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dumbQuestion
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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.
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.
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