Why is the Mandelbrot set a fractal?
5 Answers
Edward Kmett, M.S. in Mathematics from Eastern Michigan University in 2005
Answered 249w ago · Upvoted by David Joyce, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 70 answers and 533.5k answer views
John Baez, mathematical physicist trying to save the planet
The Mandelbrot set is not exactly self-similar. It's only approximately self-similar: little copies of the Mandelbrot set can be found in the Mandelbrot set, but these little copies are distorted.
The structure of the Mandelbrot set is actually quite subtle. Anders Kaseorg's answer gives a nice explanation of why it's approximately self-similar, so I'll tell you about the "Mandelbrot-Julia correspondence". There are many Julia sets, one for each complex number. Julia sets are simpler than the Mandelbrot set, and their fractal properties are easier to understand. But near any specific number, the Mandelbrot set tends to look like the Julia set for that number!
Both the Mandelbrot set and the Julia sets are defined using this function of a complex number x, which also depends on the number z:
[math]f(x)=x^2+z[/math]
If we fix z, this function defines a map from the complex plane to itself. We can start with any number x and keep applying this map over and over. We get a sequence of numbers. Sometimes this sequence shoots off to infinity and sometimes it doesn't. The set where it doesn't is called the Julia set for this number z.
On the other hand, we can start with x=0, and draw the set of numbers z for which the resulting sequence doesn't shoot off to infinity. That's called the Mandelbrot set.
The Julia set for the number z is connected precisely when that number lies in the Mandelbrot set. But the Mandelbrot-Julia correspondence says this: if you zoom into the Mandelbrot set near the number z, it tends to look at lot like the Julia set for that number!
For example, the Julia set for the number
z=−0.743643887037151+0.131825904205330i
looks like this:
![]()
while an ultra-closeup of the Mandelbrot set near that number looks like this:
![]()
For an astounding image showing many Julia sets and how they look like different regions of the Mandelbrot set, see:
http://en.wikipedia.org/wiki/Fil...
You can learn more about this stuff on David Joyce's website:
http://aleph0.clarku.edu/~djoyce...
and I especially recommend comparing the Mandelbrot set to Julia sets using this webpage:
http://aleph0.clarku.edu/~djoyce...
There are also references to read, if you want proofs, and more references here:
http://en.wikipedia.org/wiki/Jul...
http://en.wikipedia.org/wiki/Man...
The structure of the Mandelbrot set is actually quite subtle. Anders Kaseorg's answer gives a nice explanation of why it's approximately self-similar, so I'll tell you about the "Mandelbrot-Julia correspondence". There are many Julia sets, one for each complex number. Julia sets are simpler than the Mandelbrot set, and their fractal properties are easier to understand. But near any specific number, the Mandelbrot set tends to look like the Julia set for that number!
Both the Mandelbrot set and the Julia sets are defined using this function of a complex number x, which also depends on the number z:
[math]f(x)=x^2+z[/math]
If we fix z, this function defines a map from the complex plane to itself. We can start with any number x and keep applying this map over and over. We get a sequence of numbers. Sometimes this sequence shoots off to infinity and sometimes it doesn't. The set where it doesn't is called the Julia set for this number z.
On the other hand, we can start with x=0, and draw the set of numbers z for which the resulting sequence doesn't shoot off to infinity. That's called the Mandelbrot set.
The Julia set for the number z is connected precisely when that number lies in the Mandelbrot set. But the Mandelbrot-Julia correspondence says this: if you zoom into the Mandelbrot set near the number z, it tends to look at lot like the Julia set for that number!
For example, the Julia set for the number
z=−0.743643887037151+0.131825904205330i
looks like this:
while an ultra-closeup of the Mandelbrot set near that number looks like this:
For an astounding image showing many Julia sets and how they look like different regions of the Mandelbrot set, see:
http://en.wikipedia.org/wiki/Fil...
You can learn more about this stuff on David Joyce's website:
http://aleph0.clarku.edu/~djoyce...
and I especially recommend comparing the Mandelbrot set to Julia sets using this webpage:
http://aleph0.clarku.edu/~djoyce...
There are also references to read, if you want proofs, and more references here:
http://en.wikipedia.org/wiki/Jul...
http://en.wikipedia.org/wiki/Man...
