Even function-Fourier Series

[iocal]

\(\displaystyle \displaystyle F(x)=x^2,\ 0<x<2\pi,\) with period 2π.

This is an even function right? Could someone please explain to me why the resulting series is not solely a cosine series and instead the coefficient of the sine part is not zero if we integrate from 0 to 2π?

Thanks.

 

[Kavi Rama Murthy]

Not an even function!

Let me first point out that your function is not defined properly. Its value at 0, for example, has not been specified. That apart, it is not even! The value at -1 is [(2*pi)-1]^2 ant its value at 1 is 1. Before studying Fourier series make sure that the function is defined at all real numbers and that it has period 2*pi. To conclude that f is even (respectively odd) you have to ensure that f(x)=f(-x) (respectively f(x)=-f(-x)) for all real numbers x.