Is the sequence $\{n/\{ \sqrt{2}n\} \}$ uniformly distributed in $[0,1)$?

Question

A classical theorem of Kronecker says that the sequence $(\{\alpha_1 n\}, \{\alpha_2 n\},\dots,\{\alpha_d n\})$ ($n \in \mathbb{N}$) is uniformly distributed in $[0,1)^d$ provided that $1,\alpha_1,\alpha_2,\dots,\alpha_d$ are linearly independent over $\mathbb{Q}$. More generally, for a polynomial $p(x)$ (or a $d$-tuple of polynomials $p_1(x),p_2(x),\dots,p_d(x)$) a theorem of Weyl completely describes the distribution of $\{p(n)\}$ ($n \in \mathbb{N}$) (respectively, $(\{p_1(n)\}, \{p_2(n) n\},\dots,\{p_d(n)\})$). More generally still, the distribution of bounded generalised polynomials - i.e., expressions built up using polynomials, and the operations of addition, multiplication and fractional part - is completely understood thanks to work of Bergelson and Leibman. Generalised polynomials include, for instance, expressions such as $\{\alpha n \{\beta n \} \}$.

I'm interested in the behaviour of expressions that resemble generalised polynomials, but also include division. (In the application that I have in mind it is sufficient to consider relatively simple expressions of this type - such as $\{ n g(n)/h(n) \}$ where $g,h$ are generalised polynomials bounded by $1$ and $h(n) \neq 0$ for all $n \in \mathbb{N}$.) It seems to me that nothing is known about expressions like that, although I'd be delighted to be proven wrong. With this in mind, I would like to ask about the first non-trivial instance:

Is the sequence $\{n/\{ \sqrt{2}n\} \}$ uniformly distributed in $[0,1)$? Is the set of its values at least dense in $[0,1)$? What about sequences $\{\alpha n/\{ \beta n\} \}$, where $\alpha$ is non-zero and $\beta$ is irrational?

Answer 1

I'll show that $\{ n / \{ \alpha n \} \}$ is equidistributed under a certain Diophantine condition on $\alpha$ which holds generically (and in particular for $\sqrt 2$), but the proof goes through verbatim for $\{ \beta n / \{ \alpha n \} \}$.

A very reasonable guess is that $n$ and $\{ n \alpha \}$ are "independent", in the sense that for any rational function $p(x, y)$, if $\{ p(x, y) \}$ is equidistributed (as $N \to \infty$) for $0 \leq x \leq N$ and $0 \leq y \leq 1$, then the sequence $\{ p(n, \{ n \alpha \} ) \}$is equidisitributed for any irrational $\alpha$. I haven't thought about how much this proof can be adapted to this more general question.

The Diophantine condition on $\alpha$ is that $$ \liminf_{p, q \to \infty} q^{\frac{7}{3}} \left\lvert \alpha - \frac{p}{q} \right\rvert = \infty $$ For a generic number $\alpha$ this condition holds with $\frac{7}{3}$ replaced by $2 + \epsilon$.

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The general idea is this: take $q$ such that $q \alpha$ is very close to an integer (say, the denominator of a convergent in the continued fraction). Then, $\frac{n + k q}{\{ (n + k q) \alpha) \}} = \frac{n + k q}{\{ n \alpha \} + k \varepsilon}$ and so we can understand the behavior of this sequence better than the original sequence, and show that along almost all arithmetic progressions of difference $q$ the sequence is equidistributed.

Expanding the denominator using the formula for a geometric progression, it turns out that the sequence is very close to a quadratic polynomial in $k$, and these are easily shown to be equidistributed using Weyl differencing. Now for the details:

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First some definitions and preliminaries.

Let's look at the sequence of $\{ n / \{ \alpha n \} \}$ for $1 \leq n \leq N$. Take $p, q$ such that $q \alpha - p = \varepsilon$ is small (we will choose them precisely later). For now, we will require that $q = o(N)$. Let $\ell$ be the largest integer such that $q \ell \leq N$. Since $q = o(N)$ we will show equidistribution of $\{ n / \{ \alpha n \} \}$ for $1 \leq n \leq q \ell$ and this will be sufficient.

