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Author Topic: Buddha's Jewel: a special subset of the Buddhabrot, symmetrical about both axis!  (Read 2231 times)
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3dickulus
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« Reply #45 on: April 18, 2015, 08:52:26 PM »

Spectacular! ... in a cosmic sort of way  A Beer Cup
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Resistance is fertile...
You will be illuminated!

                            #B^] https://en.wikibooks.org/wiki/Fractals/fragmentarium
tit_toinou
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« Reply #46 on: April 19, 2015, 06:41:05 PM »

Hi there,

Good looking images wink .
Like I said in this post, I think the points you use to plot to the Buddhabrots are the ones that are white in my images (high derivative of the distance <=> high potential) : Mandelbrot with DE (at the end).

One thing about the symmetrical thing : the Buddhabrot IS symmetrical since as all complex polynomials have the property P(conjugate(z))=conjugate(P(z)). It just requires a lot of points to notice that (my renders of Buddhabrot don't show this because there is not enough points, but like someone said you can just plot the conjugate orbits and get something symmetrical at the end wink ).
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Chris Thomasson
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« Reply #47 on: April 19, 2015, 10:18:37 PM »

A zoom on my last Buddhabrot, centered on the main cardioid:



I think I might be falling in love with these Buddhabrot's!

;^o
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billtavis
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« Reply #48 on: April 19, 2015, 11:48:54 PM »

Good looking images wink .
Like I said in this post, I think the points you use to plot to the Buddhabrots are the ones that are white in my images (high derivative of the distance <=> high potential) : Mandelbrot with DE (at the end).
Thanks! And yeah, that seems like a good way to describe the points
One thing about the symmetrical thing : the Buddhabrot IS symmetrical since as all complex polynomials have the property P(conjugate(z))=conjugate(P(z)). It just requires a lot of points to notice that (my renders of Buddhabrot don't show this because there is not enough points, but like someone said you can just plot the conjugate orbits and get something symmetrical at the end wink ).
Well, like the Mandelbrot set, buddhabrots are symmetrical across the real axis, but not the imaginary axis. It is only a certain subset of orbits that form symmetry across the imaginary axis.
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tit_toinou
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« Reply #49 on: April 20, 2015, 12:28:56 AM »

Ok you're not talking about exact mathematical symmetry, right ? The images you have given in the first post show approximate symmetry about the imaginary axis (look at the area you called "the rolls", it is not copied on the other half of the buddha)
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billtavis
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« Reply #50 on: April 20, 2015, 01:40:26 AM »

that's correct, the images I posted are not perfectly symmetrical across the imaginary axis, although they are very close to being so. The areas which break the symmetry the most are elephant valley and seahorse valley, where higher iteration orbits are found even at the branch tips. It's not clear to me how to isolate the exact symmetry, although it is apparent that the symmetry starts decreasing rapidly for orbits with above 100 iterations.
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