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Author Topic: binary decomposition and external angles  (Read 1982 times)
Description: for quadratic polynomials
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tit_toinou
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« Reply #15 on: June 08, 2015, 09:32:37 PM »

Rather the grid is generated implicitly by iterations of z→z²+c, see http://www.mrob.com/pub/muency/binarydecomposition.html for the basic idea, you can get "local coordinates" within each cell by taking the fractional part of continuous escape time and the argument of the first iterate to escape.
Yes I've done that inspired by images like


But I can't establish the formula of the map from this space to the the Poincare Disk sad http://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model...
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claude
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« Reply #16 on: June 08, 2015, 10:19:10 PM »

Perhaps work with polar coordinates - points in the complement of the Mandelbrot set have external angle (I use Newton's method to trace rays, billtavis has an alternative using Runge-Kutta integration), and continuous iteration count heading towards infinity at the boundary of the set.  Use the continuous iteration count as the hyperbolic distance from the origin.  To convert hyperbolic distance to Euclidean distance in the Poincaré disc model, you need to invert the formula on the wiki page, that is, you have d(0,r) and want to find r, turns out

r = tanh(d/2)

(at least according to http://www.wolframalpha.com/input/?i=solve+d+%3D+acosh%281+%2B+2+x^2+%2F+%281+-+x^2%29%29 which I didn't verify by hand, and I might have made a mistake in the input...).

But tanh(20) is already rounded to 1.0 in double precision, so this is only good to 40 iterations max (and will already be horribly grainy due to insufficient precision before then), and then double precision for external angles runs out at 53 iterations max.  High precision maths is needed, no way out that I can see.
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