Claude
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Parts of the Mandelbrot set look like circles with smaller circles attached recursively: They aren't quite exact circles (except for the one centered at $-1+0i$), but the property that the radius of ...

I wrote a small Haskell program using the Diagrams library: import Diagrams.Prelude hiding (aspect) import Diagrams.Backend.SVG.CmdLine (B, defaultMain) import Data.Complex main :: IO () main = ...

$i$ is indeed the tip of a filament. However, mini-Mandelbrot copies are dense in the boundary of the Mandelbrot set, which means that for any radius $r > 0$ you can find a little Mandelbrot copy ...

You can fill the finite extruded Koch snowflake 3-D volume with a finite amount of 3-D paint. But the Koch snowflake, while topologically a 1-D line, has Hausdorff dimension $\log_3 4 \approx 1.2618\... View answer Accepted answer 7 votes fractint is one of the oldest programs, it has a formula interpreter. xaos also supports user-settable formulae. shadertoy probably has many escape-time fractal experiments on it that you can edit ... View answer 7 votes Subtract the point, rotate around origin, add the point back: $$(-(y-b)+a, (x-a) + b)$$ View answer 7 votes If$(p_x,p_y)$are the coordinates of a point off the parabola and we'd like the line from$(p_x,p_y)$to the point$(x,x^2)$on the parabola to be normal to the parabola, then we need: $$\begin{... View answer Accepted answer 7 votes The start value x_0 is determined by the coordinates of each pixel. Pixels are coloured acccording to how quickly the orbit for that pixel diverges (escape time colouring). Black pixels remained ... View answer 6 votes You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop. Points at rational internal angles (... View answer 6 votes As this graph shows, the behaviour of normalized iteration counts of points near the Mandelbrot set varies widely, indicating that attempts at a formula based on scale factors is doomed to fail. In ... View answer 6 votes Pretty sure it's an aesthetic choice, as Mark says in his comment. It arguably makes a more satisfying end to end inside a mini copy (or just before). Alternatives are to end outside the set (which ... View answer 6 votes One strategy to find some interesting (at least to my own subjective taste) places (which might not make interesting videos) semi-automatically is to spot patterns in the binary expansions of external ... View answer 6 votes https://commons.wikimedia.org/wiki/File:Multibrot_Lupanov_power-2_Q1.png suggests using the Lyapunov exponent as a stability criterion, before performing periodicity checking for colouring. The ... View answer Accepted answer 5 votes This is not a method of construction, but a brute force exhaustive search. Not mathematically elegant, but may eventually give you a maximally dispersed Latin square. Wikipedia says there are ... View answer 5 votes You want to colour the complement of the Mandelbrot set using the exterior distance estimate. For each pixel you calculate the running derivative (w.r.t. c) as well as the z iterate, then at the ... View answer Accepted answer 5 votes I think yes Consider the sequence of "westernmost" islands increasing in period: Here are some examples, you can see them increasing in hairyness / spinyness. Period 20: Period 30: Period 40: ... View answer Accepted answer 4 votes The Viper substitution tiling divides a particular triangle into 9 similar sub-triangles without reflection, in a non-trivial way. The same triangle can also be divided into 4 similar sub-triangles ... View answer 4 votes Windows of Periodicity Scaling gives a formula \beta \Lambda_p^2 for the size (and orientation, taking it as a complex number) of a minibrot island with nucleus c of period p relative to the top ... View answer 4 votes Your Julia set is similar to the one for z^2+\frac{1}{4}, as mentioned by other answerers. This Julia set is parabolic, and parabolic Julia sets are hard to compute accurately efficiently. One ... View answer 4 votes The book Indra's Pearls covers this stuff. Quoting Wikipedia: Indra's Pearls: The Vision of Felix Klein is a geometry book written by David Mumford, Caroline Series and David Wright, and published ... View answer Accepted answer 4 votes Here's a zoom into the Mandelbrot set centered on i, with a factor of 10 magnification per frame. As you can extrapolate from the first frame, your path will have to go through an infinite number ... View answer Accepted answer 3 votes Substitute x=m \cos t to get$$m^3 \cos^3 t - a m \cos t -1=0$$Let$$m=\sqrt{\frac{4a}{3}}$$then it rearranges (assuming a,m\ne 0) to$$4 \cos^3 t - 3 \cos t = \sqrt{\frac{27}{4a^3}}$$Using ... View answer Accepted answer 3 votes Consider iteration of z \mapsto c^{\sin(z)} near c:$$z \mapsto c^{\sin(c + z)} - c$$The Taylor series near z = 0 is (via Wolfram Alpha):$$ z \mapsto \left(c^{\sin(c)} - c\right) \\ + \left(\... View answer 3 votes The first image may be a zoom into a quadratic Julia set for$f_c(z) = z^2 + c$near$c = 0.270723273 + 0.575139611 i$, centered on$0$with zoom factor$5.3$(zoom factor$1$would have$\pm i$at ... View answer 3 votes For the promenade around the main cardioid, the point in the Julia set you want is a fixed point$z = f_c(z)$. By tuning (renormalization), the point you want in the "little Julia set" at ... View answer Accepted answer 3 votes Shapes can be approximated by Julia sets of rational functions with a large number of roots [1]: we are [...] interested in factored rationals of the form$$R(q) = e^C \frac{ \prod (q - t_i)^{T_i} }{... View answer Accepted answer 3 votes The first thing to note is that each of the hyperbolic components (either cardioid-like or disk-like) is associated to a period, which is a positive integer. The biggest cardioid has period$1$, the ... View answer 3 votes You may be able to calculate a graph-directed IFS for the boundary of the tile, though the rotated tiles may make this tricky - I'm not sure whether this is classed as a self-affine tile, or whether ... View answer Accepted answer 3 votes First answer: "the$\delta^3$term remains significantly smaller than the$\delta^2$term" means that$\left|C_n \delta^3\right| << \left|B_n \delta^2\right|\$, usually a few orders of magnitude ...