fractals - lecture 12 - wordpress.com€¦ · 12/01/2018 · outline 1 fractals kochsnowﬂake...

FractalsLecture 12

Justin Stevens

Justin Stevens Fractals (Lecture 12) 1 / 25

Outline

1 FractalsKoch SnowflakeHausdorff DimensionSierpinski TriangleMandelbrot Set

2 Constructions

3 Parting Shots


Recursive Koch Snowflake

Definition. The Koch snowflake can be constructed by starting with anequilateral triangle, then recursively altering each line segment as follows:

Divide the line segment into three segments of equal length.Draw an equilateral triangle that has the middle segment from theprevious step as its base and points outward.Remove the line segment that is the base of the new triangle.




Divide the line segment into three segments of equal length.

Draw an equilateral triangle that has the middle segment from theprevious step as its base and points outward.Remove the line segment that is the base of the new triangle.




Divide the line segment into three segments of equal length.Draw an equilateral triangle that has the middle segment from theprevious step as its base and points outward.

Remove the line segment that is the base of the new triangle.




Divide the line segment into three segments of equal length.Draw an equilateral triangle that has the middle segment from theprevious step as its base and points outward.Remove the line segment that is the base of the new triangle.


Koch Snowflake

Figure 1: The Koch snowflake is one of the earliest fractal curves. It is based on a1904 paper titled “On a continuous curve without tangets, constructible fromelementary geometry" by the Swedish mathematician Helge von Koch.


Perimeter of the Koch Snowflake

The number of sides of the Koch snowflakes increases by a factor of 4 eachtime, hence after j iterations, the number of sides is Nj = 4 · Nj−1 = 3 · 4j .

If the original equilateral triangle has side length S0 = s, after j iterations,Sj = Sj−1/3 = s/3j . Thus the perimeter of the snowflake is given by

Pj = Nj · Sj = 3s ·(43

)j.

As the number of iterations tends to infinity, the limit of the perimeter is:

limn→∞

Pn = limn→∞

3s ·(43

)n=∞.

Hence the Koch curve has an infinite perimeter.




If the original equilateral triangle has side length S0 = s, after j iterations,Sj = Sj−1/3 = s/3j .

Thus the perimeter of the snowflake is given by

Pj = Nj · Sj = 3s ·(43

)j.


limn→∞

Pn = limn→∞

3s ·(43

)n=∞.





If the original equilateral triangle has side length S0 = s, after j iterations,Sj = Sj−1/3 = s/3j . Thus the perimeter of the snowflake is given by

Pj = Nj · Sj = 3s ·(43

)j.


limn→∞

Pn = limn→∞

3s ·(43

)n=∞.



Area of the Koch Snowflake

In each iteration a new triangle is added on each side of the previousiteration, so the number of new triangles added in iteration j is

Tj = Nj−1 = 3 · 4j−1 = 3/4 · 4j .

If a0 is the area of the original triangle, then the area of each triangle is

aj = aj−19 = a0

9n .

Therefore, the total area of the snowflake after n iterations is

An = a0

1 + 34

n∑j=1

(49

)j

As n→∞, this is an infinite geometric series with first term 1/3:

= a0

(1 + 1/3

1− 4/9

)= 8/5 · a0.


Hausdorff Dimension of Koch Snowflake

Figure 2: 3Blue1Brown


Hausdorff Dimension

Definition. The Hausdorff dimension measures the roughness of a metricspace. If S is the scaling factor and N is the mass-scaling factor, then

N = SD ⇐⇒ D = logS(N) = logNlog S .


Recursive Sierpinski Triangle

Definition. The Sierpinski triangle starts with an equilateral triangle andrecursively divides each triangle into four smaller congruent equilateraltriangles and removes the central one.

The first five iterations are:

The Hausdorff Dimensino is given by 2D = 3 ⇐⇒ D = log2 3 ≈ 1.585.


Recursive Sierpinski Triangle

Definition. The Sierpinski triangle starts with an equilateral triangle andrecursively divides each triangle into four smaller congruent equilateraltriangles and removes the central one. The first five iterations are:

The Hausdorff Dimensino is given by 2D = 3 ⇐⇒ D = log2 3 ≈ 1.585.


Pascal’s Triangle Mod 2

Figure 3: Take Pascal’s triangle with 2n rows and color the even numbers white andthe odd numbers black


How Long is the Coast of Britain?

“How Long Is the Coast of Britain? Statistical Self-Similarity andFractional Dimension" is a 1967 paper by Benoit Mandelbrot.

Figure 4: If the coastline of Great Britain is measured using units 100 km long,then the length of the coastline is approximately 2, 800 km. With 50 km units, thetotal length is approximately 3, 400 km, approximately 600 km longer.


