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On the radius of convergence of a power series

  1. Aug 16, 2010 #1
    Hi, I'm new here. I am curious that why a power series must have a radius of convergence? I mean, even in a complex plane, there is always a so-called convergent radius for a power series. Is it possible that a power series is convergent for a certain range in one direction, and for an apparent shorter/longer range in some other direction? So far all the text books I read do not give lessons over this question. Any answer or hint or instruction will be much appreciated.
     
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  3. Aug 16, 2010 #2

    LCKurtz

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    The answer is no, it can't have a larger range of convergence in a different direction. The relevant theorem is the Cauchy-Hadamard theorem. Lots of links on the internet, one of which is:

    http://eom.springer.de/c/c020870.htm
     
  4. Aug 17, 2010 #3
    This is a really great question, though. Its answer is part of the beauty of complex analysis.
     
  5. Aug 17, 2010 #4

    HallsofIvy

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    It is, basically, an application of the "ratio test".

    If [itex]f(z)= \sum a_n(z- z_0)^n[/itex]] then the series converges, absolutely, as long as
    [tex]\lim_{n\to\infty}\frac{|a_{n+1}(z- z_0)^{n+1}|}{|a_n (z- z_0)^n|}[/tex][tex]= |z- z_0|\lim_{n\to\infty}\frac{a_{n+1}{a_n}|< 1[/tex]
    and diverges if that limit is larger than 1.

    As long as
    [tex]\lim_{n\to\infty}\frac{a_{n+1}}{a_n}= A[/tex]
    exists, then we have that the power series converges for
    [tex]|z- z_0|< \frac{1}{A}[/tex]
    and diverges for
    [tex]|z- z_0|> \frac{1}{A}[/tex]

    You can get the same result by using the root test instead of the ratio test:
    [itex]\sum a_n (z- z_0)^n[/itex] converges absolutely as long as
    [tex]\lim_{n\to\infty}\left(a_n(z- z_0)^n)^{1/n}= \left(\lim_{n\to\infty}\sqrt[n]{a_n}\right)|z- z_0|[/tex]
    is less than 1.
     
  6. Aug 17, 2010 #5
    thank you all for the replies. That does help.
     
  7. Aug 17, 2010 #6
    OK, I think I get it. Thank you.
     
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