Fourier basis functions

Question

What are Fourier basis functions? And how do I prove that Fourier basis functions are orthonormal?

Answer 1

An orthonormal basis for $L^2([0,1],\mathbb{R})$ (the space of real valued square integrable functions on the interval $[0,1]$) is $1, \sqrt{2}\cos(2\pi nx), \sqrt{2}\sin(2\pi nx)$ for $n=1,2,3,...$. These functions can be written as (convergence in $L^2$, many details omitted): $$ f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(2\pi nx)+b_n\sin(2\pi nx) $$ where $a_0=\int_{[0,1]}f(x)\,dx$, and for $n\geq 1$ $$ a_n=2\int_{[0,1]}f(x)\cos(2\pi nx)dx, \quad b_n=2\int_{[0,1]}f(x)\sin(2\pi nx)dx. $$ The orthonormality of the basis functions is established by showing that $$ \int_{[0,1]}\cos(2\pi nx)\sin(2\pi mx)dx=0, $$ $$ \int_{[0,1]}\cos(2\pi nx)\cos(2\pi mx)dx= \left\{ \begin{array}{cc} 1/2&\text{ if } n=m\\ 0&\text{ if } n\neq m\\ \end{array} \right., $$ $$ \int_{[0,1]}\sin(2\pi nx)\sin(2\pi mx)dx= \left\{ \begin{array}{cc} 1/2&\text{ if } n=m\\ 0&\text{ if } n\neq m\\ \end{array} \right., $$ so they are orthonormal with respect to the inner product $$ \langle f,g\rangle=\int_{[0,1]}f(x)g(x)dx. $$ You can learn a lot more by finding a good reference. Most differential equations books cover Fourier series to some extent to provide solutions to the heat/wave/Laplace equations (e.g. Boyce and DiPrima). Here is something random from google showing the orthogonality relations (don't know if its good).

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EDIT: Since the link is broken (although any other random google search will bring up something), here is a sample derivation of one of the orthogonality relations using integration by parts twice ($m\neq n$ both non-zero): \begin{align*} &\int_0^1\sin(2\pi nx)\cos(2\pi mx)dx\\ &=\frac{1}{2\pi m}\sin(2\pi nx)\sin(2\pi mx)\Bigg|_0^1-\int_0^1\frac{2\pi n}{2\pi m}\cos(2\pi nx)\sin(2\pi m x)dx\\ &=-\frac{n}{m}\int_0^1\cos(2\pi nx)\sin(2\pi m x)dx\\ &=\frac{n^2}{m^2}\int_0^1\sin(2\pi nx)\cos(2\pi m x)dx \end{align*} Since $n^2/m^2\neq1$, the integral(s) must be zero (i.e. $x=\lambda x\Rightarrow x=0$ for $\lambda\neq1$).

The others are left as an exercise (which might be why you looked for this answer in the first place).

A more compact way of seeing the orthogonality relations is to use complex exponentials. We have \begin{align*} \int_0^{1}e^{2\pi inx}e^{2\pi imx}dx&=\int_0^1e^{2\pi i(n+m)x}dx\\ &=\left\{ \begin{array}{cc} 1&n+m=0\\ \frac{e^{2\pi i(n+m)}-1}{2\pi i(n+m)}&\text{else}\\ \end{array} \right.\\ &=\left\{ \begin{array}{cc} 1&n+m=0\\ 0&\text{else}\\ \end{array} \right. \end{align*}