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That is, what are the possible values of a real number $\lambda$ for which there exists a nonintegral real $\alpha >1$ such that, given any $\varepsilon >0,$ all but finitely many powers of $\alpha$ lie within $\varepsilon$ of an integral multiple of $\lambda$? As an example, we may take $\lambda =\sqrt 5$ with $\alpha =\frac 12+{\frac 12} \sqrt5.$

Formally, define$$\begin{align}\Lambda &:=\\\{\lambda &\in \Bbb R_{>0}: (\exists\,\alpha\in \Bbb R_{>1}{\setminus}\Bbb N)(\forall\,\varepsilon\in\Bbb R_{>0})(\exists\,k\in\Bbb N)(\forall\, n\in\Bbb N_{>k})(\exists\,m\in\Bbb N)\;\; |\alpha^n-\lambda m|< \varepsilon\}.\end{align}$$What are the known properties of $\Lambda$? In particular, can $\Lambda$ possess a rational or a transcendental element?

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3 Answers 3

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If I am not mistaken, the class $\Lambda$ of positive reals $\lambda$ which you describe is known to be countable and include all positive rationals as well as reals $a+b\sqrt{D} \in \mathbb{Q}\left[\sqrt{D}\right]$ as well as a larger class of algebraic numbers described below. (I'm not sure if this known class is actually all algebraic numbers, but the answer might be easy, one way or the other.) It is conjectured that there are no transcendental numbers in $\Lambda.$ I am not an expert and my answer relies somewhat on claims whose references I did not personally check.


The nice example you give with $\alpha=\Phi=\frac{1+\sqrt{5}}{2}$ comes from an interesting algebraic setting and satisifes stronger conditions than the ones you give. There are theorems and conjectures depending on what conditions one sets. Other quadratic examples are $\alpha=a+\sqrt{D}$ where $(a-1)^2 \lt D \lt (a+1)^2$ and also $\alpha=\frac{a+\sqrt{D}}{2}$ when $a$ is odd, $D$ is congruent to $1$ $\mod{4}$ and $(a-2)^2 \lt D \lt (a+2)^2.$ These are the quadratic Pisot numbers. Note that a power of such a number has the same form, for example $\Phi^{10}=\frac{123+55\sqrt{5}}{2}=\frac{123+\sqrt{D}}{2}$ for $D=5\cdot 55^2=123^2-4.$ More generally, a Pisot number is a real root $\alpha \gt 1$ of a monic integer polynomial $P(x)$ such that all other roots of $P(x)$ lie strictly inside the unit circle in the complex plane. Any Pisot number $\alpha$ satisfies properties similar to $\Phi$ and it may be that they are the only reals which do. Then the $\lambda$ which work for any given (Pisot number) $\alpha$ are all algebraic numbers from the field of $\alpha$. In other words, $\lambda \in \mathbb{Z}[\alpha]$ and $\lambda$ is a root of a polynomial with integers coefficients (possibly not monic). In the quadratic cases this would include all positive numbers $a+b\sqrt{D} \in \mathbb{Z}[\sqrt{D}].$ We can get $\lambda =\frac{a+b\sqrt{D}}{k}\in \mathbb{Q}[\sqrt{D}]$ by replacing $\alpha$ with some appropriate $k\alpha^t$ where $t$ depends on $k$ and the convergence rate. The details here are a bit sketchy but I don't think it would be hard to more precise.


Let me start with the example which will illustrate why Pisot numbers have this property. The golden ratio $\Phi=\frac{1+\sqrt{5}}{2} \approx 1.618$ is one root of $x^2-x-1$, the other being $\overline{\Phi}=\frac{-1}{\Phi}\approx-0.61803$. So $\Phi$ is a Pisot number. $\Phi$ has the nice property that its powers are nearly integers and also are nearly multiples of $\sqrt{5}$.

For example $\Phi^{10}=\frac{123+55\sqrt{5}}{2}.$ The two terms in the numerator are very nearly equal $\Phi^{10}=122.991869\cdots\approx 123$ and $\frac{1}{\sqrt{5}}\Phi^{10}=55.00363\cdots\approx 55.$ One might recognize that $F_{10}=55$ is the tenth Fibonnacci number, The next is $34+55=89$ and $34+89=123.$ As is well known, these properties generalize: $\frac{\Phi^n+{\overline{\Phi}}^n}{\sqrt{5}}=F_n$ while $\Phi^n+{\overline{\Phi}}^n=F_{n-1}+F_{n+1}.$ Since $\overline{\Phi}^n$ goes to zero we can say that $\lim_{n \to \infty}\|\Phi^n\|=0$ and also $\lim_{n \to \infty}\|\frac{1}{\sqrt{5}}\Phi^n\|=0.$ Here $\|x\|$ denotes the distance from the real $x$ to the nearest integer. $\|x\|=|\ x-\lfloor x \rceil \ |.$

In fact $\overline{\Phi}^n$ goes to zero so rapidly that $\sum_1^{\infty}\|\Phi^n\| \lt \infty$ and also $\sum_1^{\infty}\|\frac{1}{\sqrt{5}} \Phi^n\| \lt \infty.$

It is an easy exercise using $\|x+y\| \le \|x\|+\|y\|$ that the statements above hold for $\| \lambda \Phi^n\|$ where $\lambda$ is any real $\lambda=\frac{a}{\sqrt{5}}+b+c\Phi$ with $a,b,c$ integers. I think that we can expand to allowing $a,b,c \in \mathbb{Q}$ if we replace $\alpha=\Phi$ with $\alpha=(k\Phi)^t$ where $k$ is the greatest common divisor of the denominators and $t$ is so large that the convergence to zero of $\left(k\overline{\Phi}^t\right)^n$ is rapid enough to make $\lfloor\ (k\Phi^t)^n \ \rceil=k^n\lfloor\ (\Phi^t)^n \ \rceil$

Consider the following conditions in increasing order of strength which might be satisfied by a pair of reals $\lambda \gt 0$ and $\alpha \gt 1$ with $\alpha$ irrational along, in the first case, with a positive $\epsilon \lt \frac{1}{2}$.

