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The question is in the title. Is there a one-dimensional rational function $f\in\Bbb R(X)$ which restricts to $\Bbb N\to\Bbb Q$, which is a surjection onto $\Bbb Q$? My guess is no.

Expanding the scope a little (at the expense of precision), are there any "nice" functions that enumerate $\Bbb Q$? Here "nice" is meant to exclude the floor function or absolute value function and related trickery. At first I thought it might work to use analytic functions here, but there is an analytic function taking any chosen values on $\Bbb N$ subject to a mild growth hypothesis (I think $f(n)\in o(n)$), by defining $f(z)=\sum_{n\in\Bbb N}a_n{\rm sinc}(\frac{z-n}\pi)$, where ${\rm sinc}(z)=\frac{\sin(z)}z$ continuously extended over zero. I will let answerers supply their own definitions of "nice" if they want to tackle the broader question.

Mario Carneiro
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  • What do you mean "one-dimensional"? – AnalysisStudent0414 Jan 25 '16 at 16:42
  • @AnalysisStudent0414 Just the obvious thing, $f$ is a function of one variable. (There are polynomial surjections $\Bbb N\times\Bbb N\to\Bbb Q$, for comparison.) – Mario Carneiro Jan 25 '16 at 16:46
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    There even exist a bijection between $\mathbb{N}$ and $\mathbb{Q}$ see https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite – stity Jan 25 '16 at 16:46
  • @stity Of course. The interesting part of this question is the "niceness" constraint. Most bijections such as the one in your link are defined piecewise; I want a smoother or simpler function. – Mario Carneiro Jan 25 '16 at 16:48
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    It seems to me that a rational function would be monotone after a certain point, so it could never induce a surjection $\mathbb{N}\to \mathbb{Q}$. But I don't know what would happen with an analytic function. – Arnaud D. Jan 25 '16 at 16:56
  • The question looks a bit confused. The interresting part in the second section indicates that you're aware that there are non-nice surjections (which there are). With that taken into consideration the first section seems confusing - if there wouldn't be a surjection there wouldn't be a nice surjection either. – skyking Jan 27 '16 at 13:10
  • @skyking I am aware there are non-nice surjections. My interest is in finding nice ones, and in the OP I consider two definitions of "nice", rational functions and analytic functions. The answer turns out to be no in the first case and yes in the second, but both for fairly trivial reasons, and I am not confident that I have correctly captured "niceness" here. – Mario Carneiro Jan 27 '16 at 14:17
  • @MarioCarneiro But the first section does not make such a restriction, it only restricts it to be "one-dimensional". I'd suggest that you update the first section to be not a question. – skyking Jan 27 '16 at 14:23
  • @skyking Note that both the title and the first section use the word "rational" or "rational function". This does not mean a function into $\Bbb Q$, it means a quotient of polynomials. (edit to clarify) – Mario Carneiro Jan 27 '16 at 14:25
  • @MarioCarneiro OK, you're correct – skyking Jan 27 '16 at 14:26
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    From your answer, we get that if "nice" means tame, then there are no nice surjections of $\Bbb{N}$ onto $\Bbb{Q}$. – Rob Arthan Jan 27 '16 at 17:17

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At least for the question of a rational function $\Bbb N\to\Bbb Q$, the answer is no. In fact, we can make a stronger claim:

If $f:\Bbb N\to\Bbb R$ is a rational function, then the range of $f$ is either bounded above or bounded below.

Proof: Separate the highest order term of $f$, writing it as $f(x)=ax^n+g(x)$ where $g\in o(x^n)$ for some $n\in\Bbb Z$. If $a\ge 0$, then we claim $f$ is bounded below (and by symmetry, $f$ is bounded above if $a\le 0$).

By the definition of little-O, there is some $N\in\Bbb N$ such that for all $x>N$, $|g(x)|\le ax^n$. Then $f(x)\ge0$, so $f(x)\ge\min\{f(0),f(1),\dots,f(N)\}$.

Edit: Alternatively, we can use Arnaud's suggestion: the derivative of $f$ is also a real rational function, which only changes sign at the roots of the numerator and denominator. Thus $f$ is monotone above the largest root of the derivative (let's say $f$ is increasing, the other case is covered by symmetry), so $f$ is eventually bounded below, and hence bounded below by the same argument as above.

The analytic function example in the OP can be extended to functions with any polynomial growth rate. Define ${\rm sincz}(z)={\rm sinc}(\frac z\pi)=\frac\pi z\sin(\frac z\pi)$. This function has the property ${\rm sincz}(0)=1$ and ${\rm sincz}(n)=0$ for all other $n\in\Bbb Z$. Additionally, since $|\sin(x)|\le 1$ on $\Bbb R$, $|{\rm sinc}(x)|\le\frac 1x$ and $|{\rm sincz}(x)|\le\frac\pi x$. To achieve even higher convergence rates, we can take ${\rm sincz}^n(x)$, which also has the same behavior on integers but is $O(x^{-n})$.

