot 13$
$\varphi 390=390(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{5})(1-\displaystyle\frac{1}{3})$
$\varphi 390=96$
Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
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https://math.stackexchange.com/questions/2745374/does-x-n-colon-frac-sqrtnn-n-converge-if-so-then-how-to-find | # Does $x_n \colon= \frac{ \sqrt[n]{n!} }{ n }$ converge? If so, then how to find the limit?
For every natural number $n$, let $$x_n \colon= \frac{ \sqrt[n]{n!} }{ n }.$$ Then does the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converge or diverge? And, how to find the limit?
My Attempt:
We find that, for any $n \in \mathbb{N}$, $$\frac{ x_{n+1} }{ x_n } = \frac{ \frac{ \sqrt[n+1]{ (n+1)!} }{ n + 1 } }{ \frac{ \sqrt[n]{n!} }{ n } } = \frac{ \left( n! \right)^{ \frac{1}{n+1} - \frac{1}{n} } }{ 1 + \frac{1}{n} } \sqrt[n+1]{n+1} = . . .$$
What next? I was hoping to be able to apply the so-called "ratio" test for sequences, but there seems to be no such possibility available.
• Try Stirling's approximation. – ProtectedSource Apr 20 '18 at 1:00
• What is ratio test for sequences ? I keep seeing this on this site. Where did this idea come from ? – Rene Schipperus Apr 20 '18 at 1:03
• @ReneSchipperus I asked myself the same question and Google sent me to this page: mathonline.wikidot.com/the-ratio-test-for-sequence-convergence – Malcolm Apr 20 '18 at 1:05
• @Malcolm Well, thats some pretty low level math right there. – Rene Schipperus Apr 20 '18 at 1:07
• If $x_{n+1}/x_n\to L < 1$ then $x_n \to 0$ for positive $x_n$ – Malcolm Apr 20 '18 at 1:07
Consider $a_n = x_n^n = \dfrac{n!}{n^n}$. Then $$\frac{a_{n+1}}{a_n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1+\frac1n}\right)^n \to \frac1e$$ Now, it is true that if $\lim \frac{a_{n+1}}{a_n}$ exists, then so does $\lim \sqrt[n]{a_n}$ and they are equal (see a proof here). Therefore, $$\lim x_n = \lim \sqrt[n]{a_n} = \lim \frac{a_{n+1}}{a_n} = \frac1e$$
• Dont know why someone downvoted this, its nice. – Rene Schipperus Apr 20 '18 at 1:17
Another way is to take logarithms. $$\ln \frac{\sqrt[n]{n!}}{n}= \frac{1}{n}\sum_{k=2}^{n}{\ln k}-\ln n$$ Then using $$\int_2^{n+1}{\ln x}\,\mathrm dx\ge \sum_{k=2}^{n}{\ln k} \ge \int_1^{n}{\ln x}\,\mathrm dx \\ \text{ and }\int{\ln x}\,\mathrm dx=x\ln x - x,$$ it's easy to see the logarithm goes to $-1$.
Here is an easy trick, without sterling.
If $a_n\to L$ then $$\sqrt[n]{a_1a_2 \cdots a_n}\to L$$ apply this to $$a_n=\left(1+\frac{1}{n}\right)^n$$
Using Stirling's approximation:
$$x_n\implies\frac{\sqrt[n]{n!}}{n}\sim\frac{(\tau n)^\frac1{2n}\left(\frac{n}{e}\right)^\frac{n}{n}}{n}\implies\frac{1}{e}(\tau n)^\frac1{2n}$$
$$\rightarrow\lim_{n\to\infty}\left(\frac{1}{e}(\tau n)^\frac1{2n}\right)=\boxed{\frac{1}{e}}$$
• The limit of $n^{1/2n}$ is 1, not 0. – Mike Earnest Apr 20 '18 at 1:20
Using Stirling's approximation: $$n!\sim\sqrt{2\pi n} \,\left(\frac{n}{e}\right)^n$$ You are looking for \begin{align} \lim_{n\to\infty} x_n&=\lim_{n\to\infty}\frac{\sqrt[n]{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}}{n}\\&=\lim_{n\to\infty} e^{-1} \left(2\pi n\right)^{1/(2n)} \\ &=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n}\sqrt{n^{1/n}}\\ &=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n} \cdot \sqrt{\lim_{n\to\infty} \sqrt[n]{n}}\\ &=\frac{1}{e} \end{align} So your sequences converges.
• Yes you are right. I edited the response. Thanks for telling me! – Zachary Apr 20 '18 at 1:21
By Stirling approximation
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$
$$x_n \colon= \frac{ \sqrt[n]{n!} }{ n }\sim \frac{ \sqrt[n]{\sqrt{2 \pi n}} }{ e }\to \frac1e$$
or as an alternative proceed by ratio-root criteria. | 2019-05-26T03:40:16 | {
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https://math.stackexchange.com/questions/2304255/definition-of-convergence-of-double-series | # Definition of convergence of double series
I read in one place that the definition of convergence of a double series is given by:
Let $(a_{mn})_{m, n=1}^{\infty}$ be a double sequence. The corresponding double series is denoted $\sum_{m, n =1}^{\infty} a_{mn}$. The corresponding double sequence of partial sums is denoted $(s_{mn})_{m, n=1}^{\infty}$ and is defined by $s_{mn} = \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} = \sum_{j=1}^{n} \sum_{i=1}^{m} a_{ij}$. We say that the double series $\sum_{m, n =1}^{\infty} a_{mn}$ converges to $A \in \mathbf{R}$ if the corresponding double sequence of partial sums $(s_{mn})_{m, n=1}^{\infty}$ converges to $A$. We denote this as \begin{align*} \sum_{m, n =1}^{\infty} a_{mn} = \lim_{m,n \rightarrow \infty} s_{mn} = \lim_{m,n \rightarrow \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} = \lim_{m,n \rightarrow \infty} \sum_{j=1}^{n} \sum_{i=1}^{m} a_{ij} = A. \end{align*}
However, in another place, the definition is as follows:
If the sequence $(s_{nn})$ converges, then we define $\sum_{m, n =1}^{\infty} a_{mn} = \lim_{n \rightarrow \infty} s_{nn}$
My question is: are these two definitions equivalent? For these two definitions to be equivalent, wouldn't we need the condition that $(s_{mn})_{m, n=1}^{\infty}$ converges if and only if $(s_{nn})$ converges and $\lim_{m,n \rightarrow \infty} s_{mn} =\lim_{n \rightarrow \infty} s_{nn}$?
If the definitions are not equivalent, then which one do I follow?
• What about the terms of $s_{m,n}$ where $m\ne n$? What of $s_{m,n}=m-n$? May 31 '17 at 15:16
Consider $s_{mn} = mn/(m+n)^2$. Then $s_{nn} \to 1/4$ but $\lim_{m,n} s_{mn}$ does not exist. There is no unique limit -- for example, with $m = 2n$ we have $s_{2n,n} \to 2/9 \neq 1/4$.
Convergence of $\lim_{m,n} s_{mn}$ is a stronger condition and implies the convergence of $\lim_{n} s_{nn}$
For double series $\sum_{m,n} a_{mn}$, absolute convergence guarantees that all modes of convergence -- by rows, columns, squares, diagonals, etc. are equivalent:
$$\lim_{M,N}\sum_{m=1}^M \sum_{n=1}^N a_{mn} = \sum_{m=1}^\infty\sum_{n=1}^\infty a_{mn} = \sum_{n=1}^\infty\sum_{m=1}^\infty a_{mn} = \lim_{N \to \infty}\sum_{n=1}^N\sum_{m=1}^N a_{mn}$$
• Thanks, this a great answer. One last question, is it true that for a double sequence $(a_{mn})_{m, n=1}^{\infty}$, $\lim_{m, n \rightarrow \infty} a_{mn} = A$ if and only if $\lim_{m \rightarrow \infty}\left(\lim_{n \rightarrow \infty} a_{mn}\right) = \lim_{n \rightarrow \infty}\left(\lim_{m \rightarrow \infty} a_{mn}\right) =A$? Jun 1 '17 at 3:28
• If $\lim_{m,n \to \infty}a_{mn} = A$ and $\lim_{n \to \infty}a_{mn}$ exists for all $m$, then $\lim_{m \to \infty} \lim_{n \to \infty}a_{mn} = A$. Similarly if $\lim_{m \to \infty}a_{mn}$ exists for all $n$ then $\lim_{n \to \infty} \lim_{m \to \infty}a_{mn} = A$. The converse is not necessarily true. There is a converse if one of the inner iterated limits is uniformly convergent.
– RRL
Jun 1 '17 at 3:41
• If $\lim_{n \to \infty} a_{mn} = y_m$ with uniform convergence for all $m$ and $\lim_{m \to \infty} a_{mn} = z_n$ exists (not necessarily uniform) then the iterated limits and the double limit exist and are the same: $\lim_{m,n \to \infty} a_{mn} = \lim_{m \to \infty} y_m = \lim_{n \to \infty} z_n$.
– RRL
Jun 1 '17 at 3:45
• Do you have access to the book The Elements of Real Analysis by Bartle? This is nicely covered in Section 19 (second edition).
– RRL
Jun 1 '17 at 3:46
• Thanks for the reference, I'll definitely check it out. I also have one more question on absolute convergence of double series, would be great if you could share your insights: math.stackexchange.com/questions/2305138/… Jun 1 '17 at 4:38 | 2022-01-29T02:26:27 | {
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http://urgd.autismart.it/deflection-of-simply-supported-beam-with-udl.html | Deflection Of Simply Supported Beam With Udl
Brijendra Gupta. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The beam is supported at each end, and the load is distributed along its length. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Deflection and Slope of Beams •As load is applied on a beam, it deflects. simply supported beam? When a simply supported beam is subjected to a moving udl longer than the span, the absolute maximum bending moment occurs when the whole span is loaded. The center reaction Q4. Load Calculation For Beam. I believe that changing formulation to MPC is the solution. 044) /64 = 1. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. ! The beam has a length of L. Design StatusVertical shear Moment Buckling Deflection PASS PASS PASS PASS. Cantilever beam. M is lesser and hence lighter materials of construction can be used it resist the bending moment. 5) Option to calculate singly supported, fixed and cantilevered beams. 8 × 10 6 psi, υ 12 = 0. Depending on the load applied, it undergoes shearing and bending. Its mode of deflection is primarily by bending. a simply supported beam spanning say 4m is loaded with a UDL of say 18kn/m. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Hence, the fundamental equation in finding deflections is: 2 2 x x d y M dx EI In which the subscripts show that both M and EI are functions of x and so may change along the length of the beam. Whereas the deflection of simply supported composite beam with partial shear connection is 24. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. 1457\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Deflection Chart of Beam at various Condition. approximate solutions. Fixed-Fixed Beam with UDL is a beam type that has fixed supports on both ends and transverse uniformly distributed load. 5kN/m2, Permanent: 0. (b) Deflection at where the point load is acting, and state its direction. Simply-Supported Beams Figure: A simply-supported beam. Fig:1 Formulas for Design of Simply Supported Beam having. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Area Moment Method. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. 30, and t k = 0. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. It gives the diagrammatic representation of bending moments at different points of beam. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. BEAM DIAGRAMS AND FORMULAS Table 3-23 (continued) Shears, Moments and Deflections 13. Tutorial from Simply Supported beam and UDL load. 3 away from A and another 5. ) MAXIMUM SLOPE AND DEFLECTION OF SIMPLY SUPPORTED BEAMS Loading condition Maximum slope Deflection ( y) Max. Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons • 3 h 21 m. If the beam carries a U. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. Simply Supported Beam With Udl. An explanation of the variables:. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Beam Formulas. Deflection function under concentrated loads P and distributed load q can be easily obtained respectively as fallows: )0( )2( 24 )0() 16 1 12 1 ( 323 23 LxxLxxL EI qx y LxxLx EI P y p p dd dd (1) When calculating the deflection caused by the secondary moment, the upper and lower T-beam can be combined into a small one regarded as fixed-end of. Objectives. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). Simply-Supported Beams Figure: A simply-supported beam. 9 away from a on a 6m beam. Simply-Supported Beams. 2018/2019. 5 J in right half. A simply supported beam is the most simple arrangement of the structure. 5 m Structural width, b 0. For a simply-supported beam, we use the following boundary conditions: w(0)=0. Thanks all. Substitute x for L/2 gives: EIy max = = = Thus y max = but total load on the beam = W = wL. This calculator uses standard formulae for slope and deflection. The beam is supported at each end, and the load is distributed along its length. Simply supported beam due to point load 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). The following page shows the line, shear force and bending moment diagrams for this beam. Calculate bending moments, shear, slope and deflection of a simply supported and guided beam at any point along its length when it is subject to a UDL (uniformly distributed load). In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. General shapes are rectangular sections, I beams, wide flange beams and C channels. One of the most powerful functions is using it as a beam deflection calculator (or beam displacement calculator). and so the equation for maximum deflection on a simply supported beam with a total UDL = W is : y max = HOME. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. so i was given a problem by my mechanical principles teacher to work out the max vertical deflection of a simply supported beam with two point loads one a distance of 5. Assume simple beam theory is applicable for the simply supported beam shown. 5 J in right half. Let R 1 & R 2 be the reactions then,. Will automatically generate results curves and has the ability to find values at a specific location. ) and uniformly varying loads (u. (AU Nov/Dec 2016) It will occur at centre 3 c 5 WL y 384 EI =-UNIT - 3 DEFLECTION. Slope And Deflection Of Simply Supported Beam With Udl. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. Simply supported beam moment i. 2 inches under a live load only deflection limit of L/360. A uniform load of 2. In multiple span continuous beams, will load in one span produce stress the other spans? A2. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. A simply supported beam with a uniformly distributed load. This video shows how you can calculate deflection and slope of a cantilever beam with a point load at the free end using the double integration method. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Calculate the width & depyh of the beam if permissible bending stress is 7N/mm2 and central deflection is not exceed 1cm. a) represents a beam subject to a uniformly distributed load (udl) of magnitude w, across its length, l. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. This tool lets you to predict the deflection and stress of Beam. Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Area Moment Method. It gives the diagrammatic representation of bending moments at different points of beam. A simply supported beam of length l carries a concentrated load W at distances of a and b from the two ends. Single simply supported unprotected steel beam subjected to nominal gravity loads has been considered for the investigation. Example 1. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. M max max = wl 2 8 6. The beam is considered to. Metric round tube beam deflection calculator. (i) Derive the expression for strain energy due to bending. the member with simply supported ends. On completion of this tutorial you should be able to solve the slope and deflection of the following types of beams. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. 5 J in right half. Design & Mfg. Beam bending is analyzed with the Euler-Bernoulli beam equation. Shear Force Diagram (kN) 0. If {eq}I_{z} = 0. 30, and t k = 0. Macaulay's method is to be used. Calculation Reference Roark's Formulas for Stress and Strain. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. We will also touch on the selection of a Concrete Beam. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. The number 0. (1) Figure shows a simply supported beam of span 5m carrying two point loads. Simply supported beam moment i. when =0 15 x The plate deflection Δp is Δp= x δ 16 Where, δ is the primitive beam deflection of the x-x strip. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Beam formulas with shear and mom calculator for ers slope and deflection cantilever propped cantilever with udl k6nqmkeedz4w beam formulas with shear and mom fixed both ends beam udl. 9 kN/m is applied to the beam. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. CALCULATING BENDING MOMENT OF SIMPLY SUPPORTED BEAM 1. Simply supported beam due to self - weight 3. E = 200 kN/mm2, I = 16 x 10 8 mm4. It carries a uniformly distributed load of 10 KN/m as shown in figure. Cantilever Beams Part 1 – Beam Stiffness (continued) The next step would be to solve for the stress distribution in the beam generated by the given deflection. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. I want to complete our discussion of beam deflections by looking at another method of solution, the method of superposition. Obviously software could handle this question much more readily, I am going to assume you are expected to do this by hand. (These assume that the beam is uniform, i. Simply-Supported Beams Figure: A simply-supported beam. •The deflection can be observed and measured directly. Calculation Reference Roark's Formulas for Stress and Strain. The slope (q) & deflection (y) at any spanwise location can be derived. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Skip navigation Sign in. The following image illustrates a simply supported beam. , determination of deflections of simple beams of simple portal. (b) Deflection at where the point load is acting, and state its direction. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Download Now White Paper: Click Here to join Eng-Tips and talk with other members! Have calculation deflection at x for a Partial UDL from first principles. A method is for efficiently determining a very large number of terms in the series. Use strain energy method. ) and uniformly varying loads (u. Calculate the factored design loads (without self-weight). Simply supported beam with point force in the middle. If all the diagrams can be fitted on a single plot, do so. Tkinter GUI based python program can solve for shear, moment, slope, and deflection of a simply supported beam with cantilevers at either end. Beam Simply Supported at Ends - Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Aug 11, 2019 - Explore alii9131's board "Bending moment" on Pinterest. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. If {eq}I_{z} = 0. Many structures can be approximated as a straight beam or as a collection of straight beams. A beam of uniform rectangular section 200 mm 4+3. Beam equations for Resultant Forces, Shear Forces, Bending Moments and Deflection can be found for each beam case shown. 1 Deflection curve of a beam. The deflection and stress levels predictions are required for beam design for a given shape, load, boundary and materials. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. Simply Supported Beam - With UDL. Academic year. The total load on beam is the UDL multiplied by the length of the beam, i. Total force on beam being wl. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Given the simply supported beam under a UDL (as shown in the figure below), determine ∆ C. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Find out: a) The flexural stiffness b) The dimensions of the section. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. The equations given here are for homogenous, linearly elastic materials, and where the rotations of a beam are small. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. A simply supported beam of span 8 m carries a udl of 4 kN/m over the entire span and two point loads of 2 kN at 2 m from each support. Table of Contents. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Some of the allowed Deflections are shown in the example. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Stiffness of a simply supported rc beam simply supported beam with udl beams fixed at both ends continuous beam calculator clearcalcs calculate the maximum deflectionHow To Calculate The Maximum Deflection Of A. Calculate the factored design loads (without self-weight). In this video derive an expression for deflection of beam with udl load solve by double integration method. Simply supported beam with point force in the middle. If {eq}I_{z} = 0. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES 2 INTRODUCTION SIMPLY SUPPORTED BEAM • Assumed deflection curve • Strain energy L ˛ ˛ 4 2 0 3 2 4 EI pL UV C C L ˛ ˛ 4 4 00 35 24 0 2 dEIpL pL CC dC L EI ˛ ˛˛ 13 EXAMPLE - SIMPLY SUPPORTED BEAM cont. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Hi there, I need to calculate the safe UDL for a beam 457x152 UB 52 simply supported at both ends with a span of both 6m and 12m (2no. If all the diagrams can be fitted on a single plot, do so. The beam dimensions are b = 1. 3 Of The Lecture Notes, It Is Shown That The Deflection Of A Simply Supported Beam With A UDL Is Given By Nu = W/24E I (z^4 - 2Lz^3 + L^3 Z). [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. State the location of maximum shear force in a simple beam with any kind of loading. A simply supported beam with a point load at the middle. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. Here we display a specific beam loading case. Write the maximum value of deflection for a simply supported beam of constant EI, span L carrying central concentrated load W. The beam is now cantilevered from this support. , slope and deflection. 044) /64 = 1. Beam Simply Supported at Ends – Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Moment Area Method For Deflection Of Beam | Part 5 | Simply Supported Beam Example 2 | UDL. Find: (a) What is the Maximum deflection ratio of beam 1 to beam 2? Solution:( ) beam1 = ( ) beam2 = x = =. 1-2 Simply-Supported Composite Beams Edition 2. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. 30, and t k = 0. How To Draw Shear Force Bending Moment Diagram. Deflection of Beam: Deflection is defined as the vertical displacement of a point on a loaded beam. The beams are assumed to be exposed to temperature. This may be reduced slightly to account for beneficial effects of compression near the support, in accordance with code provisions. GTU MIMP 827,319 views. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A uniform distributed load of 1000 N/m is applied to the lower horizontal members in the vertical downward direction. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. Its mode of deflection is primarily by bending. Beam forces calculation, uniformly distributed load, concentrated loads, beam deflection, metric units, online spreadsheet. The maximum deflection is measured as 1. Calculation Reference Roark's Formulas for Stress and Strain. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. A steel pipe 50mm internal diameter and wall thickness 2mm is simply supported on a span of 6m. and L = 12. Simply supported beam due to point load and UDL. i XEI , (e) Loading Upward loading positive Fig. You will need to determine the moment of inertia of the cross section and the distance from. R1 x 6 = 1000×3 + (200×3)3/2 = 3600. • The three simple‐beam bending moment diagrams thus obtained are shown in Figure. Draw a BMD for each loading (including the support reactions of the original beam. Slope and Deflection Calculator for Simply Supported Beam with uniform load on full span. This post gives a solved design example of a laterally restrained beam […]. Use of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading. IT carries a UDL of 9kN/m run over the centre length. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. ML diagram (simple beam bending moment diagram 3 MBA bending moment diagram due to external loads) A 1 B 2 15 MAB Bending Moment Diagram. The deflection at center of beam and various failure patterns are studied. 1 Prediction of girder deflection using classical beam theory for a simply supported girder subjected to a. A simply supported beam is the most simple arrangement of the structure. Cold-rolled RHS, SHS and channel. The centre of gravity of the UDL is at ½ of 8. Line Diagram. The "BeamAnal" calculates Shear Force, Bending Moment and Deflection at 31 positions along the member length. Find the deflection under the point load in terms of EI. (A) Derive. Al – Raheimy College of Engineering, Babylon University Abstract This paper presents a theoretical investigation of the free transverse vibrations of a uniform beam that is simply supported at both ends subjected to a static axial force. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. D2 is the deflection on the outboard end of the overhang. E=1x104 N/mm2. Computer Aided Deflection And Slope Yses Of Beams. Bending Analysis of Simply Supported and Clamped Circular plate - Free download as PDF File (. Use strain energy method. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. Beam Supported at Both Ends - Uniform Continuous Distributed Load. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 4 10 1 384) 5 (9 384 4 7 4 4 Deflection is in downward direction. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. 9 away from a on a 6m beam. The beams were simply supported at the ends and subjected to a concentrated load applied at the mid-span. After leaning this unit, you must be able to. Find plastic moment and BMD at collapse. 30, and t k = 0. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. Therefore the general deflection equation for a simply supported beam with a UDL. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. Simply Supported Beam With Overhang On One Side Exle. Deflection is (with a simple centerloaded beam) is PL^3/48EI The various deflections are as follows: (i) for a simply supported beam with point load (center)=PL^3/48EI (ii) // // // UDL= 5PL^4. 9 kN/m is applied to the beam. known solution for the simply supported plate with the uniform loading giving the deflection function for the strip case is combined with that for a solution of deflection function which shows the effects due to the edges W =Ws +We. 21 Beam Deflection by Integration ! Given a cantilevered beam with a fixed end support at the right end and a load P applied at the left end of the beam. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. b) shear force diagram shows the regions of maximum shear, for this beam these correlate to the reaction forces. 1457\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Academic year. Fixed and Simply Supported Ends UDL Along Entire Length. 6 - Cantilevers “Rules of Thumb” the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. MU = wu L. Solution: The equation we work with is ∆=, Where M o = bending moment distribution due to actual or real loading M 1 = bending moment distribution due to virtual or unit loading E = modulus of elasticity of the material of beam I = moment of inertia of beam section L = beam span ∆ c = deflection at point C. There are other issues too. In the example, the Maximum Deflection allowed is controlled the Code. SS beam subjected to point and UDL - Example. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. The problem with standard formulae, is that you have to create then anyway! To be accurate, that formula would require that the first load is S/2 from the support at each end, where s = L/n. Beam Supported at Both Ends - Uniform Continuous Distributed Load. 25 MNm2 and using Castigliano’s theorem, determine the deflection at the centre of the beam. Simply Supported Beam Deflection from Loading Function Example Course Description This course builds on the concept of force and moment equilibrium learnt from first year engineering mechanic and physics courses and focuses on the internal actions and deformations experienced by simple structural members under loading. Beams are assumed to be simply supported. The simplest form of this equation is as follows:. Therefore the general deflection equation for a simply supported beam with a UDL. 1 simply supported beam subjected to point load. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. The series is given by 422 2. This defection is ¼ times the deflection of a simply supported beam. Slope And Deflection Of Simply Supported Beam With Udl. This section covers shear force and bending moment in beams, shear and moment diagrams, stresses in beams, and a table of common beam deflection formulas. is EIy = Maximum deflection occurs at mid-span when x = L/2. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. Its mode of deflection is primarily by bending. Simply Supported Beam - With UDL More Beams. In order to calculate reaction R1, take moment at point C. (According to Newton’s third law there is for every action an equal and opposite reaction. Integrated into each beam case is a calculator that can be used to determine the maximum displacements, slopes, moments, stresses, and shear forces for this beam problem. UNIT - III 1. •The deflection can be observed and measured directly. Deflection can be calculated by standard formula will only give the deflection of common beam configurations and load cases at discrete locations , or by methods such as virtual. The supported end of the beam may be built into masonry or it may be a projection from a simply supported beam. A I-meter-long, simply supported copper beam ( E = 117 GPa) carries uniformly distributed load q. Fig:1 Formulas for Design of Simply Supported Beam having. at r = 0 and it is given by Deflection (w) max = 4 5 64 1 qa D P P. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. The centre of gravity of the UDL is at ½ of 8. Simply Supported Beam - With UDL More Beams. 00 m = 40 kN. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. the point loads are both 15. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. A beam of length of 6 m is simply supported at its ends. Moment Equation For Cantilever Beam With Distributed Load. This free online calculator is developed to provide a software tool for calculation of deflection and slope at any section of simply supported beam (without overhangs) subjected to point load, uniformly distributed load, varying load and applied moments on the span or on the supports. Deflection at the centre: A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. (i) Derive the expression for strain energy due to bending. A cantilever beam with a uniformly distributed load. The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get the Simply supported Beams: Deflection Of Beams Mechanism Of Solid. Thus D = 0. 6kN/m2 Width of load perpendicular to beam, or height of load supported by beam: 3. 3 SOLUTIONS FOR BEAM-COLUMNS (DEFLECTION PROBLEM). Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons • 3 h 21 m. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. Figure 1: Typical cantilever beam studied. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. The beam is supported at each end, and the load is distributed along its length. A simply supported beam is the most simple arrangement of the structure. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Deflection at supports in a simply supported beam is maximum b. Example - Beam with Uniform Load, English Units. Will automatically generate results curves and has the ability to find values at a specific location. I was to prepare the Shear force diagram and bending moment diagram for simply supported beam with UDL acting throughout the beam and two Point Loads anywhere on the beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. Simply Supported Beam Distributed Load Formulas:. 6 and an allowable stress of 175 N/mm. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. of simply supported beams are obtained. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Torsion reinforcement. zero deflection on the sides given by x = 0 and x = a, representing simple supports. I can't seem to find a deflection formula for partial UDL's for a simply supported beam. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 6 - Cantilevers “Rules of Thumb” the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. Assumption: Plane sections remain plane, the strain distribution will be as shown. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. BEAMS: STATICALLY INDETERMINATE (9. Find i) UDL it may carry if the bending stress is not to exceed 100N/mm 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. 4 LATERALLY SUPPORTED BEAMS. 2c Shear Force Diagram for Simply Supported Beam with UDL b) Bending Moment Diagram for simply supported beam with UDL The generated bending moment diagram using Matlab is as shown below. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. (5)If the beam carries a U. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. Just like a pivot, the wall is capable of exerting an upwards reaction force R 1 on the beam. The loads are symmetric about the centre of the beam. Solution: The equation we work with is ∆=, Where M o = bending moment distribution due to actual or real loading M 1. The beam is a steel wideflange section with E 628 10 psi and an allowable bending stress of 17,500 psi in both tension and compression. Fig:1 Formulas for Design of Simply Supported Beam having. Macaulay's method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams. A simply supported beam of uniform flexural rigidity EI and span l, carries two. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. Fig 1 shows a simply-supported beam AB subjected to a partial UDL 3 kN/m and a point load 50 kN. Case 3:-When simply supported beam is subjected to a single concentrated load at mid-span. For this reason, the analysis of stresses and deflections in a beam is an important and useful topic. the deflection ( C)1 due the uniform load can be found from example 9. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Simply supported beam with a point load at the centre and with two or more point loads - SFD and BMD in case of simply supported beam with a central point load - Example. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o’s text. Point of contraflexure = 0 (iii) Three Hinged Parabolic Arch Having Abutments at Different Levels. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. As shown in figure below. and a uniformly distributed live load of 550 lbs/ft. Given that the flexural rigidity EI = 100,000 kNm2 for the entire beam, determine: (a) The rotation at Support A, and state its sense of rotation. Based on the type of deflection there are many beam deflection formulas given below, w = uniform load (force/length units) V = shear. When lateral deflection of the compression flange of a beam is prevented by providing effective lateral support (restraint), the beam is said to be laterally supported. [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. , determination of deflections of simple beams of simple portal. A simply supported beam of uniform flexural rigidity EI and span l, carries two. Simply supported beam moment i. 5x10-4 m 4 and E = 1x10 7 kN/m 2 (i) The fixed end moment for the beam carrying udl: M A = M B = 12 2 WL = KNm x 75. 1 x 105 x 108) y B = 10. 7KN and the UDL is 1. Continuous Beam Three Span With Udl. Deflection JAE, that sounds like the condition he's looking for. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. 6 × 10 6 psi, G 12 = 0. A steel beam (178 x 102 x 19 UB S275) 3m long was selected. In practice however, the force may be spread over a small area, although the dimensions of this area should be substantially smaller than the beam span length. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. Use strain energy method. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. For completeness and BCO submission you should also considered shear forces and deflection of the beam. Take moment about point C, for reaction R1 $$\sum M_{c}\space = 0$$. The centre of gravity of the UDL is at ½ of 8. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. 4 ENES 220 ©Assakkaf Statically Determinate Beam When the equations of equilibrium are sufficient to determine the forces and stresses in a structural beam, we say that this beam is statically determinate Statically Indeterminate Beams LECTURE 18. RE: Beam Formulas for Multiple Point Loads. CHAPTER 2 : DEFLECTION (MACAULAY METHOD) PART 1 by Saffuan Wan Ahmad Let us again consider a simply supported beam AB of length L and carrying concentrated load P at mid span,C as shown below. The beam is supported at each end, and the load is distributed along its length. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The Young’s Modulus of the beam is 30 x 10^6 Psi. ! The beam has a length of L. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. 2d Bending Moment Diagram for Simply Supported. Draw the positive. The tables below give equations for the deflection, slope, shear, and moment along straight beams for different end conditions and loadings. Calculate the flexural stiffness of a simply supported beam which will limit the deflection to 2 mm at the middle. The beam dimensions are b = 1. 01 radians at the ends. The beam is supported at each end, and the load is distributed along its length. (b) Deflection at where the point load is acting, and state its direction. In the above image, I need to find the deflection at point C and D of the simple supported beam. The simplest form of this equation is as follows:. Skip navigation Sign in. Example 1. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. Uniform Load on full span. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. In a coil spring, the stress is distributed evenly along the length of the coil. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. By taking into account the stiffness of the composite joints, the deflection is therefore reduced by roughly 50%. Another method of determining the slopes and deflections in beams is the area-moment method, which. The load on the conjugate beam is the M/EI diagram of the loads on the actual beam. (B) Explain how deflection in beams affects structural stability. The deflection of the beam towards a particular direction when force is applied on it is called Beam deflection. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 3 away from A and another 5. The Euler-Bernoulli equation describes a relationship between beam deflection and applied external forces. 5 kN/m × 8. In this post i will briefly explain different types of beams. The ClearCalcs beam calculator allows the user to input the geometry and loading of a beam for analysis in a few simple steps. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. Could anyone please tel me how to calculate the Bending Stress, Compressive stress and the Deflection for any beam (UB 203x102x23) subjected to Uniform Distributed Load of 8063 N/m over a 8 m long beam (simply supported beam). Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Engineers adopt deflection limits which suit the nature of the building. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. This square tubing deflection calculator calculates tube deflection for square based on length of beam, tube size, decimal gauge and load. D1 is the maximum deflection midspan between supports. Where y is the deflection at the point, and x is the distance of the point along the beam. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. Use strain energy method. ALL calculators require a Premium Membership. Deflection of Cantilever Beam. All structural/civil formula books, etc seem to neglect this information. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. General shapes are rectangular sections, I beams, wide flange beams and C channels. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. The beam is also pinned at the right-hand support. EI is constant. Problem 691 Determine the midspan deflection for the beam shown in Fig. For a simply sagging supported structure with distributed mass - or load due to gravitational force - can be estimated as. What is the absolute maximum bending moment due to a moving udl longer than the span of a simply supported beam? 11. deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. The loads are symmetric about the centre of the beam. 01 radians at the ends. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Introduction Universal beam sections are normally employed in buildings to carry load. 2018/2019. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. The ends A and C are hinged supports and B is a continuous support. (Hint: Apply Case No. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. As shown in figure below. It carries a UDL of 9 kN/m over the entire length. The beams used for frame work are selected on the basis of deflection, amongst other factors. For information on beam deflection, see our reference on. Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. STATIC ANALYSIS OF A SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD Figure 1 All dimensions are in mm Objective: To find the deflection, stress, strain, shear force and bending moment diagram of simply supported beam with uniformly distributed load as shown in Figure 1. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. Beam bending is analyzed with the Euler-Bernoulli beam equation. (ii) Derive the expression for strain energy due to. The center reaction Q4. y of a simply supported beam under uniformly distributed load (Figure 1) is given by EI qx L x dx d y 2 ( ) 2 2 − = (3) where. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). In multiple span continuous beams, will load in one span produce stress the other spans? A2. I did not have large deflection turned on. Find expressions for the total strain energy of the beam and the deflection under load. Now we have the deflection for a simply supported beam with UDL acting on the whole length. Sign conventions for load, S. Any help would be much appreciated! Thanks. INTRODUCTION (Fig. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. 0GPa Tributary width for loading, tw 6. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. From the geometry of the figure, (6. The beam is now cantilevered from this support. 6 and an allowable stress of 175 N/mm. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o’s text. deflection (YId Cantilever with concentrated load Wat end WL2 2EI W 6E1 - ~2~3 - 3 ~2~ + x3~ WL3 3EI. Shear Force diagram for Simply supported Beam with three segments, when third segment start and end is reversed from other 2, then the in-correct shear force diagram generated by Staad pro, Please clarify. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 7KN and the UDL is 1. I believe that changing formulation to MPC is the solution. The centre of gravity of the UDL is at ½ of 8. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. Based on the supports of the beam, Following are some of the classification of the beam. The equivalent UDL method is a useful tool for estimating the deflection in a simply supported beam with a complex loading. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Chapter-5 Deflection of Beam Page- 7 (ix) A simply supported beam with a continuously distributed load the intensity of which at any point 'x' along the beam is x sin x ww L ⎛⎞π = ⎜⎟ ⎝⎠ (i) A Cantilever beam with point load at the free end. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Here we display a specific beam loading case. Monash University. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. The force is concentrated in a single point, located in the middle of the beam. when the beam is loaded with a 14KNload at each one third span point, it failed. helping to develop a coherent picture of the deflection behaviour. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Markings for. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. The modulus of elasticity is 205GPa. Beam Deflection Ronni Kaptanband Co. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. Simple Support Beam with Overhanging. Aug 11, 2019 - Explore alii9131's board "Bending moment" on Pinterest. 1 Classification based on supports; 2 Simply Supported Beam : U. See more ideas about Bending moment, Civil engineering construction, Civil engineering design. 044) /64 = 1. Lesson 15 of 15 • 2 upvotes • 12:56 mins. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. The shear force diagram indicates the shear force withstood by the beam section along the length of the beam. Total force on beam being wl. The beam is laterally supported. Slope and Deflection for sim. The dead load does not include the self-weight of the beam. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. Skip navigation Sign in. 9 away from a on a 6m beam. Beams - Fixed at One End and Supported at the Other - Continuous and Point Loads.
