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### Author Topic:  Parabolic Julia sets  (Read 3821 times)

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#### marcm200

• 3c
• Posts: 894
##### Parabolic Julia sets
« on: September 24, 2019, 09:43:19 AM »
I was computationally searching for parabolic Julia sets solving numerically the equations f(z)=z, f(f(z))=z for the quadratic case and looking at the multiplier.

I found some period-1 and period-2 parabolic seeds. They seem to be located at the boundary of the main cardioid (period 1 parabolic) and the period-2 bulb (period 2 parabolic).

Bulbs contain attracting cycles of a specific length - but do the parabolic ones also obey by that law? Since parabolic cycles can be attracting or repelling, I was quite surprised to see that.

Is this in general the case? So when I'm looking for a period-3 parabolic Julia set I only need to look in the period-3 bulbs (or close to their border?)

Below are two Mandelbrot sets showing where I currently have found parabolic seeds (only lower half was checked). Blue is interior, white are regions where a parabolic Julia seed is.

Technical note:
• In the first period-1 image, a pixel represents 32x32 underlying complex numbers, if white, at least one of them was parabolic.
• In the 2nd period-2 image, a pixel represents 8x8 underlying complex numbers.
• Images were downscaled where white took precedence over blue over black resulting in a "hairy" look.
• White pixels are drawn in double-size so they can be spotted directly without having to zoom into the image.

#### claude

• 3f
• Posts: 1736
##### Re: Parabolic Julia sets
« Reply #1 on: September 24, 2019, 11:00:37 AM »
I read in a paper (forgot which one, sorry) that when crossing the boundary from a period P component to a period P*Q bulb, the period P cycle transits from attracting to repelling with an attracting period P*Q cycle bifurcating from it (each point in the period P cycle splits into Q points); I think it is parabolic with period P at the bond point between the period P bulb and the period P*Q bulb.  I think the cusp at the root of a cardioid of period P is also parabolic with period P, though afaict the analysis of this involves "parabolic implosion" which I don't understand.

• Fractal Frogurt
• Posts: 434
##### Re: Parabolic Julia sets
« Reply #2 on: September 24, 2019, 05:25:05 PM »

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #3 on: September 25, 2019, 09:04:48 AM »
Thanks. I will focus on the meeting points between bulbs. Hopefuily I can determine the ones which have rational real and imaginary parts.

My main goal is to find parabolic Julia sets with interior in the trustworthy Julia set sense. If both real and imaginary c part are irrational I need small intervals, but for c's being the touching point of two bulbs any small square around that c will contain points exterior to the Mset, so the resulting trustworthy Julia set will always remain gray.

My other focus will be minibrots. Is there a way to symbolically calculate their main cardioid end point (the one corresponding to c=0.25+0i in the overall case)?

The c=0.25 parabolic set is currently the only one with TSA interior (it has something to do with the fact that its fixed point has only curvature on two sides, form link). So maybe the minibrot's corresponding point would work too.

#### claude

• 3f
• Posts: 1736
##### Re: Parabolic Julia sets
« Reply #4 on: September 25, 2019, 02:18:24 PM »
a minibrot (or bond point) cusp is a solution c to the system of two polynomials with integer coefficients $f_c^p(z) = z \quad\quad \frac{\partial}{\partial z} f_c^p(z) = 1$ thus it is an algebraic number, there might be a way to show that they are not rational numbers for all p above a certain number (p = 3 minibrot has rational cusp at -1.75, not sure about p = 4 or above)

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #5 on: September 26, 2019, 09:47:13 AM »
-1.75 - that's great. I found the same with my brute-force approach. And a parabolic period-3-cycle that seems to be attracting - numerically almost every interior point comes close to -1.746999285,0.

The trustworthily computed image below (level 16) does not show interior cells. And at level 21 there is still no black following the origin's orbit.

Looking at the periodic points (pixels in small colored rectangles in the crossing regions of horizontal and vertical colored lines) suggests that those might either be the cusp of a cauliflower-like object or within its interior.

