- Level: International Baccalaureate
- Subject: Maths
- Word count: 1490
The Koch Snowflake
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Introduction
Math 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 | |||
Pn | 3 | 4 | ||
An |
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:
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 side
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.
Middle
Then we can put our data into the diagram below:
The length of perimeter
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 , obviously. So we can get the geometric formula:
Pn=3
To verify the formula, we put n=0, 1,2and 3 into the formula. So we get:
P0==3 P1==4 P2== P3==
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.
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.
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 = 2.32305710-8
P16 = 299.323656 A16 = 0.692820
Then we consider the expression for Pn which:
Pn=Pn-1
We can see that Pn is always increasing in the same ratio of. So the value of Pn increase to positive infinity as n→∞.
For the expression of An which is:
An= An-1+
According to the formula of sum of geometric series to infinity:
sn
If
Thus,
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 know
P(1) is true since
Assume that n=m is true for integer t≥1.
Consider n=m+1
Am+1 =
Prove
Thus P(m+1) is true when P(m) is true. Therefore P(n) is true for all integers m≥1.
Richard ouyang
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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