Title: ln(e^(z^n)) = different??? Post by: M Benesi on October 11, 2012, 08:10:16 AM It's either because ln is multivalued, or it's a problem with the ln or exponential function in ChaosPro, but I get Mini-Mandelbrots all around the main one when I do that. In addition, it has almost burning ship type "scaffolding" around it. Anyways... here are a couple images. I better try it in another program, ehh?
(https://lh5.googleusercontent.com/-BalbZLmeSE8/UHZh2Pg6-UI/AAAAAAAABvE/HyIUiGoJ0Eg/s288/main%2520thing.jpg)(https://lh5.googleusercontent.com/-zks7U4hLPKQ/UHZh3v4YTdI/AAAAAAAABvE/UD3k4iP9aMA/s288/zoom%2520out%25204.jpg) (https://lh4.googleusercontent.com/-I5VKTLLeg18/UHZh3XlQ44I/AAAAAAAABvE/g-P3KlkRIOQ/s288/zoom%2520out%25203.jpg) (https://lh5.googleusercontent.com/-4V0A7f7OQuI/UHZh3PdiQwI/AAAAAAAABvE/X0gNtSp4SO4/s288/zoom%2520out%25202.jpg) (https://lh5.googleusercontent.com/-JZCwpf-BGp0/UHZh2HqaAFI/AAAAAAAABvE/QCij-Ip55a0/s288/weird%2520zoom%2520out.jpg)(https://lh5.googleusercontent.com/-ygNaT2soudA/UHZh2PgSZbI/AAAAAAAABvE/VPhpNnIcSLE/s288/weird%2520lightnin.jpg) And it has mini julias scattered around it (they don't have any mini-brots-- they are just julias): (https://lh6.googleusercontent.com/-7BCqmeFreV4/UHZjHIazNvI/AAAAAAAABvg/MCB5HFNurwU/s640/julia%2520outside%2520of%2520it.jpg) And stranger yet: doing ln(e^(z^n))^n, with n of 2 gives you the z^4, as expected, but in addition, you get tons of little mini z^2, which have mini z^4s around them.... wild stuff. hahahah.. that's hilarious.. a Julia with seed 0,0 has a fractal pattern of infinite circles! Title: Re: ln(e^(z^n)) = different??? Post by: lkmitch on October 11, 2012, 05:41:49 PM I think it's because of the branch cuts in the log function. If you reverse the order of log and exp, all the weirdness disappears (at least, in my Ultra Fractal formula). For more fun, also try sqrt(sqr(z^n))+c. The two attached images are from the same region in the Seahorse Valley of the Mandelbrot set, one with ln(exp(z^2))+c and the other with sqrt(sqr(z^2))+c.
Title: Re: ln(e^(z^n)) = different??? Post by: hobold on October 11, 2012, 07:14:01 PM In the complex numbers, the exponential function is not injective. That means you can find complex numbers u and v, with u != v but exp(u) == exp(v). The complex logarithm cannot reconstruct the information that was lost this way. Any of the branches of a complex logarithm function is in some sense "arbitrary" in that it picks just one of many candidates, none of which is really "the one right" choice.
Similarly, squaring a number (real or complex) is not injective either. So the square root function similarly introduces an "arbitrary" choice of one of two branches. If you like the above fractals and want to speed them up, you could simply emulate the "cut" without going trough the trouble of computing complex logarithms and exponentials. Title: Re: ln(e^(z^n)) = different??? Post by: M Benesi on October 11, 2012, 08:45:09 PM Thanks. I thought of branch cuts after I shut down my computer last night. :) Was pretty tired at the point I implemented that in 2d.
Definitely a good idea to implement the branch cuts in a "cheaty" way instead of using log and exponential function. I do like the sqrt(sqr(z^n)) + c fractal type. I think I'm going to have to play with that idea a bit. Might need to find a "cheaty" way of doing that as well. In fact, that reminds me of this recursive formula I was working on quite some time ago: f0=0 ft+1(x) = sqrt(x^2-x + ft) ---OR--- ft+1(x) = (x^n-x + ft)^(1/n) for higher n Might be interesting.... Title: Re: ln(e^(z^n)) = different??? Post by: kram1032 on October 12, 2012, 12:37:59 AM It's interesting though that removing information in the equation effectively adds visual information in the fractal.
Title: Re: ln(e^(z^n)) = different??? Post by: M Benesi on October 12, 2012, 03:07:52 AM It's weird... It obviously cancels out a certain portion.
