6 Fractal Geometry

6.1 ‘Natural’ geometry

Classical geometry typically considers continuous, mostly differentiable objects such as lines, curves,spheres, planes, cones, etc. These objects have the property of seeming flatter and less interestingas one zooms in: a differentiable curve at small scales looks indistinguishable from a line segment.The real world is somewhat different. If we look closer at a real object we tend to see more detail.A seemingly spherical orange is dimpled on close inspection. Should we say that its surface area isthat of a sphere, or greater due to the dimples? What if we zoom in further? Under a microscope, thedimples are seen to have minute cracks and fissures. With modern technology, we can ‘see’ almost tothe molecular level: what does surface area even mean at such a scale?

The Length of a Coastline In 1967 Benoit Mandelbrot asked a related question in a now-famouspaper, How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension. He arguedthat the question had no good answer.1 Should one measure by walking along the mean high tideline? But where is this? Do we ‘walk’ round every pebble? Around every grain of sand? Everymolecule? As one shrinks the scale, the measured ‘length’ becomes absurdly large. You can readmore regarding Mandelbrot’s approach here, though the process is essentially this:

• Choose a ruler of length R and measure how many N of these rulers placed end-to-end arerequired to trace round the coastline.

• Repeat with rulers of other sizes.

• Plotting log N against log(1/R) seems to give a straightline!

log N ≈ log k + D log(1/R) = log(kR−D) =⇒ N ≈ kR−D

• The number D is the fractal dimension of the coastline.

Mandelbrot’s fractal dimension2 is purely empirical and only makes sense for measurements at rea-sonable scales. Still, the fractal dimension does seem to capture something about the shape of thecoastline., Namely, the ‘bumpier’ a coastline, the greater its fractal dimension. For Britain the dimen-sion is approximately 1.25. The dimension is higher for rougher coastlines such as Norway’s.As a sanity check, consider a perfectly circular ‘coastline.’ Approxi-mate the circumference using N rulers of length R: clearly

R = 2 sinπ

N

As N → ∞, the small angle approximation for sine applies:

R ≈ 2π

N=⇒ N ≈ 2πR−1

1 R2πN

where the approximation improves as N → ∞. The fractal dimension of a circle is therefore 1.

1The official answer from the Ordnance Survey (the UK government’s mapping department) is, ‘It depends.’ The all-knowing CIA states 7723 miles, though no evidence is given as to why.

2Mandelbrot didn’t invent the concept, rather he coined the word fractal, applied earlier developments of Hausdorff,Minkowski and others, and observed how the natural world contains many examples of fractal structures.

1

We can generalize the above. The arc-length of a regular curve parametrized by r(t), t ∈ [0, 1], is

L =∫ 1

0

∣∣r′(t)∣∣ dt ≈N−1

∑k=0

∣∣∣∣r′ ( kN

)∣∣∣∣ 1N≈

N−1

∑k=0

∣∣∣∣r( k + 1N

)− r

(kN

)∣∣∣∣where we used a Riemann sum with left-endpoints and N sub-intervals, and the linear approxima-tion r(t + ε) ≈ r(t) + εr′(t) with ε = 1

N . Now suppose that the curve is parametrized such that thesegments on the right hand side are all rulers with the same length R. We conclude that

L ≈ NR =⇒ N ≈ LR−1

whence any such curve has fractal dimension 1.

6.2 Self-similarity and Fractal Dimension

The word ‘fractal’ is hard to define precisely. It is more of a catch-all term for a type of object whichexhibits one or more of several properties. Here are two important ones:

1. Greater structure becomes apparent at increasing magnification: as we’ve observed, this is incontrast to the types of objects studied in calculus and traditional geometry.

2. Self-similarity. A fractal often looks similar to a proper subset of itself. We’ll define this moreproperly later. For the present, consider the pictures. These ‘botanical’ fractals provide yetmore evidence that fractal geometry is somewhat more natural than smooth geometry. The‘tree’ seems to be similar to each of its three branches, while the ‘fern’ is similar to each of itsseven fronds.

It is somewhat trickier to explicitly describe an object with non-integer fractal dimension rather thanmerely observing that certain natural objects provide candidates. To do this, we first consider afamous fractal, indeed one of the first described.

2

Cantor’s Middle-third Set

In the late 1800’s, the following construction was proposed. Define a sequence of sets C0, C1, C2, C3, . . .

• C0 = [0, 1] is the unit interval.