Mandelbrot Definition

The Mandelbrot set is a family of complex quadratic polynomials given by

Pc : z 7→ z2 + c,

where c is a complex number.

For each c, one considers the behavior of

(0,Pc(0),Pc(Pc(0)),Pc(Pc(Pc(0))), · · · )

obtained by iterating Pc(z) starting at critical point z = 0. This eitherescapes to infinity or stays within a disk of some finite radius. TheMandelbrot set is defined as the set of all points c such that the abovesequence does not escape to infinity. Specifically, |Pn

c (0)| ≤ 2 for all n ≥ 0.




Pc : z 7→ z2 + c,

where c is a complex number. For each c, one considers the behavior of

(0,Pc(0),Pc(Pc(0)),Pc(Pc(Pc(0))), · · · )

obtained by iterating Pc(z) starting at critical point z = 0.

This eitherescapes to infinity or stays within a disk of some finite radius. TheMandelbrot set is defined as the set of all points c such that the abovesequence does not escape to infinity. Specifically, |Pn

c (0)| ≤ 2 for all n ≥ 0.




Pc : z 7→ z2 + c,


(0,Pc(0),Pc(Pc(0)),Pc(Pc(Pc(0))), · · · )

obtained by iterating Pc(z) starting at critical point z = 0. This eitherescapes to infinity or stays within a disk of some finite radius.

TheMandelbrot set is defined as the set of all points c such that the abovesequence does not escape to infinity. Specifically, |Pn

c (0)| ≤ 2 for all n ≥ 0.




Pc : z 7→ z2 + c,


(0,Pc(0),Pc(Pc(0)),Pc(Pc(Pc(0))), · · · )

obtained by iterating Pc(z) starting at critical point z = 0. This eitherescapes to infinity or stays within a disk of some finite radius. TheMandelbrot set is defined as the set of all points c such that the abovesequence does not escape to infinity. Specifically, |Pn

c (0)| ≤ 2 for all n ≥ 0.


First Mandelbrot Set

Figure 5: The first published picture of the Mandelbrot set, by Robert W. Brooksand Peter Matelski in 1978.


Modern Mandelbrot Set


Outline

1 Fractals

2 ConstructionsConstructible PolygonsFermat Numbers

3 Parting Shots


Dividing Circular Area into Equal Parts

Take curves on k and n − k parts and on m and n −m parts when m > k.If each part is the diameter of a small semicircle, the enclosed area is

π

2

(m2

2− k

2

2)+ π

2

((n − k)

2

2− (m − k)

2

2)= π

4 · n(m − k).

For m = k + 1, this is π/4 · n. The area of the circle is π(n/2)2 = π/4 · n2,which is n times the area between two successive curves as desired.




π

2

(m2

2− k

2

2)+ π

2

((n − k)

2

2− (m − k)

2

2)= π

4 · n(m − k).

For m = k + 1, this is π/4 · n.

The area of the circle is π(n/2)2 = π/4 · n2,which is n times the area between two successive curves as desired.




π

2

(m2

2− k

2

2)+ π

2

((n − k)

2

2− (m − k)

2

2)= π

4 · n(m − k).

For m = k + 1, this is π/4 · n. The area of the circle is π(n/2)2 = π/4 · n2,which is n times the area between two successive curves as desired.


x 2 Construction by Compass and Straightedge

Example. Given the unit length and a segment of length x construct x2.

1

x

P AC O

B

From the right similar triangles, we see 4OPB ∼ 4OBA, therefore

OPOB = OB

OA =⇒ OP = x2.


Heptadecagon

In 1796, Carl Friedrich Gauss proved that a heptadecagon is constructibleusing a compass and unmarked straightedge using the trigonometric identity

16 cos 2π17 =− 1 +√17 +

√34− 2

√17+

2√17 + 3

√17−

√34− 2

√17− 2

√34 + 2

√17

Gauss’ proof relies on the fact that constructibility is equivalent toexpressing trigonometric functions of common angles in terms of arithmeticoperations and square root extractions.


Heptadecagon Construction

Figure 6: Construction According to Herbert Richmond in 1893.


Gauss-Wantzel Theorem

Theorem. A regular n-gon is constructible with a straightedge andcompass if and only if n = 2kp1p2 · · · pt where k and t are non-negativeintegers, and the pi ’s (when t > 0) are distinct Fermat primes:

n = 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80,85, 96, 102, 120, 128, 136, 160, 170, 192, 204, 240, 255, 256, 257, 272, 320, 340, 384, 408, 480, 510, 512, 514, 544, 640, 680, 768, 771, 816, 960, 1020, 1024,1028, 1088, 1280, 1285, 1360, 1536, 1542, 1632, 1920, 2040, 2048, · · ·

The rationale is because φ(n) is a power of 2 for precisely these values.


