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Anonymous

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

S0.

(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

S(j+1).

(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 innity 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 innite

perimeter, in the spirit of Mandelbrot's considerations about the coastal line length of

Britain.

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 G1τD;

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 dened as

D =

logN

log s

:

Verify that if you don't remove the center triangle, the object has dimension D = 2 as

expected. What is the dimension of T1?

And the second, too:)The solutions are ready. Please ask me if something is unclear.docx and pdf files are identical

2. Koch snowflake.

2.1. The perimeter length of πΊπ .

Denote the number of sides at the πth step as π π . We know π 0 = 3 (a triangle),

and at each step we make 4 sides from any previous. This way π π+1 = 4π π .

The lengths of all sides are the same for each given step. It is π0 = 1 for the initial triangle

1

and 3 times less for each next, ππ+1 = 3 ππ .

4

π

This way the perimeter length is π(ππ ) = π π ππ and π(ππ+1 ) = π π+1 ππ+1 = 3 π π ππ = π π·(πΊπ ).

And π(π0 ) = π 0 π0 = 3.

π π

Now it is obvious that π(ππ ) = π β (π) .

Because

4

3

> 1, we know that π(ππ ) β β as π β β.

1 π

Also note that obviously π π = 3 β 4π and ππ = (3) .

2.2. The area of πΊπ .

We add one triangle to each existing side at (π + 1)-th step.

There are π π = 3 β 4π sides and π π new triangles.

1 π+1

The side length of any new triangle is ππ+1 = (3)

This way the area added to the area of ππ is 3 β 4π β

β3

4

and its area ...

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