Extending radius of convergence by analytic continuationby jackmell Tags: analytic, continuation, convergence, extending, radius 

#1
Mar112, 08:37 AM

P: 1,666

Hi,
Suppose I have an analytic function [tex] f(z)=\sum_{n=0}^{\infty} a_n z^n [/tex] the series of which I know converges in at least [itex]z<R_1[/itex], and I have another function [itex]g(z)[/itex] which is analytically continuous with [itex]f(z)[/itex] in [itex]z<R_2[/itex] with [itex]R_2>R_1[/itex] and the nearest singular point of [itex]g(z)[/itex] is on the circle [itex]z=R_2[/itex]. Can I conclude the power series has a radius of convergence [itex]R_2[/itex] and represents both f(z) and g(z) in that domain? I'm confident I can but not sure how to prove that. How about this: If f(z) and g(z) are analytically continuous, then by the Principle of Analytic Continuation, they are the same function and therefore, the power series converges up to the nearest singular point of that same function which in this case, is the singular point on [itex]R_2[/itex] and therefore, the radius of convergence of the series is [itex]R_2[/itex]. Is that sufficient? Thanks, Jack 



#2
Mar112, 01:05 PM

Sci Advisor
HW Helper
P: 2,020

I think you have the right idea but you've kind of written it up in an odd way. There's also the problem of your usage of the nonstandard term "analytically continuous". (One says that "g is an analytic continuation of f", and not that f and g are "analytically continuous".)
Basically what's going on here is that g is analytic in the disc z<R_2, so has a power series expansion there, with coefficients b_n say. But g=f in the smaller disc z<R_1, so $$b_n=\frac{g^{(n)}(0)}{n!}=\frac{f^{(n)}(0)}{n!}=a_n.$$ 



#3
Mar112, 01:50 PM

P: 1,666

Thanks for helping me. 


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