Since $\{ \alpha n \}$ is equidistributed, for the rest of the proof we will work only with the subset of $n$ such that $c \leq \{ \alpha n \} \leq 1 - c$ for some arbitrarily small positive constant $c$. This will make things easier because the function $\frac{1}{x}$ behaves nicer when $x$ is bounded away from $0$.

Now, take some $1 \leq n \leq q$ and look at $\frac{n + k q}{\{ \alpha n + k q \alpha \}} = \frac{n + k q}{\{ \alpha n \} + k \varepsilon}$ for $0 \leq k < \ell$ (we have assumed here that $\lvert \ell \varepsilon \rvert < c$). Notice that $$ \frac{n + k q}{\{ \alpha n \} + k q} = \frac{1}{\{ \alpha n \}} \cdot \frac{n + k q}{1 + k \frac{\varepsilon}{\{ \alpha n \}}} = $$ $$ \frac{1}{\{ \alpha n \}} \cdot (n + k q) \cdot \left( 1 - k \frac{\varepsilon}{\{ \alpha n \}} + k^2 \frac{\varepsilon^2}{\{ \alpha n \}^2} + \mathcal{O} \left( \ell^3 \varepsilon^3 \right) \right) = $$ $$ \frac{n}{\{ \alpha n \}} + \left( \frac{q}{\{ \alpha n \}} - \frac{\varepsilon}{\{ \alpha n \}^2} \right) \cdot k - \frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k^2 + \mathcal{O} (\ell^3 q \varepsilon^2) $$

Assume that $\ell^3 q \varepsilon^2 = o(1)$. Then, it is sufficient to show equidistribution (for $0 \leq k < \ell$) of the quadratic sequence $$ \frac{n}{\{ \alpha n \}} + \left( \frac{q}{\{ \alpha n \}} - \frac{\varepsilon}{\{ \alpha n \}^2} \right) \cdot k - \frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k^2 $$ The technique of Weyl differencing says that for a sequence $a_n$, if $a_{n + h} - a_n$ is equidistributed for every $h$ then $a_n$ is equidistributed. Because our sequence is a quadratic polynomial, and so what we need to show is that for almost all $n < q$ the sequence $$ \frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k $$ is equidistributed for $0 \leq k < \ell$.

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It is easy to see that a necessary condition for this sequence to be equidistributed for almost all $n$ is $q \varepsilon \ell \to \infty$. The following simple fact is useful:

Suppose that $q \varepsilon \ell \to \infty$.

1. If $q \varepsilon = o(1)$ then the sequence is equidistributed for all $n$.
2. If $q \varepsilon = \Theta(1)$ then the sequence is equidistributed for almost all $n$.

Proof:

1. We have an arithmetic progression with tiny step size which wraps around the interval a number of times which goes to infinity, and so this is clearly equidistributed.

2. The exceptional set of $\beta$ which are $\Theta(1)$ such that $\beta k$ is not equdistributed is a finite set of very small intervals around rational numbers of very small denominator, and from the equidistribution of $\{ \alpha n \}$ we see that a negligible proportion of $n$'s fall into this exceptional set.

Either way, we see that when $q \varepsilon \ell \to \infty$ and $q \varepsilon = \mathcal{O}(1)$ the sequence is equidistributed, so we will try to choose $q, \varepsilon$ like this.

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As an example, suppose for a moment that the continued fraction of $\alpha$ has bounded coefficients, as happens in the case of $\sqrt 2$. In this case, we have convergents of any size we want, and so we can take $q$ such that $\ell$ grows arbitrarily slowly to infinity and $q \varepsilon = \Theta (1)$, which proves equidistribution in this case by the above lemma.

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Unfortunately, sometimes there are large coefficients in the continued fraction and so we have to work a bit more. Here is the general method.

Take $t$ growing to infinity. By Dirichlet's theorem, there are some $q_0, p_0 \leq N t$ such that $\lvert \varepsilon_0 \rvert = \lvert q_0 \alpha - p_0 \rvert \leq \frac{1}{t N}$. We will take $q = r q_0, \ \ell = \frac{N}{r q_0}$ for some $r$ which we will choose now.

