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        general >> complex analysis >> Analytic continuation
        
(Message started by: Jason on Feb 20^th, 2004, 1:34pm)

Title: Analytic continuation
Post by Jason on Feb 20^th, 2004, 1:34pm

Let g(z)= [sum] [supinfty]₁z^2j/(j+1). Determine the disc of convergence for this series. Find the largest open set to which g can be analytically continued.

Title: Re: Analytic continuation
Post by Icarus on Feb 20^th, 2004, 9:21pm

The first is trivial. The second is still a topic of considerable mathematical research, and last I have heard, not an entirely settled question.

Title: Re: Analytic continuation
Post by Icarus on Feb 20^th, 2004, 9:25pm

Sorry, I was thinking of the wrong function. The second is fairly well known too. Hint: g is the composition of two more familiar functions.

Title: Re: Analytic continuation
Post by Jason on Feb 24^th, 2004, 9:53pm

As I understand the radius of convergence is 1, and
g(z)=ln(1-z²)/z². Am I correct?
But I'm confused with the largest open set which the function can be analytically continued. I see that at 0 the function has a removable singularity and also at z=1 and z=-1 the function has a branch points. So is it correct that
ln(1-z²)/z² is an analytic continuation of g(z) on the C\{1},{-1} ?

Title: Re: Analytic continuation
Post by Icarus on Feb 25^th, 2004, 5:03pm

Close but not quite. As you circle either of the poles, the natural logarithm picks up an additional 2[pi]i in value. So somewhere you have to have a line of discontinuity, over which the value makes this 2[pi]i jump. The "smallest" (depending on how you measure the size) set for this discontinuity would be a line from one pole to the other. So the maximal set to which you can analytically continue is [bbc] \ [-1, 1]

Title: Re: Analytic continuation
Post by Icarus on Feb 25^th, 2004, 5:14pm

Of course that is assuming that you are restricted to [bbc] as your domain. If you are allowed a Riemann surface, then you can get an infinitely layered one. The surface can be constructed this way: take an infinite number of copies of [bbc], one for each integer, and stack them in order. Slit all of the planes along the line segment connecting -1 & 1. Glue the upper side of the slit for each plane to the lower side of the slit in the plane above it. The points -1 & 1 from every plane are gone, but the remainder of the newly formed surface is smooth, and ln(1 - z²)/z² can be extended to all of it by analytic continuation. If the point of the domain comes from plane #n, then the value differs from that of the principle branch by 2n[pi]i/z².

Title: Re: Analytic continuation
Post by delusionCdotnet on Mar 26^th, 2004, 7:49am

doesn't that equal x? ::)

Title: Re: Analytic continuation
Post by Icarus on Mar 27^th, 2004, 7:33am

??? Doesn't what equal x? ???