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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

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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.

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  • $\begingroup$ Thanks! I think I understand it now. After thinking about your response for a little while I think I understand now. From my understanding, The derivation of a_0 as the integral is valid when n is positive, but when n=0, the integral actually is twice the coefficient, so we have to cut it in half. $\endgroup$ – Joshua French Feb 23 '15 at 5:10
  • $\begingroup$ @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. $\endgroup$ – robjohn Feb 23 '15 at 5:17

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