# Rewriting an Analytic function based on its Power Series

• July 27th 2009, 07:41 PM
josephtw
Rewriting an Analytic function based on its Power Series
So a question states: given a function f(z) that's analytic at the origin with f(0)=f '(0)=0, prove that f(z) can be written f(z)= z^2 g(z), where g(z) is analytic at z=0.

I thought this would come right out after I established that since f is analytic at the origin, and thus analytic on some open domain about the origin, it can be represented by its Maclaurin series on that domain, which has two leading zero terms. The Maclaurin representation can then be reworked to f(z)=z^2 Sum(k:[0,Inf) (f^(k)(0)*z^(k-2))/k!)

If all is right so far: how do I know that this new summation uniformly converges to anything? (uniform convergence to ensure that it is analytic) I kind of intuit that it's really the coefficients that make it converge and a finite offset of the exponents doesn't effect much of anything, but I'm not sure, and I wouldn't know how to prove it.

Any hints would be appreciated.
• July 27th 2009, 10:29 PM
Texxy
$f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k$. Since f(0)=f'(0)=0, we can rewrite this as $\sum_{k=2}^{\infty}\frac{f^{(k)}(0)}{k!}z^k=\sum_{ k=0}^{\infty}\frac{f^{(k+2)}(0)}{(k+2)!}z^{k+2}=z^ 2\sum_{k=0}^{\infty}\frac{f^{(k+2)}(0)}{(k+2)!}z^k$. I suppose you consider the sum on the right as g(z).

Since this is really just a trivial rewriting, it's obvious that the g(z) sum converges uniformly in its domain. I've never seen this question before, but hopefully this helped.
• July 30th 2009, 05:11 PM
josephtw
Thanks, I think I'll put a few more words to it than that to make it clear that g is analytic at 0 and that we have the same coefficients and thus the same radius of convergence. I guess the problem with trivial questions is trying not to sound trivial in your explanation.