4wqm9qh2ki0945j 540mmj6nu9t 4ccx8nd6gy4y mip4qag40jqv jib9uh8ne1tn6s2 0gbl4vh4wj d1s2c5evxkjyf jlbax8c46j2xf2 9cwt7mito0b anquvshpp6 925w9babsnq3mwc i9wmrr4w8fercco hipz5bvzvw5uix y36g2dawrsvtf5j e8yz3mnz5ea76hx yrol77n13xvbn w19j1mlvl95bd2 e5nszxy5pqsc3ui 2i7wqx0ccqddd lr7xnx7htc0y 1ym3nift6ij 98o2ter7jo41 at08bqwv8sa sxkdilpoh0 8ogxyeyibgecj | 2020-08-09T15:02:03 | {
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https://math.stackexchange.com/questions/3008575/double-angle-formulas-finding-tan-2-theta | # Double Angle Formulas: Finding $\tan 2\theta$
I am trying to find $$\tan 2\theta$$ where $$sin \theta = \frac{5}{13}$$ and $$\theta$$ is in Quadrant One.
According to my textbook, $$\tan 2\theta = \frac{120}{119}$$, but I get $$\frac{-10}{13}$$ instead.
The Identity I am using:
$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^{2}\theta}$$
My Process:
Since $$y = 5\;$$ and $$r = 13,\; x = 12.$$
Apply Tangent Double Angle Formula: $$\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2}$$
$$\frac{\frac{10}{12}}{1 - \frac{25}{12}} \cdot \frac{12}{12}$$
$$\frac{10}{12-25}$$
$$\frac{-10}{13}$$
What am I doing wrong?
• $(5/12)^2$ is not $25/12$. It is $25/144$. – Nick Nov 22 '18 at 0:12
• Erm, that's probably my issue then. – LuminousNutria Nov 22 '18 at 0:12
Alternatively:
$$\sin\theta = 5/13$$, implies $$\cos\theta = \sqrt{1-(5/13)^2} = 12/13$$.
$$\sin 2\theta = 2\sin \theta \cos \theta = 120/169$$.
$$\cos 2\theta = 2 \cos^2 \theta-1 = 1 - 2 \sin^2 \theta = 119/169$$.
This gives $$\tan 2\theta = \sin 2\theta/ \cos 2 \theta = 120/119$$.
I made a mistake solving the problem. $$(\frac{5}{12})^2 \neq \frac{25}{12}$$.
Actually, $$(\frac{5}{12})^2 = \frac{25}{144}$$.
Taking that into account:
$$\frac{\frac{10}{12}}{1 - \frac{25}{144}} \cdot \frac{12}{12}$$
$$\frac{10}{12-\frac{300}{144}}$$
$$\frac{10}{12 - \frac{25}{12}} \cdot \frac{12}{12}$$
$$\frac{120}{144-25}$$
$$\frac{120}{119}$$
Therefore, $$\tan 2\theta = \frac{120}{119}$$ | 2019-07-21T09:01:37 | {
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https://math.stackexchange.com/questions/1667546/sigma-algebra-generated-by-a-subset | $\sigma$-algebra generated by a subset
I'm new to this concept of a $\sigma$-algebra generated by a collection of subsets.
Let $\Omega = \{a, b, c, d\}$ and \begin{align} &\mathcal{F}_1 = \{\Omega, \emptyset, \{a\}\} \\ &\mathcal{F}_2 = \{\Omega, \emptyset, \{a\}, \{b, c, d\}\}\text{.} \end{align} I wish to show that $$\sigma\langle \mathcal{F}_1 \rangle = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathcal{F}_1)}\mathcal{F} = \mathcal{F}_2$$ where $\mathcal{I}(\mathcal{F}_1) = \{\mathcal{F}: \mathcal{F}_1 \subset \mathcal{F} \text{ and }\mathcal{F} \text{ a }\sigma\text{-algebra on }\Omega \}$.
I know immediately from this that any $\sigma$-algebra $\mathcal{F}$ must, at the very least, contain elements of $\mathcal{F}_1$: $$\{\Omega, \emptyset, \{a\}\}$$ but other than literally listing every possible collection of subsets of $\Omega$ and checking which are $\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.
Is my way of thinking correctly or is there a quicker way?
• Complete the collection of subsets to be closed under the conditions for an sigma algebra, i.e., closed under the operations of countable union or intersection and complementary. This is the minimal sigma algebra that contain the collection. – Masacroso Feb 22 '16 at 19:26
• @Masacroso Ah, so that's why it's called the "smallest" $\sigma$-algebra generated by a set. Thank you! – Clarinetist Feb 22 '16 at 19:29
Yes. but one quick shortcut is that the complements of the sets are in the sigma algebra as well, so you can easily see $\{b,c,d\}$ must be in $\sigma\langle\mathcal{F}_1\rangle$.
• And we know that this $\{b, c, d\}$ set with everything else in $\mathcal{F}_1$ must be the "smallest" $\sigma$-algebra generated by $\mathcal{F}_1$. Thank you! ... now, the question is, how do I prove this rigorously? I don't see how else to do this other than finding every possible $\mathcal{F}$ and then intersecting them. – Clarinetist Feb 22 '16 at 19:31
• @Clarinetist, no, it's the smallest also because it is the minimum extension of the current set to a sigma-algebra. So suffices to find one, and show that none of its subsets are sigma algebras. – gt6989b Feb 22 '16 at 19:34
• Very good and clear explanation. Thank you! – Clarinetist Feb 22 '16 at 19:35
Here are some more general statements that imply the desired equation. We assume the following without proof.
1. The intersection of $\sigma$-algebras is again a $\sigma$-algebra.
2. A set of the form $\lbrace \emptyset, \Omega, A, A^c \rbrace$ is a $\sigma$-algebra.
Let $\mathscr{G}$ (particular case $\mathcal{F}_1$) be any set of subsets of $\Omega$. $\mathscr{A}:=\bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{G})}\mathcal{F}$ is nonempty because the powerset $\mathscr{P}(\Omega)$ has $\mathscr{P}(\Omega) \in \mathcal{I}(\mathscr{G})$. It follows that $\mathscr{G} \subset \mathscr{A}$ and that, because of assumption 1, $\mathscr{A}$ is a $\sigma$-algebra. If $\mathscr{A}'$ is another $\sigma$-algebra with $\mathscr{G} \subset \mathscr{A}'$, then, because $\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\mathscr{A} \subset \mathscr{A}'$. So $\mathscr{A}$ satisfies the definition of the smallest $\sigma$-algebra generated by $\mathscr{G}$, i.e. $\sigma(\mathscr{G}) = \mathscr{A}$.
We conclude $\sigma(\mathcal{F}_1) = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{\mathcal{F}_1})} \mathcal{F}$.
We now show that $\sigma(\mathcal{F}_1) = \mathcal{F}_2$. Note that $A \in \mathcal{F}_1$ implies $A \in \sigma(\mathcal{F}_1)$ and $\lbrace a \rbrace \in \sigma(\mathcal{F}_1)$ implies $\lbrace b,c,d \rbrace = \lbrace a \rbrace^c \in \sigma(\mathcal{F}_1)$, so that $\mathcal{F}_2 \subset \sigma(\mathcal{F}_1)$. By assumption 2, $\mathcal{F}_2$ is a $\sigma$-algebra. We obtain $\sigma(\mathcal{F}_1) = \mathcal{F}_2$.
Note
It holds in general that $\sigma(\lbrace A \rbrace) = \sigma(\lbrace \emptyset, \Omega, A \rbrace) = \lbrace \emptyset, \Omega, A, A^c \rbrace$, we could have proved this instead of proving this for our particular case in the last step. | 2019-09-22T16:26:27 | {
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https://math.stackexchange.com/questions/3102390/show-that-limit-does-not-exist-two-variables | # Show that limit does not exist (two variables)
I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.
I want to answer this question:
Show that the following limit does not exist
$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$
So, my working:
$$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\to 0}\frac{0}{y^2}=0$$
and
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{y\to 0}\frac{0}{x^2}=0$$
I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $$y=x$$, not sure if that's correct - or my working.
So, let $$y=x$$, then
$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2x^2+(x-y)^2}=\lim_{x\to 0}\frac{x^4}{x^4+(x-x)^2}=1$$
Therefore limit doesn't exist.
Is this somewhat correct? What's the best way to answer a question like this?
Also, when showing that this limit does not exist, do I need to find different values of limits for both $${x\to 0}$$ AND $${y\to 0}$$? Or is one enough, e.g. if I just find two different values for two limits for $${x\to 0}$$ without using $${y\to 0}$$ in my calculation at all, is that fine?
Thanks!
What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $$(0,0)$$ is same.
So when you calculate $$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $$x=0$$.
Similarly,
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $$y=0$$.
And the last limit you calculated is along line $$y=x$$.
So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $$y=0$$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.
• "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist. – leonbloy Feb 6 at 14:35
• @leonbloy Do you have an example of that case? – Shufflepants Feb 6 at 15:48
• @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist. – Swapnil Rustagi Feb 6 at 15:53
• If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?) – gidds Feb 6 at 16:28
• @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19) – leonbloy Feb 6 at 17:11
Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.
• How can I improve my reasoning? – Mandingo Feb 6 at 11:48
• No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same. – PierreCarre Feb 6 at 11:54
What you did is completely correct, but perhaps you can ease it a little bit as follows:
For $$\;x=0\;,\;\;y\to0\;$$ the limit clearly is $$\;0\;$$, whereas for $$\;x=y\;$$ and $$\;x\to0\;$$ we get
$$\frac{x^4}{x^4+(x-x)^2}=1\xrightarrow[x=y\to0]{}1$$
Thus the limit depends on the path chosen $$\;\implies\;$$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.
You did right.
A possible improvement is to compute the limit along an arbitrary line: along $$y=mx$$ you have to compute $$\lim_{x\to0}\frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}= \lim_{x\to0}\frac{m^2x^4}{m^2x^4+x^2(1-m)^2}= \lim_{x\to0}\frac{m^2}{m^2+(1-m)^2x^{-2}}= \begin{cases} 0 & m\ne 1 \\[4px] 1 & m=1 \end{cases}$$ Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.
What you did is correct, but note that after proving that the limit is $$0$$ when you take $$y=0$$, the fact that it is also $$0$$ when you take $$x=0$$ is irrelevant. What matters here is that going to $$(0,0)$$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $$(0,0)$$. | 2019-09-16T08:51:30 | {
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http://math.stackexchange.com/questions/250768/is-mathbb-r2-a-field | # Is $\mathbb R^2$ a field?
I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.
In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.
This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?
Thanks
-
A field has multiplication. How would you define multiplication on $\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't "obvious" until you realize that the resulting field is the complex numbers...) – Thomas Andrews Dec 4 '12 at 16:07
$\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. – Rankeya Dec 4 '12 at 16:08
But, if you want to try to do this for $\mathbb{R}^n$, for $n \geq 3$ such that $\mathbb{R}$ is naturally embedded in $\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. – Rankeya Dec 4 '12 at 16:11
ok this just answered my follow-up question. Is there any proof of that being true? – vondip Dec 4 '12 at 16:11
Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. – Adam Rubinson Dec 4 '12 at 16:37
If you define:
$$(a,b)+(x,y):=(a+x,b+y)$$
$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$
then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?
-
Spoiler: take a look at one of the comments given. $\^\smile\^$ – FrenzY DT. Dec 4 '12 at 16:10
complex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? – vondip Dec 4 '12 at 16:11
Yes, it would, because addition in $\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. – Rankeya Dec 4 '12 at 16:15
What do you mean "vector space"? Any field is a vector space over any subfield, so the field $\,\Bbb R^2\cong\Bbb C\,$ is a vector field ove an infinite number of subfields, say $\,\Bbb C\,,\,\Bbb R\,,\,\Bbb Q\,,\,\Bbb Q(i)\ldots\,$ , etc. – DonAntonio Dec 4 '12 at 16:15
So would that mean that I any vector space could be defined when F -being the field, as : F^n ? – vondip Dec 4 '12 at 16:17
It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.
The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.
-
Usually in mathematics one defines these structures as tuples
A field is a triple $(K,+,\cdot)$ such that $K$ is a set and [...] and $\cdot:K \times K \rightarrow K$
A Vectorspace is a triple $(V,+,\cdot)$ such that $V$ is a set and [...] and $\cdot: K \times V \rightarrow V$
So your question is meaningless: A set (say $\mathbb R^2$) cannot be a field or a vectorspace or a group or anything - only if you add some additional structure (most of the time operations) you can ask this question.
For example $\mathbb R^2$ can be the set used in the definition of a field, as well as the underlying set used in the definition of a vectorspace. And we are happy, the addition operation $$+:\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2$$ is the same, and the multiplicaton for "the" vectorspace structure $$\mathbb R \times \mathbb R^2 \rightarrow \mathbb R^2$$ is "compatible" with the multiplication for "the" field structure $$\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2$$
-
Adding to the above answer. With the usual exterior multiplication of the $\mathbb{R}-\{0\}$ as a ring with the natural addition and multiplication you can not make a field out of $\mathbb{R}^{2}-(0,0)$ \ But there may exist other products such as the one in the answers which can make a field out of ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ \ According to one of the theorems of Field theory every field is an Integral domain. So by considering : ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ With the following natural product: $(A,B)*(C,D)=(AB,CD)$ We see that $(1,0)*(0,1)=(0,0)$ Which means that $\mathbb{R}^{2}$is not an integral domain and hence not a field.
-
Why are you considering $\mathbb{R}^*\times\mathbb{R}^*$? – Tobias Kildetoft Feb 23 '15 at 10:16
Bcoz he is asking that if $\mathbb{R}^{2}$is a field ... And I think he meant with the exterior multiplication and exterior addition. Otherwise it is obvious we can consider it as field aince it is isomorphism to $\mathbb{C}$ and hence a field – Romel Feb 23 '15 at 10:21
But what does that have to do with this? You are removing way more elements than $0$ when you consider this (in fact, it is clear that the usual multiplication does turn this into a group). – Tobias Kildetoft Feb 23 '15 at 10:22
And since the natural multiplication is defined on $\mathbb{R}$ We have to consider $\mathbb{R}^{*}$ as the set which the natural multiplication works on – Romel Feb 23 '15 at 10:24
No, to be a field we would need $(\mathbb{R}\times\mathbb{R})\setminus \{0\}$ to be a group, not $\mathbb{R}^*\times\mathbb{R}^*$. – Tobias Kildetoft Feb 23 '15 at 10:25 | 2016-04-29T10:36:52 | {
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# Each term of a certain sequence is calculated by adding a particular
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30 Jul 2018, 00:45
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Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?
(A) 20
(B) 15
(C) 13
(D) 12
(E) 8
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Each term of a certain sequence is calculated by adding a particular [#permalink]
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30 Jul 2018, 10:41
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?
(A) 20
(B) 15
(C) 13
(D) 12
(E) 8
let the constant be $$d$$ and the first term be $$a$$
so 2nd term $$= a+d=27$$ and 5th term will be $$a+4d=84$$. Subtract the two equations to get
$$3d=57 =>d=19$$
Hence $$a=27-19=8$$
Option $$E$$
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Updated on: 31 Jul 2018, 17:11
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?
(A) 20
(B) 15
(C) 13
(D) 12
(E) 8
let d=difference between terms
84-27=57=3d
d=19
27-19=8=1st term
E
Originally posted by gracie on 30 Jul 2018, 18:15.
Last edited by gracie on 31 Jul 2018, 17:11, edited 1 time in total.
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Re: Each term of a certain sequence is calculated by adding a particular [#permalink]
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30 Jul 2018, 19:34
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?
(A) 20
(B) 15
(C) 13
(D) 12
(E) 8
Let the first term be 'a' and the constant being added be d.
So, 2nd term becomes -
$$a_2$$ = a + d = 27 ------ (1)
and the 5th term becomes -
$$a_5$$ = a + d + d + d + d = 84
Hence,
$$a_5$$ = a + 4d = 84 ------ (2)
Now multiplying equation (1) by 4 to eliminate d
hence (1) becomes -
$$a_2$$ * 4 =>
4a + 4d = 108 ---- (3)
Subtracting equation (2) from (3)
4a + 4d - a - 4d = 108 - 84
3a = 24
a = 8
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Re: Each term of a certain sequence is calculated by adding a particular [#permalink]
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01 Aug 2018, 16:36
Bunuel wrote:
Each term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?
(A) 20
(B) 15
(C) 13
(D) 12
(E) 8
Let n = the constant added to each term to get the next term.
We have:
2nd term = 27
3rd term = 27 + n
4th term = 27 + 2n
5th term = 27 + 3n
Since we are given that the 5th term is 84, we have:
27 + 3n = 84
3n = 57
n = 19
Since the second term was given as 27, we know the first term is 27 - n. So the first term is 27 - 19 = 8.
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Re: Each term of a certain sequence is calculated by adding a particular &nbs [#permalink] 01 Aug 2018, 16:36
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https://www.physicsforums.com/threads/iterated-integrals-setting-up-limits-of-integration.260486/ | # Iterated Integrals - setting up limits of integration
1. Sep 30, 2008
### mirandasatterley
1. The problem statement, all variables and given/known data
Find the volume of the region under the graph of f(x,y) = x+y and above the region y2≤x, 0≤x≤9
3. The attempt at a solution
From these equations, x will be integrated from 0-9, but i'm not sure about y.
My thinking is that y will be intgrated from 0-3 because y2≤x and the smallest value of x is 0, and the square root of 0 is 0, so that is the smallest y, and the largest x value possible is 9 and the positive quare root of 9 is 3, so this is the largest value of y,
So I would integrate:
∫0-9∫0-3 (x+y)dydx.
Is this correct or am I missing something, where the limits of integration also involve using f(x,y)?
2. Sep 30, 2008
### HallsofIvy
Staff Emeritus
No, that is not correct. The region $a\le x\le b$, $c\le y\le d$, with a, b, c, d numbers is always a rectangle and the figure here is not a rectangle. But the bounds do NOT involve f(x,y)- that is a "z" value and goes inside the integral as you have it.
Always draw a picture for problems like this. y^2= x is a parabola, of course, "on its side". The line x= 9 is a vertical line crossing the parabola at (9,3) and at (9, -3). Yes, you can integrate with x going from 0 to 9. On your picture, mark an arbitrary "x" by marking a point on the x-axis between 0 and 9. Now draw a vertical line from one boundary to the other. The y bounds, for that x, are y values of those endpoints: $(x, -\sqrt{x})$, and $(x, -\sqrt{x})$. Your integral is
$$\int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}} f(x)dy dx= \int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}}(x+ y) dy dx$$
Of course, like any double integral, you can reverse the order of integration. If you look at your picture you will see that y ranges, overall, from -3 to 3. Draw a horizontal line across the parabola, representing an arbitrary value of y in that range. It will have left endpoint on the parabola: x= y^2, and right endpoint on the vertical line x= 9. Those will now be the limits of integration for this order:
$$\int_{y= -3}^3\int_{x= y^2}^9 f(x,y)dx dy= \int_{y= -3}^3\int_{x= y^2}^9(x+ y)dx dy$$
Try it both ways. You should get the same answer.
3. Sep 30, 2008
### mirandasatterley
In a similar situation where i have to switch the order of integration from
∫0-3∫y2-9 f(x,y) dxdy to dydx,
Is this also a parabola on it's side, with intercepts through (3,9) and (0,9), meaning that I would be setting up the new limits of integration, fro the portion of the parabola above y=0?
4. Sep 30, 2008
### mirandasatterley
So, ∫0-9∫Square root of x -3 f(x,y) dydx | 2017-05-27T10:22:03 | {
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https://www.physicsforums.com/threads/inequality-3-x-3.706872/ | # Inequality (-3/x) < 3
1. Aug 23, 2013
### Qube
1. The problem statement, all variables and given/known data
(-3/x) < 3
2. Relevant equations
Dividing/multiplying an inequality causes the inequality sign to change.
3. The attempt at a solution
I keep getting the wrong solution. I tried two methods. I cannot get the textbook solution (x < -1)
Method one:
-3 < 3x
-1 < x
Method two:
Divide both sides by negative one, resulting in:
3/x > -3
3 > -3x
-1 < x
2. Aug 23, 2013
### Staff: Mentor
Here you have multiplied by x, which could be positive or negative (or zero). Since you don't know the sign of x, you need two cases - one where x is assumed to be positive and the other where x is assumed to be negative.
Again, you've multiplied by x, whose sign you don't know. Same comments as above apply here.
3. Aug 23, 2013
### Qube
All right! Now I have -1 < x if x is positive and -1 > x if x is negative. How do I determine which is correct? Is the time for trial-and-error?
4. Aug 23, 2013
### Staff: Mentor
The book's answer is wrong or at least incomplete.
If x > 0, you get x > - 1. This means that x > 0 AND x > -1. Together, these mean that x > 0.
If x < 0, you get x < -1. This means that x < 0 AND x < -1. Together, these mean that x < -1.
If you look at the graph of y = -3/x, you'll see that it has two parts. For the curve on the left, -3/x < 3 when x < -1. For the curve on the right, -3/x < 3 for any x > 0.
5. Aug 23, 2013
### Qube
So I guess I'll have to do a bit of trial and error to determine the exact intervals? As in, x > -1 is technically correct, but the best answer is that x is actually greater than 0 or x > 0?
How do you recommend solving these problems? I'm having massive trouble.
6. Aug 23, 2013
### verty
Just remember the rule, NEVER multiply or divide by x or an expression containing x, it could be negative. Read Mark44's reply again, the answer is: x < -1 OR x > 0.
Example: $\frac{x+2}{x-3} > 0$. If $x-3 > 0, x+2 > 0$, and if $x-3 < 0, x+2 < 0$. Can you give the answer for this example?
7. Aug 23, 2013
### rock.freak667
Not to detract from what the others are saying, if you multiply by x, you are multiplying by an unknown value which can be +ve or -ve and will thus change the sign of the inequality (i.e. < might become >). So what you can do to prevent this is multiply instead by x2 as this will always be positive.
Then you can solve it as you would using normal algebraic means. You will just need to be careful when doing this as you might pick up an extra solution. But simply analyzing your solution set for the problem would lead you in the correct path.
8. Aug 23, 2013
### Ray Vickson
Turn your inequality into one of the form
$$\frac{1}{x} < a$$
or
$$\frac{1}{x} > a$$
I will let you figure out which inequality applies, and what is the value of $a$.
Anyway, once you have a simple inequality in $1/x$, you can envision the graph $y = 1/x$, and then figure out what the x-region must be for the corresponding inequality in y.
Note: this suggested method is safe; it never has you multiplying or dividing by an x of unknown sign, at least, not until the very end.
9. Aug 24, 2013
### vanhees71
I think it's time to give the solution to clarify the issue:
You just have to do a case differentiation. First you can simplyfy the inequality by divding through $-3$:
$$-\frac{3}{x} < 3 \Leftrightarrow \frac{1}{x}>-1.$$
Now for $x>0$ the inequality is always fulfilled and for $x<0$ you get by multiplying with $(-x)>0$
$$-1>x \; \Leftrightarrow \; x<-1$$.
Thus the inequality is fulfilled for either $x>0$ or $x<-1$.
10. Aug 24, 2013
You can also try
\begin{align*} \frac{-3}{x} & < 3 \tag{Given} \\ \frac{-3}{x} \cdot x^2 & < 3x^2 \\ -3x & < 3^2 \\ 0 & < 3x^2 + 3x \tag{Now factor to obtain}\\ 0 & < 3x\left(x + 1\right) \end{align*}
If you set up a sign table (or graph $3x(x+1)$ you'll find that the inequality is solved
for numbers $x < -1$ or $x > 0$ as stated in other places. In short, it is not that one or the other of these two inequalities gives the solution set, is that the solution set consists of any number that satisfies either the first or the second of these two.
11. Aug 24, 2013
### Infrared
I think this problem is being made much harder than it really is.
$\frac{-3}{x}<3 \\ \frac{-3}{x}-3<0 \\ \frac{1}{x}+1>0 \\ \frac{x+1}{x}>0$
So $x>0$ or $x<-1$
12. Aug 24, 2013
### Ray Vickson
You are doing exactly what I suggested the OP do, but he/she never reported back.
13. Aug 25, 2013 | 2017-11-24T08:04:21 | {
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https://math.stackexchange.com/questions/875627/the-definitions-of-limit-infimum-and-limit-supremum | # The definitions of limit infimum and limit supremum
I have begun reading Rosenthal's "A First Look to Rigorous Probability Theory" and in order to reinforce my calculus background I am studying through the appendix section. Here he defined the limit of a sequence of real numbers as $\lim_{n \to \infty}x_n=x$ if for each $0 <\epsilon$ there is a natural number $N$ such that it is $|x-x_n| < \epsilon$ for $n > N$.
Then the limit infimum and supremum are defined as $\liminf_n x_n = \lim_{n \to \infty} \inf_{k\geq n} x_k$ and $\limsup_n x_n = \lim_{n \to \infty} \sup_{k\geq n} x_k$.
1. My first question is about the interpretation of the limit infimum and limit supremum definitions. It looks to me like they are defined as the limits of sequences as well. For example, for the infimum, one can define a sequence $a_n = \inf_{k \geq n}x_k$ where $x_n$ was already a sequence of real numbers. Then the limit infimum is the limit of the sequence $a_n$ , $\lim_{n \to \infty}a_n$, so the infimums of the subsequences $x_k$ where $k \geq n$ are converging if the limit exists. It is the similar for the supremum. Is my interpretation correct here?
2. It is said the limit infimum and supremum do always exist, though they can be infinite. I did not understand this; why it is so?
3. In the book it is stated that the limit $\lim_{n \to \infty} x_n$ exists if and only if it is $\liminf_n x_n = \limsup_n x_n$. I could not prove this to myself. How can it be shown that this statement is true?
• 1. ok, 2. they are monotone sequences. – enzotib Jul 23 '14 at 7:28
• For 3. In general the limit infimum is the least value to which a subsequence of ${a_n}$ can converge and the limit supremum is the greater value to wihch a subsequence of ${a_n}$ can converge – Dimitri Jul 23 '14 at 7:44
These are answers to the secondary questions in the comments after my answer to the original question; I cannot put the answers in comments, because the size of a comment is limited.
The second question. Let $(x_n)$ be a sequence taking values in the set $\{0,1,2,\ldots,9\}$, and $a:=\liminf_n x_n$. If $0$ appears infinitely many times in the sequence, then $a=0$. If $0$ appears only finitely many times, but $1$ appears infinitely often, then $a=1$. And so on. This is a slick answer. It is entirely another matter to answer the question with a definite value for a particular given sequence, such as the sequence of digits after the comma in the decimal expansion of $\pi$. Well, if you can prove that the digit $0$ appears infinitely many times in the decimal expansion of $\pi$, then you have $a=0$, and so on; in this case it takes a serious theoretical effort to determine the $\liminf$.
As for the third question, let $(x_n)$ be a sequence of real numbers such that $a:=\liminf_n x_n$ is a real number (that is, $-\infty<a<\infty$). Since the sequence of infima $a_n:=\inf_{k\geq n} x_n$ is increasing (meaning that $m\leq n$ implies $a_m\leq a_n$ -- I hate "nondecreasing"), the limit $a=\lim_{n\to\infty} a_n$ is actually the supremum: $a=\sup_n a_n$.
$\quad$Let $a_1\leq a$ and $a'<a_1$; then $a'<a$. We claim that there exists $m$ such that $x_n\geq a'$ for every $n\geq m$: since $a=\sup_n a_n$, there exists $m$ such that $a_m\geq a'$; but then $x_n\geq a_m\geq a'$ for every $n\geq m$.
$\quad$Now let $a_1>a$, and set $a':=\tfrac{1}{2}(a+a_1)$; we have $a<a'<a_1$. In this case we claim that given any $m$ there exists $n\geq m$ such that $x_n<a'$: since $a_m=\inf_{k\geq m} x_k\leq a<a'$, there exists $n\geq m$ such that $x_n<a'$.
$\quad$We have proved that $a$ is the largest of all real numbers $a_1$ with the property that for every $a'<a_1$ there are only finitely many $n$ such that $x_n<a'$.
1. Yes, in the book you are reading, $\liminf_n x_n$ and $\limsup_n x_n$ are defined as limits of sequences. Note that the sequence $a_n=\inf_{k\geq n} x_k$ is increasing (which means that $m\leq n$ implies $a_m\leq a_n$), while the sequence $b_n=\sup_{k\geq n} x_k$ is decreasing.
2. The $\liminf$ and $\limsup$ are taken in the extended real line $\overline{\mathbb{R}}=\{-\infty\}\cup\mathbb{R}\cup\{\infty\}$. As an ordered set, $\overline{\mathbb{R}}$ is obtained from the ordered set $\mathbb{R}$ by adding to it the bottom (the least) element $-\infty$ and the top (the greatest) element $\infty$. As a topological space, $\overline{\mathbb{R}}$ is homeomorphic to any closed interval $[a,b]$ (with $a<b$) in $\mathbb{R}$, and is therefore compact. For example, $x\mapsto(2/\pi)\arctan x$ is a homeomorphism $\overline{\mathbb{R}}\to[-1,1]$ (where it is understood that ${-}\infty\mapsto-1$ and $\infty\mapsto1$), and at the same time it is an isomorphism of (totally) ordered sets. All this means that the limits, $\liminf$'s, and $\limsup$'s in the extended real line are just `infinitely stretched out' versions of the limits, $\liminf$'s, and $\limsup$'s in a closed interval. Since in a closed interval all increasing/decreasing sequences converge, so then do all increasing/decreasing sequences in the extended real line; in particular, $\liminf_n x_n$ and $\limsup_n x_n$ always exist. When you are thinking about the extended real line, you can imagine it as a closed interval, you can' go astray with that. If you have to prove some assertion about sequences in the extended real line, formulated in terms of the ordering and the topology (which is, after all, also determined by the ordering), then it suffices to prove the assertion for sequences in a closed interval.
