Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If a function is even or odd, it implies that there are respectively only cos and sine terms in its Fourier expansion. But is there a condition for a function to have an expansion with only odd or even harmonics, like this:

$\frac{a_0}{2} + \sum_{n=1,3,5,...}^\infty a_n \cos(nx)$

share|cite|improve this question
up vote 4 down vote accepted

Yes, you have to consider symmetries around $\pi/2$. Let me show you an example. Let $f\colon[-\pi,\pi]\to\mathbb{R}$ be an odd continuous function. Then it's Fourier series has only sine terms: $$ f\sim\sum_{n=1}^\infty b_n\sin(n\,x),\quad b_n=\frac{2}{\pi}\int_0^\pi f(x)\sin(n\,x)\,dx. $$ Suppose further that $f$ satisfies $f(\pi-x)=f(x)$ if $0\le x\le\pi$. Then for all $n\in\mathbb{N}$ $$ \int_0^\pi f(x)\sin(2\,n\,x)\,dx=\int_0^\pi f(\pi-x)\sin(2\,n\,\pi-2\,n\,x)\,dx=-\int_0^\pi f(x)\sin(n\,x)\,dx. $$ This implies that $b_{2n}=0$.

If $f(\pi-x)=-f(x)$, then $b_{2n+1}=0$. You can find similar conditions for cosine series.

share|cite|improve this answer
Thank you for the explanation ! – vkubicki Mar 5 '13 at 18:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.