Banach's fixed point theorem

Question

For the following question, we are to use Banach's fixed point theorem to show that $\{x_n\}$ converges to the root of the polynomial: $x^4 - 4x^2 - x + 4 = 0$ ; lying between $\sqrt{3}$ and $2$. Let ${x_1} = \sqrt{2}$ and $x_{n+1} = \sqrt {2 + \sqrt{x_n}}$.

I know the theorem states that if $f: X \to X$ is a contraction of a complete metric space X, then f has a unique fixed point.

I think I need to prove that f has a contraction, but I don't quite understand how to proceed with that. Is that the right approach or is there another approach that would be more efficient, per se.

Thank you in advance.

Answer 1

Guide Set $X := [\sqrt{3},2]$ as your complete metric space, and set $f(x) := \sqrt{2+\sqrt{x}}$. Use Calculus to see that $f$ is increasing, and a self-map on $X$. Moreover, $f'$ is positive and decreasing on $X$. Since $f'(\sqrt{3})<1$, the Mean Value Theorem guarantees that $f$ is a Banach contraction on $X$. Hence your sequence will converge to the unique fixed point.

Answer 2

Let $X=[\sqrt{3},2]$ and $f(x)= \sqrt {2 + \sqrt{x}}$. $X$ is a complete metric space.

Show that $f(X) \subset X$ and that there is $q \in (0,1)$ such that $|f'(x)| \le q$ for all $x \in X$. By the mean value theorem:

$|f(x)-f(y)| \le q|x-y|$ for all $x,y \in X$.

Hence there is $x \in X$ with $f(x)=x$.

But: $f(x)=x \iff x^4 - 4x^2 - x + 4 = 0$.