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

Maths HL Kochs Snowflake

Extracts from this document...

Introduction

The Koch Snowflake

image19.gif

The koch snowflake is a fractal that was identified by Helge Von Koch in 1904. It is created by starting with an equilateral triangle, removing the centre third of each side and replacing it with another, smaller equilateral triangle.

Table 1.1: Table of Values For the Koch Snowflake

Stage

Nn

Ln

Pn

An

0

3

1

3

0.433

1

12

image00.png

4

0.57735

2

48

image01.png

image02.png

0.6415

3

192

image03.png

image04.png

0.67

Finding the values of N was relatively easy as this was just a process of counting the sides. It was discovered that the Number of sides was

...read more.

Middle

. The stage 0 was equal to image07.png

 as the side length in stage 0 is equal to 1. Every stage after that was equal to the area of the previous iteration, + (Area of an added triangle) x (Number of added Triangles.).

It should be noted that the graph for Pn is divergent. The perimeter has an infinite value when n approaches infinity.

The graph of An is convergent.  A has a finite value when n approaches infinity.

Generalizations for behaviour within Graphs.

General Formula for Nn:

n

Nn

0

3

1

12

2

48

3

192

N

3(4)n

...read more.

Conclusion

ng" alt="image08.png" />

The general formula for length is one third of the previous length. The general formula is then 1/3n.

Verifications

N = 0        image09.png

 = image10.png

 = 1

N = 1        image11.png

 = image00.png

N = 2        image12.png

N = 3        image13.png

General Formula for Pn

n

Pn

0

3

1

4

2

5image00.png

3

7image01.png

n

image14.png

The general formula for perimeter is equal to the Nn multiplied by the Ln. 3(4)n x image08.png

 . The general formula is thenimage14.png

.

Verifications

N = 0         image15.png

N = 1         image16.png

N = 2        image17.png

N = 3        image18.png

...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. El copo de nieve de Koch

    la siguiente fase, gracias a la implementaci�n de este sistema y a la necesidad de crear un modelo matem�tico que nos proporcione la longitud de cada lado, surgi� el siguiente modelo, que c�mo antes mencione se toma la variaci�n de y se eleva a la n-�sima que ser� sustituida por

  2. Koch Snowflake

    Number of sides of the Koch Snowflake Iteration (n) Number of sides (Nn) 0 3 1 12 2 48 3 192 From the table above we can see that there is an increase in the number of sides at every stage by a factor of 4.

  1. Investigating the Koch Snowflake

    ((V3)/4), A1= [(V3)/4] + (4/9)0(1/3) ((V3)/4) A2= 10(V3)/27, A2= [(V3)/3] + ((V3)/27), A2= [(V3)/3] + (1/3) (4/9)1 ((V3)/4) A3= 94(V3)/243, A3= [10(V3)/27] + (4(V3)/243), A3= [10(V3)/27] + (1/3) (4/9)2 ((V3)/4) And hence we can see that the general expression is An = An-1 + (1/3)

  2. The Koch Snowflake

    To find the formula in this case we'll need to look at the similarities between each stage and perform some calculations as shown below: Area Stage 0 = . Area Stage 1 = + Area Stage 2 = + + Area Stage 3 = + + + An = An-1

  1. Math Portfolio - The Koch snowflake investigation.

    It is directly proportional. Generalizations and predictions As we know the nth term for Nn, Ln, Pn and An we can now predict the results for terms till n=6. n Nn Ln Pn An 0 3 1 3 0.433012701 1 12 4 0.577350269 2 48 0.641500298 3 192 0.670011422 4

  2. MAths HL portfolio Type 1 - Koch snowflake

    it is a geometric progression since every successive term after the 1st term is being multiplied by the common ratio 4. We can confirm this common ratio by taking two values from the table on the left. Ex. 1 N1 = 3 and N2 = 12 Ex.

  1. The Koch Snowflake

    Since when finding the perimeter the length of all sides are added together, I was able to realize that if I multiplied the answers I would get my answer for perimeter. In order to prove my idea I estimated my answer for n=2 which was,16-3..

  2. Math HL portfolio

    conjecture 1 = 1 = 1 = 1 Now lets use the example of a = 2 Like in Y=2x�-3x+1.2 (example of page 7) where a = 2, b = -3 , c = 1.2 D= - =( - )

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