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Author Topic: Perturbation for Julia  (Read 503 times)
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Kalles Fraktaler
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kallesfraktaler
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« on: October 31, 2014, 12:41:19 PM »

I wanted to try to render a Julia fractal with Perturbation, but I failed.

This is how I applied the Julia formula to master K.I.Martin's functions:
X[n+1] = X[n]^2 + C
Y[n+1] = X[n+1] + D[n+1]
D[n+1] = Y[n+1] - X[n+1]
D[n+1] = (Y[n]^2 + C) - (X[n]^2 + C)
D[n+1] = ((X[n] + D[n])^2 + C) - (X[n]^2 + C)
D[n+1] = X[n]^2 + 2*X[n]*D[n] + D[n]^2 + C - X[n]^2 - C
D[n+1] = 2*X[n]*D[n] + D[n]^2 (1)

A nice thing with Julia is that the constant C, that needs to be full precision, is eliminated.
So... now it is time for extracting the complex variables

(dr+di) = 2*(xr+xi)*(dr+di) + (dr+di)
(dr+di) = 2*(xr*dr + xr*di + xi*dr - xi*di) + dr*dr + 2*dr*di - di*di
dr = 2*(xr*dr - xi*di) + dr*dr - di*di
di = 2*(xr*di + xi*dr) + 2*dr*di

And............................. it doesn't work...........  hurt
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Kalles Fraktaler
Fractal Senior
******
Posts: 1458



kallesfraktaler
WWW
« Reply #1 on: April 23, 2015, 08:53:57 PM »

Have anyone had a look on this?

Is it because the the X and the Y is not starting with the same value for Julia, that makes K.I.Martin's equation
Dn = Yn - Xn not applicable for Julia?
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Want to create DEEP Mandelbrot fractals 100 times faster than the commercial programs, for FREE? One hour or one minute? Three months or one day? Try Kalles Fraktaler http://www.chillheimer.de/kallesfraktaler
http://www.facebook.com/kallesfraktaler
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