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

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

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!!

• No need Parseval for this. See here. Maybe you wish $\sum_{n}\frac{1}{n^4}$ ? – Surb Apr 8 at 14:45
• Why don't you post the proof that you saw in class and how you're attempting to adapt it? – cmk Apr 8 at 14:55
• Thanks @Surb. Does that last equality somehow equal $\frac{\pi^2}{6}$?? They are close with $\frac{\pi^2}{3}$ + extra terms. – HaKuNa MaTaTa Apr 8 at 15:12
• No it's not correct. But Parseval will help you to compute it. – Surb Apr 8 at 15:13
• Ahh okay okay hmm interesting... I don't see a mistake in the computations though.... I'm still a little stuck, could you walk me through this? – HaKuNa MaTaTa Apr 8 at 15:15

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 $$\sum_{n=1}^\infty\frac{1}{n^4}=\frac1{16}\left(\frac{2\pi^4}5-\frac{2\pi^4}9\right)=\frac{\pi^4}{90}$$
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$$\tfrac23\pi^3=\int_{-\pi}^\pi\theta^2 d\theta=4\pi\sum_{n\ge1}\tfrac{1}{n^2}\implies\sum_n\tfrac{1}{n^2}=\tfrac16\pi^2.$$