I'm trying to find the Fourier series for $f(\theta) = |\sin\theta|$. The function is even, so $b_n = 0 \space\forall n$. Doing the integration for $a_n$ yields

$$a_n = \frac{2}{\pi}\left(\frac{1+(-1)^n}{1-n^2}\right)$$

Then, calculating $a_0$ to be $\frac{4}{\pi}$ and plugging into the Fourier series formula gives

$$f(\theta) = \frac{2}{\pi} + \frac{2}{\pi} \sum_{n=1}^\infty \frac{1+(-1)^n}{1-n^2} \cos(n\theta)$$

I then tried graphing this function to check my answer, and realised that the first term of the series is undefined as it involves a division by $0$. I then tried starting the series from $n=2$ instead of $n=1$, and the graph looked correct. How do we deal with Fourier terms that are undefined? Can we always just ignore them?

EDIT: Formula for $a_n$:

$$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(\theta) \cos(n\theta)d\theta = \frac{2}{\pi} \int_{0}^\pi \sin\theta \cos(n\theta)d\theta$$


Let $$I = \int_{0}^\pi \sin\theta \cos(n\theta)d\theta$$

By parts:

$$I = \Bigl[-\cos\theta \cos(n\theta)\Bigr]_0^{\pi}-n\int_0^\pi\cos\theta \sin(n\theta)d\theta$$ $$=(1+(-1)^n)-n\Bigl(\Bigl[\sin\theta\sin(n\theta)\Bigr]_0^\pi-n\int_0^{\pi}\sin\theta\cos(n\theta)d\theta\Bigr)$$ $$=1+(-1)^n+n^2I$$ $$\implies I = \frac{1+(-1)^n}{1-n^2}$$ $$\therefore a_n = \frac{2}{\pi}\left(\frac{1+(-1)^n}{1-n^2}\right)$$

  • 2
    $\begingroup$ If that's really true, then your integration for $a_n$ is not valid for $n=1$. You should compute $a_1$ separately. Try writing out that specific case and you will probably find you have to handle it differently. $\endgroup$
    – MPW
    Oct 26 '17 at 21:18
  • $\begingroup$ @MPW Doing it manually gives $a_1 = 0$, so I can ignore it here. Do we always need to compute them manually? $\endgroup$
    – imulsion
    Oct 26 '17 at 21:34
  • 1
    $\begingroup$ Can you write out the formula so I can see it (for $a_n$)? You probably have a step which isn't valid if $n=1$. Just as a dumb example, the formula $\int x^n\; dx = \frac{x^{n+1}}{n+1}+C$ is not valid for $n=-1$. You just have to inspect your formula for invalid cases and handle them separately, if there are any (there needn't be any). $\endgroup$
    – MPW
    Oct 26 '17 at 21:42
  • $\begingroup$ @MPW Added the calculations in for $a_n$ $\endgroup$
    – imulsion
    Oct 26 '17 at 21:54
  • 1
    $\begingroup$ Indeed. The equation reduces to $I=1-1+I$, that is, $I=I$, when $n=1$. So another approach is needed. But it's easy since $\int\sin t\cos t \;dt = \int \frac12\sin 2t \; dt$ which can be computed directly. $\endgroup$
    – MPW
    Oct 26 '17 at 21:57

You cannot always ignore the indeterminate forms. Consider that if $f(\theta) = 1$, then $$\int_{-\pi}^\pi \cos n \theta \,d\theta = \frac {2 \sin \pi n} n.$$ $a_0$ is non-zero, but it cannot be found by computing the integral for general $n$ and then substituting $n = 0$.

If $f$ is bounded, then $a_n$ can be obtained by treating $n$ as a continuous variable and taking the limit: $$a_n = \frac 2 \pi \int_0^\pi \sin \theta \cos n \theta \,d\theta = \frac {2(1 + \cos \pi n)} {\pi(1 - n^2)}, \\ a_1 = \lim_{n \to 1} a_n = 0.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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