Firstly, we will want $\varepsilon = r \varepsilon_0$ (that is, for the closest integer to $r q_0 \alpha$ to be $r p_0$) and for this we need $r \lvert \varepsilon_0 \rvert < \frac{1}{2}$. In the proof we have required $\lvert \ell \varepsilon \rvert < c$ and $\ell^3 q \varepsilon^2 = o(1)$ which is equivalent to $$ \frac{N \lvert \varepsilon_0 \rvert}{q_0} < c \iff N \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert < c $$ and $$ \frac{N^3 \varepsilon_0^2}{q_0^2} = o(1) \iff N^3 \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert^2 = o(1) \iff N^{\frac{3}{2}} \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert = o(1) $$ The first condition obviously follows from the second. We do have one additional condition on $r$, which is $\ell \to \infty$ (which actually is a condition on $r$).

It is easy to see that $\lvert q_0 \varepsilon_0 \rvert \leq 1$. Let $r$ be the smallest positive integer such that $\lvert q \varepsilon \rvert = r^2 \lvert q_0 \varepsilon_0 \rvert \to \infty$. From the fact mentioned above, for such $r$ the sequence is equidistributed (assuming $\ell \to \infty$).

The only thing that is left now is to check that in this case $\ell \to \infty$. If $r = 1$ then obviously $\ell \to \infty$. Otherwise, $r \approx \frac{1}{\sqrt{\varepsilon_0 q_0}}$ $$ \ell = \frac{N}{r q_0} \approx \frac{N \sqrt{\varepsilon_0}}{\sqrt{q_0}} = N \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert^{\frac{1}{2}} $$ This tends to infinity iff $\ell^2 \approx N^2 \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert \to \infty$.

Thus, all that is left is to show that for every $N$ there exist $p_0, q_0$ such that $N^{\frac{3}{2}} \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert \to 0$ but $N^2 \left\lvert \alpha - \frac{p_0}{q_0} \right\rvert \to \infty$ which follows in an elementary manner from our Diophantine condition on $\alpha$ at the beginning and choosing $t \approx N^{-\frac{1}{7}}$.

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Some comments:

1. I think the Diophantine condition can be relaxed all the way to requiring $\alpha$ not to be a Liouville number by taking higher order terms in the geometric series expansion, and applying iterated Weyl differencing.

2. The Diophantine condition occurs because when $q$ is very small compared to $N$, the sequences become long and very complicated, which makes them more difficult to control. Perhaps one could divide the interval into a bunch of more short arithmetic progressions. The fact that the numbers appearing are bigger might be compensated by the fact that the denominator changes very slowly, but I'm not sure of this.

Answer 2

Here is at least a heuristic argument in favour of equidistribution.

For any equidistributed variable $x$ in $(0,1)$, the values that fall into $I(n)= (1/(n+1),1/n]$ are also equidistributed in $I(n)$ and can be written as $$\frac1{n+1}+\frac1{n\,(n+1)}\,y$$ where $y$ is equidistributed in $(0,1]$. In this interval the fractional part of the inverse is $$ \eta_n(y) = \frac{1}{\frac1{n+1}+\frac1{n\,(n+1)}\,y} -n~=~ \frac{n\,(n+1)- n\,(n+y) }{n+y}~=~ \frac{ 1-y }{1+y/n}.$$ This function is decreasing and its own inverse, such that $$p(\eta_n\leq\eta)~=~1-y(\eta ) ~=~ \frac{\eta\,(n+1)}{n+\eta}.~~~(*) $$ The probability of the equidistributed $x$ to fall into $I(n)$ is $\frac1{n\,(n+1)}$, so the overall probability of $$p(\eta'\leq \eta) ~=~ \sum \limits_{n=1}^\infty \frac1{n\,(n+1)} \, \frac{\eta\,(n+1)}{n+\eta} ~=~ \sum \limits_{n=1}^\infty \frac{\eta }{n\,(n+\eta)}.$$ From this the probability density of the inverse of an equidistributed variable in $(0,1)$ is $$ p(\eta)~=~ \frac{d}{d\eta} p(\eta'\leq \eta) ~=~ \sum \limits_{n=1}^\infty \frac{1}{ (n+\eta)^2} ~=~ \psi_1(1+\eta),$$ where $\psi_1$ is the trigamma function. Note that $(*)$ implies that $p(\eta_n\leq\eta)$ in the limit $n\to \infty$ approaches equidistribution. For a heuristical argument, rewrite the sequence $$\left\{ \frac{n}{\{ \sqrt{2} \,n\}}\right\} ~=~ \left\{ \frac{n}{ \sqrt{2} \,n -\lfloor \sqrt{2} \,n \rfloor }\right\} ~=~ \left\{ \frac{1}{ \sqrt{2} -\frac{\lfloor \sqrt{2} \,n \rfloor}{n} }\right\} $$ and note that because $\{ \sqrt{2} \,n\}$ is equidistributed in $(0,1)$ the values $ \sqrt{2} -\frac{\lfloor \sqrt{2} \,n \rfloor}{n}$ are in some sense `equidistributed' in $(0,1/n)$, which similar to above would imply that $$p_n(\eta'\leq \eta) ~=~ \sum \limits_{k=n}^\infty \frac{\eta }{k\,(k+\eta)},$$ and for the not normalized density one has $$ p_n(\eta)~=~ \sum \limits_{k=n}^\infty \frac{1}{ (k+\eta)^2} ~=~ \psi_1(n+\eta).$$ After normalizing by $$ \int \limits_0^1 p_n(\eta) \,d\eta ~=~ \frac1n,$$ one has $$ p_n(\eta)~=~ n\, \psi_1(n+\eta) \to 1 ~~{\rm for}~~n\to\infty,$$ which implies equidistribution for sufficiently large $n$. Numericial evidence displays a consistent peak in $p_n(\eta)$ at $\eta =\frac1{2\,\sqrt{2}}$ which decreases for large $n$.

