# Why does the Mandelbrot set contain (slightly deformed) copies of itself?

The Mandelbrot set is the set of points of the complex plane whos orbits do not diverge. An point $c$'s orbit is defined as the sequence $z_0 = c$, $z_{n+1} = z_n^2 + c$.

The shape of this set is well known, why is it that if you zoom into parts of the filaments you will find slightly deformed copies of the original shape, for example:

I measured some points on the Mandelbrot, and the corresponding points from one of these smaller Mu-molecules. Comparing the orbit sequences it was possible to find points on each sequence which were very close - but this experiment did not really help me to understand anything new.

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+1 for the eye-popping image! – Tom Stephens Aug 18 '10 at 13:07
All the fractal books go on about "fixed-point theory" and how the Mandelbrot set separates "attractive" from "repulsive" fixed points, as well as how it is related to the first self-similar object that was found in ancient times: the "equiangular (logarithmic) spiral". All this is a bit too dense for me. I for one would love to see a more accessible explanation to non-specialists. – J. M. Aug 18 '10 at 13:43
There is more to the Mandebrot set that containing copyes of itself: a square contains four copies of itself! – Mariano Suárez-Alvarez Aug 19 '10 at 22:52
And the logarithmic spiral looks the same if you zoom in/out properly, yes (the reason why Jakob Bernoulli had requested it be put on his tombstone). Sam's answer looks to be a rough sketch of why the Mandelbrot set is the way it is... – J. M. Aug 20 '10 at 23:13
Good question. +1. – Mehper C. Palavuzlar Oct 27 '10 at 11:35

I really like this question! I can't yet upvote, so I'll offer an answer instead. This is only a partial answer, as I don't fully understand this material myself.

Suppose that $f$ is a quadratic polynomial. Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a "quadratic-like map." Then we'll call $f$ renormalizable. (See Chapter 7 of McMullen's book "Complex dynamics and renormalization" for more precise definitions.) Now renormalization preserves the property of having a connected Julia set. Also the parameter space of "quadratic-like maps" is basically a copy of $\mathbb{C}$.

So, fix a quadratic polynomial $f$ and suppose that it renormalizes. Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$. This gives a map from a small region about $f$ to the space of quadratic-like maps. This gives a partial map from the small region to the Mandelbrot set and so explains the "local" self-similarity.

To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way. (I believe that there are issues as you approach the place where the baby is attached to the parent.) Thus renormalization explains why the baby Mandelbrot set appears.

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Incurring the risk of sounding too simplistic, isn't that one of the fundamental properties of a fractal [Wikipedia]:

Quasi-self-similarity. This is a looser form of self-similarity; the fractal appears approximately (but not exactly) identical at different scales. Quasi-self-similar fractals contain small copies of the entire fractal in distorted and degenerate forms. Fractals defined by recurrence relations are usually quasi-self-similar but not exactly self-similar. The Mandelbrot set is quasi-self-similar, as the satellites are approximations of the entire set, but not exact copies.

Update: Sorry to dig this topic out, but I was just reading about folding something into itself, and I remembered having seen something similar related to fractals. The following videos are amazing and present an incredibly intuitive notion of where the self-similarity of a fractal comes from: https://code.google.com/p/mandelstir/

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While this is true, Sam Nead's answer actually answers why this is true. – Jacques Carette Aug 21 '10 at 13:53
But why is the Mandelbrot set a fractal? Nothing in its definition says it can't be something simple like a circle. – weux082690 May 30 at 2:34

It's slightly tricky for the Mandelbrot set, because this exists in parameter space. It's easier to think about the corresponding Julia sets - although the idea is similar.

The answer that I like is this: A Julia set is a "hall of mirrors". When you look at one, you are seeing a reflection of a reflection of a reflection... In fact, with the right software, you can animate this idea in real time. This also lets you see features of the set which aren't readily apparent just by looking at the usual escape-time colouring of it. But I digress...

When you think of the fractal as being a reflection of reflections of reflections, suddenly the idea that one small part of it might look like the whole thing seems... unsurprising.

If you want to really blow you mind, try this: Did you know that certain parts of the Mandelbrot set resemble particular Julia sets? I still haven't figured out exactly why yet...

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There are some references for this business about $M$ at $c$ resembling $J_c$ here: math.stackexchange.com/questions/32062/… – Sam Nead May 2 '14 at 18:42

See this paper:

McMullen, Curtis T., The Mandelbrot set is universal. In The Mandelbrot set, theme and variations, 1–17, London Math. Soc. Lecture Note Ser., 274, Cambridge Univ. Press, 2000. MR1765082 (2002f:37081)

PDF available at the author's site.

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