• Construct Cn+1 by deleting the open‘middle-third’ of every interval in Cn.

Thus C1 = [0, 13 ] ∪ [ 2

3 , 1], etc. Cantor’s set C is es-sentially the limit of this sequence of sets:

C :=∞⋂

n=0

Cn

0 13

23 1

C0C1C2C3...

It has several strange properties, some of which you should have seen in a previous class:

• If the length of a set is the sum of the lengths of any disjoint sub-intervals of a set, then

length(Cn) =

(23

)n

since we always delete 13 of every interval each time we create a new set in the sequence. How-

ever C is a subset of every Cn, whence

length(C) ≤(

23

)n

for all n ∈N0

It follows that C has zero length. Otherwise said, the Cantor set contains no subintervals.

• The Cantor set is uncountable: there exists a bijection between C and the original interval [0, 1]!

• The Cantor set is self-similar. Abusing notation somewhat, we can write

Cn+1 =13

Cn ∪(

13

Cn +23

)where we mean that Cn+1 consists of two copies of Cn, each shrunk by a factor of 1

3 , one of thembeing shifted 2

3 to the right. The upshot is that the Cantor set itself satisfies

C = 13C ∪

(13C + 2

3

)It is similar to two disjoint subsets of itself! The animation below shows how the full set maybe doubled to produce itself.

One of Cantor’s purposes was to show that a set which appears very small (no intervals) can still beenormous (uncountable). The self-similarity of the set was of less concern at the time.

3

Fractal Dimension

Our goal is to measure self-similar objects and, in so-doing, create a new notion of dimension relatedto Mandelbrot’s. First we think about some of the standard objects of pre-fractal geometry.

Line A line segment can be viewed as N copies of itself scaled by a factorof r = 1

N .

Square A square comprises N copies of itself scaled by a factor r = 1√N

.

Cube A cube comprises N copies of itself scaled by a factor r = 13√N

.

In each case observe that N =( 1

r

)Dwhere D is the dimension of the object.

We now extend this observation with a loose definition.

Definition 6.1. A geometric figure is self-similar if it may be sub-divided into N ∈ N similar copiesof itself, each scaled by a magnification factor r < 1.

The fractal dimension of such a figure is D =log N

log(1/r)= − log N

log r

Examples

1. The Cantor set consists of N = 2 copies of itself, each shrunk by a factor of r = 13 . Its fractal

dimension is therefore D = log 2log 3 ≈ 0.6309. Observe that this is between 0 and 1, encapsulating

the notion that the Cantor set is simultaneously:

• More than a countable collection of points of dimension 0, and,

• Less than a line (interval) of dimension 1.

2. The tree fractal on page 2 contains3 N = 3 copies of itself, each scaled by a factor of r = 25 . Its

fractal dimension is D = log 3log(5/2) ≈ 1.199.

Similarly, the fern comprises seven copies of itself shrunk by a factor of r = 310 for a fractal

dimension D = log 7log(10/3) ≈ 1.616.

Both dimensions are between 1 and 2 reflecting the idea that both objects are somehow greaterthan mere lines, but do not have area. The larger fractal dimension of the fern suggests that itseems to occupy more space than the tree.

3. Look up pictures of the Sierpinski triangle (D = log 3log 2 ≈ 1.58496) and carpet (D = log 8

log 3 ≈ 1.8928)and consider how they occupy some middle ground between lines and areas.

4. The Menger Sponge (D = log 20log 3 ≈ 2.7268) is an extension of the Sierpinski carpet to R3 and and

example of a fractal which doesn’t quite manage to have volume.3Because of the construction, including translating the picture, the whole contains a little more than three copies of

itself. The bottom part of the stem and the largest two green branches are ‘extra.’ This is of no import.

4

The Koch Curve and Snowflake

The Koch curve is a generalization of the Cantor set. It is produced as the limit of a sequence ofcurves.

• Let K0 be a line segment of length 1.

• Delete the middle-third of K0 and replace it with the other two sides of an equilateral triangleto create K1.

• Delete the middle-third of each of the four line segments in K1 and replace each with the othertwo sides of an equilateral triangle to create K2.

• Repeat this process ad infinitum.

The curve is shown below along with an animation showing how it is built.

