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Fun with Fractals
Mike Whittaker
University of Glasgow
Royal Institution of Great Britain Masterclass in Mathematics
University of Glasgow
7 November 2015
Plan for today:
1. Introduction
2. Introduction to complex numbers
3. Julia sets
4. The Mandelbrot set
Romanesco broccoli
A fractal
A fern
The Barnsley Fern
Lungs
A fractal
Snowflake
The Koch Snowflake
Painting “Winter Oak” by Virginia Daley
Fractal tree
What are fractals?
Definition (Oxford English Dictionary)
A fractal is a curve or geometrical figure, each part of which has
the same statistical character as the whole. They are useful in
modelling structures (such as snowflakes) in which similar patterns
recur at progressively smaller scales, and in describing partly
random or chaotic phenomena such as crystal growth and galaxy
formation.
Fractals often come from unexpected places...
Paper folding
Take your long piece of paper and fold it once. With the fold to
the left make a 2nd fold (this isn’t really important but will make
sure we all get the same result). Now unfold the result
After 2 folds you should get
Paper folding
Refold your paper (with the first fold to the left). Now make a 3rd
fold and unfold the result. It should look like:
Paper folding
Refold your paper (with the first fold to the left). Now make a 4th
fold and unfold the result. It should look like:
Paper folding
Refold your paper (with the first fold to the left). Now make a 5th
fold and unfold the result. It should look like:
Paper folding
6th iteration:
Paper folding
7th iteration:
Paper folding
8th iteration:
Paper folding
9th iteration:
Paper folding
10th iteration:
Paper folding
11th iteration:
Paper folding
12th iteration:
Paper folding
13th iteration:
Paper folding
14th iteration:
Paper folding
15th iteration:
Paper folding
... 20th iteration:
Aside about paper folding
How many times do you think you could fold a piece of paper?
I bet you all a chocolate bar that none of you can fold a piece of
paper 10 times...
Lets work out how thick the paper will be if you could fold it 10,
15, 20, and 43 times...
Let’s assume that a standard piece of paper is 0.05mm thick (I got
this figure from the internet). If you fold the paper 10 times how
many sheets of paper thick is the stack?
Hint: Use your paper to check the numbers for 2,3,4,5 folds.
Aside about paper folding
At 10 folds your paper is 210 sheets thick. So we have
210 × 0.05mm = 1024 × 0.05mm = 51.2mm
At 15 folds your paper is 215 sheets thick. So we have
215 × 0.05mm = 32768 × 0.05mm = 1638.4mm ≈ 1.6m
At 20 folds your paper is 220 sheets thick. So we have
220 × 0.05mm ≈ 1718kms
At 43 folds your paper is 243 sheets thick. So we have
243 × 0.05mm ≈ 439804kms
Note: The distance from earth to the moon is 384400kms.
Part 2: Introduction to complex numbers
Introduction to complex numbers
Is there a solution to the polynomial equation x 2 + 1 = 0?
Solving for x we have
x2 + 1 = 0
=⇒
x 2 = −1
=⇒
√
x = ± −1.
But hold on! You can’t take the square root of a negative
number..............
In the 1500s, this annoyed Rafael Bombelli. So he came up with
an alternative:
√
Bombelli’s idea is to imagine that −1 does exist, and to declare
√
i = −1.
Introduction to complex numbers
√
Assuming that i = −1 exists we still get a mathematically
consistent framework (i.e. this crazy idea actually works!).
We call i an imaginary number since you have to imagine that it
exists.
Lets check that x = i is a solution x 2 + 1 = 0:
√
i 2 + 1 = ( −1)2 + 1 = −1 + 1 = 0
X
Introduction to complex numbers
A complex number is a number of the form a + bi where a and b
are real numbers.
We can add and multiply complex numbers as follows:
(a + bi) + (c + di) = (a + c) + (b + d)i and
(a+bi)(c +di) = ac +bci +adi +bdi 2 = (ac −bd)+(bc +ad)i.
For example
(3+2i)2 = (3+2i)(3+2i) = 9+6i+6i+4i 2 = (9−4)+(6+6)i = 5+12i
On your scrap paper try working out
(1 + i)(4 − 2i) = 4 − 2i + 4i − 2i 2 = (4 + 2) + (−2 + 4)i = 6 + 2i
The geometry of complex numbers
For any complex number a + bi we have two coordinates, the real
coordinate and the imaginary coordinate
imaginary
a + bi
b
a
real
The modulus of a complex number is the quantity
p
|a + bi| = a2 + b 2 .
Using the Pythagorean Theorem, the modulus is the length of the
blue arrow.
Part 3: Julia sets
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Julia sets
Consider the equation f (z) = z 2 + c for c a complex number.
Let f 2 (z) = f (f (z)) and f 3 (z) = f (f (f (z))) and so on.
For each fixed complex number z0 we want to look at the values
|f (z0 )|, |f 2 (z0 )|, |f 3 (z0 )|, |f 4 (z0 )|, |f 5 (z0 )|, · · ·
and determine if this sequence of numbers bounded or if is goes off
to infinity.
The (filled) Julia Set is the set of complex values where the above
sequence is bounded.
Julia set example: f (z) = z 2 − 1
Lets see what sort of sequences we get for f (z) = z 2 − 1.