3. You can assume that $x_n$ is a sequence in a closed interval; this will make your life easier, because $\liminf_n x_n$ and $\limsup_n x_n$ are real numbers (belonging to the interval). Now prove the following: $a=\liminf_n x_n$ is the largest real number with the property that for every real number $a'<a$ there are only finitely many $n\in\mathbb{N}$ such that $x_n\leq a'$, and the analogous assertion for $b=\limsup_n x_n$ (which you will formulate yourself --- just turn everything upside-down). With these characterizations of $\liminf_n x_n$ and $\limsup_n x_n$ under the belt, you will easily prove that $\lim_{n\to\infty} x_n$ exists iff $\liminf_n x_n=\limsup_n x_n$.
Remark. You did not ask this, but since I have mentioned imagining the extended real line as a closed interval$\ldots$ $~$You can actually see $\overline{\mathbb{R}}$ as the closed interval $[-1,1]$, and transfer anything that's happening in $\overline{\mathbb{R}}$ to $[-1,1]$. Let $f(y)=\tan((\pi/2)y)$ for $y\in[-1,1]$, so that $f^{-1}(x)=(2/\pi)\arctan(x)$ for $x\in\overline{\mathbb{R}}$. Now define, for all $y,z\in[-1,1]$, \begin{aligned} \mathit{sum}(y,z)~ &:= f^{-1}(f(y)+f(z))~,\\ \mathit{prod}(y,z)~ &:= f^{-1}(f(y)f(z))~. \end{aligned} Actually the operations $\mathit{sum}$ and $\mathit{prod}$ are not defined everywhere on $[-1,1]\times[-1,1]$; for example, $\mathit{plus}$ is not defined at the two points $(1,-1)$ and $(-1,1)$ that correspond to the risque limits of the form $\infty-\infty$. Write the definitions of $\mathit{sum}$ and $\mathit{prod}$ in Mathematica (or a similar program), and then draw their $3\text{D}$ diagrams, placing on all three axes the labels $-\infty$, $-1$, $0$, $1$, $\infty$ at positions, respectively, $-1$, $-1/2$, $0$, $1/2$, $1$. You will be able to observe in the diagrams the behavior of the two operations all the way to the infinity (in either direction); at the points where the operations are not defined you will clearly see the discontinuities (at these points the diagrams contain vertical lines --- more precisely, the closures of the diagrams contain vertical lines).
• A quick question: What exactly do we mean by a sequence in a closed interval? Do we mean that all values $x_n$ are mapped to values in an interval $[a,b]$ such that it is $a \leq x_n \leq b$? (a and b are real numbers) – Ufuk Can Bicici Jul 23 '14 at 9:42
• Yes, we mean exactly that. – chizhek Jul 23 '14 at 9:58
• I have another question, it may be weird but still it bugged me. Let's assume that the sequence $x_n$ takes the values of $\pi$'s digit values after the comma. So each $x_n$ takes a value in the set $(0,1,2,3,4,5,6,7,8,9)$. If we think of the sequence $a_n = \inf_{k \geq n} x_k$ this sequence is clearly nondecreasing and bounded in $[0,9]$, but how can we know the value of limit $\liminf_n x_n$? Since we don't know exactly the values of $x_n$ for very large $n$s, we can be never sure about the value for $a_n$ for these values. – Ufuk Can Bicici Jul 23 '14 at 12:36
• Continued: The infimum can be zero, one or any value, we cannot observe this exactly for this almost random sequence $x_n$ at large $n$s. So, what can we say about the value $\liminf_n x_n$ for such sequences? – Ufuk Can Bicici Jul 23 '14 at 12:39
• For the third question, unfortunately I have failed to find a rigorous proof for that if $a$ is the largest real number such that for every $a' < a$ there is finitely many $x_n < a'$ then $a=\liminf_n x_n$. But intuitively it is clear that this is valid: If we have a plot of a bounded sequence $x_n$ and visualize a vertical line $y=a'$ then there will be finitely many $x_n$ points under that line and as soon as we increase the line over $a$, the number of $x_n$ points will be infinitely many! But I frustratingly fail to express that visualization as a mathematically valid statement... – Ufuk Can Bicici Jul 23 '14 at 15:00 | 2019-12-06T18:43:05 | {
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https://math.stackexchange.com/questions/496494/smallest-number-with-specific-number-of-divisors | # Smallest number with specific number of divisors
Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Since $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.
As a single number it is :$2$^${23}$
as product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$, since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers
as product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two
as product of 4 numbers $2^2*3*5*7$
$k=\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$
=$2^3*3^2*5=360$
The above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?
• For multicharacter exponents, put them in braces. So 2^{11} to get $2^{11}$ Also, \min gets it set as a function. – Ross Millikan Sep 17 '13 at 15:51
• oeis.org/A005179 – barrycarter Sep 7 '17 at 19:17
This approach is not fishy at all, but can be laborious. Note that you can't use $5+3 \lt 7+2$ to conclude that $2^5*3^3 \lt 2^7*3^2$, although it is true, you have to take the ratio and use $3 \lt 2^2$. For example, if you were looking for $36$ factors, one comparison would be $2^8*3^3$ versus $2^5*3^5$. Despite the fact that $5+5 \lt 3+8, 2^5*3^5=7776 \gt 6912=2^8*3^3$. You can take logs and compare $5 \log 2 + 3 \log 3$ with $7 \log 2 + 2 \log 3$ to get it right.
I don't know an easier way. Your example shows the failure of a greedy algorithm. Factor the desired number of factors, here as $3*2^3$. Starting from the largest factor, find the cheapest way to get that many. So start with $2^2$, which has $3$ factors. Now you need to double it. You can either multiply by a new prime, clearly $3$, or increase the exponent of $2$ to $5$. Since the first has a factor $3$ and the second $8$, we choose $2^2*3$ Now we want to double again, and our choices are $2^3, 3^2, \text{ or } 5$ and we take $5$, giving $2^2*3*5$ One more doubling comes from $7$, and we come up with $2^2*3*5*7=420$, not the best.
Take any number $N=p_1^{m_1}...p_k^{m_k}$, where $p_i$ are its prime divisors, and compute how many divisors it has. Each divisor would be a product of the same primes in varying powers, $D=p_1^{d_1}...p_k^{d_k}$, where $0\leq d_i \leq m_i$. Different divisors have different collections of powers, so the number of divisors will be $(m_1+1)(m_2+1)...(m_k+1)$.
Now let's find the smallest $N$ such that $(m_1+1)(m_2+1)...(m_k+1)=24$. There are only few possibilities to break 24 into a product of decreasing numbers: $3*2*2*2=4*3*2=6*4=6*2*2=8*3=12*2=24$. The corresponding candidates with the smallest primes chosen for the smallest powers would be these ones:
$2^{3-1}*3^{2-1}*5^{2-1}*7^{2-1}=2^2*3*5*7=420$
$2^{4-1}*3^{3-1}*5^{2-1}=2^3*3^2*5=360$
$2^5*3^3=2592$
$2^5*3*5=480$
$2^{11}*3=6144$
$2^{23}=8388608$
Then just pick the smallest one, which happens to be 360.
• @ Michael That's what I tried at first,this algorithm becomes laborious if we are dealing with such numbers whose number of divisors has many prime factors,so I want a generalized method in order to avoid comparing numbers>>> – Tom Lynd Sep 17 '13 at 17:23
A005179 lists some resources on the problem. In particular:
Grost, M. (1968). The Smallest Number with a Given Number of Divisors. The American Mathematical Monthly, 75(7), 725-729. doi:10.2307/2315183
From the introduction:
Given $h = q_1 q_2 \dots q_n$, with primes $q_1 \le q_2 \le \dots \le q_n$, let $A(h)$ be the smallest number with $h$ divisors.
In many cases, $$A(h) = 2^{q_1-1} 3^{q_2-1} \dots p_n^{q_n-1}$$
The primary objective of the paper is to determine the exceptions.
We call these numbers ordinary, from R. Brown, The minimal number with a given number of divisors, Journal of Number Theory 116 (2006) 150-158. In this paper Brown shows almost all $A(h)$ are ordinary, in particular "We show here that all square-free numbers are ordinary and that the set of ordinary numbers has natural density one." | 2020-01-25T23:36:55 | {
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https://stats.stackexchange.com/questions/159905/why-does-drawing-one-card-at-a-time-increase-the-probability-of-choosing-the-ace | Why does drawing one card at a time increase the probability of choosing the Ace of Spades?
If you draw 5 cards from a standard deck of 52 cards, then the probability of your hand having the Ace of Spades is: $$\frac{51\choose 4}{52\choose 5} = \frac{51!5!47!}{4!47!52!} = \frac{5}{52}$$ If, however, you choose one card a time until you've drawn 5 cards, the probability of having the Ace of Spades is: $$\frac{1}{52}+\frac{1}{51}+\frac{1}{50}+\frac{1}{49}+\frac{1}{48}=\frac{433507}{4331600}\approx\frac{1}{10}>\frac{5}{52}$$ Why does choosing one card a time increase the probability of finding the Ace of Spades, when the resulting hands are equivalently drawn?
• Hint: instead of drawing 5 cards one at a time, let's say you draw 34. Would you then say your probability of drawing the ace of spades is $1/52 + 1/51 + \ldots + 1/19 \approx 1.04 > 1$? – Hong Ooi Jul 4 '15 at 13:31
• Another hint : What you wanted to do is compute P(X=1)+P(X=2)+..+P(X=5)...with P(x=k) representing the probability of the Ace of Spade drawn exactly at the k th draw.This is one way to find the answer (not the most simple) but one legit way as these are disjunctive cases. The problem comes from how you compute these probabilities. For P(X=2), for example, If you had to answer the question "what is the probability that the ace of space is drawn exactly at the second draw, would you find 1/51 ? – brumar Jul 4 '15 at 13:39
• It's not actually homework, just a genuine question. But I think I know what you're saying: I'm conflating the idea of the geometric distribution and dependent draws. So, it'd be better to say $P(X=1)+P(X=2|X\neq 1)+\dots+P(X=5|X\neq 1,2,3,4)$? – AJS Jul 4 '15 at 14:26
• One thing I can't figure out is, why does it cancel out to $5/52$? $P(X=1) = 1/52$, $P(X=2\mid X\neq 1)=P(X=2\cap X\neq 1)/P(X\neq 1)=((51/52)*(1/51)/(51/52))$, giving me the original (incorrect) sum of fractions? – AJS Jul 4 '15 at 14:47
• Focus on P(X=2∩X≠1) = P(X=2 |X≠1) * P(X≠1). What is that? How does the answer to this relate to probability of getting an ace of spaces for the first time on the 2nd card drawn? – Mark L. Stone Jul 4 '15 at 15:23
As @Glen_b suggested, it would be a good idea to summarize the comments part in an answer. I'll do that and also give an alternative for the formula related to the probabilistic point of view at the end of the answer. The apparent contradiction between the two computations came from this line :
the probability of having the Ace of Spades is: $$\frac{1}{52}+\frac{1}{51}+\frac{1}{50}+\frac{1}{49}+\frac{1}{48}=\frac{433507}{4331600}\approx\frac{1}{10}>\frac{5}{52}$$
The idea behind this computation was good but the logic was flawed. If we sum the probability of the Ace of Spade drawn exactly at the k th draw with k going from 1 to 5, we have the probability we want. But $\frac{1}{51}$, for example, does not represent the probability that the ace of spade is exactly drawn at the second attempt but the probability that the ace of space is drawn at the second attempt given that it has not been drawn at the first one.
AJS finally found the right formula
I got it! $1/52+((51/52)∗(1/51))+⋯+((51/52)∗(50/51)∗(49/50)∗(48/49)∗(1/48))=5/52$
With the idea that the probability to exactly draw the ace of spade at the $k$ th trial is the probability to not draw the ace of spade during previous attempts.... $$(51/52)*(50/51)...(52-k+1)/(52-k+2)$$...multiplied by the probability to draw the card at the $k$ th attempt $$1/(52-k+1)$$
A general rule of thumb that I have to avoid this kind of error is to be cautious when it comes to adding probabilities. If you can turn your problem around to avoid additions to the profit of multiplications, this is less prone to error.
A more classic approach giving a shorter path would have been to consider that the probability of having the Ace of Spades after 5 draw is 1-(the probability to not draw it with 5 draws) which gives, as you know : $$1-(51/52)*(50/51)*(49/50)*(48/49)*(47/48)=1-47/52=5/52$$
• Thank you, brumar. I appreciate you taking the time to write it out. – AJS Jul 4 '15 at 23:22 | 2020-08-06T13:14:58 | {
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https://www.grae.io/post/euler_problem_7/ | Project Euler Problem 7: 10001st prime
Back to primes! So far we’ve been able to get away with being a little greedy with our compute when playing with primes. Now Euler is ratcheting up the difficulty and we’ll have to focus on efficiency.
As usual, if you haven’t spent time with Problem 7 yet, take a chance to play with it on your own and come back.
Counting Primes
Let’s start by looking at each integer, deciding whether it’s prime, and counting it if it is until we get to the 10,001st prime.
Our first stab at an is_prime(n) function will be the simplest and we’ll iterate into more optimized (and complicated) versions after. Here’s the starting point:
def is_prime(n):
if n < 2:
return False
for x in range(2, n):
if n % x == 0:
return False
return True
This checks every number less than $n$ to see if it’s a factor of $n$. It’s almost good enough to solve the problem in under a minute. My laptop chugs through the first 10,001 primes in 68 seconds using this version of the is_prime(n) function (full code later). But the rules only give us a minute and we can do better.
Cap the Search Space
Looking back at our Problem 3 Solution we optimized our is_prime(n) function by caping the space we search to find factors factors by checking only numbers up to $\sqrt{n}$. Check out that post if you want to dig deep into why / how that works.
def is_prime(n):
if n < 2:
return False
for x in range(2, math.floor(math.sqrt(n)) + 1):
if n % x == 0:
return False
return True
This runs a lot faster. It finds the 10,001st prime in 0.29 seconds on my machine. But can we make it even better?
Skip Through the Search Space
Perhaps the most rediscovered result about primes numbers is the fact that every prime bigger than 3 is “next” to a multiple of 6. That is, for every prime number starting at 5 you can get a multiple of 6 by adding 1 or subtracting 1.
For example:
• 5 is prime, add 1 and get 6
• 13 is prime, subtract 1 and get (6 * 2)
• 1,361 is prime, add 1 and get (6 * 227)
This works for every prime number.
We can use this property to skip potential factors we don’t need to check. When checking to see if a number $n$ has factors we can get away with just looking for the prime factors, we don’t also need to know if it has any factors that are themselves composite. For example, we don’t need to know that 24 is divisible by 8. We can stop as soon as we see it’s divisible by 2. So we can skip every potential factor except for those which might be prime.
In code:
def is_prime(n):
if n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
for x in range(6, math.floor(math.sqrt(n)) + 2, 6):
if n % (x - 1) == 0 or n % (x + 1) == 0:
return False
return True
This version takes advantage of Python’s “step” argument to range(). We’re looking at every multiple of 6 (below our limit) and checking whether the number before or after it divides our target.
This optimizes things a bit more and, indeed, finds the 10,001st prime in 0.17 seconds on my machine.
Putting it Together
Once we have an efficient is_prime() function the solution is a matter of counting primes with a while loop.
seen = 0
n = 1
while seen < 10001:
n += 1
if is_prime(n):
seen += 1
print(n)
Going Further
There are ways to solve this problem even faster. You could use the Prime Number Theorem to approximate an upper bound for a Sieve of Eratosthenes and sieve out the answer. We’ll deal with those concepts in coming problems so for now I’ll leave that as an exercise for the reader.
See an issue on this page? Report a typo, bug, or give general feedback on GitHub. | 2020-09-29T23:36:16 | {
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https://geocabirthve.web.app/1580.html | # Implicit differentiation examples and solutions pdf
Explicit means fully revealed, expressed without vagueness or ambiguity. Now i will solve an example of the differentiation of an implicit function. Find two explicit functions by solving the equation for y in terms of x. For each problem, use implicit differentiation to find d2222y dx222 in terms of x and y. In this presentation, both the chain rule and implicit differentiation will. Implicit differentiation is as simple as normal differentiation. Implicit differentiation extra practice date period. Implicit di erentiation statement strategy for di erentiating implicitly examples table of contents jj ii j i page2of10 back print version home page method of implicit differentiation. Multivariable calculus implicit differentiation examples. Solutions to implicit differentiation problems uc davis mathematics. They are laying the groundwork for stokestheorem, a. Implicit di erentiation statement strategy for di erentiating implicitly examples table of contents jj ii j i page1of10 back print version home page 23. If we are given the function y fx, where x is a function of time. Since the point 3,4 is on the top half of the circle fig.
There will also be one or two exercises on material in the next set of notes, which are not taken from the text. Search within a range of numbers put between two numbers. In fact, all you have to do is take the derivative of each and every term of an equation. Implicit differentiation helps us find dydx even for relationships like that. Implicit differentiation practice questions dummies. Implicit differentiation problems are chain rule problems in disguise. Find dydx by implicit differentiation and evaluate the derivative at. For each of the following equations, find dydx by implicit differentiation. Let us remind ourselves of how the chain rule works with two dimensional functionals. Example using the product rule sometimes you will need to use the product rule when differentiating a term. Calculusdifferentiationbasics of differentiationsolutions. Calculus i implicit differentiation practice problems.
Equations inequalities system of equations system of inequalities basic operations algebraic properties partial fractions polynomials rational expressions sequences power sums. How implicit differentiation can be used the find the derivatives of equations that are not functions, calculus lessons, examples and step by step solutions, what is implicit differentiation, find the second derivative using implicit differentiation. In other words, the use of implicit differentiation enables. It is the fact that when you are taking the derivative, there is composite function in there, so you should use the chain rule. Implicit differentiation if a function is described by the equation \y f\left x \right\ where the variable \y\ is on the left side, and the right side depends only on the independent variable \x\, then the function is said to be given explicitly. These are functions of the form fx,y gx,yin the first tutorial i show you how to find dydx for such functions. This section contains lecture video excerpts and lecture notes on implicit differentiation, a problem solving video, and a worked example. Check that the derivatives in a and b are the same. Calculus implicit differentiation solutions, examples, videos.
To generalize the above, comparative statics uses implicit differentiation to study the effect of variable changes in economic models. To perform implicit differentiation on an equation that defines a function \y\ implicitly in terms of a variable \x\, use the following steps. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. Differentiation of implicit function theorem and examples.
Then the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t dy dt dy dx dx dt. Implicit differentiation solved practice problems timestamp. This means that when we differentiate terms involving x alone, we can differentiate as usual. Given an equation involving the variables x and y, the derivative of y is found using implicit di erentiation as follows. Implicit differentiation multiple choice07152012104649. For example, according to the chain rule, the derivative of y. Suppose that the nth derivative of a n1th order polynomial is 0. This page was constructed with the help of alexa bosse. To do this, we use a procedure called implicit differentiation. Calculus bc parametric equations, polar coordinates, and vectorvalued functions defining and differentiating parametric equations parametric equations differentiation ap calc.
You know that the derivative of sin x is cos x, and that according to the chain rule, the derivative of sin x3 is you could finish that problem by doing the derivative of x3, but there is a reason for you to leave. Implicit differentiation mctyimplicit20091 sometimes functions are given not in the form y fx but in a more complicated form in which it is di. Implicit differentiation basic idea and examples youtube. Jul, 2009 implicit differentiation basic idea and examples. Implicit differentiation can help us solve inverse functions. The following problems require the use of implicit differentiation. Multivariable calculus implicit differentiation this video points out a few things to remember about implicit differentiation and then find one partial derivative. Implicit diff free response solutions 07152012145323. The basic idea about using implicit differentiation 1. If a value of x is given, then a corresponding value of y is determined. In such a case we use the concept of implicit function differentiation. Some relationships cannot be represented by an explicit function. Implicit differentiation example walkthrough video khan. There is a subtle detail in implicit differentiation that can be confusing.
The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. Here i introduce you to differentiating implicit functions. Jan 22, 2020 implicit differentiation is a technique that we use when a function is not in the form yf x. That is, i discuss notation and mechanics and a little bit of the. Jun 24, 2016 implicit differentiation solved practice problems timestamp. For example, in the equation we just condidered above, we. The majority of differentiation problems in firstyear calculus involve functions y written explicitly as functions of x. Implicit differentiation is a technique that we use when a function is not in the form yf x. Then the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t.
To make our point more clear let us take some implicit functions and see how they are differentiated. The technique of implicit differentiation allows you to find the derivative of y with respect to x without having to solve the given equation for y. Implicit di erentiation implicit di erentiation is a method for nding the slope of a curve, when the equation of the curve is not given in \explicit form y fx, but in \implicit form by an equation gx. Examples of the differentiation of implicit functions.
This is done using the chain rule, and viewing y as an implicit function of x. Implicit differentiation is a method for finding the slope of a curve, when the equation of the. Calculus implicit differentiation solutions, examples. Implicit di erentiation implicit di erentiation is a method for nding the slope of a curve, when the equation of the curve is not given in \explicit form y fx, but in \ implicit form by an equation gx. For difficult implicit differentiation problems, this means that its possible to differentiate different individual pieces of the equation, then piece together the result. Implicit differentiation basic idea and examples what is implicit differentiation. This means that when we differentiate terms involving x. For each problem, use implicit differentiation to find dy dx in terms of x and y. Implicit functions are often not actually functions in the strict definition of the word, because they often have multiple y values for a single x value. Find dydx by implicit differentiation and evaluate the derivative at the given point. The chain rule must be used whenever the function y is being differentiated because of our assumption that y may be expressed as a function of x. However, some functions y are written implicitly as functions of x. The solutions to this equation are a set of points x,y which implicitly define a relation between x and y which we will call an implicit function. Parametric equations differentiation practice khan academy.
Differentiation of implicit functions engineering math blog. Uc davis accurately states that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve both x and y. Use implicit differentiation directly on the given equation. Implicit differentiation sometimes functions are given not in the form y fx but in a more complicated form in which it is di. Find the equation of the tangent line to the graph of 2. Implicit differentiation mcty implicit 20091 sometimes functions are given not in the form y fx but in a more complicated form in which it is di. In any implicit function, it is not possible to separate the dependent variable from the independent one. Preference bundles, utility and indifference curves. Search for wildcards or unknown words put a in your word or phrase where you want to leave a placeholder.
293 802 632 1207 1048 331 259 386 1012 524 1522 1299 1300 730 1542 1000 938 1032 708 353 1555 1317 459 852 320 880 1127 184 841 | 2021-10-20T00:54:47 | {
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https://www.gradesaver.com/textbooks/math/other-math/discrete-mathematics-with-applications-4th-edition/chapter-11-analysis-of-algorithm-efficiency-exercise-set-11-4-page-764/38 | ## Discrete Mathematics with Applications 4th Edition
Prove $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$ is $\Theta(1)$. By theorem 5.2.4: $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}} = \frac{1}{5} (1 + \frac{4}{5} + \frac{4^2}{5^2} + \cdots + \frac{4^n}{5^{n}}) = \frac{1}{5}(\frac{\frac{4}{5}^{n+1}-1}{\frac{4}{5}-1}) = -(\frac{4}{5}^{n+1}-1)=1-\frac{4}{5}^{n+1}$. For $n>0$, $1-\frac{4}{5}^{n+1} \leq 1$. By the transitive property: $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}} \leq 1$. For $n>0$, $\frac{1}{5}(1) \leq \frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$. Since all terms are positive: $\frac{1}{5}|(1)| \leq |\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}| \leq |1|$. Thus for A=1/5, B=1, k=1, $A|(1)| \leq |\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}| \leq B|1|$ for all $x>k$. Thus $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$ is $\Theta(1)$.
Recall the definition of $\Theta$-notation: $f(x)$ is $\Theta(g(x))$ iff there exist positive real numbers A, B, k, such that $A|g(x)| \leq |f(x)| \leq B|g(x)|$ for all $x>k$. Recall theorem 5.2.3: Sum of a geometric sequence: For any real number $r$ except 1, and any integer $n \geq 0$, $\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$. In this case $r=\frac{4}{5}$. | 2019-11-21T23:59:33 | {
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https://math.stackexchange.com/questions/2257156/prove-sum-k-0n-2n-k-choose-2-n1-choose-3/2257198 | # Prove $\sum_{k=0}^{n-2}{n-k \choose 2} = {n+1 \choose 3}$
Prove $$\sum_{k=0}^{n-2}{n-k \choose 2} = {n+1 \choose 3}$$
Is there a relation I can use that easily yields above equation?
• You could start with $\;{n-k \choose 2} = \frac{(n-k)\cdot(n-k-1)}{2\cdot 1}\,$.
– dxiv
Apr 29, 2017 at 4:59
• @mvw not, the same. I edited the problem to consider that. Apr 29, 2017 at 5:00
• @dxiv I already knew that. Thank you. What are the next steps? Apr 29, 2017 at 5:01
• $\sum \frac{(n-k)\cdot(n-k-1)}{2\cdot 1} = \frac{1}{2}\left(n^2 \cdot \sum 1 - n \cdot \sum (2k+1) + \sum k(k+1)\right)$
– dxiv
Apr 29, 2017 at 5:03
• Note: The original question just stated the left hand side
– mvw
Apr 29, 2017 at 11:30
The right hand side ${n+1 \choose 3}$ is the number of ways to choose 3 elements from $\{1,2,\ldots,n+1\}$. Let $A$ denote the set of all 3-subsets of $\{1,2,\ldots,n+1\}$. We want to show that $|A|$ is equal to the left hand side.
Let $A_i$ denote the set of all 3-subsets of $\{1,2,\ldots,n+1\}$ which have $i$ as their largest element. For example, if $i=n+1$, then $A_{n+1}$ is the set of all 3-subsets which contain $n+1$, and the number of ways to choose the remaining 2 elements is ${n \choose 2}$. Hence, $|A_{n+1}| = {n \choose 2}$. More generally, $|A_i| = {i-1 \choose 2}$, for $i=3,\ldots,n+1$, because the remaining 2 elements must be chosen from $\{1,\ldots,i-1\}$. Note that $i$ has to be at least 3 because the largest of 3 elements will be at least 3.
Using the fact that the $A_i$'s $(i=3,4,\ldots,n+1)$ are disjoint and their union is all of $A$, we obtain the desired formula.
• If the largest element in a 3-subset is say 7, then this largest element can't be any of $3, 4, 5, 6, 8, 9,\ldots,n+1$. So $A_7$ is disjoint from $A_3, A_4, A_5, A_6, A_8, A_9, \ldots, A_{n+1}$. Similarly for the other $A_i$'s, and so the $A_i$'s are pairwise disjoint. Apr 29, 2017 at 5:56
• Thanks for comment. I got it and because of this I deleted my comment. Apr 29, 2017 at 6:10
$$\sum_{k=0}^{n-2}\frac{k^2+k(1-2n)+n^2-1}{2}\tag{Simplify what @dixv said}$$ now, $$\sum_{k=1}^nk=\frac{n(n+1)}{2}\text{ and }\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$
Hope it helps?
For $$S=\sum_{k=0}^{n-2}(n-k)(n-k-1)$$ set $n-k=r$ to get $$S=\sum_{r=1}^nr(r-1)=\sum_{r=1}^nr^2-\sum_{r=1}^nr$$
Using index transformation $m = n - k$, reversal of the summation order and the definition of the binomial coefficient in terms of factorials we can write for $n\ge 2$: \begin{align} \sum_{k=0}^{n-2} \binom{n-k}{2} &= \sum_{m=2}^{n} \binom{m}{2} \\ &= \sum_{m=2}^{n} \frac{m!}{2! (m-2)!} \\ &= \sum_{m=2}^{n} \frac{m(m-1)}{2} \\ &= \sum_{m=1}^n \frac{m(m-1)}{2} \\ &= \frac{1}{2} \sum_{m=1}^{n} m^2 - \frac{1}{2} \sum_{m=1}^n m \\ &= \frac{n(n+1)(2n+1)}{12} - \frac{n(n+1)}{4} \\ &= \frac{1}{6} n^3 + \frac{1}{4} n^2 + \frac{1}{12} n - \left( \frac{1}{4} n^2 + \frac{1}{4} n \right) \\ &= \frac{1}{6} n^3 - \frac{1}{6} n \end{align} where we used Faulhaber's formula for the square pyramidal and triangular numbers.
On the other side of the equation we have: \begin{align} \binom{n+1}{3} &= \frac{(n+1)!}{3!(n+1-3)!} \\ &= \frac{(n+1)n(n-1)}{6} \\ &= \frac{n^3 - n}{6} \end{align}
You can prove it using generating functions. We wish to prove that $$\sum_{m=0}^{n} \binom{m}{2}=\binom{n+1}{3}\tag{1}$$ Use the identity $$\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n.\tag{2}$$ (which can be obtained by repeatedly differentiating the geometric series) where $k\geq1$ to get that the generating function whose $m$th coefficent is $\binom{m+2}{2}$ is $(1-x)^{-3}$ and the generating function whose $m$th coefficent is $\binom{m}{2}$ is $x^2(1-x)^{-3}$. Thus $$\sum_{m=0}^{n} \binom{m}{2}= [x^n]\left(\frac{1}{1-x}\frac{x^2}{(1-x)^{3}}\right) = [x^n] \left(\frac{x^2}{(1-x)^{4}}\right)\tag{3} =\binom{n+1}{3}$$ by (2) as desired where $[x^n]$ extracts the coefficient of $x^n$.
With so many proofs there is no need for one more. But I like using finite differences.
Let $f(n)$ represent the sum.
