# Taylor series vs. Fourier series

1. Nov 5, 2009

### jaejoon89

Is a Fourier series essentially the analogue to a Taylor series except expressing a function as trigs functions rather than as polynomials? Like the Taylor series, is it ok only for analytic functions, i.e. the remainder term goes to zero as n->infinity?

2. Nov 5, 2009

### jbunniii

A Taylor series has to be expanded around a specific point, and the coefficients consist of the derivatives of the function at that point: in particular, the function must be infinitely differentiable there. Convergence may be limited to a neighborhood of a certain radius around that point.

The Fourier series for a function is not dependent upon a specific point. A function need not be infinitely differentiable at any point (or even differentiable at all) to have a Fourier series. Every function that is integrable ($L^1$) has a formal Fourier series, i.e., the coefficients exist.

Mere continuity is sufficient to ensure convergence almost everywhere. More generally, if $f$ is any function in $L^p$ for $p > 1$, then the Fourier series for $f$ converges almost everywhere. (This is a very hard result that wasn't obtained until the late 1960s.) On the other hand, there exists an $L^1$ function whose Fourier series diverges at every point.

3. Nov 5, 2009

### lurflurf

The Taylor series is essentialy the Fourier series on a loop around the point of expansion.