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Does there exist a real number $a$ bigger than $1$ that's not a rational power of an integer such that $\displaystyle\lim_{n\to\infty}${$a^n$}$=0$ where values for $n$ are positive integers?

(P.S. $\{ x \} = x - \lfloor x \rfloor$ or namely the fractional part function)

I just saw this problem today. I've just started studying about the fractional part approximations so unfortunately I've made no progress solving this.
*I made a slight change in the problem to avoid any confusions about small values

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    $\begingroup$ If we are allowed to restrict the the exponents to odd integers, then $\phi$, the Golden Ratio, is an example. In fact, the sequence $$ a_n=\phi^{2n+1}-\frac1{\phi^{2n+1}}\in\mathbb{Z} $$ The sequence starts out $1,4,11,29,\dots$ and satisfies the recurrence $a_n=3a_{n-1}-a_{n-2}$. Since $\phi=\frac{1+\sqrt5}2\gt1$, we have $\lim\limits_{n\to\infty}\frac1{\phi^{2n+1}}=0$ so that $$ \begin{align} \lim_{n\to\infty}\left\{\phi^{2n+1}\right\} &=\lim_{n\to\infty}\frac1{\phi^{2n+1}}\\ &=0 \end{align} $$ $\endgroup$
    – robjohn
    Commented Jan 9, 2021 at 14:41
  • $\begingroup$ Thanks for your point. However, we are not allowed to restrict all positive integers to odd integers. If there is an even integer, sufficiently big and contradicting the foresaid condition, it is still a counterexample. So removing even numbers is not a valid move, since for even integers, the golden ratio may not work as you said $\endgroup$
    – Aryan
    Commented Jan 9, 2021 at 14:49
  • $\begingroup$ To reduce confusion and noise, I have deleted comments to earlier versions and comments that replied to comments deleted by their authors. $\endgroup$
    – robjohn
    Commented Jan 9, 2021 at 15:21
  • $\begingroup$ Do you mean $\{ x \} = x - \lfloor x \rfloor$? Sometimes $[x]$ is used for the closest-integer function which is different. $\endgroup$ Commented Jan 9, 2021 at 20:24
  • $\begingroup$ Thank you for the notation. Corrected it $\endgroup$
    – Aryan
    Commented Jan 9, 2021 at 23:00

2 Answers 2

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It is hard to find such $a$. Indeed, according to Wikipedia, the set of such $a$ is countable. (Remark that Corollary 3 of Zeldich’s attraction theorem states that a set $\{a\in (1,\infty): \{a^n\}\mbox{ is not dense in }[0,1]\}$ is meager in $(1,\infty)$.)

Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then there is no such $a$. Indeed, for a contradiction pick such $a$. The affirmative answer to Pisot-Vijayaraghavan problem implies that $a$ is a Pisot–Vijayaraghavan number. Let $a_1=a$ and $a_2, \dots, a_m$ be the conjugates of $a$. Then for each natural $n$, $\sum_{k=1}^m a_k^n$ is integer.

For each $k=1,\dots, m$ let $a_k=r_ke^{\varphi_ki}$ for some positive $r_k$ and $\varphi_k$. Let $G=\Bbb T^m=\{z\in\Bbb C:|z|=1\}$ be a (multiplicative) topological group and $g=(e^{\varphi_1i}, e^{\varphi_2i},\dots, e^{\varphi_m i})$ be an element of $G$. Since the group $G$ is compact, it is well-known and easy to show that $G$ topologically periodic, so for any neighborhood $U$ of the identity of $G$ there exists arbitrarily big natural number $n$ such that $g^n\in U$. Pick $U_0=\{(x_1,\dots,x_m)\in\Bbb T^m: \forall i (\operatorname{Re} x_i\ge 0)\}$.

Let $N>0$ be any number. Since $a$ is a Pisot-Vijayaraghavan number, we have $r_k<1$ for each $k=2,\dots, m$. So there exists $M>0$ such that $\sum_{k=2}^m r_k^M<1/2$. Pick $n>N,M$ such that $g^n\in U_0$. Then for some integer $K$ we have

$$K=\sum_{k=1}^m a_k^n=\sum_{k=1}^m r_k^ne^{n\varphi_ki} =\sum_{k=1}^m \operatorname{Re} r_k^n (e^{n\varphi_ki})= a^n+\sum_{k=2}^m r_k^n \operatorname{Re} e^{n\varphi_ki}.$$

So $a^n\le K\le a^n+\sum_{k=2}^m r_k^n\le a^n+\sum_{k=2}^m r_k^M<a^n+1/2$. Thus $\{a^n\}>1/2$, a contradiction.

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    $\begingroup$ Thanks for all the effort! It took me a while to understand it but after all it's not "a solution" since it's based on a conjecture! Thank you very much $\endgroup$
    – Aryan
    Commented Jan 19, 2021 at 9:11
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    $\begingroup$ @AryanHemmati Yes, my answer provides not a solution but a message: “if there is a not extremely hard solution then it shows that there is no such $a$”. $\endgroup$ Commented Jan 19, 2021 at 9:16
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    $\begingroup$ Actually I derived this theorem from a very unrelated problem which is again a conjecture by one of my friend that seems very much to be true. This I strongly sense that there is no such $a$, too $\endgroup$
    – Aryan
    Commented Jan 19, 2021 at 17:06
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I don't have a positive answer to OP's question; but note that it's easy to find solutions to the seemingly similar question where you would like to have $\lim_{n\rightarrow\infty}\{a^n\}=1$. For instance, $$a_m = \frac{1}{2}\left(m + \sqrt{m^2-4}\right) \in (1,\infty)$$ has this property for any integer $m > 2$. This works because the linear recurrence defined by $f_0=2,$ $f_1=m,$ and $f_{n+2}=mf_{n+1} - f_n$ can be explicitly solved to yield $$ f_n = a_m^n+b_m^n, $$ where $b_m=\frac{1}{2}\left(m - \sqrt{m^2-4}\right)\in(0,1)$. Hence $$ \{a_m^n\}=\{f_n - b^n_m\}=\{-b^n_m\}\rightarrow 1. $$ It seems as if it should be possible, perhaps with a higher-range linear recurrence, to do the same thing with the opposite sign: we would need, say, $f_n=a^n-Bb^n+...$ for some $a>1$ and $B> 0$ and $b\in(0,1)$, and for the remaining terms to be real and decaying faster than $b^n$. Some constraints are placed on this by the fact that the sum of the eigenvalues of an integer matrix is integral.

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    $\begingroup$ I am afraid that a number $a$ constructed this way be such that $\lim_{n\to\infty}\{ a^n\}=0$ would be a Pisot-Vijayaraghavan_number, that, according to my answer, leads to a contradiction. $\endgroup$ Commented Jan 21, 2021 at 18:51

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