The Koch Snowflake

IB Mathematics HL

Portfolio

(Type I)

The Koch Snowflake

IB Candidate name: Juliana Peña

IB Candidate number: D 000033 049

May 2008

Juliana Peña 000033 049

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Table of Contents

Questions . . . . . . . . . 3

Answers . . . . . . . . . 4

Answer 1 . . . . . . . . 4

Answer 2 . . . . . . . . 6

Answer 3 . . . . . . . . 7

Answer 4 . . . . . . . . 9

Answer 5 . . . . . . . . 11

Answer 6 . . . . . . . . 12

Answer 7 . . . . . . . . 14

Bibliography . . . . . . . . . 16


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Questions

(Extracted from the IB Mathematics HL Internal Assessment Teacher Support Material)


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Answers

Answer 1

Table 1

Finding the values for was simple since only counting of the number of sides in the diagram was

needed. After finding a few of the values, the pattern was clear: each value was four times the

previous value.

In each iteration of the fractal, each side is divided in three; therefore the length of one side of the

snowflake is one third of the length of the snowflake in the previous iteration:

.

was found by using a simple calculation. Since the Koch snowflake's sides are all the same size,

and perimeter is defined as the sum of the lengths of all the sides, the perimeter of the snowflake

is the same as multiplying the number of sides by the length of each side: .

was the most complicated value to find. Given that the formula for the area of an equilateral

triangle is √

, where is the length of a side of the triangle, the area when is

√

, since

. √

.

Since after each iteration a certain number of triangles are added to the previous shape, the area

of iteration is ( ) ( ).

Table 2 shows the values of the area of each added triangle, , and the number of added

triangles, , together with , for . was found by calculating √

. was found by

counting (although later it was apparent that ).

0 3 1 3 √

1 12

4

√

(

)

2 48

√

(

)

3 192

√

(

)


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Table 2

1

3 √

2

12 √

3

48 √

With these values, finding is now simple. Using , the values of to

were found.

The simplified results are shown together in Table 1.


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Answer 2

Something important to notice is that is convergent. This means that has a finite value

when . Contrastingly, is divergent, so it has an infinite value when . Therefore, the

Koch snowflake is fractal with a finite area bounded by an infinite perimeter, a characteristic quite

common amongst fractals.

0

50

100

150

200

250

0 1 2 3

Nn

n

Graph of Nn

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3

Ln

n

Graph of Ln

012345678

0 1 2 3

Pn

n

Graph of Pn

0.00.10.20.30.40.50.60.70.8

0 1 2 3

An

n

Graph of An


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Answer 3

Table 3

Generalization for :

Since each result is four times the last result, the general formula for

is , or .

Verifications:

For For For For

Table 4


Each result is a third of the previous one. The general formula is

.

Verifications:

For

For

For

For

Table 5


As mentioned before, . Therefore,

.

Verifications:

For

For

For

For

0 3

1 12

2 48

3 192

0 1

1

2

3

0 3

1 4

2

3


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Table 6


The general formula for the area must be

the sum of the area of the previous

iteration plus the area of the added

triangles. Therefore, the general formula

must be the sum of all these areas.

Table 6 shows the un-simplified sums and

the approximate results of to . The

generalized formula can then be deduced

as √

* ∑ (

)

+.

Verifications:

For √

( )

√

For √

(

)

√

(

)

√

(

)

√

For √

(

)

√

(

)

√

(

)

√

For √

(

)

√

(

)

√

(

)

√

0 √

0.43

1 √

(

)

2 √

(

)

3 √

(

)

√

[ ∑(

)

]


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Answer 4

If , the following must be true:

√

[ ∑(

)

] √

*

+

√

[

]

√

[

]

√

(

)

√

Figure 1 on the next page shows one of the three sides of the Koch snowflake on the fourth stage.

Counting the number of sides on this curve and multiplying them by four gives 768, which is . If

the snowflake were completed, the other numbers would also be shown to be true.


