20 different methods of inverting the Mandelbrot set

Or: how to get from c to 1/c, and back again.
First the cardioid of the main body of the Mandelbrot set is converted to a circle After the inversion the process is reversed so that when a second inversion is applied one gets back to the original image.
In the unlikely event that someone is interested in the formulas used, please ask and I won't show any hesitation in sharing them.

[@GawdonBennett]

I'd like to take a look at the formula...thanks

[@Arneauxtje]

Well, there's 20 of them, as the title of the video suggests, on how to get from c to 1/c.
But they all group in 3 families:
1. addition: c -> c-c*t+t/c and t=0..1
2. multiplication: c -> c/(t*(c*c-1)+1) and t=0..1
3. exponentiation: c -> c^t and t=1..-1
The first 2 are pretty elementary to work out. The 3rd makes use of the fact that:
c = a+ib = r*exp(i*phi) where r=sqrt(a*a+b*b) and phi=arctan(b/a)
then c^t = r^t*exp(t*i*phi) = r^t*[cos(t*phi)+i*sin(t*phi)]

[@GawdonBennett]

Thanks for prompt reply

[@00001__]

@@GawdonBennett Say Thanks!

[@noahdirksen3623]

@@00001__ bro?

[@user-rl9gb9ho1w]

I HAVE FORMULA

[@arthurrochamattathecreator3070]

That pop on the 3rd, 5th and 7th one 😳

[@alan2here]

You kind of want to flip the bulbs over the cardioid, and (or perhaps not) flip the feature on the right side of the set as well.

[@user-oj7dq3wr7k]

ج

[@WevordinDoesAndroidEditor]

0:30 inverted Monobar

[@Tootbird]

0:21 spider on the 6th

[@nodrones]

hi! would you please share formulas for the 3, 15 and 16 ones? counting left to right, top to bottom. thank you!

[@Arneauxtje]

Well, first there's the transformation from the cardioid of the body of the set to a circle. This is done in c-space (a+ib) as follows:
rho=sqrt(a*a+b*b)-1/4
phi=arctan(b/a)
a-new=rho*(2*cos(phi)-cos(2*phi))/3
b-new=rho*(2*sin(phi)-sin(2*phi))/3
Frame 3, 15 and 16 are based on an inversion through exponentiation, i.e.:
c -> c^t and t=1..-1
Making use of the fact that:
c = a+ib = rho*exp(i*phi) where rho=sqrt(a*a+b*b) and phi=arctan(b/a)
then c^t = rho^t*exp(t*i*phi) = rho^t*[cos(t*phi)+i*sin(t*phi)]
This is frame 3.
Frame 15 and 16 is a variant where rho^t in the equation above is substituted with rho/(t*(rho*rho-1)+1) and t=0..1
Additionally, frame 16 substitutes cos(t*phi) with cos(phi), in other words, the transition is cancelled for Re(c) but kept for Im(c).
I hope this all makes sense to you.

[@nodrones]

@@Arneauxtje it does! thank you so much for a prompt reply! it's an awesome job!

[@scribbllllll]

0:18 19th one *eclipse with glasses 15th without

[@lucaspierce3328]

Holographic Holo-Fractal Stochastic Super-Tension Shape Dynamics!.

[@WevordinDoesAndroidEditor]

0:07 Monobar

[@adammajewski6381]

I'm interested in the formulas used. Please share them.

[@Arneauxtje]

Happy to oblige. But could you be a bit more specific though?
Also, in the comments below a lot has been disclosed already. If you would like some additional explanations on some of the procedures, please let me know.

[@adammajewski6381]

@@Arneauxtje right ( I see them after expandig the comments tree ) Thx

[@truongquangduylop33yyuh34]

@@Arneauxtje i want formula for 4

[@Arneauxtje]

@@truongquangduylop33yyuh34
First there's the transformation from the cardioid of the body of the set to a circle. This is done in c-space (a+ib) as follows:
rho=sqrt(a*a+b*b)-1/4
phi=arctan(b/a)
a-new=rho*(2*cos(phi)-cos(2*phi))/3
b-new=rho*(2*sin(phi)-sin(2*phi))/3
Frame 4 is based on an inversion through exponentiation, i.e.:
c -> c^t and t=1..-1
Making use of the fact that:
c = a+ib = rho*exp(i*phi) where rho=sqrt(a*a+b*b) and phi=arctan(b/a)
then c^t = rho^t*exp(t*i*phi) = rho^t*[cos(t*phi)+i*sin(t*phi)]
This is frame 3.
For frame 4 the t in in cos(t*phi) is dropped, i.e.:
c^t = rho^t*[cos(phi)+i*sin(t*phi)]
Hope this helps.

[@afteroayfractalspro4486]

John

[@survivalizer]

Is this not just an inversion of the human body?

[@Arneauxtje]

or at least the human psyche.

[@yahoo5726]

Ok, now what do I do with my newfound confusedness

[@ConsciousRobot]

Wow!!

[@thisisachannel8472]

0:19 is this me or did it lag?

[@Arneauxtje]

I think it's you

[@DmitriiFractaler]

0:05 - 20 monobars?

[@lukatolstov5598]

A 1st what ap? At 2nd I want formulas for 1 and 2.

[@Arneauxtje]

No app, write my own code, mainly in assembly.
1: c -> c-c*t+t/c and t=0..1
2: c -> c/(t*(c*c-1)+1) and t=0..1

[@00001__]

Make 100 Diffrent Ones

[@00001__]

Or 36

[@adammajewski6381]

other method of plane transformations : commons.wikimedia.org/wiki/File:Feigenbaum_stretch_3.png

[@Arneauxtje]

Very interesting, I'll look into it. Thanks.

[@chrisrodriguezm13]

0:05 everything is monobar LOL

[@KingGreenscreenKid420]

Monobar formula:
z^2 + (c iabs c+0.25)

[@__________________________hi52]

​@@KingGreenscreenKid420 wrong it's actually
complex p;
complex t;
complex r;
complex d;
complex y;
complex o;
complex pa;
pa = 1;
y = (0.1);
d = c^w;
o = 1;
p = arg(d);
r = cos(2/o*arg(cos(p* o/2)));
d = d * r;
r = rad(d) - 0.25;
p = arg(d);
d = r * (2*exp(1i*p) - exp(2*1i*p)) / 3;
bailout = 10;
t = pa;
z = (z)^po + t * d + (1 - t) * c;

[@KingGreenscreenKid420]

​@LatinNegativeElolnoma no shit sherlock i didn't know

[@DoNotExpose5624]

Obviously not, it is not a monobar, It is a half lambda or a circle body mandelbrot

[@chrisrodriguezm13]

@@DoNotExpose5624 No, I thought that 2 years ago

[@cheems0]

ı am dızzy 😵

[@testingman1a797]

0:19 DONOT look at the 2th mandelbrot to the top

[@Arneauxtje]

Takes one to know one

[@tesscruzat5277]

0:19 DONOT look at the 2th mandelbrot to the top

[@KingGreenscreenKid420]

Uhm actually its
"1st"
"2nd"
"3rd"
"4th"
🤓

[@arthurrochamattathecreator3070]

hmmm

[@DoNotExpose5624]

​@@tesscruzat5277stop copying him