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Need help

  1. Feb 10, 2007 #1
    Anyone knows anything about the Radius on convergence in the complex plane (Complex Analysis)
     
  2. jcsd
  3. Feb 10, 2007 #2
    Yes. Somebody knows something about the radius of convergence in the complex plane.

    Hope I was helpful
     
  4. Feb 10, 2007 #3

    ranger

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    :rofl: :rofl:
     
  5. Feb 11, 2007 #4

    HallsofIvy

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    Radius of convergence in the complex plane is exactly the same as radius of convergence in the real numbers. Exactly what is your question?
     
  6. Feb 13, 2007 #5
    Great

    The problem states:
    Find the Radius of Convergence of the following Power Series:
    (a) Sumation as n goes from zero to infinity of Z^n!
    (b) Sumation as N goes from zero to infinity of (n + 2^n)Z^n

    For (a) I think the radius of convergence is 1 but I'm a bit unsure of that...
     
  7. Feb 14, 2007 #6

    HallsofIvy

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    As I said- same thing as on the real line (except now it really is a radius!). Apply the "ratio" test: a series [itex]\Sigma a_n[/itex] converges absolutely if the limit ration
    [tex]\lim_{n\rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|[/tex]
    is less than 1.
    (Diverges if that limit is greater than one, may converge absolutely or conditionally or diverge if it is equal to 1).

    In particular, for Zn, we have |Zn+1/Zn|= |Z|. That series converges absolutely for |Z|< 1. (It obviously diverges for Z= 1 or -1 and diverges for |Z|> 1)

    Try (n+ 2n)Zn yourself.
     
  8. Feb 14, 2007 #7

    StatusX

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    Is that [itex]z^{n!}[/itex]? If so, you can prove a general result that if a_n is any increasing sequence of natural numbers, then [itex]z^{a_n}[/itex] converges iff |z|<1. This is the case HallsofIvy did if a_n=n, and yours if a_n=n!. The general proof follows from the result for a_n=n (which is the smallest increasing sequence of natural numbers).
     
    Last edited: Feb 14, 2007
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