Print Page - z' = z^5 +c

Title: z' = z^5 +c
Post by: flok on November 16, 2009, 12:44:38 PM

Hi,

Here's my implementation of z' = z^5 + c:
http://www.vanheusden.com/fractals/mand_z5+c_4.0.png
(http://www.vanheusden.com/fractals/mand_z5+c_4.0.png)

The source-code van be found here: http://www.vanheusden.com/fractals/ (http://www.vanheusden.com/fractals/)

(hopefully I did not make any mistakes while expanding that formula to real/img part)

Title: Re: z' = z^5 +c
Post by: cKleinhuis on November 16, 2009, 01:00:52 PM

hello, and welcome to the forums ... ehrm, what formula you are rendering there ?!?!?
certainly not a 5th order mandelbrot : :'(

Title: Re: z' = z^5 +c
Post by: David Makin on November 16, 2009, 02:36:14 PM

Quote from: flok on November 16, 2009, 12:44:38 PM

Hi,

Here's my implementation of z' = z^5 + c:

(http://www.vanheusden.com/fractals/mand_z5+c_4.0.png)

The source-code van be found here: http://www.vanheusden.com/fractals/ (http://www.vanheusden.com/fractals/)

(hopefully I did not make any mistakes while expanding that formula to real/img part)

When you say z^5+c, which number format do you mean ?
If ordinary complex then Trifox is correct - it's definitely not correct.

To expand a complex power to reals 1st start with Pascal's triangle to the row required for your power, e.g. for z^5 start with:

Code:

               1
             1   1
           1   2   1
        1   3   3   1
     1   4   6    4   1
   1   5  10  10   5   1

So for z^5 that's "1, 5, 10, 10, 5, 1"
Now add your x/y terms starting with the highest power of x and y^0 (i.e. 1) on the left decrementing the x power and incrementing the y power across the row - remembering that for each y we also have the equivalent power of i:

x^5, 5*x^4*y*i, 10*x^3*y^2*i^2, 10*x^2*y^3*i^3, 5*x*y^4*i^4, y^5*i^5

Now consider the result of raising i to the power (i.e. giving i, -1, -i, or +1):

x^5, 5*x^4*y*i, -10*x^3*y^2, -10*x^2*y^3*i, 5*x*y^4, y^5*i

giving z^5 = x^5 - 10*x^3*y^2 + 5*x*y^4 + i*(5*x^4*y - 10*x^2*y^3 + y^5)

i.e. the Mandy iteration done longhand in Ultra Fractal terms:

x = real(z)
y = imag(z)
z = x^5 - 10*x^3*y^2 + 5*x*y^4 + flip(5*x^4*y - 10*x^2*y^3 + y^5) + #pixel

Edit: Just to add that for higher powers there are formulas for getting the coefficients for a given row of Pascal's triangle - see Wikipedia.

Title: Re: z' = z^5 +c
Post by: flok on November 16, 2009, 05:29:48 PM

Quote from: Trifox on November 16, 2009, 01:00:52 PM

hello, and welcome to the forums ... ehrm, what formula you are rendering there ?!?!?
certainly not a 5th order mandelbrot : :'(

Hmmm. Has been a while ago that I wrote that program, might have made a bug. I tried implementing z' = pow(z, 5.0) + c.
Did that the naive way by expanding into (x + iy)(x+iy)(...)... etc.
The

Code:

dummy1 = w * ln(y);
dummy2 = exp(w * ln(x))
re(z) = dummy2 * cos(dummy1);
im(z) = dummy2 * sin(dummy1);

method doesn't seem to work (all black).

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