# extending a function holomorphic in a neighborhood of 1 to be entire

I'm stuck on a qualifying exam question, not really familiar with the method of solution. It goes as follows:

Let $f(z)$ be an an entire function and $g(z)$ be analytic in a neighborhood of $z=1$ which satisfies $g^{(n)}(1)=(f^{(n)}(1))^{\alpha}/(n!)^{\alpha-1}$, where $\alpha >0$. Show that $g(z)$ can be extended to an entire function.

Any help would be greatly appreciated.

edit: fixed typo

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Yes, I apologize, that is correct. Just a typing a error on my part. $g^{(n)}(1)=$ etc. –  C.Cheshire Aug 17 '12 at 3:23
I would guess that you can show that the taylor series for $g(z)$ is well-defined everywhere based on that condition. –  JSchlather Aug 17 '12 at 3:28

Use Hadamard's formula to compute the convergence radius of $g$, it's not hard work.

Actually, you can show in a neighborhood of 1, $g$ has the expansion coefficients $g_n=(f_n)^\alpha$, where $f(z)=\sum_{n \geqslant0}f_n(z-1)^n$.

Explicitly, $f_n=\frac{f^{(n)}(1)}{n!},~g_n=\frac{g^{(n)}(1)}{n!}$, note the formula of $g^{(n)}(1)$ given by you, $g_n=(f_n)^\alpha$ is easy to check !

So that $\rho(g)=\rho(f)^\alpha$, which is $\infty$ by $\rho(f)=\infty$.

Remark. For an expansion $\sum_{n\geqslant 0} a_n(z-z_0)^n$ with $a_n$ complex numbers,

$(\mbox{the convergence radius})^{-1}=\mbox{lim sup}_{n\rightarrow\infty}\sqrt[n]{|a_n|}$