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.

Linkback: https://fractalforums.org/fractal-mathematics-and-new-theories/28/parabolic-julia-sets/3091/

https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/parabolic

images with code:
https://commons.wikimedia.org/wiki/Category:Parabolic_Julia_sets

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)

"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: 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.

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.

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.

"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!