Deriving the moebius transformation that maps 3 points to 3 other points
is explained here(section 3.4.11 Building a mobius transformation):
https://mphitchman.com/geometry/section3-4.html
Using equated cross ratios. (handling infinity is also explained.)
For turning the moebius transformation formula into a formula involving x and y,
see the 4th answer here:
https://math.stackexchange.com/questions/36542/real-and-imaginary-parts-of-the-m%C3%B6bius-transformation
Uses a, b, c, d, (real numbers with imaginary parts a2= b2= c2 = d2 = 0) ,
and alpha, beta, gamma, and delta.
After about 8 pages of tedius hand calculation with many corrected mistakes,
the moebius transformation and the x and y formulas were calculated, for the current mapping of the attractor points to -1 and +1, and also to
-2 and +2 . The simple expected result is that this scaling by 2 is equivalent to multiplying the x and y formulas by 2. So a scaling input could be added to the double doyle. Or the scaling could be done after forming the double doyle, by MoI scaling.
The original moebius transformation is V(z) = (z-1)/(z+1).
The 2 times scale moebius transformation is V(z) = 2*(z-1)(z+1).
For the x and y formulas:
The new denominator is the same as used in double doyle node.
denom = (x+1)*(x+1) + y*y.
The numerator is 2 times the original numerator.
New numer = 2*((x*x - 1) + (y*y))
- Brian
