• January 21, 2021, 08:22:43 AM

### Author Topic:  Parabolic Julia sets  (Read 3820 times)

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#### gerrit

• 3f
• Posts: 2288
##### Re: Parabolic Julia sets
« Reply #75 on: October 02, 2020, 04:48:00 AM »
about the Golden mean Siegel disk in the quadratic case $$f(z)=\lambda z + z^2$$ with $$\lambda=e^{2i\pi\cdot\alpha}$$ and $$\alpha=\frac{\sqrt{5}-1}{2}$$.
Here's a picture near that point in the M-set, pretty hard to render.

• Fractal Frogurt
• Posts: 434
##### Re: Parabolic Julia sets
« Reply #76 on: October 02, 2020, 08:08:52 AM »
Here's a picture near that point in the M-set, pretty hard to render.
Is it parameter plane ( c-plane) or dynamic plane ( z-plane) ?

#### gerrit

• 3f
• Posts: 2288
##### Re: Parabolic Julia sets
« Reply #77 on: October 02, 2020, 08:30:35 AM »
Is it parameter plane ( c-plane) or dynamic plane ( z-plane) ?
M-set is c-plane, Julia set z-plane, so former.

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #78 on: October 02, 2020, 10:17:33 PM »
Here's a picture near that point in the M-set, pretty hard to render.
Could you elaborate on being hard? Does it take long computation times or iteration count? A high precision number type? Does the image change sharply if a slightly different grid positioning was chosen?

#### gerrit

• 3f
• Posts: 2288
##### Re: Parabolic Julia sets
« Reply #79 on: October 03, 2020, 08:14:17 PM »
Could you elaborate on being hard? Does it take long computation times or iteration count? A high precision number type? Does the image change sharply if a slightly different grid positioning was chosen?
High max iteration + subpixel structure so how to color?

• Fractal Frogurt
• Posts: 434
##### Re: Parabolic Julia sets
« Reply #80 on: October 04, 2020, 10:03:52 AM »
so it is a dense set ?

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #81 on: October 08, 2020, 05:20:41 PM »
"Putting it all together"

The conformal radius associated with the Siegel disk in the classical quadratic case

$f(z)=e^{2i\pi\cdot \alpha}\cdot z + z^2$

with $$\alpha$$ being a bounded-type Brjuno number (hence there is a constant C > 0 such that all continued fraction entries ai <= C for i >= 1, is bounded with mathematical guarantee from below by

$R(\alpha) \geq e^{-69-2\cdot ln~(C+1)}$

with ln being the natural logarithm.

In particular for the Golden mean Siegel disk being of constant type ai=1 for i >= 1, C=1 can be chosen, hence the conformal radius is definitely >= e-71. By the relation between conformal and inner radius (post #61), a circle of radius 1/4 of that value can safely be marked in the TSA to detect preimages by cell-mapping.

This would technically be possible around L102R2 if one were trying to compute the whole filled in Julia set. Preimages of the Siegel center point or that circle alone can readily be computed with a sufficient number type directly.

This was quite a long and interesting journey, starting from the first idea of assisted TSA (post #11) back in October 2019 to the now reached first goal line.

The proof in the attached pdf is based on a collection of already proven crucial facts about the Brjuno sum, the Brjuno function and their relation to one another and the conformal radius (Yoccoz, Marmi, Buff), facts about continued fractions, the Fibonacci sequence and the sum of their reciprocals (Wang, DeVille) and splitting the Brjuno sum in several parts and mayoranting them (my humble input).

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #82 on: November 11, 2020, 02:29:07 PM »
"Rotation of z^2-0.75: Unsuccessful"

As rotation was successful in detecting interior in a previous case, I tried this with the classic z^2-0.75, resulting in a conjugated polynomial f(z)=(1+i)*z^2-z with the parabolic fix point at the origin.

Analyzing the four squares of the tiling of the complex plane, that have the origin at one of their corners, two seem to lie fully inside the interior (image, upper right, point-sampled overview, yellow pixel is the origin).

One starting square (pixel) A iterates to a screenRect of size 3x3, extending over the coordinate axis into a different quadrant and into the exterior.

Code: [Select]
A: -0.000244140625..0,0..0.000244140625  -> f(A): -1.4901161193847656e-007..0.00024434924125671387,-0.00024434924125671387..1.4901161193847656e-007
The red square in the image below is a zoomed version of that one pixel in the original image. Its screenRect is the blue rectangle (computed via IA), yellow is the exaggerated origin and purple are the 1-iterate images of single complex numbers from the red square.

The bounding box cannot be tightened by monotonicity (the partial derivative of the real component function, u*dy e.g. is straddling zero, and that derivative is tight). Additionally the corners are not the most extended points of the bbx. So, independent of whether the 3x3-pixel size is due to IA-overestimation or by bulging arcs connecting the corner's images, there is no chance here to get rid off of quadrant-spanning.

So I have to resort to the two remaining complicated options: ball arithmetics (however, as tiling the plane will requre overlapping balls, at the border between interior and exterior, those balls overlap and hence a bounding ball might probably overlap with exterior), or affine arithmetics with the starting set a triangle pointing towards the parabolic point and a triangular tiling (as mentioned previously).

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