5 Fractal Geometry
5.1 Natural Geometry, Self-similarity and Fractal Dimension
Classical geometry typically considers objects (lines, curves, spheres, etc.) which seem flatter and
less interesting as one zooms in: a differentiable curve at small scales looks like a line segment!
By contrast, real-world objects tend to exhibit greater detail at smaller scales. A seemingly spherical
orange is dimpled on closer inspection. Is its surface area that of a sphere, or is it greater due to the
dimples? What if we zoom in further? Under a microscope, the dimples are seen to have minute
cracks and fissures. With modern technology, we can see almost to the 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-famous
paper, How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension. His essential
point was that the question has no simple answer:
34
Should one measure by walking along the mean
high tide line? But where is this? Do we ‘walk’ round every pebble? Round every grain of sand?
Every molecule? As one shrinks the scale, the measured length becomes absurdly large. We sketch
Mandelbrot’s approach.
35
• Given a ruler of length R, measure how many N are required to trace round the coastline when
laid end-to-end.
• Plotting log N against log(1/R) for several sizes of ruler seems to give a straight line!
log N ≈ log k + D log(1/R) = log(kR
−D
) =⇒ N ≈ kR
−D
The number D is Mandelbrot’s fractal dimension of the coastline.
Mandelbrot’s fractal dimension is purely empirical, though it does seem to capture something about
the ‘bumpiness’ of a coastline: the bumpier, the greater its fractal dimension. For mainland Britain
with its smooth east and rugged west coasts, D ≈ 1.25. Given its many fjords, Norway has a far
rougher coastline and a higher fractal dimension D ≈ 1.52.
Example 5.1. As a sanity check, consider a smooth circular ‘coastline.’
Approximate the circumference using N rulers of length R: clearly
R = 2 sin
π
N
As N → ∞, the small angle approximation for sine applies,
R ≈
2π
N
=⇒ N ≈ 2π R
−1
where the approximation improves as N → ∞. The fractal dimension
of a circle is therefore 1.
1
R
2π
N
34
The official answer from the Ordnance Survey (the UK government mapping office) is, ‘It depends.’ The all-knowing
CIA states 7723 miles, though offers no evidence as to why.
35
For more detail see the Fractal Foundation’s website. Mandelbrot coined the word fractal, though he didn’t invent the
concept from nothing. Rather he applied earlier ideas of Hausdorff, Minkowski and others, and observed how the natural
world contains many examples of fractal structures.
80
Our goal is to describe self-similar objects and thus create a new notion of dimension related to
Mandelbrot’s. To begin consideration of self-similarity, we first consider some of the standard objects
of pre-fractal geometry.
Segment A segment can be viewed as N copies of itself scaled by a factor
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 =
1
3
√
N
.
In each case observe that N =
1
r
D
where D is the usual dimension of the
object (1, 2 or 3). Inspired by this, we make a loose definition.
Definition 5.2. A geometric figure is self-similar if it may be subdivided into N similar copies of
itself, each scaled by a magnification factor r < 1. The fractal dimension of such a figure is
D := log
1/r
N =
log N
log( 1/r)
= −
log N
log r
Example 5.3. The botanical pictures below offer some evidence for non-integer fractal dimension
and that self-similarity is a natural phenomenon. The ‘tree’ comprises N = 3 copies of itself, each
scaled by a factor of r = 0.4. Its fractal dimension is D = −
log 3
log 0.4
≈ 1.199.
The fern has N = 7 and r = 0.3 for a fractal dimension D = −
log 7
log 0.3
≈ 1.616.
Tree fractal D ≈ 1.199 Fern fractal D ≈ 1.616
The pictures illustrate the interpretation of fractal dimension. Both objects seem to occupy more
space than mere lines, but neither has positive area. Moreover, the fern seems to occupy more space
than the tree. The ‘trunk’ and ‘branches’ in the first picture aren’t part of the fractal, and are drawn
only to give the picture a skeleton.
81
Example 5.4 (Cantor’s Middle-third Set). This famous example dates from the late 1800s.
