# Derivation of Fourier Series?

Can someone point me to the full derivation of the Fourier Series? I'm having problems understanding how the a's and b's coeffients are worked out.

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## migrated from physics.stackexchange.comFeb 23 '11 at 1:25

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This isn't really a physics question. I'll flag for the mods to migrate it to math.stackexchange. – Mark Eichenlaub Feb 17 '11 at 20:54
I don't quite understand the question: the Fourier Series is a definition: for any integrable function on a closed interval, its coefficients are given by a particular formula: see e.g. en.wikipedia.org/wiki/Fourier_series#Definition. So what exactly is it that you want to work out? – Pete L. Clark Feb 23 '11 at 2:16
I think the OP wants to see a proof that if $f(x) = \sum a_n\cos nx + b_n\sin nx$, where the convergence is (at least) pointwise, then the coefficients $a_n$ and $b_n$ are given by those standard formulas. – Jesse Madnick Feb 23 '11 at 2:43

Imagine that $f(x) = \sum_n a_n \cos n x + \sum_n b_n \sin n x$ (for $x \in [0,2\pi]$, say). The idea for computing the $a_n$s and $b_n$s is that when you write down integrals of the form $\int_0^{2\pi} \cos m x \cos n x,$ or $\int_0^{2\pi} \sin m x \sin n x$, or $\int_0^{2 \pi} \cos m x \sin n x$, then the integrals vanish (just compute them!) unless $m = n$ and the functions coincide; and in the cases when they don't vanish, their values are easily computed.
So taking $f$, and then computing $\int_0^{2\pi} f(x) \cos n x$ or $\int_0^{2 \pi} f(x) \sin n x$, one exactly reads off $a_n$ or $b_n$ (for the value of $n$ you chose). This is where the formulas come from.
The way people normally think about this is as a kind of orthogonal projection: the functions $\cos n x$ and $\sin n x$ are like orthogonal basis vectors in a vector space, and the integral is like an inner product. So to find the coefficient of a given basis vector (i.e. an $a_n$ or a $b_n$) one takes the inner product against that particular basis vector.