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Author Topic: "Spherical Power" function?  (Read 1682 times)
0 Members and 1 Guest are viewing this topic.
JColyer
Guest
« on: December 01, 2009, 02:36:05 PM »

Hey there, I've got (thanks to fractal rebel) a power function for triplex numbers which looks like this (sans the conversion to and from spherical coordinates...)

r = Math.Pow(a.R, b);
phi = a.Phi * b;
theta = a.Theta * b;

I'm re-working another old fractal function and it's got a complex ^ complex operation, which I've coded up two ways in spherical, both seem to work (i.e. generate images that look 'right') but which one is more "correct"?

r = Math.Pow(a.R, b.R);
phi = (a.Phi + b.Phi) * b.R;
theta = (a.Theta + b.Theta) * b.R;

- or -

r = Math.Pow(a.R, b.R);
phi = (a.Phi * b.R) + b.Phi;
theta = (a.Theta * b.R) + b.Theta;

- or -

Is it something else completely??

Thanks!

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BradC
Safarist
******
Posts: 85



« Reply #1 on: December 05, 2009, 04:14:44 AM »

...or maybe it's a^b=e^{b \ln (a)}, where e^z=(1,0,0)+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots and \ln (z)=(z-(1,0,0))-\frac{(z-(1,0,0))^2}{2}+\frac{(z-(1,0,0))^3}{3}-\cdots, haha! These expressions should actually define something since we have integer powers and multiplication by a scalar defined already. The logarithm definition would have a small radius of convergence however, and I'm not sure how to get around that because \ln \left(z^2\right) probably wouldn't be equal to 2 \ln (z). Might be worth playing with; it would be cool to have transcendental functions defined somehow...

Edit: Oops, had the wrong series for ln.
« Last Edit: December 05, 2009, 04:22:26 AM by BradC » Logged
Paolo Bonzini
Guest
« Reply #2 on: December 06, 2009, 11:18:27 AM »

See http://www.fractalforums.com/theory/triplex-algebra/msg9537/#msg9537 for my little conjecture about triplex exponential...
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