[Resolução] III.6.10b
Posted: 07 Dec 2022 23:07
\(a_n=\frac{1}{\pi }\int_{\pi}^{-\pi}\left | x \right |cosnxdx
= \frac{2}{n^2\pi}cos(nx)|^\pi _0 =
\left\{\begin{matrix}
-\frac{4}{n^2\pi}&impar \\
0&par
\end{matrix}\right.
\\
|x|=\frac{\pi^2}{2}-\frac{4}{\pi} \sum_{n=1}^{\infty}\frac{cosnx}{n^2}=\frac{\pi^2}{2}-\frac{4}{\pi} \sum_{n=1}^{\infty}\frac{cos(2n-1)x}{(2n-1)^2}
\\
0 = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}
\\
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}\)
= \frac{2}{n^2\pi}cos(nx)|^\pi _0 =
\left\{\begin{matrix}
-\frac{4}{n^2\pi}&impar \\
0&par
\end{matrix}\right.
\\
|x|=\frac{\pi^2}{2}-\frac{4}{\pi} \sum_{n=1}^{\infty}\frac{cosnx}{n^2}=\frac{\pi^2}{2}-\frac{4}{\pi} \sum_{n=1}^{\infty}\frac{cos(2n-1)x}{(2n-1)^2}
\\
0 = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}
\\
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}\)