### Julia set of exponential function

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• 3f
• Posts: 1329

#### Re: Julia set of exponential function

« Reply #15 on: October 31, 2019, 01:59:17 PM »
here's an attempt for fragm with opengl 4 for double precision using dual complex numbers for automatic differentiation for newton's method

• 3f
• Posts: 1329

#### Re: Julia set of exponential function

« Reply #16 on: October 31, 2019, 02:20:44 PM »
C = 0 + pi i  (black is no period <= 16 attractor found near pixel)

#### marcm200

• Fractal Frankfurter
• Posts: 543

#### Re: Julia set of exponential function

« Reply #17 on: October 31, 2019, 03:50:51 PM »
That is not a correct escape radius, there is no escape radius, not sure if there is any algorithm to decide if an orbit escapes or not, except on the real axis.
|Re(z)|>50 as the escape criterion  is what you do if you naively enter exp(z) in some UF code and are surprised it doesn't work.
The criterion Re(z) > 50 is originally from Peitgen's book about "Science of fractal images" (50 however was an arbitrary value, as "e^50 is quite large").

Interesting is that if |z|->inf so does Re(z), so points escape in the positive x-axis (p. 158).

And an interesting tid bit of information: if 0 < lambda < 1/e, the half plane Re(z) < 1 is not in the Julia set.

(I find it confusing that the Julia set for entire functions contains all points escaping to infinity, while that of polynomials does not (p. 157)).

z_new=2*exp(z_old + c) where c=-0.1-0.1i in [-4..8] x [-6..6]*i

I used the polynomial quadratic smooth iteration count (probably mathematically not applicable here, but good for artistic purposes) and the 50-criterion, color by heat-map (black is "not escaped").

#### marcm200

• Fractal Frankfurter
• Posts: 543

#### Re: Julia set of exponential function

« Reply #18 on: November 01, 2019, 10:08:19 AM »
using interval arithmetic starting from any pixel size will probably blow up to include the whole plane for most (all?) points if the Julia set is the whole plane?

I used the kv interval arithmetics library and my rational map software to compute an exponential map: 1/128*exp(z), or more precisely its non-escaping regions.

It is quite an interesting experiment to use an IA library for the complete computation with double precision as interval end points - usually I only perform division by the library and the rest by symbolic IA bounding box calculation at an appropriate data type.

Numerical analysis showed 2 fixpoints: one attracting at ~0.00787-4.157e-017*i, and one repelling (when starting in the 16-square).

Upper right is a point-sampled version with the 50-criterion. Black is not-escaping.

Lower right is the IA image with the basin of attraction (yellow) at refinement level 10, i.e. squares are bounded and not in the Julia set (and since 0 < 1/128 < 1/e, so Re(z) < 1 was expected not to be in the Julia set and I hoped there was an attractor that could be found at decent refinement levels).

Left is the IA visitor's log (anti-buddha style), all interior points' orbits to depth 3 increase a counter per pixel visited, heat-map-colored. The white hotspot is at the image's center, where the attracting fix point resides.

• 3f
• Posts: 1900

#### Re: Julia set of exponential function

« Reply #19 on: November 02, 2019, 08:58:55 PM »
Article in pdf attached. Changed extension to "jpg" to get past security.

• Fractal Furball
• Posts: 207

#### Re: Julia set of exponential function

« Reply #20 on: November 03, 2019, 08:18:15 AM »
Not helpful in any way I can see.

Have you checked linked papers ?

Mandel also has tanscendental function ( source code ia available) and do not use arbitrary precision
in JUlia lang:
https://github.com/chakravala/Fatou.jl/wiki/Explore-Fatou-sets-&-fractals

• 3f
• Posts: 1900

#### Re: Julia set of exponential function

« Reply #21 on: November 03, 2019, 07:22:11 PM »
Have you checked linked papers ?

Mandel also has tanscendental function ( source code ia available) and do not use arbitrary precision
in JUlia lang:
https://github.com/chakravala/Fatou.jl/wiki/Explore-Fatou-sets-&-fractals
It's not hard to make "wrong" images of exp(z), as in the link, but I think impossible to do it right.
AFAIK there is no way to compute exp(z) for very large |z| because there is no way to compute sin(x) for very large x because you need as many digits as the size of x (say $$10^{10^10}$$).

