Show that if $\phi$ is an odd function on $(-l,l)$, its full Fourier series on $(-l,l)$ only has sine terms.
The full Fourier series is defined as $$\phi(x)=\frac{1}{2}A_0+\sum_{n=1}^{\infty}(A_n cos \frac{n \phi x}{l}+B_n sin \frac{n \pi x}{l})$$, for $-l <x< l$, and $$A_n=\frac{1}{l} \sum_{-l}^l \phi(x) cos \frac{n \pi x}{l} dx(n=0,1,2,…)$$ $$B_n=\frac{1}{l} \sum_{-l}^l \phi(x) sin \frac{n \pi x}{l} dx(n=0,1,2,…)$$
I also know that an odd function is like this $\phi(-x)=-\phi(x)$, but am not sure how to prove it.
Best Answer
The result holds for both the discrete version of the coefficients and the ones involving integrals. Consider the term $A_n$. We will show that it is always $0$. First of all, write: $$ {A_n} = \frac{1}{l} \sum_{x=-l}^{-1} \phi(x) \cos(\frac{n \pi x}{l}) + \phi(0) + \frac{1}{l} \sum_{x=1}^{l} \phi(x) \cos(\frac{n \pi x}{l}) $$ Since $\phi$ is odd, $\phi(0)=-\phi(0)$, which implies that $\phi(0)=0$. Moreover, recall that $\cos(\cdot)$ is an even function, so that we can write: $$ A_n = \frac{1}{l} \sum_{x=1}^{l} -\phi(x) \cos(\frac{n \pi x}{l}) + 0 + \frac{1}{l} \sum_{x=1}^{l} \phi(x) \cos(\frac{n \pi x}{l}) = 0$$ And the result is thus proved, because the coefficients of the cosines in the series are always $\mathbb{0}$. The same can be proved also for integrals, using the fact that $\phi$ is odd.