Find the Fourier series for $f(\theta)=\theta^2$ and prove $\Sigma^\infty_{n=1} \frac{1}{n^2}=\frac{\pi^2}{6}$

Question

Find the Fourier series for $f(\theta)=\theta^2$ and use Parseval's identity for $f$ to derive the identity:

$$\sum^\infty_{n=1} \frac{1}{n^2}=\frac{\pi^2}{6}$$

In addition, find the expansion for $f$ in terms of the functions $\{1, \cos(2\pi \theta), \sin(2\pi \theta), \cos(4\pi \theta),....\}$

Here is a link explaining Parsevals identity: https://en.wikipedia.org/wiki/Parseval%27s_identity

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Can somebody help me with this one? There seems to be a lot going on and a lot of these ideas are really new to me. I'm trying to follow a proof we got in class that we can use the fourier expansion $f(x)=\theta$ on $[0,1)$ to show that

$$\sum^\infty_{n=1} \frac{1}{n^4}=\frac{\pi^2}{6}$$

But the proof is sort of confusing, and I don't see how I could adapt it. I'd really appreciate some help on this one! Thanks MSE!!

Answer 1

You don't need Parseval (which will square your coefficients so it will give you something with $n^4$). From this question, $$ f(x)=\frac{\pi^2}{3}+4 \ \sum_{n=1}^{+\infty} \frac{(-1)^n}{n^2} \ \cos(nx). $$ As $f$ is differentiable everywhere, we have pointwise convergence. Then, evaluating at $\pi$, $$ \pi^2=f(\pi)=\frac{\pi^2}{3}+4 \ \sum_{n=1}^{+\infty} \frac{(-1)^n}{n^2} \ \cos(n\pi) =\frac{\pi^2}{3}+4 \ \sum_{n=1}^{+\infty} \frac{1}{n^2} . $$ Then $$ \sum_{n=1}^{+\infty} \frac{1}{n^2} =\frac14\,\left(\pi^2-\frac{\pi^2}3\right)=\frac{\pi^2}6. $$

If you were to use Parseval for this function, what you get is $$ \frac{2\pi^5}5=\int_{-\pi}^\pi (t^2)^2\,dt=\pi\,\left(2{a_0^2}+\sum_{n=0}^\infty a_n^2\right)=\pi\left(\frac{2\pi^4}{9}+\sum_{n=1}^\infty\frac{16}{n^4}\right). $$ Solving, you get $$ \sum_{n=1}^\infty\frac{1}{n^4}=\frac1{16}\left(\frac{2\pi^4}5-\frac{2\pi^4}9\right)=\frac{\pi^4}{90} $$

Answer 2

As has been noted, Parseval with this choice of $f$ sums $\tfrac{1}{n^4}$, not $\tfrac{1}{n^2}$. So let's do the latter with Parseval on $g(\theta)=\theta$ instead. From here, $g=-2\sum_{n\ge1}\tfrac{(-1)^n}{n}\sin n\theta$, so$$\tfrac23\pi^3=\int_{-\pi}^\pi\theta^2 d\theta=4\pi\sum_{n\ge1}\tfrac{1}{n^2}\implies\sum_n\tfrac{1}{n^2}=\tfrac16\pi^2.$$