I'm studying Fourier analysis and my book gives the following definitions for the Fourier series and Fourier coefficients:

Fourier series of $2\pi$-periodic function $f(\theta)$ is defined as:

$$f(\theta) = \frac{1}{2}a_0 + \sum_{1}^{\infty}a_n\cos n\theta + b_n\sin n\theta\;\;\;\;(1),$$

where:

$$a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\;d\theta,$$ $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos n\theta\;d\theta\;\;\;\;(n\geq0),$$ $$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin n\theta\;d\theta\;\;\;\;(n\geq 1).$$

The definition of $a_n$ puzzles me, because of the $(n\geq0)$. For example if I have the $2\pi$-periodic function $f(\theta)=|\theta|$, for $-\pi\leq\theta\leq\pi$, the Fourier coefficients of $a_0$ and $a_n$ are (from my book):

$$a_0 = \pi,\;\;\;\;\text{for}\;\;\; n=0$$ $$a_n=\frac{2}{\pi}\frac{(-1)^n-1}{n^2},\;\;\;\;\text{for}\;\;\; n>0$$

The author states also:

Note that the formula for $a_n$ also holds for $n=0$; this is the reason for the factor of $\frac{1}{2}$ in $(1)$.

Do you see my problem? If I now plug in $n=0$ for my example problem in $a_n$ I get a zero division? Why does the author make the statement that it also holds for $n=0$. Shouldn't then:

$$a_0=\frac{2}{\pi}\frac{(-1)^0-1}{0^2} = \frac{2}{\pi}\frac{0}{0}=\pi,$$

which of course makes no sense...?

Thank you for any help! =)

You are confusing two things. One is the definition of the coefficients, which can be valid for all values of $n$. And the other is the actual evaluation of the integral, which can give different solutions for $n=0$ and $n>0$.
$$a_0 = \frac{1}{\pi}\int_\pi^{-\pi} f(\theta) \cos(0\cdot \theta)d\theta =\\ =\frac{1}{\pi}\int_\pi^{-\pi} f(\theta)\cdot 1d\theta$$
• Thank you for your help @5xum, I noticed what you mean. But shouldn't (According to slightly conflicting definitions): $a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(0\cdot\theta)\;d\theta = \frac{2}{\pi}\frac{0}{0}=\pi$? I mean, shouldn't it NOT matter whether you plug in $n=0$ before you do the integration or after integration?... – jjepsuomi Mar 4 '14 at 7:33
• No, because the equation $$a_n=\frac{2}{\pi}\frac{(-1)^n-1}{n^2},\;\;\;\;\text{for}\;\;\; n>0$$ only holds for $n>0$ and not for $n=0$. The equation the author says holds for $n=0$ is the one I wrote in my answer, not one you think. – 5xum Mar 4 '14 at 7:42