A sequence is eventually periodic if we can drop a finite number of terms from the beginning and make it periodic.


I think this is periodic since the function is periodic and it seems to converge to 0 by iterations. The thing that confused me is that there are infinite numbers in the domain in the period of $\sin(k)$ since it's continuous. So I can't really assume that if a function is periodic -> sequence is periodic?

  • $\begingroup$ Related post: math.stackexchange.com/questions/4764/… And other posts linked there might be of interest in relation to this problem, too. $\endgroup$ – Martin Sleziak Nov 3 '15 at 13:05
  • $\begingroup$ If I understand correctly, you mean there exist two integers $u,v$ where $u-v$ is a multiple of $2\pi$? It does not seem true to me though. $\endgroup$ – cr001 Nov 3 '15 at 13:13

If you have $\sin(k+n)=\sin k$, then1 either $n$ or $2k+n$ is an integer multiple of $\pi$. Can this happen for integers $k,n\ge1$? (Knowing that $\pi$ is irrational2 might be useful.)

You should arrive to the conclusion that the sequence is not periodic.

1We know that $\sin x=\sin y$ holds if and only if $x=y+2z\pi$ or $x=\pi-y+2z\pi$ for some $z\in\mathbb Z$.

2 Wikipedia: Proof that $\pi$ is irrational

  • $\begingroup$ $sin(k+2\pi)$ will be the same as $sin(k)$ but since, we can only work with integers this is not possible, so there is no such period? So not periodic even though the function itself is periodic? $\endgroup$ – GRS Nov 3 '15 at 13:19
  • $\begingroup$ I think this is the correct solution and it does not matter whether the set is dense or not. $\endgroup$ – cr001 Nov 3 '15 at 13:19
  • $\begingroup$ @Nick It is true that $\sin(k+2\pi)=\sin k$. But you should also keep in mind that $\sin k =\sin (\pi -k) = \sin (3\pi-k)=...$. $\endgroup$ – Martin Sleziak Nov 3 '15 at 13:22
  • $\begingroup$ Thank you very much. So I need to show that since I can't make this period with integers only (because of irrationality), there is no such period and thus the sequence is not periodic. $\endgroup$ – GRS Nov 3 '15 at 13:24
  • $\begingroup$ @Nick What you wrote is correct. (The sequence is not periodic.) I have added this to my answer. (So that this fact is mentioned there explicitly.) $\endgroup$ – Martin Sleziak Nov 3 '15 at 13:26

It cannot be eventually periodic since it can be shown the set of values of $ \sin k\ (k\in \mathbf N)$ is dense in $[-1,1]$, and not equal to the whole of $[-1,1]$.

  • $\begingroup$ You probably mean $[-1,1]$ instead of $[0,1]$. math.stackexchange.com/questions/4764/… $\endgroup$ – Martin Sleziak Nov 3 '15 at 13:15
  • $\begingroup$ Actually, this is not an answer. $\endgroup$ – user147263 Nov 3 '15 at 13:17
  • $\begingroup$ @Martin Sleziak: You're right. One of my freudian slips… Thanks for pointing it. $\endgroup$ – Bernard Nov 3 '15 at 13:23
  • $\begingroup$ @Normal Human: It was implicit. I've added the dots. $\endgroup$ – Bernard Nov 3 '15 at 13:28
  • $\begingroup$ Still not an answer. "It's not periodic because it can be shown that it's not periodic" I can answer a lot of questions here in this fashion. $\endgroup$ – user147263 Nov 3 '15 at 14:03

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