# Fixed-point iterations for quadratic function $x\mapsto x^2-2$

Let $$f(x)$$ be $$x^2-x-2$$. I want to find the root using FPI in an interval where it will converge. I have chosen $$g(x)=x^2-2$$ and so $$g'(x)=2x$$. The convergence condition, $$|g'(x)|<1$$ is obviously satisfied in $$-0.5.

Problem: I have failed to find a consistent interval outside of this convergence interval up to $$\pm 1$$ for which the iteration consistently diverges.

Question: Does this convergence condition only guarantee convergence (in the bounds) but not divergence (outside the bounds)?

• Is that fixed-point iteration fixed? From $x^2=2+x$ one finds the better iteration $x_{n+1}=\sqrt{2+x_n}$ for the positive root. – LutzL Apr 16 at 16:25
• Yes, but I thought the reason it’s ‘better’ is because it satisfies abs(g’(x))<1 in some interval. But g(x) in op works just fine up to -+1. – AKubilay Apr 16 at 18:10
• $g(x)=\sqrt{2+x}$ has $g'(x)=\frac1{2\sqrt{2+x}}\le\frac25$ for all $x>0$, $g(x)=1+\frac2x$ has $g'(x)=-\frac{2}{x^2}>-1$ for $x>\frac32$, so you get intervals with $|g'(x)|<1$ for many fixed-point functions. – LutzL Apr 16 at 18:53

Let us consider the fixed point iterations associated to the function $$g: x \mapsto x^2-2$$, defined by the quadratic map $$x_{n+1} = {x_n}^2 - 2, \qquad x_0 \in \Bbb R .$$ This map has many periodic points, even with large period. The period-one fixed points $$-1$$, $$2$$ are both repelling fixed points (indices $$2>1$$ and $$4>1$$, respectively). Thus, fixed-point iterations will not converge towards these values unless the starting value $$x_0$$ is exactly equal to $$-1$$ or $$2$$. Setting $$y_n = -\frac{1}{4} x_n + \frac{1}{2}$$, the logistic map $$y_{n+1} = r y_n (1-y_n)$$ with parameter $$r=4$$ is obtained, which exact solution is bounded and exhibits chaotic behavior (see also this article). Therefore, $$x_n = 2\cos\left(2^n\cos^{-1}(x_0/2)\right)$$ is bounded and exhibits chaotic behavior too (the sequence does not diverge to infinity).