The density of the fractional part of $x^n$ in $(0,1)$?

Question

If $\{ \text{the fractional part of } x^n, n \in \mathbb{N} \}$ is dense in $(0,1)$, what is the range of $x$?

Obviously, $[-1,1]$ is not the case, and all integers are not the cases either, what will happen if $x>1$?

Answer 1

The most reasonable thing I could easily show is: The set of such $x$ is dense in $\Bbb R\setminus[-1,1]$, and the complement is also dense (at least in $\Bbb R\setminus[-2,2]$)!

I'll elaborate on $x>1$ only. If $x,y\in \Bbb R$, write $x\equiv y$ as shorthand for $x-y\in\Bbb Z$ (i.e., $x$ and $y$ have same fractional part). If $x\in\Bbb R$ and $A\subseteq \Bbb R$, let us write $x\dot\in A$ as abbreviation for: There exists $y\in A$ with $x\equiv y$.

Lemma. Let $[a_1,b_1],[a_2,b_2],\ldots$ be a sequence of closed intervals $\subset \Bbb R$. Let $q>1$ and $d>0$. Let $0<n_1<n_2<\ldots$ be a strictly increasing sequence of positive integers such that $q^{n_{k+1}-n_k}(b_k-a_k)\ge 2$ for $k\ge 1$ and $q^{n_1-1}d\ge 2$. Then for every interval $[a,b]$ with $q\le a$ and $b-a\ge d$, there exists $x\in[a,b]$ with $x^{n_k}\dot\in[a_k,b_k]$ for all $k$.

Proof. Note that $x\dot\in[u,v]$ is trivially true if $v\ge u+1$. Hence we may assume wlog that $b_k<a_k+1$ for all $k$.

Let $I_0=[a,b]$. Then $b^{n_1}-a^{n_1}>a^{n_1-1}(b-a)\ge q^{n_1-1}d\ge 2$. Then there exists $m\in\Bbb Z$ with $a^{n_1}< a_1+m<b_2+m< b^{n_1}$. Let $I_1=[\sqrt[n_1]{a_1+m},\sqrt[n_1]{b_1+m}]$. Then

• $I_1\subsetneq I_0$ and
• $x^{n_1}\dot\in[a_1,b_1]$ for all $x\in I_1$ and
• for each $y\in [a_1,b_1]$ there exists $x\in I_1$ with $x^{n_1}\equiv y$.

Assume we have found $I_0\supsetneq I_1\supsetneq \ldots \supsetneq I_k$ with the following properties:

• $x^{n_j}\dot\in [a_j,b_j]$ for all $1\le j\le k$ and $x\in I_k$
• for each $y\in[a_k,b_k]$ there exists $x\in I_k$ with $x^{n_k}\equiv y$.

Let $I_k=[u,v]$. Then $v>u\ge q$ and so $$v^{n_{k+1}}-u^{n_{k+1}}>u^{n_{k+1}-v_k}(v^{n_k}-u^{n_k})\ge q^{n_{k+1}-n_k}(b_k-a_k)\ge 2.$$ Again, we find $m\in\Bbb Z$ with $u^{n_{k+1}}<a_{k+1}+m<b_{k+1}+m<v^{n-{k+1}}$. By letting $I_{k+1}=[\sqrt[n_{k+1}]{a_{k+1}+m},\sqrt[n_{k+1}]{b_{k+1}+m}]$ we extend our nested sequence and still have the properties above. Then $\bigcap_{k\in\Bbb N}I_k$ is not empty and for each $x\in\bigcap_{k\in\Bbb N}I_k$ we have $x^{n_j}\dot\in[a_j,b_j]$ as desired. $\square$

Using the lemma and e.g. $n_k=k^2$ we can easily prescribe the intervals $[a_k,b_k]$ in such a way that the fractional part of $x^{k^2}$ follows a sequence that is dense in $[0,1]$. At least for $a>2$, we can use the lemma and $n_k=N+k$ for $N$ big enough to find $x$ such that $x^n\dot\in[0,\frac2a]$ for almost all $n$.