Lecture 6: Chaos and Fractals
Julia Collins
30th October 2013
The Koch snowflake
In the Koch snowflake, each iteration of the shape is given by deleting the middle third of
each edge and attaching an equilateral triangle whose base would have been the missing
edge.
4 iterations of the Koch snowflake
Each edge that was 3 units long becomes 4 units long after the new iteration, so the
perimeter of gets multiplied by 34 after each iteration. This means that the Koch snowflake
has infinite perimeter.
Homework: Show that the Koch snowflake has finite area. (Requires some
knowledge of infinite sums and the geometric series!)
1
Fractal dimension
We can think of dimension as being the ratio between the scale of an object and the detail
we see.
For example, take a line of 1 unit in length. If we scale that line to 13 of its original
length, then we need 3 = 31 of the new sticks to fit into the same space as the old one. So
the dimension of a line is 1.
Now suppose we have a square with sides 1 unit in length. Then if we scale the square
so that it is only a third as tall and wide, then we need 9 = 32 as many squares to fit inside
the old one. So the dimension of a square is 2.
Finally, take a cube with sides 1 unit in length. If we scale the width, height and depth
of the box to be a third of what it was before, then we need 27 = 33 as many cubes to fit
inside the old one. So the dimension of a cube is 3.
So what is the dimension of the Koch snowflake? When we start with our equilateral
triangle, we can assume that all sides are 1 unit in length. Then we take that side, divide
into thirds, remove the middle third and replace with another triangle. So the lines making
up the side of the triangle have been scaled down by 13 , but now there are 4 of them instead
of 3. Therefore the fractal dimension is the number x which solves the equation 3x = 4.
We can find this value by taking logarithms of both sides:
3x = 4 ⇒ log 3x = log 4 ⇒ x log 3 = log 4 ⇒ x =
So x = 1.2619, which is the ‘dimension’ of the Koch snowflake.
2
log 4
.
log 3
Homework solution
To find the area of the Koch snowflake, we need to know how much new area we are adding
on after each step in its construction.
The area of a triangle is 12 × base × height . The triangles in an iteration have 31 the
base AND 13 the height of the triangles in the previous iteration, meaning they have exactly
1
of the area. So after n iterations, the triangles have 91n the area of the original triangle we
9
started with.
How many new triangles are added in each step? In Step 1, 3 new triangles are added.
Each of the three ‘sides’ of the triangle now have 4 edges, so 3 × 4 = 12 new triangles are
added on in Step 2. This pattern repeats, so in Step n there are 3 × 4n−1 triangles added on.
n−1
This means that the total area added in Step n is 3×49n times the area of the original
triangle.
Let An be the area of the Koch snowflake after n iterations. So A0 is the area of the
triangle we start with.
A1 = A0 + 39 A0 = 34 A0 .
4
40
A2 = A1 + 3×4
A0 = 34 A0 + 27
A0 = 27
A0
92
An = A0 +
n
X
3 × 4k−1
k=1
n
X
9k
A0
4k−1
A0
9k−1
k=1
!
n−1 k
1X 4
= A0 1 +
3 k=0 9
1 9
= A0 1 +
(see below for explanation of this step)
3 5
8
=
A0 .
5
3
= A0 +
9
A∞
To understand this manipulation, the first thing we need to remember is that the
n
X
k=2
symbol means “the sum from when k = 2 up to when k = n”. The other thing we need
to know is the sum of an infinite series where the next number in the series is obtained by
multiplying the last number by a constant factor. This type of series is called a geometric
3
series and the formula for the sum using multiplicative factor x is
∞
X
xk =
k=0
In our calculation we needed to find
∞ k
X
4
k=0
9
∞ k
X
4
k=0
9
=
1
.
1−x
, so x = 94 . This gives us
1
1−
4
9
9
= .
5
Now we are just left with the task of finding A0 . This is the area of an equilateral triangle
with side length 1. What is the height of this triangle? To get the answer we need some
basic school trigonometry. Each angle in the triangle is 60◦ , and
sin 60 =
height
1
√
and sin 60 =
3
,
2
so this is the height of the triangle. The total area of the triangle is therefore
√
√
3
3
1
×
×1=
2
2
4
Finally, the area of the Koch snowflake is therefore
√
√
8
3
2 3
×
=
.
5
4
5
4