36
Starting with the unit interval C
0
= [0, 1], define a se-
quence of sets (C
n
) where C
n+1
is obtained by deleting the
open ‘middle-third’ of each interval in C
n
; for instance
C
1
=
0,
1
3
∪
2
3
, 1
Cantor’s set is essentially the limit of this sequence:
C :=
∞
\
n=0
C
n
0
1
3
2
3
1
C
0
C
1
C
2
C
3
.
.
.
Cantor’s set has several strange properties. . .
Zero length If the length of a set is the sum of the lengths of its disjoint sub-intervals, then
length(C
n
) =
2
3
n
since we delete
1
3
of the remaining set at each step. It follows that
∀n ∈ N
0
, length(C) ≤
2
3
n
=⇒ length(C) = 0
Otherwise said, C contains no subintervals.
Uncountable There exists a bijection between C and the original interval [0, 1]!
Self-similarity Abusing notation somewhat,
C
n+1
=
1
3
C
n
∪
1
3
C
n
+
2
3
where we mean that C
n+1
consists of two copies of C
n
, each shrunk by a factor of
1
3
and one
shifted
2
3
to the right. The upshot is that the Cantor set itself satisfies
C =
1
3
C ∪
1
3
C +
2
3
Being similar to two disjoint subsets of itself, its fractal dimension is D =
log 2
log 3
≈ 0.631. The
above image links to an animation showing how the full set may be doubled to produce itself.
The Cantor set has many generalizations. Look up the Sierpi
´
nski triangle (D =
log 3
log 2
≈ 1.585) and
carpet (Examples 5.8, D =
log 8
log 3
≈ 1.893), and the Menger sponge (D =
log 20
log 3
≈ 2.727).
36
Henry Smith discovered this set in 1874 while investigating integrability, in which context the ‘length’ of a set was later
formalized using measure theory. Cantor’s description in 1883 was more focused on topological properties. Self-similarity
was less of a concern at the time.
82
Example 5.5 (The Koch Curve and Snowflake). The Koch curve is another generalization of the
Cantor set, produced as the limit of a sequence of curves.
• Let K
0
be a segment of length 1.
• Replace the middle third of K
0
with the other two sides of
an equilateral triangle to create K
1
.
• Replacing the middle third of each segment in K
1
as before
to create K
2
.
• Repeat this process ad infinitum.
The curve is drawn, along with the Koch snowflake obtained by
arranging three copies around an equilateral triangle.
The relation to the Cantor set should be obvious in the construc-
tion. Indeed if K
0
= [0, 1], then the intersection of this with the
Koch curve is the Cantor set!
The Koch curve is self-similar in that it comprises N = 4 copies
of itself shrunk by a factor of r =
1
3
. Its fractal dimension is
therefore
log 4
log 3
≈ 1.2619.
We may also consider the curve’s length. Let s
n
be the number
of segments in K
n
, each having length t
n
. Also let ℓ
n
= t
n
s
n
be
the length of the curve K
n
. We easily see that
s
n
= 4
n
, t
n
=
1
3
n
=⇒ ℓ
n
=
4
3
n
→ ∞
from which the Koch curve is infinitely long!
Koch Curve
Koch Snowflake
Self-similarity
Exercises 5.1. 1. By removing a constant middle fraction of each interval, construct a fractal analo-
gous to the Cantor set but with dimension
1
2
.
2. Prove that the area inside the n
th
iteration of the construction of the Koch snowflake is
A
n
=
1 +
3
5
1 −
4
9
n
√
3
4
The area inside the complete snowflake is therefore
8
5
that of the original triangle.
3. Suppose r(t), t ∈ [0, 1] is a regular (smooth) curve in the plane.
(a) Use the arc-length formula L =
R
1
0
|
r
′
( t)
|
dt together with Riemann sums and the linear
approximation r(t + ϵ) ≈ r(t) + ϵr
′
( t) with ϵ =
1
N
to argue that
L ≈
N−1
∑
k=0
r
k + 1
N
−r
k
N
(∗)
(b) Suppose the curve is parametrized such that each segment on the right side of (∗) has the
same length R. Prove that L ≈ NR.
Any regular curve thus has fractal dimension 1 in the sense stated by Mandelbrot (pg. 80).
83
5.2 Contraction Mappings & Iterated Function Systems
Thus far we have only dealt with fractals where the whole consists of pieces scaled by the same
factor. In general we can mix up scaling factors. To do this it is helpful to borrow some language
from topology.
