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Find the function for this Taylor series

  1. Jul 13, 2009 #1
    Find the function that has the following Taylor series representation:


    Where s is a constant such that 0<Re(s)<1.

    Any ideas?
  2. jcsd
  3. Jul 14, 2009 #2


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    Do you have any reason to believe it is an elementary function?
  4. Jul 14, 2009 #3

    Gib Z

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    Well the first thing to do would be to consider what a Taylor series itself looks like of a function f, assuming it has one;

    [tex]f(x) = \sum_{m=0}^{\infty} f^m (0) \frac{x^m}{m!} [/tex] where f^m indicates the m-th derivative, not exponent.

    So if we use the theorem that two power series are equal if and only iff their coefficients are identical, we get;

    [tex]f^m (0) = \frac{1}{m+s} [/tex]

    I have tried for a while at seeing what that tells us exactly, and my intution leads me to believe this derivatives look like something from an exponential function, so if I were you, I would now try letting [tex]f(x) = e^{g(x)}[/tex] and seeing what I can find out about g(x) .

    EDIT: Whoops I didn't see Halls post.
  5. Jul 14, 2009 #4
    We can get a differential equation for this function, call it [tex]f(x)[/tex]. That denominator [tex]m+s[/tex] looks scary , let's get rid of it. Multiply by [tex]x^s[/tex] so that we have [tex]\sum x^{m+s}/((m+s)m!)[/tex], differentiate to get rid of the [tex]m+s[/tex]. Then multiply by [tex]x^{1-s}[/tex] and we end up with the series [tex]\sum x^m/m![/tex] which can be recognized as [tex]e^x[/tex]. So the result is...
    [tex]sf \left( x \right) +x{\frac {d}{dx}}f \left( x \right) ={{\rm e}^{x}}[/tex]
    Is this helpful? The solution is not elementary, so maybe or maybe not.
  6. Jul 14, 2009 #5
    This function is a hypergeometric function [tex]{{}_1\mathrm{F}_1(s;\,1+s;\,x)}[/tex], that's just the definition... Maple evaluates this in terms of the incomplete Gamma function, so we conclude it is:
    [tex]s \left( -x \right) ^{-s}\int _{0}^{-x}\!{{\rm e}^{-t}}{t}^{s-1}{dt}[/tex]
    at least for [tex]x < 0[/tex]
  7. Jul 22, 2009 #6
    I think it is pretty clever to construct a differential equation like that. Well done.
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