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Author Topic: Glynn Julia set  (Read 23551 times)
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David Makin
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« Reply #60 on: January 30, 2010, 03:10:39 PM »

I think you call this formula the 'flipped' one :
Code:
z= (R^(@pow/2))*sin(@pow*ph)*exp(i*@pow*th)+c1
   zz=(R^(@pow/2))*cos(@pow*ph)+cz
     R=(|z|+zz*zz)
     th=atan2(z)
     ph=atan2(zz+i*cabs(z))
     if ph>=#pi/2,ph=#pi-ph,endif

Yea - the one with the adjustment to phi smiley
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« Reply #61 on: January 30, 2010, 11:46:20 PM »

I think you call this formula the 'flipped' one :
Code:
z= (R^(@pow/2))*sin(@pow*ph)*exp(i*@pow*th)+c1
   zz=(R^(@pow/2))*cos(@pow*ph)+cz
     R=(|z|+zz*zz)
     th=atan2(z)
     ph=atan2(zz+i*cabs(z))
     if ph>=#pi/2,ph=#pi-ph,endif

Thanks Jos. Yes, that is the 'flipped' version.
To recap, the reason for the parity changes in some versions is to get equivalence between forward and inverse iteration for a given formula, which is necessary due to phi giving a double cover when iterated beyond 0 to pi.
The usual forward iterations have no parity change (of course), and to reproduce them using MIIM we have to add some parity change in the inverse method.
This flipped version originally resulted from a cosine phi inverse iteration without any parity, which turned out to be often more interesting than the one with correct roots. To duplicate it in forward iteration requires a parity change there instead.

However, that simple parity formula is only "correct" for integer powers, so you've discovered another useful mutation. If you haven't already tried it without the parity change to phi, that would give the "true" cosine version, which should be asymmetric. I haven't figured out how to reproduce this flipped Glynn inversely.
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xenodreambuie
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« Reply #62 on: January 31, 2010, 12:00:56 PM »

I have solved it, sort of. The actual inverse of Jos's formula is simple, but it has a major flaw. It has two disjoint orbits, so it needs two different seed points to produce the complete image. That's very inconvenient because it needs a modification of the MIIM routine, and also to choose two good seeds for a given set of parameters. I currently adjust a seed interactively, and to do two that way would get painful. The picture below is one of the orbits for (-0.1,0,0), which shows the patchwork pattern more clearly. The best seed points are in one of the big patches on the left. I'm guessing that deducing a formula for ideal seeds depending on parameters is difficult. Another possibility is if there is some mapping between the two orbits that could be used like an extra root choice.

The partial formula I used is:
// two roots each for phi, theta: kphi, ktheta = 0 or 1
power = 1.5
Rootlimit = 1.5pi
theta = arctan2(y,x) + 2pi*(ktheta - floor((theta+rootlimit)/2pi))
if abs(theta)<rootlimit then
  phi = arccos(z/r)
  if phi>0.75pi then phi = pi - phi
  phi = phi/power
  if kphi>0 then phi = pi - phi
  theta = theta/power
  r = r^(1/power)
  ... (standard calculations from r,theta,phi)
else // root not valid



* Glynnflipproblem.jpg (18.86 KB, 263x217 - viewed 484 times.)
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Regards, Garth
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