The curve can easily be analysed. Let sn be the number of line segments making up the curve Kn,and tn be the length of each. Also let `n = tnsn be the length of the curve Kn. Then:

s0 = 1, s1 = 4, s2 = 16, . . . sn+1 = 4sn =⇒ sn = 4n

t0 = 1, t1 =13

, t2 =132 , . . . tn+1 =

13

tn =⇒ tn =13n

`n =

(43

)n

In particular, the length of the Koch curve must be greater than that of every Kn, whence it onlymakes sense to say that it is infinitely long. We therefore have the paradoxical situation of an infinitelylong curve which has no self-intersections existing in a finite amount of space.

Self-similarity The Koch curve is self-similar in that it comprises N = 4 copies of itself shrunk bya factor of r = 1

3 . Its fractal dimension is therefore log 4log 3 ≈ 1.2619.

5

The Koch Snowflake This is formed by arranging three copies of the curve around the edges of atriangle. It is an amusing exercise to compute the area contained inside. Let A0 =

√3

4 be the area ofthe triangle at stage 0 (three copies of the straight line K0).

• To create A1, we add on 3s0 = 3 extra triangles each having area 19 A0: each extra triangle is

similar to the original but scaled by a length-factor of 13 for an area scaling of 1

9 .

• To create A2, we add on 3s1 = 12 extra triangles each having area 192 A0.

• Generally, An+1 = An +3sn

9n+1 A0.

• A little geometric series shows that

An = A0

(1 +

n

∑k=1

3sk−1

9k

)= A0

(1 +

13

n

∑k=1

(49

)k−1)

= A0

(1 +

35

[1−

(49

)n])

• The limit as n→ ∞ is 85 A0 so that the snowflake contains 8

5 the area of the original triangle. Wetherefore have a finite area inside an infinitely long boundary curve!

6

6.3 Contraction Mappings

Thusfar we have only dealt with fractals where the whole consists of N pieces each scaled by thesame factor. In general we can mix up scaling factors. To do this we need some of the language ofcontraction mappings from topology.

Example Consider the Cantor set and the functions S1, S2 : R→ R defined by

S1(x) =x3

S2(x) =x3+

23

S1, S2 are continuous and C = S1(C) ∪ S2(C). These functions encapsulate the self-similarity of theCantor set. The functions moreover satisfy an important property:

∀x, y ∈ R, |S1(x)− S1(y)| = |S2(x)− S2(y)| =13|x− y|

This makes the functions examples of contraction mappings.

Definition 6.2. A contraction mapping is a function S on a subset of Rn such that ∃c ∈ [0, 1) with

|S(x)− S(y)| ≤ c |x− y|

In essence, a contraction mapping moves points closer together. It should be clear that every con-traction mapping is continuous. The main idea of this section is that fractals may be generated byrepeatedly applying contraction mappings applied to an initial shape.

Returning to the Cantor set, at each stage of the construction, we have

Cn+1 := S1(Cn) ∪ S2(Cn)

The limit of this process is the Cantor set itself. Surprisingly, it barely matters what set we choose asour initial input C0. For example, we could start with C0 = {0}, from which

C1 = {0, 23}, C2 = {0, 1

9 , 23 , 7

9}, C3 = {0, 127 , 2

9 , 727 , 2

3 , 1927 , 8

9 , 2527}, . . .

We draw the first few iterations below. In the second picture, we start with a very different initial setC0 = [0.2, 0.5] ∪ [0.6, 0.7]. Iterating this also appears to produce the Cantor set!

0 13

23 1

C0C1C2C3C4C5C6

...

0 13

23 1 0 1

323 1

C0C1C2C3C4C5C6

...

0 13

23 1

7

It seems like the Cantor set might be independent of the initial data C0. Our major result shows inwhat sense this is the case. It relies on some heavy lifting from topology: if you’ve at least done someanalysis, then several of the concepts will be familiar, though a lot of work is needed if one wants tomake this watertight. We summarize the discussion without proof.

• A compact subset of Rm is one which is closed (contains its boundary points) and bounded (allpoints lie within some finite distance of the origin).

• The set of all compact subsets of Rn is a metric space H. This means that there is a sensibledefinition for the distance between two compact subsets of Rn: constructing this is a little tricky,look up the Hausdorff metric if you are interested.4

• Since H is a metric space, it makes sense to speak of a convergent sequence of compact sets:limn→∞ Kn → K if d(Kn, K) → 0 where d is the Hausdorff metric. It also makes sense to speakof Cauchy sequences in H. Moreover, H is complete in that every Cauchy sequence (Kn) ⊆ Hconverges to some K ∈ H.

