mike whittaker university of glasgow royal …paper folding take your long piece of paper and fold...

Fun with Fractals

Mike WhittakerUniversity of Glasgow

Royal Institution of Great Britain Masterclass in MathematicsUniversity 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 hasthe same statistical character as the whole. They are useful inmodelling structures (such as snowflakes) in which similar patternsrecur at progressively smaller scales, and in describing partlyrandom or chaotic phenomena such as crystal growth and galaxyformation.

Fractals often come from unexpected places...

Paper folding

Take your long piece of paper and fold it once. With the fold tothe left make a 2nd fold (this isn’t really important but will makesure 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 ofpaper 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 gotthis figure from the internet). If you fold the paper 10 times howmany 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 x2 + 1 = 0?

Solving for x we have

x2 + 1 = 0 =⇒ x2 = −1 =⇒ x = ±√−1.

But hold on! You can’t take the square root of a negativenumber..............

In the 1500s, this annoyed Rafael Bombelli. So he came up withan 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 itexists.

Lets check that x = i is a solution x2 + 1 = 0:

i2 + 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 bare 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+bdi2 = (ac−bd)+(bc+ad)i .

For example

(3+2i)2 = (3+2i)(3+2i) = 9+6i+6i+4i2 = (9−4)+(6+6)i = 5+12i

On your scrap paper try working out

(1 + i)(4− 2i) = 4− 2i + 4i − 2i2 = (4 + 2) + (−2 + 4)i = 6 + 2i

The geometry of complex numbers

For any complex number a + bi we have two coordinates, the realcoordinate and the imaginary coordinate

real

imaginary

b

a

a + bi

The modulus of a complex number is the quantity

|a + bi | =√a2 + b2.

Using the Pythagorean Theorem, the modulus is the length of theblue arrow.

Part 3: Julia sets

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Julia sets

Consider the equation f (z) = z2 + 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 offto infinity.

The (filled) Julia Set is the set of complex values where the abovesequence is bounded.

Julia set example: f (z) = z2 − 1

Lets see what sort of sequences we get for f (z) = z2 − 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 tothe Julia set of f (z) = z2 − 1.

Question: Is 1 in the Julia set of f (z) = z2 − 1?Solution: Yes! Since f (1) = 12 − 1 = 0, using the sequence for 0we get {0, 1, 0, 1, 0, · · · } which is bounded.

Julia set example: f (z) = z2 − 1

Lets take a look at the sequence for the complex number i :

f (i) = i2 − 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 weget is {2, 3, 8, 63, 3968, · · · }. This is an unbounded sequence, so idoes not belong to the Julia set f (z) = z2 − 1.

Question: So what’s the point?Answer: The set of complex numbers that give boundedsequences (the Julia Set) is a remarkable geometric object.

The Julia set of f (z) = z2 − 1

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Every point in the black part of the image above is in the Julia setand the coloured points are not!

The Julia set of f (z) = z2

Exercise for problem session: Can you work out (guess) the Juliaset of f (z) = z2?

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) = z2 + i

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The Julia set of f (z) = z2 + 8i/9

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The Julia set of f (z) = z2 − 1.3

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The Julia set of f (z) = z2 − 1.7549

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The Julia set of f (z) = z2 − 0.12256− 0.74486i

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The Julia set of f (z) = z2 − 1.037 + 0.17i

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The Julia set of f (z) = z2 − 0.52 + 0.57i

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The Julia set of f (z) = z2 + 0.295 + 0.55i

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The Julia set of f (z) = z2 − 0.624 + 0.435i

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The Julia set of f (z) = z2 − 0.8− 0.175i

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The Julia set of f (z) = z2 − 0.8− 0.15i

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Part 4: The Mandelbrot set

Benoıt Mandelbrot 1924–2010

The Mandelbrot set

To study Julia sets we considered complex functions f (z) = z2 + c .

What happens if we study these functions as the value c varies?

Let fc(z) = z2 + c and look at the sequence

{fc(0), f 2c (0), f 3c (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 looklike...

The Mandelbrot Set

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The Mandelbrot set

“There is nothing more to this than a simple iterative formula. It isso simple that most children can program their home computers to

produce the Mandelbrot set... Its astounding complication wascompletely out of proportion with what I was expecting.”

- Benoıt 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 thatget larger than 2 after 300 iterations and on the right after 800iterations.

Zooming in on the Mandelbrot set

https://www.youtube.com/watch?v=PD2XgQOyCCk

This video uses 10227 iterations to produce this high resolutionversion of the Mandelbrot set. To give you a sense of the scale ofthis 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 ofthe points c such that the Julia set of fc(z) = z2 + c is connected.

Connectedness is a topological property. So the Mandelbrot setcan be said to be a single object encoding interesting features ofthe 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 onthe 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 onthe 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.