Mandelbrot set inversion, 4 different methods in one image

Where first the cardioid of the main body of the Mandelbrot set is converted to a circle, in an attempt to make the inversion look more interesting. After the inversion the process is reversed so that when a second inversion is applied one gets back to the original set.

[@Adam46221]

when you use the wrong formula in math but get the correct answer:

[@Masman40777]

Bottom left one looks pretty cool, i like how it collapses in on itself

[@whatupboi7834]

0:08 circlebrot

[@lukatolstov5598]

(Monobar)!

[@user-te3ii8ru1m]

Nice

[@WevordinDoesAndroidEditor]

The 4 is my favorite

[@ioium299]

Bottom right?

[@blox1496]

the mandelbrots are digging a cave base to survive the night, and then building a normal base

[@samuelluce8286]

That's a stretch

[@luxuriousgaming8927]

0:21

[@whatupboi7834]

0:30 inverted circlebrot

[@AltrrxOfficial]

Monobar*

[@HomeworkBan5624]

call it monobar not circlebrot

[@Unlimit42]

It's circlebrot not tricorn

[@ryanmapping7944]

0:39 : mandelbrot on the bottom right, then reverse "power morph" 0:48 to 0:58 : in the bottom right we see an "power morph" of some sorts

[@H_fromDiscord_real]

c morph

[@JacobPlat]

enig!

[@Imnotanoob761]

W I D E

[@Friendlyneighborhoodalien]

is it possible that these images are the best description to show the shape of the surface tension of space time?

[@CB_Mandelbrotich_Japany_]

0:08 Mandelbrot turns into monobar

[@YOUTUBE_IS_COOL470]

Formula from 00:8 ? Monobar fractal

[@ninjanianin]

It's kind of beautiful! I can't wait to see more stuff on your Mandelbrot KZbin channel! Also, It's a bit confusing that in 0:30 it's a circle, But it's actually still great! I wonder what your deepest Mandelbrot zoom is!

[@Arneauxtje]

Thank you for your comment. It's a circle because I chose to convert the cardioid into one, just for esthetic purposes, and to see how it would play out under the inversion. I'm not really into deep zooms. It seems to me that those are just more of the same, but more importantly, I refuse to participate in the rat race of people trying to top each other with yet another video claiming to be the deepest zoom ever. It's a bit pointless to me to let your PC crunch numbers for months on end in order to make this idle claim to fame. Instead, I try to seek slightly more original things to do with this subdivision of complex analysis, admittedly not all of them very succesfully, but nevertheless.

[@michlop452]

@@Arneauxtje I agree, most mandelbrot zooms are so damn repetitive, though some channels try to be more creative and do other fractals/mandelbrot powers. Those zooms are actually pretty interesting, though they take more computing power from what I know.

[@ninjanianin]

@@michlop452 Thx :D

[@tesscruzat5277]

0:30

[@__________________________hi52]

Monobar

[@TalalThaju-gn7zj]

Monobar momento

[@pomni_tadc_real]

What is the formula for the 2nd one?

[@Arneauxtje]

Inversion is achieved by c -> c/(t*(c*c-1)+1) and t=0..1

[@pomni_tadc_real]

GOD IN MANDELBROWSER

[@featherfractal212]

0:08 monobar

[@Arneauxtje]

What is this 'monobar' people keep talking about?

[@WevordinDoesAndroidEditor]

@@Arneauxtje Man look: Deseptor Made A Fractal (Monobar) The Monobar Is Made Look Like That, A Circle-Shaped.

[@__________________________hi52]

@@WevordinDoesAndroidEditor yes inspired by it

[@__________________________hi52]

@@WevordinDoesAndroidEditor bro he's not blind, it didn't exist back then

[@AnComplexFraktal]

​​@@Arneauxtjeif youre asking what is the fractal about, it is the fractal shown in 0:09, when the mandelbrot's cardioid turns into a circle and skewed. that fractal is made by deseptor.

