Half Range Expansion

Theory

If a function $f(x)$ is defined only on a half-period $[0,L]$ , it can be extended to $[-L,L]$ as:

Exercise 1

Question

Write the half-range Fourier sine series expansion for

f(x)= \begin{cases} 0, & 0\le x\le \frac{\pi}{2},\\ 1, & \frac{\pi}{2}\le x\le \pi. \end{cases}

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Solution

For a half-range sine series on $[0,\pi]$ , we use

f(x)=\sum_{n=1}^{\infty} b_n \sin(nx),

where

b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)\,dx.

Since $f(x)=0$ on $[0,\frac{\pi}{2}]$ and $f(x)=1$ on $[\frac{\pi}{2},\pi]$ ,

b_n=\frac{2}{\pi}\int_{\frac{\pi}{2}}^{\pi}\sin(nx)\,dx.

b_n=\frac{2}{\pi}\left[-\frac{\cos(nx)}{n}\right]_{\frac{\pi}{2}}^{\pi} =\frac{2}{n\pi}\left(\cos\frac{n\pi}{2}-\cos n\pi\right).

Since $\cos n\pi=(-1)^n$ , we obtain

b_n=\frac{2}{n\pi}\left(\cos\frac{n\pi}{2}-(-1)^n\right).

Hence, the half-range sine series is

f(x)=\sum_{n=1}^{\infty} \frac{2}{n\pi}\left(\cos\frac{n\pi}{2}-(-1)^n\right)\sin(nx), \quad 0&lt;x&lt;\pi.

Exercise 2

Question

Write the half-range cosine series expansion of

f(x)=x,\quad 0&lt;x&lt;4.

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Solution

For a half-range cosine series on $[0,L]$ with $L=4$ ,

f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos\frac{n\pi x}{4},

where

a_0=\frac{2}{4}\int_{0}^{4}x\,dx, \quad a_n=\frac{2}{4}\int_{0}^{4}x\cos\frac{n\pi x}{4}\,dx.

First,

a_0=\frac{1}{2}\left[\frac{x^2}{2}\right]_{0}^{4}=4.

Next, using integration by parts,

a_n=\frac{1}{2}\int_{0}^{4}x\cos\frac{n\pi x}{4}\,dx =\frac{8(-1)^n}{n^2\pi^2}.

Thus, the half-range cosine series is

f(x)=2+\sum_{n=1}^{\infty}\frac{8(-1)^n}{n^2\pi^2} \cos\frac{n\pi x}{4}, \quad 0&lt;x&lt;4.