Expansions of Functions Analytic at Infinity

  • Thread starter Thread starter ferret123
  • Start date Start date
  • Tags Tags
    Functions Infinity
ferret123
Messages
23
Reaction score
0

Homework Statement



Prove that if f(z) is analytic at infinity, then it has expansion of the form

f(z) = \sum_{n=0}^{+\infty} \frac{a_{n}}{z^{n}}

converging outside some disk.2. The attempt at a solution

I know that for f(z) to be analytic at infinity we want to consider the composite function g(w) = f(\frac{1}{w}), then if g is analytic at 0 f will be analytic at infinity.

I've been attempting to Taylor expand g(w) around the point w=0 and then substituting the appropriate f terms in. However I'm not sure if this is a valid method as I can't see much pattern in the derivatives.

So far I have

g(w) = g(0) + g'(0)w + \frac{g''(0)w^{2}}{2} + \frac{g'''(0)w^{3}}{3!} + ...

as well as

g'(w) = \frac{-f'(\frac{1}{w})}{z^{2}}
g''(w) = \frac{2f'(\frac{1}{w})}{w^{3}} + \frac{f''(\frac{1}{w})}{w^{4}}
g'''(w) = \frac{-6f'(\frac{1}{w})}{w^{4}} - \frac{6f''(\frac{1}{w})}{w^{5}} - \frac{f'''(\frac{1}{w})}{w^{6}}

meaning I'm just gaining more of each derivative each time and I'm not sure if this is the right way to go about this. Any help or tips would be much appreciated.
 
Last edited:
Physics news on Phys.org
ferret123 said:

Homework Statement



Prove that if f(z) is analytic at infinity, then it has expansion of the form

f(z) = \sum_{n=0}^{+\infty} \frac{a_{n}}{z^{n}}

converging outside some disk.2. The attempt at a solution

I know that for f(z) to be analytic at infinity we want to consider the composite function g(w) = f(\frac{1}{w}), then if g is analytic at 0 f will be analytic at infinity.

What you need here is that if f is analytic at infinity then g is analytic at zero.

I've been attempting to Taylor expand g(w) around the point w=0 and then substituting the appropriate f terms in.

You don't need to do that. It is enough to note that if g is analytic at zero then it has a power series expansion
<br /> g(w) = \sum_{n=0}^\infty a_n w^n <br />
which converges inside some disc.
 
Ok I've really over complicated it then, thanks! So then when I change back to f I get the desired result analytic outside of a disk with radius 1/radius of the g disk.

The next part of the question is looking for an expansion of this form for for \frac{z-1}{z+1} \&amp; \frac{z^{2}}{z^{2} - 1}. I think for the first of these I can re-express as 1 - \frac{2}{z+1} and then expand this?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top