Half range Fourier series

Short description: Fourier series defined on an interval [0,L]

A half range Fourier series is a Fourier series defined on an interval [math]\displaystyle{ [0,L] }[/math] instead of the more common [math]\displaystyle{ [-L,L] }[/math], with the implication that the analyzed function [math]\displaystyle{ f(x), x\in[0,L] }[/math] should be extended to [math]\displaystyle{ [-L,0] }[/math] as either an even (f(-x)=f(x)) or odd function (f(-x)=-f(x)). This allows the expansion of the function in a series solely of sines (odd) or cosines (even). The choice between odd and even is typically motivated by boundary conditions associated with a differential equation satisfied by [math]\displaystyle{ f(x) }[/math].

Example

Calculate the half range Fourier sine series for the function [math]\displaystyle{ f(x)=\cos(x) }[/math] where [math]\displaystyle{ 0\lt x\lt \pi }[/math].

Since we are calculating a sine series, [math]\displaystyle{ a_n=0\ \quad \forall n }[/math] Now, [math]\displaystyle{ b_n= \frac{2}{\pi} \int_0^\pi \cos(x)\sin(nx)\,\mathrm{d}x = \frac{2n((-1)^n+1)}{\pi(n^2-1)}\quad \forall n\ge 2 }[/math]

When n is odd, [math]\displaystyle{ b_n=0 }[/math] When n is even, [math]\displaystyle{ b_n={4n \over \pi(n^2-1)} }[/math] thus [math]\displaystyle{ b_{2k}={8k \over \pi(4k^2-1)} }[/math]

With the special case [math]\displaystyle{ b_1=0 }[/math], hence the required Fourier sine series is

[math]\displaystyle{ \cos(x) = {{8 \over \pi} \sum_{n=1}^{\infty} {n \over(4n^2-1)}\sin(2nx)} }[/math]