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Convergence of Laurent Series

  1. Oct 21, 2006 #1
    This is more of a general question than a specific question.
    Find the annulus of convergence for the Laurent series
    [tex]\sum_{n=-\infty}^{-1} \left( \frac{z}{2} \right)^n + \sum_{n=0}^\infty \frac{z^n}{n!} [/tex]

    I know what to do for the second series, but I am not sure about the first series. The main thing that is bugging me is the [itex]-\infty[/itex] to -1. In general, how do I deal with such a series? Is there something special that needs to be done, or do I treat it the same as a series from 0 to [itex]\infty[/itex]? Thanks!
  2. jcsd
  3. Oct 21, 2006 #2


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    An "annulus" is the region between two concentric circles. You are really looking for two "radii of convergence". It should be clear to you that if |z| is large, there will be no problem with thefirst series (where z has negative exponent) but might with the second. The series with the positive exponents will give you the outer boundary of convergence but converges everywhere inside it. (Actually, you have probably noticed that the second series is just the standard series for ez and converges for all z.)
    It is the first series, with negative exponents, that will not converge for z= 0 but will converge for |z| large that will give you the "inner" radius.

    Do this: replace z with x-1 so that
    [tex]\sum_{n=-\infty}^{-1} \left( \frac{z}{2} \right)^n[/tex]
    [tex]\sum_{n= 1}^{\infty} \left(2x \right)^n[/tex]

    What is the radius of convergence of that? Now convert back to z= x-1.
  4. Oct 21, 2006 #3
    Awesome idea, thanks!

    The region of convergence for the new series is |x| < 1/2 => |z| > 2

    Hence the region of convergence for the whole series is |z| > 2 since the other series converges everywhere.
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