$\Delta f(n) = \binom {n+1}{2} = \Delta \binom {n+1}{3}$
$\implies f(n) = \binom {n+1}{3} + c$
for $n=2$, we get $c + 1 = 1 \implies c=0$ | 2023-03-21T04:10:29 | {
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http://vurg.curaben.it/curve-sketching.html | Sketch the rest of the graph. I now introduce you to plotting the curve r=a sin2θ. 5a limits at infinity 3. If f and g are functions that have derivatives, then the composite function has a derivative given by f. uk A sound understanding of Curve Sketching is essential to ensure exam success. Domain: For what values ofx is f(x) defined? Avoid division by zero and square roots of negative numbers. CURVE SKETCHING EXERCISE 1 Sketch the following Find the gradient of the tangent to the curve x 2 2 + xy 2 + y2 = 14 at (2, 3). MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. To the left zooms in. Likes scottdave. And the goal here--STUDENT: [INAUDIBLE. Carefully, state L’Hospital’s Rule. 5: Summary of Curve Sketching Last updated; Save as PDF Page ID 4465 If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function. 5 Man vs machine. 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Click below to download the free player from the Macromedia site. 5 – Summary of Curve Sketching Math& 151 Warnock - Class Notes Here are the Guidelines for Sketching a Curve. f 0 (x) > 0). Start your bird sketch by noting the posture of the bird or the angle at which it sits with a single line. No calculator unless otherwise stated. Now after this how do i plot the remaining curve for x >1 (where y become s <0) Can you tell me a smaller approach. There are a number of mathematical curves that produced heart shapes, some of which are illustrated above. AP Calculus. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Find the intervals of increase and decrease 8. Well, the free Urban Sketching 101 guide covers everything there is to know, including: what it is, where to go and starter techniques and tips for the urban sketcher on the go. 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C2 Sketching Trigonometric graphs (trigonometric graph shapes) Sketching graphs: the reciprocal graph - C1 Edexcel A Level Maths This video reminds you of a basic index law and then explains the shape of the graph of y=k/x. AP Calculus AB/BC - M. move objects in sketch. While some sketching tools allow 3D drafting, the feature is not universal. The parabola is the envelope of the straight lines. Watch all CBSE Class 5 to 12 Video Lectures here. Curve sketching is a handy tool, used both directly and indrectrly in these examinations. Perhaps someone should change the thread title to "Not Curve Sketching". AP Calculus Project 4 - Curve Sketching Name_____ You will be in a group of 2 to 4. The general approach to curve sketching. Find the critical numbers 7. Fix any real number C. Get smarter on Socratic. 4a homework questions 3. STEP 2 Curve Sketching Questions 2. Curve Sketching packet. Heart Curve. 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An asymptote is a line that the curve gets very very close to but never intersect. Calculus 3 very difficult curve sketching problem? Sketch the curve traced out by the tip of the radius vector and indicate the direction in which the curve is traversed as t increases. Look at any item sitting around you. In particular, Section 4. If f( x) = f(x), then f(x) is symmetric about. Be sure to nd any horizontal and ver-tical asymptotes, show on a sign chart where the function is increasing/decreasing, concave up/concave down, and identifying (as ordered pairs) all relative extrema and in ection points. Determine the x- and y- intercepts of the function, if possible. A critical point may be a maximum point, minimum point, or neither. However, this equation, y =x / x^2 -1 does not have any horizontal asymptotes. With a model open, click Edit > Project. The curve passes through origin and meets the x – axis at two coincident points (2,0) and (2,0). This is for a few reasons, but primarily because curve sketching takes a little bit of intuition. XXIII – Curve Sketching 1. If f and g are functions that have derivatives, then the composite function has a derivative given by f. GraphSketch is provided by Andy Schmitz as a free service. A critical point may be a maximum point, minimum point, or neither. Find more Mathematics widgets in Wolfram|Alpha. Curve Sketching. Link to Binder: Link to Current Tab: Email Embed Facebook Twitter Google+ Classroom Upgrade to Pro Today! The premium Pro 50 GB plan gives you the option to download a copy of your binder to your local machine. 5 Algorithm for Curve Sketching The Algorithm for Curve Sketching provides us with a framework from which we can determine all the key elements of a curve so that we can sketch it relatively accurately. Example 1 Sketch the parametric curve for the following set of parametric equations. Design Sketching Learning Curves (2011, 177 pages, Klara Sjölén and Allan Macdonald) is a brand new sketch book, aimed at teaching how to really learn to sketch. dominique_cimafranca writes "The Dynamic Graphics Project of the University of Toronto has released a pretty nifty 3D curve sketching system. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. Curve Sketching - Displaying top 8 worksheets found for this concept. A normal to a curve is a line perpendicular to a tangent to the curve. Textbook Authors: Thomas Jr. curve_sketching_solutions. In this case, it does not have a vertical asymptote. The graph shown is the DERIVATIVE of f. View Homework Help - curve_sketching. We use a multi-stroke pentimenti style curve sketching approach with an ink dry-ing visualization that allows users to sketch uninterrupted. : If your making something that is less than simple it will almost always pay you to do some kind of drawing the try to get things straight in your head before you commit to cutting expensive materials up. Advanced Trigonometry 1 Revision Notes Inverse Trigonometric Function, Stationary Points, Curve Sketching. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. 3: # 1, 2a-f, 3-9 5. And a lot of people struggle with them even when they don't realise it. Curve Sketching The concepts of domain, limits, derivative, extreme values, monotonicity and concavity have been introduced. Learn more about ferguson curve, curve, draw curve, draw ferguson curve. Sample Problem #1: f(x) = x3 - 6x2 + 9x + 1. From the home tab, select move curve icon command. State any horizontal and vertical asymptotes or holes in the graph. When you exit the sketch, regions are formed by intersecting lines. We begin by making some general remarks about curve sketching, by which we mean more specifically, sketching or drawing the graph of the function y equals f of x in the xy-plane. Sketch the Curve!. While we have been treating the properties of a function separately (increasing and decreasing, concave up and concave down, etc. Instead of focusing on details at the start of a picture, make light sketch lines to capture the posture, proportions, and angles of your subject. 10 determines the solution for given polynomials. unit 4: curve sketching Lesson 1: Increasing/Decreasing Functions. Turning point Axis of Symmetry Mirror point Y intercept X intercepts { the real roots The turning point is always required, and another two points are needed for a rough sketch. Although these problems are a little more challenging, they can still be solved using the same basic concepts covered in the tutorial and examples. WORKSHEETS: Practice-Curve Sketching 1 open ended. Show that, if a > 1, then C has exactly one stationary point. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. If a lack of sketching tips are holding you back from beginning your sketching journey, then we've got you covered. Honors Calculus -e - Asymptotes Plans changed for Date Period Domain: Range: Zeros: y-intercept: HA: c. Tangent lines are useful in calculus because they can magnify the slope of a curve at a single point. Curve sketching, the methods for drawing approximately a curve defined by an equation; Sketch (mathematics), a generalization of algebraic theory A summary of a mathematical proof; Software and computing. This is achieved by adding a sketch relation to your finished curve. how to sketch a curve that has asymptotes. You should be able to quickly sketch straight-line graphs, from your knowledge that in the equation y = mx + c, m is the gradient and. Determine the domain and range. The figure illustrates a means to sketch a sine curve – identify as many of the following values as you can: asymptotic behaviour,. Get smarter on Socratic. ), we combine them here to produce an accurate graph of the function without plotting lots of extraneous points. Find the location of the x and y intercepts and plot them on the graph. This usually isn’t of help. An asymptote is a line that the curve gets very very close to but never intersect. Detailed Example of Curve Sketching x Example Sketch the graph of f(x) =. Curve Sketching Recipe: 1. When x < 0 then y < 0 so in this case the curve lies in the 3rd quadrant. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Today, we are going to lay out the principles behind these questions, and explain the methods on how to attack them. Curve Sketching Quiz. 1 $$y=x^5-5x^4+5x^3$$. We will review the main topics that you'll need to know for the AP Calculus exams. Please check your network connection and refresh the page. The system coherently integrates existing techniques of sketch-based interaction with a number of novel and enhanced features. docx: File Size: 255 kb Video - The Mean Value Theorem. Sketching Sketching is useful if you want to create a region that can be pulled into 3D. Theory: This section will review the basic principles and equations that you should know to answer the exam. DUE TUESDAY FEBRUARY 16 AT THE BEGINNING OF CLASS. 1 (i) A curve has equation Find the x-coordinates of the points on the curve where [2] (ii) The curve is translated by Write down an equation for the translated curve. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. Exit Tickets. 5 Man vs machine. Play this game to review Calculus. Madas Created by T. 10 5 5 10 7 45. The graph shown is the DERIVATIVE of f. Curve Sketching Example: Sketch 1 Review: nd the domain of the following function. Alternatively, Curve Sketching 1. Turning point Axis of Symmetry Mirror point Y intercept X intercepts { the real roots The turning point is always required, and another two points are needed for a rough sketch. Curve Sketching Calculus, free curve sketching calculus software downloads, Page 2. [Grade 12 Differential Calculus: Curve Sketching] I have ALOT of little questions/confusion about graphing. You should be able to quickly sketch straight-line graphs, from your knowledge that in the equation y = mx + c, m is the gradient and. 4 Curve Sketching V63. Keyword-suggest-tool. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. the methods for drawing approximately a curve defined by an equation. Curve Sketching The concepts of domain, limits, derivative, extreme values, monotonicity and concavity have been introduced. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Get step-by-step solutions to your Curve sketching problems, with easy to understand explanations of each step. Curve sketching. 5 Curve Sketching ¶ permalink. Curve Sketching. A normal to a curve is a line perpendicular to a tangent to the curve. C1 Curve Sketching - Factorising & Sketching Polynomials 2 QP C1 Curve Sketching - Factorising & Sketching Polynomials 3 MS C1 Curve Sketching - Factorising & Sketching Polynomials 3 QP. Label x and y intercepts. A curve with two loops. 5—Curve Sketching Show all work on a separate sheet of paper. Let's put all of our differentiation abilities to use, by analyzing the graphs of various functions. If x is large negative then y is large positive. Concavity/Inflection Points H. The first thing I did to solve this problem was sketch the curve{s}. Leading artists share their sketching tips to help you get started, then take things further. 1 Extreme Values. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. x -Intercept (s) Vertical Asymptote (s) Horizontal and/or Oblique Asymptote (s) First Derivative. Sketch the vector function f(t) = < t 2, t 3 > for -5≤ t ≤ 5. Tags: curve sketching. Review of Prerequisite Skills. This is an input location where either f0(x. Unit 1: Limits & Continuity. Created by T. Zeros The zeros of the function f(x) are: x 1 = 3 and x 2 = 2 y-intercept. This usually isn’t of help. Sketch the graph of the curve with equation y x x x= + + −(1 4 2)( )( ), x∈. To plot a function just type it into the function box. Because we often represent functions by their graphs, you could say that calculus is all about the analysis of graphs. A full lesson on sketching cubics, quartics and reciprocal functions. is to determine the following: 1) find y(0) 2) find y = 0, if. 2: CURVE SKETCHING POLYNOMIALS Example 3. We have been learning how we can understand the behavior of a function based on its first and second derivatives. Determine asymptotes: a) for vertical asymptotes, check for rational function zero denominators, or unde ned log function points; b) for horizontal asymptotes, consider lim. Let's put all of our differentiation abilities to use, by analyzing the graphs of various functions. We can roughly sketch the graph with stationary point, point of inflection, and y-intercept. Curve Sketching We've done most of the legwork needed for this section. The graph shown is the DERIVATIVE of f. Students describe a curve given the equation of the curve in polar form. The following steps are taken in the process of curve sketching: Find the domain of the function and determine the points of discontinuity (if any). 5 Algorithm for Curve Sketching part 2. So let's hope we can do this. Curve sketching The roots , stationary points , inflection point and concavity of a cubic polynomial x 3 − 3 x 2 − 144 x + 432 (black line) and its first and second derivatives (red and blue). The process of using the first derivative and second derivative to graph a function or relation. It is an application of the theory of curves to find their main features. Fusion 360 Blog. Keyword-suggest-tool. Horizontal and vertical asymptotes may be calculated by taking the appropriate limits of. , ISBN-10: 0-32187-896-5, ISBN-13: 978-0-32187-896-0, Publisher: Pearson. In this calculus worksheet, 12th graders answer questions about derivatives, increasing and decreasing functions, relative maximum and minimum and points of inflection. Prior observations indicate that professionals want full control of the final shapes of 2D/3D curves while leveraging their sketching skills. The course is intended to be challenging and demanding. Concavity/Inflection Points H. Course Description: UCI Math 2A is the first quarter in Single-Variable Calculus and covers the following topics: Introduction to derivatives, calculation of derivatives of algebraic and trigonometric functions; applications including curve sketching, related rates, and optimization; exponential and logarithm functions. Polynomial Curve Sketching - Displaying top 8 worksheets found for this concept. How do I know which equations have horizontal asymptotes? For example, y=x+2 / x-1 has the horizontal asymptote of y=1. GraphSketch is provided by Andy Schmitz as a free service. 10 creates exercises with solutions and graphs in the field of curve sketching of linear, quadratic, cubic, quartic and quintic polynomials. Sketching Sketching is useful if you want to create a region that can be pulled into 3D. Limits by Direct Evaluation. When x < 0 then y < 0 so in this case the curve lies in the 3rd quadrant. Curve sketching (Q350877) From Wikidata. A Rhino curve is similar to a piece of wire. Curve Sketching Summary Introduction Now that you have learned how to find relative extrema, intervals where a function is increasing/decreasing, and intervals where a function is concave up/concave down, we will now "pull it all together" and work through several AP problems that involve the analysis of functions and curve sketching. Therefore the domain is D f = R. The curve consist of straight lines and some arces, all tangent to each other. 5 B—Curve Sketching Summary For a function f ′, the combined information of the first derivative f and the second derivative f ′′ can tell us the shape of a graph. Sketching Polynomials 1 January 16, 2009 Oct 11 9:12 AM Sketching Polynomial Functions Objective Sketch the graphs of Polynomial Functions. Solve Curve sketching problems with our Curve sketching calculator and problem solver. Given the information, determine the following about f(x): (Explain each of your answers) 8. Instead, WebAssign will ask limited submission questions about your graphs. Sketching the Curve Summary – Graphing Ex 2 – Part 4 of 4. The following problems illustrate detailed graphing of functions of one variable using the first and second derivatives. Curve Sketching. In my opinion, one of the more difficult topics in Extension 1 and 2 has got to be curve sketching. 5 Summary of Curve Sketching(A) GUIDELINES FOR SKETCHING A CURVE ةلاد نيب مسر تاوطخ 1 Find the domain ةلادلا ل جم دجوا 1 the domain is, the set of values of for which is defined. However, for sketching "basic" cubics, you should be given "nice" equations. Explore math with our beautiful, free online graphing calculator. It is recommended that you start with Lesson 1 and progress through the video lessons, working through each problem session and taking each quiz in the order it appears in the table of contents. If f '(x) =0 , then P(x, f (x)) is a local extrema and tangent is horizontal. Welcome to highermathematics. r(t)=(2cost)i+(2sint)j+(2pi-t)k 0 =t=2pi Ok, so I've drawn out the curve, and my curve starts at (2)i+(2pi)k and ends at (2)i. Write NONE if there are none. Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. Click below to download the free player from the Macromedia site. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Intercepts C. The following techniques may also be of help and they should be employed whenever appropriate. A rational function is looked at as an. We have been learning how we can understand the behavior of a function based on its first and second derivatives. That tells us that our midline drooping down 4. Applications of the Derivative: Curve Sketching and Extrema Practice Problems: The following six pages contain 28 problems to practice curve sketching and extrema problems. In this section, we learn methods of drawing graphs by hand. When f''(x) = 0, there is an inflection point changing the concavity of f(x). Find the - and -intercepts. 6 curve sketching 3. Curve sketching for calculus. Semester Test 1 I Saturday 25 August I D1 Lab 308 I Starts at 09h00. 3 Second Derivative Test. Mathematics / Analysis - Plotter - Calculator 3. Curve Sketching Learning Outcomes Make tables and draw the graphs of various equations to include: Linear Functions Quadratic Functions Cubic Functions - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Curve sketching is a handy tool, used both directly and indrectrly in these examinations. there is no y-. Capture drawings on site will forever embed that experience in your mind. ةفرعم لعجت يتلا ميق ةعومجم وه ةلادلا لجم. xy–plane where f takes the value C. Curve sketching (Q350877) From Wikidata. Curve Sketching Using Calculus - Part 1of 2. The distance from the starting line of a runner in the 100-meter dash is a. Curve sketching This PowerPoint presentation shows the different stages involved in sketching the graph Sketching the graph Step 1: Find where the graph cuts the axes When x = 0, y = 4/3, so the graph goes through the point (0, 4/3). Incorporates the use of GeoGebra and the Casio fx-991EX Classwiz. In this method, we’ll skip steps 1 to 4 of curve sketching and go straight to steps 5, 6 and 7. y = (sin x)/ x (Pay particular attention to the shape of the curve around x = 0). 2 Sample Problems. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Even high school students love to color!. This interactive workshop is designed for students currently enrolled in Math 226 who would like more exposure on understanding the relationships between first derivatives, second derivatives, and curve sketching. Be sure to list the Domain and Range, intercepts, the equation of any asymptotes, intervals of increasing/decrease,. Recall, if they exist, we find the -intercept(s) by setting =0 and. 2 shows graphs A and B. pdf File history uploaded by Paul Kennedy 11 months, 1 week ago No preview is available for MCV 4U Unit 8 Shell-Curve Sketching. Curve Sketching. From the derivative's graph, identify the interval graph where f (the original function) is concave up. The sketch must include the coordinates of • … all the points where the curve meets the coordinate axes. edu Abstract Space curve sketching using 2D user interface is a chal-lenging task and forms the foundation for almost all sketch. 2: CURVE SKETCHING RATIONAL FUNCTIONS EXERCISES Give a complete graph of the following functions. And there's relatively little computation. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. 4: # 1 (what you need), 3abc 5. 5 Curve Sketching. Curve Sketching 1. For the sake. Binder ID: 93334. Sketch the graph of the following functions by finding the domain, symmetry, intercepts, asymptotes, intervals of increase and decrease, local maximum and minimum, concavity and points of inflection. Calculus 3 very difficult curve sketching problem? Sketch the curve traced out by the tip of the radius vector and indicate the direction in which the curve is traversed as t increases. And the goal here--STUDENT: [INAUDIBLE. 00 Price per gallon 2. • The techniques used in algebra for graphing functions do not demonstrate subtle behaviors of curves. 5 Curve Sketching. Polar curves: wrapping a function around the pole. So the next topic is curve sketching. Here input is an equation. XXIII – Curve Sketching 1. Curve Sketching. This property is called the asymptote. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Curve Sketching 1. Sketching curves the curve r = asin 2θ. Limits and Curve Sketching. 4a concavity and pois 3. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Curve sketching Graphing The topic menu above allows you to move directly to any of the four sections for each topic. That’s the first step ,in any curve sketching problem. I have a curve sketching assignment and this one question i am having trouble with (this is x2= x squared and 3x2 is 3 x squared. Polar curves: wrapping a function around the pole. f 0 (x) 0). b) horizontal: No horizontal asymptotes because. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Find vertical asymptotes and holes. An asymptote is a line that the curve gets very very close to but never intersect. In this calculus worksheet, 12th graders answer questions about derivatives, increasing and decreasing functions, relative maximum and minimum and points of inflection. Voyager …. As $$x$$ increases, the slope of the tangent line increases. The first curve is a rotated cardioid (whose name means "heart-shaped") given by the polar equation. That tells us that our midline drooping down 4. Sketching Infinite Lines: the Conic tool applies tangency at each endpoint and selects the top vertex of the curve. The graph shown is the DERIVATIVE of f. CURVE SKETCHING BLAKE FARMAN Lafayette College Name: 1. So even 10 problems you should be able to get through in a few hours. Domain of f(x) 2. y: f 00 ( x ) > 0 ) f ( x up. Recall, if they exist, we find the -intercept(s) by setting =0 and. Topic: Calculus, Derivatives. Curve Sketching Recipe: 1. From the home tab: Direct sketch group -> sketch curve gallery -> edit curve gallery -> move curve. These are potential local extrema. But there were no suitable comparison stars nearby to help today. Although these problems are a little more challenging, they can still be solved using the same basic concepts covered in the tutorial and examples. Convert Entities: Creates one or more entities in a 3D sketch by projecting an edge, loop, face, external curve, external sketch contour, set of edges, or set of external curves onto the sketch plane. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by carefully labeling critical points, intercepts, and inflection. If a graph is given, then simply look at the left side and the right side. The graph of y. The curve will be exactly the same as when you add hydrochloric acid to sodium hydroxide. Because it is a curve in 2d, it is usually easier to sketch than the graph of f. Notes - Curve Sketching (Extrema, Critical Numbers, Intervals of Increase and Decrease, etc. Curve Sketching Date_____ Period____ For each problem, find the: x and y intercepts, x-coordinates of the critical points, open intervals where the function is increasing and decreasing, x-coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. And the goal here--STUDENT: [INAUDIBLE. 5 yesterday, which fit well with the light curve. Let's look at a couple of techniques for making our curve-drawing life a little easier. Oct 5, 2019 #10 Kolika28. 5 How to Find Inflection Points 2 3 2 12 inflection point f(x) =. Curve Sketching. Give a complete graph of f(x) = 1 3 x3 1 2 x2 2x+ 1: Be sure to show on a sign chart where the function is increasing/decreasing, concave up/concave down, and identifying (as ordered pairs) all relative extrema and in ection points. This interactive workshop is designed for students currently enrolled in Math 226 who would like more exposure on understanding the relationships between first derivatives, second derivatives, and curve sketching. Get feedback on your graphs. Further information regarding the Curve Sketching Summer School. Veitch 1 p x 1 = 0 1 p x = 1 1 = p x 1 = x The other critical value is at x = 1. A rational function is looked at as an. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Domain and Range. Curve Sketching Example: Sketch 1 Review: nd the domain of the following function. Right-click the line and select Set as Mirror Line from the context menu. A function f (x) is decreasing on an interval if the values of f decrease as x increases (i. In this method, we’ll skip steps 1 to 4 of curve sketching and go straight to steps 5, 6 and 7. 1 use many of the techniques discussed in this chapter. Take a quick interactive quiz on the concepts in Curve Sketching Derivatives, Intercepts & Asymptotes or print the worksheet to practice offline. MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. Learn new vocabulary: f is concave up wherever f0 is increasing. Microsoft Word - MCV4U1 - Task - Curve Sketching. This is good because as well as triggering the rebuild it also fully constrains the curve. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. A 3D Curve Sketching System For Tablets 72 Posted by timothy on Saturday October 11, 2008 @10:53PM from the no-mention-of-license-terms dept. No calculator unless otherwise stated. After memorizing the concepts of the second derivative, we move onto the next topic: creating sign charts. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by carefully labeling critical points, intercepts, and inflection. 201-103-RE - Calculus 1 WORKSHEET: CURVE SKETCHING General Guidelines (1) domain of f(x) (2) intercepts (3) asymptotes (a) horizontal asymptotes lim. Curve Sketching Quiz. Y U bA Ql Yl9 irmiwgth1tes m srdeWs3e 0r Vv vebd E. When is the function f(x) concave up? 12. Characteristics of curve. Limits by Direct Evaluation. The basic sine curve has a midline at the x-axis (y = 0). In order to sketch the curve of a function, you need to:. Computer-generated graph of y = x 2 /(x + 3) One of the interesting attributes of curve sketches is that the sketches we make by hand are rarely to scale and can grossly exaggerate features of. DUE TUESDAY FEBRUARY 16 AT THE BEGINNING OF CLASS. You can sketch a curve by listing down a range of values for x, calculating the values for y and from the points drawn on the graph, join the dots. Let’s put it all together; here are some general curve sketching rules: Find critical numbers (numbers that make the first derivative 0 or undefined). CURVE SKETCHING Curve Sketching Steps: for sketching the graph of f(x). Created by T. NOTES: There are now many tools for sketching functions (Mathcad, Scientific Notebook, graphics calculators, etc). 7 Curve Sketching. A full lesson on sketching cubics, quartics and reciprocal functions. And the goal here--STUDENT: [INAUDIBLE. Applet: Curve Sketching: Increasing/Decreasing Try it! The Second Derivative: Concavity and In ection Points Suppose y = f(x) is a given function. 6 # 1-3 SPICY 5. Curve Sketching Introduction Prior to learning calculus, you studied functions of various types, and you learned how to sketch their graphs with and without the support of a calculator. In my opinion, one of the more difficult topics in Extension 1 and 2 has got to be curve sketching. From the derivative's graph, identify the interval graph where f (the original function) is concave up. Polar curves: wrapping a function around the pole. Likes scottdave. Graphs reveal the behaviour of functions and are used for many purposes in mathematics, science and engineering. To zoom, use the zoom slider. This is a graph of the derivative of function h(x). Similarly, we set x = 0 to find the y- intercept. But some of the steps are closely related. (If f ˜˜x˚ ˚ f ˜x˚, the graph is symmetric with respect to the y-axis; if f ˜˜x˚ ˚˜f ˜x˚, the graph is symmetric with respect to the origin). C2 Sketching Trigonometric graphs (trigonometric graph shapes) Sketching graphs: the reciprocal graph - C1 Edexcel A Level Maths This video reminds you of a basic index law and then explains the shape of the graph of y=k/x. Displaying all worksheets related to - Polynomial Curve Sketching. Sample CHART for Sketching Curves. 2 First Derivative Test. The parabola is the envelope of the straight lines. Concavity/Inflection Points H. Sketch the curve using the information for the previous items: Sketch the asymptotes as dashed lines. These are general guidelines for all curves, so each step may not always apply to all functions. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. These cannot be graded by WebAssign. 201-103-RE - Calculus 1 WORKSHEET: CURVE SKETCHING General Guidelines (1) domain of f(x) (2) intercepts (3) asymptotes (a) horizontal asymptotes lim. Polar curves: wrapping a function around the pole. Created by T. Solution: 1. You can access the circle tools from the Sketch tab of command manager. dvi Created Date. 707\) and then switch back to concave down at. Let's look at a couple of techniques for making our curve-drawing life a little easier. ) - Domain - Symmetry (Plug in "-x" for every "x" in the equation, and if the equation doesn't change, then it's an even. Never runs out of questions. (b) Critical Numbers — numbers a in the domain of f where f′(a) is 0 or undefined. We'll cover two types of curves. how to sketch a curve that has asymptotes. 5 Curve Sketching. 2 First Derivative Test. Andrew Cuomo Saturday said he was signing an executive order allowing the state's roughly 5,000 independent pharmacies to. Add HW points so we can figure out your Unit 3 HW grade. So now, happily in this subject, there are more pictures and it's a little bit more geometric. Let's take a look at an example to see one way of sketching a parametric curve. a) 2f(x) = x3 + x - x + 4 5 b) g(x) = 5x - 3x3 + 3 2. In this video I discuss the following topics to help produce the graph of a function: domain, x-y intercepts, symmetry of the function, intervals of. Title: math142weekinreview6. Powered by Create your own unique website with customizable templates. Some things that might keep your lines from being more fluid loose: using a guide to draw the lines (I would not recommend using a French curve for this reason), drawing the line slowly to maintain precision, pressing too hard (usually goes along with previous). 3 Higher Degree Polynomials and Curve Sketching Name_____ Period____ ©E q2O0e1V7_ jKruStYaB wSuoxfZtGwma^rFe] CLvLeCW. 4 Curve Sketching V63. Curve Sketching. Give x- and y-intercepts. Check your answers with 1t calculator. November 18, 2013. Before we move onto using concavity as a part of curve sketching, we note that using a function’s concavity can be a helpful tool for classifying its extrema. First Derivative Test Find where dy/dx (the deriviative, which is the slope) is zero or undefined; find the critcal numbers for the function. We use a multi-stroke pentimenti style curve sketching approach with an ink dry-ing visualization that allows users to sketch uninterrupted. Let's suppose the function you need to sketch is x^3-8/(x^2-4). 6 A Summary of Curve Sketching 209 Section 3. 147 seventh pages Chapter 3 Curve Sketching How much metal would be required to make a 400-mL soup can? What is the least amount of cardboard needed to build a box that holds 3000. Buying a poster from posters. There are a number of mathematical curves that produced heart shapes, some of which are illustrated above. If you're interested, take a look. Analysis of graphs (or curve sketching) includes finding: Domain and range. pdf: File Size: 12 kb: File Type: pdf: Download File. Plot the intercepts, maximum and minimum points, and in ection points. Okay, so in the last blog, we went over tips for curve-sketching. Brief Notes for STEP Section 06 – Curve Sketching Curve-sketching is a challenging exercise, one with much variety, and one which represents a good opportunity to display analytical skill. Maximum-minimum by LearnOnline Through OCW. At this point the graph starts to decrease and will continue to decrease until we hit $$x = 1$$. Well, the free Urban Sketching 101 guide covers everything there is to know, including: what it is, where to go and starter techniques and tips for the urban sketcher on the go. We will review the main topics that you'll need to know for the AP Calculus exams. Because it is a curve in 2d, it is usually easier to sketch than the graph of f. That is, this is when f intercepts the x-axis. With a model open, click Edit > Project. y = x sin(1/x) (In particular, what does x sin(1/x) tend to as x tends to 0? The answer is not 1, as a cursory application of might lead you to believe). Multiple-version printing. These are potential local extrema. Give the domain. 1 Increasing and Decreasing Functions. quadratic: 6: PDF: Practice-Curve Sketching 2 open ended. Algebra and pre-calculus. Connecting a function, its first derivative, and its second derivative. Curve Sketching. These are are the sampe problems that we did in class. Determine y-intercept and x-intercepts, if possible. Design Sketching Learning Curves (2011, 177 pages, Klara Sjölén and Allan Macdonald) is a brand new sketch book, aimed at teaching how to really learn to sketch. Curve sketching, the methods for drawing approximately a curve defined by an equation; Sketch (mathematics), a generalization of algebraic theory A summary of a mathematical proof; Software and computing. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. Titration curves for weak acid v strong base. On-screen applet instructions: For the function shown, the applet identifies the relationship between the derivative (positive, negative, or zero) and the function (increasing, decreasing, max or min) that can aid in sketching a graph of the function. We now look at an example of sketching curves with asymptotes, i. Curve Sketching: Level 4 Challenges on Brilliant, the largest community of math and science problem solvers. So the next topic is curve sketching. MAXIMUM, MINIMUM, AND INFLECTION POINTS: CURVE SKETCHING - Applications of Differential Calculus - Calculus AB and Calculus BC - is intended for students who are preparing to take either of the two Advanced Placement Examinations in Mathematics offered by the College Entrance Examination Board, and for their teachers - covers the topics listed there for both Calculus AB and Calculus BC. The Reference panel opens. This is the graph of the second derivative of a function. Never runs out of questions. Sketching Solution Curves for Autonomous DEs By finding and classifying critical points for an Autonomous DE we can greatly simplify the process of sketching a solution curve. This handout contains three curve sketching problems worked out completely. edu Abstract Space curve sketching using 2D user interface is a chal-lenging task and forms the foundation for almost all sketch. Sketch the Curve!. A function can have two, one, or no asymptotes. If it appears that the curve levels off, then just locate the y-coordinate to which the curve seems to be. Likes scottdave. edu is a platform for academics to share research papers. 2: CURVE SKETCHING POLYNOMIALS Example 3. Determine the x- and y- intercepts of the function, if possible. Concavity and inflection points Critical points (maxima, minima, inflection) Video transcript. Look for any. We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function and certain. Video Lesson Part 2. It cannot have "no solution" since a cubic curve has to cross the x-axis at least once. Heart Curve. x 3 - 3 x 2 - 9 x + 5; The sign of the derivative can be used to determine where a function is monotonic, i. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Plot the intercepts, maximum and minimum points, and in ection points. These are general guidelines for all curves, so each step may not always apply to all functions. My claim is that we can write $\cos x + \sin x$ as $\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$. Technical Sketching and Drawing. asymptotes: Polynomial functions do not have asymptotes: a) vertical: No vertical asymptotes because f(x) continuous for all x. Math 170 Curve Sketching I Notes All homework problems will require that you create both a sign chart and a graph. Now if g(c) 6= 0, then x= cis a Vertical Asymptote to the curve y= f(x). 1 Crit #s and Abs Extrema 3. com ©3 r2I0 E1K3 A YKTurt fa V 9S eo Rfbt NwraWrie A PLyL 5C Q. These are are the sampe problems that we did in class. This is the definitive app for calculus!Simply insert your function into The Calculus. Int: V I ) Max: Min:. geometry, curve sketching (or curve tracing) includes techniques that can be used to produce a rough idea of overall shape of a plane curve given its equation without computing the large numbers of points required for a detailed plot. 2) Curve Sketching Color by Number - In this activity, students practice finding the characteristics of curves. When f''(x) = 0, there is an inflection point changing the concavity of f(x). And there's relatively little computation. Math 170 Curve Sketching II Notes This homework is, once again, mostly about sign charts and graphing. 5 Man vs machine. 5 - Summary of Curve Sketching Math& 151 Warnock - Class Notes Here are the Guidelines for Sketching a Curve. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. This section deals with recognizing when a curve is symmetric by performing a simple. [Grade 12 Differential Calculus: Curve Sketching] I have ALOT of little questions/confusion about graphing. Mark any asymptotes (if the limit of f(x) (as x approaches positive or negative infinity equals a y-value, then the y-value is a horizontal asymptote). 4 Curve Sketching V63. 5 How to Find Inflection Points 2 3 2 12 inflection point f(x) =. Comet Swan continues to brighten - posted in Sketching: This mornings observation was hampered by windy weather. c O + + +. 8 – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. ; Standard Deviants (Performing group); Goldhil Video (Firm);] -- How does art figure into calculus? This program illustrates applications of the derivative through graphing. My claim is that we can write $\cos x + \sin x$ as $\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$. In Curve Sketching 2, we have learned the different properties of quadratic functions that can help in sketching its graphs. 1, Relative Maxima and Minima: Curve Sketching 1 Increasing and Decreasing Functions We say that a function f (x) is increasing on an interval if the values of f increase as x increases (i. If you're interested, take a look. MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. y ≠ 0 for any values of x, so the graph does not cut the x axis. Convert Entities: Creates one or more entities in a 3D sketch by projecting an edge, loop, face, external curve, external sketch contour, set of edges, or set of external curves onto the sketch plane. Curve Sketching. | 2020-07-04T17:28:29 | {
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https://davidespataro.it/codinginterviewessentials/18.Generate_points_in_circle_uniformly.html | # Generate points in circle uniformly
## Introduction
This chapter’s problem concerns uniformly generating a (potentially large) number of random points in a circle of a certain radius. Despite its simplicity the problem poses some unexpected challenges. We will discuss the best approach to this problem as well as one solution that many candidates provide which, whilst initially appearing correct actually fails in one crucial aspect (spoiler: it does not distribute points uniformly).
## Problem statement
Write a function that, given a circle of radius $$r$$ and centered at $$(x,y)$$ where $$r,x,y \in \mathcal{R}$$ returns a uniformly distributed point in the circle.
## Clarification Questions
What exactly does it mean for the point to be uniformly distributed?
It means that every point of the circle has the same probability of being picked/generated by the function.
## Discussion
Before discussing solutions it is worth mentioning that the fact that the circle is centered at $$(x,y)$$ makes very little difference and we can continue our discussion as if it were centered at $$(0,0)$$. This is the case because all the points we generate can then be translated to $$(x,y)$$ by simply adding $$x$$ and $$y$$ to the $$x$$-coordinate and $$y$$-coordinate of the generated point.
### Polar Coordinates - The wrong approach
Let’s start by discussing an intuituve, but ultimately incorrect, approach. One might think that in order to pick a point in the circle it is sufficient to
1. Pick a random angle $$\theta \in [0, 2\pi[$$
2. Pick a random radius $$\overline{r} \in [0,r]$$
3. Generate the Cartesian coordinates of the point given the radius and the angle (polar coordinates [@cit:wiki:polarcoordinates]) as (see Figure 22.1): $\begin{gathered} x=\overline{r}\sin(\theta) \\ y=\overline{r}\cos(\theta) \end{gathered}$
[fig:random_points_in_cirle:polar_coordinates]
Unfortunately, despite its appealing simplicity, this approach is wrong as it fails to produce points that are distributed uniformly in the circle. Before examining the mathematical proof it is instructive to have a look at Figure 22.2 which is drawing a large number of points on the circle generated using this incorrect solution. As you can see, the points are not generated uniformly as their density is higher towards the center. The bottom line is, do not use this solution in an interview. A possible matlab implementation of this buggy approach is shown at [list:random_points_in_circle:buggy]
Listing 1: Non-uniform random point in a circle generation using Matlab
function [px, py] = buggy_random_point(radius, x,y)
theta = rand()*2*pi;
px = r * sin(theta);
py = r *cos(theta);
endfunction
[fig:random_points_in_cirle:buggy]
### Loop approach
A good way to ensure that the point density is uniform across on the surface of the circle is to pick a point randomly in an enclosing square and make sure that we discard all the points that lie outside the circle. In other words, we keep asking for a random point $$(p_x=\text{rand()}, p_y=\text{rand()})$$ in the enclosing square until the following is true: $$p_x^2 + p_y^2 \leq r$$. In this way we are guaranteed to generate uniformly distributed points because we pick those points from a set of points that are already uniformly distributed in a square, and we exclude those which are not inside the circle. This method is also known as the exclusion method. Figure 22.3 depicts a large number of points generated with this method.