Anders Kaseorg, MIT S.B. in Math ’08; MIT PhD student in CS ’14–
Updated 236w ago · Upvoted by David Joyce, Ph.D. Mathematics, University of Pennsylvania (1979) and Alon Amit, PhD in Mathematics; Mathcircler. · Author has 556 answers and 7.6m answer views
The word “fractal” is only loosely defined, but everyone agrees that fractals share two general kinds of properties:
Fractals have more exotic behavior. For example, if you scale up the boundary of the Koch snowflake by a factor of 3, you can arrange the result into 4 identical copies of the original boundary—so the measure has increased by a factor of 4, not 3. In general, if you scale it up by [math]k[/math], the measure increases by a factor of [math]k^{\log_3 4}[/math]. Thus we say that the Koch snowflake’s boundary has dimension [math]\log_3 4 \approx 1.262[/math].
The Mandelbrot set is an extreme case of this behavior. Its boundary is actually 2-dimensional: if you scale up the Mandelbrot set by a factor of two, the boundary measure increases by four times. This is the largest possible dimension for a plane figure, and equal to the dimension of the Mandelbrot set itself. The Mandelbrot is set is so “bumpy” that we can talk about its boundary having an area, not a length—and whether that area is nonzero remains an open question to this day!
Let [math]f_c(z) = z^2 + c[/math], and consider the Taylor series of [math]f_c^n[/math] ([math]f_c[/math] iterated [math]n[/math] times) for some positive integer [math]n[/math]:
[math]f_c^n(z) = f_c^n(0) + \tfrac12(f_c^n)''(0)z^2 + O(z^4)[/math].
(There are no [math]z^1[/math] or [math]z^3[/math] terms because [math]f_c[/math] is an even function.)
If we scale everything by a factor of [math]k = \tfrac12(f_c^n)''(0)[/math], we end up with
[math]k f_c^n\bigl(\tfrac z k\bigr) = \tfrac12(f_c^n)''(0)f_c^n(0) + z^2 + O(z^4)[/math]
[math]= f_{g_n(c)}(z) + O(z^4)[/math],
where [math]g_n(c) = \tfrac12(f_c^n)''(0)f_c^n(0)[/math].
That is, for small [math]z[/math], taking [math]mn[/math] iterations of [math]f_c[/math] works very similarly to taking [math]m[/math] iterations of [math]f_{g_n(c)}[/math]. So we can expect that [math]g_n(c)[/math] is in the Mandelbrot set iff [math]c[/math] is in something that looks very similar to the Mandelbrot set.
Let’s test this theory. On the left are the points [math]c[/math] with [math]g_4(c)[/math] in the Mandelbrot set, compared with the real Mandelbrot set on the right:
![]()
Not too bad. But the approximation becomes much better at smaller scales. Let’s zoom in on one of the roots of [math]f_c^4(0) = 0[/math], in this case the one near [math]-0.156 + 1.032i[/math] (the tiny black spot near the top of the above image):
![]()
We’ve found a baby Mandelbrot set, and indeed it looks very close to the transformed version of the whole Mandelbrot set!
In general, each baby Mandelbrot sets occurs at a root [math]c = c_0[/math] of [math]f_c^n(0) = 0[/math]; it’s approximately sent back to the whole Mandelbrot set under the transformation [math]g_n[/math].
This transformation, like any analytic function, is approximately a scaled rotation near [math]c_0[/math], also thanks to Taylor series:
[math]g_n(c) = g_n(c_0) + g_n'(c_0)(c - c_0) + O(c - c_0)^2[/math]
[math]\approx g_n'(c_0)(c - c_0)[/math] for [math]c \approx c_0[/math].
(Multiplication by [math]g_n'(c_0)[/math] scales the complex plane by [math]|g_n'(c_0)|[/math] and rotates by [math]\arg g_n'(c_0)[/math].)
- Fractal dimension: A fractal has a fractal (Hausdorff) dimension exceeding its topological dimension. Roughly speaking, this means that a fractal is so infinitely “bumpy” that scaling it up causes its measure to increase faster than you would expect from a non-fractal.
- Self-similarity: Zooming in to a fractal yields pictures that are similar (whether exactly or somehow approximately) across widely varying scales.
