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Author Topic: Bounds on the Mandelbox  (Read 279 times)
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rudi
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Posts: 1


« on: April 03, 2014, 08:05:31 PM »

Hi Fractalforums folks!

As part of a project in a pure math course on fractals, I found formulas for the size of the (standard) Mandelbox. Nothing very exciting, but it gives a useful minimum bound which can be used for rendering.

The proofs are on my projects page https://rudi-c.github.io/mandelbox370/ but here are the main results :

Let s be the scale parameter.
For s > 1, points with abs(any coordinate) > 2(s+1)/(s-1) are not in the mandelbox.
For s < -1, points with abs(any coordinate) > 2 are not in the mandelbox.
For 1 < s < 2, points with abs(all coordinates) < 2(s^2-s-2)(1-s) are in the mandelbox (i.e., the box has a solid region in the middle)
For -2 < s < -1, points with abs(all coordinates) < 4+2s are in the mandelbox (idem).

I still need to have someone verify those proofs, but renderings suggest that the results are correct.

I also have a conjecture that if s > 4sqrt(n) + 1 where n is the number of dimensions of the mandelbox, then the Mandelbox is empty (has no points inside - escape time rendering should still give something to look at, but no details after zooming in) except for the origin. This is about 6.65 for the 2D mandelbox. For smaller values, the Mandelbox is non-empty (it will at least have some isolated fixed points).

Hopefully someone will find this useful.

- Rudi
« Last Edit: April 03, 2014, 08:10:54 PM by rudi » Logged
knighty
Fractal Iambus
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Posts: 819


« Reply #1 on: April 03, 2014, 10:23:17 PM »

Hi Rudi!
Awesome work! and thank you for sharing.

I have once found the same formulas for the size of the mandelbox but without any proof, just by experimentation and guessing. embarrass

Regarding you conjecture, it also happens with kaleidoscopic IFS and kaliset. For example, with kaliset, when one of the coordinates of c is negative the fractal is empty.
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