Consider the function $f(x)=\sin(\cos(x))$. What are the attractive fixpoints for $f(x)$, ie. the set of values of $x$ for which $f \circ f \circ f \circ \cdots \circ f(x) = x$ in the limit?
Let $$F_n(x)= \begin{cases} F_{n-2}(\sin(\cos(x))&\text{ if }n\geq 2\\ \cos(x)&\text{ if }n=1\\ x&\text{ if }n=0 \end{cases}$$
Does $\lim_{k\rightarrow\infty} F_{2k}(x)$ exist?
I'm empirically getting a fixpoint close, but not exactly equal to to $\ln(2)$, but I guess that could just be a fluke.
The fixed point(s) are where $f(x)=x$. They are attractive when $|f'(x)|<1$ (equal to 1 is more complex but not relevant here)
But why is the fixed point near $\ln 2$? $\ln 2$ is the solution of $e^x-2=0$.
Instead of the roots of $f(x)-x$, consider the roots of $g(x)=-\cos(x)+\arcsin (x)$. It turns out that the series expansion of $e^x-2$ and $g$ around 0 are the same for the first few terms.
The first disagreement is at the fourth derivative (1 vs -1).