Find all $x\in\mathbb{R}$ such that $$\lim_{n\to\infty}|x^n-\langle x^n\rangle|=0$$ where $\langle t\rangle$ is the integer nearest to $t$ (eg. $\langle\frac{1}{3}\rangle=0$, $\langle\frac{8}{3}\rangle=3$, $\langle k+\frac{1}{2}\rangle$ is not defined for $k\in\mathbb{Z}$).

I found this somewhere in the internet (today I searched again in IMO shortlists and didn't found, so it's probably not from there), tried to solve for a long time, but without nontrivial results (numbers $x\in\mathbb{Z}$ and $x\in(0,1)$ satisfy this, but I have no idea how to examine e.g. $x=\sqrt{2}$).

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    $\begingroup$ Probably you have already solved the problem. The case $x=\sqrt{2}$ is not a solution, as the limit of the resulting oscillating function is undefined. By similar considerations, for example any real number of the form $j^{1/k}$, with $j$ and $k$ positive integers $>1$, is not a solution. $\endgroup$
    – Anatoly
    Jul 17, 2020 at 8:38
  • $\begingroup$ I found this your question, comments to which by Jack D'Aurizio suggest that the present question can be hard. $\endgroup$ Jul 18, 2020 at 4:34
  • $\begingroup$ I believe it has to be true that $x^n$ either converges to an integer or always is an integer. This only happens when $-1 < x < 1$ (in which case $x^n$ converges to $0$) or $x$ is an integer. $\endgroup$ Jul 18, 2020 at 5:55

2 Answers 2


Update. (2021, January 15). According to Wikipedia’s article, there are only countably many such $x$ with $|x|>1$, and among them no transcendental numbers are known. Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then the set of such $x>1$ is coinsides with the set of Pisot-Vijayaraghavan numbers.

In 2011, at a conference in my institute, I met Mark Zel’dich, in whose thesis “On one analogue of the Poincaré recurrence theorem” (in Ukrainian) is announced the following

Attraction theorem. Let $f(t):\Bbb R\to\Bbb R$ be a continuous function, strictly increasing for $t\ge a$ for some $a>0$ and unbounded for $t\to+\infty$ and $A$ be an open unbounded subset of $\Bbb R$. Then a set $$\{t\in (0,\infty): \{n: f(nt)\in A \} \mbox{ is finite}\}$$ is meager, that is a union of a countable many nowhere dense sets.

The equality $x^n=e^{n\ln x}$ (for $x>1$) and Attraction theorem applied to a function $f(t)=e^t$ and any open neighborhood $A$ of the set $\Bbb Z+1/2$ imply that a set of $x>1$ (and so also of $x<-1$), satisfying the claim, is meager.

On the other hand, following this answer by Ewan Delanoy, we can prove the next

Proposition. For each $0\le s<t<\ell$ a set $$X=\left\{x>1: \frac{s}\ell \le\left\{\frac{x^n}\ell\right\}\le \frac{t}\ell \mbox{ for each natural }n\right\}$$ contains a subset parameterized by an infinite binary tree with $2^{\aleph_0}$ nodes, that is $|X|=2^{\aleph_0}$.

Given $0\le s<t<\ell$, let a finite (or infinite) sequence of integers $(a_1,a_2,a_3,\ldots,a_n)$ (or $(a_i)_{i\geq 1}$) is good if $a_1\ge\tfrac{2\ell}{t-s}$, $$\left(a_i\ell+s\right)^{\frac{i+1}{i}} \leq a_{i+1}\ell+s< (a_{i+1}+1)\ell+t\le \left(a_i\ell+t\right)^{\frac{i+1}{i}}\label{1}\tag{1}$$ for every $i$ between $1$ and $n-1$ (or every $i$ if the sequence is infinite).

Lemma 1. If $(a_i)$ is good, then $a_i\ell+s\ge \left(\tfrac{2\ell}{t-s}\right)^i$ for every $i$.

Proof. Use induction on $i$ and $a_i^{\frac{i+1}{i}} \leq a_{i+1}$.

Lemma 2. If $(a_i)$ is good then $\left(a_i\ell+t\right)^{\frac{i+1}{i}}-\left(a_i\ell+s\right)^{\frac{i+1}{i}}\ge 2\ell$ for every $i$.

