# Fourier Cosine Coefficients for Even Function over Symmetric Range

## Theorem

Let $f \left({x}\right)$ be an even real function defined on the interval $\left[{-l \,.\,.\, l}\right]$.

Let the Fourier series of $f \left({x}\right)$ be expressed as:

$\displaystyle f \left({x}\right) \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \left({a_n \cos \frac {n \pi x} l + b_n \sin \frac {n \pi x} l}\right)$

Then for all $n \in \Z_{\ge 0}$:

$a_n = \displaystyle \frac 2 l \int_0^l f \left({x}\right) \cos \frac {n \pi x} l \, \mathrm d x$

## Proof

As suggested, let the Fourier series of $f \left({x}\right)$ be expressed as:

$\displaystyle f \left({x}\right) \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \left({a_n \cos \frac {n \pi x} l + b_n \sin \frac {n \pi x} l}\right)$

By definition of Fourier series:

$a_n = \displaystyle \frac 1 l \int_{-l}^{-l + 2 l} f \left({x}\right) \cos \frac {n \pi x} l \, \mathrm d x$
$\cos a = \cos \left({-a}\right)$

for all $a$.

By Even Function Times Even Function is Even, $f \left({x}\right) \cos \dfrac {n \pi x} l$ is even.

Thus:

 $\displaystyle a_n$ $=$ $\displaystyle \frac 1 l \int_{-l}^{-l + 2 l} f \left({x}\right) \cos \frac {n \pi x} l \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 1 l \int_{-l}^l f \left({x}\right) \cos \frac {n \pi x} l \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 2 l \int_0^l f \left({x}\right) \cos \frac {n \pi x} l \, \mathrm d x$ Definite Integral of Even Function

$\blacksquare$