Fourier Series/Absolute Value of x over Minus Pi to Pi

Theorem

For $x \in \openint {-\pi} \pi$:

$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

Proof 1

By definition, the absolute value function is an even function:

$\size {-x} = x = \size x$

Thus by Fourier Series for Even Function over Symmetric Range, $\size x$ can be expressed as:

$\ds \size x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$

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

$a_n = \ds \frac 2 \pi \int_0^\pi \size x \cos n x \rd x$

On the real interval $\openint 0 \pi$:

$\size x = x$

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

$a_n = \ds \frac 2 \pi \int_0^\pi x \cos n x \rd x$

Thus Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$ can be applied directly.

So for $x \in \openint {-\pi} \pi$:

$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

$\blacksquare$

Proof 2

By Fourier Series for Absolute Value Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:

$\forall x \in \openint {-\lambda} \lambda: \map f x = \size x$

has a Fourier series:

$\map f x \sim \dfrac \lambda 2 - \ds \dfrac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$

Substituting for $\lambda = \pi$ gives:

$\size x = \dfrac \pi 2 - \ds \dfrac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

as required.

$\blacksquare$