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http://www.gibney.de/does_anybody_know_this_fractal

Is this some known kind of fractal?

Update: This one got a lot of great feedback from around the net. I summarized it here: http://www.gibney.de/does_anybody_know_this_fractal_-_part_2

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  • $\begingroup$ In case anyone is wondering about the number of upvotes, this is linked from Hacker News. $\endgroup$ – Chris Taylor Oct 16 '12 at 22:07
  • $\begingroup$ What are the limits of your plot? For example, what is the complex number in the upper right corner of the plot? Is it 4000 + 4000i? $\endgroup$ – Chris Taylor Oct 16 '12 at 22:09
  • $\begingroup$ Chris: My original question contained a link to the page where I describe this thing and give more informations. For some reason this question got edited heavily and the link got dropped. Here it is: gibney.de/does_anybody_know_this_fractal $\endgroup$ – no_gravity Oct 16 '12 at 22:31
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    $\begingroup$ I edited the question to make it clearer, but I didn't remove the link - it's still there. I've now edited to include the limits of the plot as well (FYI, we prefer questions to be as complete as possible - links to external sites often go dead, especially on a timescale of years). $\endgroup$ – Chris Taylor Oct 16 '12 at 22:34
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    $\begingroup$ Your image appears to be the inversion w.r.t. the origin of a regularly repeating pattern; the latter reminds me of a higher-dimensional variation of Thomae's function. $\endgroup$ – Rahul Oct 17 '12 at 0:32
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Probably a part of what we see is due to rounding in your program: note that the value at $c$ is the same as that at $ic$ so that the image should be symmetric under rotation by 90 degrees. This roughly is the case, apart from the deformed squares all over the place. In fact, they all occupy one quadrant like region with one vertex at a point with a high value, probably due to rounding. Without the squares you would have a better approximation to the real picture.

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The intensity as defined is $0$ for almost all values of $z$, so it is unclear how the visualization relates to the definition. If you post the code that generated the images you will have a better chance of getting a satisfactory answer, and that would also address @doetoe's observation of what are apparently artifacts.

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  • $\begingroup$ You are right. The check if the result of a division is a gaussian integer is actually checking if the result is near a gaussian integer. Thats why the intensity is higher then 0 for most values. $\endgroup$ – no_gravity Oct 23 '12 at 15:57

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