The convergence rate is necessary to ensure that $\sum_{n\in\Bbb N}a_n{\rm sincz}^m(z-n)$ converges. If $a_n$ is $O(n^k)$, then $a_n{\rm sincz}^m(z-n)$ is $O(n^{k-m})$, so the sum converges for $m>k+1$. Then as a uniformly convergent sum (uniformly because it is using a global bound on ${\rm sincz}$) of analytic functions, it is itself analytic, and takes the value $f(n)=a_n$ at each integer. Needless to say, most constructed bijections $\Bbb N\to\Bbb Q$ have polynomial growth, usually $O(n^{1/2})$, so this shows that there is an analytic function whose restriction to the natural numbers is a bijection to $\Bbb Q$, although it is quite contrived and does not really "simplify" the expression.

Community
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Mario Carneiro
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  • I think the gist of the idea here is okay, but the details have a lot of gaps - what's the 'highest-order term' of $1/(3x-2)$, for instance? You have to prove some results on boundedness of rational functions that are relatively straightforward but still need fleshing-out. – Steven Stadnicki Jan 25 '16 at 17:12
  • @StevenStadnicki The Laurent expansion of $\frac1{3x-2}$ near $\infty$ is $\frac1{3x}+\frac2{9x^2}+\frac4{27x^3}+\dots$, so the highest order term is $\frac1{3x}$ and the remaining piece is $O(x^{-2})$. The easy way to find the highest order term of a rational function is to divide the highest order terms of the numerator and denominator polynomials. – Mario Carneiro Jan 25 '16 at 17:44
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    Here's an alternative argument (which still needs some fleshing out): Write your rational function as $p(x) + \frac{q(x)}{r(x)}$ where $p,q,r$ are polynomials with $\deg q < \deg r$. Then your function is asymptotic to $p$. But $p$ is bounded above or below on $\mathbb{N}$, and there are only finitely many elements of $\mathbb{N}$ for which your rational function is more than $1$ away from $p(x)$, so your rational function must also be bounded above or below. – Jason DeVito - on hiatus Jan 25 '16 at 19:44
  • @StevenStadnick: alternatively, you can assume that the rational function is $p/q$ with $\deg(p) \ge \deg(q)$. – Rob Arthan Jan 27 '16 at 12:16
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This is an alternative argument based on Arnaud's suggestion. Without loss of generality, you can assume the rational function $f$ is $\frac{p}{q}$ where $n = \deg(p) \ge \deg(q) = m$ (otherwise trade $f$ in for $\frac{1}{f}$). But then (using $(\frac{p}{q})' = \frac{p'q - pq'}{q^2}$) you have: $$ f'(x) = \frac{(n-m)p_nx^{m+n-1} + \mbox{terms of lower degree}}{x^{2m} + \mbox{terms of lower degree}} $$ where, without loss of generality, I have assumed $q$ is monic and where $p_n \neq 0$ and $n > 0$ (if $n = 0$, $f$ is constant). So as $x$ tends to $\pm\infty$, $f'(x)$ tends to $\pm\infty$, if $n > m+ 1$, and to $p_n \neq 0$ if $n = m + 1$. So $f(x)$ is monotonic whenever $|x|$ is sufficiently large if $n > m$. If $m = n$, then as $x$ tends to $\pm\infty$, $f(x)$ tends to $\pm p_n$, so $f(x)$ is bounded. But if $f(x)$ is either monotonic for large enough $|x|$ or bounded, then $f$ cannot map $\Bbb{N}$ onto $\Bbb{Q}$.

Rob Arthan
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  • The $x^2m$ part looks like it came out wrong, but I'm not sure what you wanted to write. (Why is an exponent of the numerator a coefficient in the denominator?) (Edit: oops, I see you took a derivative. But then the $2$ is wrong...) – Mario Carneiro Jan 27 '16 at 14:19
  • It's fixed. On reflection, your argument is much better in that it shows that no $f$ that is first-order definable in the language of a ring will do for $f$ (all you are using is O-minimality of the zero set of $f$. Unfortunately, I can't tick it for you! – Rob Arthan Jan 27 '16 at 16:52
  • I think it really is fixed now $\ddot{\frown}$. – Rob Arthan Jan 27 '16 at 17:08
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If nice is taken to mean analytic function there sure is. There exists a result from complex analysis that states that if $\lim_{n\to\infty} c_n=\infty$ and $a_n$ is a sequence of complex number then there's an analytic functionn $f$ such than $f(c_n) = a_n$. Now let $c_n = n$ and $a_n$ be an enumeration of $\mathbb Q$ then this result guarantees the existence of such a function.

skyking
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