If you calculate bailout BEFORE adding in your pixel components, you get infinite arms for the julias (they go on forever). I'm having problems implementing a simpler formula. I can take: log (sin( IM (z ^n ))) <IM = Imaginary part> and assign the result to the imaginary part of z. This doesn't really improve calculation times at all. What should work, at least AFAI can tell, is IM(z) % (pi*2), right? But it doesn't!!@$!@ Title: Re: ln(e^(z^n)) = different??? Post by: kram1032 on October 12, 2012, 04:48:21 AM It would be something like
x=x²-y²+a y=ArcTan(Tan(2xy))+b But I doubt that would be any faster. for the sqrt version you end up with something like: abs = x²+y² arg = 1/2 atan(tan(4 atan2(x+y i))) x=abs*cos(arg)+a y=abs*sin(arg)+b but that also probably makes it slower rather than faster.... I guess the reason for, let's say "increased fractality" is the quick change of values in close proximity to the branch cut. There, the results depend a lot more on tiny changes, increasing the overall observed chaotic behavior of the system. Makes me wonder what would happen if you went for the triple-branch-cut productlog. Something like z=ProductLog(ez2z²) Or some even more branch-torn function. Title: Re: ln(e^(z^n)) = different??? Post by: M Benesi on October 12, 2012, 05:26:20 AM Gee. Now I have to try and implement that. Thanks.
Title: Re: ln(e^(z^n)) = different??? Post by: M Benesi on October 13, 2012, 04:25:35 AM Well, I found that there are very interesting Julia (0,0) for certain multiple branch cut fractal types. For z= log [ sqrt( sqr(z)^z^n) ] You get nice webs around n=3. Here's some webs from n=3 (click to enlarge): (https://lh4.googleusercontent.com/-hO2Hi2IFBE0/UHjPpfAa9aI/AAAAAAAABwE/Lg5B80GJHdM/s400/other%2520exp%25207%2520type%2520switch%25201.jpg) (https://lh4.googleusercontent.com/-hO2Hi2IFBE0/UHjPpfAa9aI/AAAAAAAABwE/Lg5B80GJHdM/s0/other%2520exp%25207%2520type%2520switch%25201.jpg) For z= log [ (z^z^n)^b ]^(1/b) } n=2 b=7 Julia (0,0) you get nice ferns: (https://lh4.googleusercontent.com/-1HiKd-YrV7M/UHjPpSwylKI/AAAAAAAABwE/QGiy0HbrFz0/s400/nice%2520fern%2520fractal.jpg) (https://lh4.googleusercontent.com/-1HiKd-YrV7M/UHjPpSwylKI/AAAAAAAABwE/QGiy0HbrFz0/s0/nice%2520fern%2520fractal.jpg) Here is the first formula, Mandelbrot style. (https://lh6.googleusercontent.com/-Oy6gkPlj45A/UHj61YFA1NI/AAAAAAAABwc/GRcK3d23uJs/s640/blue%2520and%2520black.jpg) (https://lh6.googleusercontent.com/-Oy6gkPlj45A/UHj61YFA1NI/AAAAAAAABwc/GRcK3d23uJs/s0/blue%2520and%2520black.jpg) And this is the distorted mini mandy for n=3: (https://lh6.googleusercontent.com/-F7KX07lao9M/UHkHFYtz1yI/AAAAAAAABw0/xNgIlosz9BA/s640/mini%2520mandelbrot%2520in%2520n%25203.jpg) (https://lh6.googleusercontent.com/-F7KX07lao9M/UHkHFYtz1yI/AAAAAAAABw0/xNgIlosz9BA/s0/mini%2520mandelbrot%2520in%2520n%25203.jpg) A seahorse valley shot... a bit broken up. The mini Mandy was after a continuity break- the section it is in has many breaks in continuity, so this image does as well. I like this stuff, it's old school 2d stuff. Can't find many references to z^z^n types of fractals though. Wouldn't mind reading about them a bit. (https://lh3.googleusercontent.com/-1XmRIyUN2JE/UHkPaCfZo0I/AAAAAAAABxM/gNTZVnVWMvU/s400/seahorse%2520valley%2520minibrot.jpg) (https://lh3.googleusercontent.com/-1XmRIyUN2JE/UHkPaCfZo0I/AAAAAAAABxM/gNTZVnVWMvU/s0/seahorse%2520valley%2520minibrot.jpg) |