21

Is $\{\sin n^m \mid n \in \mathbb{N}\}$ dense in $[-1,1]$ for every natural number $m$?

Progress

For $m=1$, I can prove this using the fact that $\sin$ is continuous and $a+b\pi$ is dense in the real line, for integer $a$ and $b$. However, this approach breaks down for $m>1$; what method should be used then?

Community
  • 1
Human Learning
  • 1,033
  • 3
    Perhaps you could elaborate on what you've tried in answering the question so that we can best direct our assistance. – Emily Dec 29 '14 at 20:29
  • 14
    @Arkamis The whole 'what have you tried' business, in my unsophisticated opinion, only makes sense for questions that are not obviously non-trivial. Sure, the OP should have included that he knows the statement for $m=1$ and maybe a link to a solution, but other than that the question is obviously interesting and non-trivial. @ OP Note that it would suffice that $a+b^2\pi$ is dense but I think problems like that usually boil down to questions about approximations of $\pi$ that are known to be 'out-of-reach' hard. – Myself Dec 29 '14 at 20:44
  • 4
    @Myself I disagree. Asking what the poster has tried isn't just a matter of making them put more effort into the question, but also guides possible responses in such a way that they present useful information to the OP. A great example is the bevy of linear algebra answers that invoke the Cayley-Hamilton theorem. The answers are great -- assuming the OP knows (and can use) that theorem. – Emily Dec 29 '14 at 20:47
  • Can we extend solution for $m=1$ to general case? we can project (continous function) $(a+b\pi)^m$ to $a$ and use similar statement as for $m=1$. – Human Learning Dec 29 '14 at 20:50
  • @Myself To be fair, this is a fairly standard exercise in equidistribution theory, so this particular one isn't out-of-reach :). It's still unclear whether the OP has any background in Weyl's method, but it's looking less likely after the followup comments. In fact one does extend the $m=1$ solution to higher $m$ by Weyl differencing, but I can't make any sense of the phrase "project $(a+b\pi)^m$ to $a$" – Erick Wong Dec 29 '14 at 22:33
  • 4
    Define a (projection) function that sends every $(a+b\pi)^m$ to $a$. It's continuous and hence sends dense set to $a^n^ dense. – Human Learning Dec 30 '14 at 04:42
  • 6
    @user180168 Thanks, that makes much more sense as a concept, but it is a deeply flawed approach. The $\sin()$ is a crucial part of the density of $a$. Extricating that while preserving any notion of continuous function will prove to be subtle if not impossible. For instance, the map $n \mapsto \pi n$ is continuous (in either direction) but I think you'll agree that ${\sin n}$ and ${\sin \pi n}$ behave very differently. – Erick Wong Dec 30 '14 at 16:58

1 Answers1

1

The answer is yes.

Given any real number $x_0$, the map $T_{x_0} : (x_1,\ldots,x_d) \mapsto (x_1+x_0,x_2+x_1,\ldots,x_d+x_{d-1})$ from $(\mathbf{R}/\mathbf{Z})^d$ to itself preserves the Haar probability measure. When $x_0$ is an irrational number, it is ergodic. This can be checked by using Fourier series.

A recursion shows that for every integer $n \ge 0$ and $(x_1,\ldots,x_d) \in (\mathbf{R}/\mathbf{Z})^d$, $$T_{x_0} ^n(x_1,\ldots,x_d) = \Big(x_1+x_0~,~x_2+nx_1+{n \choose 2}x_0~,~\ldots~,~\sum_{k=0}^d {n \choose k} x_{d-k}\Big).$$ Applying Birkoff ergodic theorem to any integrable fonction depending on the last variable $x_d$ only yields the following statement.

If $P$ is a non-constant polynomial whose leading coefficient is irrational, then the sequence $(P(n))_{n \ge 0}$ is uniformly distributed modulo 1. https://en.wikipedia.org/wiki/Equidistributed_sequence

In particular, the sequence $(n^m/(2\pi))_{n \ge 0}$ is uniformly distributed modulo 1, yielding the positive answer to your question.

Christophe Leuridan
  • 8,223