Definition 5.6. A contraction mapping is a function S on a subset of R
n
such that ∃c ∈ [0, 1) with
|
S(x) − S(y)
|
≤ c
|
x −y
|
A contraction mapping therefore 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 by
repeatedly applying contraction mappings to an initial shape. We have already seen a example:
Example (5.4, mk. II). Consider the following functions S
1
, S
2
: R → R
S
1
(x) =
x
3
S
2
(x) =
x
3
+
2
3
These are certainly contraction mappings
∀x, y ∈ R,
|
S
1
(x) − S
1
( y)
|
=
|
S
2
(x) − S
2
( y)
|
=
1
3
|
x −y
|
with scale factor c =
1
3
. More importantly, these functions define the Cantor set: at each stage of its
construction, we have
C
n+1
:= S
1
(C
n
) ∪S
2
(C
n
)
As the limit of this process, the self-similarity of the Cantor set can be expressed in the same manner:
C = S
1
( C) ∪ S
2
( C).
Surprisingly, it barely seems to matter what initial set C
0
we choose. For example, we could start
with the singleton set C
0
= {0}, from which
C
1
= {0,
2
3
}, C
2
= {0,
2
9
,
2
3
,
8
9
}, C
3
= {0,
2
27
,
2
9
,
8
27
,
2
3
,
20
27
,
8
9
,
26
27
}, . . .
We draw the first few iterations below. In the second picture, we start with a very different initial set
C
0
= [0.2, 0.5] ∪[0.6, 0.7]. Iterating this also appears to produce the Cantor set!
0
1
3
2
3
1
C
0
C
1
C
2
C
3
C
4
C
5
C
6
.
.
.
0
1
3
2
3
1
0
1
3
2
3
1
C
0
C
1
C
2
C
3
C
4
C
5
C
6
.
.
.
0
1
3
2
3
1
84
Iterated Function Systems
It certainly seems as if the Cantor set might be generated by the contraction maps S
1
, S
2
indepen-
dently of the initial data C
0
. The following result shows in what sense this is the case, though it relies
on some heavy lifting from topology. If you’ve done some analysis, then several of the concepts will
be familiar. We summarize the discussion without proof.
• A subset of R
m
is compact if it is closed (contains its boundary points) and bounded (all points lie
within some ball centered at the origin).
• The set of all compact subsets of R
m
is a metric space H. This means that the distance d(X, Y)
between two compact sets X, Y ∈ H may sensibly be defined, though it is a little tricky. . .
37
• Since H is a metric space, we can discuss convergent sequences (K
n
) of compact sets
lim
n→∞
K
n
= K ⇐⇒ lim
n→∞
d(K
n
, K) = 0
It also makes sense to speak of Cauchy sequences in H. Moreover, H is complete in that every
Cauchy sequence (K
n
) ⊆ H converges to some K ∈ H.
• The Banach Fixed Point Theorem now applies.
If S : H → H is a contraction mapping on a complete metric space H, then S has a
unique fixed point (some F ∈ H such that S(F) = F). Moreover, if F
0
∈ H is any
initial value, then the sequence defined iteratively by F
k+1
:= S(F
k
) converges to F.
This powerful result has applications throughout mathematics.
Theorem 5.7. Let S
1
, . . . , S
n
be contraction mappings on R
m
with ratios c
1
, . . . , c
n
. Define
S : H → H by S(D) =
n
[
i=1
S
i
(D)
1. S is a contraction mapping on H, with contraction ratio c = max{c
i
}.
2. S has a unique fixed set F ∈ H given by F = lim
k→∞
S
k
(E) for any non-empty E ∈ H.
Part 1 is not 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 is Banach’s theorem.
The upshot is this: if we take any non-empty compact set E and repeatedly apply contraction map-
pings, the process will converge to a limit which is independent of E! We call the limit set F for fractal.
Such fractals are often called attractors: being limit-sets, they ‘attract’ data towards themselves.
