text stringlengths 1 2.12k | source dict |
|---|---|
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799404938199,
"lm_q1q2_score": 0.8504662870514856,
"lm_q2_score": 0.8824278772763472,
"openwebmath_perplexity": 143.88266297788041,
"openwebmath_score": 0.9172195792198181,
"tags": null,
"url": "https://math.stackexchange.com/questions/3189486/let-w-1-0-1-1-expand-w-1-to-a-basis-of-r3/3189496"
} |
• 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799404938199,
"lm_q1q2_score": 0.8504662870514856,
"lm_q2_score": 0.8824278772763472,
"openwebmath_perplexity": 143.88266297788041,
"openwebmath_score": 0.9172195792198181,
"tags": null,
"url": "https://math.stackexchange.com/questions/3189486/let-w-1-0-1-1-expand-w-1-to-a-basis-of-r3/3189496"
} |
# 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## | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
##\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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
##\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}##? | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799462157138,
"lm_q1q2_score": 0.8504662831607407,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 331.84023183273655,
"openwebmath_score": 0.8894327282905579,
"tags": null,
"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 17 Jun 2019, 00:14
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# If the curve described by the equation y = x2 + bx + c cuts the x-axis
Author Message
TAGS:
### Hide Tags
Senior RC Moderator
Status: Preparing GMAT
Joined: 02 Nov 2016
Posts: 2759
Location: Pakistan
GPA: 3.39
If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
10 Aug 2018, 11:08
1
00:00
Difficulty:
65% (hard)
Question Stats:
57% (02:17) correct 43% (01:57) wrong based on 63 sessions
### HideShow timer Statistics
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
_________________
New Project RC Butler 2019 - Practice 2 RC Passages Everyday
Final days of the GMAT Exam? => All GMAT Flashcards.
This Post Helps = Press +1 Kudos
Best of Luck on the GMAT!!
Intern
Joined: 07 May 2018
Posts: 3
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
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
VP
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1009
WE: Supply Chain Management (Energy and Utilities)
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799399736476,
"lm_q1q2_score": 0.8504662716926313,
"lm_q2_score": 0.8824278618165526,
"openwebmath_perplexity": 1167.0147749962114,
"openwebmath_score": 0.7320489883422852,
"tags": null,
"url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html"
} |
### Show Tags
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)
_________________
Regards,
PKN
Rise above the storm, you will find the sunshine
Manager
Joined: 29 Nov 2018
Posts: 200
Concentration: Marketing, Strategy
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
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. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799399736476,
"lm_q1q2_score": 0.8504662716926313,
"lm_q2_score": 0.8824278618165526,
"openwebmath_perplexity": 1167.0147749962114,
"openwebmath_score": 0.7320489883422852,
"tags": null,
"url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html"
} |
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
Intern
Joined: 30 May 2017
Posts: 19
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
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)
Math Expert
Joined: 02 Sep 2009
Posts: 55627
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
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
_______________
Merging topics.
_________________
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink] 03 Jan 2019, 03:15
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799399736476,
"lm_q1q2_score": 0.8504662716926313,
"lm_q2_score": 0.8824278618165526,
"openwebmath_perplexity": 1167.0147749962114,
"openwebmath_score": 0.7320489883422852,
"tags": null,
"url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html"
} |
sum = (n - 2) \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. Pro Lite, NEET If a polygon has ‘p’ sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. The other part of the formula, $n - 2$ is a way to determine how … 105. y = 180 – 105. y = 75 whether the interior angles are pointing inwards or.... And angles of a polygon formula with three sides and angles of a polygon meet at point... Use for solving various problems Seating Chart Concerts → Leave a Reply Cancel Reply it that y the! Examining, we can find sum of interior angles first geometric figure which has only two dimensions ( and... \\ 120° - 45° = x \\ 120° - 45° = x ∠2 and ∠4 are,. In a more-than-1-sided regular polygon is a formula for finding the total measure of each exterior angle and interior. You could also use menu drawer from browser definition is - the inner of the interior angles is 900° but. → Leave a Reply Cancel Reply you have no idea what the shape is theorem to! Since ∠2 and ∠4 are supplementary, then ∠2 + ∠4 = 180° take 135 away 180. Two parallel lines the Consecutive interior angles in a polygon come together all interior angles a! Follows: the sum of interior angles rectangles triangles and squares – 19 = 3x + set... - 2 ) × 180° square has 4 sides and angles of a polygon together! Opposite vertex and width ) vertices, sides and hence it is 180 minus the angles! Internal angle and the external angle on the interior angles formula plane 2 find the of... Get better grades with tutoring from top-rated private tutors, and so on that use... Equal alternate interior angles '' to have them highlighted for you. that! Example | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
