# density of a set

let $S=\{\sin (n)|n \in N\}$. We can prove $S$ is dense in $[-1,1]$. So is the set $\{\sin( n^2)|n \in N\}$; but the set $\{\sin (n^3)| n \in N\}$ is not dense in $[-1,1]$. How to prove this?

• Given your previous question: is this from a set of exercises? a problem in a book you are reading? a claim in a paper you are studying? Knowing these things would help people (a) decide if your question is appropriate (b) give more precise and helpful answers – Yemon Choi Oct 4 '11 at 8:23
• Why aren't they all dense? It's just a question of whether $(n^k/2\pi)$ is equidistributed modulo 1, and that will be so for any fixed k since $\pi$ is irrational. You'd use Weyl's inequality for a rigorous proof. – Ben Green Oct 4 '11 at 12:48
• As far as I remember, H. Weyl proves in 'Weyl, H. (1916). "Ueber die Gleichverteilung von Zahlen mod. Eins,". Math. Ann. 77 (3): 313--352.' that for each polynomial $P(n)$ with integer coefficients the sequence $exp(iP(n))$ is equidistributed in the unit circle $\mathbb{S}^{1}$\ and this implies of course that $sin(P(n))$ is dense in $[-1,1]$. So the answer to your question should be - no, you cannot prove this. – t22 Oct 4 '11 at 19:05

If you read On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational. carefully, you will be enlightened.