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

Table of Contents

Questions

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Answers

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4

Answer 1

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4

Answer 2

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6

Answer 3

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Answer 4

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9

Answer 5

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Answer 6

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Answer 7

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16

Bibliography

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Juliana Peña 000033 049

Questions

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

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Answers Answer 1 Table 1

0

3

1

12

2

48

3

192

1

√

3 √

4 √ √

(

)

(

)

(

)

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, triangles,

, together with

, for

.

, and the number of added

was found by calculating

counting (although later it was apparent that

Page 4 of 16

).

√

.

was found by

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Table 2

1

3

2

12

3

48

With these values, finding were found.

is now simple. Using

The simplified results are shown together in Table 1.

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√ √ √

, the values of

to

Juliana Peña 000033 049

Answer 2

Graph of Ln

Graph of Nn

Nn

250

1.2

200

1.0 0.8

150

Ln 0.6

100

0.4

50

0.2

0

0.0 0

1

2

0

3

1

Graph of Pn

An

1

2

3

Graph of An

8 7 6 5 4 3 2 1 0 0

3

n

n

Pn

2

2

3

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

1 n

n

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.

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Answer 3 Table 3

0

3

1

12

2

48

3

192

Generalization for : Since each result is four times the last result, the general formula for is , or . Verifications: For For For For Table 4

Generalization for

:

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

.

0

1

1

Verifications: For

2

For

3

For For Table 5

Generalization for 0

3

As mentioned before,

1

4

Verifications:

2 3

:

For For For For

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. Therefore,

.

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Table 6

Generalization for : 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.

0 1

√

*

∑

(

√

2

Table 6 shows the un-simplified sums and the approximate results of to . The generalized formula can then be deduced as

√

√

3

√

0.43

(

)

(

)

(

) √

)+.

∑(

[

)]

Verifications: For

√

(

For

√

(

For

√

(

For

√

(

)

√

)

√

( )

) √

√

( )

√

( ) )

√

(

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√

√

( ) )

√

(

)

√

Juliana Peña 000033 049

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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Juliana Peña 000033 049

Answer 5 Using spreadsheet software, Table 7 was produced. Table 7

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

As shown on Table 7, the values of places. Hence, the value of where

0.433013 0.577350 0.641500 0.670011 0.682683 0.688315 0.690818 0.691930 0.692425 0.692645 0.692742 0.692786 0.692805 0.692813 0.692817 0.692819 0.692820 0.692820

when and are equal up to six decimal up to six decimal places is 16.

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Juliana Peña 000033 049

Answer 6 Table 8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

3.000000 4.000000 5.333333 7.111111 9.481481 12.641975 16.855967 22.474623 29.966164 39.954885 53.273180 71.030907 94.707875 126.277167 168.369556 224.492742 299.323656 399.098207

The spreadsheet-generated table above shows values of

0.433013 0.577350 0.641500 0.670011 0.682683 0.688315 0.690818 0.691930 0.692425 0.692645 0.692742 0.692786 0.692805 0.692813 0.692817 0.692819 0.692820 0.692820

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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)]

Juliana Peña 000033 049

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

Therefore, since the first term of ∑ √

As

approaches infinity,

(

(

) is

)

√

with first term

and ratio is

,

(

)

√

(

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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Juliana Peña 000033 049

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 ( true. ( )

√

) must be proved true given that ( ) is

√

∑(

[

, √

√

√

[

)]

( )

√

Since

∑(

[

√

]

√

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[

[

]

]

)]

Juliana Peña 000033 049

So ( ) holds true. ( ) √

∑(

[

√

∑(

[

√

[

)

∑(

)

√

)]

∑(

[

√

So (

)]

(

√

√

(

(

√

)]

)]

√

)

√

[

[

[

∑(

∑(

[

∑(

)]

∑(

)]

)]

) also holds true.

Since ( ) and (

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

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)]

Juliana Peña 000033 049

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).

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