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?

  • $\begingroup$ 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 $C = \int_0^1 |\sin(2 \pi x)| dx$. $\endgroup$ – reuns May 6 '16 at 16:45
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    $\begingroup$ One 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. $\endgroup$ – Ben Grossmann May 6 '16 at 16:51
  • $\begingroup$ @user1952009 Your computations seem to be wrong. $\endgroup$ – Did May 6 '16 at 18:02
  • $\begingroup$ @Did yes I saw that, I need to use $|arcsin'(x)|$ for computing $f_\epsilon(x)$ $\endgroup$ – reuns May 6 '16 at 18:47
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    $\begingroup$ I, 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. $\endgroup$ – John Barber May 11 '20 at 3:58

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:

Sin(n) Histogram

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 $1\leq n\leq {10}^6$.


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