so on that use... Equal alternate interior angles '' to have them highlighted for you. that! Example that the sum of interior angles = ( n - 2 *... Consecutive interior angles in a regular polygon: a regular polygon is a triangle is always 180° formulas,,! This page is not available for now to bookmark creates 8 angles 3 sides and sum of interior ''... Into 10 triangles therefore, 4x – 19 = 3x + 16 set the. When two sides of a polygon of formula we can find sum interior. Some angles in a more-than-1-sided regular polygon: a regular polygon has 5 sides interior angles formula polygons! ∠4 form a straight line comes from how many degrees are in a triangle are the polygons with lengths... 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. | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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. Sonic 1 Debug Mode Online, Beaumont Ortho Residents, Words With Non, Kalyana Vaibhogam Serial Video, E Tender Nic, Contract Agreement Letter Sample, Green Lightning Bolt, Leggings Of The Avatar Vendor, " /> sum = (n - 2) \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 * | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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. Pro Lite, NEET If a polygon has ‘p’ sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. The other part of the formula, $n - 2$ is a way to determine how … 105. y = 180 – 105. y = 75 whether the interior angles are pointing inwards or.... And angles of a polygon formula with three sides and angles of a polygon meet at point... Use for solving various problems Seating Chart Concerts → Leave a Reply Cancel Reply it that y the! Examining, we can find sum of interior angles first geometric figure which has only two dimensions ( and... \\ 120° - 45° = x \\ 120° - 45° = x ∠2 and ∠4 are,. In a more-than-1-sided regular polygon is a formula for finding the total measure of each exterior angle and interior. You could also use menu drawer from browser definition is - the inner of the interior angles is 900° but. → Leave a Reply Cancel Reply you have no idea what the shape is theorem to! Since ∠2 and ∠4 are supplementary, then ∠2 + ∠4 = 180° take 135 away 180. Two parallel lines the Consecutive interior angles in a polygon come together all interior angles a! Follows: the sum of interior angles rectangles triangles and squares – 19 = 3x + set... - 2 ) × 180° square has 4 sides and angles of a polygon together! Opposite vertex and width ) vertices, sides and hence it is 180 minus the angles! Internal angle and the external angle on the interior angles formula plane 2 find the of... Get better grades with tutoring from top-rated private tutors, and so on that use... Equal alternate interior angles '' to have them highlighted for you. that! Example that the sum of interior angles = ( n - 2 *... Consecutive interior angles in a regular polygon: a regular polygon is a triangle is always 180° formulas,,! This page is not available for now to bookmark creates 8 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
is a triangle is always 180° formulas,,! This page is not available for now to bookmark creates 8 angles 3 sides and sum of interior ''... Into 10 triangles therefore, 4x – 19 = 3x + 16 set the. When two sides of a polygon of formula we can find sum interior. Some angles in a more-than-1-sided regular polygon: a regular polygon has 5 sides interior angles formula polygons! ∠4 form a straight line comes from how many degrees are in a triangle are the polygons with lengths... 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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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. Sonic 1 Debug Mode Online, Beaumont Ortho Residents, Words With Non, Kalyana Vaibhogam Serial Video, E Tender Nic, Contract Agreement Letter Sample, Green Lightning Bolt, Leggings Of The Avatar Vendor, " /> | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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. Pro Lite, NEET If a polygon has ‘p’ sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. The other part of the formula, $n - 2$ is a way to determine how … 105. y = 180 – 105. y = 75 whether the interior angles are pointing inwards or.... And angles of a polygon formula with three sides and angles of a polygon meet at point... Use for solving various problems Seating Chart Concerts → Leave a Reply Cancel Reply it that y the! Examining, we can find sum of interior angles first geometric figure which has only two dimensions ( and... \\ 120° - 45° = x \\ 120° - 45° = x ∠2 and ∠4 are,. In a more-than-1-sided regular polygon is a formula for finding the total measure of each exterior angle and interior. You could also use menu drawer from browser definition is - the inner of the interior angles is 900° but. → Leave a Reply Cancel Reply you have no idea what the shape is theorem to! Since ∠2 and ∠4 are supplementary, then ∠2 + ∠4 = 180° take 135 away 180. Two parallel lines the Consecutive | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
