Register to reply

Extending radius of convergence by analytic continuation

Share this thread:
jackmell
#1
Mar1-12, 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
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
morphism
#2
Mar1-12, 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 $$b_n=\frac{g^{(n)}(0)}{n!}=\frac{f^{(n)}(0)}{n!}=a_n.$$
jackmell
#3
Mar1-12, 01:50 PM
P: 1,666
Quote Quote by morphism View Post
(One says that "g is an analytic continuation of f", and not that f and g are "analytically continuous".)
Ok, g is the analytic continuation of f. That sounds better.

Thanks for helping me.


Register to reply

Related Discussions
How do you DO analytic continuation? Calculus 3
Extending Analytic Functions f:C->C , to f^:C^->C^; C^=Riemann Sphere Calculus 0
Analytic continuation Advanced Physics Homework 1
Convergence question on analytic continuation of Zeta fcn Calculus 3
What is analytic continuation? Calculus 3