# Show that function has asymptotically stable 2-cycle

dynamical systemsstability-theory

I need to show that the function

$$F: \frac{19}{6}x(1-x)$$

has an an asymptotically stable 2-cycle.

I'm completely lost on this math problem. Do I need to use Lyapunov functions or something similar?

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 $$Q$$.