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What are the complex fourier coefficients of the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by the $2\pi-periodic$ continuation of $f(x)=\pi-x$ , for x $0\le x < 2\pi$ ?

And how can one use that fact together with the Bessel function to show that $$\sum \frac{1}{k^2} = \frac{\pi^2}{6}$$?

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  • $\begingroup$ what is the Bessel equation? $\endgroup$ – user50336 Nov 28 '12 at 21:31
  • $\begingroup$ @Thomas Andrews I dont see what to try.... @ cassandrao it was a typo, i meant the bessel function $\endgroup$ – bakabakabaka Nov 28 '12 at 21:34
  • $\begingroup$ @bakabakabaka How do you compute the Fourier coefficients? Can you do those integrals yourself? $\endgroup$ – Thomas Andrews Nov 28 '12 at 21:39
  • $\begingroup$ did you mean the Parseval identity? $\endgroup$ – user50336 Nov 28 '12 at 21:41
  • $\begingroup$ @Thomas Andrews yea i can compute these integrals, but I have troubles knowing what to do $\endgroup$ – bakabakabaka Nov 28 '12 at 22:20
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The fourier coefficients for nonzero $n$ are

$$\begin{eqnarray} \hat f(n) &=& \frac{1}{2 \pi} \int_0^{2 \pi} (\pi-t) e^{-i n t} dt \\ &=& \frac{1}{2} \int_0^{2 \pi} e^{-i n t} dt - \frac{1}{2 \pi} \int_0^{2 \pi} t e^{-i n t} dt \\ &=& - \frac{1}{2 \pi} \left(\left[t \frac{e^{- i n t}}{- i n}\right]_0^{2 \pi} - \int_0^{2 \pi} \frac{e^{- i n t}}{- i n} dt \right) \\ &=& \frac{1}{-in} \\ \end{eqnarray}$$

and $\hat f(0) = 0$.

Thus applying Parseval's identity we have

$$\sum_{n=-\infty,n\neq 0}^\infty \frac{1}{n^2} = \frac{1}{2\pi}\int_{-\pi}^\pi (\pi - t)^2 \, dt = \frac{1}{2\pi}\int_{0}^{2\pi} t^2 \, dt = \frac{\pi^{2}}{3}$$

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    $\begingroup$ Pretty +1. Just in the last line's sum one has to take care that $\,n\neq 0\,$ $\endgroup$ – DonAntonio Nov 28 '12 at 23:16
  • $\begingroup$ @DonAntonio, thank you I don't know to do the latex. for that $\endgroup$ – user50336 Nov 29 '12 at 10:41
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    $\begingroup$ Just did it for you, @cassandrao $\endgroup$ – DonAntonio Nov 29 '12 at 11:43

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