Lowest / optimal bailout values for the Mandelbrot sets

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Dinkydau

• Fractal Furball
• Posts: 271

Lowest / optimal bailout values for the Mandelbrot sets

« on: February 04, 2019, 09:05:31 PM »
I want to know the lowest possible bailout values for all the Mandelbrot sets with power n, so with the formula z^n + c.

I think for even n it's:
$\sqrt[(n-1)]{2}$

It seems to work well with the 4th and 8th power M-sets (that I tested) and for the 2nd power M-set the formula is just 2 like we're used to.

The reasoning behind that bailout value formula is based on that M-sets with even power have a spike on the real axis. I tried to figure out how long it is for n=4.
For a point to not escape, c needs to be negative enough to compensate for z^n all the time. The higher n, the larger z^n is. Substracting c once can compensate for something weaker than multiplying by two, so I think if z^4 is more than multiplying by 2, a point escapes. So I wonder: when is
z^4 > z*2 ? That's when
z^3 > 2, so
z > $$\sqrt[3]{2}$$
z > 1.2599210498948731647672106072782
This is also the length of the spike when measured by zooming in on it in a fractal program.

For higher even powers you can do the same thing to arrive at the general formula. So I have some evidence that it makes sense for even powers. But it's not at all clear to me why it would work well for uneven powers. For n=3 the bailout of $$\sqrt[2]{2}$$ looks large enough, but I'm not sure if it's the lowest possible. It's hard to measure or calculate how far out the longest branches reach. They're not straight like a spike, and my reasoning doesn't work because z^n is not necessarily positive.

• 3f
• Posts: 1546

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #1 on: February 04, 2019, 10:10:36 PM »
I think you're right (for all positive integer powers).  Here's my reasoning:

The assumption that R is an escape radius means $z_m > R \ge c$
implies $|z_{m+1}| = |z_m^n + c| > |z_m|^n - |c| > R^n - R = R (R^{n-1} - 1)$

Taking the limit as $$|z_m| \to R$$ from above gives
$\frac{|z_{m+1}|}{|z_m|} > R^{n-1} - 1 \ge 1$
=>
$R^{n-1} \ge 2$
=>
$R \ge \sqrt[n-1]{2}$

which is what you claimed.  The existence of a c in the set with |c|=R is proof of minimality for some powers, but the tips where $$z_3 = z_2$$ for $$z \to z^3 + c$$ have magnitude $$\sqrt[4]{3}$$ which is a little bit less (at least, I think they are the relevant tips..).  But I think the smallest escape radius just has to be >= the largest c, equality is not guaranteed...

EDIT I suggest checking R=1.36 (which is between the two escape values I calculated) for the cubic Mandelbrot and inspecting carefully for any problems... maybe I'm talking nonsense...

EDIT2 these are not the cubic tips you are looking for
« Last Edit: February 04, 2019, 10:41:44 PM by claude, Reason: error »

Dinkydau

• Fractal Furball
• Posts: 271

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #2 on: February 05, 2019, 12:16:53 AM »
But I think the smallest escape radius just has to be >= the largest c, equality is not guaranteed...
Yeah, that too.

So for even powers there's always a c in the set that's on the escape radius. Maybe for uneven powers there's no such c, but the formula is still optimal. The space may be used differently. There could be a c in the set well within the escape radius that then comes very close to it and moves away again and doesn't escape. Lowering the escape radius would be problematic for such c.

I made some changes to my own simple fractal program to try to find c in the 3rd power M-set that need the high escape radius but I couldn't (easily) find them. Either they escape or they reach at most about 1.355 absolute value. That also the largest distance of a c inside the set from 0 that I could find. It doesn't rule out the possibility that the escape radius can be lowered.

lkmitch

• Fractal Fanatic
• Posts: 38

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #3 on: February 05, 2019, 05:31:42 PM »
If you want a tighter restriction, keep in mind that the minimum radius varies with |c|. Doing that will give images where the iteration contours come much closer to the tips of the set. I implemented this in Ultra Fractal several years ago in the "elliptical bailout" formulas.

Dinkydau

• Fractal Furball
• Posts: 271

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #4 on: February 05, 2019, 08:58:26 PM »
I don't understand. I want the minimum radius that works for all c.

greentexas

• Fractal Phenom
• Posts: 53

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #5 on: February 06, 2019, 03:54:02 AM »
These are not Mandelbrot sets, but I do know that for fractals such as the Burning Ship, Buffalo, Perpendicular, etc. the western spike of the Nth order fractal ends at -(2^(1/(n-1))). For SOME fractals (e.g. the Mandelbrot, Quasi Heart), it appears to be that the fractal is always bounded in the circle with that radius. I might experiment around soon and see which fractals break the bounding rules.

