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Author Topic: Julia-sets for continous functions  (Read 420 times)
Description: a numerical approach to compute the Julia-sets for every continous function
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SamTiba
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« on: January 25, 2017, 06:36:12 PM »

Since I did not found it anywhere here (even though I thought this would be a known approach), here is a method to compute Julia-sets for continous functions:

Known definitions from the Julia- and Fatou-set (taken from the german wiki-page and translated):

Fatou-Set
The initial values from the Fatou-set iterated over a function lead to a static dynamic behaviour. So if the initial value changes a bit, the dynamic flow of the iterated function shows a similar behaviour.
Julia-Set
The points from this set lead to instable dynamics: Every inifinitesimal small change of the initial value leads to a completely different dynamic.

To further explain this behaviour, I am going to show this on an example later on.

To make a numerical approach we have to balance the complex plane, since functions can also have this static behaviour when converging to ∞, but we can't describe this using the normal complex numbers.
A possible way of doing so is using the Riemann Sphere. We use following moebius-transformation to get the height of every point z = a+bi on the Riemann Sphere:
   Zriem = (abs(z)^2 - 1)/(abs(z)^2 + 1) = (a^2+b^2 - 1) / (a^2+b^2 + 1)
After transformation we have the Complex plane + ∞ as the unit sphere. All further definitions are applied on this sphere.

Numerical approach to qualify a point
After a certain number of iterations n0 the following behaviours occur for points Z of the Fatou- and Julia-set.
For every number of iterations n greater then n0 in relation to its neighbourpoint Zp we have for the Fatou-set:

And for the Julia-set:


d(Zn,Zpn) is not beeing calculated on the complex plane. We have a sphere so we would have to use spherical distances.

Numerical approach:
We calculate the distance by the subtraction of the angles of the z-values. Therefore we have a maximum distance for j of pi (from d(∞,0)).

Through this method we get two escape values, one for each set, and therefore can use two different color gradients.
It is quite slow, but produces a lot of new patterns and the possibility to extend it to quaternion numbers.

This approach produces nice Julia-fractals for all functions, especially the continous ones. (Polynom with positive and negative exponent, sin cos tan (sinh cosh tanh) and exp).
For multivalued functions we get non-exact results.

Comparison to known examples from the quadratic Julia-set
Julia-set for z^2 + c with c = -0.8 +0.2i (in red, Fatou-set in blue)


Julia-set for z^2 + c with c = -0.2 +0.3i (in red, Fatou-set in blue)




Some complex and beautiful examples from my gallery:
Painted Julia fractal
Red Butterfly
Smooth rational Julia

Credits to Ijon (german site with mathematical explanation of the properties) for the mathematical backround.
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macawscopes
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« Reply #1 on: February 05, 2017, 04:05:48 AM »

This is really cool... I'm gonna have to read up on the math, as I hadn't ever heard of this kind of fractal.  What is meant by "continuous"?  I'm having trouble understanding this.  As opposed to "discrete"?
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SamTiba
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« Reply #2 on: February 05, 2017, 09:01:36 AM »

Yes cheesy

The math behind it is much more complex then my approach to visualize it.
If you're into it it's definetly worth a shot, but unfortunately I can't provide you the underlying theories in english.
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Adam Majewski
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« Reply #3 on: February 05, 2017, 09:41:50 AM »

Compare with :
https://commons.wikimedia.org/wiki/File:Quadratic_Julia_set_with_Internal_binary_decomposition_for_internal_ray_1_over_3.ogv

HTH
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SamTiba
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« Reply #4 on: February 05, 2017, 10:20:16 AM »

never heard of a decomposition-method or a parabolic-method?
it seems to be very different
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