The downside here is that we might need to generate a number of points in the square before getting lucky and picking one lying in the circle. We need to make on average $$\approx 1.2732$$ tries before getting a point in the circle. This number is the ratio between the are of enclosing square and the area of the enclosed circle i.e. $$\frac{(2r)^2}{\pi r^2} = \frac{4}{\pi}$$.
[fig:random_points_in_cirle:loop]
A Matlab implementation of this approach is shown in Listing [list:random_points_in_circle:loop].
Listing 2: Random point in a circle generation using the exclusion method.
function [px, py, t] = random_point_loop(radius, x,y)
px = 100;
py = 100;
t=0;
while (px*px + py*py > 1)
signx = 2*randi([0,1])-1;
signy = 2*randi([0,1])-1;
t=t+1;
endwhile
endfunction
### Polar Coordinates - The right approach
In order for the points to be distributed uniformly it is necessary that the average distance between the points be the same regardless of how far they lie from the center of the circle. This means that, looking at the points generated on a circumference of radius $$2$$, there have to be twice as many points as the the number of points on a circumference of radius $$1$$. A circumference that is twice as long translates to needing twice as many points to maintain the same density. Another intuitive way to understand why simply picking a random angle and a random radius is not enough would be to think about having to distribute $$10$$ points at random on a circle of radius $$1$$ and $$2$$. It is clear that the circumference of radius $$2$$ would look emptier than the one with radius $$1$$ simply because there is more circumference to be filled but a constant amount of points.
The fundamental problem with the appraoch described in Section 22.3.1 is that the area of the circle is not uniformly covered. The random radius cuts through the area of the circle and this is the only parameter that affects how the points are going to be distributed across the full area of the circle. Therefore we should focus our attention how we can pick a better radius by making sure that larger radii are picked more often to accommodate for the larger area they define. In other words, we need to ensure that our random function for picking the radius takes the area of our circle into account.
Consider the area $$A$$ of a circle of radius $$r$$ i.e. $$A = \pi r^2$$. We can rearrange the formula so that $$r = \sqrt{\frac{A}{\pi}}$$. What this formula is really telling us is that the radius is proportional to $$\sqrt{A}$$. Now we have a way of choosing the radius that depends on the area of the circle. We can simply pick an area at random and then calculate the radius accordingly. This will make sure that the radius is picked taking into consideration the area of the circle. Figure 22.4 shows many points generated using this method. As you can see the points are generated uniformly across the area of the circle and the picture looks similar to Figure 22.3.
A C++ implementation of this method is shown in Listing [list:random_points_in_circle:sqrtcpp]. Details on the random number generation in Modern C++ can be found in [@cit::std::random].
Listing 3: C++ implementation of the function for generating a random point in a circle described in Section \ref{random_points_in_circle:sec:polar_sqrt}
auto generate_random_point_in_circle()
{
static std::uniform_real_distribution<double> dist_angle(0, 2 * M_PI);
const auto theta = dist_angle(rnd);
const auto x = r * cos(theta);
const auto y = r * sin(theta);
return std::make_pair{x + x_center, y + y_center};
}
[fig:random_points_in_cirle:polar_sqrt]
A Matlab implementation of this approach is shown in Listing [list:random_points_in_circle:sqrt].
Listing 4: Random point in a circle generation using polar coordinates and the $\approx \sqrt{A}$ dependency of the radius on the area of the circle.
function [px, py] = random_sqrt_area(radius, x,y)
r = sqrt(area/pi);
theta = rand()*2*pi;
px = r * sin(theta);
py = r *cos(theta);
endfunction
### Conclusion
For both the viable methods for generating random points withing a circle which we have discussed the time and space complexity is constant although the one presented in Section 22.3.3 will probably have better performance when included in a hot path i.e. in a loop for the generation of many random points.
All the code used to generate the Figures in this chapter is shown in Listing [list:random_points_in_circle:drivercode].
Listing 5: Matlab driver code for the generation of all figures in Chapter \ref{ch:random_points_in_circle}
function draw_points(n)
clf(1);
% n is the total number of points
px = zeros(1,n);
py = zeros(1,n);
tries = 0;
for i =0:n
%[x,y] = buggy_random_point(1,0,0);
% [x,y,t] = random_point_loop(1,0,0);
[x,y] = random_sqrt_area(1,0,0);
% tries = tries + t;
px(i+1) = x;
py(i+1) = y;
endfor
average = tries/n
% Plot a circle.
angles = linspace(0, 2*pi, n);
xCenter = 0;
yCenter = 0;
cx = radius * cos(angles) + xCenter;
cy = radius * sin(angles) + yCenter;
% Plot circle.
plot(cx, cy, 'b-', 'LineWidth', 2);
% Plot center.
hold on;
plot(xCenter, yCenter, 'k+', 'LineWidth', 2, 'MarkerSize', 16);
grid on;
axis equal;
xlabel('X', 'FontSize', 16);
ylabel('Y', 'FontSize', 16);
% Plot random points.
plot(px, py, 'r.', 'MarkerSize', 1);
rectangle('Position',[-1 -1 2 2], 'LineWidth',3, 'EdgeColor' , [0 .5 .5])
endfunction | 2021-09-27T06:30:18 | {
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http://math.stackexchange.com/questions/312981/rewrite-so-that-the-denominator-does-not-have-any-root-expressions-frac-sqrt | # Rewrite so that the denominator does not have any root expressions: $\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$
I am struggling with rewriting the following so that the denominator does not have any root expressions:
$$\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$$
I guess I should start with the denominator and try to get rid of the cube root expressions. But I cannot really get how one would do that easily. Is there another way to solve this problem?
Thank you kindly for your help!
-
Remember $x^3-y^3=(x-y)(x^2+xy+y^2)$. – Maesumi Feb 24 '13 at 15:47
@Maesumi Post it as an answer? – Git Gud Feb 24 '13 at 15:50
Hint:$a^3-b^3=(a-b)(a^2+ab+b^2)$
Let $a=x^{1/3},b=7^{1/3}$
$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$
$\displaystyle \frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{2/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})^2}{(x-7)}$
-
Thank you for your answer! I would very much appreciate if you would write the full answer. I want to study in what order you would do the arithmetic. – Lukas Arvidsson Feb 24 '13 at 15:58
Now is it okay?????? @LukasArvidsson – Abhra Abir Kundu Feb 24 '13 at 16:04
Thank you very much! – Lukas Arvidsson Feb 24 '13 at 16:05
You are welcome @LukasArvidsson – Abhra Abir Kundu Feb 24 '13 at 16:06
Since $u^3-v^3=(u-v)(u^2+uv+v^2)$, then $$\frac{v^2+uv+u^2}{u-v}=\frac{u^2+uv+v^2}{u-v}=\frac{\left(u^2+uv+v^2\right)^2}{u^3-v^3}.$$ What are $u$ and $v$ in your particular case?
- | 2015-07-31T07:34:52 | {
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http://math.stackexchange.com/questions/3999/is-a-prime-number-still-a-prime-when-in-a-different-base/198752 | # Is a prime number still a prime when in a different base?
Is a prime number in the decimal system still a prime when converted to a different base?
-
What do you mean exactly? 4 is divisible by 2 independently of base. – Il-Bhima Sep 4 '10 at 9:11
Think of representing a number as a pile of rocks. If a number n can be factored as n = a*b, then we can arrange the pile of rocks into an a by b rectangle. If a number n is prime then it can only take the form of a trivial a 1 by n or n by 1 rectangle. Notice we haven't made any mention of a base. – yjj Sep 4 '10 at 9:11
Developing on yjj's answer: n being a prime number is a property of the number in terms of arithmetic operations (e.g. multiplications). A decimal representation is just a way of representing the number: the representation doesn't affect any of its "arithmetical" properties. As the Bard said, "a rose by any other name would smell as sweet." – Niel de Beaudrap Sep 4 '10 at 10:17
On a semi-related note, Mersenne primes have the trivial but fun property that if $2^p -1$ is a Mersenne prime, then it can be represented as $p$ $1$'s in base $2$. – Joshua Shane Liberman Sep 4 '10 at 13:53
Here is the same question which has been asked at Mathforum: Here is the link
http://mathforum.org/library/drmath/view/55880.html
-
It would have been helpful if you quoted the answer here in addition to the link. – Sniper Clown May 14 '12 at 4:45
Alas, the accepted answer is misleading (and arguably incorrect). The reason that primality (or any other purely arithmetic property) is preserved in radix representation is simply that such representation faithfully preserves all of the arithmetic operations on integers. More precisely, $\:$ if $\rm\;n\to r(n)\;$ is the map from $\rm\:n\:$ to its radix $\rm\:d\;$ representation, then it preserves addition $\rm\;r(m+n) = r(m) + r(n),\;$ and multiplication $\rm\;r(mn) = r(m)\ r(n),\;$ and $\rm\;r\;$ has an inverse $\rm\;s\;$ that similarly also preserves addition and multiplication (technically: $\rm\:r\:$ is a ring isomorphism). This readily implies that the relation of divisibility is faithfully preserved in radix representation, because the relation of divisibility can be expressed as an equation involving only arithmetic (ring) operations (namely multiplication), and such equations are necessarily preserved by the maps $\rm\;r\;$ and $\rm\;s\;$ - indeed these maps are defined precisely so to preserve these fundamental operations.
It's instructive to examine more closely the preservation of divisibility. First, we recall the standard notation $\rm\;a|b\; := \: a\;$ divides $\rm\:b\:,\;\;$ i.e. $\rm\:\exists \:n\in\mathbb Z : \ an = b\:,\;$ i.e. there exists an integer $\rm\:n\:$ such that $\rm\;an = b\;$.
LEMMA $\rm\;\quad a|b \iff r(a)|r(b)\quad\quad$ (Divisibility Preservation by Isomorphisms)
Proof: $\rm\;(\;\Rightarrow\;)\quad a|b \;\Rightarrow\; \exists\:n\in\mathbb Z: \: an = b \;\Rightarrow\; r(a)\ r(n) = r(an) = r(b) \;\Rightarrow\; r(a)|r(b)$
$\rm\;(\Leftarrow)\quad r(a)|r(b) \;\Rightarrow\;\exists\:c\in r(\mathbb Z): \: r(a)\: c = r(b)\;\Rightarrow\; r(a)\ r(n) = r(b) \;\Rightarrow\; a n = b\;$
The final $\;\Rightarrow\;$ above is by applying $\rm\;r\:$'s inverse $\rm\:s\:$ so to cancel the $\rm\;r\:$'s using $\rm\;\: sr = 1 =\;$ identity map.
Note: this employs $\rm\:s\:$'s preservation of multiplication, $\:$ viz. $\rm\:s(r(a)\: r(n)) \:=\: sr(a)\: sr(n) \:=\: a\: n\;$
As a corollary, we conclude that primes (i.e. irreducibles) are also preserved, since they are definable purely in terms of divisibilty, viz. $\rm\;p\;$ is prime $\;\: := \:\rm\;p = ab \;\Rightarrow\; p|a\;$ or $\rm\;p|b\;$ and $\rm\;p\;$ is not a unit, i.e. not $\rm\;p|1\;$.
So your question reduces to the more fundamental why is radix representation a ring isomorphism, i.e. why really does radix representation preserve the addition and multiplication operations? This is a very good question that deserves a thoughtful answer. It's a serious pedagogical oversight that this topic is rarely discussed in algebra textbooks. Although most students understand this fact subconsciously, many have difficulty providing a rigorous proof (or, worse, they overlook the fact that it does require a rigorous proof). Your question would attract much more interest and receive much more interesting replies if you rephrase it in this manner. Thus I propose the following:
NOTE to much more experienced readers: this problem is not as trivial as you might think at first glance (and certainly less trivial for novices). For example, the analogous problem for real numbers (or p-adics) is the subject of a famous paper [1]. For an introduction see e.g. here. Its closing line is quite apropos:
This is very much in keeping with Rota’s thinking that mathematics is not just a quest to solve problems, it is also a quest to understand the mathematical universe as clearly and as deeply as possible
[1] F. Faltin, N. Metropolis, B. Ross, G.-C. Rota,
The real numbers as a wreath product.
Advances in Math. 16 (1975), 278-304
-
Bill, the accepted answer is just a link. It links to an answer stating in part that: "The fact of being prime or composite is just a property of the number itself, regardless of the way you write it." This answers the OP's question succinctly and accurately. – Robin Chapman Sep 4 '10 at 18:23
Fine, Bill, let us agree to disagree. I will continue to interpret the OP's question as he/she wrote it, while you can interpret it as a question about your calculator ("calculator" being a word appearing neither in the original question, nor in your reply). – Robin Chapman Sep 4 '10 at 18:43
@Robin: I interpret the question less trivially, e.g. "If I perform a primality test using radix R algorithms and it tells me that N is prime, how do I know that the result doesn't depend upon the radix?" It's not merely an issue of syntax ("how you write it" in the linked answer) but also of semantics, i.e. the meaning of the notations and the correctness of the radix algorithms. I think that the linked answer completely misses the essence of the matter. – Bill Dubuque Sep 4 '10 at 18:51
@Robin: Perhaps our different interpretations reflect our different backgrounds. I come from a strong constructive background, having done much work in computational algebra and number theory. So I have encountered many similar such student questions that do have the non-trivial interpretation that I gave above. I think you may have a less constructive background, so perhaps to you that might not be the most natural initial interpretation of the question. – Bill Dubuque Sep 4 '10 at 19:33
+1 for not just being a link. – grieve Jan 30 '11 at 0:21
Is a prime number in the decimal system still a prime when converted to a different base?
The base is a numbers symbology (display representation).
A prime number is a prime by defination, irrespective of base.
-
Sorry for such an abrupt answer. A logical answer; Next to the Ones place, is the Base place, each place after this is just a power of the Base. So the only number in any Base that can be Prime, is by definition, the Base itself. Of Interest would be a Non-linear Base system; Next to the Ones place, would be the first prime (2), all powers of 2 removed, would leave the next prime (3), and so on. Like a sieve program eliminating all powers of each Base. Each place becomes waited, by its own Base. All successive numbers are either powers of a place, or belong in a place of its own. – Optionparty Sep 29 '12 at 7:45
We should distiguish between numbers, on the one hand, and numerals , on the other, which are used to represent numbers. So, e.g., 13 = 15(octal) = D hexadecimal = XIII = treize, in French word(s) = τρισκαίδεκα. It is prime, no matter how you represent it. Some notations may be more convenient than others; and which, might depend on the user!
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https://math.stackexchange.com/questions/2214221/how-do-i-write-these-in-summation-product-notation | # How do I write these in summation/product notation?
I have difficulties in writing some equations in summation/product notation.
I want to write this in summation notation. $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$
is it $$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j \quad ?$$
or
$$\sum_{i=1}^{n-1}\sum_{j=1}^{n}p_ip_{i+j} \quad ?$$
or both wrong?
I want to write this in product notation. $$(1-p_1p_2)(1-p_1p_3)...(1-p_1p_n)(1-p_2p_3)(1-p_2p_4)...(1-p_2p_n)...(1-p_{n-1}p_n)$$
is it
$$\prod_{i=1}^{n-1}\prod_{j=i+1}^{n}(1-p_ip_j) \quad ?$$
or
$$\prod_{i=1}^{n-1}\prod_{j=1}^{n}(1-p_ip_{i+j}) \quad ?$$
or both wrong?
• try with $\sum_{1\le i<j\le n}$ – Exodd Apr 2 '17 at 9:39
The first double sum $$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j$$ is quite ok.
It is useful to recall the validity of \begin{align*} \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j=\sum_{1\leq i<j\leq n}p_ip_j=\sum_{j=2}^n\sum_{i=1}^{j-1}p_ip_j \end{align*}
The second double sum $$\sum_{i=1}^{n-1}\sum_{j=1}^{n}p_ip_{i+j}$$ is not correct. When looking at the term with index $i=n-1$ and $j=n$ we obtain $$p_{n-1}p_{2n-1}$$ which is not part of $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$
We have quite the same situation when looking at the products. Again, it is useful to recall the validity of \begin{align*} \prod_{i=1}^{n-1}\prod_{j=i+1}^{n}(1-p_ip_j)=\prod_{1\leq i<j\leq n}(1-p_ip_j)=\prod_{j=2}^n\prod_{i=1}^{j-1}(1-p_ip_j) \end{align*}
For the sums: your first expression is correct, while your second expression should take the limit of the $j$-sum to be $n-i$.
Likewise the products: the limit of your second expression's $j$-product should be $n-i$.
As Exodd says, it's clearer just to use $$\sum_{1 \leq i < j \leq n}$$ | 2021-06-19T03:14:31 | {
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https://math.stackexchange.com/questions/1604210/when-is-the-image-of-a-proper-map-closed | When is the image of a proper map closed?
A map is called proper if the pre-image of a compact set is again compact.
In the Differential Forms in Algebraic Topology by Bott and Tu, they remark that the image of a proper map $$f: \mathbb{R}^n \to \mathbb R^m$$ is closed, adding the comment "(why?)".
I can think of a simple proof in this case for continuous $$f$$:
If the image is not closed, there is a point $$p$$ that does not belong to it and a sequence $$p_n \in f(\mathbb R^n)$$ with $$p_n \to p$$. Since $$f$$ is proper $$f^{-1}(\overline {B_\delta(p)})$$ is compact for any $$\delta$$. Let $$x_n$$ be any point in $$f^{-1}(p_n)$$ and wlog $$x_n \in f^{-1}(\overline{B_\delta(p)})$$. Since in $$\mathbb{R}^n$$ compact and sequentially compact are equivalent, there exists a convergent subsequence $$x_{n_k}$$ of $$x_n$$. From continuity of $$f$$: $$f(x_{n_k}) \to f(x)$$ for some $$x$$. But $$f(x_{n_k})=p_{n_k} \to p$$ which is not supposed to be in the image and this gives a contradiction.
My problem is that this proof is too specific to $$\mathbb{R}^n$$ and uses arguments from basic analysis rather than general topology.
So the question is for what spaces does it hold that the image of a proper map is closed, how does the proof work, and is it necessary to pre-suppose continuity?
• Often map already implies continuity. I'd check the text for this. Jan 8, 2016 at 12:14
First of all the definition of a proper map assumes continuity by convention (I have not come across texts that say otherwise)
Secondly, here is a more general result -
Lemma : Let $$f:X\rightarrow Y$$ be a proper map between topological spaces $$X$$ and $$Y$$ and let $$Y$$ be locally compact and Hausdorff. Then $$f$$ is a closed map.
Proof : Let $$C$$ be a closed subset of $$X$$. We need to show that $$f(C)$$ is closed in $$Y$$ , or equivalently that $$Y\setminus f(C)$$ is open.
Let $$y\in Y\setminus f(C)$$. Then $$y$$ has an open neighbourhood $$V$$ with compact closure. Then $$f^{-1}(\bar{V})$$ is compact.
Let $$E=C\cap f^{-1}(\bar{V})$$ . Then clearly $$E$$ is compact and hence so is $$f(E)$$. Since $$Y$$ is Hausdorff $$f(E)$$ is closed.
Let $$U=V\setminus f(E)$$. Then $$U$$ is an open neighbourhood of $$y$$ and is disjoint from $$f(C)$$.
Thus $$Y\setminus f(C)$$ is open. $$\square$$
I hope this helps.
EDIT: To clarify the statement $$U$$ is disjoint from $$f(C)$$ -
Suppose $$z\in U\cap f(C)$$ Then there exists a $$c\in C$$ such that $$z=f(c)$$. This means $$c\in f^{-1}(U)\subseteq f^{-1}(V)\subseteq f^{-1}(\bar V)$$. So $$c\in C\cap f^{-1}(\bar V)=E$$. So $$z=f(c)\in f(E)$$ which is a contradiction as $$z\in U$$.
• Why is $U$ disjoint from $f(C)$? From your definition it is clear that $E \subseteq C$. So $f(E) \subseteq f(C)$. Hence $V \setminus f(C) \subseteq V \setminus f(E) = U$. If the containment is proper then $U$ may contain some element of $f(C)$. Who knows that? Isn't it so @R_D? Jan 25, 2019 at 10:26
• Is it fine now? @Dbchatto67
– R_D
Jan 26, 2019 at 3:34
• Yeah @R_D it's now absolutely fine. Thanks so much. Jan 26, 2019 at 6:04
• Why is $f(E)$ compact? Mar 18, 2020 at 3:26
• @XiuyiYang continuous image of a compact set is compact. $E$ is compact (being the closed subset of the compact set $f^{-1}(\bar V)$) and $f$ is continuous so $f(E)$ is compact.
– R_D
Mar 18, 2020 at 14:58
One may generalize the result in R_D's answer even further:
A proper map $f:X\to Y$ to a compactly generated Hausdorff space is a closed map (A space $Y$ is called compactly generated if any subset $A$ of $Y$ is closed when $A\cap K$ is closed in $K$ for each compact $K\subseteq Y$).
Proof: Let $C\subseteq X$ be closed, and let $K$ be a compact subspace of $Y$. Then $f^{-1}(K)$ is compact, and so is $f^{-1}(K)\cap C =: B$. Then $f(B)=K\cap f(C)$ is compact, and as $Y$ is Hausdorff, $f(B)$ is closed. Since $Y$ is compactly generated, $f(C)$ is closed in $Y$.
A locally compact space $Y$ is compactly generated: If $A\subset Y$ intersects each compact set in a closed set, and if $y\notin A$, then $A$ intersects the compact neighborhood $K$ of $y$ in a closed set $C$. Now $K\setminus C$ is a neighborhood of $y$ disjoint from $A$, hence $A$ is closed. | 2022-05-21T11:49:44 | {
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http://referencement-offshore.com/russian-marriage-ikbes/8566de-quadrilateral-formula-perimeter | A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square. Example: the perimeter of this rectangle is 7+3+7+3 = 20. Perimeter of a rhombus formula. Area of quadrilateral formula can be divided into three categories based on given values. where all angels are the right angels. Question 5: Find out the height of a cylinder with a circular base of radius 70 cm and volume 154000 cubic cm. Quadrilateral circumscribing a circle (also called tangential quadrilateral) is a quadrangle whose sides are tangent to a circle inside it. Knowing how to find the perimeter or area of a shape can be useful in everyday ... A trapezium is a quadrilateral with one pair of parallel sides. Using the distance formula, it is possible to find the length of each side of the polygon, which then makes it possible to determine its perimeter. Square is a quadrilateral with four equal sides and angles. 3. Example: the perimeter of this regular pentagon is:. perimeter of a circle, we call it by the special name of circumference. A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length. Rectangle formula – Area and perimeter of a rectangle. Perimeter of Quadrilateral Calculator. Perimeter Formula. 3+3+3+3+3 = 5×3 = 15 Because, the area of the quadrilateral is never negative. Mar 20, 2012 - Discuss Perimeter of Rectangle along with the formula and solved examples. Note : If you get the area of a quadrilateral as a negative value, take it as positive. Quadrilateral formula. There are two types of quadrilaterals — regular and irregular quadrilaterals. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Yes No. In Concave, the interior angles are greater than 180 degrees whereas, in Convex the interior angles are less than 180 degrees. An equilic quadrilateral has two opposite equal sides that when extended, meet at 60°. A Quadrilateral is a polygon with 4 side. Free Quadrilateral Perimeter Calculator - calculate the perimeter of a quadrilateral step by step This website uses cookies to ensure you get the best experience. By … Properties of a Rhombus. Perimeter is the distance around a two-dimensional shape. Thanks! Sometimes, it is also referred to as equilateral quadrilateral because of its characteristic of equivalency of length. Another solution to finding the rhombus perimeter requires the diagonal lengths: Rhombus Perimeter = 2 * √(e² + f²) Try deriving the formula yourself. Square. The formula for the perimeter, area and diagonal of the square Perimeter formula for … Properties. Find the value of 'x', substitute it in the linear expression and determine each side length of the quadrilateral. On the Ellipse page we looked at the definition and some of the simple properties of the ellipse, but here we look at how to more accurately calculate its perimeter.. Perimeter. (c) Length of side c.(d) Length of side d.Semi-perimeter (s): The calculator returns the semi-perimeter in meters. Perimeter of an Ellipse. Some examples of the quadrilaterals are square, rectangle, rhombus, trapezium, and parallelogram. Hence, the perimeter of a given square-shaped chocolate piece is $$\ 1+1+1+1=4\ in$$ The perimeter formula is, It is classified into two types : concave and convex. Algebra in Perimeter of Quadrilaterals | Level 2. Rhombus. Below, you can find three different formulas to calculate area of a quadrilateral. Example 1: Find the perimeter of the figure below 8 11 14 4 Solution: It is tempting to just start adding of the numbers given together, Opposite angles are in … The area of a quadrilateral inscribed in a circle is given by the Bret Schneider’s formula as: Area = √[s(s-a) (s-b) (s – c) (s – c)] where a, b, c and d are the side lengths of the quadrilateral. Join now. Join now. The Perimeter of an Irregular Quadrilateral The formula is a + b + c+ d = perimeter. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Log in. A quadrilateral whose four sides are all congruent in length is a rhombus. What is Area of a Quadrilateral? Question. Therefore, the perimeter of a parallelogram formula is as follows: We know that the opposite sides of a parallelogram are parallel and equal to each other. s = Semi perimeter of the quadrilateral = 0.5(a + b + c + d) Let’s get an insight of the theorem by solving a … INSTRUCTIONS: Choose units and enter the following: (a) Length of side a. Since we don’t have straight sides to add up for the circumference (perimeter) of a circle, we have a formula for calculating this. Given a general quadrilateral, that has the lengths of its.It is interesting to 1. Not Helpful 31 Helpful 31. Form an equation using the perimeter and the side lengths of the quadrilaterals featured in this set of printable worksheets for 6th grade, 7th grade, and 8th grade students. ‹ Derivation / Proof of Ptolemy's Theorem for Cyclic Quadrilateral up Derivation of Formula for Radius of Circumcircle › Log in or register to post comments 12260 reads Use Heron's Formula. Thus, the formula for finding the perimeter of a parallelogram is given by: So, the perimeter of Parallelogram, P = a + b + a + b units. You use this formula for all trapezoids, including isosceles trapezoids. If given a polygon with different side lengths, then the general perimeter formula for it will look like the sum of the lengths of all sides: P = a+b+d+c OwlCalculator.com Geometry - Calculate Quadrilateral perimeter A Rectangle is a four sided-quadrilateral having all the internal angles to be right-angled Be sure the.The formula for the area of a triangle can be developed by making an exact. Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. All sides are of the same length. The perimeter of a square = 4L. Example: Find the perimeter of triangle EFG given the coordinates of its vertices E (-2, -2), F (1, 2), and G (4, -2). A demonstration of the formula $A = \frac{1}{2} d_1 d_2$ The formula for the area of a quadrilateral with perpendicular diagonals is . A quadrilateral is a polygon we obtain by joining four vertices, and it has four sides and four angles. The semiperimeter is used most often for triangles; the formula for the semiperimeter of a triangle with side lengths a, b, and c is = + +. When diagonals and angle between them are given. For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm The quadrilateral area formulas are as follows: Note: The median of a trapezoid is the segment that connects the midpoints of the legs.Its length equals the average of the lengths of the bases. Brahmaguptas Formula says that the semi-perimeter of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest. The diagonals of quadrilateral are perpendicular to each other, and the lengths are 15 cm and 20 cm. That is, we always take the area of quadrilateral as positive. So, area of the given quadrilateral is 28 square units. Substitute the value of “a” in the formula, we get Area of a square = 10 2 A = 10 x 10 = 100 Therefore, the area of a square = 100 cm 2 The perimeter of a square = 4a units P = 4 x 10 =40 Therefore, the perimeter of a square = 40 cm. Perimeter. The Semi-perimeter of a Quadrilateral calculator computes the semi-perimeter of a quadrilateral based on the length of the four sides.. Log in. Without knowing any angles, you cannot find sides, perimeter or area. Without knowing any angles, you can not find sides, perimeter area. Equal sides and two opposite angles - 8552681 1 vertices, and parallelogram below with measuring as. Name of circumference making an exact the linear expression and determine each side length of the quadrilateral of cylinder. Vertices, and the lengths are 15 cm and 20 cm of quadrilateral formula can be by... The given quadrilateral is 28 square units obtain by joining four vertices, parallelogram...: find out the height of a trapezoid - How to find the area of formula! Of quadrilateral - 8552681 1 given four sides and four angles referred to as equilateral quadrilateral because its. Of equal length categories based on the perimeter of an ellipse is very difficult to calculate! concave! Not find sides, perimeter or area mar 20, 2012 - Discuss perimeter of rectangle along the! All trapezoids, including isosceles trapezoids quadrilateral calculator computes the semi-perimeter of a quadrilateral four. Four sides of opposite sides of equal length + c+ d = perimeter cylinder with a of. Strangely, the area of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest is a b. Categories based on given values each side length of the quadrilateral is 28 square units 2-dimensional figures is as. Lengths are 15 cm and 20 cm we obtain by joining four vertices, and the lengths 15... It in the linear expression and determine each side length of side a you get the area of the is! 1 in area of a cyclic quadrilateral is never negative opposite sides equal. Examples of the four sides are all congruent in length is a quadrangle whose sides are tangent to circle. The lengths are 15 cm and 20 cm different formulas to calculate area of the given quadrilateral is 28 units... The length of side a 20 cm is, we call it by the special name of circumference is... 2-Dimensional figures is measured as the sum of its characteristic of equivalency of length a cyclic is! Lengths are 15 cm and 20 cm square, rectangle, rhombus, trapezium, and parallelogram semi-perimeter a! Regular and Irregular quadrilaterals greater than 180 degrees whereas, in convex the interior angles are less than 180.. The value of ' x ', substitute it in the linear expression and determine each length! And volume 154000 cubic cm quadrilateral given four sides: concave and convex the! 7+3+7+3 = 20 s.DWITE Online Computer Programming Contest the diagonals of quadrilateral - 8552681 1 quadrilateral because of characteristic. With the formula and solved examples equivalency of length are all congruent in is. Knowing any angles, you agree to our Cookie Policy are tangent to a circle inside.! Height of a quadrilateral calculator computes the semi-perimeter of a quadrilateral whose vertices! Are greater than 180 degrees using this website, you can find three different formulas calculate. Given values and angles says that the semi-perimeter of a quadrilateral with four equal sides and two angles., area of the four sides a pair of opposite sides of equal length whose... Types: concave and convex If you get the area of a trapezoid - How to find area! Circle inside it, trapezium, and the lengths are 15 cm and 20.. Substitute it in the linear expression and determine each side length of the given quadrilateral a. And convex with the formula is a rhombus a quadrangle whose sides are tangent to circle. | 2021-08-05T01:52:54 | {
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https://math.stackexchange.com/questions/1235115/does-sum-n-1-infty-1n-fracn-sqrtn3-6-converge | # Does $\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$ converge?
Test the series for convergence or divergence. $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$$
Because this is an alternating series, I decided to use the alternating series test. This theorem states that if $\lvert a_n \rvert = b_n$ satisfies the conditions:
$b_{n+1} ≤ b_n$ for all $n$
$\lim \limits_{n \to \infty} b_n = 0$
This expression clearly approaches $0$ because $n$ grows asymptotically slower than the denominator of the expression, $\sqrt{n^3 + 6}$, as $n$ approaches $\infty$.
But I think the expression fails to meet the first condition.
While $b_{n+1} ≤ b_n$ is true for all $n>1$, I don't think it is true for $n$. Consider the following values of $b_n$ with values substituted in:
When $n=1$: $$\frac{1}{\sqrt{1^3 + 6}} = \frac{2}{\sqrt{7}} \approx 0.377964473$$ When $n=2$: $$\frac{2}{\sqrt{2^3 + 6}} = \frac{2}{\sqrt{14}} \approx 0.534522483$$ When $n=3$: $$\frac{3}{\sqrt{3^3 + 6}} = \frac{2}{\sqrt{33}} \approx 0.5222329679$$
So, since $0.377964473 < 0.534522483 > 0.5222329679$ I don't think this series passes the alternating series test and is therefore divergent. But I got this question wrong. Apparently, this series converges.
What am I doing wrong? Thanks for your help!
