Fourier Cosine Coefficients for Even Function over Symmetric Range

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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:

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


From Cosine Function is Even:

$\cos a = \cos \left({-a}\right)$

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

$\blacksquare$


Also see


Sources