Find the Fourier Series of the function:
Here we have
We obtain b_2n = 0 and
b 2 n + 1 = 2 2 n + 1 {\displaystyle b_{2n+1}={\frac {2}{2n+1}}}
Therefore, the Fourier series of f(x) is
f ( x ) = π 2 + 2 ( s i n ( x ) + s i n ( 3 x ) 3 + s i n ( 5 x ) 5 + . . . ) {\displaystyle f(x)={\frac {\pi }{2}}+2(sin(x)+{\frac {sin(3x)}{3}}+{\frac {sin(5x)}{5}}+...)}
Fourier Series: Basic Results