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Let $P$ be the set of all prime numbers. Is $\sin(P)$ dense is $[-1,1]?$ How could we approach such a problem?

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    +1 Cool question. What have you tried? – draks ... Feb 15 '12 at 22:40
  • On distributional principles it sure seems true. Very cool question. – Brian B Feb 15 '12 at 22:53
  • @draks Well, I know the proof that $\sin(\mathbb N)$ is dense. Actually, I just remembered it from my first course in analysis and thought about this problem. The proof of the case with $\mathbb N$ doesn't seem to generalize, and it would be strange if it did I think. But I have simply no idea how to find another approach. –  Feb 15 '12 at 22:57
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    I don't know the $\sin(\mathbb{N})$ proof (can you provide a link?), but would it help to think of $\mathbb{N}$ as sum of all primes, semi-primes, k-almost primes? – draks ... Feb 15 '12 at 23:09
  • @draks There's a question about it on this site. There's a link to a paper with a proof there but I can't access it from my house so I'm not sure what's in it. –  Feb 15 '12 at 23:12
  • Maybe this could help... there it is proven, that $\sum \cos p_n$ does oscillate/diverge. What would happen if $\sin(P)$ is not dense? I think it would clearly diverge. What do you think? – draks ... Feb 15 '12 at 23:35
  • @draks I only see a proof (a nice one!) that it would follow from the TPC there. Am I missing something? –  Feb 15 '12 at 23:56
  • The Green-Tao theorem guarantees infinitely long arithmetic progressions within the set $P$ of primes. However, this fact is not enough yet to conclude that the set $sin(P)$ is dense in $[-1,1]$; and I don't see any extra argument to establish density from the Green-Tao theorem. – Jonas Kibelbek Feb 16 '12 at 01:35
  • @Jonas, arbitrarily long arithmetic progressions. It's not hard to show that infinitely long arithmetic progressions are impossible. – Gerry Myerson Feb 16 '12 at 04:05
  • @Gerry: Thanks for correcting that. I can't believe I wrote that and reread it without noticing. Of course, the image under sine of any infinitely long arithmetic progression of integers is dense in [-1,1] by the same arguments that work for $\sin(\mathbb{Z})$. – Jonas Kibelbek Feb 16 '12 at 05:59

1 Answers1

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According to the Wikipedia article about the discrepancy of a sequence:

The sequence of all multiples of an irrational $\alpha$ by successive prime numbers, $2 \alpha$, $3 \alpha$, $5 \alpha$, $7 \alpha$, $11 \alpha$, ... is equidistributed modulo 1. This is a famous theorem of analytic number theory, proved by I. M. Vinogradov in 1935.

With $\alpha = \frac{1}{2 \pi}$, this implies that $P$ is equidistributed modulo $2 \pi$. Using this, and the continuity of the sine function, I think it is straightforward to show that $\sin(P)$ is dense in $[-1,1]$ (although not equidistributed).

Dan Brumleve
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    Also, the distribution of $sin(P)$ can be derived from the fact of the equidistribution of $P$ modulo $2\pi$. – Michael Lugo Feb 16 '12 at 05:37
  • Dan, could you post a link or a reference to Vinogradov's proof? –  Feb 16 '12 at 08:11
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    @ymar, I couldn't find one with a search, but here it is claimed to be "a byproduct of [Vinogradov's work on] the odd Goldbach conjecture". It is said to have been proven in 1935 so probably the document can be narrowed down on that basis. – Dan Brumleve Feb 16 '12 at 08:34
  • How about $P^{it}$? Is this dense over the unit circle? – draks ... Jul 16 '12 at 14:40
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    @draks The phase of $P^{it}$ varies extremely slowly for any fixed $t$, so there's no question that it is dense on the unit circle. One only needs $\frac{p_{n+1}}{p_n} \to 1$, which is slightly stronger than Bertrand's postulate and weaker than the Prime Number Theorem. – Erick Wong Jan 06 '13 at 07:40