1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted



Similar Discussions: Inverting a Taylor series
  1. Taylor Series (Replies: 4)

  2. Taylor series (Replies: 2)

  3. Taylor series (Replies: 7)

  4. Taylor series (Replies: 5)

Loading...