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Consider the $2\pi$ periodic function $f(x) = x^2$ defined over the interval $x\in (0,2\pi].$

The Fourier expansion of $f$ is as follows: $$f(x) = \frac{4\pi^2}{3}+\sum_{n\geq 1}\left(\frac{4}{n^2}\cos(nx)-\frac{4\pi}{n}\sin(nx)\right).$$

Now I am asked to compute the sum of the series

$$\sum_{n\geq 1}\frac{1}{n^2}$$

using the expansion given above. I guess we have to set $x= 0,$ but when I do that, I do not get the correct answer ($\frac{\pi^2}{6}$). Where am I going wrong?

Student
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    Set $x=2\pi\quad\quad$ – DINEDINE May 08 '19 at 13:53
  • $x=0$ is not in the domain of definition… – Dirk May 08 '19 at 13:58
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    I believe you want to consider what happens when you "consider the $2\pi$ periodic function $f(x)=x^2$" since this function is not inherently periodic. By taking a periodic extension of this function, you will have a jump discontinuity at $x=0$, and something interesting happens to the value of Fourier series at jump discontinuities that you may want to consider. – TM Gallagher May 08 '19 at 13:59
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    I think you can get away with change of variable and reworking the problem on the interval (-pi,pi]. This will kill off your sine terms. – TurlocTheRed May 08 '19 at 14:14

1 Answers1

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At $ x=0$ the Fourier series converges to $\frac {f ({-0})+f ({+0})}{2}=\frac {(2\pi)^{2}+0^{2}}{2}=2\pi^2$

Plug that into your fourier series and you get the desired result.

ibnAbu
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