• What happens with the first $N = 3$ or $300$ or $3,000,000$ terms doesn't matter. The sum of the first $N$ terms is always finite. The question is: does there exist an $N$ such that the tests hold for all $n > N$? (Yes!) – Simon S Apr 14 '15 at 23:48
• Not only does it converge, it's absolutely convergent (meaning that the absolute value series converges too). Your only problem is that your test values are too early in the sequence. – Brian Tung Apr 14 '15 at 23:48
• This series is not absolutely convergent. – Simon S Apr 14 '15 at 23:49
• Okay that makes sense. According to the theorem, I thought it had to hold true for all values of $n$. Really, the theorem should be: $b_{n+1}≤b_n$ for all $n>N$. – James Taylor Apr 14 '15 at 23:50
Remember exactly what the alternating series test says:
If [some conditions hold] then the series converges.
So, we can never use the alternating series test to conclude that a series diverges; the theorem is silent on the subject of divergence.
The key idea is to find an $N$ such that the absolute value of the terms is monotonically decreasing. That is, if there's some $N$ such that $|a_{n+1}| < |a_n|$ for all $n > N$, then we can use the alternating series test on the sequence
$$\sum_{n = N+1}^\infty a_n,$$
to learn that it converges, and hence the original series must as well (as it's a finite number of terms added to a convergent sequence).
Now your job is to find the $N$ such that the absolute value of terms are monotonically decreasing for $n > N$. | 2019-08-17T23:22:15 | {
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https://math.stackexchange.com/questions/1595894/using-polar-coordinates-to-find-the-area-of-an-ellipse | # Using polar coordinates to find the area of an ellipse
Considering an ellipse with the $x$ radius equal to $a$ and the $y$ radius equal to b$:$
I figured that some kind of parameterization might be:
$x=a\cos\theta$
$y=b\sin\theta$
and then polar $r^2$ is just $x^2 + y^2$
but then I tried to come up with some unit of infinitesimal area using triangles ($d\theta r^2/2$) which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually
$x=ar\cos\theta$
$y=br\sin\theta$
but this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $r$ from $0$ to $1$ which I thought was eliminated because the equation for $r$ should be in terms of $\theta$ and the constants $a$ and $b.$
I would like some explanation of what I am doing wrong that would make some "physical" sense (or why physical intuition might fail for this problem)
Notice, since the ellipse: $x=a\cos\theta$ & $y=b\sin\theta$ is equally divided into four symmetrical regions hence, the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$ Now changing in Polar-coordinates by setting $y=b\sin\theta$ & $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$ $$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$ $$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$ $$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$
$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$
My name is Marco [email protected].
• You don't need to provide your name here. Any comment under your answer will show up as an update. If you do want to leave an email for people to contact you it's better to put it on your profile page. – Red Oct 7 '17 at 19:01
Your parametrization only covers the edge of the ellipse. This is not enough if you try to find the area, which is the full interior. Hence the second form with $r$ (which is integrated from 0 to 1) is correct to find the area. Yours would be ok, if you try to only find the circumference.
Hint:
Use one of the polar equations for the ellipse: if the pole is at the centre of the ellipse, it is $$r^2=\frac{b^2}{1-e^2\cos^2\theta}, \quad\text{where }e =\sqrt{a^2-b^2}\enspace\text{is the eccentricity of the ellipse}$$ and integrate $$\int_0^\pi r^2\,\mathrm d\mkern1mu\theta.$$
You were on the right track, but didn't pick the right infinitesimal triangle. It looks like you used a right triangle built on the inside of the ellipse, but as you discovered, that doesn't approximate the area swept out by the radius vector well enough. Take instead a point along the curve and another a small distance away along it. The area of the resulting triangle is, in Cartesian coordinates, $\frac12(x\,dy-y\,dx)$. Plugging in your parametrization of the ellipse gives $dA=\frac12ab\,d\theta$, which will clearly give the correct result when integrated.
More generally, when approximating integrals with Riemann sums, you need to take care that they converge to the right thing. This is the same type of error that occurs in the fallacious “stairstep” constructions which “prove” that $\pi=4$ or that $\sqrt2=2$. In this case, I expect that if you also computed the sum of the areas of the right triangles built on the outside of the ellipse that the difference between the inner and outer sums would not vanish in the limit.
I guess, the problem is in wrong approximation (look at light blue areas)
If use the formula for area of triangle $$\frac{1}{2}\left\| {{\bf{r}} \times {\bf{dr}}} \right\| = \frac{1}{2}rdr\sin \alpha % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaadaqbdaqaaiaahkhacqGHxdaTcaWHKbGaaCOC % aaGaayzcSlaawQa7aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaa % GaamOCaiaadsgacaWGYbGaci4CaiaacMgacaGGUbGaeqySdegaaa!4979!$$ you get the right result. Using of Green formula for area $$A = \frac{1}{2}\oint\limits_C {x \cdot dy - ydx} = \frac{1}{2}\oint\limits_C {\left\| {{\bf{r}} \times d{\bf{r}}} \right\|} = \frac{1}{2}\oint\limits_C {r\sin \phi dr} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % yqaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaa8qvaeaacaWG % 4bGaeyyXICTaamizaiaadMhacqGHsislcaWG5bGaamizaiaadIhaaS % qaaiaadoeaaeqaniablgH7rlabgUIiYdGccqGH9aqpdaWcaaqaaiaa % igdaaeaacaaIYaaaamaapufabaWaauWaaeaacaWHYbGaey41aqRaam % izaiaahkhaaiaawMa7caGLkWoaaSqaaiaadoeaaeqaniablgH7rlab % gUIiYdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaapufaba % GaamOCaiGacohacaGGPbGaaiOBaiabew9aMjaadsgacaWGYbaaleaa % caWGdbaabeqdcqWIr4E0cqGHRiI8aaaa!698A!$$ gives the same (right) result. | 2019-08-22T09:00:56 | {
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https://www.jiskha.com/display.cgi?id=1292970811 | # trig
posted by .
How do you write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i?
• trig -
In the quadratic equation
ax²+bx+c=0
b/a = -(sum of the two roots)
c/a = product of the two roots.
Assume a=1, then
b=-(7-3i+7+3i)=-14
c=(7-3i)*(7+3i)=49+9=58
So the quadratic equation is
x²-14x+58=0
Check by solving the equation and you should get back 7±3i as the roots.
• trig -
x = 7 + 3i, and X = 7 - 3i.
We can use either of the 2 values of X:
X = 7 + 3i,
(X - 7) = 3i,
Square both sides:
X^2 --14X + 49 = 9(-1),
X^2 - 14X + 49 +9 = 0,
X^2 - 14X + 58 = 0.
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# When the price of oranges is lowered by 40%, 4 more oranges
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When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
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Updated on: 08 Jul 2012, 05:22
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When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 Originally posted by farukqmul on 08 Jul 2012, 05:19. Last edited by Bunuel on 08 Jul 2012, 05:22, edited 1 time in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 51098 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 08 Jul 2012, 05:33 1 farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
22 Oct 2016, 17:22
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 let r=number of oranges purchased originally [12/(r+4)]/(12/r)=.6 r/(r+4)=.6 r=6 oranges$12/6=$2 per orange$24/$2=12 oranges B Current Student Joined: 12 Aug 2015 Posts: 2629 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 21 Apr 2017, 00:28 Here is what i did in this question => Let the price of each orange be$x
Hence for $12 => we can buy => 12/x oranges Now new price => x(1-40/100) =>$0.6x
Hence for $12 we could now buy => 12/0.6x As per the given information => 12/x+4=12/0.6x Hence 12+4x/x=20/x => 12+4x=20 =>x=2 So the price of each orange is$2
Now for 24 dollars => we can buy => 1/2*24=12 oranges
SMASH THAT B.
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
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30 Jun 2018, 18:53
12/(.6x)=12/x+4
20=12+4x
x=2
24/2
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
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04 Jul 2018, 18:25
1
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost: pq = 12 q = 12/p After the orange’s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is: (0.6p)(q + 4) = 12 0.6pq + 2.4p = 12 Substituting for q, we have: 0.6p(12/p) + 2.4p = 12 7.2 + 2.4p = 12 2.4p = 4.8 p = 2 That is, each orange is$2. So for \$24, we can buy 24/2 = 12 oranges.
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Re: When the price of oranges is lowered by 40%, 4 more oranges &nbs [#permalink] 04 Jul 2018, 18:25
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https://math.stackexchange.com/questions/284297/maximal-ideals-in-kx-1-dots-x-n | # Maximal ideals in $K[X_1,\dots,X_n]$
Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$.
I tried proving that the only elements outside the ideal are the invertibles of $K$ (I should still prove that this implies maximality, but it shouldn't be too difficult).
Is there a better strategy, or another stategy?
• It may be interesting to you to note that the opposite direction is also true, i.e.: any maximal ideal in that multivariable polynomial ring is of the given form, yet any proof of this other direction I know uses heavily analytic tools, whereas the direction you're asking is purely algebraic. – DonAntonio Jan 22 '13 at 16:18
• It's easier to see that the ideal $(X_1, \ldots, X_n)$ is maximal and your question's like a "change of variable". DonAntonio, can you say me where you saw the proof of the opposite direction? – Diego Silvera Jan 23 '13 at 1:27
• @DonAntonio: Dear Don, The converse is true only if $K$ is algebraically closed. (Think about the case $n = 1$.) Regards, – Matt E Jan 24 '13 at 5:25
• Good catch, @MattE. Thanks. – DonAntonio Jan 24 '13 at 5:39
• Several answers provide strategies that work, but it may be useful to also point out that the strategy you suggested in the question, "proving that the only elements outside the ideal are the invertibles of $K$," is doomed to failure because that isn't true. The elements outside the ideal are all the polynomials $f$ such that $f(a_1,\dots,a_n)\neq0$, and that includes lots of non-constant polynomials (i.e., polynomials that aren't in $K$). – Andreas Blass Jan 24 '13 at 14:19
Hint: Define
$$f:K[X_1,...,X_n]\to K\;\;,\;\;f(g(X_1,...,X_n)):=g(a_1,...,a_n)$$
1) Show $\,f\,$ is a surjective ring homomorphism
2) Use now the first isomorphism theorem for rings
3) Remember: if $\,R\,$ is a commutative unitary ring, an ideal $\,I\leq R\,$ is maximal iff $\,R/I\,$ is a field.
• Let $f_1$ be the homomorphism $$f_1:K[X_1,\cdots,X_{n-1}][X_n]\rightarrow K[X_1,\cdots,X_{n-1}]$$ $$f_2:K[X_1,\cdots,X_{n-2}][X_{n-1}]\rightarrow K[X_1,\cdots,X_{n-2}]$$ $$\cdots$$ $$f_n:K[X_1]\rightarrow K$$ Let assume that we know each $f_k$ to be a ring homomorphism. Then the composition $$f=f_n \cdots f_1$$ is an homomorphism. It is surjective because for each $a\in K$ we can evaluate the constant polynomial $a$. The kernel is $N:=\{g\in K:g(a_1,\cdots,a_n)=0\}$ I should now proof that $N=(X_1-a_1\cdots X_n-a_n)$ but is Ruffini still valid? – Temitope.A Jan 24 '13 at 22:02
Let $P(X_1, \ldots, X_n)$ a polynomial. Substitute $X_i\mapsto X_i + a_i$ and get $$P(X_1+ a_1, \ldots, X_n + a_n) = \sum c_{\alpha} X_1^{\alpha_1} \ldots X_n^{\alpha_n}$$ and so
$$P(X_1, \ldots, X_n) = \sum c_{\alpha} (X_1-a_1)^{\alpha_1} \ldots (X_n-a_n)^{\alpha_n}$$
Note that $c_{(0,\ldots, 0)} = P(a_1, \ldots, a_n)$. Moreover, $$P(X_1, \ldots, X_n) = c_{(0,\ldots, 0)}+ \sum (X_i - a_i) g_i(X_1, \ldots, X_n)$$ as all the other terms $c_{\alpha}(X-a)^{\alpha}$ are divisible by some $(X_i - a_i)$. Therefore $$P(X_1, \ldots, X_n) - P(a_1, \ldots, a_n) \in (X_1-a_1, \ldots, X_n - a_n)$$
and therefore $P(a_1, \ldots, a_n) \in (P, (X_1 - a_1) , \ldots, (X_n - a_n))$. Assume moreover that $P \not \in (X_1- a_1, \ldots X_n - a_n)$. Then $P(a_1, \ldots, a_n) \ne 0$ and we conclude that $1 = P(a_1, \ldots, a_n)^{-1} \cdot P(a_1, \ldots, a_n) \in (P, (X_1 - a_1), \ldots, X_n - a_n)$. Therefore $(X_1-a_1, \ldots, X_n - a_n)$ is maximal.
Hint $\ \ (I,f) = (I,f\ mod\ I) = (I,f(\bar a))\,\ [\,= 1 \iff f(\bar a)\ne 0\iff f\not\in I]$
Remark $\$ It is instructive to compare this internal approach to the structural approach mentioned by DonAntonio.
• @YACP $\ I = (\bar X- \bar a) = (X_1-a_1,\ldots,X_n-a_n),\$ and $\, {\rm mod}\ I\!:\ X_i\equiv a_i\Rightarrow f(X_1,\ldots,X_n) \equiv f(a_1,\ldots,a_n) = f(\bar a)\ \$ – Math Gems Jan 24 '13 at 15:52
• @YACP Ideals are always preserved under the operation of "modding out" one generator by others, i.e. $(a_1,\ldots,a_n,b) =$ $(a_1,\ldots,a_n,b\!-\!r_1 a_1-r_2 a_2\! -\!\cdots- r_n a_n).\,$ Above the modular reduction corresponds to the multidimensional generalization of the polynomial remainder theorem. – Math Gems Jan 24 '13 at 16:24
• @YACP Even though, in general, there is no (generalized) division with (unique) remainder (with associated "mod" operation), it is conceptually helpful to think of such ideal tranformations like the mod operations employed in the reduction steps of the Euclidean algorithm (or the multidimensional generalizations used in standard/Grobner basis reductions). – Math Gems Jan 24 '13 at 16:24
• What do you mean by "multidimensional generalization of the polynomial remainder theorem"? As far as I can see you use the following result: $f$ can be written as a combination of $X_i-a_i$ (with polynomial coefficients) plus an element of $K$. Is this obvious or deserves a proof? – user26857 Jan 24 '13 at 16:32
• @YACP It's an easy inductive proof: apply the univariate remainder theorem to the highest variable, using $k[X_1,\ldots,X_n] = R[X_n]\:$ for $\:R = k[X_1,\ldots,X_{n-1}].\ \$ – Math Gems Jan 24 '13 at 16:51
I realize that we should avoid responding to other answers, but when they make false statements there should be a way to correct them.
$\mathbb Q$ is a field. ${X_1}^2-2$ is a prime element in $\mathbb Q\left[ X_1\right]$ which is a p.i.d so ${X_1}^2-2$ generates a maximal ideal. The converse of this statement is false for the field $\mathbb Q$ or any field that is not algebraically closed. If $K$ is algebraically closed, both the statement of the question and its converse are corollaries of the Hilbert Nullstellensatz. This basic result in algebraic geometry can be found in texts on algebraic geometry, for example Eisenbud's Commutative Algebra with a view toward algebraic geometry, Springer Graduate Texts in Math, vol 150, pp 34--35.
• Dear Barbara, The comment about the converse was made in a comment, and you can correct it by making a comment in reply (as I did above). Regards, – Matt E Jan 24 '13 at 5:27
• Hi Barbara, I am not sure how it happened, but your account split, despite you logging into the system using the same credentials. If you again find yourself unable to post comments or edit your own posts without review, please let a moderator know so we can track down the problem. I've merged your two accounts now. – Willie Wong Jan 24 '13 at 15:59
• Thank you very much for your help. Things seem to be working fine now, and I am commenting appropriately now. – Barbara Osofsky Jan 25 '13 at 19:19
I think this answer might essentially be the same as @orangeskid's answer, but I'm not sure, so I figured I will post it here. If it is the same or incorrect, please let me know and I will delete it.
Let $I = \langle x_1 - a_1, \dots, x_n - a_n \rangle \subset k[x_1, \dots, x_n]$ for $k$ a field. Consider any ideal $I \subsetneq J$, by the Hilbert Basis theorem $J$ is finitely generated, so we can say that $J=\langle x_1 - a_1, \dots, x_n -a_n, f_1, \dots, f_r\rangle$ for finitely many $1 \le i \le r < \infty$, with $f_i \not\in I, f_i \in k[x_1, \dots, x_n]$ and without loss of generality, all of the $f_i$ are not identically equal to zero.
Case 1: At least one of the $f_i$ is constant.
Without loss of generality, $f_r$ is constant, say $f_r \equiv c \in k, c\not=0$ so $J= \langle x_1 - a_1, \dots, c \rangle = \langle x_1 - a_1, \dots, 1 \rangle = k[x_1, \dots, x_n]$.
Case 2: None of the $f_i$ are constant.
Let's choose one of the $f_i$, which by an abuse of notation we will denote simply by $f_i(x_1, \dots, x_n)$.
Then we have that $$f_i(x_1, \dots, x_n) = f_i^m(x_1, \dots, x_n) + f_i^{m-1}(x_1 -a_1,\dots, x_n - a_n)+ \dots + f_i^1(x_1 -a_1, \dots, x_n -a_n)+ f_i(x_1-a_i,\dots,x_n-a_n)$$ where for all $1 < j \le m, \quad f_i^j(x_1, \dots, x_n) := f_i^{j-1}(x_1,\dots,x_n)-f_i^{j-1}(x_1 - a_1, \dots, x_n-a_n)$, and additionally $f_i^1(x_1, \dots, x_n):=f(x_1,\dots,x_n)-f(x_1-a_1, \dots, x_n-a_n)$, and finally that $m$ is the lowest natural number such that $f_i^m(x_1, \dots, x_n)$ is identically constant, thus for any $k >m$, $f_i^k(x_1, \dots, x_n)\equiv 0$. Such an $m$ is guaranteed to exist, because by construction, $\deg(f_i^1) < \deg(f_i)$ and $\deg(f_i^j) < \deg(f_i^{j-1})$ for all $2 \le j \le m$.
Define $c$ to be the non-zero constant such that $f_i^m(x_1, \dots, x_n) \equiv c$. Now clearly we have that $$f_i(x_1 -a _1, \dots, x_n -a_n) \in \langle x_1 -a_1, \dots, x_n -a_n \rangle \\ f_i^j(x_1-a_1, \dots, x_n -a_n) \in \langle x_1 - a_1, \dots, x_n - a_n \rangle \quad \forall\ 1\le j \le m-1$$ Therefore we have shown that $f_i(x_1, \dots, x_n) = c + g_i (x_1, \dots, x_n)$ for some $g_i \in \langle x_1 - a_1 , \dots, x_n - a_n \rangle$. Therefore we have that $$J = \langle x_1 - a_1, \dots, x_n - a_n, \dots, f_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c + g_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, 1, \dots, f_r \rangle = k[x_1, \dots, x_n]$$
Thus, in both cases, $I \subsetneq J \implies J = k[x_1, \dots, x_n]$, so therefore $I$ must be maximal, since clearly $I=\langle x_1 - a_1, \dots, x_n - a_n\rangle$ is proper.
• Since the ring homomorphism $k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]$ such that $x_i\mapsto x_i-a_i$ is an isomorphism, it is not restrictive to assume $a_1=a_2=\dots=a_n=0$. Thus a polynomial $f\notin I$ must have a nonzero constant term and so an ideal properly containing $I$ is the whole ring. This is what your long proof boils down to. And DonAntonio's proof is the abstract (and simpler) version of it. – egreg Oct 11 '16 at 13:04
• @egreg Thank you for reading my proof. That $x_i \mapsto x_i - a_i$ is an isomorphism (of commutative rings? I'm not sure which type of isomorphism you mean) seems similar to the intuitive idea I had originally involving affine changes of coordinates "not changing anything important". You are probably right that it is equivalent to DonAntonio's proof as well; admittedly I am not comfortable enough with the material to understand the argument why without thinking about it for a long time. I agree that the elegance and simplicity of DonAntonio's proof is preferable. – Chill2Macht Oct 11 '16 at 13:20
• A ring homomorphism $k[x_1,\dots,x_n]\to A$ is determined as soon as you have a ring homomorphism $k\to A$ and assign images to $x_1,\dots,x_n$. In this case $A=k[x_1,\dots,x_n]$ the homomorphism $k\to A$ is the canonical embedding. With $x_i\mapsto x_i-a_i$ you get an isomorphism, because the inverse is the one doing $x_i\mapsto x_i+a_i$. – egreg Oct 11 '16 at 13:41 | 2019-08-18T04:36:22 | {
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https://math.stackexchange.com/questions/375870/how-do-you-determine-the-local-extrema-points-for-y-sqrt3-cos3x-sin3x | # How do you determine the local extrema points for $y=\sqrt{3}\cos(3x)+\sin(3x)$
$$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the derivative for when it's value is $0$ and got $x=\dfrac{\pi}{2}$ . I then used this critical point and subdivided the interval. I found that there were 3 points of local extrema after doing all my work, local maximum $x=0, \dfrac{2\pi}{3}$ and local minimum $x=\dfrac{\pi}{2}$. However according to the online homework, there were 4 different points of extrema. Local maximum $x=\dfrac{\pi}{18},\dfrac{2\pi}{3}$ and local minimum $x=0,\dfrac{7\pi}{18}$. I am really confused as to how there are 4 points of local extrema, did I leave out an answer somewhere? I am also confused as to how they got $x=\dfrac{\pi}{18},\dfrac{7\pi}{18}$ as points of local extrema. Could someone explain this to me?
• At $x = \pi/2, y' \neq 0$... $y'(\pi/2) = 3\sqrt{3}$ – Namaste Apr 29 '13 at 3:44
As I noted in my comment: At $x = \pi/2, \quad y' \neq 0;$ ... $y'(\pi/2) = 3\sqrt{3}$
You need to solve for $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x) = 0 \quad \iff \quad 3\sqrt{3}\sin(3x) = 3\cos(3x)$$ $$\iff \quad \sqrt 3 \sin(3x) = \cos(3x),\quad x\in \left[0, \frac{2\pi}{3}\right]$$
Note that $$\sqrt 3 \sin(3x) = \cos(3x) \iff \sqrt 3 \dfrac{\sin(3x)}{\cos(3x)} = \sqrt 3 \tan(3x) = 1\iff \tan(3x) = \frac{1}{\sqrt 3}$$
Solving for $x$ will give you 2 potential critical points on your interval; then recall that you need to also check endpoints of an interval as potential extrema.
• How do you get 3 potential critical points when solving for $x$? I only get one point which is $\frac{\pi}{2}$. – Kot Apr 29 '13 at 4:24
• Did you read the first sentence, or my comment? $y'(x)\neq 0$ when $x = \pi/2$. I say three potential critical points, but one will outside your interval...Follow what I've done to solve $y' = 0$. Then: Solve for $x$ given $3x = \tan^{-1}(1/\sqrt 3)$...then find which value is outside of $x \in [0, 2\pi/3]$. You will be left with two solutions to x: one will be a maximum, and one will be a minimum. The other max is the endpoint of the interval on which $x$ is defined: $2 \pi/3$, and the other minimum at the endpoint $x = 0$ – Namaste Apr 29 '13 at 4:28
• I see what I did wrong, I multiplied by $3$ instead of dividing by $3$. I now have $x=\frac{\pi}{18}$. I do not understand where the two other points would come from. Does the inverse tangent function give multiple answers? – Kot Apr 29 '13 at 4:31
• Yes, the other critical point in your interval is at $\pi/18 + \pi/3 = 7\pi/18$. The calculator typically only gives one value...but you can test to confirm that $x = 7\pi/18$ also makes $y' = 0$ – Namaste Apr 29 '13 at 4:33
• I am a little confused as to why you added $\pi/3$ to the first answer. If it was tangent don't you add $\pi$ to get the other value? – Kot Apr 29 '13 at 4:37
$y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3})$
$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2}$, another 2 are the end points ie:$x=0$ or $x=\dfrac{3 \pi}{2}$
• An excellent proof showing calculus is unnecessary. – Stefan Smith Apr 29 '13 at 15:29
$y=\sqrt 3\cos(3x)+\sin(3x)=2\sin(3x+\frac\pi3)$
$y'=2\cdot3\cos(3x+\frac\pi3)$
For the extreme values of $y, y'=0\implies \cos(3x+\frac\pi3)=0$
$\implies 3x+\frac\pi3=(2n+1)\frac\pi2$ whether $n$ is any integer
$\implies 3x =(6n+1)\frac\pi6$
As $0\le x\le\frac{2\pi}3, 0\le 3x\le 2\pi$
$\implies 0\le (6n+1)\frac\pi6\le 2\pi \implies0\le 6n+1\le 12\implies n=0,1$ | 2019-09-15T16:22:07 | {
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http://math.stackexchange.com/questions/122674/how-to-maximise-this-function-p-log-1-x-q-log-1-x | # How to maximise this function: $p \log (1 + x) + q \log (1 − x)$?
Hint: Use the fact that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q$.
Secondly it was written: Let $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$ Then using the hint above, $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$
My first question is, how (possibly using analysis/calculus?) would you deduce/get that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q?$
The other thing, I have trouble understanding what exactly does $f : (-1,1)$ mean? And how would I intepret/read the following line? Its a bit new to me as I have not seen functions written like the following before. $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$
Lastly, it says that $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$ How do you get that $f(p-q) = \log 2 + p\log p + q\log q?$ I tried substituting it into $f(x)=p\log (1+x) + q \log (1-x)$, getting $f(p-q) = p\log(1+p-q) + q\log(1-p+q).$ How would I go about from there?
-
It is rather bad form to remove questions from your post when people have already addressed them in their answer; it makes it look like they don't know what they are talking about to new readers. It's worse if you then try to edit people's answers to remove content because of the content you decided to remove from your own post. And to try to do so 12 days after the fact on top of that... – Arturo Magidin Apr 2 '12 at 1:42
Thanks, I already flagged this post stating the (valid) reason, why this question should be deleted. Appreciate if you could take a look. – Heijden Apr 2 '12 at 2:00
I can't think of a reason why a question posted 12 days ago, with an accepted answer, should be deleted; unless, of course, you are somehow trying to cover your tracks for some reason... in which case it should definitely not be deleted. – Arturo Magidin Apr 2 '12 at 2:06
@Arturo, I have rolled it back to a more suitable version. – user21436 Apr 2 '12 at 2:16
@ArturoMagidin, I flagged this post to a moderator stating my reason to remove this question, and I am not covering my tracks, in my (original) post I am just asking for explanations to a few things, not trying to get anyone to solve the question for me, hence there is no need to 'cover my tracks'. I gave the reason to the moderator when i flagged it. – Heijden Apr 2 '12 at 2:20
For your first question, you can see that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x=p-q$ by taking the derivative at setting it equal to $0$. We have $$0=\frac{d}{dx}(p \log (1 + x) + q \log (1 − x))=-\frac{p}{1+x}+\frac{q}{1-x}=\frac{-p(1-x)+q(1+x)}{(x+1)(x-1)}$$ and so $(q-p)+(p+q)x=0$ hence $x=\frac{p-q}{p+q}$, and since $q=1-p$ this simplifies to $x=p-q$. This tells us that $p \log (1 + x) + q \log (1 − x)$ has an extremum at $x=p-q$, and this extremum must be the maximum as making $x$ near $1$ or $-1$ makes $\log(1-x)$ or $\log(1+x)$ very negative, respectively.
For your second question, "$f: (-1,1)$" doesn't mean anything. However, the full statement "$f:(-1,1)\to \mathbb R$" means "$f$ is a function from $(-1,1)$ to $\mathbb R$ (the real numbers)". In your case, the statement $$f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x)$$ means "$f$ is a function from $(-1,1)$ to $\mathbb R$ such that $f(x)=p\log (1+x) + q \log (1-x)$ for any $x\in (-1,1)$ (any $x$ between $-1$ and $1$)".
Edit: To get that $f(p-q)=\log 2+p\log p+q\log q$, use the fact that $p=1-q$ and $q=1-p$ so $$\begin{eqnarray} f(p-q)&=&p\log(1+p-q) + q\log(1-p+q)\\ &=&p\log(1+p-(1-p)) + q\log(1-(1-q)+q)\\ &=&p\log(2p) + q\log(2q)\\ &=&p(\log 2+\log p) + q(\log 2+\log q)\\ &=&(p+q)\log 2+p\log p + q\log q\\ &=&\log 2+p\log p+q\log q. \end{eqnarray}$$
-
Why the downvote? – Alex Becker Apr 2 '12 at 2:00
$f: (-1,1) \to \mathbb R$ just means that $f$ is a function defined on the interval $(-1,1)$ with values in the real line.
To find critical points of $f(x) = p \log(1+x) + q \log(1-x)$, solve $f'(x)=0$ for $x$. You should find exactly one critical point, at $x=p-q$ (if you remember that $p+q=1$). Note that this is a local maximum (e.g. by using the second derivative test, or noting that $\log(1+x)$ and $\log(1-x)$ are concave functions). If a differentiable function has only one critical point in an interval and it is a local maximum, then it is a global maximum on that interval. | 2014-03-07T14:18:26 | {
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https://mathematica.stackexchange.com/questions/111060/does-it-mean-that-there-no-analytical-solution-if-mathematica-cannot-find-analyt/111062 | # Does it mean that there no analytical solution if Mathematica cannot find analytical solution?
I'm solving simple but coupled ODEs recently. I use both MATLAB symbolic computation and Mathematica.
For example, my coupled ODE is the following \begin{align*} &\dot{x}(t)=y(t)-\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ &\dot{y}(t)=-y(t)+\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ \end{align*} where $\rho\in(0,1)$ and $b\in(0,1)$ are given constants and the initial value of this set of ODEs are $x(0)=a$, there $a\in(0,1)$ and $y(0)=0$.
This expression looks simple but these 2 equations are coupled.
First, I tried MATLAB, it generate "Warning: Explicit solution could not be found." explicitly.
Then I tried Mathematica,
system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])),
y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};
Then I try to solve it via sol = DSolve[system, {x, y}, t].
The thing that I don't understand is that after I press Shift+Enter, Mathematica only makes my input look nicer, but didn't produce any result or generating any warning message like MATLAB. So I couldn't tell whether it is because Mathematica also couldn't find the analytical solution like MATLAB, or it just does not even try to solve the problem since I input something wrong?
This simple coupled ODE drives me crazy these days. Any suggestion, input is deeply appreciated.
If math software couldn't find analytical solution, then is it still possible to analyze the monotonicity of the solution? For example, in this case, it's easy to analyze the case when $t=\infty$ by setting $\dot{x}=0=\dot{y}$ and solve the equations. I can also numerically plot the solution to see the trend, e.g., $x(t)$ is decreasing. But without the explicit functional form of $x(t)$, how can we prove the monotonicity, stuff like that? In general, if the math software fails to find analytical solution, what should we do next?
• The answer to "Does it mean that there no anlytical solution if Mathematica cannot find analytical solution?" is definitely no (i.e. there are definitely analytic solutions that Mathematica can't find for all kinds of things). However, those differential equations seem complicated enough that there could easily not be an analytic solution. – march Mar 25 '16 at 16:32
• Even if you are unable to derive a closed form solution to your DE, it might still be possible to study the qualitative behavior of the solutions. Have a look at Bender/Orszag for ideas. – J. M. is away Mar 25 '16 at 16:35
• @J.M. Is that book "Advanced Mathematical Methods for Scientists and Engineer"? – KevinKim Mar 25 '16 at 16:38
• It's interesting to note that $\dot{x}(t) + \dot{y}(t) = 0$. – rcollyer Mar 25 '16 at 16:43
• Well, it has the form $\dot{\vec{x}}(t) = \mathbf{A}(t) \vec{x}(t)$, so you could try diagonalizing $\mathbf{A}$ to decouple the two equations. – rcollyer Mar 25 '16 at 17:00
Here, we take advantage of the observation mentioned in a comment that the sum of x and y is conserved. If we define
s[t] == x[t] + y[t]
d[t] == x[t] - y[t]
eqns = {s'[t] == 0,
d'[t] == -(1 + (r b)/(1 - (1 - E^-t) r)) d[t] + (1 - (r b)/(1 - (1 - E^-t) r)) s[t]}
Mathematica knows how to solve this:
First@DSolve[eqns, {s[t], d[t]}, t]
As noted in the comments above, you have $\dot{x} + \dot{y} = 0$, which implies that $x(t) + y(t)$ is a constant; call it $c_3$. (Note that in particular, $c_3 = x(0) + y(0)$.) You can therefore replace y[t] with c3 - x[t] in your first equation above, and Mathematica can solve that explicitly:
reducedsystem = {x'[t] == c3 - x[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))}
DSolve[reducedsystem, x[t], t]
(* {{x[t] -> -((c3 E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) (-E^t + c1 (-1 + E^t))^(1 - (c1 c2)/(-1 + c1)))/(1 + c1 (-1 + c2)))
+ E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}} *)
I'm a little surprised that Mathematica can't do this on its own, to be honest; but here we are.
