• April 23, 2021, 03:40:47 PM

### Author Topic:  hybrid formulas  (Read 1769 times)

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#### FractalAlex

• Fractal Frankfurter
• Posts: 541
• Experienced root-finding method expert
##### Re: hybrid formulas
« Reply #30 on: November 06, 2020, 04:50:47 PM »
There's something I also wanted to say: Is there a way to recreate the False Quasi Perpendicular/Heart fractals as well as the Quasi Perpendicular/Hearts with the hybrid formula system? And would it be possible to include non-integer powers? I think the latter is not possible...
"quotquotI am lightning, the rain transformed."quotquot
- Raiden, Metal Gear Solid 4: Guns of the Patriots

#### claude

• 3f
• Posts: 1839
##### Re: hybrid formulas
« Reply #31 on: November 06, 2020, 05:22:14 PM »
Looking at it I can't see if the Quasi formulas can be decomposed into the simpler operations needed for the KF formula designer.

How do you perturb arbitrary powers?  I guess you start from $$Z \to \exp(p \log(Z))$$, which I think works out (using perturbation algebra) as $z \to 2 \sinh(p\, \mathrm{log1p}(z / Z) / 2) \exp(p \log(Z (Z + z)) / 2)$
So it's possible, but whether it is worth the effort to implement sinh log1p exp log for complex numbers, with all the real (dual, floatexp) functions that they will need also, plus OpenCL versions of all of that, plus all the algorithms for Newton zooming (period checking, root finding, size estimate, skew calculation), it adds up to a surprisingly large amount of work...

EDIT the branch cuts ruin eveything by lack of continuity...
« Last Edit: February 03, 2021, 04:57:15 PM by claude »

#### FractalAlex

• Fractal Frankfurter
• Posts: 541
• Experienced root-finding method expert
##### Re: hybrid formulas
« Reply #32 on: November 06, 2020, 05:27:16 PM »
Oh, through factoring? I'll have a look at it.

#### FractalAlex

• Fractal Frankfurter
• Posts: 541
• Experienced root-finding method expert
##### Re: hybrid formulas
« Reply #33 on: November 06, 2020, 07:05:38 PM »
Just a question: how do I turn something like $$z = z^3 + c$$ into something like this?
$y = y(3x^2 - y^2) + b$
$x = x(x^2 - 3y^2) + a$

Where x and y are the real and imaginary components of z, and a and b, the real and imaginary components of c.

#### hobold

• Fractal Frogurt
• Posts: 420
##### Re: hybrid formulas
« Reply #34 on: November 06, 2020, 08:23:57 PM »
If I understand the question correctly, you start with the two correspondences you mentioned:

z = (x + i*y),
c = (a + i*b)

And then you plug that into, say, z^3 + c simply by substituting:

z_n+1 := (x_n + i*y_n)^3 + (a + i*b)

(meaning the (n+1)th z is derived from the nth x and y)

You have to multiply these out (remembering that i*i = -1), and finally re-group to isolate i again, to arrive at

z_n+1 = (x_n+1 + i*y_n+1) with the appropriate expressions for x_n+1 and y_n+1.

#### FractalAlex

• Fractal Frankfurter
• Posts: 541
• Experienced root-finding method expert
##### Re: hybrid formulas
« Reply #35 on: November 06, 2020, 08:36:56 PM »
Well, I tried, but I got a different result:
$x = x^3 - 2xy^2 - xy^2 - y^3 + a, y = 3x^2y + b$

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