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Does a irrational number $x > 1$ exist such that $\{x^n \} \le \frac{1}{2}$ for all positive integers $n$ ?

$x=1+ \sqrt 2$ holds for $n$ odd, but not in even

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    $\begingroup$ For what it's worth, an example of $x>1$ such that $\{x^n\} \geq \frac{1}{2}$ for all $n\geq 1$ is easily found: for example, $x = 2+\sqrt{3}$ satisfies this. $\endgroup$
    – Gro-Tsen
    Apr 15, 2023 at 22:18
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    $\begingroup$ @Gro-Tsen -- indeed all socalled "strong Pisot numbers" satisfy $\{x^n\}\rightarrow 1$ for $n\rightarrow\infty$ (theorem 4 in Ref. 2 cited below). $\endgroup$ Apr 16, 2023 at 10:53
  • $\begingroup$ I don't understand why the answer of Carlo Beenakker (sorry Carlo!) got the +50 bounty. The other answer of @user42355 explicitly gives an algorithm as well as one explicit example of an irrational number $x>1$ such that $\{x^n\}\leq 1/2$, as asked, though without proof. Numerics suggest that there are infinitely many of such numbers $x$ between every two integers $(m,m+1)$ for $m\geq 2$, most of them at $x=m+\epsilon$ with $0<\epsilon\ll 1$. Am I missing something? $\endgroup$
    – Fred Hucht
    Apr 24, 2023 at 14:47

3 Answers 3

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The OP asks for an instance of what Dubickas [1] has called a ${\cal Z}$-number: A real number $x>1$ for which there exists a real $\xi\neq 0$ such that $\{\xi x^n\}<1/2$ for every integer $n$.

An example [2] of an irrational ${\cal Z}$-number is $x=\tfrac{1}{2}(7+\sqrt{41})$, when $\{\xi x^n\}$ with $\xi=\tfrac{1}{4}(2+3x)$, converges to $1/4$ for $n\rightarrow\infty$.

The OP asks for an irrational ${\cal Z}$-number with $\xi=1$. It is known [3] that there exists no algebraic non-integer $x>1$ such that $\{x^n\}$ converges to 0. A stronger result (theorem 4 of [2]) indicates that a non-integer ${\cal Z}$-number with $\xi=1$ cannot be a Pisot number. I am not aware of any results for non-algebraic numbers.

  1. Even and odd integral parts of powers of a real number, A. Dubickas (2006).
  2. On the limit points of the fractional parts of powers of Pisot numbers, A. Dubickas (2006).
  3. On a question of G. Kuba, F. Luca (2000).
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    $\begingroup$ $+1$ but why was this accepted? It is indeed a very helpful and useful remark. But it does not provide an answer to the OP's question (nor, e.g., a concrete reference where the problem is stated to be open). $\endgroup$ Apr 16, 2023 at 2:31
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    $\begingroup$ I see nothing wrong with the answer being accepted. $\endgroup$ Apr 16, 2023 at 8:01
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For $x \in \mathbb{R}_{\ge 1}$ let $\phi(x) = (\lfloor x^n \rfloor + \frac{1}{2})^{\frac{1}{n}}$, where $n$ is the smallest positive integer such that $(\lfloor x^n \rfloor + \frac{1}{2})^{\frac{1}{n}} < x$; if there is no such $n$, then let $\phi(x) = x$. Clearly $\phi(x) \le x$ for every $x$, and $\phi(x) = x$ if and only if $\{x^n\} \le \frac{1}{2}$ for every $n \ge 1$. So the set $S$ of fixed points of $\phi$ are exactly the special numbers we are looking for. Every positive integer is a fixed point of $\phi$. If $m \in \mathbb{Z}_{\ge 1}$ and $x \in (m,m+1)$, then $\phi(x) \in (m,m+1)$. Note that $\phi(x) \le \phi(y)$ for $1 \le x \le y$. Let $\phi^k$ denote the $k$th iteration of $\phi$ for $k \in \mathbb{Z}_{\ge 0}$, and let $\phi^{\infty}(x) = \lim_{k \to \infty} \phi^k(x)$. It is easy to see that $\phi^{\infty}(x)$ a fixed point of $\phi$.

We have $\phi((\frac{3}{2})^{\frac{1}{k}}) = (\frac{3}{2})^{\frac{1}{k+1}}$ for $k \ge 2$. So $\phi^{\infty}(x) = 1$ for every $x \in [1,2)$. However for larger $x$'s it seems we actually get non-integer fixed points too. E.g., for $x \in [\frac{5}{2},3)$ we seem to get $$\phi^{\infty}(x) \approx 2.488183157043128473812375937641611512972854767026801209322612522.$$ (Iterating $\phi$, one can calculate this numerically to hundreds of digits, the convergence rate seems to be fast (exponential in $n$). To be clear, I do not have a proof for $\phi^{\infty}(x) > 2$.) This number should be $\max(S \cap (2,3))$. Similarly one could study $\max(S \cap (m,m+1))$ for $m \in \mathbb{Z}_{\ge 2}$. I suspect that $S \setminus \{1\}$ is a perfect subset of $\mathbb{R}$ (i.e., it is closed and has no isolated points). If this is true, then $S$ has continuum many elements, so in particular there are irrational and even transcendental numbers in $S$.

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    $\begingroup$ This is an extremely nice answer! Here are a few Mathematica lines to check it: Phi[x_]:=Module[{r,n=1,xn=1}, Until[r<x, r=(Floor[xn=x xn]+1/2)^(1/n); n++]; r]; Nest[Phi, 2.489, 300] which gives (19099715638144966649411967754404585379550888675163857213264337986739620780445974268881417007581482431498308210695025482412510525929947181011469848332740147329410787147897399517/2)^(1/442), being correct to $\approx 180$ digits. $\endgroup$
    – Fred Hucht
    Apr 17, 2023 at 9:46
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In the book Selected Problems in Real analysis page 164 have the proof of following result: For $0\leq p < q\leq1$, there exists a number $x \in \mathbb{R}$ such that $\{x^n\} \in [p,q]$ for all $n \in \mathbb{Z_{>0}}$

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