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[tex]

\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n^{2}}

[/tex]

I need to show it equals pi^2 / 12. Not sure where to begin :\ Tried plugging in values but non cancel out which I can see :\

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In summary, you are looking for a function that y=x^2 is defined on (-pi, pi). You find the Fourier series for it and then plug x=0 to get the formula you're looking for.

- #1

- 655

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[tex]

\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n^{2}}

[/tex]

I need to show it equals pi^2 / 12. Not sure where to begin :\ Tried plugging in values but non cancel out which I can see :\

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Does this problem arise for you in a Fourier Series course, or if not that, where?

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thomas49th said:

How did I guess? Well, given that it has a pi in the answer, it isn't likely you are going to find the sum with simple algebra; it is going to take something more advanced. And this problem is the type typically asked in Fourier Series chapters where you are learning about how the FS converges.

You didn't state what FS you calculated, but if you evaluate it at the magic point and use what you know about what the series converges to at continuity points or finite jumps (whichever is relevant to your problem), the result will drop out.

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the magic point? How do I find that?

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thomas49th said:the magic point? How do I find that?

You say "presto - change-o" and try the simplest point you can think of. Be brave...

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I shall try n=1. Therefore -1 pops out. What do I know about series convergence. Not allow. Some series convergences to a single value. I recall something about continuous and jumpy series. If your series jumps around you can get Gibbs Phenomena. Lovely stuff but don't think that is what you were after? Also in discontinuous series the Fourier series always goes through the middle point or average of the discontinuity between each peroid.

Am I any closer?

Thomas

Am I any closer?

Thomas

Last edited:

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thomas49th said:

Am I any closer?

Thomas

No, no... Look, you have a function f(x) and you have expanded it in a FS, which I will call s(x)

So you have f(x) = s(x), hopefully, for most values of x (you have a theorem about that). Put the simplest x you can think of in both sides of that equation and see what happens.

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set t = 0

as adding 1 to the exponent of (-1)^n is the same as multiplying by -1 we find that we get pi^2 / 12

THANKS!

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[tex] \sum_{n=1}^{\infty} \frac{1}{n^2} [/tex]

If you do, then what you have to find is half of it.

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thomas49th said:

set t = 0

as adding 1 to the exponent of (-1)^n is the same as multiplying by -1 we find that we get pi^2 / 12

THANKS!

You're welcome. Glad you found the magic point.

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bigubau said:

[tex] \sum_{n=1}^{\infty} \frac{1}{n^2} [/tex]

If you do, then what you have to find is half of it.

On a second thought, there's a method of finding exactly the sum you're looking for

Assume that y=x^2 is defined on the interval (-pi, pi). Find the Fourier series for it and then plug x=0. You'll get exactly the formula you're looking for.

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