Enunciado:
Seja a série de Fourier da função \(2\pi\) periódica \(f(x) = x^2\) para \(x \in [-\pi, \pi]\). Mostre que \(\sum_{n=1}^\infty\ \frac{1}{n^4} = \frac{\pi^4}{90}\)
Solução:
Sabendo que \(f(x) = x^2 = S_f(x) = \frac{\pi^2}{3} + \sum_{n=1}^\infty\ (-1)^n\frac{4 cos(nx)}{n^2}\) para \(x \in [-\pi, \pi]\)
\(a_0 = \frac{2\pi^2}{3}\)
\(a_n = \frac{4 (-1)^n}{n^2}\)
Pela Identidade de Parseval sabemos que:
\(\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \,dx = \frac{a_0^2}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2)\)
\(\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \,dx = \frac{1}{\pi} \int_{-\pi}^{\pi} x^4 \,dx = \frac{1}{\pi} \frac{x^5}{5}|_{-\pi}^{\pi} = \frac{1}{\pi} (\frac{2\pi^5}{5}) = \frac{2\pi^4}{5}\)
\(\frac{2\pi^4}{5} = \frac{\frac{4\pi^4}{9}}{2} + \sum_{n=1}^\infty [\frac{4(-1)^n}{n^2}]^2\)
\(\frac{2\pi^4}{5} = \frac{2\pi^4}{9} + \sum_{n=1}^\infty \frac{16}{n^4}\)
\(\frac{2\pi^4}{5} - \frac{2\pi^4}{9} = 16 \sum_{n=1}^\infty \frac{1}{n^4}\)
\(\frac{8\pi^4}{45} = 16 \sum_{n=1}^\infty \frac{1}{n^4}\)
\(\frac{8\pi^4}{45 \cdot 16} = \sum_{n=1}^\infty \frac{1}{n^4}\)
\(\frac{\pi^4}{90} = \sum_{n=1}^\infty \frac{1}{n^4}\)
