Half-Range Fourier Series/Identity Function

Theorem

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

Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:

$\forall x \in \openint 0 \lambda: \map f x = x$

The half-range Fourier series of $f$ over $\openint 0 \lambda$ can be given in the following forms:

Cosine Series for Identity Function

The half-range Fourier cosine series for $\map f x$ can be expressed as:

\(\ds \map f x\)

\(\sim\)

\(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda\)

\(\ds \)

\(=\)

\(\ds \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \paren {\cos \dfrac {\pi x} \lambda + \frac 1 {3^2} \cos \dfrac {3 \pi x} \lambda + \frac 1 {5^2} \cos \dfrac {5 \pi x} \lambda + \dotsb}\)

Sine Series for Identity Function

The half-range Fourier sine series for $\map f x$ can be expressed as:

\(\ds \map f x\)

\(\sim\)

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

\(\ds \)

\(=\)

\(\ds \dfrac {2 \lambda} \pi \paren {\sin \dfrac {\pi x} \lambda - \frac 1 2 \sin \dfrac {2 \pi x} \lambda + \frac 1 3 \sin \dfrac {3 \pi x} \lambda - \dotsb}\)