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Analytic Extension on a Complex Function

  • Thread starter moo5003
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Homework Statement


Prove there cannot be an analytic extension containing the unit disk of:

f(x) = Series on n from 1 to infinity: x^(n!)


Homework Equations



Unique Extension theorem, no real explicit equations I can think of.


The Attempt at a Solution



So far I've proved the range of convergence is 1 and that it diverges for all |z|=1. There is an accumilation point for the coincidence set of f(x) which is all |z|=1 thus the extension must be unique. Since the extension must contain the unit disk it follows that all z such that |z|=1 must be defined on the extension in which it agrees with f(x) for all |z|<1. My professor gave us the hint to show that the roots of unity are dense along the unit circle... though I'm unsure how to apply this to the problem itself. I was hoping one of you could help me or give me some direction on the problem.

I'm guessing I a)Need to show that the extension cannot be defined on a root of unity without a contradiction (No clue how to show this). b)Since the roots of unity are dense along the unit circle its impossible for any domain containing the unit circle to not be defined on a root of unity, thus the extension cannot exist.

It seems like this is the steps I need to take although I'm unsure how to prove or start proving them. I supposed for a, I can just state F(e^(2ipim/n) = k for some nth root of unity, but I dont know how to derive a contradiction from that.

Thanks for any help you guys can provide!
 

Answers and Replies

  • #2
Dick
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Judging by the hint he gave, I think he is looking for a proof along the lines of showing that the function diverges as you approach a dense set of points on the unit circle. For example, show it diverges as r->1 for r real. If that's clear then think about what what happens to f(r*u) where u is a root of unity as r->1.
 

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