How do I show that the 2-cycle {-1,0} is attracting for the function $F(x)=x^2-1$
The textbook says that an attracting point is one that has an interval around it in which $F^n(x)$ converges to it. And that a 2-cycle is a when $F(0)=-1$ and $F(-1)=0$. I need to prove that there is some interval in which $F^n(x)$ converges to two limits -1 and 0 but how do I do this? (The only way I can think of is to try to find a formula for $F^n(x)$ in terms of x but all I can think of is the trivial recursion.)
$F(-1)$ is not $0$ for that function. Maybe $F(x)=-x^2+1$ or something.
If so, let $G(x) = F(F(x))$. I think you can show $G^n(x)$ converges to each limit in some interval.
[p.s.]
Then $G(x) = x^4-2x^2, G(0)=0, G(-1)=-1, G'(0)=0, G'(-1)=0$.
Let I=[-c,c]. We can satisfy $|G'(x)|<1/2$ for all $x\in I$ by choosing c small enough. Then $\frac{G(x)-G(0)}{x-0}=\frac{G(x)}{x}<1/2$ from mean-value theorem, which means $G^n(x)$ converges to $0$. Similarly for x=-1.