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I am trying to work on a problem in complex analysis. Although I know how to solve it, I am only stuck at one point.

The problem asks if there exist a holomorphic function $f$ on the unit disk such that $f(\frac{1}{n})=\frac{1}{n!}$.

Here, the approach will be to consider another function $g$ that coincides with $f$ on a discrete set (mainly $\{\frac{1}{n};n \in \mathbb{N}$}) and use the uniqueness theorem to show that they coincide with each other everywhere on the unit ball.

Now, if our function $g$ is not analytic on the unit ball, we will get what we need.

I cannot find such function $g$. It is easy to do it when $f(n)=\frac{1}{n+1}$ but here the factorial is making it a little bit difficult.

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  • $\begingroup$ Do you know a complex function that extends the positive integers factorial? I'm thinking $f(x) = \frac{1}{\Gamma(x+1)}$ $\endgroup$ Feb 29, 2020 at 19:48
  • $\begingroup$ I do not. I can read about it. It is not analytic on the u it disk? $\endgroup$
    – m96
    Feb 29, 2020 at 19:49
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    $\begingroup$ Your logic seems flawed to me: if $g$ is not analytic, then you have no reason to claim that it coincides with $f$ on the unit ball. For instance, you could just define $g(\frac{1}{n})=\frac{1}{n!}$ and $g(x)=0$ if $x$ is not the reciprocal of a positive integer; but it is clear that this proves nothing. $\endgroup$
    – TonyK
    Feb 29, 2020 at 19:59
  • $\begingroup$ Yeah you are totally right. I think this would have worked if f analytic on a bigger disk and we only need them to coincide on the smaller one. $\endgroup$
    – m96
    Feb 29, 2020 at 20:03

2 Answers 2

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The idea is simply that such a function, to exist, would have to be "too flat". More formally: let $f$ be an hypotetical holomorphic function that satisfies your hypothesis. It would then satisty:

$$f(0)=\lim f(\frac 1n)=\lim \frac 1{n!}=0\\ f'(0)=\lim \frac{f(\frac 1n)}{\frac{1}{n}}=\lim \frac{n}{n!}=0\\ f^{(k)}=k!\lim \frac{f(\frac 1n)}{n^k}=k!\lim \frac{n^k}{n!}=0 $$

Thus $f=\sum \frac{f^{(n)}(0)z^n}{n!}=0$, which contradicts $f(\frac 1n)=\frac 1{n!}$

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There is no such function. If there was, its Taylor series centered at $0$ would be of the form$$a_kz^k+a_{k+1}z^{k+1}+\cdots,$$for some $k\in\mathbb N$ and $a_k\neq0$. But then$$\lim_{n\to\infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\left\lvert\frac{a_k}{n^k}\right\rvert}=1.$$In particular$$\lim_{n\to\infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\frac1{n^k}}\neq0.$$But$$\lim_{n\to\infty}\frac{\frac1{n!}}{\frac1{n^k}}=0.$$

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  • $\begingroup$ Thank you. Both answers make sense. Do you think there is an approach using the uniqueness theorem without going through the gamma function? $\endgroup$
    – m96
    Feb 29, 2020 at 20:00
  • $\begingroup$ I don't think so. $\endgroup$ Feb 29, 2020 at 20:01

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