# Question on Fixed Point Iteration and the Fixed Point Theorem.

The Question: Let's approximate the root $p \in [0,1]$ by applying fixed point iteration. Consider the iteration function $g(x) = 1 - x^{2}.$ Can you find an interval which the fixed point theorem can be applied ?

The Attempt: I have tried using the Bisection Method to figure out the root of the function $h(x) = 1 - x - x^{2}$. However, when I do this, I am not getting any values that belong to the intervals when I compute for the iterations. Is there some other way I can find an interval that I can apply the fixed point theorem to?

Thank you for the help!!

• You want a bounded fixed point method? – ja72 Sep 6 '16 at 20:31
• If you iterate, $g(x)=1-x^2$, you'll quickly get stuck in an attractive 2-cycle - not what you want. You can write the equation $x=1-x^2$ as $\sqrt{1-x}=x$, however, which has the same positive root. Thus, you can find the root by iterating $f(x)=\sqrt{1-x}$. That kind of rewrite is a common trick in this area. – Mark McClure Sep 6 '16 at 20:33
• Okay. Can you explain again how you got $f(x) = \sqrt(1-x)$ ? – user211962 Sep 6 '16 at 22:08

Hint: If I have understood the statement correctly the answer is no. The reason being that at the fixed point the derivative of $g$ is smaller than $-1$.
• I don't understand why we cannot use it because the fixed point of the derivative is less than $-1$. Can you please elaborate on that more? – user211962 Sep 7 '16 at 1:26
• The fixed point method, (I suppose you are talking about: $x_{n+1}=g(x_n)$), requires a strict Lipschitz contraction of an interval $[a,b]$. If this is possible to find, then at the fixed point $a=0.6180340...$ the Lipschitz contraction of $g$ would imply $|g'(a)|=2a<1$ which is false. – H. H. Rugh Sep 7 '16 at 5:20