Fractional part of rational power arbitrary small

Question

I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).

Answer 1

I think that the wolfram link cited the first result (Vijayaraghavan 1941) incorrectly where it reports that

$\{ (3/2)^n \}$ has infinitely many accumulation points in both $[0, 1/2)$ and $[1/2, 1]$.

The first result actually is

$\{ (3/2)^n \}$ has infinitely many accumulation points in $[0,1]$.

The second result cited by wolfram is by Flatto, Lagarias, Pollington (1995) which is available here: http://matwbn.icm.edu.pl/ksiazki/aa/aa70/aa7023.pdf. In the introduction of this paper by them, it reports that

Vijayaraghavan later remarked that it was striking that one could not even decide whether or not $(3/2)^n$ mod $1$ has infinitely many limit points in $[0, 1/2)$ or in $[1/2, 1)$.

This paper proved that

$$\limsup\{ (3/2)^n \} - \liminf \{(3/2)^n\} \geq \frac13. $$

A few later results are mentioned (Dubickas 2006, 2008) in Yann Bugeaud's book "Distribution Modulo One and Diophantine Approximation" (for example, p. xi, p. 67 and p. 68).

$\{ (3/2)^n \}$ has a limit point in $[0.238117 . . . , 0.761882 . . .]$ which has length $0.523764 . . .$.

$\{(3/2)^n\}$ has a limit point in $[0, 8/39] \cup [18/39, 21/39] \cup [31/39, 1]$, of total length $19/39$.

Answer 2

I will show that if $a = 1+\sqrt{2}$ then the limit points of $\{a^n\}$ are $0$ and $1$.

I know that this doesn't tell anything about rational $a$, but this might be of use.

Note that this method can show this for $a$ and $b$ roots of $x^2-2ux-v = 0$ where $u$ and $v$ are positive integers such that $v < 2u+1$. This case is $u=v=1$; $u=1, v=2$ also works.

If $a = \sqrt{2}+1$ and $b = 1-\sqrt{2}$ then $ab = -1$ and $a+b = 2$.

Therefore $a$ and $b$ are the roots of $x^2-2x-1 = 0$.

If $u_n = a^n+b^n$, then $u_0 = 2, u_1 = 2$.

Since $a^{n+2} =a^n(a^2) =a^n(2a+1) =2a^{n+1}+a^n$ and similarly for $b$,

$\begin{array}\\ u_{n+2} &=a^{n+2}+b^{n+2}\\ &=2a^{n+1}+a^n+2b^{n+1}+b^n\\ &=2u_{n+1}+u_n\\ \end{array} $

Therefore $u_n$ is a positive integer for all $n$.

Since $|b| < 1$, $|b^n| \to 0$.

Since $b < 0$, $b^{2n} > 0$ and $b^{2n+1} < 0$.

Therefore, since $a^n = u_n - b^n$, $\{a^{2n}\} =\{u_{2n}-b^{2n}\} =1-b^{2n} $ so $\{a^{2n}\} \to 1 $ and $\{a^{2n+1}\} =\{u_{2n+1}-b^{2n+1}\} =\{u_{2n+1}+|b^{2n+1}\} =|b^{2n+1}| $ so $\{a^{2n+1}\} \to 0 $.

Therefore the limits points of $\{a^n\}$ are $0$ and $1$.

Answer 3

Edit: sorry, I now realize that the question asks whether 0 is a limit point of $\{a^n\}$. My answer below is not a paraphrase of the question but is something weaker. But I still hope it helps.

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Let me rephrase the question: Let $a$ be a rational greater than 1, is it true that $\lim_{n \to \infty}\{a^n\} = 0$?

When $a$ is an integer, yes, $\lim_{n \to \infty}\{a^n\} = \lim_{n \to \infty} 0 = 0$.

However, When $a$ is a noninteger rational, $\lim_{n \to \infty}\{a^n\} \ne 0$.

Proof by contradiction, assume $\lim_{n \to \infty}\{a^n\} = 0$. let $$ be the denominator of $\{a\}$, we can find an $$ such that $∀ n > $, $\{x^n\} < \frac{1}{aq}$.

Let $ > $, we have

$$\{a^m\} < \frac{1}{aq} \tag{1}\label{1}$$ and

$$\{a^{m+1}\} < \frac{1}{aq} < \frac{1}{q} \tag{2}\label{2} $$ where

\begin{align} \{a^{m+1}\} &= \{ (⌊a^m⌋ + \{a^m\}) (⌊a⌋ + \{a\}) \} \\ & = \{ ⌊a^m⌋⌊a⌋ + ⌊a^m⌋\{a\} + \{a^m\}(⌊a⌋ + \{a\}) \} \\ &= \{⌊a^m⌋\{a\} + \{a^m\}(⌊a⌋ + \{a\})\} \\ &= \{⌊a^m⌋\{a\} + \{a^m\}a\} \\ & = \{\{⌊a^m⌋\{a\}\} + \{a^m\}a\} \end{align}

where the last step comes from the observation that { + } = {{} + }.

Note $\{⌊a^m⌋\{a\}\}$ is one of $1/, 2/, 3/... (-1) / $, in particular

$$ \frac{1}{q} \le \{⌊a^m⌋\{a\}\} \le \frac{q-1}{q} \tag{3}\label{3}$$

We have $1/ < \{⌊a^m⌋\{a\}\} + \{a^m\}a < 1$ by \eqref{1} \eqref{3}, which implies $\{a^{m+1}\} = \{\{⌊a^m⌋\{a\}\} + \{a^m\}a\} > 1/$ which contradicts \eqref{2}.