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I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the trigonometric Fourier series expansion of $x^{2}/4$. How can one prove the same result using the complex Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?

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@VFG: Did you calculate the Fourier coefficients? Can you see any similarity to $\zeta(2)$? How can you manipulate the Fourier series? –  AD. Aug 27 '10 at 20:41
    
AD: Yes. One of the difficulties I have is dealing with a complex rather than a trigonometric Fourier series. –  Américo Tavares Aug 27 '10 at 20:56
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I believe you are supposed to be using Parseval's identity. –  Qiaochu Yuan Aug 27 '10 at 21:16
    
Qiaochu Yuan: That is a good hint, thanks. Anyhow could you please elaborate a little bit? –  Américo Tavares Aug 27 '10 at 21:37
    
I just want to point out this process can be generalized to give you $\zeta(2n)$ for all $n\in\mathbb{N}$. In this case dealing with $[0,1]$ is simpler than $[-\pi,\pi]$, so $f(z)=\sum_{n=-\infty}^{\infty}c_ne^{2\pi inz}$ where $c_n=\int_0^1 f(z)e^{-2\pi inz}dz$. The coefficients of $f(x)=x^{2n}$ will give you $\zeta(2n)$. However, if you instead use the Bernoulli polynomials, the integration by parts turns out much nicer (the $uv|_0^1$ terms all go away). en.wikipedia.org/wiki/Bernoulli_polynomials –  Riley E Jun 2 '11 at 12:55

3 Answers 3

up vote 5 down vote accepted

Use the definition:

Say $f$ is defined on $[-\pi, \pi]$.

If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$

then

$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$

If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?

To integrate, you can try integration by parts.

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thanks for the solution! –  Américo Tavares Aug 27 '10 at 23:18
    
@VFG: I am curious though, was this homework? It is ok to ask I believe, as long as you say it is homework and show what you have tried so far. –  Aryabhata Aug 27 '10 at 23:29

Extending off from Aryabhatta answer:


For our situation: We have $f(x)=x ~~~{\text{ for }} 0\leq x\leq 1$

$2L=1,\Rightarrow L=\frac{1}{2}$

So restating we have:

$f(x) = \displaystyle\sum_{n=-\infty}^{\infty} {c_{n} e^{inx}}, \text{ where }c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{f(x)e^{-inx}} ~\mathrm{d}x,~~~~~~n=0,~\pm 1,~\pm 2, \cdots~ $

$ \Rightarrow~~ c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{xe^{-inx}}~\mathrm{d}x $

After integrating the complex Fourier coefficient we see that we get the following:

$\Rightarrow~~~~\displaystyle c_n=i\left(\frac{\cos(\frac{n}{2})}{2\pi n}-\frac{\sin(\frac{n}{2})}{\pi n^2}\right),~~~\text{for }n \in \mathbb{R}$

Lastly plugging back $c_n$ into $f(x)$ we then get our desired result for $n=0,~\pm 1,~\pm 2, \cdots~$.

Please update if you see any mistakes with any of the work. It has been quite some time since I work with Fourier Series and went off from my head. Feel free to edit mistakes as necessary if willing.

Thanks.

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It seems strange that you have included Aryabhatta's answer word for word within your own. –  Jonas Meyer Jun 2 '11 at 7:36
    
@Jonas: His really was not needed to be restated, I was just putting so people could follow along without having to scroll back and forth between the two. His was just for some generic interval as I was trying to ask the question at hand. Could remove if seems strange. That is why I added the note above before proceeding. –  night owl Jun 2 '11 at 7:42

This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.

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