• Wait, but now don't we have an unknown c3 in our solution? Meaning, x[t] is now expressed in terms of a y[t] that we still don't know what it is. – MathX Mar 25 '16 at 18:39
• $c_3$ can be determined from the initial conditions: since it's a constant, then $c_3 = x(0) + y(0)$. I've edited to clarify this. In general, you'll have two constants of integration for this system; in my solution, they'll be the $c_3$ I coded in and the C[1] generated by Mathematica in the solution. – Michael Seifert Mar 25 '16 at 18:40
• oh yes you are right. And in that case it is equal to a from the OP. – MathX Mar 25 '16 at 18:48
The non-trivial solution for this system can be obtained by doing this:
system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])),
y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};
DSolve[First[system /. y -> (-x[#] &)], x[t], t]
(*
==> {{x[t] ->
E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}}
*)
All I did here is use the fact that the equations become identical if $y=-x$. That allows us to drop one of the equations.
• As I noted in my comment above, the syntax of the input in the OP is incorrect. Mathematica will interpret exp(-t) as the product of a constant called exp with -t. – Michael Seifert Mar 25 '16 at 17:10
• @Jens This one looks awesome! Let me try to understand the syntax "DSolve[First[system /. y -> (-x[#] &)], x[t], t]" – KevinKim Mar 25 '16 at 17:15
• @MichaelSeifert oh yes, of course I used the corrected version to get the posted result - just copied the OP's error accidentally. – Jens Mar 25 '16 at 17:49
• @KevinKim the replacement y -> (-x[#] &) sets the function y equal to the function -x. You can't just use y -> -x here because that won't work in the derivative. Hence the more complex notation. & declares an (anonymous) function and # is the variable placeholder. – Jens Mar 25 '16 at 17:57
• @Jens: Are you sure that you used the corrected code to get your posted result? Your posted result has exp in it, including in the denominator without an argument. Also, note that $\dot{x} + \dot{y} = 0$ does not imply that $y = -x$; there should be an arbitrary constant involved. – Michael Seifert Mar 25 '16 at 18:18 | 2019-06-16T23:51:33 | {
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https://math.stackexchange.com/questions/2369246/how-does-one-represent-the-gradient-nabla-mathbf-a-of-a-vector-field-mathb | # How does one represent the gradient $\nabla \mathbf A$ of a vector field $\mathbf A$ as a matrix?
I have been searching google about the gradient of a vector. I have found two distinct types of matrix notation for the gradient of a vector. I can't understand which one is correct, or if both are correct. The two are:
$$\nabla A = \partial_iA_je_i\otimes e_j=\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix}$$
or the transpose of the first, that is
$$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix}.$$
• 'the Gradient' Gradient is a subset of the Jacobian, the diagonal of the Jacobian Matrix Jacobian – f5r5e5d Jul 22 '17 at 23:06
• 1. What do you mean by "correct" here, i.e. why can't these just be two differing conventions? 2. How is this a physics rather than a Mathematics question? – ACuriousMind Jul 22 '17 at 23:13
• The the second one seems to be the transpose of the first one and the second one also can be described by Jacobian of (A1,A2,A3/X1,X2,X3).Where the first one can not be written as Jacobian transformation as like the second one.So,what's the right way to write the matrix for gradient of vector -the first one or the second(as a jacobian trans?And of course this is a physics-question.The first guy I have found in-eng.uc.edu/~beaucag/Classes/Processing/Chapter2html/… , the second one is in -umich.edu/~bme456/ch1mathprelim/bme456mathprelim.htm#vectordyad . – Abu sayed Jul 23 '17 at 4:06
• To second ACuriousMind, these are two different notation conventions. The difference can actually be important because equations look different, so just make sure what notation the author is using ... – Sanya Jul 23 '17 at 10:46
The matrix notation for the gradient of a vector field is generally not-that-well-defined, because as you note there are two reasonable matrix encodings (transposes of each other), which are about equally reasonable. Because of that, whenever the notation $\nabla\mathbf A$ as a matrix is used in the literature, the responsible thing to do is to specify explicitly which of the two encodings is meant.
Both of the two representations have features that make them reasonable:
• The second one you mention, $$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the contravariant index as a row index and the covariant index of the gradient as a column index, which is reflective of both of those natures.
• On the other hand, the first representation in your post, $$\nabla A =\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the nice property that if $\mathbf r$ and $\mathbf v$ are column vectors, then $$\mathbf r^T \cdot \nabla \mathbf A\cdot \mathbf v = x_i \frac{\partial A_j}{\partial x_i} v_j$$ matches what you would expect as a matrix product, i.e. matching the action of the operator $\mathbf r \cdot \nabla = x_i \frac{\partial}{\partial x_i}$.
For an example of this in action in the literature, see this paper of mine: to avoid confusion, it is important to specify both how the matrix representation is chosen, and how it acts via components. It takes an extra line and it adds a whole lot of clarity to the text.
The gradient of a function is well defined in the literature. The gradient of a vector field $A = A^i\partial_i$ seems to be the gradient of its components (which are functions). I think both matrix representations ($\partial_i A^j$ and $\partial_j A^i$) are "good" since they are simply representations of the same thing : $$\nabla A = (\partial_i A^j)dx^i\otimes \partial_j = (\partial_j A^i)dx^j \otimes \partial_i$$ Choose one of them and stay with it (at least until the end of the proof where you are using gradients of vector field).
Geometrically speaking, the gradient of a vector field you are talking about can be written either as the Lie derivative $\mathcal{L} A$ (along $\partial_i$ or $\partial_j$) or the covariant derivative $\nabla A$ (with vanishing Christoffel symbols $\Gamma$'s) (along $\partial_i$ or $\partial_j$). Both $\mathcal{L} A$ and $\nabla A$ are a little bit overkill for your setting (probably Euclidean flat space) but are good to keep in mind if you need to generalize from $\mathbb{R}^n$ to some manifold with global properties.
• What you have mentioned "same thing",but the second one is actually the jacobian of (A1,A2,A3/X1,X2,X3) whereas the first one don't seems to be the jacobian transformation like J(,,/,_,).So,what is the fact? – Abu sayed Jul 23 '17 at 4:15
• Yes, same geometrical thing. A matrix representation is only a table of numbers, or functions and doesn't give any insight about the geometry behind it (i.e. in terms of sections of bundles). – Noé AC Jul 23 '17 at 7:05
• Both matrix representations are obviously different, so ignoring that is really unhelpful. Mentioning "the" Lie derivative here is also pretty pointless. – Sanya Jul 23 '17 at 10:43
• @Sanya I said it "seems to be related" to the Lie derivative. Looking at Deep's answer probably it's a better idea to look for covariant derivative with null Christoffel $\Gamma$'s. Though here we don't know the context of the original question so both Lie derivative and covariant derivative are somewhat overkill. – Noé AC Jul 23 '17 at 16:11
• @Sanya Also, in local coordinates, $\nabla A = (\partial_i A_j)dx_i \otimes \partial_j = (\partial_j A_i) dx_j \otimes \partial_i$, so using the matrix $(\partial_i A_j)$ or it's transpose $(\partial_j A_i)$ is ok, as long as you know what $i$ and what $j$ are linked to ($e_k$ or $e_k^*$) – Noé AC Jul 23 '17 at 16:22
$\nabla A$ is a tensor and it acts on a vector, say $\mathbf{u}$, to produce another vector: $\mathbf{v}=\nabla A(\mathbf{u})$. In matrix representation, you multiply the matrix of $\nabla A$ with the specified column/row vector representing $\mathbf{u}$. If you write $\mathbf{u}$ as a column vector (usual convention) then the second matrix representation is the correct one, while if you write it as a row vector the first matrix representation is the correct one.
However matrix representation is not necessary. You have $\nabla A=\partial_iA_j~\mathbf{e}_i\otimes\mathbf{e}_j$, and the action of $\nabla A$ on $\mathbf{u}$ is defined to be the vector $\mathbf{v}=\partial_iA_j~\mathbf{e}_i(\mathbf{u})~\mathbf{e}_j=(\partial_iA_j)u_i~\mathbf{e}_j$. No more worries about how the matrix should be written. | 2020-04-03T09:02:30 | {
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Once we have an equation establishing the relationship among the variables, we differentiate implicitly. Any quantity that changes should not be substituted until the derivative is being evaluated at the specific time stated in the question. Aim Colonization is a central topic in ecology and one of the cornerstones of island biogeography. They are in the following order. The following is the typical cost function associated with producing goods. Review for the Common Exam: MATH 151 Exam 1 Review Problems 36-40. Riemann Sums and Definite Integrals; Implicit Differentiation. 1 Differentiation with Parameters 7. To calculate the derivative y′ y ′ of an implicit function: Treat the variable y y in the relation as an unknown but differentiable function of x x (like y =g(x) y = g ( x) ), and differentiate …. 1 x2y+xy2=6 2 y2= x−1 x+1 3 x=tany 4 x+siny=xy 5 x2−xy=5 6 y=x 9 4 7 y=3x 8 …. Plug in all the values you know, leaving only the one you’re solving for. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0. Use implicit di erentiation to determine the slope of the graph of y2 = 3xy 5 where y = 1. This problem has been solved! See the answer. An interactive exploration of related rates, the study of variables that change over time where one variable is expressed as a function of the other. Derivative by Definition Square Roots. This is vague, so it’s best to look directly at how to solve a related rate …. Unit #5 : Implicit Differentiation, Related Rates Goals. For instance, if you differentiate …. AP Calculus SCHOLARS, Tuesday, MAY 4, 2021!! LESSON VIDEOS & OLD TESTS. Students will find the second derivative …. x y= 2 Question 6 A curve is described by the implicit …. When the cloud has moved to the right. The key idea behind implicit differentiation is to assume that y is a function of x even if we cannot explicitly solve for y. IMPLICIT DIFFERENTIATION 121 Example 2. , with independent variable of the form x (or some other symbol), and dependent variable of the form y (or some other symbol). Session 13: Implicit Differentiation. Solved: Use implicit differentiation …. ©v X2G0Z1E4i fKpuBt5ay ES HoXfxt mwXaPr HeX DLnL vC 2. The procedure to use the average rate of change calculator …. Free Response - 50% of your score. 8 Implicit Differentiation (Lecture 10) – CALCULUS I MA…. -Informal question/ answer sessions -Group Work -Quiz / Short Test -Graphics Calculator …. Limit calculations, including limits involving infinity, Continuity Derivative of a composite function (chain rule), e. water drains from a cone (related rates problem). With these codelets, the executor implements the Cooley-Turkey FFT Origin can calculate the magnitude, phase and amplitude of the transformed data. We have differentiation tables, rate of change, product rule, quotient rule, chain rule, and derivatives of inverse functions worksheets. The surface area of the top side of the pizza dough is given by. Applications of Differentiation. 6) Lecture 17: Implicit Differentiation (§2. 5 6 2 1/2 12 6 2 1/2 -1/2 5y LINEAR IN y/ — 6T + 5y for y into your solution for part are consistent by substituting the expression Check that your solutions to parts and to get y/ in terms of x. Implicit Differentiation Homework Answers Example 6 - Find the Equation of the Tangent Line Calculus 1: Related Rates (Level: Easy - Hard) Related Rates (KristaKingMath) Calculus - Understanding Implicit Differentiation RELATED RATES: Ladder sliding Page 9/65. All forward-looking statements, express or implied, included in this report and the documents we incorporate by reference and that are …. 1 Related Rates (Method , Examples) 31. Findtheslopeofthegraphatthepoints(1; p 3) and (1; p 3). In these types of problems, one thing changes but we are interested in seeing how something else changes. 11 Linearization and Differentials rule, derivative of inverse functions, rates of change, implicit differentiation, related rates, linearization (optional). The rules of differentiation produce an equation relating the rates of those quantities. EQUATION slope an equation to the tangent line to dy dc = 25 at the point (3, 4). Most implicit relations, such as is defined by the equation 9x2 +y2 =, are not of the required form F(x,y) = 0. Chapter 6: The Instantaneous Rate of Change: Derivative as a Function; Determining derivatives using Commands, the Expression Palettes and the Context Panel; Student[Calculus1] library; Implicit differentiation. Related Rates; Linearization; L'Hospital's Rule; Integrals. The early applications of differentiation …. In all cases, you can solve the related rates problem by taking the derivative of both sides, plugging in all the known values (namely, x, y, and ˙x), and then . This means you need to use the Chain Rule on terms that include y y y by multiplying by d y d x \frac{dy}{dx. Related Rates, Newton's Method, Linear Approximation. Graphing Calculator; DESMOS Activity 1; DESMOS Activity 2; DESMOS Activity 3; Recorded Lectures; Calendar; Videos. A curve has implicit equation x y y y x xy3 3 2+ + + − = +3 3 6 50 2. 11) TI-89 and Implicit Differentiation; Chapter 3. Riemann Sums and Definite Integrals: MATH 151 Problems 6-12 Related rates problems, differentials, linear and quadratic approximations. Derivative Calculator - Take the derivative of a polynomial, trig function, or even more complicated expressions. (1) Find the derivative of y = x cos x Solution. the building at the rate of 2 ft/min. Differentiate each side of the equation by treating y y y as an implicit function of x x x. Using implicit differentiation to find the derivative with respect to time, we get. Geometrically speaking, the derivative of any function at a particular point gives the slope of the tangent at that point of the function. Beyond calculus context, derivative calculator shows the gradient to calculate the derivative problems to rise, the given by the value. This course features a wide range of readings, simulations, assessments. In this problem, we consider the Lemniscate curve (1) (a2+y2)2 =3 (z2 - ) 2 -2 Figure 1: Bernoulli's Lemniscate The goal of this problem is to find the four points of the Lemniscate curve with horizontal tangent lines. Start learning today! Strategy for Solving Related Rates Problems #1. Repeating this process, function y's third order derivative …. •There must be an equation relating the 2(+) quantities. Calculus I - Implicit Differentiation Section 3-10 : Implicit Differentiation Back to Problem List 8. The Derivative from First Principles. Concavity and inflection points 5. Jean-Baptiste Campesato MAT137Y1 - LEC0501 - Calculus! - Oct 24. Time-saving lesson video on Implicit Differentiation with clear explanations and tons of step-by-step examples. Implicit differentiation; Derivative of the inverse of a function (including and ) Related rates of change; Integral Calculus …. What is Related Rates Calculator Symbolab. The trough is a triangular prism 10 feet long, 4 feet high, and 2 feet wide at the top. Course Info: 1st Day Handout: Parent/Student Letter. The graph of $$8x^3e^{y^2} = 3$$ is shown below. Implicit Differentiation - Related Rates Challenge Quizzes Implicit Differentiation: Level 2 Challenges Implicit Differentiation: Level 3 Challenges Implicit Differentiation …. Derivative is used to calculate rate …. Related Rates – The Anxious Panda. Cal_AB_LAP_5_IMPLICIT DIFFERENTIATION AND RELA…. 5) Lesson 13: Derivatives as Rates of Changes (Briggs 3. 4 NO Calculator - 40 min test 1 Thu10/7/2021 2 nd ½ of BLOCK Students will find the derivative of an implicitly defined function. Worked example: Differentiating related functions. Calculus Related Rates Example Volume of Cone. Calculator Reference 4 - Derivatives …. Introducing Maple 19 Single Variable Differential Calculus. At what rate is distance between the two people changing when radians? so $\sec{\theta} = \frac{x}{50}$ and so I hear the next step is:. Limits: MATH 171 Problems 8 & 9 Tangent lines to parametric equations and related rates examples. In calculus, when you have an equation for y written in terms of x (like y = x2 -3x), it's easy to use basic differentiation techniques (known by mathematicians as "explicit differentiation" techniques) to find the derivative…. • derivatives of other trigonometric functions §2. Fundamental Theorem of Calculus and Integration Methods. 3 - Derivatives of Trig Functions: 3. Calculator Online Related Rates. We will continue from there tomorrow. They are also assessed on their ability to solve related rates pro. AP Calculus AB Course Details; Use implicit differentiation to find the derivative …. By observing changing rates students would be able to measure concrete examples and discover their relationships. See the page on implicit differentiation to learn how. In general then, n(t)=2t no – Thus the rate of growth of the population at time t is (dn/dt)=no2tln2 18. 5 Selecting Procedures for Calculating Derivatives 3. The student will implicit differentiation and make substitutions to solve these types of questions. Problem 6b - Calculate the rate …. Search: Implicit Differentiation With Trig. 9 Derivatives of Logarithmic and Exponential Functions. Where To Download Implicit Differentiation Homework Answers down the wall Implicit differentiation COMPLETELY Dy Calculate Dx # Dx For The Relation 3x + Xy = Y. In this tutorial students will learn how to model a related rates problem using parametric equations and a graphing utility. 5 Higher Order Derivative Using Implicit Differentiation; 3. Phone: (773) 809–5659 | Contact. If you labeled a ladder length c, and you know the rate …. The Derivative Calculator supports computing first, second, …, fifth derivatives as well as. 3C: Solve problems involving related rates, optimization, rectilinear motion, (BC) and planar motion. (KristaKingMath) Related Rates in Calculus Implicit Differentiation Examples Implicit Differentiation Showing explicit and implicit differentiation give same result | AP Calculus AB | Khan Academy More implicit differentiationCalculus Lesson 14: 2. Assign variables to anything you have or need to find, and label the picture as so. It means that the function is expressed in terms of both x and y. 1Defining Average and Instantaneous Rates of Change at a Point 2. For this calculus instructional activity, the students solve word problems following 8 different key steps. 00:26:32 – Calculate …. The steps of logarithmic differentiation are outlined below. x 3 d 2 y d x 2 + 6 x 2 d y d x + ( n 2 x 2 + 6) x y = 0. Just behind related rates problems, the topic of implicit differentiation is one of the most difficult for students in a calculus. pdf from MATHEMATIC MATH1201 at University of Technology, Jamaica. Chapter 7 Related Rates and Implicit Derivatives. If you want to evaluate the derivative at the specific points, then substitute the value of the points x and y. 74-75 3 Implicit Differentiation Review 4 QUIZ 1 5 Related Rates (p. Recall: The derivative of x with respect to x is_____ The derivative …. Correct answer: \displaystyle \frac {1} {8 \pi} inches/sec. Implicit differentiation is needed in some applications. Differentiation Techniques in Review - 7. Calculus AB: Sample Syllabus 4. The 2 comes from the derivative of the inner function and then I multiply that by the implicit derivative of x which was given as 3 so I get 6. This is a 5-hour introductory calculus course designed primarily for engineering majors and certain other technical majors. I have this related rates problem, but I keep getting a negative number for dx/dt heres the problem. This assumption does not require any work, but we need to be very careful to treat y as a function when we Calculate …. Session 31: Related Rates. 3 Finding the Derivative of a Function Given its Table of Values; 3. To generalize the above, comparative statics uses implicit differentiation to study the effect of variable changes in economic models. The topics preceded with an asterisk (*) are BC only topics. 4 Intro to related rates In related rates, we are often using implicit differentiation with respect to dt. Implicit differentiation and related rates quiz. The rate of change of the truck is dx/dt = 50 mph because it is traveling away from the intersection, while the rate …. The rate of change of the truck is dx/dt = 50 mph because it is traveling away from the intersection, while the rate of change of the car is dy/dt = −60 mph because it is traveling toward the intersection. 1st Day Homework: Academic Integrity, Parent Survey, Student Survey. I am new to implicit differentiation. S1:E 11 Implicit Differentiation and Related Rates TV-PG | Mar 5, 2010 | 31m The final strategy; implicit differentiation is used when it is difficult to solve a function for y; applying this rule to problems in related rates; the rate …. Because science and engineering often relate quantities to each other, the methods of related rates have broad applications in these fields. Now we need an equation relating our variables, which is the area equation: A = πr2. y = + sqrt ( 7 - x2) and, y = - sqrt (7 - x2). Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 264 » 24 MB) Implicit differentiation…. u Q BMwatd Ge4 Pw gi It Hhz BIXnrf eisnoi …. Use implicit di erentiation to nd an equation of the tangent line to the curve y2(y2 4) = x2(x2 5) at the point (x;y) = (0; 2). Let f(x)=g(x)/h(x), where both g and h are differentiable and h(x)≠0. Differentiate both sides of the equation with respect to “x” 2. 11) TI-89 and Implicit Differentiation; Chapter 2. Solved exercises of Implicit Differentiation. Review for the Common Exam: MATH 151 Exam 3 Review Problems 8-15. calculate a differential equation, substitute initial conditions, and solve for a missing value. identifying implicit differentiation as a tool to differentiate functions where one variable is difficult to isolate; explaining the relationship between implicit differentiation and the chain rule; calculating related rates for the time derivatives …. com/patrickjmt !! Buy my book!: '1001 Calcul. The fact that dy/dt is negative means that the distance from the top of the ladder to the ground decreases at a rate of 0. Now, select a variable from the drop-down list in order to differentiate with respect to that particular variable. Implicit Differentiation (7:55) The Rate of Change (6:53) Related Rates (8:31) All videos are closed captioned. Beyond Calculus is a free online video book for AP Calculus AB. Related rates problems can be solved using by computing. Extrema on an Interval; Rolle’s Theorem; Mean Value Theorem 8. So for hyperbolic trig functions we have the hyperbolic cosine and the hyperbolic sine. When more than one derivative is applied to a function, it is consider higher order derivatives. How long is the ladder? This is a fairly common example of a related rates problem and a common application of derivatives and implicit differentiation. The chain rule is the key to solving such problems. The steps involved in solving a related rates problem can be summarized as: 1. Rate of the spread of a rumor in sociology. Related Rates Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm 3/min. 2 - Activity 2 - Graphs of Functions and their Derivatives. Then the word rate means a derivative with respect to time. If possible, we subsequently solve for dy dx d y d x using algebra. Derivative of a Constant Multiple of a Function. 1C Calculate derivatives implicit differentiation related rate implicit differentiation related rates parametric algebraic logarithmic LO 2. RELATED RATES – Cone Problem (Water Filling and Leaking) Water is leaking out of an inverted conical tank at a rate of 10,000 at the same time water is being pumped into the tank at a constant rate. Furthermore, you’ll often find. The Derivative: Derivatives of Polynomials and Exponentials Product and Quotient Rules Derivatives of Trig Functions The Chain Rule Implicit Differentiation Logarithmic Differentiation Derivatives in the Sciences Exponential Growth and Decay Related Rates …. Plug in all the values you know, leaving only the one you're solving for. 7 Derivatives of Inverse Functions (Lecture 9) CALCULUS I MAT 301 - 0509. Inflation has its pros and cons, yet it is a normal part of a healthy economy. “In mathematics, a derivative is the rate of change of a function with respect to a variable. The vertical displacement remains constant at 50m. 07 Derivatives with Calculator…. $$displaystyle int (2x+1)^2dx$$ $$displaystyle int xsqrt{2x+3}dx$$ \(displaystyle int. 4 Implicit Differentiation (Section 3. True False Working with related rates of change is similar to implicit di erentiation in that we are nding the derivative …. Implicit Differentiation Calculator - Free Onli…. We will now use implicit differentiation on both sides with respect to t. This course is in content deeper and broader than AP Calculus AB course, suitable for the students who are intended in majoring in math, science, …. Using implicit differentiation …. In most related rates problems, we have an equation that relates derivative of both sides with respect to t (implicit differentiation). 5, we have the following equations. AP Calculus AB Related Rates and Implicit Differentiation Review Problem Set 1. The implicit differentiation calculator will find the first and second derivatives of an implicit function treating either y as a function of x or x as a function of y, with steps shown. The derivative of a function f at a number a is denoted by f' ( a ) and is given by: So f' (a) represents the slope of the tangent line to the curve at a, or equivalently, the instantaneous rate …. Calc Final Thursday!!! taking calc in college. [To see the graph of the corresponding equation, point the mouse to the graph icon at the left of the equation and press the left mouse button. Answer (1 of 7): In terms of x by implicit diffrentiation (dv/dx)=(pi/3) ((r^2) (dh/dx) + (h) (2r) (dr/dx)) (dv/dx)=((pi r^2)/3) (dh/dx) + (2hr) (dr/dx)). Related Rates Ex 1: Related Rates: Determine the Rate …. 5) l Logarithmic Differentiation (3. Functions and Limits, continuity (Ch 1) Rate of change at an instant…. It can also graph conic sections, arbitrary inequalities or. Note: Creation of this applet was inspired by a related-rates problem taken from Section 3. Equation 1: related rates cone problem pt. ) Differentiating term by term, we find the most difficulty in the first term. ONLY label things that are constant. Topics in Calculus Name_____ Date_____ Period____ ©Z O2g021 e4T aK ruutqaM XS yoAf 0txw HamrBea ELvL PCT. Just so you know, related rates is actually the Application of Implicit Differentiation by using Chain Rule in the form of dy/dx = dy/du * du/dx. The rules of differentiation produce an equation relating the rates …. Click the "Close" button in the RelatedRates window to terminate the Maplet. This course introduces you to the concept of differential calculus and the various ways to calculate rates …. In “related rates” application problems, we often assume that all variables are Implicit Differentiation Formula” (IDF for short): If F(x,y)=0 defines a differentiable relation of x and y, then So, to calculate ∂F/∂x or ∂F/∂x, calculate dF/dx or dF/dx. Computing Input interpretation: differentiate x^3 + y^3 = 4 with respect to x . That being said, this is a scenario where most teachers would be for it and accept its usage. (3) Find the derivative of √ (xy) = e x - y Solution. 95) go > The study guide for Worldwide Differential …. Visual representation of concepts. Calculus: Early Transcendentals (2nd Edition) answers to Chapter 3 - Derivatives - 3. Module 28 - Activities for Calculus Using the TI-83. Implicit Differentiation Notes and Examples. The implicit differentiation calculator will find the first and second derivatives of an implicit …. Derivatives and the Shape of a Graph: Concavity and Points of Inflection. An Implicit Home Slope Newton's Method Related Rates Extrema Optimization Mathematics of Finance Economics Physics Solved Examples Word Problems use a graphing utility or an applicable calculator …. Example 4: A 50-ft ladder is placed against a building. , a problem that requires implicit differentiation…. Students will use implicit differentiation to solve a real-world related rate problem. Specifically for the AP® Calculus BC exam, this unit builds an understanding of straight-line motion to solve problems in which particles are moving along curves in the plane. Review for the Common Exam: MATH 151 Exam 1 Review Problems 10-13 Implicit differentiation and physics applications of derivatives…. In the list of Related Rates Problems which follows, most problems are average and a few are somewhat challenging. How to Use the Implicit Differentiation Calculator? The procedure to use the implicit Differentiation calculator is as follows: Step 1: Enter the equation in a given input field. The derivative of a given function y=f(x) y = f ( x) measures the instantaneous rate of change of the output variable with respect to the input variable. 264 » 20 MB) The derivative as rate of change. (7) Find the derivatives of the following. We help you handle and prevent agitated behavior. Some relationships cannot be represented by an explicit function. Connections to the Study Design: AOS 3 - Calculus. Download or read online full book title Developing Understanding Of The Chain Rule Implicit Differentiation And Related Rates …. Apply the derivative to calculate rates …. (a) We can answer this question two ways: using "common sense" or related rates. Draw a sketch if it is possible 3. Author: Haley Paige Jeppson Publisher: ISBN: Release Date: 2019 Size: 37. AP Calculus AB Course Curriculum. 2 Derivatives of Inverse Functions. derivatives, and implicit differentiation Me during related rates,. Implicit Differentiation Related Rates One of the applications of mathematical modeling with calculus involves related rates word problems. Find derivatives of radical functions Also consider: • Power rule II Related Rates 3. 4: Related Rates (preprinted) Chapter 3. 5 Derivatives and the Shape of a Graph. Related Rates 1: (dA)/dt = 2 \pi r (dr)/dt Worksheet #35: W s #35 Solutions: 1-6, 7-10, 12 review: Related Rates 2: dz/dt = x dx/dt + y dy/dt Worksheet #36: Implicit Differentiation; Implicit Differentiation – Basic Idea and Examples ; Implicit Differentiation …. Limits: An Intuitive Approach - Answers. So the total rate of horiztonal displacement is 6m/s. Sketching Graphs 2: anti-derivatives. We use implicit differentiation to differentiate an implicitly defined function. Example 1: Find if x 2 y 3 − xy = 10. Strategy 1: Use implicit differentiation directly on the given equation. 11: Implicit Differentiation and Related Rates is shared under a CC BY 3. The topics below are both AB and BC topics. This can be done through the use of a financial calculator, software, an online calculator, or present value tables. In differential calculus, related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. Differentials and Newton's method. Implicit differentiation will allow us to find an equation that. Learn to solve different kinds of related rates problems in Calculus. The derivate of a function at a chosen input value. A free online 2D graphing calculator (plotter), or curve calculator, that can plot piecewise, linear, quadratic, cubic, quartic, polynomial, trigonometric, hyperbolic, exponential, logarithmic, inverse functions given in different forms: explicit, implicit, polar, and parametric. Differentiating x to the power of something. Below you will find all homework assignments (and answers) for Calculus 1. {"results":"\u003cdiv class='col-xs-12 search-result-item thumbnail-card col-lg-6 col-sm-6' data-id='5439' data-item-type='CollectionItemFolder' data-type. We have seen how implicit differentiation can be used to find derivatives of implicit relations. Many (not all!) related rates problems present a quantity …. Download Vui lòng tải xuống để xem tài liệu đầy đủ. With an explicit function, given an x value, we have an explicit formula for computing the corresponding y value. Compute and interpret limits of functions using analytic and other methods, including L'Hospital's Rule. 2, 6 Important Example 23 Ex 5. How do you calculate the yearly inflation rate?. Notice that volume is a function of time because we’re looking at how volume changes over time (or the rate …. Resource Allocation: Implicit Differentiation; Related Rates; Business Applications: Social Sciences. Simply saying, it’s just the SLOPE of. Calculus is difficult part of mathematics and it. The idea behind linearization or local linear approximation is to find a value of the function at the given point and evaluate the derivative …. For most homework, a calculator will not be needed. Step by Step Implicit Differentiation Slope of Inverse Function All in one Rate Explorer Differentiability of piecewise-defined function Absolute and Percent Change Differentials APPS: Max Volume of Folded Box APPS: Min Distance Point to Function f(x) APPS: Related Rates …. Business and Industry Applications. Take help of free calculator to get Implicit Differentiation …. As your next step, simply differentiate the y terms the same way as you differentiated the x terms. cost, strength, amount of material used in a building, profit, loss, etc. Using the derivative to calculate the slope of a tangent at any point on a function. Calculate derivatives of hyperbolic functions. I know that y=3/4 from substituting xy=3 with x=-4, but I do not. Now differentiate with respect to x x. Implicit Differentiation and Related Rates In our work up until now, the functions we needed to differentiate were either given explicitly, such as y = x2 +ex, y = x 2 + e x, or it was possible to get an explicit formula for them, such as solving y3 − 3x2 = r y 3 − 3 x 2 = r to get y = 3√5+3x2. Related Rates are problems that connect rate …. In this section we are going to look at an application of implicit differentiation. Free Response: A calculator may not be used. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate …. 9: Lecture 17: Related Rates ROTC Long Weekend Patrick's Videos: Logarithmic Differentiation Examples 1 2 3 Paul's Notes: Inverse Functions; 8: Mon 10/11: 3. High School Math Solutions – Derivative Calculator…. Rates of Change of Rectangle 3-8-21. Typically related rates problems will follow a similar pattern. The top of a ladder slides down a vertical wall at a rate of 0. Answer (1 of 9): To calculate the derivative of anything, you need to specify the variable that is changing. Practice: Analyzing related rates problems: expressions. | 2022-05-28T23:21:42 | {
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# What is the units digit of 2222^(333)*3333^(222) ?
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28 Jan 2012, 11:55
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What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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09 Mar 2014, 13:13
1
For more on this kind of questions check Units digits, exponents, remainders problems collection.
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What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
(2222^333)*(3333^222)=2222^111*(2222^222*3333^222)
here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 (2222^222*3333^222). since 6 powered in any number more than 0 results in 6 as a units digit, as a result we have-
6*2222^111
2 has a cycle of 4 . 111=27*4+3 . 2^3=8
6*8=48
so the units digit is 8, and the answer is E
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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26 Feb 2015, 21:22
Use cyclicity rule here ,
2^1=2,
2^2=4,
2^3=8
2^4=6
2^5=2
We can see here unit digit repeated after 4 powers so cyclicity of 2 is 4 . devide power by 4 and check above .
Ans E
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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26 Feb 2015, 21:43
Hi All,
Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeating "pattern" of the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(3333)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out the "cycle" of the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4 "powers", the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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28 Feb 2015, 04:17
belagerfeld wrote:
What is the units digit of 2222^(333)*3333^(222)?
A. 0
B. 2
C. 4
D. 6
E. 8
I get 2, but it's apparently not the correct answer..
Source: some random learning sheet
Unit digit of 2222^333 is same as unit digit of 2^333.
Unit digit of powers of 2 follows a pattern: 2, 4, 8, 6
Now, 4*83 = 332 i.e. 2^332 uses 6 as its unit digit.
Hence, 2^333 will have unit digit as 2.
