Question

Obtain the Fourier series for the following periodic function which has a period of 2π: f(x)=x for−π≤x≤π

Answer 1

$f(x)=x,[-\pi,\pi]$

The function f(x) is not even, so a₀=0, a_n=0.

The numbers a₀, a_n and b_n (n=1,2,...) are called the Fourier coefficients of the function f.

The Fourier series of the function f(x) on the interval (0;T) in terms of the sines of multiple arcs is called the series:

$f(x)=\sum b_n*sin(\frac{\pi*n*x}{T})$

$b_n=\frac{2}{T}\displaystyle\int_0^Tf(x)*sin(\frac{\pi*n*x}{T})dx$

For our data:

$b_n=\frac{2}{\pi}\displaystyle\int_0^\pi f(x)*sin(\frac{\pi*n*x}{\pi})dx=\frac{2}{\pi}\displaystyle\int_0^\pi x*sin(n*x)dx=$

$\frac{2}{\pi}(-\pi*\frac{cos(\pi*n)}{n}+\frac{sin(\pi*n)}{n^2}-0)=-2*\frac{(-1)^n}{n}$

Finally

$f(x)=0.0+\sum(-2*\frac{(-1)^n}{n})*sin(n*x)$