It is a classical result that $$ \limsup_n \sin(n) = 1 $$ Even more, the set $\{\sin(n):n\in\mathbb{N} \}$ is dense in $[-1,1]$. I was wondering if it is possible to say something about the distribution of the sequence in the interval $[-1,1]$. I have no idea if one should expect equidistribution or not. Is there any result about this question?
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show that $n / (2 \pi) - \lfloor n / (2 \pi) \rfloor$ is equidistributed, and consider for $x \in \ ]-1,1[$ : $ \displaystyle f_\epsilon(x) = \lim_{N \to \infty} \frac{# { \ |\sin(n) - x| < \epsilon \ \mid \ 1 \le n \le N}}{N}$, you should get that when $\epsilon \to 0$, $f_\epsilon(x) \to C |\sin(2 \pi x)|$ with $C = \int_0^1 |\sin(2 \pi x)| dx$. – reuns May 06 '16 at 16:45
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2One should not expect equidistribution in this case: the distribution will be "denser" near $1$ and $-1$. We can derive the exact result using equidistribution on the circle and considering the $y$-coordinate. – Ben Grossmann May 06 '16 at 16:51
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@user1952009 Your computations seem to be wrong. – Did May 06 '16 at 18:02
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@Did yes I saw that, I need to use $|arcsin'(x)|$ for computing $f_\epsilon(x)$ – reuns May 06 '16 at 18:47
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why isnt john barber's answer accepted? – mathworker21 May 08 '20 at 17:33
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I did not know how the site worked back then, I have accepted it now. – Felipe Pérez May 09 '20 at 01:27
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1I, likewise, didn't know how to add an image to an answer back then. I have added a plot comparing the computed and predicted pdfs. – John Barber May 11 '20 at 03:58
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Assuming $n \mod 2\pi$ is distributed uniformly on $[0,2\pi]$, we can model the distribution of $\sin(n)$ as having the same distribution as $\sin(\frac{\pi}{2}u)$, where $u$ is uniformly distributed on $[-1,1]$. $u$ thus has the probability density function (pdf) $$ f(u) = \frac{1}{2} $$ on $u\in [-1,1]$ and zero everywhere else. What we want is the pdf $g(x)$ of the quantity $x = \sin(\frac{\pi}{2}u)$. This is given by $$ g(x) = \left|\frac{du}{dx}\right| f(u(x)). $$ Since $u(x) = \frac{2}{\pi} \sin^{-1}(x)$ and $f = \frac{1}{2}$ in the region of interest, this becomes: $$ g(x) = \frac{1}{\pi\sqrt{1 - x^2}}. $$ I've checked this by looking at the distribution of $\sin(n)$ for the first 1,000,000 integers $n$, and I find that it matches this prediction perfectly.
Edited 4 years later to include a comparison plot:
Here the black curve is the function $g(x)$ given above, and the red dots are the computed pdf of $\sin(n)$ obtained by looking at $1\leq n\leq {10}^6$.
John Barber
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