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Sequence of complex polynomials

  1. Mar 6, 2009 #1
    This is problem 13.3 from Rudin's Real and Complex analysis. It is not homework.

    Is there a sequence of polynomials {Pn} such that Pn(0) = 1 for n = 1,2,3,... but Pn(z) -> 0 for all z != 0 as n -> infinity?

    My guess here is no. Sketch of proof: Assume such a sequence existed. Then we should be able to contradict the maximum modulus theorem for any disk around 0 since all Pn(z) for |z| = r will be approaching 0 for large enough n, but Pn(0) = 1.

    Is this correct?

    thanks
     
  2. jcsd
  3. Mar 8, 2009 #2

    lurflurf

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    Homework Helper

    It depends what you mean by converges.
    clearly if you fix n lim x->infinity pn(x)=infinity
    but it may be that if you fix x lim n->infinity pn(x)=0
    consider
    f=(1-x^2/n^n)^n
     
  4. Mar 8, 2009 #3
    Can you clarify that? Say we fix an "r", and take the circle |z| = r. Then for any finite n, and z on this circle, with your f, f(z) will either be approching 0, or approaching infinity. It doesn't seem like all the values on the circle can be driven to 0 at the same time for a given n.

    Your function will work fine for real x. But I can't seem to make it work for the complex case. What am I missing?

    Also, I could be blowing the limit computation, but doesn't that f converge to 1 for all finite x as n goes to infinity?

    And if we go with:
    f = (1-x^2/n)^n, that'll converge to exp(-x^2), which has the same problem that I can't drive all the values on the circle to 0.

    thanks for your time.
     
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