# Dynamics in the integers - Floor function

Let $\alpha$ be an irrational with $0<\alpha<1$. Consider the function given by \begin{align*} f: &\mathbb{N}\longrightarrow \mathbb{N}\\ &x\longmapsto [ \alpha\cdot x]\end{align*} where $[r]$ is the largest integer that is less than or equal to $r$ for $r\in\mathbb{R}$.

Let $N=\left[\frac{1}{\alpha}\right]$ be the integer part of $\frac{1}{\alpha}$, and consider the following numbers:

$A_1(n)=\#\{k\leq n : f(k)<f(k+1)\}$

$A_2(n)=\#\{k\leq n : f(k)=f(k+N)\}$

The questions are the following:

1) Is it true that $\displaystyle{\lim_{n\to\infty} \dfrac{A_1(n)}{n}=\alpha}$?

2) What is the value of $\displaystyle{\lim_{n\to\infty} \dfrac{A_2(n)}{n}}$, if it exists?

The problem above came out when I was trying to find examples of classes of finite ordered structures that satisfies strong conditions on their definable sets (for my Ph.D. Thesis). I translated the problem in simple terms, and I would really appreciate any help.

(If it is too easy, a hint will be good. If the problem is related with something in number theory, combinatorics, ergodic theory or anything else, I would also appreciate if you tell me how)

• For the second part use Weyl's equidistribution theorem which shows that the limit is $1-\alpha N$. Aug 27, 2014 at 0:57
• @Lucia: I just wrote the same. But you probably beat me by a minute or so. Aug 27, 2014 at 1:00

The answer to the first question is: yes. The function $$f(x)$$ jumps $$0$$ or $$1$$ at every integer (because $$0\leq\alpha\leq 1$$), and also $$f(0)=0$$, hence $$A_1(n)=f(n+1)$$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1).$$
Regarding the second question: the limit exists and equals $$1-N\alpha$$. Indeed, $$f(k+N)=[k\alpha+N\alpha]$$, hence $$f(k)=f(k+N)$$ means that the fractional part of $$k\alpha$$ is less than $$1-N\alpha$$. Here we used that $$0. As $$\alpha$$ is irrational, the fractional parts of $$k\alpha$$ are equidistributed in $$(0,1)$$ by Weyl's theorem, hence the density of $$k$$'s with $$f(k)=f(k+N)$$ equals the length of the interval $$(0,1-N\alpha)$$ which is $$1-N\alpha$$.
More generally, if $$1\leq M\leq N$$ is fixed, then the density of $$k$$'s with $$f(k)=f(k+M)$$ equals $$1-M\alpha$$. In particular, the case $$M=1$$ yields an alternative proof for the first paragraph, since $$1-(1-\alpha)=\alpha$$.
• The sequence $[n\alpha]$ is known as a Beatty sequence; the indicator function of a Beatty sequence (with irrational $\alpha$) is known as a Sturmian word. There are hundreds of papers working out properties of Beatty sequences. Nov 19, 2014 at 6:03