Fourier Example

Find the Fourier Series of the function:

f(x)={\begin{cases}0,&-\pi \leq x<0\\\pi ,&0\leq x\leq \pi \\\end{cases}}

Solution

Here we have

$a_{0}={\frac {1}{2\pi }}(\int _{-\pi }^{0}0\ dx+\int _{0}^{\pi }\pi \ dx)={\frac {\pi }{2}}$

$a_{n}=\int _{0}^{\pi }\pi cos(nx)\ dx=0,n\geq 1,$

and

b_{n}=\int _{0}^{\pi }\pi \sin(nx)\,dx={\frac {1}{n}}(1-cos(x\pi ))={\frac {1}{n}}(1-(-1)^{n})

We obtain $b_{2n}$ = 0 and

$b_{2n+1}={\frac {2}{2n+1}}$

Therefore, the Fourier series of f(x) is

$f(x)={\frac {\pi }{2}}+2(sin(x)+{\frac {sin(3x)}{3}}+{\frac {sin(5x)}{5}}+...)$