Mandelbrot and Captive Spirals

[element90]

Two version of a Mandelbrot island in a fractal other than the standard Mandelbrot set, the spirals contain areas where the orbits haven't escaped hence "captive spirals".


https://copy.com/3cQMgY6cDJ0k

https://copy.com/KPBCtLjmyCZY

The formula for the fractal is:

z = (alpha(2z - z^2) + beta)^2

alpha = the location in the complex plane
beta = -0.5
initial z = 1

and it looks like this:


https://copy.com/qKvIsjCwCdY1

[Pauldelbrot]

Actually, technically that's only half of the Mandelbrot for this. Your mapping has this derivative:

4(alpha(2z - z^2) + beta)(alpha(1 - z))

which has a zero at z = 1, the critical point you studied to get these images, but also has zeroes at -alpha z^2 + 2 alpha z + beta = 0. That can be solved using the quadratic formula:

z = [(-2 alpha) +/- sqrt(4(alpha^2) - 4(-alpha)(beta))]/(-2 alpha)

which can be simplified:

z = 1 +/- sqrt((alpha + beta)/alpha)

So, when (alpha + beta)/alpha != 0 there are two additional critical points. However, since these are zeroes of alpha(2z - z^2) + beta they both go to zero on the first iteration.

The upshot is that you can skip the annoying square roots -- the other half of this thing's M-set can be seen by using initial z of 0. Also, there only is one other half, rather than two other thirds. And I predict that the first two images will be near the center of a big round bulb of M-material when initial z is set to 0, which will be the front bud of a big Mandelbrot figure.

Interesting find, and hope this helps you find more interesting stuff in this formula. The Julia sets should be interesting, too, with up to two finite attractors as well as infinity; the Julia sets from inside the minibrot in the first two images you posted should have small "jewel" basins of one attractor enclosed by loops and whorls of the other, similar to how the minibrot is situated in your images.

[Pauldelbrot]

Oh, and as for the behavior when (alpha + beta)/alpha = 0? Then beta = - alpha, and the iteration becomes z -> (z - 1)^4. So you should see quartic Mandelbrot shapes then with either initial z (0 goes to 1 in one iteration in this instance).

You might get a real bonanza of stuff when beta is close to, but not quite equal to, alpha. Stay away from alpha = 0 though, as z -> beta^2 is a constant and nothing interesting happens.

[element90]

Pauldebrot, I know that the solutions of the derivative of the fractal formula give the critical points that can be used as the initial values of the Mandelbrot algorithm, however, I always got stuck when the result includes the parameter for the location in the complex plane.

For convenience I'll call the location in the complex plane c so the roots of the quadratic part become:

root 1 = 1 + sqrt((c-1)/c)
root 2 = 1 - sqrt((c-1)/c)

So for a root to be 0 then (c - 1)/c must equal 1 which also leads to a root of 2. Checking the pictures for initial values of 0 and 2 show that the first contains well formed Mandelbrot islands and the second does not, this indicates that the root 0 is indeed a critical point but 2 is not, currently I don't see why zero is a critical value.

Determining critical values for a fractal formula can lead to a wide variety of results, I've done this for cubics see http://element90.wordpress.com/?s=cubic+observations and http://element90.wordpress.com/2012/11/25/more-cubic-observations/, so determining critical values from root formulae that contain c values would be useful. Depending on the complexity of derivative of the fractal formula finding some roots is easy but others require much more effort.

Going back to the formulae for root 1 and 2, c is the location in the complex plane so values of root 1 and 2 are different for every location calculated by the Mandelbrot algorithm.

Here is a picture of the fractal when using zero as the initial value:


https://copy.com/TJxJSRTxekUg

I have a generalised form of the fractal formula used for the pictures that started this thread:

z = (alpha(beta*z^gamma + delta*z^epsilon) + zeta)^eta

as it has multiple powers it is relatively slow.

So using this formula as a base I contrived a formula with i as a critical value:

z = (c(3z + z^3)  + 1)^2

which produces:


https://copy.com/TBhgmUJnBPoE

Using other parameters for the generalised formula I came across a strange phenomenon of an unexpected need to use multi-precision calculation, see http://www.fractalforums.com/general-discussion/unexpected-need-for-multi-precision/.

That's it for now.
 

[Pauldelbrot]

When the critical points depend on the image coordinate c, then you can't have initial z be a constant -- it needs to be a function of c. In UF you'd have not

init:
  z = 0

or

init:
  z = 1

or whatever, but rather

init:
  z = sqrt(#pixel)

or whatever the function was.

In the instant example, it ended up not mattering since both the additional critical points went, c-independently, to z₁ = 0.

[element90]

Does that mean that when the the root formula is used as the initial value and the result is zero then zero can be used as the critical value? Can the value be something other than zero?

Currently my programs only allow for a fixed initial value or the location in the complex plane. 

[Pauldelbrot]

If the first iterate of a critical point is some complex number w, then you can use w as the initial value and get essentially the same results. You've simply pre-performed one iteration.

[element90]

Thanks for the help. It took a while to understand I've got it now.