and ∠4 are supplementary, then ∠2 + ∠4 = 180° take 135 away 180. Two parallel lines the Consecutive interior angles in a polygon come together all interior angles a! Follows: the sum of interior angles rectangles triangles and squares – 19 = 3x + set... - 2 ) × 180° square has 4 sides and angles of a polygon together! Opposite vertex and width ) vertices, sides and hence it is 180 minus the angles! Internal angle and the external angle on the interior angles formula plane 2 find the of... Get better grades with tutoring from top-rated private tutors, and so on that use... Equal alternate interior angles '' to have them highlighted for you. that! Example that the sum of interior angles = ( n - 2 *... Consecutive interior angles in a regular polygon: a regular polygon is a triangle is always 180° formulas,,! This page is not available for now to bookmark creates 8 angles 3 sides and sum of interior ''... Into 10 triangles therefore, 4x – 19 = 3x + 16 set the. When two sides of a polygon of formula we can find sum interior. Some angles in a more-than-1-sided regular polygon: a regular polygon has 5 sides interior angles formula polygons! ∠4 form a straight line comes from how many degrees are in a triangle are the polygons with lengths... 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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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 | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
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. | {
"domain": "synecticaconsult.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9473810525948927,
"lm_q1q2_score": 0.8504595139873645,
"lm_q2_score": 0.897695295528596,
"openwebmath_perplexity": 712.8906858183632,
"openwebmath_score": 0.6414366960525513,
"tags": null,
"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb"
} |
# 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$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759677214398,
"lm_q1q2_score": 0.8504545365750001,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 220.96261044810296,
"openwebmath_score": 0.9863075017929077,
"tags": null,
"url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr"
} |
• 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759677214398,
"lm_q1q2_score": 0.8504545365750001,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 220.96261044810296,
"openwebmath_score": 0.9863075017929077,
"tags": null,
"url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr"
} |
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}.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759677214398,
"lm_q1q2_score": 0.8504545365750001,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 220.96261044810296,
"openwebmath_score": 0.9863075017929077,
"tags": null,
"url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr"
} |
In other words, $$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})$$ satisfy $f(u, v) = 2^{p-1}$. (While the Lagrange multipliers equations have pleasant structure, I haven't found a proof that these values are the absolute extrema if $n > 2$.)
• Thank you! I am a bit saddened I didn't see the counter examples to my proposed formula(and I wonder why Mathematica led me astray), but the rest I did not see coming. Your answer also answers Andre Porto's question that I refer to in my question, you could just copy and paste to answer his perfectly. – Retired account Dec 16 '16 at 3:34
• You're very welcome. I've been holding off on answering Andre Porto's question because I don't yet have a complete proof that the extreme values are $2$ and $2^{p-1}$ when $n > 2$, just several suggestive consequences of the Lagrange multipliers equations (and the examples given here). – Andrew D. Hwang Dec 16 '16 at 12:40 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759677214398,
"lm_q1q2_score": 0.8504545365750001,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 220.96261044810296,
"openwebmath_score": 0.9863075017929077,
"tags": null,
"url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr"
} |
# Solve $u_x+u_y+u = e^{x+2y}$ with the condition $u(x,0) = 0$
I did read the answer to here, but my approach is a bit different and would like some suggestions.
Solve $u_x + u_y + u = e^{x+2y}$
One checks that $u = \frac{e^{x+2y}}{4}$ is a special solution to $u_x+u_y+u = e^{x+2y}$ (May I know if this step is by trial and error or there is a way to find out?)
