I have a question that I can't manage to answer myself. It comes from some work in PDE theory, but it is related to analytic number theory.

Let us say that we have an irrational number $\xi$. The question is basically the following: is it true that the numbers $n \{n \xi\}$, where $\{\cdot\}$ stands for the fractional part, are dense in $\mathbb{R}^+$ as $n \in \mathbb{N}$?

If I make some numerical experiments, the answer seems to be affirmative. I know that $\{n\xi\}$ is uniformly distributed in $[0,1]$ and hence everywhere dense, but I need to know if any positive numer can be approximated by terms of the sequence $n \mapsto n \{n \xi\}$.

Any help or reference is really welcome.


As I said in the comments, I am interested in the behavior of the sequence $$ n \mapsto \left| 2\{n\xi\}-1 \right| n. \tag{1} $$ If I plot $$ n \mapsto \left( 2\{n\xi\}-1 \right)n,\tag2 $$ it seems to me that the sequence is uniformly distributed in $(-\infty,+\infty)$. Now (1) is just the absolute value of (2). Can we say anything about the absolute value of a u.d. sequence? I've found nothing in the most popular books.

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    $\begingroup$ The sequence $\{n\{n\xi\}\} = \{n^2 \xi\}$, mentioned in the title, is uniformly distributed mod $1$. As for $n\{n\xi\}$ in $\mathbb{R}^+$, I think the answer depends on $\xi$. For example if $\xi = \sqrt{2}$ then you have an estimate of the form $\{n\xi\} \geq c/n$, so you cannot approximate points of $\mathbb{R}^+$ near $0$. However I would guess that it's true for random $\xi$. $\endgroup$ – Sean Eberhard Apr 15 '16 at 10:42
  • $\begingroup$ Such an estimate would imply Littlewood's conjecture right? It is actually much stronger, so the answer is no, as Sean demonstrated. Nevertheless, given such BA numbers (say quadratic irrationals), you can construct some non-trivial examples to Littlewood's, so some closely related analogs do exist. $\endgroup$ – Asaf Apr 15 '16 at 11:24
  • $\begingroup$ @SeanEberhard Thank you for your comment. Do you have any reference for estimates like $\{n \sqrt{2}\} \geq c /n$? I am totally new to number theory... $\endgroup$ – Siminore Apr 16 '16 at 8:05
  • $\begingroup$ This is a basic fact about quadratic irrationals, and it follows from the observation that $|p+q\sqrt{2}|\cdot |p-q\sqrt{2}| = |p^2 - 2 q^2| \geq 1$ for all pairs of integers $p,q$. Look up continued fractions and diophantine approximation for more information and context. Did you really need the result for all irrational $\xi$ or would almost all be enough? $\endgroup$ – Sean Eberhard Apr 16 '16 at 11:37
  • $\begingroup$ @SeanEberhard To say the truth, I am investigating the behaviour of the sequence $n \mapsto |2n\{n\xi\}-n|$, where $0<\xi<1/2$ is an irrational number. Almost all would be much more than I know ;-) $\endgroup$ – Siminore Apr 16 '16 at 11:58

For a random process that produces a countable sequence of real numbers, there are two possibilities:

1) With probability one, the sequence is dense in $\mathbb R$.

2) There is some fixed interval such that the probability that there are any elements of the sequence in the interval is less than $1$.

This is obvious because it is sufficient to work with intervals with rational endpoints, and then we can apply countable additivity of measure.

So if you want to show the sequence is dense, you just have to rule out case 2. Fix an interval, and consider the random variable that is the number of $n \leq N$ such that the $n$th element of your sequence is in that interval. We want to show that the probaiblity this random variable vanishes goes to $0$ as $N$ goes to $\infty$.

You seem to care about two different sequences, $n\{n \xi\}$ and $2n\{n \xi\}-n$. For either one, the expected value of this variable is proportional to $\log n$ and goes to $\infty$. Just a little bit more information on this variable could be sufficient to upper bound the probability of vanishing. One natural approach is to try to show the variance is $o\left((\log n)^2\right)$.

To do that, we need to upper bound the probability that simultaneously $|\{n \xi\} -a/n | < b/n$ and $|\{m \xi \} - a/m | < b/m$. I'm not yet sure how to do this.


Just some loosely connected plots of interest. First of all $\sum_{n < N} \{ n \xi\}^2 = \frac{1}{12}N + O(1)$ which can be seen from the mean ergodic theorem without the $O(1)$ error term.

enter image description here

If we change one letter and plot $\sum_{n < N} \{ n^2 \xi\}^2 = \frac{1}{12}N + O(?)$ a much noisier plot emerges.

enter image description here

Then I tried to plot your last equation $\sum_{n \leq N} 2n \{ n \xi\} - n $ and a pleasant figure emerges. Personally I don't understand why those partial sums are bounded (almost everywhere) in $\xi$.

enter image description here

  • $\begingroup$ Actually I do not need the partial sums of my sequence. Anyway, the sequence $n \mapsto 2n\{n \xi\}-n = \left(2\{n \xi\} -1 \right)n$ is (probably?) uniformly distributed in $\mathbb{R}$. Now the question is: what can we say about the absolute values of a u.d. sequence? $\endgroup$ – Siminore Apr 16 '16 at 15:51
  • $\begingroup$ I am curious as to why your number sequence arises in PDE, all I can think of is KAM theory. For example Resonances and Small Divisors by Etienne Ghys $\endgroup$ – john mangual Apr 16 '16 at 16:01
  • $\begingroup$ It is somehow related to the distribution of zeroes of some special function that solves a differential equation. $\endgroup$ – Siminore Apr 17 '16 at 8:34
  • $\begingroup$ @Siminore there is not really such a thing as uniform distribution in $\mathbb R$, as far as I know. What do you mean? $\endgroup$ – Will Sawin May 16 '16 at 17:21

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