Period $3$ orbit of the logistic map $x_{n+1}=r \cdot x_n(1-x_n)$

Question

One can proof, that the logistic map has an stable orbit of period three for $r=1+2\sqrt{2}$. This can be done by looking at the third iterated of $f$ and investigate it for stable fixed points. For the value $r=1+2\sqrt{2}$ the graph of the third iterated of $f$ has three intersections with the function $x$, hence three (stable) fixed points. See the following graphics:

$f$ with three intersections" />

The three intersections represent the values of the attractor of period $3$, so the values of the orbit of period $3$.

So this means that the values of $f^{3k}(x)$ for $k \in \mathbb{N}$ are all equal, the same for all the values of $f^{3k+1}(x)$ and $f^{3k+2}(x)$ due to the orbit of period three.

So this for I`d expect that if I plot the graphs of $f^3(0.5), f^6(0.5), f^9(0.5),...,f^{36}(0.5)$ dependent of the value $r$, that they'd be identical at $r=1+2\sqrt{2}$, because there starts the orbit of period three. But as one can see in the following graphics, the graphics of the functions $f^3(0.5), f^6(0.5), f^9(0.5),...,f^{36}(0.5)$ are not the same at $r=1+2\sqrt{2}$, this value is represented by the vertical black line. Why is that? Has anyone an explanation for this?

$f^{3k}(0.5)$ for $k \in \{1,...,12\}$ and a vertical line at $r=1+2\sqrt{2}$." />

Answer 1

According to Mathematica (see commands below), the stable fixpoints of $f^3$ for $r = 1 + 2 \sqrt{2}$ are approximately 0.159928818446256, 0.514355277061990 and 0.956317841973624.

f[x_] := (1 + 2 Sqrt[2]) x (1 - x)
g[x_] := Nest[f, x, 3]
N[Solve[g[x] == x, x], 15]

Since the point 0.5 isn't one of them, there's no reason for it to be a fixpoint of $f^{3k}$ either. (Notice that although the graph $y=f^3(x)$ has an even symmetry with respect to the point 0.5, the line $y=x$ hasn't, so the curves don't touch at 0.5 but a little bit further to the right.)

But if you plot $f^{3k}(0.514355277061990)$ instead of $f^{3k}(0.5)$, then you should get your expected equality to a good approximation.

Answer 2

For anyone who is interested: Instead of $x=0.5$ I started the iteration of $f^{3k}(x^*), k \in \{1,...,12\}$ with $x^*=0.514355277061990$ wich is a fixed point of $f^3$ and adjusted the graphics from above.

One can see, that all the iterates of $f$ have one intersection wich represent the value of that specific attractor of the orbit. The vertical black line has the value $r=1+2\sqrt{2}$.