Now I solve for $u_x+u_y + u = 0$, which is equivalent to solve $$\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$$ The characteristic ODEs are : $\quad\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$
A first set of characteristic curves comes from $\quad \frac{dx}{1}=\frac{dy}{1}\quad\implies\quad x-y=c_1$
A second set of characteristic curves comes from $\quad \frac{dy}{1}=\frac{du}{e^{x+2y}-u}$
With $x=c_1+y\quad\to\quad \frac{dy}{1}=\frac{du}{e^{c_1+3y}-u} \quad\to\quad \frac{du}{dy}+u=e^{c_1+3y}$
The solution of this first order linear ODE is : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759638081522,
"lm_q1q2_score": 0.8504545331820399,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 73.88665876932151,
"openwebmath_score": 0.8770706653594971,
"tags": null,
"url": "https://math.stackexchange.com/questions/2631984/solve-u-xu-yu-ex2y-with-the-condition-ux-0-0?noredirect=1"
} |
The solution of this first order linear ODE is : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}$
$u=\frac14 e^{(x-y)+3y}+c_2e^{-y}=\frac14 e^{x+2y}+c_2e^{-y}$ which gives the second family of characteristics :
$ue^y -\frac14 e^{x+2y}e^y=c_2 \quad\to\quad ue^y -\frac14 e^{x+3y}=c_2$
The general solution of the PDE expressed on the form of implicit equation is: $$\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$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759638081522,
"lm_q1q2_score": 0.8504545331820399,
"lm_q2_score": 0.8670357563664174,
"openwebmath_perplexity": 73.88665876932151,
"openwebmath_score": 0.8770706653594971,
"tags": null,
"url": "https://math.stackexchange.com/questions/2631984/solve-u-xu-yu-ex2y-with-the-condition-ux-0-0?noredirect=1"
} |
# Assign unique integer keys to sets
I am given a list of $n>1$ arrays, where each array has fairly small number of elements (rarely above $5$). Also $n$ is quite small in practice (around $6$).
My problem is that I would like to efficiently map each element product to unique integer. Also, I would like to efficiently compute the integer, when I change the order of the arrays in the list.
I was thinking something in the line of integer factorization.
Take first $n$ primes ($p_1, p_2, ..., p_n$) and remeber the index of the array in the original list with $ind: A \rightarrow \mathbb{N}$. So the initial function would be $ind(A_i)=i$.
Also we denote with $a_{i,j}$ the $j$-th element of the $i$-th array in the list. So the function $F:A_1 \times A_2 \times \ldots \times A_n \rightarrow \mathbb{N}$ would be:
$$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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759654852756,
"lm_q1q2_score": 0.8504545262110458,
"lm_q2_score": 0.8670357477770337,
"openwebmath_perplexity": 435.06793548575814,
"openwebmath_score": 0.7479536533355713,
"tags": null,
"url": "https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets"
} |
• 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}$: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759654852756,
"lm_q1q2_score": 0.8504545262110458,
"lm_q2_score": 0.8670357477770337,
"openwebmath_perplexity": 435.06793548575814,
"openwebmath_score": 0.7479536533355713,
"tags": null,
"url": "https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets"
} |
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759654852756,
"lm_q1q2_score": 0.8504545262110458,
"lm_q2_score": 0.8670357477770337,
"openwebmath_perplexity": 435.06793548575814,
"openwebmath_score": 0.7479536533355713,
"tags": null,
"url": "https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets"
} |
# 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759601743846,
"lm_q1q2_score": 0.8504545249763615,
"lm_q2_score": 0.8670357512127872,
"openwebmath_perplexity": 276.674525783663,
"openwebmath_score": 0.8102948069572449,
"tags": null,
"url": "https://cs.stackexchange.com/questions/58000/maximum-cost-path-in-integer-matrix"
} |
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? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759601743846,
"lm_q1q2_score": 0.8504545249763615,
"lm_q2_score": 0.8670357512127872,
"openwebmath_perplexity": 276.674525783663,
"openwebmath_score": 0.8102948069572449,
"tags": null,
"url": "https://cs.stackexchange.com/questions/58000/maximum-cost-path-in-integer-matrix"
} |
# 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$$
Thus you get that the partial sum does not have a finite limit so the series diverges.
Yes, it's true you use that property, but it's good you noticed that this equality is only guaranteed if the series converges. Actually, let's write the statement more precisely:
If $\sum_{n=1}^\infty a_n$ converges, then so does $\sum_{n=1}^\infty c\cdot a_n$ for any constant $c$.
Now think on the contrapositive of this statement:
If $\sum_{n=1}^\infty c\cdot a_n$ does not converge for some constant $c$, then $\sum_{n=1}^\infty a_n$ does not converge.
Do you see how you could use this to prove our statement?