First look at 0. We get
f (0) = 02 − 1 = −1
f 2 (0) = f (f (0)) = f (−1) = (−1)2 − 1 = 1 − 1 = 0
f 3 (0) = f (f 2 (0)) = f (0) = −1
f 4 (0) = f (f 3 (0)) = f (−1) = 0,
and the pattern repeats. So the sequence we get is
{1, 0, 1, 0, 1, 0, · · · }. This is a bounded sequence, so 0 belongs to
the Julia set of f (z) = z 2 − 1.
Question: Is 1 in the Julia set of f (z) = z 2 − 1?
Solution: Yes! Since f (1) = 12 − 1 = 0, using the sequence for 0
we get {0, 1, 0, 1, 0, · · · } which is bounded.
Julia set example: f (z) = z 2 − 1
Lets take a look at the sequence for the complex number i:
f (i) = i 2 − 1 = −1 − 1 = −2
f 2 (i) = f (f (i)) = f (−2) = (−2)2 − 1 = 4 − 1 = 3
f 3 (i) = f (f 2 (i)) = f (3) = 32 − 1 = 8
f 4 (i) = f (f 3 (i)) = f (8) = 82 − 1 = 63,
and we get a sequence that grows very large. So the sequence we
get is {2, 3, 8, 63, 3968, · · · }. This is an unbounded sequence, so i
does not belong to the Julia set f (z) = z 2 − 1.
Question: So what’s the point?
Answer: The set of complex numbers that give bounded
sequences (the Julia Set) is a remarkable geometric object.
The Julia set of f (z) = z 2 − 1
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Every point in the black part of the image above is in the Julia set
and the coloured points are not!
The Julia set of f (z) = z 2
Exercise for problem session: Can you work out (guess) the Julia
set of f (z) = z 2 ?
Hint: Plug in some complex numbers and see what you get!
1.0
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You get a circle of radius 1.
The Julia set of f (z) = z 2 + i
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The Julia set of f (z) = z 2 + 8i/9
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-1.0
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0.0
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1.0
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The Julia set of f (z) = z 2 − 1.3
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0.4
0.2
0.0
-0.2
-0.4
-0.6
-2
-1
0
1
2
The Julia set of f (z) = z 2 − 1.7549
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0.0
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-2
-1
0
1
2
The Julia set of f (z) = z 2 − 0.12256 − 0.74486i
1.0
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The Julia set of f (z) = z 2 − 1.037 + 0.17i
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The Julia set of f (z) = z 2 − 0.52 + 0.57i
1.0
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The Julia set of f (z) = z 2 + 0.295 + 0.55i
1.0
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0.0
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-1.0
-1.0
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0.0
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1.0
The Julia set of f (z) = z 2 − 0.624 + 0.435i
1.0
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0.0
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-1.0
-1.5
-1.0
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0.0
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1.0
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The Julia set of f (z) = z 2 − 0.8 − 0.175i
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The Julia set of f (z) = z 2 − 0.8 − 0.15i
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Part 4: The Mandelbrot set
Benoı̂t Mandelbrot 1924–2010
The Mandelbrot set
To study Julia sets we considered complex functions f (z) = z 2 + c.
What happens if we study these functions as the value c varies?
Let fc (z) = z 2 + c and look at the sequence
{fc (0), fc2 (0), fc3 (0), · · · }
(1)
for different values of c.
The Mandelbrot Set consists of the values of c where the sequence
(1) is bounded.
For example, in the Julia set section, we saw that the sequence (1)
is bounded for f0 and f−1
So what does the collection of all points that are bounded look
like...
The Mandelbrot Set
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The Mandelbrot set
“There is nothing more to this than a simple iterative formula. It is
so simple that most children can program their home computers to
produce the Mandelbrot set... Its astounding complication was
completely out of proportion with what I was expecting.”
- Benoı̂t Mandelbrot.
Zooming in on the Mandelbrot set
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-0.40
The complexity of the Mandelbrot set is absolutely astounding.
Zooming in on the Mandelbrot set
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-0.740
-0.735
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-0.740
-0.735
-0.730
-0.725
These two images are both the same region in the Mandelbrot set.
On the left we have coloured points black that have sequences that
get larger than 2 after 300 iterations and on the right after 800
iterations.
Zooming in on the Mandelbrot set
https://www.youtube.com/watch?v=PD2XgQOyCCk
This video uses 10227 iterations to produce this high resolution
version of the Mandelbrot set. To give you a sense of the scale of
this number:
The universe is estimated to contain between 1078 and 1082
atoms.
it took 4 weeks for a really fast computer to produce.
Another definition of the Mandelbrot set
There is an alternative definition: The Mandelbrot set consists of
the points c such that the Julia set of fc (z) = z 2 + c is connected.
Connectedness is a topological property. So the Mandelbrot set
can be said to be a single object encoding interesting features of
the collection of all Julia sets.
Lets see this with some more pictures...
Finding Julia sets in the Mandelbrot set
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The red dot in the Mandelbrot set shows where this Julia set on
the left comes from in within the Mandelbrot set.
Finding Julia sets in the Mandelbrot set
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The red dot in the Mandelbrot set shows where this Julia set on
the left comes from in within the Mandelbrot set.
Some Julia sets of the Mandelbrot set
Resources
The Dragon Curve - by Numberphile video:
https://www.youtube.com/watch?v=wCyC-K_PnRY
Mandelbrot Set - by Numberphile video:
https://www.youtube.com/watch?v=NGMRB4O922I
Filled Julia Sets - by Numberphile video:
https://www.youtube.com/watch?v=oCkQ7WK7vuY
Mandelbrot zoom video:
https://www.youtube.com/watch?v=PD2XgQOyCCk
The images were produced by Mathematica.