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Expansion of a function f(x) with poles

  1. Feb 5, 2010 #1
    If a function f(x) have SIMPLE POLES , could in principle this f(x) be expanded into

    [tex] f(x)= \sum_{r}a_{r} (x-r)^{-1} [/tex]

    where 'r' are the poles on the complex plane of the function

    Another question, would it be possible to relate using the Euler-Mac Laurin resummation, a series of the form [tex] \sum_{n=0}^{\infty}(-1)^{n}f(n) [/tex] to the integral

    [tex] \int_{0}^{\infty}f(x)dx [/tex]
     
  2. jcsd
  3. Feb 5, 2010 #2
    No. But (assuming a finite sum, or a convergent sum) if you subtract this from f(x), you get something with no poles, perhaps reducing to a simpler problem.
     
    Last edited: Feb 5, 2010
  4. Apr 9, 2010 #3
    The answer to the first question is YES: see the Mittag-Leffler theorem.

    I would substitute [tex]x=\sin\left(\cos(n\pi) x\right)[/tex] in the integral. Differentiate several times under the integral and add the terms. Substitute [tex]n=1[/tex] in the final step. That's how I would go about and see if something happens. See if it is related to the Fourier series of some other function.
     
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