37
This is the Hausdorff metric. Given Y ∈ H, and x ∈ R
n
, define d
Y
(x) = inf
y∈Y
||
x −y
||
to be the distance from x to the
‘nearest’ point of Y. Define d
X
(y) similarly. The Hausdorff distance between X and Y is then
d(X, Y) := max
(
sup
x∈X
d
Y
(x), sup
y∈Y
d
X
(y)
)
Roughly speaking, find x ∈ X which is as far away d
Y
(x) as possible from anything in Y, and find y ∈ Y similarly; d(X, Y)
is the larger of these distances.
85
Examples 5.8. 1. (Cantor set Ex. 5.4) Theorem 5.7 shows that we may let C
0
be any closed bounded
subset of R. Repeatedly applying the contraction mappings S
1
and S
2
will always result in the
same set C.
A nice application is that one can easily find all sorts of interesting points in the Cantor set. For
instance, suppose x, y ∈ R are a pair such that y = S
1
(x) and x = S
2
( y): otherwise said
y =
1
3
x and x =
1
3
( y + 2)
Since E = {x, y} is a compact set satisfying E ⊆ S(E), it follows that E ⊆ lim S
k
(E) = C, from
which x, y both lie in the Cantor set! However, we can easily solve to see that (x, y) = (
3
4
,
1
4
).
This seems paradoxical:
1
4
does not lie at the end of any deleted interval (denominators of the
form 3
n
) but yet the Cantor set contains no intervals. How does
1
4
end up in there?!
2. (Koch curve, Ex. 5.5) Define four mappings S
i
: R
2
→ R
2
,
each with scale factor c =
1
3
.
Mapping Effect
S
1
(x, y) =
x
3
,
y
3
Scale
1
3
S
2
(x, y) =
1
6
x −
√
3
6
y +
1
3
,
√
3
6
x +
1
6
y
Scale
1
3
, rotate 60°, translate
S
3
(x, y) =
1
6
x +
√
3
6
y +
1
2
,
√
3
6
x −
1
6
y +
√
3
6
Scale
1
3
, rotate −60°, translate
S
4
(x, y) =
x
3
+
2
3
,
y
3
Scale
1
3
, translate
Applied to the Koch curve, the image of each map corresponds by color. The picture links to a
series of animated constructions of the curve starting with different initial sets E.
3. (Sierpi
´
nski carpet) Eight contraction mappings produce this fractal,
each reducing the whole by a (length-scale) factor of
1
3
.
As with the Koch curve, the image links to several alternative con-
structions using different initial starting sets.
4. (A Fractal Fern) This is built from three contraction mappings:
S
1
: Scale by
3
4
, rotate 5° clockwise, and translate by (0,
1
4
)
S
2
: Scale by
1
4
, rotate 60° counter-clockwise, and translate by (0,
1
4
)
S
3
: Scale by
1
4
, rotate 60° clockwise, and translate by (0,
1
4
)
86
Fractal Dimension Revisited
Since Theorem 5.7 permits several different contraction factors, we need a new
approach to computing fractal dimension. We ask how many disks of a given
radius ϵ are required to cover a set. In the picture, the unit square requires four
disks of radius ε = 0.4. For smaller ε, we will plainly need more disks. . .
Definition 5.9. Let A be a compact subset of R
m
.
1. If ε > 0, the closed ε-ball centered at x ∈ A consists of the points at most a distance ε from x:
B
ϵ
(x) = {y ∈ R
m
: d(x, y) ≤ ε}
2. The minimal ε-covering number for A is
N(A, ε) = min
(
M : ∃x
1
, . . . , x
M
∈ A with A ⊆
M
[
n=1
B
ϵ
(x
n
)
)
3. Given a compact set A ⊆ R
m
, its fractal dimension is the limit
D = lim
ε→0
log N(A, ε)
log( 1/ε)
We don’t claim to prove that D must exist, though a simple example should at least convince you
that the definition is reasonable!
Example 5.10. Let A = [0, 1] be the interval of length 1. It is not hard to see that
ε ≥
1
2
⇐⇒ N(ε) = 1, and
1
4
≤ ε <
1
2
⇐⇒ N(ε) = 2
etc. More generally, N and ϵ are related via
1
2N
≤ ϵ <
1
2(N −1)
The dimension of the line (1) may therefore be recovered via the squeeze theorem
D = lim
ϵ→0
log N
log( 1/ε)
= 1
Thankfully an easier-to-use modification is available using boxes.
