Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Is there a faster approach for finding the Fourier series of $$\sin(x/2)~,\cos(x/2)~~~\text{etc}$$ other than the usual approach?

share|cite|improve this question
Fourier series generally apply to period function a period $2\pi$. These are not quite so periodic. You could define though a function equal to these on the interval $[0,2\pi]$, and define for periodicity outside that interval. – Thomas Andrews Mar 7 '13 at 2:21
The usual approach is pretty fast, isn't it? You have to calculate integrals like $\int\sin nx\sin(x/2)\,dx$, and you can use trig identities to transform that product of sines into a simple sum, which sum is easy to integrate. – Gerry Myerson Mar 7 '13 at 2:29

I'll give a non-standard derivation for $\cos (x/2)$ on $[-\pi,\pi]$. Faster or not, you tell.

The function being even, we work with cosines. Observe that if $$f(x)\sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx\tag1$$ then $$f''(x)\sim -\sum_{n=1}^\infty n^2 a_n \cos nx\tag2$$ I am writing $\sim $ instead of $=$ here to avoid getting arrested by the convergence police. Naturally, we want to plug (1) and (2) into $f''= (-1/4)f$ and equate coefficients. This does not work.

The problem is, $f''$ is not exactly $(-1/4)f$. The values of $f'$ at the endpoints don't match, and this discontinuity contributes a point mass to $f''$. The correct formula is $$f''=(-1/4)f+\delta_{\pi}+\delta_{-\pi}\tag3$$ It's easy to find the coefficients for the delta function: $$\delta_\pi +\delta_{-\pi}= \frac{c_0}{2}+ \sum_{n=1}^\infty c_n\cos nx, \quad c_n= (-1)^n\frac{2}{\pi}\tag4$$ because "integrating" against a delta function amounts to evaluation. Now plug into (3). $$-n^2 a_n = (-1/4)a_n + c_n\tag5$$
which yields $$ a_n = \frac{-c_n}{n^2-1/4}\tag6$$ with $c_n$ as in (4).

share|cite|improve this answer
+1 for the convergence police. – nbubis Mar 7 '13 at 6:48
Convergence police is a good term. Also, why not substitute the number $\tfrac{1}{2}$ in $\cos{(x/2)}$ with a general $k$. – NikolajK Mar 7 '13 at 8:36
@Nick One could do that, although the easy case of integer $k$ must be treated separately. I "leave this as an exercise for the interested reader." – user53153 Mar 7 '13 at 12:44

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.