Answer 3

The set of the values of $\{n/\{\sqrt 2 n\}\}$ is dense in $[0,1)$.

The choice of $n$ is as follows: $n=ka_p+1$, where $a_p$ is the $p$th term of A001541. For fixed $k \in \mathbb N^+$, we let $a_p\rightarrow \infty$, and the value of $\{n/\{\sqrt 2 n\}\}$ tends to $\{-\frac {\sqrt{2}} 2 (3k^2-2)\}$.

Proof: Let $b_p=\sqrt{2a_p^2-2}$. By the properties of A001541, $b_p$ are all integers, and $b_p-\sqrt2 a_p=O(1/a_p)$.

Thus, for fixed $k$ and sufficiently large $p$, $\{\sqrt2 n\}=\{\sqrt2 ka_p+\sqrt{2}\}=\sqrt{2}ka_p-kb_p+\sqrt2-1$.

Since $\sqrt{2}ka_p-kb_p$ is $O(1/a_p)$, we may compare $n/\{\sqrt 2 n\}$ to $n/(\sqrt 2 - 1)$:

$n/\{\sqrt 2 n\}=\frac{ka_p+1}{\sqrt{2}ka_p-kb_p+\sqrt2-1}=\frac{ka_p+1}{\sqrt2-1}-\frac{(\sqrt{2}ka_p-kb_p)(ka_p+1)}{(\sqrt2-1)(\sqrt{2}ka_p-kb_p+\sqrt2-1)}$.

The first term equals to $(\sqrt2+1)ka_p+(\sqrt2+1)$, which is mod-1 equal to $\sqrt2ka_p-kb_p+\sqrt{2}$. It mod-1 converges to $\sqrt2$. So we only need to prove that the second term, without the minus sign, mod-1 converges to $\frac{3\sqrt2}2k^2$.

To see this, notice that the leading term of the second term is $ \frac{(\sqrt{2}ka_p-kb_p)ka_p}{(\sqrt2-1)^2}$. As $b_p^2=2a_p^2-2$, the term can be rewritten as $k^2\frac{2a_p}{\sqrt2a_p+b_p}(3+2\sqrt2)$. The term tends to $k^2\frac{\sqrt{2}}{2}(3+2\sqrt2)$, which is mod-1 equivalent to $\frac{3\sqrt2}2k^2$.

By the theorem of Weyl as you mentioned, when $k$ runs over $\mathbb N^+$, the set $\{-\frac {\sqrt{2}} 2 (3k^2-2)\}$ is dense in $[0,1)$.

Thus the set of the values of $\{n/\{\sqrt 2 n\}\}$ is dense in $[0,1)$.