• The Banach Fixed Point Theorem is what drives the main result.

If S : H → H is a contraction mapping on a complete metric space H, then S has a uniquefixed point (some F ∈ H such that S(F) = F). Moreover, if F0 ∈ H is any initial value, thenthe sequence defined iteratively by Fk+1 = S(Fk) converges to F.

This powerful result has applications throughout mathematics.

Theorem 6.3 (Iterated Function System). Let S1, . . . , Sn be contraction mappings on Rm with contractionratios c1, . . . , cn. Define a transformation S onH by

S(D) =n⋃

i=1

Si(D)

1. S is a contraction mapping onH, with contraction ratio c = max{ci}.

2. S has a unique fixed set F ∈ H given by F = limk→∞

Sk(E) for any non-empty E ∈ H.

Part 1 is not so difficult to prove if you’re willing to work with the definition of the Hausdorff metric(try it if you’re comfortable with analysis!). Part 2 follows immediately from Banach’s theorem.

The upshot is this: if we take any initial non-empty compact set E as our data and repeatedly apply acombination of contraction mappings, then the resulting limit is independent of E! We call the limit setF, for fractal. Such fractals are often referred to as attractors: as limit-sets, they ‘attract’ data towardsthemselves.

4Given Y ∈ H, and x ∈ Rn, define dY(x) = infy∈Y ||x− y|| to be the distance from x to the ‘nearest’ point of Y. One candefine dX(y) similarly. The Hausdorff metric is then

d(X, Y) = max

{supx∈X

dY(x), supy∈Y

dX(y)

}

Roughly speaking, find points x ∈ X and y ∈ Y such that x is a far as possible from Y and vice versa: d(X, Y) is the largerof these distances.

8

The Cantor Set (again)

Theorem 6.3 shows that we may let C0 be any closed bounded subset of R. Repeatedly applying thecontraction mappings S1 and S2 will always result in the same set C.

A nice application of this approach is that one can easily find all sorts of interesting points in theCantor set. For example, suppose that x, y ∈ R are a pair such that y = S1(x) and x = S2(y). That is

y =13

x and x =13(y + 2)

Since E = {x, y} is a compact subset of R which satisfies E ⊆ S(E), it follows that E ⊆ limk→∞

Sk(E) = C.

Thus x, y both lie in the Cantor set. However we can easily solve to see that (x, y) = ( 34 , 1

4 ). This isalso paradoxical: 1

4 does not lie at the end of any deleted interval (denominators of the form 3n) butyet the Cantor set contains no intervals. How does 1

4 end up in there?!

The Koch curve

We can also see the contraction mapping theorem at work with the Koch curve. Define four contrac-tion mappings Si : R2 → R2, each with factor c = 1/3.

Mapping Effect

S1(x, y) =( x

3,

y3

)Scale by 1

3

S2(x, y) =

(16

x−√

36

y +13

,

√3

6x +

16

y

)Scale by 1

3 , rotate by 60° and translate

S3(x, y) =

(16

x +

√3

6y +

12

,

√3

6x− 1

6y +

√3

6

)Scale by 1

3 , rotate by -60°and translate

S4(x, y) =(

x3+

23

,y3

)Scale by 1

3 and translate

In the animations below, we start with two different initial compact sets and apply the contractionmappings. Compare with the original construction of the curve.

9

Sierpinski carpet

The Sierpinski carpet (upper left) is produced by eight contraction mappings, each reducing thewhole by a factor of 1

3 . We start with three different initial compact sets in R2: as per the Theorem, alllead to the same fractal.

10

A Fractal Fern

We construct a fractal fern. This is built from the following list of contraction mappings:

S1: Scale by 34 , rotate 5° clockwise and translate by (0, 1

4 )

S2: Scale by 14 , rotate 60° counter-clockwise and translate by (0, 1

4 )

S3: Scale by 14 , rotate 60° clockwise and translate by (0, 1

4 )

11

6.4 Fractal Dimension Revisited

Recall that if a fractal is composed of N copies of itself, eachshrunk by a factor of r, then its fractal dimension is D = log N

log(1/r) .However, in the previous section we created fractals using severalcontraction maps with different scales. How can we make senseof the fractal dimension in this situation?The trick is to consider how difficult it is to cover over a set usingsmall disks of a given radius ε > 0. In the picture we are try-ing to cover the blue unit square with disks of radius ε = 0.4. Itisn’t hard to believe that we need at least four disks. If we makeε smaller, we will need more disks. It is the relationship betweenε and the number N of required disks that will allow us to definedimension in a new way.Definition 6.4. Let A be compact and ε > 0 be given. Define the closed ε-ball centered at x ∈ A by

B(x, ε) = {y ∈ A : d(x, y) ≤ ε}This is simply all the points in A which are at most a distance ε from x.

