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Let $f(x)$ be a rational function which is a ratio of two integral polynomials, and $n \in \mathbb Z$. Then the sequence of iterates $n, f(n), f(f(n)), f(f(f(n)), ...$ will be an infinite sequence of rational numbers, except in the rare cases where some iterate is a pole of $f$.

In the special case that $f(x) = \frac{a}{x^m}$ it happens that, when $n \ne 0$ is divisible by $a$ (also a nonzero integer), the iterates of $n$ under $f$ are integers infinitely often and non-integers infinitely often.

Are there any other examples where $n, f(n), f(f(n)), f(f(f(n)), ...$ contains both infinitely many integers and infinitely many non-integers?

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    $\begingroup$ Functions such as $f(x) = \frac{1}{1 - x}$ are of finite order, so taking e.g. $n = 3$ gives a "cheating" example. These are also easily generalizable. $\endgroup$ – WhatsUp Sep 3 '17 at 14:14
  • $\begingroup$ @WhatsUp Thanks---perhaps I should have asked for infinitely many distinct integers and non-integers. Does that avoid all easy generalizations of your periodic examples? $\endgroup$ – Kimball Sep 3 '17 at 16:18
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    $\begingroup$ This can be answered from Silverman's theorem "Siegel's theorem for orbits". $\endgroup$ – Pasten Sep 3 '17 at 18:24
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As Pasten suggested in the comments, the key tool here is Siegel's theorem, and this was already done by Silverman, see "Theorem A" in

Joseph H. Silverman: Integer points, Diophantine approximation, and iteration of rational maps, Duke Math. J. 71 (1993) #3, 793--829.

Proposition. Fix a rational function $f \in {\bf Q}(x)$, and some $n_0 \in {\bf Q} \cup \{\infty\}$; for $i=1,2,3,\ldots$, define $n_i$ inductively by $n_i = f(n_{i-1})$. Then if $n_i \in \bf Z$ for infinitely many $i$, then either

(i) $\{n_i\}$ is periodic, or

(ii) $f$ has the form $f(x) = c + a/(x-c)^m$ for some $a,c \in \bf Q$ (with $a \neq 0$) and $m>1$, or

(iii) $f$ is a polynomial.

In each case it is also possible to have $n_i \notin \bf Z$ for infinitely many $i$. Note that case (ii) contains Kimball's example, and is in fact equivalent to it under conjugation by $x \mapsto x+c$.

The point is that if $\{n_i\}$ is not periodic then for each $j=1,2,3,\ldots$ the $j$-th iterate $f^j$ satisfies $f^j(x) \in \bf Z$ for infinitely many distinct $x \in \bf Q$, whence $f^j$ has at most two distinct poles By Siegel's theorem on integral points. Silverman uses this to show that $f^2$ is polynomial, and thence that $f$ is either polynomial or of the form exhibited in (ii) by citing a result from

A. Beardon: Iteration of Rational Functions (GTM 132), New York: Springer 1991

($\S$4.1), which he describes as "elementary" and "well-known", and also proves as Proposition 1.1 of his paper (pages 798--799).

A simple example of a polynomial whose iterates can take both integer and non-integer values is $f(x) = x + a$ for non-integral $a \in \bf Q$, say $f(x) = x+\frac12$. More complicated examples such as $f(x) = 2x^2 + \frac12$ can be obtained as $f(x) = P(cx)/c$ for suitable polynomials $P$ of degree $2$ or greater (here $P(x) = x^2+1$ and $c=2$). One can even construct examples such as $f(x) = x^2 + \frac{x}{2}$ for which it is probably true that the iterates of every integer include both integers and non-integers but this is very hard to prove (this example encodes the behavior of the parity of iterates of $x \mapsto (x^2+x)/2$).

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