Function f(1/n)=1/n!

Question

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.

Do you know a complex function that extends the positive integers factorial? I'm thinking $f(x) = \frac{1}{\Gamma(x+1)}$ — Jonas Gomes, Feb 29 '20 at 19:48
I do not. I can read about it. It is not analytic on the u it disk? — , Feb 29 '20 at 19:49
Related: holomorphic function on punctured disk satisfying $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ has an essential singularity at $0$ — Martin R, Feb 29 '20 at 19:54
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. — TonyK, Feb 29 '20 at 19:59
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. — , Feb 29 '20 at 20:03

score 9 · Answer 1 · answered Feb 29 '20 at 19:58

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 $f(\frac 1n)=\frac 1{n!}$

score 5 · Accepted Answer · answered Feb 29 '20 at 19:51

5

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$$\lim_{n\to\infty}\frac{\frac1{n!}}{\frac1{n^k}}=0.$$

answered Feb 29 '20 at 19:51

José Carlos Santos

427,504

Thank you. Both answers make sense. Do you think there is an approach using the uniqueness theorem without going through the gamma function? – Feb 29 '20 at 20:00
I don't think so. – José Carlos Santos Feb 29 '20 at 20:01

Function f(1/n)=1/n!

2 Answers2