We say thatM⋃

n=1B(xn, ε) is an ε-covering of A if A ⊆

M⋃n=1

B(xn, ε): that is,

∀x ∈ A, ∃n : x ∈ B(xn, ε)

The next result follows from the definition of compactness (this needs a little topology).

Lemma 6.5. Given A compact and ε > 0, there exists a minimum number N (A, ε) of ε-balls required tocover A.

Definition 6.6. Let A be compact. The minimal ε-covering number for A is

N (A, ε) = min

{M : A ⊆

M⋃n=1

B(xn, ε)

}For example, if A is a square with side length 1 and ε = 0.4, then N (A, 0.4) = 4.

As ε decreases, we expect N (A, ε) to increase. We should have the same type of relationship as fordimension.

Example Consider an interval [0, 1] of length 1. It is easy to convince yourself that ε andN([0, 1], ε

)are related as follows:

range of ε N ([0, 1], ε)12 ≤ ε 1

14 ≤ ε < 1

2 216 ≤ ε < 1

4 318 ≤ ε < 1

6 41

10 ≤ ε < 18 5

12m ≤ ε < 1

2(m−1) m

12

Observe thatlogN

log(1/ε)−→ε→0

1 is the dimension of the line. A similar trick happens in other dimen-

sions, although it is harder to compute explicitly.

Definition 6.7. Given a compact set A ⊆ Rn, its fractal dimension is the limit

D = limε→0

logN (A, ε)

log(1/ε)

Thankfully an easier to use modification is available using boxes.

Theorem 6.8 (Box-counting). Let A be compact and cover Rm by boxes of side length 12n . Let Nn(A) be the

number of boxes intersecting A. Then

D = limn→∞

logNn(A)

log 2n

Theorem 6.9. Let {Sn}Mn=1 be an iterated function system (finite collection of contraction mappings) where

each Sn has scale factor cn ∈ [0, 1) and let F be its attractor (limiting fractal from Theorem 6.3). Supposethat at each stage of the construction, portions of the fractal generated by each contraction map meet only atboundary points. Then the fractal dimension is the unique number D such that

M

∑n=1

cDn = 1

Note that D exists and is unique:

Existence Apply the intermediate value theorem: f (D) = ∑Mn=1 cD

n is continuous on [0, ∞), f (0) =M ≥ 1 and limD→∞ f (D) = 0 (since each ck ∈ [0, 1)).

Uniqueness f is a strictly decreasing function of D since each cDk is strictly decreasing.

Examples

1. If all scale-factors cn = r are the same, then MrD = 1 =⇒ D = − log Mlog r = log M

log(1/r) recovers thesame formula as before.

2. Recall the fractal fern from page 11. It is generated from three contraction maps with scalefactors 3

4 , 14 , 1

4 . Its dimension is the solution to the equation(34

)D

+

(14

)D

+

(14

)D

= 1 =⇒ D ≈ 1.3267

3. You are very unlikely to be able to solve such an equation exactly and will instead rely onnumerical approximations. When you can, it is usually because of some trick. For instance,suppose that a fractal is created from two copies of itself scaled by 1

2 and three copies scaled by14 . Then

2(

12

)D

+ 3(

14

)D

= 1

Writing α =( 1

2

)Dwe have the quadratic equation

2α + 3α2 = 1 =⇒ α =13

=⇒ D = log2 3 ≈ 1.5849

13

Other methods of creating fractals

The contraction mapping approach is not the only way to create fractals. Fractal structures appearnaturally all over mathematics. Two famous examples are the logistic map (related to numerical ap-proximations to non-linear differential equations) and the Mandelbrot set. If you’re interested, readabout them! The Mandelbrot set is particularly famous. It arises from a construction in the complexplane. Consider the function

f (z) = z2 + c

where c is a constant complex number. Apply this function to c repeatedly. If the result remainsbounded forever, we place the point c in the Mandelbrot set. Below is a very low-res version of thefamous picture: much better versions can be found online.

−1 1

i

−i

14