# Fourier Series/Identity Function over Symmetric Range

## Theorem

Let $\lambda \in \R_{>0}$ be a strictly positive real number.

Let $\map f x: \openint {-\lambda} \lambda \to \R$ be the identity function on the open real interval $\openint {-\lambda} \lambda$:

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

The Fourier series of $f$ over $\openint {-\lambda} \lambda$ can be given as:

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

## Proof

From Identity Function is Odd Function, $\map f x$ is a odd function.

$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$

where:

 $\ds b_n$ $=$ $\ds \frac 2 \lambda \int_0^\pi \map f x \sin \frac {n \pi x} \lambda \rd x$ $\ds$ $=$ $\ds \frac 2 \lambda \int_0^\pi x \sin \frac {n \pi x} \lambda \rd x$ Definition of $\map f x$ $\ds$ $=$ $\ds \frac {2 \lambda} {n \pi} \paren {-1}^{n + 1}$ Half-Range Fourier Sine Series for Identity Function

Substituting for $b_n$ in $(1)$:

$\ds x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$

as required.

$\blacksquare$