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:


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

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

By Fourier Series for Odd Function over Symmetric Range, we have:


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

where:

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


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

as required.