if you have to write Fourier serie for f(x)=x^2, 0

Question

if you have to write Fourier serie for f(x)=x^2, 0<x<pi.....do you have to calculate it for -pi<x<pi?

anonymous · Answer

or why should i have $$a0=2/\pi \int\limits_{0}^{\pi} x ^{2}dx$$

anonymous · Answer

So we want the Fourier series for F(x)=x^2, for 0<x<pi.

You don't have to calculate it for -pi<x<pi. You can use the domain given to you.

So the Fourier series would be

$$f(x)=\frac{1}{2}a_o+\sum_{n=1}^{\infty}a_ncos(\frac{n \pi x}{L})+b_nsin(\frac{n \pi x}{L})$$ for -L<x<L

where $$a_0=\frac{1}{L}\int\limits_{-L}^{L}F(x)dx$$ $$a_n=\frac{1}{L}\int\limits_{-L}^{L}F(x)cos(\frac{n \pi x}{L})dx$$   $$b_n=\frac{1}{L}\int\limits_{-L}^{L}F(x)sin(\frac{n \pi x}{L})dx$$

anonymous · Answer

So for F(x)=x^2 0<x<pi, 

the L value will be pi/2.

$$a_0=\frac{1}{\frac{\pi}{2}} \int\limits_{0}^{\pi} x^2 dx$$
       $$=\frac{2}{\pi} \int\limits_{0}^{\pi} x^2 dx$$

anonymous · Answer

and if you have -pi<x<pi, you use "normal" formula?