II.
(10)
Let $f \in L^2(a, b)$ such that $f = \sum_{1}^{\infty} (f, \phi_n)\phi_n$ converges in norm. Then Parseval's equation states that
$\|\|f\|\|^2 = \sum_{1}^{\infty} (f, \phi_n)^2$
Evaluate the following series by applying Parseval's equation to certain of the Fourier expansions in Table
1 of §2.1.
$\sum_{1}^{\infty} \frac{1}{n^4}$
Answer. We use Entry 16 from Table 1:
$f(x) = x^2 = \frac{\pi^2}{3} + 4 \sum_{1}^{\infty} \frac{(-1)^n}{n^2} \cos nx$
We have $\phi_n(x) = \frac{1}{\sqrt{\pi}} \cos(nx)$, $(f, \phi_n) = \frac{4(-1)^n}{n^2}\sqrt{\pi}$, for $n \ge 1$ and $(f, \phi_0) = \sqrt{\frac{2}{3}}\pi^{5/2}$. See Maple
worksheet. Applying Parseval's equation we get
$\|\|f\|\|^2 = \|\|x^2\|\|^2 = \frac{2\pi^5}{5} = \frac{2\pi^5}{9} + \sum_{1}^{\infty} \frac{16\pi}{n^4}$
So we have that
$\frac{2\pi^5}{5} - \frac{2\pi^5}{9} = \sum_{1}^{\infty} \frac{16\pi}{n^4}$
That is,
$\frac{8\pi^5}{45} = \sum_{1}^{\infty} \frac{16\pi}{n^4}$
namely,
$\sum_{1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$
