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I'm not so sure I'm using the right terms in this question, but I will try to explain: Say I have a 3d surface, x axis and y are positives, z goes from 0-1. That surface has many "basins"(??), i.e. areas where all points go to a minima, and "mountains" i.e areas where all points go to a maxima. My questions are:

If I understand correctly, the minima of such a "pit" is the attractor, and the area around it where all points are starting to "move" toward this minima is the basin of attraction.

A. Is that correct ?

B. I want to find the radius of these basins, and their depth, How can I do that ?

I hope it makes sense.... Thanks, matlabit.

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Hum... what is your dynamical system on this surface ? is it smooth ? do you have an explicit formula ? to what data do you hope to relate the radii and depths to ? – Glougloubarbaki Oct 26 '12 at 12:23
Also, there is no particular link a priori between the "height" (z coordinate) of a point on your surface and it being an attractor or repulsor – Glougloubarbaki Oct 26 '12 at 12:24
I don't have explicit formula (this is why I wasn't sure if what I'm looking at can be interpreted as basin of attraction) , it is smooth, (gaussinan kernel), I want to relate the radii to "amount" of smoothing needed , such that all points that are within that radii represent an area on that surface that is meaningful to me. I didnt understand what you are saying in the second comment... thanks. – matlabit Oct 26 '12 at 12:35
I'm not sure I understand your problem. Is it : a. I have a function $f :S \rightarrow S$ mapping the surface into itself and I'm concerned with the behaviour of iterates $S^n(x)$, $n \in \N$ and $x \in S$ (that is what a dynamical system is). OR b. I have a fonction $f : S \rightarrow \mathbb{R}$ and I'm concerned with the extremum of $f$ on $S$ (that's what optimization is). – Glougloubarbaki Oct 26 '12 at 12:40
I have a function f(x,y) = z , f:S->R. I can find the optimums, (I can do it with gradient descent for example), but, I also want the radii of the area where the optimum starts to evolve. So, for your question, I think optimization ? – matlabit Oct 26 '12 at 12:49

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