• 3f
• Posts: 1546

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #6 on: February 06, 2019, 05:07:12 PM »
The radius I calculated might not be the minimum possible after all.  It's calculated starting from $$|Z_{m+1}| > |Z_m|$$ i.e. the point of no return after which Z keeps increasing forever, independent of C.  But the relevant Z each depend on their own C, so that might allow tighter bounds to be achieved.

However, mathematically proving that a lower radius is possible for cubic Mandelbrot set (for example) is beyond my complex-dynamics skills.

I think the problem of minimal escape radius is equivalent to finding the bounding circle of the anti-Buddhabrot for the formula?

Dinkydau

• Fractal Furball
• Posts: 271

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #7 on: February 18, 2019, 02:02:09 PM »
Even though $$\sqrt[n-1]{2}$$ is maybe not the lowest escape radius for the odd powers, it still has a beautiful property: it's the smallest such that the iteration bands are connected. The first iteration band is the only one that is infinitely thin anywhere, and it is so at n-1 points. With a larger escape radius, iteration bands are not infinitely thin anywhere.

I lowered the escape radius to 1.4 (slightly less than $$\sqrt{2}$$) for the 3rd power Mandelbrot set and you can see in the image that the iteration bands are not connected anymore.

With the escape radius set to $$\sqrt{2}$$, the first and the second iteration bands' boundaries touch because:
After 1 iteration, z_1 = c, so points c that escape after 1 iteration have abs(c) > escape_radius.
After 2 iterations, z_2 = c^3 + c, so points c that escape after 2 iterations have abs(c^3 + c) > escape_radius.
Now fill in $$\sqrt{2}*i$$ for c in both cases (this c is the "highest"/most imaginary/most northern value of c that is not larger than the escape radius).
$$abs(\sqrt{2}*i) = \sqrt(2)$$, so this c is at the boundary of the first iteration band.
$$abs((\sqrt{2}*i)^3 + \sqrt{2}*i) = abs(-2*(\sqrt{2}*i) + (\sqrt{2}*i)) = \sqrt(2)$$ as well, so it's also at the boundary of the second iteration band.

After 3 iterations the expression turns into $$(c^3 + c)^3 + c = (-\sqrt(2)*i)^3 + \sqrt(2)*i = 3*\sqrt(2)*i$$, which is larger than the escape radius, so only the first and second iteration bands' boundaries touch.

With even powers, the iteration bands all need to go around the spikes and are infinitely thin there, so they're all infinitely thin at n-1 points (the number of spikes).

The connectedness of iteration bands is something that I'm used to and I like it. I also wanted to know the lowest escape radii mainly because I didn't like an arbitrary choice. I was surprised that no fractal programs had a good choice built in. Choosing the escape radius to be the lowest so that iteration bands are still connected is a good justification for the choice in my opinion so I can live with that. it turns out the thing I asked for was not what I was looking for because of the unexpected fact that the best escape radii for odd powers are not the lowest like for even powers.

It would still be interesting to know the lowest possible values. Because of the crazy shape of the branches of the 3rd power Mandelbrot set I'm afraid it's not even a nice expression but an infinite sum or something.

I think the problem of minimal escape radius is equivalent to finding the bounding circle of the anti-Buddhabrot for the formula?
I don't know what that is.
« Last Edit: February 18, 2019, 08:09:27 PM by Dinkydau »

• 3f
• Posts: 1546

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #8 on: February 18, 2019, 05:50:36 PM »
A nice property you have discovered!

I don't know what that is.
The Buddhabrot plots all the Z_n for points that escape, for iterations of $$Z_n \to Z_n^2 + C$$, for all C.  The anti-Buddhabrot plots all the Z_n for points that don't escape.  These can both be generalized to other formulas.  The bounding circle of the anti-Buddhabrot is the largest |Z_n| among points that don't escape.

• 3f
• Posts: 1546

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #9 on: March 20, 2019, 01:19:45 AM »
The difference is bigger with higher powers, perhaps obviously.  Here's power 5 Mandelbrot:

high bailout

bailout=2

low bailout (new feature in next KF, not yet released)

Dinkydau

• Fractal Furball
• Posts: 271

Re: Lowest / optimal bailout values for the Mandelbrot sets

« Reply #10 on: March 20, 2019, 01:18:50 PM »
low bailout (new feature in next KF, not yet released)
Nice

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