Using n (number of iterations) as a variable

TheRedshiftRider

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Using n (number of iterations) as a variable

« on: June 12, 2014, 05:18:36 PM »

A few weeks ago I read the topic about iterating the mandelbrot with random powers, I liked the idea but I thought it was a little too random so I was thinking about a formula with increasing powers using the variable n.

N starts with 1 and adds one when an iteration is calculated.

I tried ot with the mandelbrot formula:

z^n+c

After that I tried some other formulas related to this one and I found some other patterns:

z^n/10+c

z^n/-1+c

z^sqrt(n)+c

z^sqrt(n+10)+c

z^n+c+c

z^n/z+c

z^2+0.1*n + c

(z^10*n+c)/(z^n+c)

(z^100*n+c)/(z^n+c)

Some of them dont really look interesting or complex.
Can the variable n be used to create nice patterns? Are there any other nice fractals that can be created with this variable?


« Last Edit: June 12, 2014, 08:36:29 PM by Toofgib »	Logged

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David Makin

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Re: Using n (number of iterations) as a variable

« Reply #1 on: June 12, 2014, 11:03:55 PM »

See Ultra Fractal mmf.ufm:Transformation (Mandelbrot or Julia) this has a choice of base formulas including z^iter+c

Also see mmf.ufm:Transpoly - this is a more sophisticated formula allowing a greater choice of base formulas which are all produced from generative recursion that give a rising degree - implemented as either use as a fixed degree i.e. Iterate Poly of degree n + c or as Poly of degree iter + c i.e. such that the power rises with iteration.

To get smooth iteration colouring on the ones with rising degree per iteration will probably require bailout-time estimation of the degree of divergence (at least if using the standard Vepstas smooth iteration colouring method).

The formulas can be found in the ultrafractal formula database under my name (David Makin) here:

http;//formulas.uitrafractal.com/

And all those in mmf.ufm will obviously work in UF but also should convert perfectly for use in ChaosPro with the UF formula import facility.


« Last Edit: June 12, 2014, 11:06:23 PM by David Makin »	Logged

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David Makin

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Re: Using n (number of iterations) as a variable

« Reply #2 on: June 12, 2014, 11:07:56 PM »

You can also use:

start:
z = pixel

loop:
z=z^pixel +pixel

(for instance)


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kram1032

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Re: Using n (number of iterations) as a variable

« Reply #3 on: June 13, 2014, 01:57:20 PM »

If you keep increasing n with the iteration count, I'd think that the final appearance critically depends on which iteration you end on. I might be wrong though. Could you try varying the iteration count by +/-1?


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TheRedshiftRider

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Re: Using n (number of iterations) as a variable

« Reply #4 on: June 14, 2014, 12:38:51 PM »

Quote from: David Makin on June 12, 2014, 11:03:55 PM

Thanks, I will have a look at those. I dont use ultra fractal but I do use chaospro so I'll try to import those formulas.

Btw, I didnt use chaospro to make those images, I used a program made by a shy member of fractalforums.


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TheRedshiftRider

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Re: Using n (number of iterations) as a variable

« Reply #5 on: June 14, 2014, 01:51:14 PM »

Quote from: kram1032 on June 13, 2014, 01:57:20 PM

Its actually the opposite, I tried some things and It actually depends on the number you start with:

If you use the formula 2^n+c and start with 1 and add one every time, the series of formulas that is calculated will be:

z^1+c, z^2+c, z^3+c, z^4+c, z^5+c...

If you start with 2 it will be:

z^2+c, z^3+c, z^4+c, z^5+c, z^6+c...

This will make that they are both much different.

Here are some other examples:

z^(n+10) + c

z^(n-10) + c

z^(n-50) + c

It is also important to change how the variable has impact on the shape, you can make it increase/decrease slower or faster:

z^(10+n*0.01) + c

z^(2+n*0.01) + c

z^(2+n*0.001) + c

z^(2-n*0.014) + c


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youhn

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Re: Using n (number of iterations) as a variable

« Reply #6 on: June 14, 2014, 04:49:00 PM »

This follows what seems to be a rule (or even The Definition?) of all chaos theory. Seemingly small choices or moves at the start, dictate the end result much more than what happens next. Holds also for deep zooms in the Mandelbrot set of Burning Ship fractal.

I really like the image at (z^100*n+c)/(z^n+c). love


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TheRedshiftRider

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Re: Using n (number of iterations) as a variable

« Reply #7 on: June 15, 2014, 01:01:54 PM »

Quote from: youhn on June 14, 2014, 04:49:00 PM

I just tried some things. But you're right about that.


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kram1032

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Posts: 1863

Re: Using n (number of iterations) as a variable

« Reply #8 on: June 15, 2014, 01:46:08 PM »

Here'd be one which always is quadratic in behavior.

$z=\frac{z^{2+n}}{{z^*}^n}$

or, as expressed in real/imaginary form:

$\begin{array}{c} x=\left(x^2+y^2\right) \cos (2 (n+1) \, \arg (x+i y)) \\ y=\left(x^2+y^2\right) \sin (2 (n+1) \, \arg (x+i y)) \end{array}$

more generally,
$z=\frac{z^{k}}{{z^*}^n}$
would be
$\begin{array}{c} x=\left(x^2+y^2\right)^{\frac{k-n}{2}} \cos ((k+n) \, \arg (x+i y)) \\ y=\left(x^2+y^2\right)^{\frac{k-n}{2}} \sin ((k+n) \, \arg (x+i y)) \end{array}$


« Last Edit: June 15, 2014, 02:02:37 PM by kram1032 »	Logged

lkmitch

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Posts: 238

Re: Using n (number of iterations) as a variable

« Reply #9 on: June 16, 2014, 05:04:02 PM »

z^(2+tanh(iter/100))+c. The tanh() function is < 1 (for real inputs), so the power starts at 2 and never goes beyond 3.

2014-06-16-a.jpg (59.61 KB, 800x600 - viewed 344 times.)
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TheRedshiftRider

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Re: Using n (number of iterations) as a variable

« Reply #10 on: June 21, 2014, 12:45:32 PM »

I also tried it with the burning ship:


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TheRedshiftRider

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Re: Using n (number of iterations) as a variable

« Reply #11 on: June 24, 2014, 07:04:31 PM »

I found another one that looks very interesting:

(z*(z^(1+n))-z)*c

The julia-sets of this one look very nice.


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