# Why does a fourier series have a 1/2 in front of the a_0 coefficient

I am reading up on the fourier series, and I keep seeing it as being defined as:

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

where

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

and

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

I understand the derivation of the coefficients using trig integral identities, but I can't find a clear explanation of why $\frac{1}{2}$ is in front of $a_0$. Can anyone help show my why this is the case? Why can't we just have $a_0$ with no number in front of it. Thanks!

edit: corrected summation term

Because $$\frac1\pi\int_0^{2\pi}\cos^2(n\theta)\,\mathrm{d}\theta=\left\{\begin{array}{}2&\text{if }n=0\\1&\text{if }n\ne0\end{array}\right.$$ Also, the summation should start at $n=1$.
We have defined $$a_n=\frac1\pi\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta$$ If we write $$g(\theta)=\sum_{n=0}^\infty a_n\cos(n\theta)+\sum_{n=1}^\infty b_n\sin(n\theta)$$ then $$\frac1\pi\int_0^{2\pi}g(\theta)\cos(n\theta)\,\mathrm{d}\theta=\left\{\begin{array}{}\color{#C00000}{2}a_0&\text{if }n=0\\a_n&\text{if }n\ne0\end{array}\right.$$ So we have to use $\frac12a_0$ to compensate.
• @JoshuaFrench: Yes. We either have to divide the definition of $a_0$ by $2$ or divide $a_0$ by $2$ when we use it to reconstruct $f(\theta)$. We have chosen the latter. – robjohn Feb 23 '15 at 5:17