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how to prove $\sup \{ \sin n \mid n\in \mathbb N \} =1$

Michael Hardy
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Leitingok
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3 Answers3

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If $\alpha$ is an irrational number, then $\{ \operatorname{fractional part}(n\alpha) : n \in\mathbb{N} \}$ is dense in $(0,1)$.

Let $\alpha = 1/(2\pi)$. Then for every open interval about $1/4$, no matter how small, there is some $n$ such that $\operatorname{fractional part}(n\alpha)$ is in that interval. When $n/(2\pi)$ is close to $1/4$, then $n$ is close to $\pi/2$, so (since the sine function is continuous) $\sin n$ is close to $1$.

How does one prove the assertion in my first paragraph above? I think Dym & McKean's book on Fourier transforms gives a proof using Fourier transforms.

There's also the question of how to prove that $\pi$, and hence $1/(2\pi)$, is irrational. http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational

In epsilon-delta language: Let $\varepsilon>0$. Since sine is continuous, there exists $\delta>0$ such that if $\pi/2-\delta<x<\pi/2+\delta$, then $1 \ge \sin x > 1-\varepsilon$ (the first inequality does not come from continuity, but from known behavior of the sine function). Since $\{ \operatorname{fractional part}(n\alpha) : n \in\mathbb{N} \}$ is dense in $(0,1)$, there exists $n$ such that $n/(2\pi)$ is between $1/4 \pm \delta/(2\pi)$. So $n$ is between $\pi/2\pm\delta$ and so $\sin n > 1-\varepsilon$.

So for every $\varepsilon>0$, the desired sup is $>1-\varepsilon$ but $\le 1$.

Michael Hardy
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Edit: Based on the suggestion of one of the comments, and the fact that this answer is bogus, I'd like to point out here in black and white that this answer is terrible. If anyone is searching for an answer to this question, look at the other ones. This one should never have been accepted.

Not a proof exactly, but remember $\forall_{\epsilon>0}$ you can find an element $\frac{p}{q}\in\mathbb{Q}$ such that $|\frac{p}{q}-\frac{\pi}{2}|&lt\epsilon$

So we can also find an element in $\mathbb{N}$ that is arbitrarily close to $\frac{\pi}{2}\mod 2\pi$. Since sine is continuous, if we find a sequence of natural numbers $a_n$ that approaches $\frac{\pi}{2}\mod 2\pi$ as a limit, then $\sin{a_n}$ approaches 1. Thus $$\sup \{ \sin n |\ n\in \mathbb \{a_{m\in\mathbb{N}}\} \} =1$$ and since $\{a_{m\in\mathbb{N}}\}\subset\mathbb{N}$ then the supremum over all of N is at least as big. Since sine is bounded from above by 1, we are done.

Lots of handwaving, but you should be able to collect the intermediate bits.

Tim Seguine
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  • how to find $n\in \mathbb N$ arbitrarily close to $\frac{π}{2}\mod{2π}?$ – Leitingok Sep 11 '11 at 12:17
  • That's the part that was a little shaky. It's possible but I need to think about a better way to explain it. – Tim Seguine Sep 11 '11 at 12:36
  • I find out about that. thanks. – Leitingok Sep 11 '11 at 12:41
  • @Tim: It's not just a little shaky, I'm afraid. It's a crucial gap in the argument. All you can deduce is that $p$ is within $q\epsilon$ of $\frac{\pi}{2}$ mod $2\pi$. And you have not shown any kind of bound on $q$. – TonyK Sep 11 '11 at 12:53
  • @TonyK Yeah, I know. I read back through it and I realized my reasoning was wrong. You can find an integer with the property that is needed, but my reasoning was bull. q is of course unbounded so the bound was useless. – Tim Seguine Sep 11 '11 at 13:10
  • @Leitingok Michael Hardy's answer is better than mine. – Tim Seguine Sep 12 '11 at 12:57
  • You could add an explicit mention of the fact that there is a serious problem with this reasoning, at the beginning of your post. – Did Feb 10 '12 at 06:10
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You can use Dirichlet's theorem. Given $\alpha\in \mathbb R$ and a positive integer $N$ there integers $p,q$ with $1\leq q \leq N$ such that $$ |q\alpha-p|< \frac1N. $$ This can be easily proved using pigeonholing. See

https://proofwiki.org/wiki/Dirichlet%27s_Approximation_Theorem

Francesco Di Plinio
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