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Or put differently, does

$$\lim_{n \to \infty}\big(\max \{\sin 1, \sin 2, \ldots ,\sin n\}\big) = 1?$$

My intuition says yes, but how can one prove this?

Diego Silvera
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Little miss sunshine
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    @Easy's response answers your question. As an aside, the values of $\sin(n)$ are dense in $[-1,1]$. This means you can arbitrarily approximate any point in this interval by choosing the right sequence of integers. – muzzlator Mar 26 '13 at 03:57
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    No, Easy's answer doesn't answer the question, because it is not clear how density proves that $\sin n$ is dense. It is not enough to know that $p/q$ can be made arbitrarily close to $\pi/2$. @muzzlator – Thomas Andrews Mar 26 '13 at 04:03
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    In fact, we can prove this by a simple application of the pigeonhole principle. – Sangchul Lee Mar 26 '13 at 05:08
  • @sos440 Could you show us how in an answer? – Julien Mar 26 '13 at 05:16
  • @julien, I also want to post a complete answer, but I'm outside and iPad is a terrible tool when it comes to writing an answer. I will post my answer as soon as possible. – Sangchul Lee Mar 26 '13 at 05:22
  • @sos440 Is my answer approximately what you were referring to? – Mark McClure Mar 26 '13 at 05:33
  • @MarkMcClure, that's exactly what I was going to figure out, with some calibration. Thanks! – Sangchul Lee Mar 26 '13 at 05:37
  • @ThomasAndrews Ah, I didn't actually read Easy's answer properly. I just wanted to make my comment – muzzlator Mar 26 '13 at 06:29
  • of course, if $\pi$ were rational the answer to the question would be "no" :-) – mau Mar 26 '13 at 14:13

3 Answers3

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If $\alpha$ is any irrational number, then $\{a+b\alpha: a,b\in\mathbb Z\}$ is dense in $\mathbb R$.

In particular, with $\alpha=2\pi$, we can find $a+2b\pi$ which is arbitrarily close to $\frac{\pi}2$, and thus $\sin a$ can be made arbitrarily close to $1$.

The above claim, of course, requires proof, but it is true.

The key is that if you take the continued fraction expansion of $\alpha$ and write the $m$th convergent as $p_m/q_m$ then $|p_m-\alpha q_m|<\frac{1}{2q_{m-1}}$. Then given any $x\in\mathbb R$, define $$d_m = \left\lfloor \frac{x}{|p_m-\alpha q_m|}\right\rfloor$$. Then $$d_m|p_m-\alpha q_m| < x < (d_m+1)|p_m-\alpha q_m|$$ But the difference between the right and left side is less than $\frac{1}{2q_{m-1}}$, and we're done.

As Julien commented, you actually need $a>0$. That's not a hard condition to resolve. We can just pick even convergents, so that $0<p_{2m}-\alpha q_{2m}<\frac{1}{2q_{2m-1}}$. Then do the above computation without the absolute values. Ultimately, your $n=d_{2m}p_{2m}$.

Thomas Andrews
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  • I think you need $a\geq 0$, given the question. But $\mathbb{N}-\alpha\mathbb{N}$ is dense for $\alpha$ irrational, so that's fine. But I think you really need this stronger fact. Excuse me if I'm wrong. – Julien Mar 26 '13 at 04:22
  • Ah, yes, in which case, my argument can be done with the even convergents of the continued fraction, which always gives $0<p_{2n}-\alpha q_{2n}<\frac{1}{2q_{2n-1}}$. @julien – Thomas Andrews Mar 26 '13 at 04:29
  • Interesting, +1. I had never seen an explicit proof of this fact with continued fractions. Yet fairly natural. – Julien Mar 26 '13 at 04:38
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    Just to make sure I got it right, the first two values I got were $d_2p_2=190$ and $d_4p_4=3872$. $\sin 3872 = 0.999916207545327$. – Thomas Andrews Mar 26 '13 at 04:46
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    The next value is $d_6p_6=18498340$, and $\sin 18498340 = 0.999999999409637$. – Thomas Andrews Mar 26 '13 at 04:49
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In fact, the magic element of the proof is Hurwitz's theorem : there exists a (universal) constant $c$ such that for every irrational $\zeta$ there are infinitely many rational approximants $\frac{m}{n}$ with $\left|\zeta-\dfrac{m}{n}\right| \lt \dfrac{c}{n^2}$; multiplying this through by $n$ we get that there are infinitely many $\frac{m}{n}$ with $\left|n\zeta-m\right|\lt\frac{c}{n}$. (More accurately, this needs a slight modification of Hurwitz's theorem which says there are infinitely many such approximants with $n$ odd - but this is actually an easy consequence of the usual proof of the theorem via continued fractions.) Now, apply this with $\zeta=\frac{\pi}{2}$; we then get that the integer $m$ is less than $\frac{c}{n}$ away from an odd multiple of $\pi/2$; if $\sin m$\lt 0 (i.e., if we have one of the 'wrong' multiples) then we can just replace $m$ with $-m$ to get a positive value for sin. Finally, this approximation lets us bound the value of $\sin m$ from below by $1-\frac{c}{n}$.

Steven Stadnicki
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Here is a proof of denseness of the sequence $k\mod\pi$ in the interval $[0,\pi]$ using the pigeon hole prinicple - possibly, the one alluded to in the comments above.

Let $\varepsilon>0$ and choose $n\in{\mathbb N}$ such that $\pi/n<\varepsilon$. By the pigeon hole principle, at least one of the $n$ intervals in the set $\cal I$ defined by

$${\cal I}=\{[m/n,(m+1)/n]\}_{m=0}^{n-1}$$

must contain two of the first $n+1$ terms from the sequence $k\mod\pi$. Let's call those terms $i\mod\pi$ and $j\mod\pi$. We then have $(i-j)\mod\pi<1/n$ so that each interval in $\cal I$ contains a number of the form $k(i-j)\mod\pi$.

Mark McClure
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