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## Homework Statement

Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8

## Homework Equations

## The Attempt at a Solution

No idea how to start. Pls guide.

Thanks

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- Thread starter Outrageous
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- #1

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Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8

No idea how to start. Pls guide.

Thanks

- #2

Ray Vickson

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## Homework Statement

Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8

## Homework Equations

## The Attempt at a Solution

No idea how to start. Pls guide.

Thanks

You need to show your work, first. Anyway, what is the context? Is this a problem in a course? If so, what is the course subject (analytic function theory, differential equations, Fourier analysis...)?

- #3

arildno

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## Homework Statement

Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8

## Homework Equations

## The Attempt at a Solution

No idea how to start. Pls guide.

Thanks

You won't be able to prove it, since the result is wrong. The sum converges to [itex]\frac{\pi^{2}}{6}[/itex] instead

- #4

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You need to show your work, first. Anyway, what is the context? Is this a problem in a course? If so, what is the course subject (analytic function theory, differential equations, Fourier analysis...)?

You won't be able to prove it, since the result is wrong. The sum converges to [itex]\frac{\pi^{2}}{6}[/itex] instead

I got that from a manual solution, I don't know where should I ask my question so I put it here .

This answer is for quantum mechanics problem. I don't think that is Fourier series.

- #5

arildno

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Well, in that thumbnail, it isn't 1/n^2 for n the integers, is it?

- #6

berkeman

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## Homework Statement

Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8

## Homework Equations

## The Attempt at a Solution

No idea how to start. Pls guide.

Thanks

Check your PMs -- you must show your work to receive tutorial help (and please do not use txt speak here like "pls")

- #7

Ray Vickson

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I got that from a manual solution, I don't know where should I ask my question so I put it here .

This answer is for quantum mechanics problem. I don't think that is Fourier series.

You gave us the wrong problem; it should be

[tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]

I would be willing to bet this can be obtained from an appropriate Fourier series, but I will leave the fun of discovery to you.

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- #9

Dick

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Sorry for my mistakes. Thanks for replying.

Do I need to know f(x)= ? for solving this problem? Like f(x)=x for showing 1 - (1/3) + (1/5)....= ∏/4 .as shown below

Use the same function, but instead of evaluating the series at a point, use Parseval's identity.

- #10

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Use the same function, but instead of evaluating the series at a point, use Parseval's identity.

f(x)= x .Range? -∏ <x< ∏ ?

What do you mean by evaluating the series at a point? Fourier series evaluating at a point

Complex fouries series and Parseval's theorem ?

Why do we need two types?

- #11

Dick

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f(x)= x .Range? -∏ <x< ∏ ?

What do you mean by evaluating the series at a point? Fourier series evaluating at a point

Complex fouries series and Parseval's theorem ?

Why do we need two types?

Yes, exactly the same function, exactly the same series. Did you look up Parseval's theorem? Evaluating the series gives you things like 1/n, Parseval's gives you things like 1/n^2. Just try it.

- #12

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Yes, exactly the same function, exactly the same series. Did you look up Parseval's theorem? Evaluating the series gives you things like 1/n, Parseval's gives you things like 1/n^2. Just try it.

I got the answer by using parseval's theorem when f(x)=x over the interval -∏<x<∏.

Thanks

But when do I know I need to use complex Fourier series and parseval's theorem? Is that because there is a square? (2m+1)^2

[tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]

Can I say Fourier series is used when my conjugate ,g(x) is not complex. Complex conjugate of g(x) = g(x), Eg. ∫g(x)f(x)dx= c

But when g(x) is complex , complex conjugate of g(x) is not equal to g(x),then I should use complex Fourier series?

This is my own conclusion after reading . Please Correct if it's wrong.

Use the same function, but instead of evaluating the series at a point, use Parseval's identity.

I am still don't understand what do you mean evaluate the series at a point, is that meant f(x)= x , the x should not be a point, so I have to integrate by using parseval's , but how do you know we can't treat it as a point?

- #13

Dick

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I got the answer by using parseval's theorem when f(x)=x over the interval -∏<x<∏.

Thanks

But when do I know I need to use complex Fourier series and parseval's theorem? Is that because there is a square? (2m+1)^2

[tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]

Can I say Fourier series is used when my conjugate ,g(x) is not complex. Complex conjugate of g(x) = g(x), Eg. ∫g(x)f(x)dx= c

But when g(x) is complex , complex conjugate of g(x) is not equal to g(x),then I should use complex Fourier series?

This is my own conclusion after reading . Please Correct if it's wrong.

I am still don't understand what do you mean evaluate the series at a point, is that meant f(x)= x , the x should not be a point, so I have to integrate by using parseval's , but how do you know we can't treat it as a point?

Yes, the square is a hint you might need Parseval. But you didn't really finish the proof you presented in post 8. Could you do that? Then you might see why evaluation of the Fourier series at a point can give you the sum of some series.

- #14

Ray Vickson

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I got the answer by using parseval's theorem when f(x)=x over the interval -∏<x<∏.

Thanks

But when do I know I need to use complex Fourier series and parseval's theorem? Is that because there is a square? (2m+1)^2

[tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]

Can I say Fourier series is used when my conjugate ,g(x) is not complex. Complex conjugate of g(x) = g(x), Eg. ∫g(x)f(x)dx= c

But when g(x) is complex , complex conjugate of g(x) is not equal to g(x),then I should use complex Fourier series?

This is my own conclusion after reading . Please Correct if it's wrong.

I am still don't understand what do you mean evaluate the series at a point, is that meant f(x)= x , the x should not be a point, so I have to integrate by using parseval's , but how do you know we can't treat it as a point?

You ask "...how do you know...?" You don't! You can try 9 methods that lead to failure, while the 10th method works. You may generate dozens of pages of scrap paper full of failed attempts, and you may spend many hours on "wasted" efforts (not really wasted, though---they teach you something). Be assured, all the posters here that are professors, etc., have gone through this type of experience many times in the past.

However, there are some lessons you can take with you. Often (not always!) when we have sums involving π or π

- #15

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Yes, the square is a hint you might need Parseval. But you didn't really finish the proof you presented in post 8. Could you do that? Then you might see why evaluation of the Fourier series at a point can give you the sum of some series.

Understood~

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