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Taylor series vs. Fourier series

  1. Nov 5, 2009 #1
    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. jcsd
  3. Nov 5, 2009 #2

    jbunniii

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    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 ([itex]L^1[/itex]) has a formal Fourier series, i.e., the coefficients exist.

    Mere continuity is sufficient to ensure convergence almost everywhere. More generally, if [itex]f[/itex] is any function in [itex]L^p[/itex] for [itex]p > 1[/itex], then the Fourier series for [itex]f[/itex] converges almost everywhere. (This is a very hard result that wasn't obtained until the late 1960s.) On the other hand, there exists an [itex]L^1[/itex] function whose Fourier series diverges at every point.
     
  4. Nov 5, 2009 #3

    lurflurf

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