Now there are some interesting possibilites:

If all periodic points lie in the interior of a cauliflower, it would in principle be possible to detect interior cells.

If all are the cusp and at a dyadic rational x-coordinate, the points lie on a corner of a square at some refinement level and beyond and like in the c=0.25 case, interior cells will emerge.

Any combination of the above would also result in black cells.

In all other cases (irrational or non-dyadic coordinates), at least one point is at any refinement level on one edge of two adjacent squares, but not on the corner. Then any such square extends away from its cauliflower and above or below the x-axis, hence in the exterior region. Then this cell would remain gray forever. And since it is an attracting cycle everything else would too.

Quite a complicated but very interesting situation here - but too demanding for me to address it mathematically.

If I limit myself to minibrots on the x-axis it is possible to use small intervals for the real part of claude's solutions (I guess multivariate Newton then comes into play) since [-2..+0.25] x 0i is completely in the Mset. Hoping that the intersection of all interior of those c values is not empty, maybe there will be black emerging.

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #6 on: September 30, 2019, 01:53:17 PM »
"Proven futile" - a very long experiment report
(or, as hobold put it in a Meet&Greet post, the occasional wall of text inbetween the imagery )

Figueiredo et al mentioned in their paper "Rigorous bounds for polynomial Julia sets" that parabolic ones are notoriously difficult to render reliably. Well, I can again corroborate that.

I've read in a text by Paul Blanchard ("Complex analytic dynamics on the Riemann sphere", 1984) about the formula z²+1*z (fig. 3.16), which is by design creating a Julia set with a parabolic fix point at the origin. I computed that - it looked like a shifted z²+0.25 set. Other values (-1*z, i*z, -i*z) had their parabolic points on the boundary and the curvature argument came into play. So that was not working.

So I tried some other values with length 1. And the first thing that comes to mind when seeing x²+y²=1 is the Pythagorean triple 3,4,5. So I tried z²+z*(-3/5-4/5*i) which - point-sampled - showed a feature I have been looking for but never found so far: A parabolic periodic point in its interior, i.e. the Fatou set and not the Julia set itself (see image 1 below, point-sampled, 20000 maxit, __float128 precision, originally 4-square, cropped out).

So no curvature, I started computing it with the TSA, using small intervals as -0.6 and -0.8 are unfortunately not representable as dyadic fractions.

Calculations went on - but no interior cells at L16 (see image 3 below, it's an intermediate result at level 17, trustworthily downscaled of the 4-square, then cropped out).

So I took a look at the parameter-space and it turned out -0.6-0.8i was - again - on the boundary (see image 2 below, point-sampled [-3..5] x i*[-4..4]).

Not unexpected but I hoped for it to be in its interior, because using small complex intervals (hence a rectangle) on a point at the boundary of a curved object does not work. There are always seeds that lie on the outside, so since the black region of an interval-c-value Julia set is the intersection of all individual black, well, there won't be any then.

Then I thought, maybe there are other values whose coordinates are dyadic fractions and can be used directly.

So I took a closer look on $$x²+y²=1$$ which as a dyadic fraction would be $$a/2^n, b/2^n$$ (brought to the same denominator), hence $a²/2^{2n} + b²/2^{2n} = 1 = 2^{2n}/2^{2n}$ which is $$a²+b²=2^{2n}$$.

But that looked suspiciously like Fermat's theorem a²+b²=c², and for natural numbers there's only one solution: 3,4,5 (unless one is 0, but that's trivial).

So no dyadic fractions to find besides the numbers I used (+-1, +-i) which are rationally indifferent, so, as I read lately, their parabolic periodic points are in the Julia set so I would need to look at the curvature again.

(I kind of hope someone spots an error in that line of argumentation, since I'd really like to have a 2nd parabolic set with interior besides z²+0.25).

So that leaves me now with two options:

• Look for a different formula where the parabolic points are in the Fatou set, the parameter used however is either non-trivially dyadic or not at the boundary of the corresponding parameter-space - if such a value exists (from my last futile experiments I currently doubt that).

