• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21

MAths HL portfolio Type 1 - Koch snowflake

Extracts from this document...

Introduction

Practice Maths Portfolio – Fractals

Introduction

A fractal is generally "a rough or fragmented geometric shape that can be split into parts, each of which is (at least approximately) a reduced-size copy of the whole”, a property called self-similarity. The term was coined by Benoît Mandelbrot in 1975 and was derived from the Latin fractus meaning "broken" or "fractured." A mathematical fractal is based on an equation that undergoes iteration, a form of feedback based on recursion.

In 1904, Helge Von Koch gave a more geometric definition of a similar function, which is now called the Koch snowflake.

One can imagine that it was created by starting with a line segment, then recursively altering each line segment as follows:

  1. Divide the line segment into three segments of equal length.
  2. Draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
  3. Remove the line segment that is the base of the triangle from step 2.

image00.jpg

In this Portfolio I am primarily going to see if there is any pattern in this fractal, and for that I am going to get the values for number of sides, length of each side, perimeter of the shape and the area of the shape.

...read more.

Middle

2

image65.png

3

image02.png

We can see from the table on the right that it is a geometric progression since every successive term after the 1st term is being multiplied by the common ratio.image66.pngimage66.png .

General formula for Pn

image68.png

image69.png

We can confirm this common ratio by taking two values from the table on the left.

Ex. 1   P1 = 3 and P2 = 4

image70.png

Ex. 2   L2 = 4 and L3 = image71.pngimage71.png

image72.png

The general formula can be verified by putting in values from the table

image73.png

image74.png

image75.pngimage75.png

Using Graphamatica, I have plotted and drawn this Graph of n vs. Pn

image76.png

Scale

X-axis – 1 step = 0.5units

Y-axis – 1 step  = 0.1 units

Relationship between n and Area of the shape:

An

Area of the shape

n (stage)

f(n)

0

image54.png

1

image67.png

2

image77.png

3

image10.png

We can see from the table on the left we can clearly see that the initial number a is image78.pngimage78.png. Now since the denominator goes from 4 to 3 and then goes on increasing in odd powers of three I hypothesised the common ratio to be image79.pngimage79.png which turned out to be correct. After subtracting image78.pngimage78.png and dividing the answers by image80.pngimage80.png the remainder was image81.pngimage81.png and so I multiplied it by the same factor.

General formula for An

image82.png

The general formula can be verified by putting in values from the table.

image58.png

image83.png

image84.png

image44.png

image85.png

image86.png

Stage 4

image87.png

For the diagram above I have used a software called Dr. Bill’s software of the Von Koch snowflake simulation which I found online.

To investigate what happens at stage four, we will have to apply all of the above equations and substitute n as 4. we will have to substitute in these for

image88.png

image89.png

image90.png

image91.png


Stage (n)

image92.png

image93.png

image94.png

image95.png

4

image96.png

image97.png

image98.png

image99.png

...read more.

Conclusion

 is true,

image13.pngimage17.pngimage17.png     is true,

Part 5.

A

B

C

D

E

1

Stage

No. of Sides

Length of a side

Perimeter

Area

2

0

3

1

3

0.4330127

3

1

12

0.333333333

4

0.5773503

4

2

48

0.111111111

5.333333333

0.6415003

5

3

192

0.037037037

7.111111111

0.6700114

6

4

768

0.012345679

9.481481481

0.6826830

7

5

3072

0.004115226

12.64197531

0.6883149

8

6

12288

0.001371742

16.85596708

0.6908179

9

7

49152

0.000457247

22.47462277

0.6919304

10

8

196608

0.000152416

29.96616369

0.6924248

11

9

786432

5.08053E-05

39.95488493

0.6926445

12

10

3145728

1.69351E-05

53.2731799

0.6927422

13

11

12582912

5.64503E-06

71.03090654

0.6927856

14

12

50331648

1.88168E-06

94.70787538

0.6928049

15

13

201326592

6.27225E-07

126.2771672

0.6928135

16

14

805306368

2.09075E-07

168.3695562

0.6928173

17

15

3221225472

6.96917E-08

224.4927416

0.6928190

18

16

12884901888

2.32306E-08

299.3236555

0.6928197

19

17

51539607552

7.74352E-09

399.0982074

0.6928201

20

18

2.06158E+11

2.58117E-09

532.1309432

0.6928202

21

19

8.24634E+11

8.60392E-10

709.5079242

0.6928203

22

20

3.29853E+12

2.86797E-10

946.0105656

0.6928203

To obtain the above table, various formulas were used in the spreadsheet software (Office ‘07).

