Show that the logistic map $x_{n+1}=Ax_n(1-x_n)$ has stable $2$-cycle for all $A>3$.
I am a newbie in non-linear dynamics and chaos theory. All I have been able to find out about the map is that it has equilibrium points at $x^*=0$ and $x^*=1-\frac{1}{A}$. At $x^*=0$, the system attains asymptotic stability and for $1<A<3$, also the system is asymptotically stable. But how to geometrically/graphically analyze about the stable $2$-cycle for values of $A$ above $3$?
I will show you here a numerical example of how to determine a 2-cycle in iterative maps. Showing that for $A>3$ 2-cycles are obtained should follow smoothly then.
Start from the basics: a 1-cycle is a fixed point. Meaning, that after a single iteration, the value obtained from the map is the same. Here, to find a fixed point, we solve:
(1): $x = Ax(1-x)$
and this is how you obtain the first two answers in your problem definition.
Next, a 2-cycle means that every two iterations you will repeat the same value. Hence, what you want is:
Define $f(x) = Ax(1-x)$, and the root solving problem is:
(2): $f(f(x)) = x$
you can solve that analytically (I believe) or show graphically that for $ A> 3 $ a point satisfying (2) without satisfying (1) exists. Actually, these will be pairs of points, I will let you deduce the reasoning for that.
I have written a small MATLAB script to illustrate the manner of thinking:
In figure 1, intersection of yellow with blue solves (2). Intersection of yellow with orange solves (1).
Here, I have tried A = 5. And you see a pair of points solving equation (2) specifically.
The value of one of the points is 0.95 (approx). I have tried this value in the mapping and, voila, a two cycle is found: