Let $ \ x_1 <x_2 < ... < x_8 \ $ be the eight fixed points of $ \ G^3(x) \ $ where $ \ G(x)= 4x(1−x) \ $.
Clearly $ \ x_1 = 0 \ $. Then,
$ (a) \ $ For which $ \ i \ $ is $ \ x_i = \frac{3}{4} \ $?
$(b) $ Find The Group of remaining six points into two orbits of three points each.
Answer:
$ G(x)=4x(1-x) \\ \Rightarrow G^2(x)=G(G(x))=G(4x(1-x))=16x(1-x)[1-4x(1-x)] \\ \Rightarrow G^3(x)=64x(1-x)[1-4x(1-x)] * (1-16x(1-x)[1-4x(1-x)]) \\ $
The fixed points of $ G (x) \ $ are $ \ x_1=0 \ $ and $ \ x=1 \ $
The fixed points of $ G^2(x) \ $ are $ \ x_1=0 \ $ , $ x=1 \ $ , $ x=\frac{1}{2} \ $ , $ x=\frac{1}{2} \ $
But $ \ \frac{1}{2} \nless \frac{1}{2} $ while $ \ x_1 <x_2 < ... < x_8 \ $
So something wrong is there.
Also how to find the fixed points of $ G^3(x) \ $ ?
I know that the $ 4 \ $ fixed pints of $ G^2(x) \ $ are also fixed points of $ G^3(x) \ $ but how to find the rest $ 4 \ $ fixed points of $ G^3(x) \ $ ?
Help me out to answer the question .
The fixed points of $x\mapsto G(x):=4x(1-x)$ $(\>x\in{\mathbb R})$ are the real solutions of the equation $G(x)=x$, namely $0$ and ${3\over4}$. There are $4$ different fixed points of $x\mapsto G^{\circ2}(x)$, determined by the equation $G^{\circ2}(x)=x$. Of course the fixed points $0$ and ${3\over4}$ of $G$ are among these. A fixed point of $G^{\circ2}$ which is not a fixed point of $G$ need not be a fixed point of $G^{\circ3}$, but each fixed point of $G^{\circ2}$ is a fixed point of $G^{\circ4}=\bigl(G^{\circ2}\bigr)^{\circ2}$.
There are $8$ different fixed points $x_i$ of $G^{\circ3}$, two of them being $0$ and ${3\over4}$. The latter is $x_6$, if the $x_i$ are ordered increasingly. The following figure shows a graph of the function $x\mapsto G^{\circ3}(x)-x$.