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Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$

a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$

$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$

$b_n = \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$

$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$

The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?

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    $\begingroup$ Do you know $\sum \frac{1}{n^2}=\frac{\pi^2}{6}$? $\endgroup$ Jul 6, 2017 at 23:29
  • $\begingroup$ No the student does not. You have to use a Fourier series. A piecewise linear function that goes from $(-\pi,0)$ to $(0,1)$ and then to $(\pi,0)$, extended periodically, should work. $\endgroup$ Jul 6, 2017 at 23:37
  • $\begingroup$ The Fourier sine series for $f(x) \equiv 1$ on $(0, \pi)$ seems to have the right coefficients to apply the theorem (drawing a blank on its name) $\lVert f \rVert_2^2 = C \sum_{n=1}^\infty |c_n|^2$. $\endgroup$ Jul 6, 2017 at 23:52
  • $\begingroup$ @Jeffrey Do you know Parseval's identity? $\endgroup$ Jul 6, 2017 at 23:52
  • $\begingroup$ I think the minis after $\frac{\pi}{8}$ should be a plus. $\endgroup$ Jul 7, 2017 at 0:38

3 Answers 3

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You can prove $$\sum \frac{1}{n^2}=\frac{\pi^2}{6}$$ using Fourier series. Hence, $$\sum\frac{1}{(2n)^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$ Therefore, $$\frac14 \sum \frac{1}{n^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$ This shows $$\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$

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First of all your $b_k$ are wrong, they should be:

$$b_k= \frac{(-1)^{k+1}}{2k}$$

Not that it matters beacause of the following. Second of all notice that $f(x)$ is continuos at zero, which is to say $f(0^+)= f(0^-)=0$. Once you get the expansion right is not that hard, just make $x=0$, easy peasy.

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This would be one way of going about it but using a different fourier series: $$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$ Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just use odd numbers instead with $(2n+1)$. So we can make a new formula with only odds and we can do this by letting $n =(2n+1)$, and because we don't want the even results we can resolve the numerator to be equal to $2$. $$f(x)\ =\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\cos\left(n\pi x\ +\frac{\pi x}{2}\right)$$ Then setting $x = 0$ gives: $$0=\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$ After rearranging: $$\frac{\pi^{2}}{8}=\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$

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