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Is $\{n \sin n | n \in \mathbb{N}\}$ dense on the real line?

If so, is $\{n^p \sin n | n \in \mathbb{N}\}$ dense for all $p>0$?

This seems much harder than showing that $\sin n$ is dense on [-1,1], which is easy to show.

EDIT: This seems a bit harder than the following related problem, which might give some insight:

When is $\{n^p [ \sqrt{2} n ] | n \in \mathbb{N}\}$ dense on the real line, where $[\cdot]$ is the fractional part of the expression?

I am thinking that there should be some probabilistic argument for these things.

EDIT 2:

Ok, so plotting a histogram over $n \sin n$ is similar to plotting $n \sin(2\pi X)$ where $X$ is a uniform distribution on $[-1,1].$ This is not surprising, since $n$ mod $2\pi$ is distributed uniformly on $[0,2\pi].$

Now, the pdf of $\sin(2\pi X)$ is given by $f(x)=\frac{2}{\pi \sqrt{1-x^2}}$ in $(-1,1)$ and 0 outside this set.

The pdf for $n \sin(2\pi X)$ is $g_n(x)=\sum_{k=1}^n \frac{1}{nk} f(x/k)$ so the limit density is what we get when $n \rightarrow \infty.$ (This integrates to 1 over the real line).

Now, it should be straightforward to show that for any interval $[a,b],$ $\int_a^b g_n(x) dx \rightarrow 0$ as $n \rightarrow \infty.$

Thus, the series $g_n$ is "too flat" to be able to accumulate positive probability anywhere. (The gaussian distribution on the other hand, has positive integral on every interval).

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    A related question: [Does $|n^2 \cos n|$ diverge to $+\infty$?](http://math.stackexchange.com/q/126583)2012-10-25
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    It's unlikely that it's true for all $p\gt0$, since for $p\gt1$ the "probability" of hitting a short interval is proportional to $n^{-p}$, and the total probability of being hit is bounded from above by the sum of the probabilities, which for $p\gt1$ converges to a finite limit $\lt1$ for sufficiently small intervals.2012-10-25
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    @Camilo: $\{q\sin q\mid q\in\mathbb Q\}$ is dense simply because $x\sin x$ is continuous.2012-10-25
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    It feels like we need effective statements about the distribution of {sin n|0 <= n < N}, at least looking at the problem as finding n given epsilon to satisfy: |sin n - x/n| < epsilon/n2012-10-25
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    It might be easier to investigate the situation with a triangle wave of period $2\pi$. That should give the same end result, since if we take any bounded interval on the $y$ axis, whether it is hit for large enough $n$ depends only the value of the sine very close to its zero crossings where a triangle wave approximates it well.2012-10-25
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    It is not obvious: looking at $n$ from 0 to 10,000,000 it seems only 17 values of $n\sin n$ lie in the interval $[-1,1]$ and only three of these have $n$ above $3195$2012-10-25
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    Henry: Well, this is not surprising, since $\ln 10000000 \approx 17$...2012-10-26
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    @Paxinum, so are you saying the statement doesn't hold? Do you think we could prove for a specific natural number that this sequence avoids it?2012-10-28
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    @sperner: No, he's saying that 17 is what one would expect even if it is dense. If we replace $n\sin n$ by "a random value in $[-n,n]$", then about $\ln N$ of the first $N$ values would be in $[-1,1]$, _and_ the sequence would almost surely be dense in $\mathbb R$ (because $\ln N\to\infty$). So Henry's empirical finding does not indicate that $n\sin n$ is _not_ dense.2012-10-28
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    Exactly. I have done some more calculations, and the limit distribution has limit 0. However, the number of actual points in any interval goes to infinity, probabilistically speaking.2012-10-28

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