If there is a trick I haven't found it in any publication. Since all the pubs show the distorted images I'm guessing no trick is known.

#### marcm200

• Fractal Frankfurter
• Posts: 543

#### Re: Julia set of exponential function

« Reply #22 on: November 03, 2019, 08:10:49 PM »
Article in pdf attached. Changed extension to "jpg" to get past security.
I find this termination circle in fig. 5 very interesting and tried to recreate it but unsuccessful. Did someone succeed? I got the protruding structure but no circle (image below)

I am not sure I understood the iteration correctly, I used z_new := exp(z_old)-1, although the formula from the paper (4) said

Fr+1 (z) = exp {Fr(rz)} — 1; F0(z) = z

I was not sure about the (rz) part, I assumed it was a typo and they meant zr and not (r*z), but maybe that's wrong and there is a multiplication in every iteration step? (But a quick test did not show a circle).

• 3f
• Posts: 1900

#### Re: Julia set of exponential function

« Reply #23 on: November 04, 2019, 12:13:21 AM »
3 render attempts of Julia set of exp(z). Top normal double precision escape R = 1e40, middle same with additional 100 digits of precision, bottom 100 extra digits of precision R=1e120.

Bright green = escaping. There should be no green regions. You see green stuff has shrunk a bit in bottom picture.
If I could run with $$10^{10^{10^{10}}}$$ digits of precision maybe it would be correct.

#### marcm200

• Fractal Frankfurter
• Posts: 543

#### Re: Julia set of exponential function

« Reply #24 on: November 05, 2019, 10:37:38 AM »
Those termination cycles would have been a great discovery, but as gerrit's images suggest they are probably an artifact due to using the escape radius concept. So I guess for the exponential function one is stuck with determining regions that are definitely periodic and the rest remains uncertain (forever?).

• 3f
• Posts: 1329

#### Re: Julia set of exponential function

« Reply #25 on: November 05, 2019, 02:47:52 PM »
another variation, this time layering all the periods found instead of just the lowest

• 3f
• Posts: 1603

#### Re: Julia set of exponential function

« Reply #26 on: November 06, 2019, 03:13:32 PM »
Julia set of $$z \mapsto (\frac{1}{2} + i)e^z$$. All points are trapped. White to green pulsations to black = orbit wanders farther to the right before eventual convergence.

#### marcm200

• Fractal Frankfurter
• Posts: 543

#### Re: Julia set of exponential function

« Reply #27 on: November 26, 2019, 01:27:02 PM »
An interesting article about the parameter space of $$\lambda \cdot exp$$, characterizing regions there with (things I don't understand) dynamic rays and addresses.

And the author states in fig.2 "... in the exponential case it is impossible to test whether the singular orbit escapes to inf". so gerrit's question was definitely unanswered in 2003.

Code: [Select]
Annales Academiae Scientiarum Fennicae MathematicaVolumen 28, 2003, 3-34Attracting dynamics of exponential mapsDierk Schleicher

• 3f
• Posts: 1900

#### Re: Julia set of exponential function

« Reply #28 on: November 26, 2019, 06:05:38 PM »
An interesting article about the parameter space of $$\lambda \cdot exp$$, characterizing regions there with (things I don't understand) dynamic rays and addresses.

And the author states in fig.2 "... in the exponential case it is impossible to test whether the singular orbit escapes to inf". so gerrit's question was definitely unanswered in 2003.

Code: [Select]
Annales Academiae Scientiarum Fennicae MathematicaVolumen 28, 2003, 3-34Attracting dynamics of exponential mapsDierk Schleicher

#### v

• Fractal Fanatic
• Posts: 35

#### Re: Julia set of exponential function

« Reply #29 on: December 03, 2019, 05:25:03 PM »
I remember trying to plot these kinds of exp(z) variant fractals by assigning a colour to the 'dominant frequency' of series z_0, z_1, ..., z_n by taking its DFT and finding the peak hoping it would expose more detail. Never really got anywhere with it, or understand what I'm looking it.  A superficial escape time for |z|>100000 or some large number would show solid colour regions that would then fill up with 'flower petals' with this method, sometimes exposing detail that wasn't there. Here is a Julia for exp(z)+c and (1/2 + i)*exp(z), looks similar to above picture
« Last Edit: December 06, 2019, 10:14:04 PM by v »

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