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Figure 1

Figure 1 was made using turtle graphics drawn by a Python program with the following source

code:

import turtle iteration = 4 set = "F" for i in range(iteration):

set = set.replace("F","FLFRFLF") turtle.down() for move in set:

if move is "F": turtle.forward(100.0/3**i) if move is "L": turtle.left(60) if move is "R": turtle.right(120)

input () ## This program source code is a slightly modified version ## of the one freely available on Wikipedia (see Bibliography, source 1)


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Answer 5

Using spreadsheet software, Table 7 was produced.

Table 7

0 0.433013

1 0.577350

2 0.641500

3 0.670011

4 0.682683

5 0.688315

6 0.690818

7 0.691930

8 0.692425

9 0.692645

10 0.692742

11 0.692786

12 0.692805

13 0.692813

14 0.692817

15 0.692819

16 0.692820

17 0.692820

As shown on Table 7, the values of when and are equal up to six decimal

places. Hence, the value of where up to six decimal places is 16.


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Answer 6

Table 8

0 3.000000 0.433013

1 4.000000 0.577350

2 5.333333 0.641500

3 7.111111 0.670011

4 9.481481 0.682683

5 12.641975 0.688315

6 16.855967 0.690818

7 22.474623 0.691930

8 29.966164 0.692425

9 39.954885 0.692645

10 53.273180 0.692742

11 71.030907 0.692786

12 94.707875 0.692805

13 126.277167 0.692813

14 168.369556 0.692817

15 224.492742 0.692819

16 299.323656 0.692820

17 399.098207 0.692820

The spreadsheet-generated table above shows values of and until .

As mentioned in Answer 2, is divergent and is convergent. As gets very large, the

perimeter only becomes larger by a larger amount each time. However, at the same time, is

getting larger, but by a smaller amount every time. Eventually, will reach a limit when .

But what is this limit?

On Answer 3, it was stated that

√

[ ∑(

)

]


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The sum shown in this formula, ∑ (

)

, is the partial sum of a geometric series with a ratio

. The formula for the sum of an infinite geometric series with first term and ratio is

Therefore, since the first term of ∑ (

)

is

,

√

(

) √

(

) √

(

)

√

As approaches infinity, gets nearer and nearer to a maximum value of √

.

, however, does not have a finite upper limit, as it is divergent. As approaches infinity, so does

.

Restating the deduction in Answer 2, a Koch snowflake at its infinite iteration will have a finite

area of √

enclosed within an infinite perimeter.


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Answer 7

From Answer 1, it was stated that

( ) ( )

which equals

√

From Answer 3, it was stated that

√

[ ∑(

)

]

Let us prove this by induction.

( ) states that

√

√

[ ∑(

)

]

To prove this statement by induction, ( ) and ( ) must be proved true given that ( ) is

true.

( ) ( )

√

√

[ ∑(

)

]

Since √

,

√

√

√

[

]

√

[

]

√

[

]


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So ( ) holds true.

( ) ( )

√

[ ∑(

)

] √

√

[ ∑(

)

]

√

[ ∑(

)

] √

(

)

√

[ ∑(

)

]

√

[ ∑(

)

(

)]

√

[ ∑(

)

]

√

[ ∑(

)

] √

[ ∑(

)

]

So ( ) also holds true.

Since ( ) and ( ) both hold true, ( ) is true for every positive integer .


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Bibliography

1. Wikipedia contributors. "Koch snowflake". Wikipedia, The Free Encyclopedia.

http://en.wikipedia.org/w/index.php?title=Koch_snowflake&oldid=132082916 (accessed

June 2, 2007).

2. Riddle, Larry. "Koch Snowflake Area." Agnes Scott College.

http://ecademy.agnesscott.edu/~lriddle/ifs/ksnow/area.htm (accessed June 2, 2007).

3. Riddle, Larry. "Koch Snowflake." Agnes Scott College.

http://ecademy.agnesscott.edu/~lriddle/ifs/ksnow/ksnow.htm (accessed June 2, 2007).

http://en.wikipedia.org/w/index.php?title=Koch_snowflake&oldid=132082916

http://ecademy.agnesscott.edu/~lriddle/ifs/ksnow/ksnow.htm

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