Repeated roots: $f^{1992}(x) = x$ for $f(x) = 4x(1-x)$ Given a function $f : [ 0 , 1 ] \to \ [ 0 , 1 ]$ defined by $f(x) = 4x(1-x)$ for all $x \in [ 0 , 1 ]$. How many distinct roots does the equation $f^{1992}(x) = x$ has. ($f^n(x) = f\left(f^{n-1}(x)\right)$)
I've already know that $\deg\left(f^{1992}(x)\right) = 2^{1992}$. So, it has $1992$ roots. But, I don't know how to deal with repeated roots. Can anyone give me some hint or a solution to this problem? Thanks in advance!
 A: This is a famous dynamic system, the $r = 4$ case of the logistic map.
In the above wiki page, a solution is given to this problem, and a link to the related OEIS sequence, and even a simple formula: $\frac{2^k - 2}k$. Note however that this is the number of "primitive" cycles.
You may gather informations from that page and use them to solve your problem.
A: It is possible to solve $f_n(x) = x$, $ \ \ f_n(x)=f(f_{n-1}(x))$ in a general way, for any integer $n$:
$ $
$$f_{n+1}(x)=4 f_{n}(x)(1-f_{n}(x))$$
Solving recursively for $f_n(x)$:
$$f_{n}(x)=\frac{1}{2}-\frac{1}{2}\cos(2^n c_x)$$
$$f_1(x)=\frac{1}{2}-\frac{1}{2}\cos(2 c_x)=4x(1-x).$$
solving for $c_x$ and $x\in [0,1]$:
$$c_x=\pm\frac{1}{2} \arccos \big(1+8x(1-x)\big)+2k\pi $$
$$f_{n}(x)=\frac{1}{2}\left(1-\cos\left(2^{n-1}\arccos\left(1+8x\left(x-1\right)\right)\right)\right)$$
$$f_{n}(x)=\sin^2\left(2^{n-2}\arccos\left(1+8x\left(x-1\right)\right)\right).$$
For $\theta \in[-\frac{\pi}{2},\frac{\pi}{2}]$, $x=\sin(\theta)^2$
$$\arccos\left(1+8\sin(\theta)^2\left(\sin(\theta)^2-1\right)\right)=\arccos\left(\cos(4 \theta)\right)=4\theta$$
therefore we have that $f_n(x)=x$ is equivalent to $\sin(\theta)^2=\sin(2^n \theta)^2$
$$\theta=\left\{\frac{\pm  \pi+4k }{1+2^n},\frac{2  \pi(\pm 1+2 k)}{1+2^n},\frac{\pm 4  \pi k}{\pm 1+2^n},\frac{\pi(1 \pm 4 k)}{\pm (1-2^n) }\right\}$$
$k$ such that
$$-\frac{\pi}{2}\leq\left\{\frac{\pm  \pi+4k }{1+2^n},\frac{2  \pi(\pm 1+2 k)}{1+2^n},\frac{\pm 4  \pi k}{\pm 1+2^n},\frac{\pi(1 \pm 4 k)}{\pm (1-2^n) }\right\}\leq \frac{\pi}{2}$$
