Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function?
[Math] Is a real power series that maps rationals to rationals defined by a rational function
ca.classical-analysis-and-odesnt.number-theoryreal-analysis
Best Answer
No. In fact, $p(x)$ can be a complex analytic function with rational coefficients that takes any algebraic number $\alpha$ in an element of $\mathbb{Q}(\alpha)$. (And everywhere analytic functions are not rational unless they are polynomials).
The algebraic numbers are countable, so one can find a countable sequence of polynomials $q_1(x), q_2(x), \ldots \in \mathbb{Q}[x]$ such that every algebraic number is a root of $q_n(x)$ for some $n$. Suppose that the degree of $q_i(x)$ is $a_i$, and choose integers $b_i$ such that $$b_{n+1} > b_{n} + a_1 + a_2 + \ldots + a_n.$$
Then consider the formal power series:
$$p(x) = \sum_{n=0}^{\infty} c_n x^{b_n} \left( \prod_{i=0}^{n} q_i(x) \right),$$
By the construction of $b_n$, the coefficient of $x^k$ for $k = b_n$ to $b_{n+1} -1$ in $p(x)$ is the coefficient of $x^k$ in $c_n x^{b_n} \prod_{i=1}^{n} q_i(x)$. Hence, choosing the $c_n$ to be appropriately small rational numbers, one can ensure that the coefficients of $p(x)$ decrease sufficiently rapidly and thus guarantee that $p(x)$ is analytic.
On the other hand, clearly $p(\alpha) \in \mathbb{Q}[\alpha]$ for every (algebraic) $\alpha$, because then the sum above will be a finite sum.
With a slight modification one can even guarantee that the same property holds for all derivatives of $p(x)$.
I learnt this fun argument from the always entertaining Alf van der Poorten (who sadly died recently).