# 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.

-
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

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

This question came from our site for active researchers, academics and students of physics.

If you already believe that any function can be written as an infinite series of sines and cosines with appropriate coefficients, and all you are wondering about is how we go about actually computing the coefficients, then you're most of the way there. The tricky part is proving that the Fourier series works at all; the derivation of the coefficients is pretty straightforward. It's still not exactly intuitive and involves some steps that you would never think of, but it's easy enough to follow. A good derivation is here: http://cnx.org/content/m10733/latest/

-
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.