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Zeroes of analytic functions

  1. Aug 29, 2009 #1
    Suppose I have an analytic function, [itex]f[/itex], with radius of convergence [itex]1[/itex]. From http://planetmath.org/encyclopedia/ZerosOfAnalyticFunctionsAreIsolated.html" [Broken] it follows that any [itex]0[/itex] of [itex]f[/itex] in [itex] \{ z : |z| < 1 \}[/itex] is isolated. But is it possible that the zeroes have a limit point on the boundary of convergence? For example, is possible that an analytic function with radius of convergence 1 has zeros exactly at [itex]1-\frac{1}{n}[/itex] for all [itex]n \in Z^{+}[/itex]?
    [Edit: Added question marks.]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 29, 2009 #2
    Take a look at the Mittag-Leffler Theorem which may be over kill. I do not really understand these issues and would be willing to go through the proofs with you.
     
    Last edited by a moderator: May 4, 2017
  4. Aug 29, 2009 #3

    Hurkyl

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    (It's hard to read that without question marks)


    Can you write down an analytic function with infinitely many zeroes?

    Can you apply a change of variable to move those zeroes to the unit circle?
     
  5. Aug 29, 2009 #4
    Let me try, [tex]\frac{1}{\Gamma(z)}[/tex] is entire and has zeroes at [itex] -1,-2, \dots [/itex], so [tex]\frac{1}{\Gamma(\frac{-1}{1-z})}[/tex] would have 0's at [itex]1-\frac{1}{n}[/itex] and since the pole closest to 0 is at 1, the radius of convergence is 1.

    Does that look right?

    [Added later]
    [itex]f(\frac{1}{1-z})[/itex] where [itex]f[/itex] is an entire function with zeroes exactly at the positive integers will do in general.
     
    Last edited: Aug 29, 2009
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