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

  • 1
    $\begingroup$ 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... $\endgroup$ – Pietro Majer Aug 3 '10 at 12:42
  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Aug 3 '10 at 13:19
  • 4
    $\begingroup$ 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. $\endgroup$ – Dan Piponi Aug 3 '10 at 21:47
  • $\begingroup$ @sigfpe That wasn't a Hungarian-drafted problem sheet was it? $\endgroup$ – 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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Pietro Majer Aug 3 '10 at 13:24
  • 1
    $\begingroup$ 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 $\mathbb R / \mathbb Q$. $\endgroup$ – Petya Aug 3 '10 at 15:21
  • $\begingroup$ 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 $\sum 1/i! \sin(n!\pi x)$... $\endgroup$ – Willie Wong Aug 3 '10 at 20:36
  • $\begingroup$ Thank you! I also stared on your comment for a few minutes! But now I understand. $\endgroup$ – 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.

  • $\begingroup$ Does someone know if $e^2$ is in $G$? $\endgroup$ – Ashutosh Nov 22 '13 at 0:43
  • 2
    $\begingroup$ It is an exercise in Rudin's real and complex analysis book that the set of $x$ for which the sequence $sin(n!x)$ converges has measure zero. $\endgroup$ – Venkataramana Dec 26 '16 at 2:36
  • $\begingroup$ Link in the post seems to be dead, here is a snapshot from the Wayback Machine. $\endgroup$ – Martin Sleziak Jul 22 '20 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.