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

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I did. But I am looking for proof of the fact too. – Brahadeesh Apr 13 '11 at 0:01
You have to compute the integrals that define the inner products. See – lhf Apr 13 '11 at 0:03
@Brahadeesh: I think you're missing what stackexchange is about. We're here to enhance each other's understanding of mathematics, not to do what is likely other people's homework for them. – Alex Becker Apr 13 '11 at 0:33
@Alex I am sorry if I gave that impression. The question was not my homework. But I encountered it when I was studying Digital Image reconstruction. As I did not know where else to go, I asked in the Math forum. I was looking for someone to provide an insight that I can use for understanding this topic. – Brahadeesh Apr 13 '11 at 0:39
up vote 3 down vote accepted

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): $$ f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx) $$ where $a_0=\int_{[0,1]}f(x)\,dx$, and for $n\ge 1$ $$ a_n=\frac12\int_{[0,1]}f(x)\cos(2\pi nx)dx, \quad b_n=\frac12\int_{[0,1]}f(x)\sin(2\pi nx)dx. $$ The orthonormality of the basis functions is established by showing that $$ \int_{[0,1]}\cos(2\pi nx)\sin(2\pi mx)dx=0, $$ $$ \int_{[0,1]}\cos(2\pi nx)\cos(2\pi mx)dx=0 \text{ if } n\neq m, \ 1/2 \text{ if } n=m,$$ $$ \int_{[0,1]}\sin(2\pi nx)\sin(2\pi mx)dx=0 \text{ if } n\neq m, \ 1/2 \text{ if } n=m,$$ 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 (eg Boyce and DiPrima). Here is something random from google showing the orthogonality relations (don't know if its good).

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