# The Koch Snowflake

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Introduction

Jade Okba IB1

IB Mathematics HL Investigation

The Koch Snowflake

n | Nn | Ln | Pn | An |

0 | 3 | 1 | 3 | ( 3)/4 |

1 | 12 | 1/3 | 4 | ( 3)/3 |

2 | 48 | 1/9 | 5 1/3 | 10 ( 3)/27 |

3 | 192 | 1/27 | 7 1/9 | 94 ( 3)/243 |

Number of Sides

3 x 4 = 12

12 x 4 = 48

48 x 4 = 192

Each successive term is a result of multiplying the previous one by 4. Therefore, this is a geometric sequence and the common ration is 4. The equation for this sequence is as follows:

n = stage no. N = number of sides r = common ratio

Nn = N0 x rⁿ

Length of Sides

1 ÷ 3 = 1/3

1/3 ÷ 3 = 1/9

1/9 ÷ 3 = 1/27

Each successive term is a result of dividing the previous term by 3. This shows that it is a geometric sequence and the common ratio is 3. Therefore the equation for this sequence is:

L = length of side n = stage no. r = common ratio

Ln = N0/rⁿ

e.g. L2 = 1/3² = 1/9

Perimeter

4 ÷ 3 = 1.333…

5 1/3 ÷ 4 = 1.333…

7 1/9 ÷ 5 1/3 = 1.333…

Therefore each successive term is the result of multiplying the previous term by 1.333… which is equal to 4/3.

The perimeter is increasing in a geometric sequence, by 1/3 of its value each time. It is also increasing by a larger amount each time, and so is a divergent series.

This

Middle

3 x 4 = 12 12 x 4 = 48

IT MULTIPLIES BY 4 EVERY SUCCESSEIVE TERM.

1/9 x 1/9 = 1/9² 1/9² x 1/9 = 1/9³

IT DIVIDES BY 9 EVERY SUCCESSIVE TERM.

This is a geometric series with the first added term being: (3 x A/9) = A/3, the common ratio being 4/9 and the number of added terms (and the index of 9) are equal to n.

In order to find the rule for this geometric sequence I must use this equation:

= a [(1-r ⁿ)/ (1-r)]

n = number of terms r = common ratio a = first term

Area = A/3 [(1-(4/9) ⁿ)/ (1-(4/9))]

This is equal to:

Area = 3A/5 (1-(4/9) ⁿ)

I must now adapt this equation in order for it to suit my sequence. The first shape’s area of ( 3)/4 is irregular, because earlier in my project, I used an odd result to calculate this number. Due to this, I must use the first added term as my first term. My common ratio is 4/9 as I worked out earlier. These have been substituted into the equation below:

Area = 3( 3)/20 (1-(4/9)ⁿ)

n | Area using equation | Real Area |

1 | ( 3)/12 | ( 3)/3 |

2 | 13( 3)/108 | 10( 3)/27 |

3 | 133( 3)/972 | 94( 3)/243 |

This equation is incorrect. However, I will now add on ( 3)

Conclusion

Proof by Induction

An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4

Firstly. I will prove that this general equation is true for n = 1

A1 = 3( 3)/20[1-(4/9)^1] + ( 3)/4

= 15( 3)/180 + ( 3)/4

= ( 3)/12 + ( 3)/4

= ( 3)/12 + 3( 3)/12

= 4( 3)/12

= ( 3)/3

The answer is correct.

Now, I will assume that it is true for n = k

Ak = 3( 3)/20 [1-(4/9) ^k] + ( 3)/4

Then, I must show that it is true for n = k + 1

A (k+1) = 3( 3)/20 [1-(4/9) ^ (k+1)] + ( 3)/4

This shows the (k+1)th term in the form of:

An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4

( 3)/4 → The initial area

( k+1) → The +1 is the inductive step

Instructions to draw fourth stage of Koch snowflake in Microsoft Windows Logo

to side :x :y → Type into commander box

if :y=0 [fd :x stop] → enter into To Mode box the click ok (applies to each of the next instructions)

side :x/3 :y-1

lt 60 side :x/3 :y-1

rt 120 side :x/3 :y-1

lt 60 side :x/3 :y-1

end →do no type this – click cancel to end

to vonkoch2 (Same process as above)

;superimpose 6 stages

cs pu bk 300 lt 90 fd 200 rt 90 pd

vonkoch2 500 0

vonkoch2 500 1

vonkoch2 500 2

vonkoch2 500 3

vonkoch2 500 4

end

to vonkoch2 :x :y (same process as above)

;draws single curve size :x stage :y at current cursor position

repeat 3 [side :x :y rt 120]

end

Now just type vonkoch2 in the commander box

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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