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Hi, I'm interested in the behaviour of the sequence $(\sin(n!\pi x))$, when $x$ is irrational, as $n$ tends to infinity.

1) Is the sequence dense in $(-1,1)$?


2) Is it possible that for some irrational $x$, $\sin(n!\pi x)$ tends to $0$ as $n$ tends to infinity?

Any reference would be appreciated,

Thank you

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The case $x=e$ is a popular riddle. A variation is $n\sin(2\pi en!)\to2\pi$ (whence $e\notin\mathbb{Q}$). Apart from these elementary cases, it sounds like a question immediately going into wild open problems... – Pietro Majer Aug 3 '10 at 12:42
However, I guess it should not be difficult to buid an x for which 1) holds. This immediately produces a dense set $x+\mathbb{Q}$; and in fact I imagine that one can also prove that 1) holds generically. – Pietro Majer Aug 3 '10 at 13:19
Homework question. Seriously! When I was an undergraduate at Cambridge this was on one of the first problem sheets in the first analysis course. I can't help but think it was intended to scare people away. It generated much discussion and I think the consensus was that there are no nice answers to this question. – Dan Piponi Aug 3 '10 at 21:47
@sigfpe That wasn't a Hungarian-drafted problem sheet was it? – Yemon Choi Aug 3 '10 at 23:26

For $x=e=\sum 1/{{i}!}$, the sequence $sin(n!\pi x)$ tends to zero since the sequence of fractional parts {$n!x$} tends to zero. Hence generally answer on your question is negative.

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also note that the sequence $\sin(\pi xn!)$ only changes by finitely many terms if a rational is added to $x$; so in particular 2) holds in a dense set. – Pietro Majer Aug 3 '10 at 13:24
Moreover, it easily generalizable to $\sum_{i\in I}1/i!$, where $I$ is any infinite subset of $\mathbb N$. That gives us a continuum points in $\mathbb R / \mathbb Q$. – Petya Aug 3 '10 at 15:21
Petya, I hope you don't mind my minor edit. I stared at your post for a full five minutes before I understood that you didn't mean to define $e$ as $\sum 1/i! \sin(n!\pi x)$... – Willie Wong Aug 3 '10 at 20:36
Thank you! I also stared on your comment for a few minutes! But now I understand. – Petya Aug 4 '10 at 22:01

Denote by G the set of all x for which $sin(n! \pi x)$ approaches 0 as n approaches infinity. Every real number $0 < x < 1$ can be uniquely written as $ \sum_{n \ge 2} \frac{x(n)}{n!}$ where $0< x(n) < n$. From this it can be immediately seen that $x \in G$ iff $\frac{x(n)}{n}$ approaches 0 or 1. This immediately shows that there are continuum many points of G in every interval.

It can also be shown that G is an F-sigma-delta (countable union of countable intersections of open sets) additive subgroup of reals but it's neither G-delta nor F-sigma. One may iterate this construction to obtain additive groups of reals at arbitrarily high finite levels of Borel hierarchy in the following way: Let $G_0 = G$. Let $G_{n+1}$ be the set of all x such that the fractional part of $(n!x)$ converges to a point in $G_n$. Then $G_n$'s form an increasing chain of additive subgroups of reals and their union is a Borel additive subgroup of reals which is not at any finite level of Borel hierarchy.

I wrote a note on this here.

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Does someone know if $e^2$ is in $G$? – Ashutosh Nov 22 '13 at 0:43

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