Theorem 5.11 (Box-counting). Let A be compact and cover R
m
by boxes of side length
1
2
n
. Let N
n
(A)
be the number of boxes intersecting A. Then
D = lim
n→∞
log N
n
(A)
log 2
n
87
We finish with a formula satisfied by the dimension of an iterated function system (Theorem 5.7).
Theorem 5.12. Let {S
n
}
M
n=1
be an iterated function system with attractor (limiting fractal) F and
where each contraction S
n
has scale factor c
n
∈ (0, 1) . At each stage of the construction, suppose
portions of the fractal generated by each contraction map meet only at boundary points. Then the
fractal dimension is the unique D satisfying
M
∑
n=1
c
D
n
= 1
Examples 5.13. 1. If all scale-factors are identical c
n
= r, we recover Definition 5.2,
Mr
D
= 1 =⇒ D =
−log M
log r
=
log M
log( 1/r)
2. The fractal fern (Examples 5.8) is generated by three contraction maps with scale factors
3
4
,
1
4
,
1
4
.
Its dimension is the solution to the equation
3
4
D
+
1
4
D
+
1
4
D
= 1 =⇒ D ≈ 1.3267
3. Numerical approximation is usually required to solve for D, though sometimes an exact solu-
tion is possible. For instance, if c
1
= c
2
=
1
2
and c
3
= c
4
= c
5
=
1
4
, then
2
1
2
D
+ 3
1
4
D
= 1
Writing α =
1
2
D
yields the quadratic equation
2α + 3α
2
= 1 =⇒ α =
1
3
=⇒ D = log
2
3 ≈ 1.584
Other methods of creating fractals
The contraction mapping approach is one of many ways
to create fractals. Two other famous examples are the logis-
tic map (related to numerical approximations to non-linear
differential equations) and the Mandelbrot set (pictured).
The Mandelbrot set arises from a construction in the com-
plex plane. For a given c ∈ C, we iterate the function
f
c
( z) = z
2
+ c
If f ( f ( f (··· f (c) ···))) remains bounded, no matter how
many times f is applied, then c lies in the Mandelbrot set.
Much better pictures and some trippy videos can be found
online. . .
−1
i
−i
88
Exercises 5.2. 1. Let S
1
(x) =
1
3
x and S
2
(x) =
1
3
x +
2
3
be the contraction mappings defining the
Cantor set and suppose x, y, z ∈ R satisfy
y = S
1
(x), z = S
2
( y), x = S
2
( z)
Show that x, y, z lie in the Cantor set, and find their values.
2. The construction of a Cantor-type set starts by removing the open intervals (0.1, 0.2) and
(0.6, 0.8) from the unit interval.
(a) Sketch the first three iterations of this fractal.
(b) This construction may be described using three contraction mappings; what are they?
(c) State an equation satisfied by the dimension D of the set and use a computer algebra
package to estimate its value.
3. A variation on the Koch curve is constructed using the following contraction mappings. Each
is built by first scaling the whole picture by a factor c, rotating the picture through an angle
counter-clockwise, and then translating the picture by adding a constant. The resulting fractal
is drawn.
map scale rotate translate (add (x, y))
S
1
1
2
0 0
S
2
1
4
90° (
1
2
, 0)
S
3
1
4
0 (
1
2
,
1
4
)
S
4
1
4
−90° (
3
4
,
1
4
)
S
5
1
4
0 (
3
4
, 0)
(a) Suppose you start with the straight line segment from (0, 0) to (1, 0). Draw the first two
iterations of the fractal’s construction.
(b) The dimension of the fractal is the unique solution D to the equation
1
2
D
+
1
4
D
+
1
4
D
+
1
4
D
+
1
4
D
= 1
By observing that
1
4
=
1
2
2
, convert this to a quadratic equation in the variable α :=
1
2
D
.
Hence compute the dimension of the fractal.
(c) The dimension computed in part (b) is larger than the dimension
log 4
log 3
of the Koch curve.
Explain what this means.
4. Verify the details of Example 5.10, including the computation of the limit.
5. In Theorem 5.12, prove that D exists and is unique.
(Hint: You’ll need the intermediate value theorem from calculus)
89
