Question

please solve fourier series of function f(x)=x^2 0 13 years ago

Answer 1

\[f(x)=\begin{cases}x^2&\text{for }0<x<\pi\\-x^2&\text{for }-\pi<x<0\end{cases}\]
The Fourier series for \(f(x)\) is
\[\frac{1}{2}a_0+\sum_{n=1}^\infty \bigg(a_n\cos(nx)+b_n\sin(nx)\bigg),\]
where
\[a_0=\frac{1}{\pi}\int_{-\pi}^\pi f(x)~dx,\\a_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx)~dx,~\text{ and }~b_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx)~dx\]