This problem refers to the presentation on chaos and fractals.
The Koch snow
ake, named after the Swedish mathematician Helge von Koch, is a domain
obtained through the following procedure.
(i) Start with an isosceles triangle, in which each edge has the length 1. Call this object
(ii) Divide each edge in three parts. Remove the middle part, and glue an isosceles
triangle in place of it. This object is S1.
(iii) Iteratively, given Sj , repeat the above process for every edge to get the next object
(iv) The Koch snow
ake is the limit object S1 as j ! 1:
Find the rule giving the perimeter length of Sj+1, given the perimeter length of Sj . Using
this rule, nd a formula for the length of the perimeter of the jth iteration Sj . Using this
rule, show that the length of the perimeter goes to in nity as j ! 1.
Also, nd the rule for computing the area of Sj , and show that the area remains nite
as j ! 1. Hence, Koch’s snow
ake is an example of a set with nite area but in nite
perimeter, in the spirit of Mandelbrot’s considerations about the coastal line length of
In his article on coastline of Britain, Mandelbrot refers to the empirical rule of Lewis Fry
Richardson, stating that the coastline length seems to follow the law
L(G) = M G1 D;
where M > 0 is some constant, G > 0 is the yardstick length used in the approximation
of the coastline, and D 1 is the dimension of the curve. If you assume that Sj is an
approximation of S1, and the yardstick length to measure j@Sj j is G = (1=3)j , what
would the dimension of the Koch snow
ake’s boundary be?
3. The Koch snow
ake is a construction in which an object of dimension larger than one is
obtained from a one-dimensional starting point. One can proceed the other way, starting
with a two-dimensional object. The following process describes the construction of the
Sierpinski triangle, named after the Polish mathematician Wac law Sierpinski.
(i) Start with an isosceles triangle, denoted by T0.
(ii) Divide each side in two, connect the midpoints, thus dividing the triangle in four
isosceles triangles. Remove the triangle in the middle, leaving you with three trian-
gles. This object is called T1.
(iii) Iteratively, Tj consisting of several isosceles triangles, repeat the above process to
each one of the triangles, giving you Tj+1.
(iv) The Sierpinski triangle is the limit object T1 as j ! 1.
Like the Koch snow
ake, the Sierpinski triangle is a self-similar fractal: It looks the same
no matter how much you zoom in. Find the formula for the area of Tj , show that it goes
to zero as j ! 1.
The Sierpinski triangle is a self-similar object: At each step Tj ! Tj+1, we generate a
given number N of similar objects, that have the size s times the size of the original object,
where the size is measured in terms of length, the fractal dimension can be de ned as
Verify that if you don’t remove the center triangle, the object has dimension D = 2 as
expected. What is the dimension of T1?