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!

  • $\begingroup$ Keep in mind that you can also write $x=1+\frac{x^3}{6}-x^5$ - you don't necessarily have to show that you have a contraction mapping over all of $\mathbb{R}$, but just over an interval where you know that there must be a root... $\endgroup$ – Steven Stadnicki Mar 8 '13 at 22:10
  • $\begingroup$ @StevenStadnicki Thanks for the suggestion. It seems that a $[0;1]$ interval could do here which is pretty good. But how did you realize that particularly this expression of $x$ is advantageous? $\endgroup$ – Pranasas Mar 8 '13 at 22:17
  • $\begingroup$ @Pranasas, if you let $F(x)=1+\frac{x^3}{6}-x^5$, then the problem is to find a fixed point of $F$. Thats why steven used this expression. $\endgroup$ – Tomás Mar 8 '13 at 22:42
  • $\begingroup$ @Pranasas For a few reasons: (a) it was the one reorganization of the original equation you hadn't looked at; (b) it's the simplest - it remains polynomial, rather than being an awkward root expression; and (c) over the region near $x=0$ then the powers of $x$ are going to be smaller than $x$ is, so it seemed plausible that it might be a contraction mapping. $\endgroup$ – Steven Stadnicki Mar 8 '13 at 23:32
  • $\begingroup$ @StevenStadnicki So what interval would you suggest? The fixed point is at $x \approx 0.78$. I realized $[0;1]$ is no good because for example $f(x)=1+x^3/6-x^5$ at $x= \sqrt{0.1}$ is not within $[0;1]$. Another option $[0;2]$ is not good either. $[0;1.1]$ would pretty much require a calculator to prove and I hate it. $\endgroup$ – Pranasas Mar 9 '13 at 11:40

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 $q(x)=6x^5-x^3+6x-6$.

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