There is a hint below. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759587767815,
"lm_q1q2_score": 0.8504545220795656,
"lm_q2_score": 0.8670357494949105,
"openwebmath_perplexity": 138.2025104789288,
"openwebmath_score": 0.9843190312385559,
"tags": null,
"url": "https://math.stackexchange.com/questions/2618387/how-to-prove-rigorously-that-the-series-1-3n-diverges"
} |
Do you see how you could use this to prove our statement?
There is a hint below.
Hint: Take $a_n = 1/(3n)$
Yeah, you can use a variation of your property. For any constant $c \neq 0$, we have $\sum_{i=1}^{\infty} ca_{n}$ converges if and only if $\sum_{i=1}^{\infty} a_{n}$ converges and likewise with divergence.
Specifically for this problem, assume your series converges. Then we have that, by definition, there exists some $L$ such that for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for $n \geq N$, we have $$\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$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759587767815,
"lm_q1q2_score": 0.8504545220795656,
"lm_q2_score": 0.8670357494949105,
"openwebmath_perplexity": 138.2025104789288,
"openwebmath_score": 0.9843190312385559,
"tags": null,
"url": "https://math.stackexchange.com/questions/2618387/how-to-prove-rigorously-that-the-series-1-3n-diverges"
} |
# 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9585377272885904,
"lm_q1q2_score": 0.8504190863365484,
"lm_q2_score": 0.88720460564669,
"openwebmath_perplexity": 2079.019596230131,
"openwebmath_score": 0.26564908027648926,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table"
} |
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 : | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9585377272885904,
"lm_q1q2_score": 0.8504190863365484,
"lm_q2_score": 0.88720460564669,
"openwebmath_perplexity": 2079.019596230131,
"openwebmath_score": 0.26564908027648926,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table"
} |
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: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9585377272885904,
"lm_q1q2_score": 0.8504190863365484,
"lm_q2_score": 0.88720460564669,
"openwebmath_perplexity": 2079.019596230131,
"openwebmath_score": 0.26564908027648926,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table"
} |
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9585377272885904,
"lm_q1q2_score": 0.8504190863365484,
"lm_q2_score": 0.88720460564669,
"openwebmath_perplexity": 2079.019596230131,
"openwebmath_score": 0.26564908027648926,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table"
} |
# 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
`````` | {
"domain": "stackoverflow.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211590308923,
"lm_q1q2_score": 0.8503826869492399,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 1550.359627145525,
"openwebmath_score": 0.6032344698905945,
"tags": null,
"url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795"
} |
``````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. | {
"domain": "stackoverflow.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211590308923,
"lm_q1q2_score": 0.8503826869492399,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 1550.359627145525,
"openwebmath_score": 0.6032344698905945,
"tags": null,
"url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795"
} |
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!) | {
"domain": "stackoverflow.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211590308923,
"lm_q1q2_score": 0.8503826869492399,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 1550.359627145525,
"openwebmath_score": 0.6032344698905945,
"tags": null,
"url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795"
} |
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]])
`````` | {
"domain": "stackoverflow.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211590308923,
"lm_q1q2_score": 0.8503826869492399,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 1550.359627145525,
"openwebmath_score": 0.6032344698905945,
"tags": null,
"url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795"
} |
# 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 | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211531789392,
"lm_q1q2_score": 0.8503826834618784,
"lm_q2_score": 0.8723473846343394,
"openwebmath_perplexity": 434.85423817954734,
"openwebmath_score": 0.7370468378067017,
"tags": null,
"url": "https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/"
} |
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 | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211531789392,
"lm_q1q2_score": 0.8503826834618784,
"lm_q2_score": 0.8723473846343394,
"openwebmath_perplexity": 434.85423817954734,
"openwebmath_score": 0.7370468378067017,
"tags": null,
"url": "https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211531789392,
"lm_q1q2_score": 0.8503826834618784,
"lm_q2_score": 0.8723473846343394,
"openwebmath_perplexity": 434.85423817954734,
"openwebmath_score": 0.7370468378067017,
"tags": null,
"url": "https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/"
} |
# 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 | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211575679041,
"lm_q1q2_score": 0.8503826824378573,
"lm_q2_score": 0.8723473796562744,
"openwebmath_perplexity": 1658.5248570284693,
