Re: [Fractint] Triternions
From: "Gerald K. Dobiasovsky" <gerald.dob@aon.at> Reply-To: fractint@mailman.xmission.com To: <fractint@mailman.xmission.com> Subject: Re: [Fractint] Triternions Date: Wed, 29 Jan 2003 01:29:59 +0100
Russ Walsmith wrote:
In the following frm files, the default imaginary values for Y are changed by imag(pixel)*(0,-1) to real.
CP4 (XAXIS){ x=real(pixel), y=imag(pixel)*(0,-1), a=b=0:
Russ, this seems to be a small misunderstanding regarding Fractint's parser. The imag()-function returns the imaginary component of a number already as a real! That is, from a + i*b you get b + i*0, not 0 + i*b.
Regards, Gerald
Thanks Gerald for calling that to my attention... but I admit that I had to run about six versions of TEST { y=imag(pixel): y^2 > 1 } before I became convinced. Good thing I didn't realize that, I guess, else I never would have rectified the imagined imaginary value in Mr T's formula, and TMan would not have come to life. Then Jim Muth couldn't have turned him sideways to create the fraternal twin, TMan1, and if TMan1 didn't exist, he could not have been so swiftly dissected, his strange internal organs expertly laid bare in front of God and everybody on the Fractint list. And if that hadn't happened, from whence would have come the inspiration for the haunting, surreal video 'Mandelbrot Rising', now playing at http://www.sonic.net/~jot/ufvp.html So hey, it's all good, and to quote Morgan Owens, "...the important thing here are the visual results..." Indeed, those and other happy outcomes are accruing to the bottom line. But if the imag(pixel) theory doesn't hang, then what is going on? I must say that the situation is way more curious than I initally supposed. There is much to ponder in this realm; e.g., I've noticed that CP8 { c1=real(pixel),c2=imag(pixel)*p3,c3=p1,c4=p2 z1=z2=z3=z4=0: t1=z1*z1-z3*z3-2*z2*z4 t2=2*z1*z2-2*z3*z4 t3=2*z1*z3+z2*z2-z4*z4 t4=2*z1*z4+2*z2*z3 z1=t1+c1,z2=t2+c2,z3=t3+p1,z4=t4+p2 z=sqrt(t1^2+t2^2+t3^2+t4^2) z < 4 } seems to be completely indifferent as to whether the value chosen for p3 is real or imaginary. Do you have any insights here? Regards, Russell _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail
Russ Walsmith wrote:
Thanks Gerald for calling that to my attention... but I admit that I had to run about six versions of TEST { y=imag(pixel): y^2 > 1 } before I became convinced.
Shows how memory can be misleading. I was _sure_ this is documented in the online-help (it is not), but found it in my old "Fractal Creations" book (1st edition, german translation) instead: "imag() - replaces the real component by the imaginary one, sets the imaginary part to zero and by that produces a real number."
So hey, it's all good, and to quote Morgan Owens, "...the important thing here are the visual results..."
An example for this is even hardwired into Fractint as a fractal type: tim's_error.
Indeed, those and other happy outcomes are accruing to the bottom line. But if the imag(pixel) theory doesn't hang, then what is going on? I must say that the situation is way more curious than I initally supposed. There is much to ponder in this realm; e.g., I've noticed that CP8 { c1=real(pixel),c2=imag(pixel)*p3,c3=p1,c4=p2 z1=z2=z3=z4=0: t1=z1*z1-z3*z3-2*z2*z4 t2=2*z1*z2-2*z3*z4 t3=2*z1*z3+z2*z2-z4*z4 t4=2*z1*z4+2*z2*z3 z1=t1+c1,z2=t2+c2,z3=t3+p1,z4=t4+p2 z=sqrt(t1^2+t2^2+t3^2+t4^2) z < 4 } seems to be completely indifferent as to whether the value chosen for p3 is real or imaginary. Do you have any insights here?
How I would love to be able to provide a clever proof showing it _has_ to be that way :-) But why shouldn't it be possible for the fractal's cross- section for c2=real to look identical to the cross-section for c2=imag? If you rotate p3 from (1,0) only part of the way towards (0,1) the image certainly changes - try 0.707106781 for both components of p3 for a well-known fractal to appear (only in part, though). Another situation where switching p3 from (1,0) to (0,1) does change the resulting image is in having c3 and/or c4 to be non-zero (say 0.1 for example)... Regards, Gerald
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Gerald K. Dobiasovsky -
Russ Walsmith