Proof. $$\left(a_i\ell+t\right)^{\frac{i+1}{i}}-\left(a_i\ell+s\right)^{\frac{i+1}{i}}=$$ $$\left(a_i\ell+s\right)^{\frac{i+1}{i}}\left(\left(1+\frac{t-s}{ a_i\ell+s }\right)^{\frac{i+1}{i}} -1\right) \stackrel{\mbox{(by Bernoulli's inequality)}}\ge$$ $$\left(a_i\ell+s\right)^{\frac{i+1}{i}}\left(\frac{i+1}{i}\cdot\frac{t-s}{ a_i\ell+s }\right)> \left(a_i\ell+s\right)^{\frac{1}{i}}(t-s)\stackrel{\mbox{(by Lemma 1) }}\ge 2\ell.$$

Lemma 3. If a finite sequence $(a_k)_{1\leq k\leq i}$ is good, then there are least two good sequences of length $i+1$ that extend it.

Proof. Consider the interval $I=\left[\left(a_i\ell+s\right)^{\frac{i+1}{i}},\left(a_i\ell+t\right)^{\frac{i+1}{i}}\right]$. By the preceding lemma it has length at least $2\ell$, so there are at least two integers $a_{i+1}$ satisfying inequality \eqref{1}.

Lemma 4. There are $2^{\aleph_0}$ infinite good sequences.

Proof. Iterate the preceding lemma a countable number of times, and starting from the good sequence $\left(a_1\right)=\left(\tfrac{2\ell}{t-s}\right)$ we built an infinite binary tree, that will contain $2^{\aleph_0}$ nodes.

Finally, if $a=(a_i)_{i\geq 1}$ is an infinite good sequence, by construction the sequences $u_i=\left(a_i\ell+s\right)^{\frac{1}{i}}$ and $v_i=\left(a_i\ell+t\right)^{\frac{1}{i}}$ are “adjacent” : $(u_i)$ is nondecreasing, $(v_i)$ is nonincreasing, and $u_i \leq v_i$ for every $i$. Then, we will have a real number $x_a \in X$ such that $u_i \leq x_a \leq v_i$ for every $i$.

Now, we have $x_a\neq x_b$ if the sequences $a$ and $b$ are different. Indeed, if $i$ is the smallest index such that $a_i \neq b_i$, then the intervals $I_a=[a_i\ell+s, a_i\ell+t]$ and $I_b=[b_i\ell+s, b_i\ell+t]$ are disjoint, the first contains $x_a^i$ and the other contains $x_b^i$.

Remark. I tried to modify the construction to obtain $2^{\aleph_0}$ many $x$ such that a sequence $\left\{\frac {x^n}\ell\right\}$ is convergent, but it doesn’t work.

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    $\begingroup$ Hey, @Alex Ravsky is the paper available in English? I want to read it, but I don't know Ukrainian :( $\endgroup$ Jul 18, 2020 at 8:08
  • $\begingroup$ @RalphClausen I think it will not help much. It is an announcement, not a full paper. It essentially consists of a few definitions, a short introduction, a formulation of the Poincaré recurrence theorem, a formulation of the attraction theorem, and seemingly very easy corollaries. That is it contains claims, but no proofs. It seems I tried to find a full paper some years ago, but failed. Maybe it is unpublished, because M.V. Zel’dich (or Zeldich?) told me that he is mainly interested in business rather than science and so he attended this conference in order to please his university heads. $\endgroup$ Jul 18, 2020 at 8:30
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    $\begingroup$ Also in 2012 I wrote a letter to him, but he didn’t answer. On the other hand, maybe the proof of the attraction theorem can be based on the Baire theorem and not so hard. $\endgroup$ Jul 18, 2020 at 8:30
  • $\begingroup$ Got it @Alex Ravsky :) $\endgroup$ Jul 18, 2020 at 8:40
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    $\begingroup$ Very nice explanation +1. $\endgroup$
    – Sebastiano
    Jul 18, 2020 at 11:39

A Pisot number will certainly work. For instance, $\sqrt{2}+1$.


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