If someone knows of Julia sets of any degree where the parabolic periodic points are in the Fatou set, I'd appreciate a short notice. If I understood all correctly, that would mean those sets are irrationally indifferent, but not all irrationally indifferent Julia sets's periodic parabolic points are in the Fatou set.

• Try combining arbitrary precision with the interval aiorthmetics used in the TSA. I only need it once when computing the bounding box, afterwards the determined intersection with the grid squares will start the following calculations again with a dyadic number.

#### claude

• 3f
• Posts: 1736
##### Re: Parabolic Julia sets
« Reply #7 on: September 30, 2019, 02:22:30 PM »
There are an infinite number of Pythagorean triples. Whether any have the desired form with the largest item a power of two I don't know.

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #8 on: September 30, 2019, 04:18:18 PM »
@claude: You're right as always. I misrecalled the result of Fermat (nothing for power 3 or above).

So there's still hope. I guess it's time for a brute-force approach (the denominator is set to constant 2^32 or 2^25 in my program, maybe it's tractable.

EDIT: I just did a quick search for natural numbers a,b which when squared and added result in an even power of 2. I did not find any up to 2^60 as the sum. So I used an incorrect argument in my last post, but unfortunately no different outcome here.

But interestingly, it works for base 25 or 125 (and further): (7/25)2 + (24/25)2 = (252)/(252)
(44/125)2 + (117/125)2 = 1252/1252 and more.

But I won't go that road for now - simulating base 5n numbers.

« Last Edit: September 30, 2019, 05:09:28 PM by marcm200, Reason: further results »

#### claude

• 3f
• Posts: 1736
##### Re: Parabolic Julia sets
« Reply #9 on: September 30, 2019, 05:25:05 PM »
Because an odd square is congruent to 1 modulo 4(*), there can be no primitive Pythagorean triple a^2+b^2=c^2 with c a power of 2 (a and b must both be odd, so the left hand size is 2 (mod 4), while the right hand side is 0 (mod 4) because c is even).  This means further that there can be no triple at all with c a power of 2, because otherwise you could divide out factors of 2 to get a primitive triple with c' a power of 2.

(*) fact found on Wikipedia https://en.wikipedia.org/wiki/Pythagorean_triple#Proof_of_Euclid's_formula, its proof by induction is straightforward I suppose.

#### pauldelbrot

• 3f
• Posts: 2422
##### Re: Parabolic Julia sets
« Reply #10 on: September 30, 2019, 08:27:04 PM »
Mapping $$z^2 + ze^{\frac{c}{\pi}}$$ should have rational (parabolic) indifferent points at rational pure imaginary coordinates (including all c = ix/2n). Irrationally indifferent points produce Siegel disks rather than parabolic attracting cycles.

The Julia set image shown for z2 + z(-3/5 - i4/5)was for a Siegel disk case.
« Last Edit: October 31, 2019, 03:54:01 PM by pauldelbrot »

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #11 on: October 31, 2019, 09:57:42 AM »
"Assisted detection"

I was interested to see what the TSA detects if one marks some cells as interior at the start of the algorithm. I hoped to circumvent the no-interior-cells-problem of parabolic Julia sets.

Here is the z²-0.75 case which will not show interior cells by the Figueiredo algorithm (as stated by the authors).

If I mark a region around the origin (18576 cells, red) as interior, the algorithm detects at level 11 a further ~100,000 new black cells. So some cells have paths to only the manually introduced interior cells and could therefore be judged as interior as well.

Now two questions arise:

• How to prove that a specific region is actually in the filled-in Julia set to get an initial black starting set?
• Is there an initial set of black cells so that in the process of refinement the whole interior of the filled-in set will be found? Or will it be just a portion, finding roughly the same (relative) points in ever smaller getting components of similar shape?

Does anyone know of proofs for specific (subsets of) points being in the filled-in Julia set for the parabolic case?

