Answer to Question #223175 in Math for sunday

Question

$$Question2 A \, periodic \,function\, f(x) \,with \,period \,2 π \,is \,defined\, by :\\ f(x)= [ (x+π)/2 , \, -π< x<0, \\(x-π)/2 , \, 0<x<π]\\ Determine \,the \,Fourier \,series \,for \,the \,periodic\, function.$$

Answer 1

$$a_0=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx$$

$$=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}dx+\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}dx$$

$$=\dfrac{2}{\pi}[\dfrac{x^2}{4}+\dfrac{\pi x}{2}]\begin{matrix} 0 \\ -\pi & d \end{matrix}+\dfrac{2}{\pi}[\dfrac{x^2}{4}-\dfrac{\pi x}{2}]\begin{matrix} \pi \\ 0 \end{matrix}$$

$$=\dfrac{\pi}{2}-\dfrac{\pi}{2}=0$$

$$a_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\cos{nx}dx$$

$$=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\cos{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\cos{nx}dx$$

$$=-\dfrac{\cos{\pi n}-1}{2\pi n^2}+\dfrac{\cos{\pi n}-1}{2\pi n^2}=0$$

$$b_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\sin{nx}dx$$

$$=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\sin{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\sin{nx}dx$$

$$=-\dfrac{1}{2n}-\dfrac{1}{2n}=-\dfrac{1}{n}$$

$$f(x)=-\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n}\sin{nx}$$