Radius on convergence in the complex plane

  • Thread starter racland
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  • #1
racland
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Anyone knows anything about the Radius on convergence in the complex plane (Complex Analysis)
 

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  • #2
DeadWolfe
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Yes. Somebody knows something about the radius of convergence in the complex plane.

Hope I was helpful
 
  • #3
ranger
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Yes. Somebody knows something about the radius of convergence in the complex plane.

Hope I was helpful

:rofl: :rofl:
 
  • #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?
 
  • #5
racland
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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...
 
  • #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.
 
  • #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).
 
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