Investigating the Koch Snowflake
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Introduction
Mathematics HL Portfolio
Omar Nahhas.
Class 12 “IB” (C).
The Koch snowflake is also known as the Koch island, which was first described by Helge von Koch in 1904. Its building starts with an equilateral triangle, removing the inner third of each side, building another equilateral triangle with no base at the location where the side was removed, and then repeating the process indefinitely.
The first three stages are illustrated in the figure below
Each step in the process is the repeating of the previous step hence it is called iteration.
If we let Nn = the number of sides, ln = the length of a single side, Pn = the length of the perimeter, and An = the area of the snowflake, all at nth stage, we shall get the following table for the 1st three iterations.
Table no.1: the value of Nn, ln, Pn, and An, at the stage zero and the following three stages.
n | Nn | ln | Pn | An |
0 | 3 | 1 | 3 | (√3)/4 |
1 | 12 | 1/3 | 4 | (√3)/3 |
2 | 48 | 1/9 | 48/9 | 10(√3)/27 |
3 | 192 | 1/27 | 192/27 | 94(√3)/243 |
Note: assume that the initial side length is 1.
We can see from the above table that the number of sides isn multiplied by four at each iteration.
Middle
= 3(16), N3 = 3(64)
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N0= 3(4)0, N1=3(4)1, N2=3(4)2, N3= 3(4)3
Hence, Nn= 3(4)n
Length of a single side (ln)
As we proceed in each stage the length of any side is 1/3 the length of the side from the preceding stage. If we begin with an equilateral triangle with side length 1, then the length of a side in nth iterations is
ln = 1/ (3)n
For stage 0 to 3, ln = 1, 1/3, 1/9 and 1/27.
Or as in the above it can from the value of ln from the zero stage till the third stage
l0=1, l1=1/3, l2= 1/9, l3= 1/27
l0=1/ (3)0, l1=1/ (3)1, l2= 1/ (3)2, l3= 1/ (3)3
And hence ln = 1/ (3) n
Perimeter (Pn)
Since the lengths of every side in every iteration of the Koch Snowflake are the same, then perimeter is simply the number of sides multiplied by the length of a side
Pn = (Nn) (ln)
Pn = (3(4) n) (1/ (3) n)
For the nth stage.
Again, for the first 4 steps (0 to 3) the perimeter is 3, 4, 16/3, and 64/9.
As we can see, the perimeter increases by 4/3 times each iteration so we can rewrite the formula as
Pn = 3(4/3) n
Or it can be derived from the value of Pn from the zero stage till the third stage
P0= 3, P1= 4, P2= 48/9,P3= 192/27
P0
Conclusion
As n increases the area converges to a certain value where that the 1st six decimals are equal in the successive terms, and such patterns beings to appear at stage number seventeen where that A17= 0.691693219 and A18= 0.691693719, so we can see that An+1 = An to six places of decimals at n=17. For the other values as n get larger there is no value of n where that ln = ln+1 to six places of decimals.
The perimeter as n → ∞ the perimeter becomes very large going to infinity, and the area converges to 0.692820323 which is equal almost to (8/5) ((√3)/4) which is (8/5) the area of the original triangle at step 0.
Hereby the general expression for An will be proved by induction
A0= (√3)/4, A1= (√3)/3
An = An–1 + (1/3) (4/9) n–1((√3)/4) prove it is true for n=1
A1 = A1–1 + (1/3) (4/9) 1–1((√3)/4)
A1 = A0+ (1/3) (4/9)0((√3)/4)
A1 = (√3)/4+ ((√3)/12)
(√3)/3 = (4(√3)/12) = (√3)/3
Then Left hand side equals Right hand Side
Now assume it is true for n=k that is
Ak = Ak–1 + (1/3) (4/9) k–1((√3)/4)
Now we want to prove it is true for n = k+1
Ak+1 = A (k+1)–1 + (1/3) (4/9) (k+1)–1((√3)/4)
Ak+1 = A k+ (1/3) (4/9) k((√3)/4)
A1 = A0 + (1/3) (4/9)0((√3)/4)
(√3)/3 = [(√3)/4] + (1/3) ((√3)/4) = (4(√3)/12)
= (√3)/3
Since An is true for n=1, so it true for n=k, then it true for n= k+1, then is true for all values of n
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