Prove that $\sin (n\beta)$ can get arbitrarily close to any value in [0, 1] if $\frac{\beta}{\pi}$ is irrational

Question

I saw the following statement and I'm not sure how to prove it:

Given a constant value $\beta \in \mathbb{R}$, if $\frac{\beta}{\pi}$ is irrational, then for some value $\alpha \in [0, 1]$, $\forall\epsilon \in \mathbb{R}:\epsilon > 0$, $\exists n \in \mathbb{N} : |\sin(n\beta) - \alpha| < \epsilon$

The modern proof of this result is with the unique ergodicity of $x \to x + \beta$ on $S^1 = \mathbb{R}/\mathbb{Z}$ for $\beta$ irrational. — anomaly, Jan 04 '17 at 00:28

score 0 · Answer 1 · answered Jan 04 '17 at 00:25

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Let $\alpha=\sin \theta$ then given $\epsilon$ there is (by a well known theorem) an $n$ such that $n\beta =k\pi+\gamma$ with $0\leq \gamma<\pi$ and $|\gamma-\theta|<\epsilon$ now use continuity of the $\sin $ function.

answered Jan 04 '17 at 00:25

Rene Schipperus

39,526

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It sounds like someone doesnt understand my proof. – Rene Schipperus Jan 04 '17 at 00:37
Did you mean $n\beta = 2k\pi + \gamma$, and also, what is the name of the theorem, or do you have a link to it? – Sully Chen Jan 04 '17 at 05:20

score 0 · Answer 2 · answered Jan 04 '17 at 05:38

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I found out that this principle is demonstrated in Kronecker's Approximation Theorem.

answered Jan 04 '17 at 05:38

Sully Chen

149
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Prove that $\sin (n\beta)$ can get arbitrarily close to any value in [0, 1] if $\frac{\beta}{\pi}$ is irrational

2 Answers2