You're given a function f from [0,L]. Extend f into (L,2L) so that is it symmetric about x = L; that is to say it satisfies f(2L - x) = f(x) for x in [0, L). Let the resulting function be extended into (-2L,0) as an odd function ans elsewhere as a periodic function of period 4L. Show that this function has a Fourier series of:
f(x) = Σ(n = 1 to inf) b_n sin((2n - 1)πx) /(2L)
b_n = (2 / L) ∫[0,L] f(x) sin((2n - 1)πx) /(2L) dx.
Since f has an odd extension I understand why we have a fourier sine series; however, I'm not sure why all the even terms vanish so that we only consider the odd terms (2n - 1). I also thought the limits of the integration for the coefficient b_n should be from 0 to 2L because we have a period of 4L, why does 0 to L suffice?