Is there an entire function with $f(\mathbb{Q}) \subset \mathbb{Q}$ and a non-finite power series representation having only rational Coeffitients

Question

I'm trying to answer the following question:

Is there an entire function $f(z) := \sum \limits_{n=0}^\infty c_nz^n$ such that

1. $f(\mathbb{Q}) \subset \mathbb{Q}$

2. $\forall n: c_n \in \mathbb{Q}$

3. $f$ is not a polynomial

?

I'm trying to show that no such function exists. Here's why I think so:

Assuming such a function existed. We would get $f(10^k) \in \mathbb{Q}$ for all $k \in \mathbb{Z}$. So the decimal representation of $f(10^k)$ either cuts at some digit or consists of repeating digits. Now my gut is telling me that if this is true for $f(10^n)$ with $n \in \mathbb{N}$, it won't be for $f(10^{-n}).$ (e.g. for $c_n$ with a finite digit representation: that's because the number of zeroes between each non-zero digit would increase indefinitely)

But, is this correct at all? And if so, how do I show it rigorously?

Answer 1

As @Chris Culter pointed out in their comment, such function exists. Here is a modified version of the example in the reference Remmert Theory of Complex Functions p.254:

Enumerate $\Bbb{Q} = \{r_n : n \geq 1\}$ and define

$$ f(z) = \sum_{n=1}^{\infty} \frac{z^{n(n+1)/2}}{n!\left(\frac{n(n+3)}{2}\right)!} \prod_{k=1}^{n} \frac{z - r_k}{1 + |r_k|}. \tag{1} $$

Now, each $m \geq 1$ is uniquely written as $m = \frac{n(n+1)}{2} + k$ for some $n \geq 1$ and $0 \leq k \leq n$, and the coefficient of $z^m$ in $f(z)$ satisfies

\begin{align*} |[z^m]f(z)| &= \Bigg|\frac{(-1)^{n-k}}{n!\left(\frac{n(n+3)}{2}\right)!} \sum_{1 \leq j_1 < \cdots < j_{n-k} \leq n} \frac{r_{j_1}\cdots r_{j_{n-k}}}{(1+|r_1|)\cdots(1+|r_n|)} \Bigg| \\ &\leq \frac{1}{n!m!} \sum_{1 \leq j_1 < \cdots < j_{n-k} \leq n} \frac{|r_{j_1}\cdots r_{j_{n-k}}|}{(1+|r_1|)\cdots(1+|r_n|)} \\ &\leq \frac{1}{n!m!} \binom{n}{k} \\ &\leq \frac{1}{m!}. \end{align*}

So $\text{(1)}$ defines an entire function. (And of course, $f$ is not a polynomial because the coefficient of $z^{n(n+3)/2}$ in $f(z)$ is non-zero for each $n$.) Moreover, for each $r \in \Bbb{Q}$ we can find $N$ such that $r = r_N$, thus

$$ f(r) = \sum_{n=1}^{N-1} \frac{r^{n(n+1)/2}}{n!\left(\frac{n(n+3)}{2}\right)!} \prod_{k=1}^{n} \frac{r - r_k}{1 + |r_k|} \in \Bbb{Q}. $$

Answer 2

If you allow meromorphic functions (and as a consequence, finite radius of convergence), you have $$ \frac{1}{1-x}=1+x+x^2+x^3+\cdots $$