# Fun with Fractals - University of Glasgow

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 2 nd 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 3 rd fold and unfold the result. It should look like:

Paper folding

Refold your paper (with the first fold to the left). Now make a 4 th fold and unfold the result. It should look like:

Paper folding

Refold your paper (with the first fold to the left). Now make a 5 th fold and unfold the result. It should look like:

Paper folding

6 th iteration:

Paper folding

7 th iteration:

Paper folding

8 th iteration:

Paper folding

9 th iteration:

Paper folding

10 th iteration:

Paper folding

11 th iteration:

Paper folding

12 th iteration:

Paper folding

13 th iteration:

Paper folding

14 th iteration:

Paper folding

15 th iteration:

Paper folding

... 20 th 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 2

10 sheets thick. So we have

2

10

× 0.05mm = 1024 × 0.05mm = 51.2mm

At 15 folds your paper is 2

15 sheets thick. So we have

2

15

× 0.05mm = 32768 × 0.05mm = 1638.4mm ≈ 1.6m

At 20 folds your paper is 2

20 sheets thick. So we have

2

20

× 0.05mm ≈ 1718kms

At 43 folds your paper is 2

43 sheets thick. So we have

2

43

× 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 x

2

+ 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 b a + bi a real

The modulus of a complex number is the quantity

|a + bi | = p a

2

+ 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 z

0 we want to look at the values

|f (z

0

)|, |f

2

(z

0

)|, |f

3

(z

0

)|, |f

4

(z

0

)|, |f

5

(z

0

)|, · · · 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) = 0

2

− 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

Solution: Yes! Since f (1) = 1

2

− 1?

− 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) = 3

2

− 1 = 8 f

4

(i ) = f (f

3

(i )) = f (8) = 8

2

− 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!

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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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The Julia set of f (z) = z

2

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0 1 2

The Julia set of f (z) = z

2

− 1.7549

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-2 -1

0 1 2

The Julia set of f (z) = z

2

− 0.12256 − 0.74486i

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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

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The Julia set of f (z) = z

2

+ 0.295 + 0.55i

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The Julia set of f (z) = z

2

− 0.624 + 0.435i

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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

Benoˆıt 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 f c

(z) = z

2

+ c and look at the sequence

{f c

(0), f c

2

(0), f c

3

(0), · · · } (1) for different values of c.

The Mandelbrot Set consists of the values of c where the sequence

For example, in the Julia set section, we saw that the sequence (1)

is bounded for f

0 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.”

- Benoˆıt Mandelbrot.

Zooming in on the Mandelbrot set

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The complexity of the Mandelbrot set is absolutely astounding.

Zooming in on the Mandelbrot set

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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 10

227 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 10

78 atoms.

and 10

82 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 f c

(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.

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