Checking when the fixed point iteration converges

Question

Let the iteration $x_{n+1} = \frac{1}{2}(x_n^2 + c)$, where $c\in (0,1)$.
I already showed the fixed points are: $0<\xi_1 = 1-\sqrt{\frac{4-c}{4}} < 1 < x_2 = 1+\sqrt{\frac{4-c}{4}}$.

Now, I was asked

1. To show that For $x_0 \in [0, \xi_2)$, the iteration converges to $\xi_1$.
2. To check what happens for other $x_0$.

So basically, for $x_0 \in [0,1]$ the iteration converges since $g'(x) = x$ and $g'([0,1])\subseteq [0,1]$ (We've got a theorem for that)

I think that it can be shown that for every $x_0 \in \mathbb{R}$ the iteration converges (since it is monotonic and should be bounded if I'm not mistaken)

so the conclusion is that the iteration converges to $\xi_1$ for every $x_0 \in\mathbb{R}$?

Answer 1

Your fixed point interation does not converge for every $x_0 \in \mathbb{R}$: Let $x_0=2$, then

$$x_1=\frac{1}{2}(4+c) = 2+\frac{c}{2}$$ $$x_2=\frac{1}{2}((2+\frac{c}{2})^2+c)=\frac{1}{2}(4+2c+\frac{c^2}{4}+c) > \frac{1}{2} (4+3c) = 2+\frac{3}{2}c > 2+c$$ Similarly it can be shown that $x_3 > 2+2c$ and so on. Hence $x_n > 2+(n-1)c$ and $x_n \to \infty$ as $n \to \infty$.