"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.