Answer to Question #196716 in Calculus for Ichigo

Question

Obtain the Fourier series for the following function

f(x)=0 when -π<x<0

f(x)=sinx when 0<x<π

Answer 1

First we calculate the constant 

$$a_0=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx=\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi} \sin xdx$$

$$=\dfrac{1}{\pi}\big[-\cos x\big]\begin{matrix} \pi \\ 0 \end{matrix}=-\dfrac{1}{\pi}(\cos \pi-\cos 0)=\dfrac{2}{\pi}$$

Find the Fourier coefficients for $n\not=0$

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

$$=\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi} \sin x\cos(nx)dx$$

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

$$=\dfrac{1}{2\pi}\big[-\dfrac{\cos( (n+1)x)}{n+1}+\dfrac{\cos( (n-1)x)}{n-1}\big]\begin{matrix} \pi \\ 0 \end{matrix}$$

$$=\dfrac{1}{2\pi}\big((-1)^n\dfrac{1}{n+1}-(-1)^n\dfrac{1}{n-1}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\big)$$

$$=\dfrac{1}{2\pi}\big(-\dfrac{2}{2k+1+1}+\dfrac{2}{2k+1-1}\big)$$

$$=\dfrac{1}{2\pi}\big(\dfrac{1}{k}-\dfrac{1}{k+1}\big)$$

$$=\dfrac{1}{2\pi k(k+1)}$$

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

$$=\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi} \sin x\sin(nx)dx$$

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

$$=\dfrac{1}{2\pi}\big[\dfrac{\sin( (n-1)x)}{n-1}-\dfrac{\sin( (n+1)x)}{n+1}\big]\begin{matrix} \pi \\ 0 \end{matrix}$$

$$=\dfrac{1}{2\pi}\big(0-0-0+0\big)=0$$

Thus, the Fourier series is

$$f(x)=\dfrac{1}{\pi}+\dfrac{1}{2\pi k(k+1)}\cos((2k+1)x)$$