I am asked to provide an iterative algorithm which would lead to finding a real root of this polynomial:
$$6x^5-x^3+6x-6=0$$
It is required to rely on the contraction mapping principle and Banach fixed-point theorem.
At the moment I think I can rewrite $$f(x) = \sqrt[5]{x^3/6-x+1}$$
and try to prove that in a complete metric space $\langle \, \mathbb{R}, d \, \rangle$ I have a contraction mapping $f:\mathbb{R} \rightarrow \mathbb{R}$ and consequently a unique fixed-point is my solution. I already see it's going to be messy (contraction mapping proof part) and the idea of it makes me sick…
Besides those 6'es in the initial polynomial temps to rewrite $$ x = \sqrt[3]{6} \sqrt[3]{x^5+x-1}$$ but I see no good coming out of it.
Since I'm very new to functional analysis and metric spaces I decided to ask you for suggestions about the most suave way to do it.
The final answer (fixed point) will be $x \approx 0.78$ so I'd love to get the contraction mapping over $[0;1]$, $[-1;1]$, $[0;2]$, $[-2;2]$ or something like that which would be easy to prove (contractility that is) without calculator… But I can't find it!
Best Answer
Taking the fifth root was a good idea: It helps to make derivatives small.
Consider the polynomial $$p(x):={x^3\over 6}-x+1$$ on the interval $X:=[0,1]$. From $p'(x)={x^2\over2}-1<0$ $\ (x\in X)$ it follows that $$ p(x)\geq p(1)={1\over6}\qquad(x\in X)\ .\tag{1}$$ Now define $$f(x):=\bigl(p(x)\bigr)^{1/5}\qquad(x\in X)\ .$$ Then $$f'(x)={1\over5}\bigl(p(x)\bigr)^{-4/5}\biggl({x^2\over2}-1\biggr)\ .\tag{2}$$ It follows that $f'(x)<0$ for all $x\in X$. Therefore $$0<{1\over 6^{1/5}}=f(1)\leq f(x)\leq f(0)=1\qquad(x\in X)\ ,$$ which proves that $f$ maps the complete metric space $X$ into itself.
To finish the proof we need an estimate of $|f'(x)|$. From $(1)$ and $(2)$ it follows that $$|f'(x)|\leq{1\over5}6^{4/5}<0.84\qquad(x\in X)\ .$$ Therefore we may conclude that in the interval $X$ there is exactly one point $\xi$ such that $f(\xi)=\xi$. It follows that $p(\xi)=\xi^5$, or that $\xi$ is a root of the polynomial $q(x)=6x^5-x^3+6x-6$.