Question

Draw the function f(x) = pi/L x defined in the interval -L < x < L and obtain its Fourier series expansion

Answer 1

Fourier series for any function f(x) defined in the interval -L < x < L is

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_ncos\left(\frac{\pi nx}{L}\right)+b_nsin\left(\frac{\pi nx}{L}\right)\right]$

where   $a_n=\frac{1}{L}\int_{-L}^{L}{f(x)cos\left(\frac{\pi nx}{L}\right)dx}$ , $b_n=\frac{1}{L}\int_{-L}^{L}{f(x)sin\left(\frac{\pi nx}{L}\right)dx}$

For given function f(x) = πx/L

$a_n=\frac{\pi}{L^2}\int_{-L}^{L}xcos\left(\frac{\pi nx}{L}\right)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}ycos\left(ny\right)dy=0$

$a_n=\frac{\pi}{L^2}\int_{-L}^{L}xsin\left(\frac{\pi nx}{L}\right)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}ysin\left(ny\right)dy=-\frac{2}{n}cos(\pi n)+\frac{2}{\pi n^2}sin(\pi n)=\frac{2{(-1)}^{n+1}}{n}$

So the Fourier series expansion for given function is

$\frac{\pi x}{L}=2\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n}sin\left(\frac{\pi nx}{L}\right)}$