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I encountered a difficult problem in Problems in Mathematical Analysis - Piotr Biler. It was problem 2.92 read as follows.

Show that if for reals $a_1, a_2, \cdots, a_k$, $$\displaystyle \lim_{n \rightarrow \infty} \sin \left (na_1 \right )\sin \left (na_2 \right )\cdots \sin \left (na_k \right ) = 0$$ then at least one of the $a_j$ is a multiple of $\pi$.

I referred to the answers and found I was left with a hint: consider separately the factors with $\frac{a_j}{\pi}$ rational or irrational. In the second case use the equipartition theorem.

However I don't know what is addressed as equipartition theorem and it occurred to me that $\sin \left( \alpha n \right)$ is almost always dense in the interval $\left ( -1 ,1 \right)$ with few exceptions. But sine is nonlinear and the distribution is therefore not uniform. I got dizzy, could anyone elaborate on this problem please?

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GendoTendoLendo
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  • I don't think there is a sensible notion of "uniformness" here, since there are no random variables. I've also never heard of the term "equipartition theorem". If you follow this argument then I think you can show that if any of the $a_j/\pi$ are irrational, then your sequence is dense in $(-1,1)$. That would imply that it does not tend to zero. – preferred_anon Oct 21 '20 at 15:39
  • If $a_j/\pi$ is rational but not an integer, then I think $\sin(na_j)$ is a periodic sequence with period $2\text{denom}(a_j)$ in lowest terms. For example $\sin(n\pi/2)$ is the sequence $0, 1, 0, -1, ...$. If you take the product of several such sequences, then that will have a period which is divisible by all the periods of the factors. Such a sequence also cannot tend to zero. – preferred_anon Oct 21 '20 at 15:42
  • @preferred_anon Thanks for your answer. I have proved that sine is dense in $(-1, 1)$, but I'm worried about whether those products, dense respectively, are remaining dense when they are put together. – GendoTendoLendo Oct 21 '20 at 15:58
  • I think that they are, but I haven't proved it. I expect, for example, that given an $x \in [-1,1]$ and $\epsilon> 0$, that we can find an infinite subset of $\mathbb{N}$ which has $\sin(a_1 i) \in (x-\epsilon, x+\epsilon)$ for all $i$ in the subset. Then I think you should be able to take an infinite subset of that where $\sin(a_2 i)$ is near $1$, or also in $(x-\epsilon, x + \epsilon$), I'm not sure. If you get one for all the $a_j$ then your sequence is dense in $[-1,1]$. – preferred_anon Oct 21 '20 at 16:14

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