These are as following: -

  1. A3: f(x)= A2+1
  2. B2: f(x)image18.pngimage18.png
  3. C2: f(x)=image19.pngimage19.png
  4. D2: f(x)=B2 x C2
  5. E2: f(x)=0.5*SIN(60*PI()/180)]         (=√3/4)                 (fixed)
  6. E3: f(x)=E2 +(B2*(0.5)*(C3^2)*SIN(60*PI()/180))

Part 6

The geometric pattern of the perimeter of the shape is a divergent series as the common ratio is > 1. Hence the perimeter at each stage increases exponentially. This is also proved by the graph.

image20.png

From the graph we cqn see that as,

image21.png

Area of the shape

Formula

image22.png

For the geometric progression in the formula above we can see that the common ratio is lesser than 1 and so this can be called a convergent series. When n increases towards ∞ , image24.pngimage24.png becomes so small a value that the highlighted part of the equation becomes  equal to one.

image25.png

image26.png

Part 7

The following iterative formula was derived: -

image27.png

We can now substitute the general formula derived in this equation: -

image28.png

To prove this equation by induction...

Step 1: Assume statement true for n=1,

image29.png

image30.png

image31.png

Therefore, the statement is proven true for n=1.

Step 2: Assume statement true for n=k,

image32.png

image34.png

image35.png

image36.png

image37.png

image38.png

image31.png

image39.png

We have already proven the statement true for n=1 and since image40.pngimage40.png, we can say the statement must be true for n=2, 3, 4,... n.

...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. Math IA- Type 1 The Segments of a Polygon

    The conjecture has been proved on the following page. In to order to able to test the validity of this conjecture, another triangle was produced using the geometer's sketch pad and the conjecture is validated if the ratio produced from GSP matches the value of the ratio provided by the conjecture.

  2. The Koch Snowflake

    since the gradient of the graph is -3 (1/3) and the first term is 1 In graph 3, which shows the perimeter of the different stages, the formula for calculating the perimeter at the nth stage is Nn x Ln .

  1. Math Portfolio Type II Gold Medal heights

    Using technology a line of best fit of exponential nature was generated. The default equation of exponential function is however the parameters of the logarithmic function were already ladled so they have to be renamed in order to avoid confusion.

  2. Math Portfolio - The Koch snowflake investigation.

    n Ln 0 1 1 ? 1 = 2 ? = 3 ? = From the table above, I can determine the nth term as: From the graph above, we can observe that the length of the side decreases gradually as the number of sides increase.

  1. Investigating the Koch Snowflake

    3, P1= 4, P2= 48/9, P3= 192/27 P0= 3, P1= 3(4/3), P2= 3 (16/9), P3= 3 (64/27) P0= 3 (4/3)0, P1= 3(4/3)1, P2= 3(4/3)2, P3= 3(4/3)3 And hence Pn= 3(4/3) n The area of the snowflake (An) In each iteration we take the area of the previous snowflake or shape,

  2. Shadow Functions Maths IB HL Portfolio

    function: Where Giving Allowing us to express in terms of and . I will now investigate similar cubic functions to see if the relationship between and remains the same. Let us consider and Graphed: The shadow generating function is given by the equation , Expressing in terms of and : Which is the same relationship as with our previous cubic.

  1. Math IA Type 1 Circles. The aim of this task is to investigate ...

    Figure 12. At this point it can be confusing, since the square of OP and two times OP are the same in the case of r = 2. Therefore another value must be tested. In the next case r= 6 and OP = 3.

  2. Maths HL Kochs Snowflake

    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 The general formula for Nn is 3(4)n. This is because the number of sides is equal to 4 times the previous iterations number of sides.

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