• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The Koch Snowflake

Extracts from this document...

Introduction

image38.pngMath Portfolio          

The Koch snowflake          

Introduction: The Koch snowflake fractal is built by starting with an equilateral triangle, and removing the inner third of each side, building another equilateral triangle at the location where the side was removed.  When n = 0, 1, 2and 3 each value of Nn, ln, Pn and An can be shown as the following table:

n

0

1

2

3

Nn

3

12

48

192

ln

1

image00.png

image01.png

image43.png

Pn

3

4

image47.png

image54.png

An

image56.png

image59.png

image65.png

image71.png

And the process I use to get the value are shown below:

The number of the side:

To get the next snowflake, we can found that, each side of the triangle will break into 4 new sides, hence the number of each side will always 4 times than previous one. Thus we get

Nn=4Nn-1

 Thus we assume that Nn is the single of the number of sides and get the process below:

image66.png

                     Shape1: we get a triangle. N1=3

                                            Shape2: We break into 4 new sides. So the N2=3×4        

                                            Shape 3: Each side then brake into 4 new sides, So N3=3×4 ×4

                                             Shape 4:We repeat the same step, thus we get N4=3×4×4×4

And then we graph our data:

The number of the sideimage02.png

We can found that the shape of the graph is a geometric series. Thus, if we want to get the next shape, we should time the same number.

...read more.

Middle

  ×image55.pngimage55.png=image57.png                          P3=3×image55.pngimage55.png  ×image55.pngimage55.png  ×image55.pngimage55.png=image54.png

Then we can put our data into the diagram below:

The length of perimeterimage58.png

We found that if we link all point together, we may get an up-sloping curve which is a geometric series. Thus, if we want to get the next shape’s perimeter, we should time the same number. Thus it’s a geometric series which has a first term of 3 and a common ratio of image55.pngimage55.png, obviously. So we can get the geometric formula:

                                                   Pn=3image60.pngimage60.png

To verify the formula, we put n=0, 1,2and 3 into the formula. So we get:

         P0=image61.png=3         P1=image62.png=4         P2=image63.png=image57.png        P3=image64.png=image54.png

The results are already proved above.

The area of the snowflake:

            From the observation, we can found a rule between the number of the side and the number of new triangle appeared.  image66.png

             Firstly, Nn=3

           Secondly, Nn=12

                            The number of new triangle appeared=3

            Thirdly, Nn=48

                            The number of new triangle appeared=12

Fourthly, Nn=192

                              The number of new triangle appeared=12

We can found that, the number of new equilateral triangle appeared in the shape is equal to the number of the side in the previous triangle (Nn-1).

And then we show the area of the each addition one.

...read more.

Conclusion

n

An

Nn

ln

Pn

0

0.433013

3

1

3

1

0.57735

12

1

4

2

0.6415

48

0.333333

5.333333

3

0.670011

192

0.111111

7.111111

4

0.682683

768

0.037037

9.481481

5

0.688315

3072

0.012346

12.64198

6

0.690818

12288

0.004115

16.85597

7

0.69193

49152

0.001372

22.47462

8

0.692425

196608

0.000457

29.96616

9

0.692645

786432

0.000152

39.95488

10

0.692742

3145728

5.08E-05

53.27318

11

0.692786

12582912

1.69E-05

71.03091

12

0.692805

50331648

5.65E-06

94.70788

13

0.692813

201326592

1.88E-06

126.2772

14

0.692817

805306368

6.27E-07

168.3696

15

0.692819

3221225472

2.09E-07

224.4927

16

0.69282

12884901888

6.97E-08

299.3237

17

0.69282

51539607552

2.32E-08

399.0982

18

0.69282

2.06158E+11

7.74E-09

532.1309

To find out the value for n where An+1 equals An to six places of decimals. We get the help from the Excel software to work out the first 18 term to check if we get the number we need.

From the result we can see that the An for n=16 is equal to for n=17 to six decimal places. So the value of n is 16 where An is equal to An+1 to six decimal places.

So when n = 16:

N16 = 12884901888   l16 = 2.323057image21.png10-8

P16 = 299.323656    A16 = 0.692820

Then we consider the expression for Pn which:

                     Pn=image22.pngPn-1

We can see that Pn is always increasing in the same ratio ofimage22.png. So the value of Pn increase to positive infinity as n→∞.

For the expression of An which is:

                            An= An-1+image23.png

According to the formula of sum of geometric series to infinity:

               snimage24.pngimage24.png

Ifimage25.pngimage25.png

       Thus,image26.png

image28.png

Thus we know: in this case as n→∞ we can have an infinity length of perimeter in a limited area.

we use the math induction to prove this conclusion:

As we knowimage29.png

P(1) is true since image30.png

Assume that n=m is true for integer t≥1.

image31.png

Consider n=m+1

Am+1 =  image32.pngimage32.png

                        Prove image33.png

image34.png

image35.png

image36.png

Thus P(m+1) is true when P(m) is true. Therefore P(n) is true for all integers m≥1.

