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Question Description

I’m studying for my Mathematics class and need an explanation.

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 G1τ€€€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

D =


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?

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Final Answer

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 𝑺𝒏 .
Denote the number of sides at the 𝑛th step as 𝑠𝑛 . We know 𝑠0 = 3 (a triangle),
and at each step we make 4 sides from any previous. This way 𝑠𝑛+1 = 4𝑠𝑛 .
The lengths of all sides are the same for each given step. It is 𝑙0 = 1 for the initial triangle
and 3 times less for each next, 𝑙𝑛+1 = 3 𝑙𝑛 .


This way the perimeter length is 𝑃(𝑆𝑛 ) = 𝑠𝑛 𝑙𝑛 and 𝑃(𝑆𝑛+1 ) = 𝑠𝑛+1 𝑙𝑛+1 = 3 𝑠𝑛 𝑙𝑛 = πŸ‘ 𝑷(𝑺𝒏 ).
And 𝑃(𝑆0 ) = 𝑠0 𝑙0 = 3.
πŸ’ 𝒏

Now it is obvious that 𝑃(𝑆𝑛 ) = πŸ‘ βˆ™ (πŸ‘) .


> 1, we know that 𝑃(𝑆𝑛 ) β†’ ∞ as 𝑛 β†’ ∞.
1 𝑛

Also note that obviously 𝑠𝑛 = 3 βˆ™ 4𝑛 and 𝑙𝑛 = (3) .

2.2. The area of 𝑺𝒏 .
We add one triangle to each existing side at (𝑛 + 1)-th step.
There are 𝑠𝑛 = 3 βˆ™ 4𝑛 sides and 𝑠𝑛 new triangles.
1 𝑛+1

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

This way the area added to the area of 𝑆𝑛 is 3 βˆ™ 4𝑛 βˆ™


and its area ...

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UC Berkeley

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