Timeline for Is the existence of $\lim_{n\to\infty}\cos(n!\pi x)$ for given arbitrary irrational $x$ an open problem?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 29, 2016 at 17:32 | vote | accept | CommunityBot | moved from User.Id=14319 by developer User.Id=322028 | |
| Dec 26, 2016 at 21:05 | answer | added | GH from MO | timeline score: 5 | |
| Dec 26, 2016 at 19:20 | comment | added | user21349 | @Jack: I don't see much connection between the linked question and mine. Your question seems to differ from that one in two ways: (1) you changed sine to cosine, and (2) you seem to be asking for an algorithm or decision procedure that determines convergence for a given $x$. What is the motivation for changing both of these things at the same time? Re #2, it's not clear to me what it would mean to have a decision procedure that took an "arbitrary irrational" as an input, since only countably many irrationals can even be described. This question seems to lack motivation and research effort. | |
| Dec 26, 2016 at 17:59 | history | edited | user14319 | CC BY-SA 3.0 |
added 29 characters in body; edited title
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| Dec 26, 2016 at 17:57 | comment | added | user14319 | @BenCrowell: Thanks for your comment. The answers you mentioned addressed the existence of irrational $x$ that $\lim_{n\to\infty}\sin(n!\pi x)=0$ while I'm interested in the existence of the limit itself for any irrational. Due to my ignorance, I don't see much connection between the linked question and mine. | |
| Dec 26, 2016 at 15:31 | comment | added | user21349 | @Jack: For the version of the problem using the sine function, we have answers by Petya and Ashutosh demonstrating a particular technique. Have you tried applying that technique to the version using the cosine? If so, then tell us what happened. If not, then it seems premature to post the question. | |
| Dec 26, 2016 at 15:29 | comment | added | Anthony Quas | For any $a\in[-1,1]$, there exist values of x such that $cos(n!x)\to a$. This argument is well known in ergodic theory and is essentially due to Pollington. The point is that $n!$ is a lacunary sequence (I.e. The ratio between successive terms is bounded away from 1) and in this case converges to infinity. | |
| Dec 26, 2016 at 6:21 | comment | added | Vladimir Dotsenko | @T.Amdeberhan depends what $y$ is; in full generality it is not true, of course. (If $y_n=\frac{\pi}2+\pi n$, then the limit of $\cos(y_n)$ exists and is $0$, while $\sin(y_n)$ obviously has no limit.) | |
| Dec 26, 2016 at 3:10 | comment | added | T. Amdeberhan | @VladimirDotsenko: The limit for $\cos(y)$ exists iff the same holds for $\sin(y)$. Do you agree? | |
| Dec 26, 2016 at 2:50 | comment | added | Vladimir Dotsenko | However, $(\sin(n!\pi x))^2+(\cos(n!\pi x))^2=1$, which relates those two sequences quite closely. | |
| Dec 26, 2016 at 2:50 | comment | added | Vladimir Dotsenko | @T.Amdeberhan $\sin(\frac{\pi}2-n!\pi x)$ potentially may behave slightly different from $\sin(n!\pi x)$, right? So maybe your suggestion is not fully clear. | |
| Dec 26, 2016 at 2:01 | comment | added | T. Amdeberhan | You can translate one into the other via $\cos(y)=\sin(\frac{\pi}2-y)$. So, follow the other MO question on $\sin(n!\pi x)$. | |
| Dec 26, 2016 at 1:32 | history | asked | user14319 | CC BY-SA 3.0 |