What is the cardinality of Maclaurin series that are entire that induce a bijection between ${\mathbb Z}$ and itself?

Question

So the cardinality of linear polynomials that induce a bijection from ${\mathbb Z}$ to itself is countable, because it is simply the set of linear polynomials of the form $x + n$ where $n$ is an integer. But what about the cardinality of arbitrary Maclaurin series that are entire that induce a bijection between ${\mathbb Z}$ and itself? It seems like this should be uncountable but I'm having a hard time seeing how to show it.

Answer 1

Indeed, there are $\mathfrak c=|\mathbb C|$ many such series, which is largest possible, since there are only $\mathfrak c$ continuous functions from $\mathbb C$ to itself.

The key is Weierstrass factorization theorem, from which one an prove the following interpolation theorem (see the stronger Theorem 3.4.1 in Berenstein-Gay Complex Variables. An introduction):

Given any discrete sequence of points $(z_n)_{n\in\mathbb N}$ and any complex numbers $(a_n)_{n\in \mathbb N}$, there is an entire function $g$ such that $g(z_n)=a_n$ for all $n$.

One can in fact give explicit formulas (see their Exercise 3.4.3) from which it follows that we can further require that if $z_n=n$ and $a_n\in\mathbb R$ for all $n$, then the restriction of $g$ to $\mathbb R$ is real analytic: We can take $g$ of the form $$ g(z)=\sum_{n\ge 0}\frac{a_n}{f'(z_n)}\frac{f(z_n)}{z-z_n}\left(\frac z{z_n}\right)^{q_n} $$ where we list the $z_n$ so that $|z_n|\le |z_{n+1}|$ for all $n$, $f$ is an entire function with simple zeros (precisely) at the $z_n$, and the $q_n$ are suitable nonnegative integers. Of course, at the $z_n$ we interpret the expression above as its limit. As for $f$ itself, there are explicit formulas for it, as listed on the Wikipedia link above.

This gives the result, since there are $\mathfrak c$ permutations of $\mathbb Z$, and each of them gives us a different entire function using the result just quoted.