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Author Topic: Area of Julia Set  (Read 1556 times)
Description: research
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hgjf2
Fractal Phenom
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Posts: 456


« on: June 20, 2013, 10:02:56 AM »

Yet , neither scientist could find a formula for area of JULIA SET , just by formula {z|P(P(P...P(z)...))<>infinity, P(z) = z*z+c}.
 snore hrmm
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hgjf2
Fractal Phenom
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Posts: 456


« Reply #1 on: June 20, 2013, 10:09:04 AM »

Yet , neither scientist could find a formula for area of JULIA SET , just by formula {z|P(P(P...P(z)...))<>infinity, P(z) = z*z+c}.
 snore hrmm

But I surprised that for c=-1 from formula {z|P(P(P...P(z)...))<>infinity, P(z) = z*z+c}, area of this Julia Set seem that give 1,4142135623... , and I know is equal with sqrt2.
 surprised surfing
I knew after I used VISUAL BASIC and C# for calculate this area with MONTE CARLO METHOD and filling space and comparing method because I don't know neither math formula for JULIA SET quadrating and calculate area. But I don't stand how the area have given sqrt2 for P(z) = z*z-1.
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Pauldelbrot
Fractal Senior
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Posts: 2592



pderbyshire2
« Reply #2 on: June 20, 2013, 11:06:09 AM »

This colors the Mandelbrot set interior by the areas of Julia sets. The UF gradient is mapped to the range of areas from 0 to pi (the area of the Julia set at (0,0), which seems to be close to the maximum attained).

Code:
JuliaArea {
final:
  int ct = 0
  int p = 0
  int seed = 385472
  WHILE (p < @julpoints)
    int i = 0
    seed = random(seed)
    float r = 2*(seed/#randomrange)
    seed = random(seed)
    complex x = r + (0,2)*(seed/#randomrange)
    WHILE (i < @juliters && |x| < 2)
      x = sqr(x)+#pixel
      i = i + 1
    ENDWHILE
    IF (|x| < 2)
      ct = ct + 1
    ENDIF
    p = p + 1
  ENDWHILE
  #index = (ct*16)/(#pi*@julpoints)
default:
  title = "Julia Area"
  param julpoints
    caption = "Number of samples"
    default = 3000
    min = 1
  endparam
  param juliters
    caption = "Julia maximum iterations"
    default = 30
    min = 1
  endparam
}



* julareamandel.png (84 KB, 640x480 - viewed 349 times.)
« Last Edit: June 20, 2013, 11:07:40 AM by Pauldelbrot » Logged

matsoljare
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Posts: 215



WWW
« Reply #3 on: June 20, 2013, 10:59:32 PM »

Yeah, that is pretty much the same as the "Supermandelbrot" rendering that i proposed a couple of years ago, not much have been done of it though:

http://www.fractalforums.com/new-theories-and-research/supermandelbrot-and-superjulia-imaging
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hgjf2
Fractal Phenom
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Posts: 456


« Reply #4 on: June 21, 2013, 10:43:20 AM »

Nice rendering on MANDELBROT SET graphic where the color show value of area of Julia set at each point.
If making same algorithm on cubic Julia Set  {z|P(P(P...P(z)...))<>infinity;P(z) = z*z*z+z*z+c}, the graphic cover two cubic Mandelbrot sets have at base formula fc(z)=z*z*z+z*z+c with starting from z=0 and z=-22/27 whick is simetrical , one is another mirrored by the ax {z=a-(1/3)i;\/a<-R}
   
  LEGEND OF MATH SYMBOLS: \/  anything to be ;    <- apartening ;  <> different; R  set of real numbers
 
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Pauldelbrot
Fractal Senior
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Posts: 2592



pderbyshire2
« Reply #5 on: June 21, 2013, 12:07:58 PM »

Nice rendering on MANDELBROT SET graphic where the color show value of area of Julia set at each point.

Thanks.
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