  1. $\|\lambda \alpha^n\| \lt \epsilon$ for all $n$ (or all large enough $n$)
  2. $\lim_{n \to \infty}\|\lambda \alpha^n\|=0$
  3. $\sum_1^{\infty}\|\lambda \alpha^n\|^2 \lt \infty$
  4. $\sum_1^{\infty}\|\lambda \alpha^n\| \lt \infty$

Note that condition 3 would allow $\|\lambda \alpha^n\|=\frac{1}{n}$ while condition 4 would not.

Note also that if $m \lambda$ is the integer multiple of $\lambda$ closest $\alpha^n$ then $m=\lfloor\frac{1}{\lambda} \alpha^n\rceil$ and $\|\alpha^n-m\lambda\|=\lambda\|\frac{1}{\lambda}\alpha^n\|$ so one goes to zero if and only if the other does. This might mean that for the problem as stated I should have said I was describing $\frac{1}{\lambda}$ for any particular $\alpha$ but in the end the set $\Lambda$ is the same.

According to the Wikepedia article (whose sources I have not personally checked)

  • Only countably many $\alpha$ satisfy condition 2 for some $\lambda$ and vice versa.

  • If $\alpha$ is real and condition 3 is satisfied then $\alpha$ is a Pisot number and $\lambda$ is an algebraic number in the field of $\alpha.$

  • If $\alpha$ is algebraic and condition 2 holds then the same conclusion holds.

  • It is conjectured, but not known, that the condition algebraic can be ommitted.

One could investigate condition 1 either with $\lambda=1$ or some other $\lambda.$ The idea is to leave $\alpha$ indefinite and consider the sequence of integers $m_n=\lfloor\frac{1}{\lambda}\alpha^n \rceil.$ Each choice limits the interval which could contain $\alpha$ and the potential values for $m_{n+1}$. It might be that at some point there is no possible value for $m_{n+1}$. Alternately, we might manage to get a possible initial sequence $m_i$ and confine $\alpha$ to a small enough interval $((m_n-\epsilon)^{1/n},(m_n+\epsilon)^{1/n} )$ that we might (or might not) be able to identify it. I did not have great luck with $\lambda=1$ but my experiments with $\lambda=e , \epsilon=0.1$ and $\alpha \approx 10e$ all ended with no way to continue.

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Thank you very much for this answer. Just a few queries concerning your 4th paragraph: (1) I can't read "$\sum_1^{\infty}\|\frac{1][\sqrt{5}} \Phi^n\| \lt \infty$"; (2) I guess the equality sign in "$\|x=y\| \le \|x\|+\|y\|$" should be a plus sign; (3) By "$\lambda$ is any real $\lambda=\frac{a}{\sqrt{5}}+b+c\Phi$" do you mean "$\lambda$ is any real of the form $\lambda=\frac{a}{\sqrt{5}}+b+c\Phi$ with integral $a$, $b$, and $c$"? –  John Bentin Mar 26 '13 at 8:45
    
Thanks. I fixed it. –  Aaron Meyerowitz Mar 26 '13 at 13:56

This is a hard problem in general. For example, one might expect that the powers of $e$ have more or less randomly distributed fractional parts. However, I believe we can't even disprove that the fractional parts of powers of $e$ approach $0$. In other words, we can't disprove that $1$ is in your set $\Lambda$ (with $\alpha=e$). (For all I know, we can't disprove that any $\lambda$ is in your set, even with $\alpha=e$.)

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I got the quantifiers wrong in my previous answer. Thanks to John Bentin and Robert Israel for helping me to see that.

Claim: $\mathbb{Q}^+ \subset \Lambda$

Proof: Since any multiple of $p$ is a multiple of $p/q$, it suffices to prove that the natural numbers are in $\Lambda$. Consider the Lucas numbers $1,3,4,7, ...,L_n =\phi^n+(-\phi)^{-n},..$. Since $|-1/\phi| \lt 1$, $\phi^n$ is close to the $n$th Lucas number so $1 \in \Lambda$. (See Pisot numbers.)

To show that $p\in \Lambda$ for a natural number $p$, find $m$ so that $|(-1/\phi)^m| \lt 1/p$. Then take $\alpha = p \phi^m$. $(p\phi^m)^n + (p(-1/\phi)^m)^n = p^n L_{mn}$, a multiple of $p$ for $n \ge 1$. Since $|p(-1/\phi)^m| \lt 1$, powers of $p \phi^m$ get closer to multiples of $p$ exponentially rapidly.

Passing to a subsequence of $\phi^n$, then rescaling, also shows that $\sqrt{5}~\mathbb{Q}^+ \subset \Lambda$.

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