Expansions of Functions Analytic at Infinity

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SUMMARY

The discussion centers on proving that if a function f(z) is analytic at infinity, it can be expressed as a series expansion of the form f(z) = ∑(a_n / z^n), converging outside a certain disk. The key approach involves analyzing the composite function g(w) = f(1/w), which is analytic at zero if f(z) is analytic at infinity. The participants clarify that a Taylor series expansion of g(w) around w=0 suffices to derive the desired result, simplifying the initial complex approach.

PREREQUISITES
  • Understanding of complex analysis concepts, specifically analytic functions.
  • Familiarity with Taylor series expansions and their convergence properties.
  • Knowledge of the relationship between functions at infinity and their behavior at zero.
  • Experience with power series and their applications in complex functions.
NEXT STEPS
  • Study the properties of analytic functions at infinity in complex analysis.
  • Learn about the convergence of power series and their implications for analytic functions.
  • Explore the method of substitution in complex functions, particularly using g(w) = f(1/w).
  • Investigate specific examples of function expansions, such as f(z) = (z-1)/(z+1) and f(z) = z^2/(z^2-1).
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in the properties of analytic functions and their expansions at infinity will benefit from this discussion.

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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.
 
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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?
 

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