Consider the logistic map $G(x) = 4x(1-x)$. Let $q_0=0<q_1<q_2<q_3...<q_7$ be the eight points left fixed by $G^3$. Determine which are the two fixed points and which other points are grouped together into period-3 orbits. Hint: Remember that the fixed points are $0$ and $3/4$.
I'm able to answer a similar question for $G^2$, but I've got no idea how to answer this question for $G^3$. I know that by my calculations $G^2(x) =4[4x(1-x)][1-4x(1-x)]$ and $G^3(x)=4[4(4x(1-x))(1-4x(1-x)][1-4(4x(1-x))(1-x)].$ How can I use this equation to determine which are the fixed points and which other points are grouped together into period-3 orbits?
Solve: $G^3(x)=4[4(4x(1-x))(1-4x(1-x)][1-4(4x(1-x))(1-x)] = x$ using Wolfram Alpha.
( Solve[64 (1 - x) x (1 - 4 (1 - x) x) (1 - 16 (1 - x) x (1 - 4 (1 - x) x)) == x, {x}] ) should work .
The two fixed points are $0$ and $\frac{3}{4} $ , There are in fact two three cycles which you can find by iterating the other roots using $G(x)$ .