- #1

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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

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

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