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I've got an image of a segment of the Mandelbrot set that I generated with an iPhone app a long time ago (I use it as my background image). I now have no idea where in the set the image came from.

enter image description here

Are there mathematical tricks I could use to help find the location in the set, which would make the search faster than brute-force?

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  • $\begingroup$ Could you link us to a picture of the image? Do you want to find the location of the specific image or are you only interested in a general method? $\endgroup$ – Seth Jan 6 '15 at 21:21
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    $\begingroup$ My gut instinct says no, because the Mandelbrot set is a fractal and fractals display self-similarity everywhere. However there are many clever users on this site and I have no idea if someone has a neat trick to do this. $\endgroup$ – graydad Jan 6 '15 at 21:21
  • $\begingroup$ The self-similarity technically only applies to the border, so the colouring used for the exterior might be useable to determine the location. The colourings I have seen can rule out some of the copies, but probably not all. We would need to see the picture to say anything definite, but my guess would be no. $\endgroup$ – Henrik supports the community Jan 6 '15 at 21:54
  • $\begingroup$ I tried uploading the image here, but it looks like SE compresses images, which isn't very helpful. Here is the full uncompressed version. $\endgroup$ – asmeurer Jan 6 '15 at 22:24
  • $\begingroup$ @graydad: Strictly speaking you are probably correct, but depending on what the OP wants to do with the desired information, it may be sufficient to find a location in the set that looks like this rather than the location. $\endgroup$ – Wrzlprmft Jan 7 '15 at 20:18
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In Mathematica V10, we can generate an image very much like yours with the following command:

MandelbrotSetPlot[
  {-0.7092 + 0.24445 I,
   -0.712 + 0.2487 I},
  MaxIterations -> 250,
  ImageResolution -> 2000,
  ColorFunction -> "CherryTones"
]

enter image description here

Of course, it might make sense to increase the iteration count. Here's the same image with MaxIterations set to 1500.

enter image description here

The key geometric feature that allows us to find this is the spiral pattern. There are 13 arms spiraling in to the Misiurewicz point near the top of the figure. That tells us that we are located near a period 13 bulb. Counting counter-clockwise from the main spike, the arms get smaller until we get to the $7^{\text{th}}$ arm, indicating we could look at the $6/13$ or $7/13$ bulb, which I'll describe in a moment. Also, the first prominent spiral (which isn't really spiraling much) has four arms; this will be useful when we try to figure out which of the smaller components off of the period bulb we should examine.

The main cardioid of the Mandelbrot set can be parametrized via $$c(t) = \frac{1}{2} e^{2 i \pi t}-\frac{1}{4}e^{4 i \pi t}.$$ There is a bulb attached to this cardioid at each rational number $t$. The denominator of that rational number tells you the period associated with the bulb, i.e. every parameter $c$ chosen from the part of that bulb directly connected to the main cardioid yields a function $f_c(z)=z^2+c$ with the property that the critical point zero is attracted to an orbit whose period is that denominator. This also happens to be the number of arms that spiral in near here.

The numerator of the fraction $t$ tells you how the points in the periodic orbit are stepped through. This is easiest to see with an example of the resulting Julia sets.

enter image description here

This also happens to mirror the changes in the sizes of the spiral arms of the Mandelbrot set.

Taking all this into account, I decided to examine the period $6/13$ bulb attached to the main cardioid, i.e. the point $$c(6/13) = \frac{1}{2} e^{\frac{12 i \pi }{13}}-\frac{1}{4} e^{-\frac{2 i \pi }{13}} \approx -0.706835 + 0.235839 i.$$

This point is shown in red in the image below. The image also includes a sketch of the cardioid as determined by the parametrization $c(t)$.

enter image description here

Zooming in, we see something like so:

enter image description here

Now, at this scale, all of the smaller disks that we see attached to the main $6/13$ disk have similar spiral patterns. This is where we use the fact that the first prominent spiral has four arms. This leads us, finally, to the portion outlined in blue in that final image.

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    $\begingroup$ wow -- very nice :) $\endgroup$ – Sheldon L Jan 7 '15 at 21:28
  • $\begingroup$ Note that the linked image (which I guess you edited out of my question) is higher quality, because SO compresses the image when you insert it, but the original is uncompressed png. $\endgroup$ – asmeurer Jan 8 '15 at 23:35
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    $\begingroup$ @asmeurer I didn't edit anything out of your original post, I just incorporated the link you included in one of your comments; that comment is still there. Also, I misspoke when I said your image was insufficient resolution. The resolution is fine but the quality is low, since the maximum iteration count is too low to determine when many points that are close to the set are not actually in it. I've included a sharpened version to illustrate. I also finished the response a bit more algorithmically. I'm glad you like it, by the way - I use sympy often. $\endgroup$ – Mark McClure Jan 9 '15 at 0:48
  • $\begingroup$ @MarkMcClure: +1. Excellent detective work! $\endgroup$ – Yiannis Galidakis Jan 9 '15 at 0:52
  • $\begingroup$ Shows how much I don't know about the Mandelbrot set. $\endgroup$ – marty cohen Jan 9 '15 at 1:03

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