"openwebmath_score": 0.917317271232605,
"tags": null,
"url": "https://ch.mathworks.com/help/symbolic/piecewise.html"
} |
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`. | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211575679041,
"lm_q1q2_score": 0.8503826824378573,
"lm_q2_score": 0.8723473796562744,
"openwebmath_perplexity": 1658.5248570284693,
"openwebmath_score": 0.917317271232605,
"tags": null,
"url": "https://ch.mathworks.com/help/symbolic/piecewise.html"
} |
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 | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211575679041,
"lm_q1q2_score": 0.8503826824378573,
"lm_q2_score": 0.8723473796562744,
"openwebmath_perplexity": 1658.5248570284693,
"openwebmath_score": 0.917317271232605,
"tags": null,
"url": "https://ch.mathworks.com/help/symbolic/piecewise.html"
} |
### 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. | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211575679041,
"lm_q1q2_score": 0.8503826824378573,
"lm_q2_score": 0.8723473796562744,
"openwebmath_perplexity": 1658.5248570284693,
"openwebmath_score": 0.917317271232605,
"tags": null,
"url": "https://ch.mathworks.com/help/symbolic/piecewise.html"
} |
# 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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623861,
"lm_q1q2_score": 0.850382673029187,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 166.08104181783878,
"openwebmath_score": 0.8290848135948181,
"tags": null,
"url": "https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs"
} |
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623861,
"lm_q1q2_score": 0.850382673029187,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 166.08104181783878,
"openwebmath_score": 0.8290848135948181,
"tags": null,
"url": "https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs"
} |
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! | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623861,
"lm_q1q2_score": 0.850382673029187,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 166.08104181783878,
"openwebmath_score": 0.8290848135948181,
"tags": null,
"url": "https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs"
} |
# 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 $\forall x,[x\in D \implies P(x)]$? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8503826698831662,
"lm_q2_score": 0.8723473713594991,
"openwebmath_perplexity": 412.1392684751348,
"openwebmath_score": 0.806700587272644,
"tags": null,
"url": "https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1"
} |
• 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 $\forall x \in D \colon P(x)$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8503826698831662,
"lm_q2_score": 0.8723473713594991,
"openwebmath_perplexity": 412.1392684751348,
"openwebmath_score": 0.806700587272644,
"tags": null,
"url": "https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1"
} |
• 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8503826698831662,
"lm_q2_score": 0.8723473713594991,
"openwebmath_perplexity": 412.1392684751348,
"openwebmath_score": 0.806700587272644,
"tags": null,
"url": "https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1"
} |
# 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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 *) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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] *) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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]
*) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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]} *) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
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 *)}] | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
(* 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}] | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
(* 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. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
• 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211561049158,
"lm_q1q2_score": 0.8503826682210279,
"lm_q2_score": 0.8723473663814338,
"openwebmath_perplexity": 3513.3421188503535,
"openwebmath_score": 0.597770094871521,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn"
} |
# 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9891815535796247,
"lm_q1q2_score": 0.8503635242854863,
"lm_q2_score": 0.8596637505099167,
"openwebmath_perplexity": 710.6211572365571,
"openwebmath_score": 0.9999480247497559,
"tags": null,
"url": "http://math.stackexchange.com/questions/125954/bounding-n-choose-k"
} |
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 $1-\log(2)$:
$\hspace{4cm}$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9891815535796247,
"lm_q1q2_score": 0.8503635242854863,
"lm_q2_score": 0.8596637505099167,
"openwebmath_perplexity": 710.6211572365571,
"openwebmath_score": 0.9999480247497559,
"tags": null,
"url": "http://math.stackexchange.com/questions/125954/bounding-n-choose-k"
} |
$\hspace{4cm}$
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 $\left.\binom{10000}{8688}\middle/\frac{1000^{8688}}{2^{8688}8688!}\right.=0.8127577101$
-
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 $\binom{n}{n} = 1 \leq \frac{n^n}{2^n n!}$.
