Physical interpretation for this? (dynamics of Newton's method)

[Unkraut]

Hello!
I'm a math student, currently trying to write my diploma thesis. My field of study is complex dynamics (iteration of holomorphic/meromorphic functions, Julia sets and stuff).
It's a farfetched idea, but currently I'm curious about a potential physical interpretation of the things I'm looking at.
Specifically I'm studying Newton's method for a certain class of functions.
Remember, Newton's method for a function f is given by N(z)=z-\frac{f(z)}{f'(z)}=z-\frac{1}{(\log\circ f)'(z)}.
For example, when f is a polynomial, f(z)=\prod_{k=1}^n(z-z_k), then Newton's method is N(z)=z-(\sum_{k=1}^n\frac{1}{z-z_k})^{-1}
Each root of f is an attracting fixed point for N and has some "basin of attraction" of starting values which will converge to said fixed point under iteration of N.
The term \sum_{k=1}^n\frac{1}{z-z_k} could be interpreted as an electric force field (at the point z) in the complex plane emanated by equal point charges at the points z_k, except that the force direction is altered by complex conjugation (\sum_{k=1}^n\frac{1}{\overline{z-z_k}} would be the proper electric field).
But we have the inverse (\sum_{k=1}^n\frac{1}{z-z_k})^{-1}, so the direction of the force is alright again, but now its absolute value is inverted.

Okay, mathematical constructs don't always have any physical interpretation. It's a long shot, but I'm just asking if maybe someone has any neat physical interpretation for this kind of "discrete motion" that is given by iteration of N, i.e. moving some "particle" at location z in discrete steps in direction of the "force field" given by (\sum_{k=1}^n\frac{1}{z-z_k})^{-1}. (I know, forces act by acceleration, there's inertia and stuff, but I'm just asking if someone sees any way to make "phyiscal sense" of this kind of dynamics.)

And ideas most welcome.

Kind regards
Unkraut

 

[JFerreira]

I like to play with some iterations functions to produce fractals. And I also like to think about the dynamics of this process. This means that this questions is very interesting to me, and I really want to know if you find some thing on this direction.

I like to think on Newton's method in a way that I describe in what follows (please see this therad). Given the equation ${\bf f}({\bf x})=0$, whe write it as
$$0=g(\textbf{a})=\min_{\textbf{x}\in A}{g(\textbf{x})},\qquad {g(\textbf{x})}=\frac{1}{2}\|{\bf f}({\bf x})\|^2,$$ to some continuously differentiable function ${\bf f}:A→\mathbb{R}^p$, where $A$ is an open set of $\mathbb{R}^m$ containing $a$. Now, if you have some differentiable curve ${\bf u}:(a,b)→A$ you can apply the chain rule to obtain
$$\frac{d\, g({\bf u}(t))}{dt}= \left\langle {\bf u}'(t), \nabla g({\bf u}(t))\right\rangle= \left\langle {\bf u}'(t),[J{\bf f}({\bf u}(t))]^*{\bf f}({\bf u}(t))\right\rangle=\left\langle J{\bf f}({\bf u}(t)){\bf u}'(t),{\bf f}({\bf u}(t))\right\rangle,$$ in which $\langle\cdot,\cdot\rangle$ means the inner product.

Now, if you choose the curve ${\bf u} (t)$ as a solution to the initial value problem $$\left\{\begin{array}{rrl}J{\bf f}({\bf u}(t)){\bf u}'(t)&=&-\alpha {\bf f}({\bf u}(t))\\ {\bf u}(0)&=&{\bf u}_0\end{array}\right., $$ to some $\alpha>0$, then you can see that $$\frac{d\, g({\bf u}(t))}{dt}= -2\alpha g({\bf u}(t))\leq 0.$$

This means that, if ${\bf u}(t)$ describe the trajectory of a particle, then $g({\bf u}(t))$ is non increasing.

Newton's method can be viewed as a Euler method to solve numerically the previous initial value problem.

I hope that this gives some glimpse on the question.

Obs: I search on this subject on SearchOnMath and find this tread.