Show that function has asymptotically stable 2-cycle

Solution 1:

You solve the equations and $F(x)=x$ and $F(F(x))=x$. The former is an algebraic equation of degree 2. The roots $x_1,x_2$ are easy to compute. The latter is an algebraic equation of degree 4. Since $x_1$ and $x_2$ are roots, you have for every $x \in [0,1]$, $F(F(x))-x = (F(x)-x)Q(x)$, where $Q$ is polynomial of degree 2 (that you compute by division). Hence, you get the other two roots $x_3$ and $x_4$.

Since $F(x_3) \ne x_3$ and since $F(x_3)$ is a solution of the equation $F(F(x))=x$, you have $F(x_3)=x_4$ and in the same way, $F(x_4)=F(x_3)$, so $(x_3,x_4)$ is a 2-cycle of $F$.

To prove its stability, you check that the the derivative $(F \circ F)'(x_3) = F'(x_3) \times F'(x_4) = (F \circ F)'(x_4)$ is in $]-1,1[$. Computing $F'(x_3)F'(x_4)$ is easy since the $x_3+x_4$ and $x_3x_4$ are known because $x_3$ and $x_4$ are the roots of $Q$.