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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 [itex]N(z)=z-\frac{f(z)}{f'(z)}=z-\frac{1}{(\log\circ f)'(z)}[/itex].

For example, when f is a polynomial, [itex]f(z)=\prod_{k=1}^n(z-z_k)[/itex], then Newton's method is [itex]N(z)=z-(\sum_{k=1}^n\frac{1}{z-z_k})^{-1}[/itex]

Each root of [itex]f[/itex] 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 [itex]\sum_{k=1}^n\frac{1}{z-z_k}[/itex] could be interpreted as an electric force field (at the point z) in the complex plane emanated by equal point charges at the points [itex]z_k[/itex], except that the force direction is altered by complex conjugation ([itex]\sum_{k=1}^n\frac{1}{\overline{z-z_k}}[/itex] would be the proper electric field).

But we have the inverse [itex](\sum_{k=1}^n\frac{1}{z-z_k})^{-1}[/itex], 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 [itex](\sum_{k=1}^n\frac{1}{z-z_k})^{-1}[/itex]. (I know, forces act by acceleration, there's inertia and stuff, but I'm just asking if someone seesanyway to make "phyiscal sense" of this kind of dynamics.)

And ideas most welcome.

Kind regards

Unkraut

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

Can you offer guidance or do you also need help?

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