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  • Level: GCSE
  • Subject: Maths
  • Word count: 1686

The Koch Snowflake

Extracts from this document...

Introduction

Jade Okba IB1

IB Mathematics HL Investigation

The Koch Snowflake

n

Nn

Ln

Pn

An

0

3

1

3

(  3)/4image04.png

1

12

1/3

4

(   3)/3image04.png

2

48

1/9

5 1/3

10 (           3)/27image04.png

3

192

1/27

7 1/9

94 ( 3)/243image04.png

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

...read more.

Middle

image01.png

3 x 4 = 12        12 x 4 = 48 image02.png

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:image04.png

Area = 3( 3)/20 (1-(4/9)ⁿ)image04.png

n

Area using equation

Real Area

1

    (    3)/12image04.png

 (  3)/3image04.png

2

13(          3)/108image04.png

 10(                3)/27image04.png

3

133(            3)/972image04.png

94(                     3)/243image04.png

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

...read more.

Conclusion

Proof by Induction

An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4image04.pngimage04.png

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

A1 = 3( 3)/20[1-(4/9)^1] + ( 3)/4image04.pngimage04.png

      = 15( 3)/180 + ( 3)/4image04.pngimage04.png

image04.pngimage04.png

      = ( 3)/12 + ( 3)/4

      = ( 3)/12 + 3( 3)/12image04.pngimage04.png

      = 4( 3)/12image04.png

      = ( 3)/3image04.png

The answer is correct.

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

image04.png

Ak = 3( 3)/20 [1-(4/9) ^k] + ( 3)/4image04.png

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

A (k+1) = 3( 3)/20 [1-(4/9) ^ (k+1)] + ( 3)/4image04.pngimage04.png

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

An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4image04.pngimage04.png

image04.png

( 3)/4 → The initial area

( k+1) → The +1 is the inductive step

image08.png

image09.png

image10.png

image11.png

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

image05.png

image06.png

...read more.

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