[@chrisrodriguezm13]

0:08 everything is monobar

[@WevordinDoesAndroidEditor]

0:29 is monobar inverted

[@chrisrodriguezm13]

@@WevordinDoesAndroidEditor you are right

[@WevordinDoesAndroidEditor]

@@chrisrodriguezm13 Thank :)

[@blue6556-n9g]

why do i see you in every fractal vid

[@chrisrodriguezm13]

@@blue6556-n9g Oh, I actually liked fractals so much back then

[@khaledk13]

bottom right... IM A BLACK HOLE!!! later... GAS GAS GAS IM GONNA STEP ON THE GAS TONIGHT and so on

[@sinx2247]

how do you get that circle shape at 0:30?

[@Arneauxtje]

The cardioid of the main body can be expressed in polar coordinates, and in that form you can easily map it back to a circle. At approx. 0:08 this circle is visible. When the mandelbrot-set is inverted, you get a shape like a hanging waterdrop tilted 90 degrees. But under the same transformation you also get a circle, which makes sense, because a circle is invariant under inversion. If you need formulas please let me know and I'll post them.

[@sinx2247]

I see. but why not just invert the whole plane directly? why do you need to transform it into a circle, invert the plane, then undo the circle transformation?

[@Arneauxtje]

I wanted to see what the inversion would look like with the cardioid transformed to a circle. It seemed to me that the whole inversion process would then look a lot more symmetric. And this turned out to be the case, as can be seen when you compare the first half of the animation with the second.

[@sinx2247]

Cool cool cool. thanks for the info. I tried making my own Mandelbrot set inversion animation where I have the exponent go from 1 to -1 through the unit circle. gfycat.com/gifs/detail/WhoppingAntiqueHawk

[@truongquangduylop33]

@@Arneauxtje Did the monobar? If so, whats the formula for the monobar you did

[@haroonrashid3297]

hello Sir can you please share the code by which you have been able to get this inverse of mendalbrot

[@Arneauxtje]

Sure. 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 Then, there's the 4 different ways on how to get from c to 1/c, that 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)] The top left is method 1, top right method 2, and the bottom 2 are variants of method 3. Hope this helps somewhat.

[@haroonrashid3297]

Sir Thank you so much i will try to use it Sir can you please share your email address so i can contact you for any further help?

[@Arneauxtje]

@@haroonrashid3297 Just ask here, and I'll try to assist as good as I can. It might be helpful for others too.

[@truongquangduylop33]

@@Arneauxtje what are the formulas

[@Arneauxtje]

@@truongquangduylop33 The 4 different ways on how to get from c to 1/c shown in this video 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)] The top left is method 1, top right method 2, and the bottom 2 are variants of method 3.

[@pomni_tadc_real]

In 0:30 it is a ○?

[@honeyobando883]

Yes

[@stopthanawat3563]

z=z^2 + (1/c)

[@BlackXxScopez]

wtf? can you tell me what it means for the mandelbrot set to be inverted? thanks

[@Arneauxtje]

It's turned inside out. A normal projection is on the c-plane, i.e. x-axis real and y-axis imaginary. Inverted projection is on the 1/c-plane. So points close to 0 go to infinity and vice versa. The actual Mandelbrot-set is in black. When it's inverted, the black is on the outside.

[@BlackXxScopez]

so if I were to have a function that took in a real and imaginary input and gave me back the iterations for the inverse mandelbrot point, it would just invert both the real and imaginary points and feed it into a regular mandelbrot function?

[@BlackXxScopez]

Arneauxtje took me a while to get the complex number math right, but I got it. Thanks for a the quick response.

[@Arneauxtje]

It's simpler than that. The inversion is just in the mapping. The iteration itself remains unchanged. So the coordinates of a point on the screen are changed to inverted ones, and then they're iterated to determine if the point is part of the set (black) or not (colorized depending on escape velocity).

[@__________________________hi52]

Z^ -2 + c

[@DekDM]

Monobar

[@Arneauxtje]

Where? What's a monobar anyway?

[@user-oo7dj3uh6w]

A

[@Arneauxtje]

Mmmmokay?

[@AltrrxOfficial]

B

[@Thisis_phil1234]

@@AltrrxOfficial C

[@KingGreenscreenKid420]

​@@Thisis_phil1234D

[@sebastiangalang2468]

​@@KingGreenscreenKid420E