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Inverting a Taylor series

  1. Apr 13, 2006 #1
    of course if we know f(x) the function and is analytic near x=0 with this we could construct its Taylor series:

    [tex] f(x)=a_{0}+a_{1}x+a_{2}x^{2}+............... [/tex]

    but the problem is..what happens if we know the [tex] a_{n} [/tex] but not f(x)?..well using Cauchy,s formula definition of the a we have:

    [tex] a_{n}=\frac{n!}{2i\pi}\int_{C}dzf(z)z^{-(n+1)} [/tex]

    where z is a closed curve..if we choose C to be a circle or radius 1 the above formula becomes:

    [tex] a_{n}=\frac{n!}{2i\pi}\int_{-\pi}^{\pi}dwf(e^{iw})e^{-iwn} [/tex]

    wich is nothing but a Fourier transform so we could invert it to obtain [tex] f(e^{iw}) [/tex] and hence the function f(x)..but this is "correct"?..thanks.:redface: :redface: :redface:
    Last edited: Apr 13, 2006
  2. jcsd
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