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I have seen that the set of all cluster point of the sequence $\{\sin(n)\}$ or $\{\cos(n)\}$ is $[-1,1]$ and the proof of this problem depends on the fact that the set $\{n+2 \pi k \mid n,k \text{ are integers}\}$ is dense in $\Bbb{R}$. There is another set $\{m/2^n \mid m,n \text{ are integers}\}$ which is also dense in $\Bbb{R}$. So is it possible to determine the cluster set of $\{\cos(2^n)\}$?

Theo Bendit
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Rintu93
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    It is probably much harder, since the sequence being non-periodic is equivalent to $\pi$ being irrational. The problem is equivalent to the following: determine all the real numbers $x$ between $0$ and $1$ such that for every $N > 0$, if $x=0.x_1 \cdots x_N \cdots$ in binary, then $x_1 \cdots x_N$ appears infinitely many times in the binary expansion of $1/\pi$. – Aphelli Sep 16 '19 at 08:22

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The sequence satisfies $u_0=\cos(1)$ and $u_{n+1}=2u_n^2-1$, hence $v_n=2u_n$ satisfies $v_{n+1}=f(v_n)$ where $f:[-2,2]\to [-2,2], x\mapsto x^2-2$.

By the homeomorphism $\tau:[0,1]\to [-2,2], x\mapsto -4x+2$, $f$ is topologically conjugate to $g:[0,1]\to [0,1], x\mapsto 4x(1-x)$. (meaning $\tau \circ g = f\circ \tau$).

$g$ is known as the logistic map with parameter $4$. It is well-known that $g$ is choatic on $[0,1]$, hence $f$ is chaotic on $[-2,2]$. This implies that the set of periodic points of $f$ is dense, hence $f$ has dense orbits.

In other words, $(v_n)$ is dense in $[-2,2]$, hence $(u_n)$ is dense in $[-1,1]$, i.e its cluster set is $[-1,1]$.

Gabriel Romon
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    Do you ever use any properties about $u_0$ here? As far as I can tell, this argument is not complete - the fact that $x\mapsto 4x(1-x)$ is chaotic does not imply that any particular orbit is dense - and there are even orbits which are aperiodic, but fail to have a dense image. All you can say is that almost every starting point has a dense orbit - and I don't think it's known whether $u_0$ is one of the rare exceptions. (In fact, I suspect that this question is open, contingent open questions like whether $1/\pi$ is a normal number in binary) – Milo Brandt Oct 14 '19 at 20:47
  • @MiloBrandt Let me think about it, I'll get back to you. – Gabriel Romon Oct 16 '19 at 09:12