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# MAths HL portfolio Type 1 - Koch snowflake

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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. 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.

Middle

2 3 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.  .

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

Ex. 1   P1 = 3 and P2 = 4 Ex. 2   L2 = 4 and L3 =   The general formula can be verified by putting in values from the table    Using Graphamatica, I have plotted and drawn this Graph of n vs. Pn Scale

X-axis – 1 step = 0.5units

Y-axis – 1 step  = 0.1 units

Relationship between n and Area of the shape:

 AnArea of the shape n (stage) f(n) 0 1 2 3 We can see from the table on the left we can clearly see that the initial number a is  . 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  which turned out to be correct. After subtracting  and dividing the answers by  the remainder was  and so I multiplied it by the same factor.

General formula for An The general formula can be verified by putting in values from the table.      Stage 4 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    Stage (n)    4    Conclusion

is true,   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)  3. C2: f(x)=  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. From the graph we cqn see that as, Area of the shape

Formula 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 ∞ ,  becomes so small a value that the highlighted part of the equation becomes  equal to one.  Part 7

The following iterative formula was derived: - We can now substitute the general formula derived in this equation: - To prove this equation by induction...

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

Step 2: Assume statement true for n=k,        We have already proven the statement true for n=1 and since  , we can say the statement must be true for n=2, 3, 4,... n.

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

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