Does three iteration are sufficient to use Newton's method

Question

Introduction:

Find a such family of polynomial then we can always use Newton's method provided we have a procedure to start.

We introduce a rational function where $b_i$ are real numbers:

$f(x)=x\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)-C=0\tag{I}$

For the sake of simplicity we call it a primitive basis of the problem since it could be generalized .

We rewrite $I$ :

$x\left(b_{0}\left(x+1\right)+b_{1}\cdot\frac{1}{x+1}+b_{2}\frac{x+1}{x+2}+b_{3}\cdot\frac{x+2}{2x+3}+\cdot\cdot+\frac{b_{n}\left(F_{n-2}x+F_{n-1}\right)}{F_{n-1}x+F_{n}}\right)-C=0$

Where $F_n$ are the Fibonacci's numbers

Now we isolate the root since the target is the numerator which is a polynomial in $I$.

$$x=\frac{C}{\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)}=g(x), g(x)\neq\pm \infty,x\geq 0,|g'(0)|<1,C>0$$

Provided that $x_{root}$ exists as a strictly positive real and is the roots of $f(x)$ and so $f(x)$ have only one positive real root other negative and imaginary.

In other word the numerator in $I$ have the form :

$$\sum_{n=0}^{M}a_{n}x^{n},a_0<0,a_n>0$$

Now : $g(g(g(\cdots (0)\cdots)))=x_{positive-root}$

Then we can stop iteration and use Newton's method for the positive root.

Question :

Provided that $|g'(0)|<1/2$ even three iteration are sufficient to use Newton's method ?

To go further :

See https://en.m.wikipedia.org/wiki/Centralizer_and_normalizer

https://en.m.wikipedia.org/wiki/Isomorphism_theorems

Answer 1

Some tought .

A solution to a starting value is done here Choosing the initial values (Newton's method)

The condition exclude large roots so we can say : $$x_{root}<M,M>0$$

Currently numerical evidence say $M<10$ .

We can use HmGm to give an sufficiently accurate starting value remains to compare with @ClaudeLeibovici answer.

Using Hm-Gm at zero iteration $x\geq 0$ and with strictly positive coefficient:

$$x_{positive-root}<\lim_{x\to 0}\frac{C}{M+1}\left(\frac{F_{M-2}x+F_{M-1}}{x+1}\right)^{\frac{1}{M+1}}\left(\prod_{n=0}^{M}b_{n}\right)^{-\frac{1}{M+1}}\tag{J}$$

Using one iteration $g(g(0))$ we have a lower bound .

As SimplyBeautifulart mentioned if the staring value is in $[1,2]$ or $[0,1]$ we need to show the positive root lies in this domain with the bounds .As we cannot choose too small $b_n$ in comparison of $C$ otherwise there exits $y>0$ such that $g(y)=\infty$ it should be a start using $J$ evaluating at $x=0$ that :

$$x_{positive-root}<\frac{C}{M+1}\left(F_{M-1}\right)^{\frac{1}{M+1}}\left(\prod_{n=0}^{M}b_{n}\right)^{-\frac{1}{M+1}}\tag{J}$$

If we assume $C<M$ and the product $\prod_{n=0}^{M}b_{n}>1$ we have the desired conclusion that three iteration are sufficient .

Example :

If the strictly positive coefficients verify : $$b_{i+1}-b_i=1>0,C=1$$

Then J is :

$$x_{positive-root}<\frac{1}{M+1}\left(F_{M-1}\right)^{\frac{1}{M+1}}\left(\prod_{n=0}^{M}(b_{0}+n)\right)^{-\frac{1}{M+1}}\tag{J'}$$

See also https://en.m.wikipedia.org/wiki/Hurwitz_polynomial