22
$\begingroup$

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?

$\endgroup$
2
  • 4
    $\begingroup$ A more realistic question: is it true that if $p(x)$ maps all rationals to rationals, then all coefficients of $p$ are rational numbers? $\endgroup$
    – user6976
    Oct 17, 2010 at 6:57
  • 2
    $\begingroup$ I'd say no. The $f(x)$ that I wrote below may have (by rearrangement) only irrational coefficients in its power series expansion, the reason being that we just have a smallness constraint in choosing the rationals $\epsilon _n$, while the coefficient of the power series expansion are given by certain series in which all the $\epsilon_n$ enter (so we can easily make them all trascendent). $\endgroup$ Oct 17, 2010 at 8:12

2 Answers 2

40
$\begingroup$

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).

$\endgroup$
2
  • 6
    $\begingroup$ Tribute to Alf by Jeff Shallit at recursed.blogspot.com/2010/10/… $\endgroup$ Oct 17, 2010 at 7:18
  • $\begingroup$ What is that "slight modification" to have derivatives with same property? $\endgroup$
    – Somnium
    Sep 28, 2022 at 16:15
18
$\begingroup$

You can map $\mathbb{Q}$ in itself by plenty of entire functions that are not rational functions. Let fix an enumeration of the rationals, $\mathbb{Q}=\{q _ j\ : j=1,2,\dots \} $. Consider a series $$f(x):=\sum_{n=1}^\infty\ \epsilon _n \ \prod _{j=1}^n (x-q _j)$$ If $\epsilon _n$ is a sequence of rationals converging to 0 with sufficient velocity, the series converges uniformly on bounded sets to an entire function.

(edit) rmk. of course with some more small care we can even make an entire function $f(x)=\sum_{n=0}^\infty p _n(x)$ invertible over $\mathbb{R}$, and a bijection between two assigned countably infinite dense subsets $A$ and $B$. Start the series with the identity $p_0(x)=x$, then add inductively only odd degree polynomials $p _n$, that vanish on the (finitely many) already settled points, and do not destroy the invertibility on $\mathbb{R}$ (say, keeping all partial sums of the series with derivative greater than $1/2$). The bijectivity between $A$ and $B$ is to be ensured by a standard ping-pong argument.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.