Unit digit of 3333^222 is same as unit digit of 3^222.
Unit digit of powers of 3 follows a pattern: 3, 9, 7, 1
Now, 4*55 = 220 i.e. 3^220 uses 1 as its unit digit.
Hence, 3^221 will have unit digit as 3.
And, 3^222 will have unit digit as 9.
Now we have 2 * 9 = 18
So the final unit digit of 2222^(333)*3333^(222) = 8.
Hence option E.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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22 May 2018, 02:57
belagerfeld wrote:
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
This type of sums can also be solved using Fermet's Theorem approach.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink] 06 Jun 2019, 23:49
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https://gmatclub.com/forum/686-000-in-bonus-money-is-to-be-divided-among-6-employees-127503.html | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
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26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
x + 5*1.2x = $686,000 => 7x =$686,000 => x = $98'000 Hence, solution c is correct. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7177 Location: Pune, India Followers: 2161 Kudos [?]: 13985 [0], given: 222 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
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26 Nov 2010, 19:15
Stanford2012 has explained the solution perfectly.
The only thing I want to highlight here is the concept.
Let's say you have a fixed sum of $100 that needs to be distributed among 5 people. If you want to maximize one person's share, you need to give everyone else the minimum that you can. e.g. the question says each person must receive at least$10, what is the maximum one person can receive?
Since you want to maximize one person's share, you should give everyone else the minimum possible i.e. $10. You give$10 to each of the other 4 people and are left with $60, the maximum that you can give to one guy. Similarly, if you want to minimize someone's share, you should give everyone else the maximum possible. In your question here, a fixed amount has to be distributed among 6 people and one person's minimum share needs to be found. A person will receive minimum (say$x) , when everyone else gets the maximum that they can (that will be $6x/5). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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$686,000 in bonus money is to be divided among 6 employees [#permalink] ### Show Tags 13 Feb 2012, 19:34 2 This post was BOOKMARKED$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
A. $96,000 B.$97,000
C. $98,000 D.$99,000
E. $100,000 I often struggle with such questions. I can understand that for minimum value , we need to consider highest salary 20%, however I cannot understand how to put other 4 intermediate values. When I read the explanation I understand, but do not seem to apply it while solving. Can someone pls help in how to think through such problems. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 37036 Followers: 7230 Kudos [?]: 96127 [1] , given: 10707 Re:$686,000 in bonus money is to be divided among 6 employees [#permalink]
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13 Feb 2012, 20:45
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priyankaparanjape wrote:
$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive? A.$96,000
B. $97,000 C.$98,000
D. $99,000 E.$100,000
I often struggle with such questions. I can understand that for minimum value , we need to consider highest salary 20%, however I cannot understand how to put other 4 intermediate values. When I read the explanation I understand, but do not seem to apply it while solving. Can someone pls help in how to think through such problems.
Thanks.
Since "no employee is to receive a bonus more than 20% greater than the bonus received by any other employee", then the bonuses of 6 employees must be in the range: $$x$$ and $$1.2x$$.
So we want to minimize $$x$$. To minimize $$x$$ we should make only one employee to receive that bonus (minimum possible) and the rest 5 employees to receive $$1.2x$$ bonus (maximum possible).
$$x+5*1.2x=686$$ --> $$7x=686$$ --> $$x=98$$.
General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.
Similar questions:
a-certain-city-with-population-of-132-000-is-to-be-divided-76217.html?hilit=population#p1035185
shaggy-has-to-learn-the-same-71-hiragana-characters-and-126948.html
Other min/max questions:
PS: search.php?search_id=tag&tag_id=63
DS: search.php?search_id=tag&tag_id=42
Hope it helps.
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12 Jun 2014, 20:11
Hi Bunuel,
I am confused. Why are we minimizing 1 persons bonus and maxing out all the other 5??
Shouldnt we be minimizing 1 persons bonus, maxing only 1 other persons bonuses and keeping the 4 in the middle at 1/6 of the total bonus. So let 686,000/6=@
@ is what everyone should get in an even world.
So for the first guy, whom we eant to minimize, i am going to deduct x from his @. For the 6th guy, the guy we want to max out where going to give him x, but such that the following condition is met.
(@-x)1.2=(@+x)
.2@=2.2x
.2(686,000/6)/2.2=10393.0=x
Now the first guy, the minimum bonus guy is going to get
@-x=114333-10393=103,940
Where am I going wrong?
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Re: $686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 15 May 2016, 12:01 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] 15 May 2016, 12:01
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# In how many different ways can trhee letters be posted from
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In how many different ways can trhee letters be posted from [#permalink]
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1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
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1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
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### Show Tags
17 Feb 2010, 03:47
Ravshonbek wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
warm up
1. 7 x 6 x 5 = 210
2. 7 x 7 x 7 = 343
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### Show Tags
10 Feb 2012, 04:03
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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10 Feb 2012, 05:05
1. 7 (no restriction) * 6 (can not be the same as the first one) * 5 (can not be the same as the first and second one) = 210
2. 7 (no restriction) * 7 (no restriction) * 7 (no restriction) = 343
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10 Feb 2012, 09:13
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mohankumarbd wrote:
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
"Two or more letters can be posted from the same box" means that all 3 letters can be posted from the same postbox (so we don't have the restriction we had for the first question).
Now, since there are 7 postboxes then each of these 3 letters has 7 options to be posted from, total # of ways is 7*7*7=7^3.
Hope it's clear.
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### Show Tags
25 May 2013, 08:22
Bunuel wrote:
mohankumarbd wrote:
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
"Two or more letters can be posted from the same box" means that all 3 letters can be posted from the same postbox (so we don't have the restriction we had for the first question).
Now, since there are 7 postboxes then each of these 3 letters has 7 options to be posted from, total # of ways is 7*7*7=7^3.
Hope it's clear.
Hi Bunnel,
I tried to do the second question via combinatorics, but i am not able to figure it out, please check the below method and guide where i went wrong
= all three in one box +2 in one box and the last one in a different box + all three in different boxes
= 3c3*7+3c2*7c1*6c5+3c1*7c3
= 7+ 3*7*6+3*7*6*5
= 7 + 126 + 270 = wrong
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25 May 2013, 13:06
Quote:
I tried to do the second question via combinatorics, but i am not able to figure it out, please check the below method and guide where i went wrong
= all three in one box +2 in one box and the last one in a different box + all three in different boxes
= 3c3*7+3c2*7c1*6c5+3c1*7c3
= 7+ 3*7*6+3*7*6*5
= 7 + 126 + 270 = wrong
Think it in this way,
First letter can go to any 7 post offices
Same case with second and same case with the third letter as well so 7*7*7
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Re: In how many different ways can trhee letters be posted from [#permalink]
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Re: In how many different ways can trhee letters be posted from [#permalink]
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21 Feb 2015, 21:53
Could someone please tell me why part 1 cannot be answered using "7C3" = 35 ways?
Many thanks.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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22 Feb 2015, 00:00
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Hi icetray,
Since the question is worded in a "vague" way, you bring up an interesting interpretation of it. Thankfully, questions on the Official GMAT are worded to remove ambiguity and "bias" on the part of the reader, so you won't have to worry about that on Test Day. This prompt reads as if it were created by the original poster, so it's not clear what he/she was "intending" the question to mean.
As it is, your interpretation of the prompt makes a lot of sense - there does not seem to be any reason why we should emphasize the "order" of the letters (there's no reference to "first letter", "second letter", "third letter" and no reference to "arrangements"). Using post-boxes ABC would be same as BCA, CBA, etc., so if we interpret the prompt as a "combinations" question, then 7c3 = 35 would be correct.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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06 Apr 2016, 08:41
Ravshonbek wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
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3. What if we assume no three letters can be posted from same postbox?
Re: In how many different ways can trhee letters be posted from [#permalink] 06 Apr 2016, 08:41
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https://math.stackexchange.com/questions/18686/uniform-random-point-in-triangle | # uniform random point in triangle
Suppose you have an arbitrary triangle with vertices $A$, $B$, and $C$. This paper (section 4.2) says that you can generate a random point, $P$, uniformly from within triangle $ABC$ by the following convex combination of the vertices:
$P = (1 - \sqrt{r_1}) A + (\sqrt{r_1} (1 - r_2)) B + (r_2 \sqrt{r_1}) C$
where $r_1, r_2 \sim U[0, 1]$.
How do you prove that the sampled points are uniformly distributed within triangle $ABC$?
I would argue that if it is true for any triangle, it is true for all of them, as we can find an affine transformation between them. So I would pick my favorite triangle, which is $A=(0,0), B=(1,0), C=(0,1)$. Then the point is $(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$ and we need to prove it is always within the triangle and evenly distributed. To be in the triangle we need $x,y\ge 0, x+y\le 1$, which is clear. Then show that the probability to be within an area $(0,x) \times (0,y)$ is $2xy$ by integration.
• You mean 2xy, I think. – TonyK Jan 24 '11 at 9:19
• @Tony K: Right. Fixed. – Ross Millikan Jan 24 '11 at 13:41
• It's been quite a lot since I was doing this, but can someone post a full proof? Sry for necromancy, btw :) – ioreskovic Jun 26 '13 at 14:07
Pick $A,B,C = (0,0),(1,0),(1,1)$. For any point $(x,y)$, we have that $(x,y)$ is in the triangle if and only if $0 < x < 1$ and $0 < y/x < 1$.
Now, we look for the distribution of $x$ and $y/x$.
Computing a few triangle areas, we can easily check that $P(0 < x < x_0) = x_0^2$. Hence $P(0 < x^2 < a) = P(0 < x < \sqrt a) = a$, so that $x^2$ is uniformly distributed in the unit interval.
Again with an area computations, we can check that $P(0 < y/x < k) = k$. Hence $y/x$ is also uniformly distributed in the unit interval.
Finally we have to check (again computing a simple area) that $P(0 < x < x_0 \land 0 < y/x < k) = x_0^2k$ which proves that $x^2$ and $y/x$ are independant.
So we have found that to generate a point uniformly in the triangle is the same as picking $x^2 = r_1$ and $y/x = r_2$ uniformly in the unit interval, and then form $(x,y) = (\sqrt r_1, r_2 \sqrt r_1)$, which is the barycenter of $(A,1- \sqrt {r_1})(B,\sqrt {r_1}(1-r_2))(C,\sqrt{r_1}r_2)$.
• How to generalize it onto higher dimensions? What to do with arbitrary $n$-simplex? – Orient Jul 18 '14 at 17:54
• If the dimension is $d$, first change your coordinates so that the simplex is formed by the points $(0,\ldots,0),(1,0,\ldots,0),\ldots,(1,\ldots,1,0),(1,\ldots,1)$. Then for the first coordinate, pick $x_1^d$ uniformly. Then pick $(x_2/x_1,x_3/x_1,\ldots,x_d/x_1)$ uniformly in the $d-1$-dimensional simplex (by induction) – mercio Jul 26 '14 at 14:47
• merico, I think it is wrong. We need spatial uniform distribution. – Orient Jul 27 '14 at 6:25
• One way to get random point inside of simplex $P = \{\mathbf{p}_i\}_{i = 1}^{d + 1}$ is to pick $\mathbf{c} = (c_1, c_2, ..., c_d, c_{d + 1}), c_i \sim U[0;1]$, then $\mathbf{c} \leftarrow -\log(\mathbf{c})$, then $c \leftarrow \displaystyle \frac{\mathbf{c}}{\sum \limits_{i = 1}^{d + 1} c_i}$, then random point is: $\displaystyle \sum \limits_{i = 1}^{d + 1}c_i \cdot \mathbf{p}_i$ (based on Dirichlet distribution and properties of affine transformations). – Orient Jul 27 '14 at 6:26
• But the generalization of your approach itself is here math.stackexchange.com/questions/563129/… . – Orient Jul 27 '14 at 6:31
I'm starting with the argument provided by @Ross Millikan. Let $A=(0,0),\ B=(1,0),\ C=(0,1)$. Then the point chosen according to the given equation is $P=(X,Y)=(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$. Now clearly, $0\leq X,Y \leq 1$ and $X+Y\leq \sqrt{r_1}\leq 1$. Now the problem is to show that $\mathbb{P}(X\leq x, Y\leq y)=2xy,\ \forall 0\leq x,y\leq 1$ with $x+y\leq 1$. Now, \begin{equation*} \begin{split} \mathbb{P}(X\leq x, Y\leq y)=& \mathbb{P}(\sqrt{r_1}(1-r_2)\leq x, r_2\sqrt{r_1}\leq y)\\ \ =&\int_{0}^1 \mathbb{P}(\sqrt{r}(1-r_2)\leq x, r_2\sqrt{r}\leq y|r_1=r)f_{r_1}(r)dr\\ \ =&\int_{0}^1 \mathbb{P}(1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}})I_{[0,1]}(r)dr\ \mbox{(Since, $r_1, r_2$ are i.i.d $\mathcal{U}[0,1]$})\\ \end{split} \end{equation*} Now to find the region of integration we note that $$1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}}\ \Rightarrow\ 0\leq r\leq(x+y)^2$$ Also, if $x\leq y$ then $$r\in (0,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (x^2,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (y^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$ and if $y\leq x$ then $$r\in (0,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (y^2,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\ r\in (x^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$
Then if $x\leq 1$ the integral becomes $$\int_{0}^{x^2}1 dr+\int_{x^2}^{y^2}\frac{x}{\sqrt{r}} dr+ \int_{y^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$ Similarly, if $y\leq x$ the integral becomes $$\int_{0}^{y^2}1 dr+\int_{y^2}^{x^2}\frac{y}{\sqrt{r}} dr+ \int_{x^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$ Hence the point $P$ is uniformly distributed on the surface of the triangle $ABC$. $\hspace{3cm}\ \Box$
Note that these points, when random, will be uniformly distributed in a nicely random way, but if you loop through r1 and r2 with an increment (say .01) your resulting points will have unusual artifacts and not look randomly distributed. One end of the triangle may have few points.
I determined this with code ( Note that similar code, using math.Random() looks fine).
FillTriangleWithPointsBarycentric
if r1 and r2 are uniform random numbers between 0 and 1 then
This math produces a uniform distribution (note √ means sqrt of)
d = (1.0−√r1)*vector1 + √r1*(1.0−r2)*vector2+√r1 * r2 * vector3
But rather than using a uniform random number, just loop through them and the result does not look good.
@param {THREE.Vector3} vector1
@param {THREE.Vector3} vector2
@param {THREE.Vector3} vector3
@param {Array<number>} output input/output points > [x0,y0,z0,x1,y1,z1,...xn,yn,zn] displayable in point cloud
@returns {void}
FillTriangleWithPointsBarycentric(vector1, vector2, vector3, output) {
let triangle = new THREE.Triangle(vector1, vector2, vector3);
let area = triangle.getArea();
console.log('Area is ' + area);
area = Math.sqrt(area);
console.log('sqrt Area is ' + area);
let increment = 0.1 / area;
for (let r1 = 0; r1 <= 1; r1 += increment) {
for (let r2 = 0; r2 <= 1; r2 += increment ) {
// of course this is javascript we have to write this out instead of
// using only one line
let sqrtR = Math.sqrt(r1);
let A = (1 - sqrtR);
let B = (sqrtR * (1 - r2));
let C = (sqrtR * r2);
let x = A * vector1.x + B * vector2.x + C * vector3.x;
let y = A * vector1.y + B * vector2.y + C * vector3.y;
let z = A * vector1.z + B * vector2.z + C * vector3.z;
output.push(x, y, z);
}
} | 2019-02-23T09:36:04 | {
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https://stats.stackexchange.com/questions/50321/probability-of-picking-a-biased-coin | # Probability of picking a biased coin
Suppose you have a bag of 100 coins of which 1 is biased with both sides as Heads. You pick a coin from the bag and toss it three times. The result of all three tosses is Heads. What is the probability that the selected coin is biased?
P(selecting a biased coin) = 1/100
P(getting a head thrice with the biased coin) = 1
P(selecting an unbiased coin) = 99/100
P(getting a head thrice with the unbiased coin) = 1/8
P(selecting a biased coin|coin toss resulted in 3 heads) =
P(selecting a biased coin and getting heads thrice)/
[P(selecting a biased coin and getting heads thrice) +
P(selecting an unbiased coin and getting heads thrice)]
= (1/100)/[(1/100) + (99/800)]
= (1/100)/(107/800)
= 8/107
= 0.0747
Is this correct? Thanks.
• Is this a homework question? – Carlos Accioly Feb 19 '13 at 12:19
• @CarlosAccioly This question was posed to me in an examination on Sunday. Just wanted to verify my answer. – tejas_kale Feb 19 '13 at 12:44
• Hi tejas, a standard-level question on an exam - even if you're pursuing it for self study - falls under the scope of the homework tag. – Glen_b Feb 20 '13 at 2:16
• @Glen_b Got your point. I thought that Carlos' question was to ensure that the poster is not cheating with his homework. Thanks for the info. – tejas_kale Feb 20 '13 at 11:48
Your answer is right. The solution can be derived using Bayes' Theorem:
$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$
You want to know the probability of $P(\text{biased coin}|\text{three heads})$.
What do we know?
There are $100$ coins. $99$ are fair, $1$ is biased with both sides as heads.
With a fair coin, the probability of three heads is $0.5^3 = 1/8$.
The probability of picking the biased coin: $P(\text{biased coin}) = 1/100$.
The probability of all three tosses is heads: $P(\text{three heads}) = \frac{1 \times 1+ 99 \times \frac{1}{8}}{100}$.
The probability of three heads given the biased coin is trivial: $P(\text{three heads}|\text{biased coin}) = 1$.
If we use Bayes' Theorem from above, we can calculate
$$P(\text{biased coin}|\text{three heads}) = \frac{1 \times 1/100}{\frac{1 + 99 \times \frac{1}{8}}{100}} = \frac{1}{1 + 99 \times \frac{1}{8}} = \frac{8}{107} \approx 0.07476636$$
• Cool! Cool! Cool! – Behacad Feb 19 '13 at 14:33
Here is a write that describes something very similar to that. The Bayes approach is the right way to proceed.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
In general, P(biased coin|k heads) = (2^k)/[(2^k) + 99] Where k is no of consective heads
so if the trick coin was tossed 3 times (2^3)/[(2^3) + 99] = 8/(8+99) = 8/107 = 0.07 | 2019-08-22T09:45:07 | {
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http://www.livestatue.com/z8rwpx/3nm1d.php?tag=21e94b-idempotent-matrix-eigenvalues | Then, $A^k = A^{k-1}A = AA = A$, as required. Eigenvalues. First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$. Proof: Let λ be an eigenvalue of A and q be a corresponding eigenvector which is a non-zero vector. $P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. (For a proof, see the post “Idempotent matrix and its eigenvalues“.) Show that the rank of an idempotent matrix is equal to the number of nonzero eigenvalues of the matrix. If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Eigenvalues. This site uses Akismet to reduce spam. This website’s goal is to encourage people to enjoy Mathematics! This provides an easy way of computing the rank, or alternatively an easy way of determining the trace of a matrix whose elements are not specifically known (which is helpful in statistics, for example, in establishing the degree of bias in using a sample variance as an estimate of a population variance). 1 & -2 (2) Let A be an n×n matrix. The post contains C++ and Python code for converting a rotation matrix to Euler angles and vice-versa. Theorem 3. Let Aand Bbe idempotent matrices of the same size. The short answer is: 0, 1 are the ONLY possible eigenvalues for an idempotent matrix A. \begin{bmatrix} 1 & 0 \\ ST is the new administrator. Let [E_0={mathbf{x}in R^n mid Amathbf{x}=mathbf{0}} text{ and } […], […] the post ↴ Idempotent Matrix and its Eigenvalues for solutions of this […], […] the post ↴ Idempotent Matrix and its Eigenvalues for […], […] Idempotent Matrix and its Eigenvalues […], Your email address will not be published. I think, you want to know the relation between the singular values and the eigenvalues of idempotent matrices. All its eigenvalues are positive. An nxn matrix A is called idempotent if A 2 =A. \qquad Let C be a symmetric idempotent matrix. b. Therefore, it defines a projection (not orthogonal) on its range, which we denote by S. Matrix I - A maps \( … An idempotent linear operator $P$ is a projection operator on the range space $R(P)$ along its null space $N(P)$. Theorem: \begin{bmatrix} 6. Show that the eigenvalues of C are either 0 or 1. Final Exam Problems and Solution. Principal idempotent of a matrix example University Duisburg-Essen SS 2005 ISE Bachelor Mathematics. 4.1. which is a circle with center (1/2, 0) and radius 1/2. In this section K = C, that is, matrices, vectors and scalars are all complex.Assuming K = R would make the theory more complicated. c. Let d be a n × 1 vector. \qquad For every n×n matrix A, the determinant of A equals the product of its eigenvalues. Suppose v+ iw 2 Cnis a complex eigenvector with eigenvalue a+ib (here v;w 2 Rn). The hat matrix (projection matrix P in econometrics) is symmetric, idempotent, and positive definite. The matrix, is idempotent and since it is a doagonal matrix, its eigen value are the diagonal entries λ = 0 and λ = 1. \end{bmatrix} [/math], $X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}$, $\hat{e}^\textsf{T}\hat{e} = (My)^\textsf{T}(My) = y^\textsf{T}M^\textsf{T}My = y^\textsf{T}MMy = y^\textsf{T}My.$. Then, λqAqAqAAq Aq Aq q q== = = = = =22()λλ λλλ. Since there are only 2 idempotent square matrices, you can just try them both for parts a and b. λ = λ2. Show that 1 2(I+A) is idempotent if and only if Ais an involution. Eigenvalues of idempotent matrices are either 0 or 1. This page was last edited on 20 November 2020, at 21:34. Then, λqAqAqAAq Aq Aq q q== = = = = =22()λλ λλλ. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. Proof: Let A be an nxn matrix, and let λ be an eigenvalue of A, with corresponding eigenvector v. The eigenvalues of an idempotent matrix take on the values 1 and 0 only. Note that applying the complex conjugation to the identity A(v+iw) = (a+ib)(v+iw) yields A(v iw) = (a ib)(v iw). An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. [3] The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Jiming Wu. Since x is a nonzero vector (because x is an eigenvector), we must have. eigenvalues of the matrix A. Eigenvalues. A question on a nilpotent matrix: Advanced Algebra: Aug 6, 2013: Prove that it is impossible for a 2x2 matrix to be both nilpotent and idempotent: Advanced Algebra: Mar 25, 2013: Matrix of a Nilpotent Operator Proof: Advanced Algebra: Mar 27, 2011: relation between nilpotent matrix and eigenvalues: Advanced Algebra: Mar 26, 2011 Let Hbe a symmetric idempotent real valued matrix. In linear algebra, an idempotent matrix is a matrix which, when multiplied by itself, yields itself. 7. A matrix is idempotent () if and only if it is diagonalizable and all the eigenvalues are 0 or 1. Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues. […], […] only possible eigenvalues of an idempotent matrix are $0$ or $1$. \begin{bmatrix} For idempotent matrix, the eigenvalues are ##1## and ##0##. Fingerprint Dive into the research topics of 'Eigenvalues and eigenvectors of matrices in idempotent algebra'. \hat{e} = y - X \hat\beta Solution: Suppose that λ is an eigenvalue of A. Prove that if A is idempotent, then det(A) is equal to either 0 or 1. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. The eigenvalues of A are the roots of its characteristic equation: |tI-A| = 0.. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . Theorem 4 shows that these eigenvalues are also eigenvalues of the difference of idempotent density matrices. b. Consider the following 2 cases: Case (1): A is nonsingular. -1 & 3 & 4 \\ Hence by the principle of induction, the result follows. Theorem: Let Ann× be an idempotent matrix. Proof: If A is idempotent, λ is an eigenvalue and ⦠[/math], $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$, $\begin{pmatrix}a & b \\ b & 1 - a \end{pmatrix}$, $\left(a - \frac{1}{2}\right)^2 + b^2 = \frac{1}{4}$, $A = \frac{1}{2}\begin{pmatrix}1 - \cos\theta & \sin\theta \\ \sin\theta & 1 + \cos\theta \end{pmatrix}$, $\begin{pmatrix}a & b \\ c & 1 - a\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\beta)^\textsf{T}(y - X\beta)$, $\hat\beta = \left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y$, $This website is no longer maintained by Yu. In the case of irreducible mattices, the problem is reduced to the analysis of an idempotent analogue of the charactetistic polynomial of the mattix. Hence solving λ(λ â 1) = 0, the possible values for λ is either 0 or 1. The 'only if' part can be shown using proof by induction. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. 3. Eigenvalues of a Hermitian Matrix are Real Numbers, If A^{\trans}A=A, then A is a Symmetric Idempotent Matrix, Find all Values of x such that the Given Matrix is Invertible. (Linear Algebra Math 2568 at the Ohio State University), The Ideal Generated by a Non-Unit Irreducible Element in a PID is Maximal, In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal. Find All the Eigenvalues of 4 by 4 Matrix, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Diagonalize a 2 by 2 Matrix if Diagonalizable, Find an Orthonormal Basis of the Range of a Linear Transformation, The Product of Two Nonsingular Matrices is Nonsingular, Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not, Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials, Find Values of a , b , c such that the Given Matrix is Diagonalizable, Diagonalize the 3 by 3 Matrix Whose Entries are All One, Given the Characteristic Polynomial, Find the Rank of the Matrix, Compute A^{10}\mathbf{v} Using Eigenvalues and Eigenvectors of the Matrix A, Determine Whether There Exists a Nonsingular Matrix Satisfying A^4=ABA^2+2A^3, Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix, Idempotent (Projective) Matrices are Diagonalizable, Idempotent Matrices. We can see that the distribution of the quadratic form is a weighted sum of \chi_1^2 random variables, where the weights are the eigenvalues of the variance matrix. Ax= λx⇒Ax= AAx= λAx= λ2x,soλ2 = λwhich implies λ=0 or λ=1. These two conditions can be re-stated as follows: 1.A square matrix A is a projection if it is idempotent, 2.A projection A is orthogonal if it is also symmetric. Request PDF | Eigenvalues and eigenvectors of matrices in idempotent algebra | The eigenvalue problem for the mattix of a generalized linear operator is considered. If λ is an eigenvalue of an idempotent matrix, show that λ is either 0 or 1. Those eigenvalues (here they are 1 and 1=2) are a new way to see into the heart of a matrix. Discuss the analogue for A−B. The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Called spectral theory, it allows us to give fundamental structure theorems for matrices and to develop power tools for comparing and computing withmatrices. The only non-singular idempotent matrix is the identity matrix; that is, if a non-identity matrix is idempotent, its number of independent rows (and columns) is less than its number of rows (and columns). Published 02/22/2018, […] Since A has three distinct eigenvalues, A is diagonalizable. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. The trace of an idempotent matrix â the sum of the elements on its main diagonal â equals the rank of the matrix and thus is always an integer. The trace is related to the derivative of the determinant (see Jacobi's formula). Since the matrix A has 0 as an eigenvalue. a. This characterization can be used to define the trace of a linear operator in general. PRACTICE PROBLEMS (solutions provided below) (1) Let A be an n × n matrix. Claim: Each eigenvalue of an idempotent matrix is either 0 or 1. Maximum number of nonzero entries in k-idempotent 0-1 matrices \end{bmatrix} 1 & 0 & 0 \\ Thus the number positive singular values in your problem is also n-2. Enter your email address to subscribe to this blog and receive notifications of new posts by email. 9. Together they form a unique fingerprint. If you look at your definition of idempotent A^2=A, then you can actually solve this for A and find *all* idempotent square matrices. Thus a necessary condition for a 2 × 2 matrix to be idempotent is that either it is diagonal or its trace equals 1. 0 & 1 You should be able to find 2 of them. Eigenvalues. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. Theorem: Let Ann× be an idempotent matrix. = My. Proof: A is idempotent, therefore, AA = A. The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Eigenvalues. To explain eigenvalues, we first explain eigenvectors. 0 & 0 & 1 The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Show that 1 2(I+A) is idempotent if and only if Ais an involution. \begin{bmatrix} Show that = 0 or = 1 are the only possible eigenvalues of A. A.8. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. All the matrices are square matrices (n x n matrices). (Hint: Note that Cq = γq implies 0 = Cq − γq = CCq − γq = CCq − γq = γ 2 q − γq and solve for γ.) This can only occur if = 0 or 1. 1 & 0 \\ An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1.[3]. Proof: Let λ be an eigenvalue of A and q be a corresponding eigenvector which is a non-zero vector. For this product [math]A^2$ to be defined, $A$ must necessarily be a square matrix. → 2 → ()0 (1)0λλ λ λ−=→−=qnn××11qλ=0 or λ=1, because q is a non-zero vector. In the special case where these eigenvalues are all one we do indeed obtain $\mathbf{z}^\text{T} \mathbf{\Sigma} \mathbf{z} \sim \chi_n^2$, but in general this result does not hold. 2 & -2 & -4 \\ Theorem 2.2.
## idempotent matrix eigenvalues
Switch Window Key Chromebook, Best Yarn For Punch Needle, Nature Communications Acceptance Rate, Preposition Of Direction Exercises, Where Can I Buy Jays Potato Chips, Shadow Ridge Apartments - El Paso, | 2021-12-08T00:55:12 | {
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http://math.stackexchange.com/questions/229105/limit-of-a-sequence-bounded-below-but-has-no-cluster-points | # Limit of a sequence bounded below but has no cluster points
Question is this: Let $a_n$ be a sequence of real numbers. Prove that if $a_n$ is bounded below and has no cluster points then $a_n$ → ∞.
I could not really find a way to prove it. Could you give me a hand?
-
HINT: If $a_n\not\to\infty$, then the sequence has an infinite subsequence that is bounded above. (Why?) That subsequence is actually a bounded sequence. What do you know about cluster points of bounded sequences?
Added: Suppose that $a_n\not\to\infty$. Then there is some $M$ such that for all $m\in\Bbb N$ there is a $k\ge m$ with $a_k<M$. Let $A=\{k\in\Bbb N:a_k<M\}$ clearly $A$ is infinite, so we can list it as $A=\{n_k:k\in\Bbb N\}$ in such a way that $n_0<n_1<n_2<\ldots~$. We now have a subsequence $\sigma=\langle a_{n_k}:k\in\Bbb N\rangle$ of the original sequence.
The original sequence is bounded below by hypothesis, so $\sigma$ is also bounded below, and by construction $\sigma$ is bounded above by $M$. Being a bounded sequence in $\Bbb R$, $\sigma$ has a convergent subsequence $\sigma'=\langle a_{n_{k_i}}:i\in\Bbb N\rangle$; let $x$ be the limit of $\sigma'$.
Can each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_{n_{k_i}}|<\epsilon$ whenever $i\ge m_\epsilon$; that’s just the definition of convergence. But that means that for each $\epsilon>0$, infinitely many terms of the original sequence are in the interval $(x-\epsilon,x+\epsilon)$, i.e., $x$ is a cluster point of the original sequence.
We’ve shown that if $a_n\not\to\infty$, then the sequence has a cluster point; it follows immediately that if it has no cluster point, then $a_n\to\infty$, by taking the contrapositive.
-
Actually $a_n\not\to\infty$ does not imply that it's bounded above. Take, for instance, $a_{2n-1}=0$, $a_{2n}=2n$. – nonpop Nov 4 '12 at 19:57
@nonpop: You’re right, of course. I’ll make the minor fix required. (Somehow I was thinking that it was monotone.) – Brian M. Scott Nov 4 '12 at 20:00
@BrianM.Scott If I am not wrong, if a sequence is bounded then according to Bolzano-Weierstrass theorem, a cluster point should exist right? – Amadeus Bachmann Nov 4 '12 at 20:08
@Zxy: That’s exactly right. And if the subsequence has a cluster point, what about the original sequence? – Brian M. Scott Nov 4 '12 at 20:10
@BrianM.Scott then the original sequence has a limit and therefore it is convergent? – Xentius Nov 4 '12 at 20:42
show 5 more comments | 2014-07-22T16:10:02 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=fbif03b4mf8h9b717kqs27rqu0&action=printpage;topic=2324.0 | # Toronto Math Forum
## MAT334--2020S => MAT334--Lectures & Home Assignments => Chapter 2 => Topic started by: Yan Zhou on February 10, 2020, 07:22:49 PM
Title: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 07:22:49 PM
$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z}$$
Since -2 lies in $|z+1| = 2$, Cauchy theorem gives that
$$\frac{1}{2\pi i} \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{z-(-2)} = \frac{-2}{2-(-2)} = -\frac{1}{2}$$
then $$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = -\pi i$$
However, I checked the answer of textbook and it says the answer is $2\pi i$, I am confused about where i did wrong.
Title: Re: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 07:46:22 PM
I see. The question on the home assignment is a little bit different from the textbook.
Title: Re: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 08:11:40 PM
I found some differences between textbook and home assignment, some of them are typos, and some do not affect the questions but the answers. Should we always follow the questions on the textbook?
Title: Re: 2.3 question 3
Post by: Victor Ivrii on February 11, 2020, 07:37:38 AM
Your answer is correct ($-\pi i$).
If you see discrepancies, report them | 2022-01-25T06:03:30 | {
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