Fermat Numbers

Example. Fermat numbers are of the form fn = 22n + 1 for natural n.1 f0 = 3, f1 = 5, f2 = 17, f3 = 257, and f4 = 65537 are primes. Prove

232 + 1 ≡ 0 (mod 641).2 Prove that if 2n + 1 is prime for positive n, then it is a Fermat number.

Proof.1 Observe that 216 = 65536 ≡ 154 (mod 641).

Therefore,

232 +1 = 655362 +1 ≡ 1542 +1 = 23717 = 641 ·37 ≡ 0 (mod 641).

2 Suppose n = 2r · s for some odd integer s > 1. Then,

2n + 1 =(22r)s

+ 1 =(22r + 1

) (22r ·(s−1) − · · ·+ 22r ·2 − 22r + 1

),

so 2n + 1 is composite. Thus n cannot have an odd factor and n = 2r .


Fermat Numbers

Example. Fermat numbers are of the form fn = 22n + 1 for natural n.1 f0 = 3, f1 = 5, f2 = 17, f3 = 257, and f4 = 65537 are primes. Prove

232 + 1 ≡ 0 (mod 641).2 Prove that if 2n + 1 is prime for positive n, then it is a Fermat number.

Proof.1 Observe that 216 = 65536 ≡ 154 (mod 641). Therefore,

232 +1 = 655362 +1 ≡ 1542 +1 = 23717 = 641 ·37 ≡ 0 (mod 641).

2 Suppose n = 2r · s for some odd integer s > 1. Then,

2n + 1 =(22r)s

+ 1 =(22r + 1

) (22r ·(s−1) − · · ·+ 22r ·2 − 22r + 1

),

so 2n + 1 is composite. Thus n cannot have an odd factor and n = 2r .


Fermat Number Identity

Example. Prove that for all n ≥ 1, fn − 2 = fn−1fn−2 · · · f1f0.

Proof by Induction.When n = 1, f1 − 2 = 5− 2 = 3 = f0. Assume the identity holds for n = k:

fk − 2 = fk−1fk−2 · · · f1f0. (Hypothesis)

Using difference of squares on the next Fermat number,

fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .




Proof by Induction.When n = 1, f1 − 2 = 5− 2 = 3 = f0.

Assume the identity holds for n = k:



fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .







fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .





fk − 2 = fk−1fk−2 · · · f1f0.

(Hypothesis)


fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .







fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .







fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)

= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .







fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)

= fk (fk−1fk−2 · · · f1f0) .







fk+1 − 2 = 22k+1 + 1− 2 = 22k+1 − 1

=(22k + 1

) (22k − 1

)= fk (fk − 2)= fk (fk−1fk−2 · · · f1f0) .


GCD of Fermat Numbers

Example. Prove that distinct Fermat numbers are relatively prime.

Proof.Let the numbers be fi and fj with i > j . From the identity, fi ≡ 2 (mod fj).By the Euclidean Algorithm and parity, gcd(fi , fj) = gcd(fj , 2) = 1.

Example. Prove that there are infinite primes using Fermat numbers.

Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor.

Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true.

Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1.

Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk .

If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.

Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1.

Since there areinfinitely many Fermat numbers, there are also infinitely many primes.






Proof by Strong Induction.Let P(n) be the proposition every fn contributes a new prime factor. Clearly,P(0) is true. Assume P(n) is true for every n ≤ k − 1. Let q be a primefactor of fk . If q | fi for some 0 ≤ i ≤ k − 1, this contradicts gcd(fi , fk) = 1.Thus q is different from the prime factors of f0, f1, · · · , fk−1. Since there areinfinitely many Fermat numbers, there are also infinitely many primes.


Outline

1 Fractals

2 Constructions

3 Parting Shots


Relevant Links

HMC: Python Turtles!HMC: Creating the Mandelbrot Set3Blue1Brown: Fractals are typically not self-similarWikipedia: List of fractals by Hausdorff DimensionBenoit Mandelbrot at TED: Fractals and the art of roughnessSquare Root Construction by Compass and StraightedgeNumberphile: Heptadecagon and Fermat Primes


https://www.cs.hmc.edu/twiki/bin/view/CS5Fall2016/Lab2Turtle

https://www.cs.hmc.edu/twiki/bin/view/CS5Fall2016/Lab8

https://www.youtube.com/watch?v=gB9n2gHsHN4

https://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension

https://www.ted.com/talks/benoit_mandelbrot_fractals_the_art_of_roughness

http://www.cs.cas.cz/portal/AlgoMath/Geometry/PlaneGeometry/GeometricConstructions/SquareSquareRootConstruction.htm

https://www.youtube.com/watch?v=oYlB5lUGlbw

fractals - lecture 12 - wordpress.com€¦ · 12/01/2018 · outline 1 fractals kochsnowﬂake...

Documents