Fourier exponential series of odd functions does not have even coefficients

Question

Consider the exponential fourier series

$$
f(t)= \sum\limits_{n = – \infty }^\infty {c_n e^{i n\omega _0 t} } $$

if $f$ is odd $f(-t)=-f(t)$, then how is this visible in the series? Are there only the odd $n$ values? i.e. $c_n=0$ for even $n$?

Answer 1

In that case,$$f(-t)=\sum_{n=-\infty}^{\infty}c_n e^{-i\omega_0nt}=\sum_{n=-\infty}^{\infty}c_{-n} e^{i\omega_0nt}\\-f(t)=\sum_{n=-\infty}^{\infty}-c_n e^{i\omega_0nt}$$which leads to$$c_{-n}=-c_n$$therefore also $c_n$ is odd and we can write$$f(t)=\sum_{n=1}^{\infty}2i c_n \sin n\omega_0 t$$