"Flowers, petals and Dv"

The polynomial z-z^5 has a parabolic fix-point with derivative 1 at the origin and 4 petals (upper row, image I, 2-square, escape time, images marked with PS are point-sampled, non-rounding controlled, double precision; images marked R are reliably computed, interval arithmetics, a pixel represents a complex square).

The point-clouds

\[

D_v(d) = \{ z \in C : |z^v - d| < d \}

\]

(v in the above case is 4) form flowers (upper row, image II).

If d is sufficiently small, these D_v sets can be taken as attracting petals (M Abate, Fatou flowers and parabolic curves, 2016. Equation 1.3). Unfortunately, the author did not state how one can arrive at or prove a value to be sufficiently small. Does anyone know of such a way? Maybe the proofs of the Leau-Fatou-Flower theorem provide a constructive method, but I haven't understood those to that extent.

For small enough petals, the polynomial maps a petal to itself (AF Beardon, Iteration of Rational Functions, theorem 6.5.4). So I thought the same might happen with the Dv sets and performed the following algorithm:

1. Calculate the flower set D_v for some value d that is definitely too large, here d=4 in a reliable manner (image II, black: pixel's square lies fully in the flower; gray, pixel's square straddles the flower's boundary; white, fully outside).

2. Iterate each black square A once under f (via IA) and judge the image f(A) whether it lies fully in the flower in general (it was not tested, if it is fully in the originating flower petal). If it is inside, A is kept (turned blue), otherwise it is discarded (set to white) (image III) and the value d judged as too large. For illustration purposes, the boundary of the flower is kept (gray).

3. Overlaying the judged image and the original parabolic one shows that it is indeed too large (IV, green pixels lie both in the flower and are (point-sampled using one corner) interior; red are within the flower, but lie in the point-sampled exterior), and the flower itself is way too large for the Fatou component it intersects with.

4. Choose a smaller d, e.g. d/2 and repeat this process until no pixel A is discarded. For d=0.25 this was the case (VI). Overlaying the flower after this f(A) test with the point-sampled version, shows it is fully in the interior (image VII).

I wonder, whether this is a general way to arrive at a sufficiently small d in the above petal sense, or at least one whose Dv set lies fully in the interior of the filled-in Julia set? Any proof ideas or counter-examples?

Starting from blue does not necessarily mean landing in blue, as f(A) might intersect with a gray pixel (so the pixel itself is not fully inside), but the part of f(A) intersecting with it, still is (even in axis-parallel IA overestimation). I think this will occur near the fix point.

Such a definite d value and its Dv set could then be used as a starting set of interior squares for the assisted TSA to detect (large?) portions of the parabolic interior.