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Author Topic: Simonbrot nth  (Read 2556 times)
Description: Does anybody have experience zooming into a Simonbrot with powers not 4 or 6?
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simon.snake
Fractal Bachius
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Experienced Fractal eXtreme plugin crasher!


simon.fez SimonSideBurns
« Reply #15 on: April 19, 2017, 12:12:30 AM »

He's upside down but apart from that, yes he's right.
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LionHeart
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« Reply #16 on: April 19, 2017, 12:20:35 AM »

Goody Simon,

I'll add him to ManpWIN smiley

Thanks for the tip.
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Paul the LionHeart
simon.snake
Fractal Bachius
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Experienced Fractal eXtreme plugin crasher!


simon.fez SimonSideBurns
« Reply #17 on: April 19, 2017, 12:23:31 AM »

If you change the initial condition to z=c=0-pixel then he goes the right way up.
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LionHeart
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« Reply #18 on: April 19, 2017, 12:32:37 AM »

Yep, thanks Simon.

I did indeed. I'll add it to ManpWIN that way.
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Paul the LionHeart
greentexas
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« Reply #19 on: May 11, 2017, 02:52:17 AM »

In April of this year, I discovered something:

My original formula for the Simonbrot 6th was wrong. The symmetry for the formula resembled the Simonbrot 6th (which is why I tricked myself), but it was so stupid of me for considering this to be the formula for the Simonbrot 6th:

z = (z^3 * |z|^2) +z0

This formula is obviously a 5th order fractal. z * z * z * abs(z) * abs(z) obviously has 5 zs. However, I will take credit for this formula as the 5th GT Quadratic Simonbrot. (GT stands for greentexas.) As an even bigger bonus, here is the code for the fractal (for you programmers):

ozx = zx;
zx = zx*zx*zx*zx*zx - 4*zx*zx*zx*zy*zy - (6*abs(zx*zx*zx))*zy*abs(zy) + 2*abs(zx)*abs(zy)*zy*zy*zy + 3*zx*zy*zy*zy*zy + cx;
zy = 3*ozx*ozx*ozx*ozx*zy - 4*ozx*ozx*zy*zy*zy - 6*ozx*abs(ozx)*abs(zy*zy*zy) + 2*ozx*ozx*ozx*abs(ozx)*abs(zy) + zy*zy*zy*zy*zy + cy;

The nth GT Quadratic Simonbrot has a formula of:

(z(n - 2) * abs(z)2) + z0

However, I used to think the formula for this fractal (and the normal Simonbrot) was:

(z(n / 2) * abs(z)2) + z0

It's difficult to explain when you have to clean up an ambiguous mess like this.

For an nth order Simonbrot, I discovered the formula is:

z = (zn/2 * abs(z)n/2) + z0

This is the true formula.

I was also mistaken by claiming the 0th power Simonbrot looked exactly like the 2nd power Burning Ship. It is just a boring circle.

Thank you, LionHeart, for showing us what the nth order Simonbrot looked like!
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vinecius
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« Reply #20 on: June 14, 2017, 05:10:07 PM »

It would also be nice to see Simonbrot 3rd, 5th, and 7th, but the formulas for these fractals are z1.5 * |z|2 + pixel, z2.5 * |z|2 + pixel, and z3.5 * |z|2 + pixel.

Due to the fact that the powers of some of the terms are fractions, this could be difficult to implement. In a roundabout way, there is a power zero Simonbrot on Kalles Fraktaler, because the power zero Simonbrot is actually the Burning Ship.

how are these?  questioning the non-symmetry in some of the regions (actually normal if you look at evolution of multibrot from z^2, the transitions are smooth but non-integer exponents introduce non-symmetries)

my understanding is that |z|^2 is a purely real quantity and has little effect on the original mandelbrot fractal other than the slight 'scaling'


* sim.png (70.99 KB, 800x2400 - viewed 229 times.)
« Last Edit: June 14, 2017, 05:37:09 PM by vinecius » Logged
vinecius
Forums Freshman
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Posts: 15


« Reply #21 on: June 14, 2017, 05:25:38 PM »

1,1.5,2,2.5,3,3.5,10 not all that different from the multibrot



* sim.png (93.81 KB, 800x4800 - viewed 232 times.)
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greentexas
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« Reply #22 on: June 16, 2017, 12:58:22 AM »

how are these?  questioning the non-symmetry in some of the regions (actually normal if you look at evolution of multibrot from z^2, the transitions are smooth but non-integer exponents introduce non-symmetries)

my understanding is that |z|^2 is a purely real quantity and has little effect on the original mandelbrot fractal other than the slight 'scaling'

That quote I made was a mistake. I made it a while back, but the formulas are all wrong! (Now I know the correct Simonbrot formulas.)
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