Fractal dimension
The boundary of a typical smooth shape in the plane is 1-dimensional. This means that a circle of radius [math]kr[/math] is [math]k^1[/math] times as long as a circle of radius [math]r[/math]—and same for a square, triangle, or any other shape you can probably think of.Fractals have more exotic behavior. For example, if you scale up the boundary of the Koch snowflake by a factor of 3, you can arrange the result into 4 identical copies of the original boundary—so the measure has increased by a factor of 4, not 3. In general, if you scale it up by [math]k[/math], the measure increases by a factor of [math]k^{\log_3 4}[/math]. Thus we say that the Koch snowflake’s boundary has dimension [math]\log_3 4 \approx 1.262[/math].
The Mandelbrot set is an extreme case of this behavior. Its boundary is actually 2-dimensional: if you scale up the Mandelbrot set by a factor of two, the boundary measure increases by four times. This is the largest possible dimension for a plane figure, and equal to the dimension of the Mandelbrot set itself. The Mandelbrot is set is so “bumpy” that we can talk about its boundary having an area, not a length—and whether that area is nonzero remains an open question to this day!
Self-similarity
The Mandelbrot set is littered with tiny approximate copies of itself. I’ll show you why this is true, but it’s going to get a little bit technical, so scroll down if you’re just looking for pretty pictures.Let [math]f_c(z) = z^2 + c[/math], and consider the Taylor series of [math]f_c^n[/math] ([math]f_c[/math] iterated [math]n[/math] times) for some positive integer [math]n[/math]:
[math]f_c^n(z) = f_c^n(0) + \tfrac12(f_c^n)''(0)z^2 + O(z^4)[/math].
(There are no [math]z^1[/math] or [math]z^3[/math] terms because [math]f_c[/math] is an even function.)
If we scale everything by a factor of [math]k = \tfrac12(f_c^n)''(0)[/math], we end up with
[math]k f_c^n\bigl(\tfrac z k\bigr) = \tfrac12(f_c^n)''(0)f_c^n(0) + z^2 + O(z^4)[/math]
[math]= f_{g_n(c)}(z) + O(z^4)[/math],
where [math]g_n(c) = \tfrac12(f_c^n)''(0)f_c^n(0)[/math].
That is, for small [math]z[/math], taking [math]mn[/math] iterations of [math]f_c[/math] works very similarly to taking [math]m[/math] iterations of [math]f_{g_n(c)}[/math]. So we can expect that [math]g_n(c)[/math] is in the Mandelbrot set iff [math]c[/math] is in something that looks very similar to the Mandelbrot set.
Let’s test this theory. On the left are the points [math]c[/math] with [math]g_4(c)[/math] in the Mandelbrot set, compared with the real Mandelbrot set on the right:
We’ve found a baby Mandelbrot set, and indeed it looks very close to the transformed version of the whole Mandelbrot set!
In general, each baby Mandelbrot sets occurs at a root [math]c = c_0[/math] of [math]f_c^n(0) = 0[/math]; it’s approximately sent back to the whole Mandelbrot set under the transformation [math]g_n[/math].
This transformation, like any analytic function, is approximately a scaled rotation near [math]c_0[/math], also thanks to Taylor series:
[math]g_n(c) = g_n(c_0) + g_n'(c_0)(c - c_0) + O(c - c_0)^2[/math]
[math]\approx g_n'(c_0)(c - c_0)[/math] for [math]c \approx c_0[/math].
(Multiplication by [math]g_n'(c_0)[/math] scales the complex plane by [math]|g_n'(c_0)|[/math] and rotates by [math]\arg g_n'(c_0)[/math].)
Tom Lowe, B.Sc. Computer Science, University of Warwick (1999)
Good question, the problem is that fractal has two meanings.
The first is something with a fractal dimension, usually excluding Euclidean shapes with integer fractal dimension. This describes almost all fractals.
The second is a self-similar shape, which is scale-symmetric or approximately so, these are solid:
These shapes may have a border which is fractal in the first sense, or non-fractal as with certain 3D tree shapes.
These two definitions are independent, the definition of fractal dimension does not require the shape to be self-similar, and a self-similar shape does not always have a fractal dimension for its size or its border.
Given this, the Mandelbrot set is not a fractal in the first sense, but it is a fractal in the second sense. However, I do think it would be less confusing if we used ‘fractal’ for the first sense (e.g. the Mandelbrot set’s border is a fractal) and ‘self-similar shape’ for the second sense.
Máté Kovács, math aficionado