Existence of holomorphic function $f$ such that $f^{(n)}(0) = n^{2n}$

Question

I need help with the following problem:

Is there a holomorphic function $f$ in an open disk around $0$ such that \begin{align} f^{(n)}(0) = n^{2n} \end{align} for all $n \in \mathbb{N}$?

I thought quite long about this and came up with the following idea:

Suppose there exists such a function. Then we can write it as \begin{align} f(z) = \sum_{n=0}^{\infty} \frac{n^{2n}}{n!}z^n. \end{align} We can compute the radius of convergence $R$ of this power series: \begin{align} R = \lim_{n \to \infty} \left| \frac{n^{2n}}{n!} \cdot \frac{(n+1)!}{(n+1)^{2n+2}} \right| = \lim_{n \to \infty}\left( \frac{n}{n+1}\right)^{2n} \cdot \frac{1}{n+1} \leq \lim_{n \to \infty} \frac{1}{n+1} = 0. \end{align}

This contradicts the assumption that $f$ is holomorphic in an open disk around $0$ or with other words: $R > 0$.

I am not very confident looking at my solution. (Is it true?) Apart from that I was wondering whether or not there is an easier way to find an answer to this question. The problem reminds me of the Identity Theorem or even the generalized Cauchy Integral Formula. Still I failed to see if (and how) those two could have been used to solve this. In order to gain a better understanding of problems of this kind I would be glad if someone has another good idea for this question!

Answer 1

Your solution is fine. Alternatively you can use the Cauchy inequalities for the derivatives (a consequence of Cauchy's integral theorem): If $f$ is holomorphic in $B_R(0)$, $0 < r < R$, and $M = \sup_{|z|=r} |f(z)|$, then $$ \frac{f^{(n)}(0)}{n!} \le \frac{M}{r^n} $$ In our case that gives $$ M \ge \frac{r^n n^{2n}}{n!} \ge r^n n^n $$ for all $n$, which is not possible.

Answer 2

Yes, if a function is holomorphic in an open region $\Omega$ containing $0$, then it equals its Taylor power series expansion in some open disk around $0$.

More precisely, the radius of convergence of the series is the supremum of all $r$ such that the open disk with center $0$ and radius $r$ is contained in $\Omega$.

Since the power series formed with the required coefficients has radius of convergence $0$, you can rightly conclude that no such analytic function exists.