Let $a_n=\cos(a_{n-1}), L=[a_1,a_2,...,a_n,...].$ Is there an $a_0$ such that $L$ is dense in$[-1,1]?$

I've been experimenting with recursive sequences lately and I've come up with this problem:

Let $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$

Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$

I know of $3$ ways of examining whether a set is dense:

$i)$The definition, that is, whether its closure is the set on which it is dense, in our case this means if: $\bar L=[-1,1]$.

ii)$(\forall x \in [-1,1])(\forall \epsilon>0)(\exists b \in L):|x-b|<\epsilon$

$iii)$ $(\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow x$

So far I haven't been able to use these to answer the question. I tried plugging in different values of $a_0$ and see where that leads but I have not found any corresponding promising "pattern" for $a_n$. Any ideas on how to approach this?

Solution 1:

Ok, if you study recursive sequences, then you probably heard about "Lamere Ladder".

According to Lamere Ladder for $\cos(x)$ (which is also a contraction because of MVT), this function has a fixed (stationary) point. So, regardless of $a_0$, $a_1$ will end in between $[-1, 1]$ and from there on, the sequence $\{a_n\}$ will tend to the fixed point of $\cos(x)$. Which makes $L$ a converging sequence, so $L$ can't be dense, because the only point satisfying ii) (in your question) is its limit.

On another note, Kronecker's approximation theorem is quite an useful tool too. For example $\{n+ m \cdot 2 \cdot \pi \space | \space m,n \in \mathbb{Z} \}$ is dense on $\mathbb{R}$ and $cos(x)$ is a continuous function, making $\{cos(n)\}_{n \in \mathbb{Z}}$ dense on $[-1,1]$.