From: Jim Muth <jamth@mindspring.com> Reply-To: fractint@mailman.xmission.com To: fractint@mailman.xmission.com CC: philofractal@lists.fractalus.com Subject: [Fractint] FOTD 30-06-02 (Algae [7]) Date: Sun, 30 Jun 2002 10:45:15 -0400
frm:MandelbrotMix4 {; Jim Muth a=real(p1), b=imag(p1), d=real(p2), f=imag(p2), g=1/f, h=1/d, j=1/(f-b), z=(-a*b*g*h)^j, k=real(p3)+1, l=imag(p3)+100, c=fn1(pixel): z=k*((a*(z^b))+(d*(z^f)))+c, |z| < l }
Question: what does this little critter actually do??? As far as I can tell, it iterates z = k(az^b + bz^f)+c (a strange choice of variable names, but hey...) I notice from the way the formula works that both the coefs and exponents have to be purely real. Then there's just the small matter of z0... I'm not sure, but isn't (-abgh)^j the same as the (f-b)th root of (-ab)/fd? And if so... why?!? I find it hard to belive that it just seemed like a nice expression to initialise the orbit to... Another question: when E is not a positive real integer, Z^E is multivalued. How about if E is an imaginary positive integer? Is it multivalued or single valued? Whimsical observation: Z^0.5 is double-valued; Z^0.25 has 4 values, z^0.1 has 10 values, z^0.01 has 100 values... by induction, z^0 has infinity values ;-) [of course, since the solutions to z^0.5, etc get progressively closer together, at z^0 all the infinity of solutions are THE SAME ONE: "1"] Andrew. "More scienide..." _________________________________________________________________ Send and receive Hotmail on your mobile device: http://mobile.msn.com
Dear Fractalists: Did anyone else on the list take note of Ronaldo's fractal haircut during the latter stages of WC Korea\Japan 2002? Better yet, can anyone provide a par formula? PDBod
participants (2)
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Andrew Coppin -
David Boddington