#### pauldelbrot

• 3f
• Posts: 2422
##### Re: Parabolic Julia sets
« Reply #12 on: October 31, 2019, 03:56:47 PM »
Quote
s there an initial set of black cells so that in the process of refinement the whole interior of the filled-in set will be found?

Such a set will need to include the parabolic point itself, while avoiding exterior points. Your rectangle doesn't do that; however, a triangle with base through the origin and point at the parabolic point might get the job done. If the base of the triangle is not wide enough there will be gaps of grey between the preimages and if the base is too wide there will be misclassified exterior points.

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #13 on: November 01, 2019, 10:15:53 AM »
Such a set will need to include the parabolic point itself, while avoiding exterior points. Your rectangle doesn't do that; however, a triangle with base through the origin and point at the parabolic point might get the job done.
As I'm working with complex intervals (squares) and the parabolic point is at the corner of 4 of them at any refinement level, I think any triangle ending in such a box encompassing the parabolic point would also have exterior points, so I cannot color it black manually. But using different shapes as initial black seeds and coloring near the parabolic point is a good idea I will follow on.

I just started reading about parabolic petals. Maybe those would be interesting regions to color black - I suspect its members having paths to the parabolic point (?) and could therefore not be colored black by the TSA, but setting them to interior manually might break that and maybe part of the basin of attraction for that petal will get determined as well? Just speculating though, the petal theme is quite complicated for me.

#### marcm200

• 3c
• Posts: 894
##### Re: Parabolic Julia sets
« Reply #14 on: December 16, 2019, 05:01:55 PM »
"Plane tiling with triangles"
(long text, but I need a bit of notation first)
(based on pauldelbrot's triangle suggestion a few posts back)

Note, all images in this post are point-sampled.

Recap
Picture 1 shows why cell mapping with squares does not detect interior in parabolic sets (except c=0.25 as of now).

Notation beforehand
Black = interior of Julia set z²-0.75, white=escaping
Red rectangles = square tiling of the complex plane
Blue object marked "S" (right half of picture): source square, blue pixels are starting complex numbers for one iteration
Blue object marked "T" (left half of picture): image object of S
Yellow rectangle around T: bounding box as computed by interval arithmetics
Exaggerated point near T: parabolic fix point -0.5+0i

Both the yellow bounding box and the red squares it intersects contain interior and exterior of the Julia set - and will at all finite sizes. As the fix point iterates to itself, i.e. in IA all squares containing the parabolic point have paths to interior and exterior and can never be black. Neither can squares that have a path to the parabolic point (so basically everything afaik).

The source of the problem is that the Julia set towards the parabolic point has curvature - but bounding box and tiling squares are parallel to the axis.

Triangle tiling
But what would happen if one tiles the complex plane with e.g. triangles? (picture 2)

Now there are image objects that lie completely in the interior but touch the parabolic point. If one could calculate a bounding polygon (maybe a triangle itself?), this lies completely in the interior as well and hence allows in principle bounded orbits.

Is the image object of a triangle curved inwards to the interior (under z²+c)? If so, one could just connect the endpoints and have a triangle. In the other case one needs to construct the convex hull.

However, there are a number of issues here:

First, the parabolic point belongs to multiple triangles, and since some of them do contain both interior and exterior, it will never be marked is interior, neither will any source triangle, so still no interior detection possible.

But would it be tractable to tile the complex plane with disjoint triangles? I.e assigning the parabolic point to only one (carefully chosen?) triangle, and doing the same with shared edges, leaving the other triangles open sets.

Secondly, the curvature in the y>0-half plane is opposite to the south part, so maybe I need to flip the triangles orientation there.

This goes a bit in the direction of affine arithmetics which exploits the fact, that the bounding box usually is not filled up completely by image points (Figueiredo et. al, Affine Arithmetic: Concepts and Applications, 2004), however I have not yet fully understood that. Does anyone have experience with other error-free computation models than interval arithmetics?

Maybe I'll make this a project for 2020.

Any thoughts are welcome!

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