Deepest Mandelbrot Set Zoom Animation ever - a New Record! 10^275 (2.1E275 or 2^915)

Music is "Research Lab" by Dark Flow ( http://itunes.apple.com/us/album/subu... , http://amzn.com/B001U9YCG8 )

Read more geeky details and download the full-resolution video at http://fractaljourney.blogspot.com

Details:
The final magnification is 2.1x10^275 (or 2^915). I believe that this is the deepest zoom animation of the Mandelbrot set produced to date (January 2010).

Each frame was individually rendered at 640x480 resolution and strung together at 30 frames per second. No frame interpolation was used. All images were lovingly rendered by 12 CPU cores running 24/7 for 6 months.

Self-similarity (mini-brots) can be seen at 1:16, 2:30, and at the end 5:00.

• Category

  Science & Technology

[TheDeadSource]

Did you find any formations you didn't expect by rendering this far inwards?

[nosro1]

Not sure how to answer that. Every area of the Mandelbrot is different. Ergo, every zoom is different. Sometimes the differences are subtle. (See my other video on Seahorse valley, which shows how a basic shape evolves while panning one particular location.) That's the fun in exploring the Mandelbrot.

[cacahahacaca]

Holy fuck wow!!! Thanks so much, really amazong.  Would love it at even more at high res ;)

[Morgoth Bauglir]

And by harnessing this power we shall control time. That or create the best acid trips in the universe.

[Dhiego Magalhães]

Did you use a supercomputer?

[Liviu Chircu]

10^275 is a formidable result! How do you even store such small floating point values? Not to mention the computational power required! (the quality seems to be very good, which means the arrays of complex numbers tested for each frame of the animation are pretty big) This must surely take entire weeks to compute, even on a supercomputer!

[masonery123]

Yeah. I'm gonna like it.

I'm in geometry, I have an algebra II textbook I'm studying at home, I'm getting extra trig work at school, I'm studying pre-calc with khanacademy, and I'm reading "calculus made easy" by Silvanus P. Thompson (best book ever) all simultaneously, so I've got most of the bases covered. Stats, anyone? I think I'm going to like college math too. I suppose my first comment didn't really reveal my background, despite it revealing my age. And I still have time for the net.

[James Slaughter]

Right! Very well done, I think you're going to like college math.

If it is in the set, the pixel is black in the video. I think that the color is related to how long it takes the computer to determine that that point would make an unbounded sequence, but I don't know how this was programmed.

[masonery123]

Thanks a million. I never understood this thing!

[masonery123]

Okay, so I did some research, and now I understand. Set c to any complex number lying on the complex plane. Use the sequence to determine wheter that value of c is bounded or unbounded. For example, your point, i, is bounded, as it iterates 0,-i,-1+i,-i,-1+i, and so on. Thus, (0,1) is found in the set. However, (1,0) is not found in the set. This is because it iterates 1,2,5,26, and so on, unbounded. This is only a guess: The number of iterations before c is determined bound is the color.

[masonery123]

Oh! That makes sense, it is simply a recursive sequence. And then you plot those complex numbers on the complex plane? That doesn't seem to make sense, since you would only, as you represented, get 3 different points until they would reappear. z(5) gives -i, am I wrong? And that doesn't account for the clearly purposeful coloring, which I believe has something to do with an amount of iterations before some boundary is passed.

Of course, I don't really know what those could be.

[James Slaughter]

More about sequences: So let's take i=sqrt(-1) (so that i^2 = -1) as "c". So "n" starts at 0, then each iteration of "n" goes + 1:

I tried to put some math here, but YouTube thinks it's ASCII art and wont let me post it. Look here for the math:go to pastebin dot com and add Ycf6hA9s to the end

So it will just keep going in this loop: (-1+i), -i... that means that for all n, the sequence z(n+1)=z(n)+c, where c=i, is bounded, it's not getting bigger. So i is in the set, make a dot at i.