Is the starting value $k$ ensure convergence for almost all polynomials?

Question

It's a follow up of Does three iteration are sufficient to use Newton's method

Come back on the previous question :

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

Or:

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

Where $F_n$ are the Fibonacci's numbers

Now let $b_0\geq 1,C>2$,$b_i\in R$

Using Newton's method on $f(x)$ provided the polynomial (numerator) have only one positive real root over $R^+$

Is the starting value $k>x_{positive-root}$ with the Newton's methods sufficient to ensure the convergence to the positive and unique root ?

Have you a proof or a counterxample ?

We can use the lower bound as Cauchy's inequality see https://en.m.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots — Miss and Mister cassoulet char, Mar 29 '24 at 09:49

Miss and Mister cassoulet char · Answer 1 · 2024-03-29T14:54:53.847

0

Some thoughts:

Using the Kantorovitch's theorem we need to show $x,y,k>0$:

$$|f'(x)-f'(y)|<k|x-y|\tag{I}$$

But by definition of $f(x)$ we have :

$$\lim_{x\to \infty}f'(x)-ax+b=0,a>0$$

So taking $k$ sufficiently large for all $x>M>0$ we have $I$.

If so taking a sufficiently large initial value we have convergence .

Conclusion :

Excluding the too small value invoking divergence we have for sufficiently large one as application of the Kantorovitch's theorem the convergence is reached.

Reference :

https://en.m.wikipedia.org/wiki/Kantorovich_theorem https://arxiv.org/abs/2001.09791 Remark :

@MartinR we need to show that the derivative of $f(x)$ in absolute value doesn't vanish AFTER the positive real root of $f(x)$ for that we need a bound . See Under what conditions a rational function has bounded derivative?

To be continued.

Eurêka as there is only one positive real root we can apply Gauss-Lucas theorem.

And so all the starting value $k>x_{positive-root}$ are OK

edited Mar 29 '24 at 14:54

answered Mar 29 '24 at 10:16

Miss and Mister cassoulet char

3,511

The Kantorovich theorem requires that any $x \in X$ has a neighborhood such that $(I)$ holds for $x, y \in I$ (with some constant $k$). It does not suffice that the condition holds for large $x, y$. – Martin R Mar 29 '24 at 13:36
@MartinR I agree numerical evidence says but a more accurate explanation is needed as bound the derivative for example . Do you want to make an answer instead of just push remark without any help ? – Miss and Mister cassoulet char Mar 29 '24 at 13:53
@MartinR I have finished – Miss and Mister cassoulet char Mar 29 '24 at 14:55
The Gauss-Lukas theorem states that for a polynomial $P$, the roots of $P'$ are inside the convex hull of all roots of the roots of $P$ (including the non-real roots). So even if $P$ has only a single positive real root $x_0$ it is still possible that $P'(x) = 0$ for some $x > x_0$. – Martin R Mar 29 '24 at 15:00
@MartinR again I agree but here all the variable are real which is sufficient. – Miss and Mister cassoulet char Mar 29 '24 at 15:05
$P(x) = (x-2)^3-x+4$ has a single real root $x_0 \approx 0.479 > 0$, and the derivative has two zeros at approx $1.423$ and $2.578$. – Martin R Mar 29 '24 at 15:14
@MartinR I believe you but here the function is convex over$R_+$ so I stop here to answer you . – Miss and Mister cassoulet char Mar 29 '24 at 15:33
For convex functions the situation is quite simple: https://math.stackexchange.com/q/2616498/42969. If your function is convex then we can close this question as a duplicate. – Martin R Mar 29 '24 at 15:51
@MartinR nothing to do with a duplicate since I introduce $f(x)$ as rational function .Cohn art. – Miss and Mister cassoulet char Mar 29 '24 at 16:29
@MartinR the function if convex it works if concave and next convex it's worst it's negative . – Miss and Mister cassoulet char Mar 29 '24 at 17:49
If proved we have a local inversion we can speak about Hasse principle and Jacobian conjecture with quadratic. – Miss and Mister cassoulet char Mar 30 '24 at 08:47

Is the starting value $k$ ensure convergence for almost all polynomials?

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