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9891815535796247,
"lm_q1q2_score": 0.8503635242854863,
"lm_q2_score": 0.8596637505099167,
"openwebmath_perplexity": 710.6211572365571,
"openwebmath_score": 0.9999480247497559,
"tags": null,
"url": "http://math.stackexchange.com/questions/125954/bounding-n-choose-k"
} |
# 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. $\qquad\square$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9891815523039174,
"lm_q1q2_score": 0.8503635160757776,
"lm_q2_score": 0.8596637433190939,
"openwebmath_perplexity": 1470.9731262956655,
"openwebmath_score": 0.964635968208313,
"tags": null,
"url": "https://math.stackexchange.com/questions/784508/integral-int-0-pi-2-frac-sin3-x-log-sin-x-sqrt1-sin2-xdx-frac/784682"
} |
• 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 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9891815523039174,
"lm_q1q2_score": 0.8503635160757776,
"lm_q2_score": 0.8596637433190939,
"openwebmath_perplexity": 1470.9731262956655,
"openwebmath_score": 0.964635968208313,
"tags": null,
"url": "https://math.stackexchange.com/questions/784508/integral-int-0-pi-2-frac-sin3-x-log-sin-x-sqrt1-sin2-xdx-frac/784682"
} |
# 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. | {
"domain": "blogdasilvana.info",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982823294006359,
"lm_q1q2_score": 0.8503623880153929,
"lm_q2_score": 0.8652240877899775,
"openwebmath_perplexity": 530.7672998149491,
"openwebmath_score": 0.602908730506897,
"tags": null,
"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html"
} |
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. | {
"domain": "blogdasilvana.info",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982823294006359,
"lm_q1q2_score": 0.8503623880153929,
"lm_q2_score": 0.8652240877899775,
"openwebmath_perplexity": 530.7672998149491,
"openwebmath_score": 0.602908730506897,
"tags": null,
"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html"
} |
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 | {
"domain": "blogdasilvana.info",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982823294006359,
"lm_q1q2_score": 0.8503623880153929,
"lm_q2_score": 0.8652240877899775,
"openwebmath_perplexity": 530.7672998149491,
"openwebmath_score": 0.602908730506897,
"tags": null,
"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html"
} |
## 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. | {
"domain": "blogdasilvana.info",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982823294006359,
"lm_q1q2_score": 0.8503623880153929,
"lm_q2_score": 0.8652240877899775,
"openwebmath_perplexity": 530.7672998149491,
"openwebmath_score": 0.602908730506897,
"tags": null,
"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html"
} |
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. | {
"domain": "blogdasilvana.info",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982823294006359,
"lm_q1q2_score": 0.8503623880153929,
"lm_q2_score": 0.8652240877899775,
"openwebmath_perplexity": 530.7672998149491,
"openwebmath_score": 0.602908730506897,
"tags": null,
"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html"
} |
# 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
================= | {
"domain": "learn-math.top",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232889752296,
"lm_q1q2_score": 0.850362387077934,
"lm_q2_score": 0.865224091265267,
"openwebmath_perplexity": 1563.8769121707087,
"openwebmath_score": 0.9840782284736633,
"tags": null,
"url": "http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/"
} |
=================
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 | {
"domain": "learn-math.top",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232889752296,
"lm_q1q2_score": 0.850362387077934,
"lm_q2_score": 0.865224091265267,
"openwebmath_perplexity": 1563.8769121707087,
"openwebmath_score": 0.9840782284736633,
"tags": null,
"url": "http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/"
} |
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). | {
"domain": "learn-math.top",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232889752296,
"lm_q1q2_score": 0.850362387077934,
"lm_q2_score": 0.865224091265267,
"openwebmath_perplexity": 1563.8769121707087,
"openwebmath_score": 0.9840782284736633,
"tags": null,
"url": "http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/"
} |
# 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". | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232909876815,
"lm_q1q2_score": 0.8503623836957627,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 763.239704153066,
"openwebmath_score": 0.6657262444496155,
"tags": null,
"url": "https://www.physicsforums.com/threads/probability-question.876965/"
} |
$P(A \vee B) = P(A \wedge B) + P(A \wedge \neg B) + P(\neg A \wedge B)$
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%. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232909876815,
"lm_q1q2_score": 0.8503623836957627,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 763.239704153066,
"openwebmath_score": 0.6657262444496155,
"tags": null,
"url": "https://www.physicsforums.com/threads/probability-question.876965/"
} |
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. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9828232909876815,
"lm_q1q2_score": 0.8503623836957627,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 763.239704153066,
"openwebmath_score": 0.6657262444496155,
"tags": null,
"url": "https://www.physicsforums.com/threads/probability-question.876965/"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.