                                                                          Richard ouyang

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    -�+�5�n�Fq3�x"���7�{�#��|#3/4?��ů"M�.�?!(tm)p"PK�"<"��ii�hth�hBh�h�h.� �1/4��J$�D{"��I,#^" ��hÉ´'�F������i{h��~%'H�$}' )�T@j �&1/2&� c�"�� �� "�k���BO �7 �C�B_B����A"���Â�P����a...'�Q�qc8�a�F�{�sdY"lB�'�kȷ��L(&!&#&?�}L�LL3�Xf1f ��|�f�a�E2�2� KK%�M-)V"("k�Q�+��?�x� ����.�����s�� �boa�"���"#"�8G;�+N4�$�=g"�)��.f.-.?(r)C\W�^p��'� �(c)�5�C�+<1/4<f<Q<�<�yxYy�yCx�y"y��t��|�|��>����-��/ p � � �Xt�l|%D#�.(T,�'�(�'l#1/4W��� ���H�H(c)�]'UQ1QW���sb�bb)b�^�"�ģ�"ÅH`%�%B%�$F$aI�`�J�GR�"�U�J�4FZC:B�Z�(tm) ���L��(tm)����ֲٲ�_�"�<��Ý�'W'"���T +X*d+t),+J*�)V*>Q")(tm)*e(u(-)K) (�R�PaR�Q9 ҧ�(c)���zQu^MX�G��3ufu;��� C� �k���q�W4��'� �jÔ�� �(r)Õ�ԡ��(tm)����=�;�' G�"�{�/��_�?k ab�d��P�0ư�p�H�(ͨ�elf|�x�"l�lRa��T�4���(tm)�Y�Y�9�����3 - ?��EK5�4�~+Z+G" "w�'�1�]6���M'�K[��]`-�(r)�]�����(r)�c���+�?8(8�u������������Ӥ��s1/4s� 1/2��K�˪"�k��"��[��wNw�{� ��ţ�ce���"g<U<s=�1/2�1/4'1/4������71/27����է�g���RMY��=�"�g�W��_߿�>@' 0`6P'�0p.H'�(h>X/�$x�jD� ....��� Y �z>t;�5�%��A����L�|%�5�}"z1�*�.����cF'áx��ot*~$�$^MbL�HJ-L�K�M1M9-�N�K��+�7k��4����P�oz_�PFN�L�Yf}MVh��l����o�\�u���d�L�7�!-.7&�����R �)��m�?t?_>�$���G�"�.,>�z��1ì±c����2�N���*�v��1/2�'Ó¥4��Se�e�����7*�+�VV���>(tm)wr�ʿj�"�(c)��yN�y�zf��ٶj��'lMBÍZ-�"���5�q���m��8?U�P�� �����x�|!��|"g�H�qs�E(tm)�g[X[�/�K�->^��<~��J�U�"��\;���z� jKn[ln��p�x�i��ץ��z]���7*o��<�MÓ�1/2}+��JOT�BoP�t�w��m��O�� ��޹}���A���4�u�W��@�AÛ�P�C���ê�m��-u�h�t=�~�=�7�;f<v��" Om�>-w-�x�lj�b�y�� /�'3_b^-z���5���7oZ�T�n3/45~;��������7fr>�>"���6�)�Ý7�����̧�O� ��?��"���_� -�-�,�,m/���7�o}+v+���__=��G����Ý(r)?g�7pe��][V[/�÷��(1"_� i�@-�#5";R;�@C�"��Å"+Â` !�-6'��a� �A�0

  2. Investigating the Koch Snowflake

    and then add the area of all of the new smaller triangles added. And the general expression can be derived from the values of An from the zero stage till the third stage A0 = (V3)/4, A1= (V3)/3, A1= [(V3)/4] + (1/3)

  1. Math Portfolio - The Koch snowflake investigation.

    Area scale factor Extra triangles 0 3 - - 1 12 3 2 48 12 3 192 48 A0: ? 12 = : A0 (1 + new triangles ? ) : (1 + 3 ? ) A1: (1 +) Stage 2 has 12 new triangles and also the scale factor becomes square of the previous scale factor.

  2. The Koch Snowflake

    Nn = 3 x 4n and Ln= 1 x (1/3)n for the reasons previously explained, therefore the formula for the perimeter is 3 x (4/3)n . In graph 4, which shows the area of each shape, we can see that as we progress in stage the area increases.

  1. Koch Snowflake

    The general formula for a geometric progression is as follows - Un = U0*Rn Where R = ratio between two successive terms, U0 = 1st term, and Un = nth term of the sequence. Substituting Un with Nn can give us the general formula for the number of sides of

  2. The Koch Snowflake

    I than counted added up the length of every side with my calculator using the drawing of stage 2. And I was able to get the same answers proving my way of quickly getting the perimeter of a fractal. A= In order to calculate the area of each stage I

  1. MAths HL portfolio Type 1 - Koch snowflake

    We can confirm this common ratio by taking two values from the table on the left. Ex. 1 L1 = 1 and L2 = Ex. 2 L2 = and L3 = General formula for Ln The general formula can be verified by putting in values from the table Using

  2. Stellar Numbers. In this task geometric shapes which lead to special numbers ...

    used two more examples: --> Using n=2 6n2 +6n + c = 121� 6(2)2 +6(2) + 1 = 121 --> Using n=6: 6n2 +6n + 1 = 121 6(6)2 +6(6) + 1 = 121 With enough proof that the general statement works I conclude that it is: 6n2 +6n +

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work