🚀 go-pugleaf

          My Mandelbrot Flower
          (C) by Jay Hill, 1993

  She loves me, she loves me not
    Will it be no or will it be yes?
  From my flower of Mandelbrot
    The answer will be anyone's guess.

  It's the area of Mandelbrot,
    Quadratic, cubic and quartic petal
  Whose total is dearly sought.
    But each component is transendental.

  1.17809724509205 is so simple,
    The period 1 cardioid, you see.
  0.19634954084867 for a circle,
    0.05654241812768 for period three.

  With petal on bud and bud on petal
    In an organized infinity.
  The solution will never settle
    Until the clock strikes eternity.


That is three petals down, infinity to go.  I want to 
thank Yuval Fisher for formulas without which I could
have computed the area in period 3 components.  This
is a numerical/analytical result.  Next, period four.

Yuval and I have modified the GIPPP program so it now
computes an upper and lower limit to the MSet area.
The code is based in part on work of Marc Parmets and
information in Appendix D of the Science of Fractal Images.
It depends on accurate determination of the period at
any sample point.  The program will be ready shortly.
We plan to have about 120 segments, each square, which
when added together, will give the desired area estimates.
Preliminary results show the upper bound is 1.60986378
and the lower bound is 1.49713877. Since we know the 
contribution for each period and the precise value for
periods 1, 2, and 3, the lower bound can be adjusted
to 1.50403787.  The difference was taken from the 
'inbetween' area which the distace estimator method
always has.  Please watch for Yuvals posting when it
is ready. We figure each subarea will require about
12 to 24 hours on a workstation.

Warmly,
Jay Hill

-- 
{ hilljr@jupiter.saic.com } begin writeln(3*ln(640320)/sqrt(163):17:15) end.
void main(){double sqrt(), y=1/sqrt(2.), a=.5, m=1,z; int n=0;
for(;m*=2,z=sqrt(1-y*y),y=(1-z)/(1+z),a=a*(1+y)*(1+y)-m*y,n<3;n++);
printf("%17.15lf\n",1/a);}