# Question about Laurent Series Expansion

1. Nov 2, 2012

### dumbQuestion

Say we have a simple function like

f(z)=4z/[(z-1)(z-3)2]

I'll use this example to demonstrate my undertanding of the motivation behind and usefulness of Laurent series: if we examine f(z), we see it is analytic except where z = 1 and where z = 3, which means it can expanded in a Taylor series, but for example if we center the Taylor series at 0, it's only valid for a radius of 1 because it will hit the singularity z=1 beyond that. We want to be able to expand the function in a power series on a much larger radius of convergence.

Usefulness: in this example, we can use a laurent series and expand on the disk |z|<1, on the annulus 1<|z|<3, and on the region |z|>3. Here is the part where I'm confused. I understand the significance of the Laurent series - we have gone from being able to represent f(z) as a power series only on a very small disk, to being able to represent it as a power serires almost everywhere. But what about for example, if we want to represent it as a power series on the boundary of one of these disks on a point where it's analytic? Take for example the point z=i. f(z) is analytic here. We know this because we can construct some small open set around i and f(z) will be differentiable at all those points. So f(z) is analytic at i, but because i is distance 1 from 0, its on the boundary of these regions we're expanding on so our Laurent series isn't valid for the point z=i.

I suppose for example, I could construct just a regular Taylor series centered at z=i, and it would be valid for all z within √2 distance from i (since 1 is √2 distance from i). But so do we have to do this constructing an individual series centered at all of these "boundary points" that f is analytic at? Or is there some general all encompassing series that will be valid at EVERY point f is analytic at? I guess it seems like Laurent series is great but we still miss out on a bunch of points the function is analytic at.

Last edited: Nov 2, 2012
2. Nov 4, 2012

### jackmell

Not sure what precisely you're asking but I'll try and summarize:

All (integer) power series have a radius of convergence equal to the distance to the nearest singular point. The series converge at least inside this radus but can also converge on the radius of convergence for some or all the point on the radius.

See:

http://en.wikipedia.org/wiki/Region_of_convergence

which gives examples of series converging on the boundary.

But there is no single power series representation of a function with singular points by virtue of what I said above. If for example, we wanted the power series representation of points where the function is analytic and on the boundary of some power series representation where the series does not converge, we would have to move the center of expansion of the series so that these points on the boundary are now inside of this new power series radius of convergence or where it converges on this new boundary.