The Koch snowflake is an infinitely jagged "fractal" curve obtained as a limit of polygonal curves (it is continuous but has no tangent line at any point). Begin with and equilateral Triangle (stage 0) and produce stage 1 by replacing each edge with 4 edges one third of the length, arranged in figure 7 continue the process: at the nth stage replace each edge with four edge one third the length.
a.) Show that the perimeter Pn of the polygon at the nth stage satisfies
Pn = 4/3P_(n-1). Prove that lim(n to infinity) Pn = infinity. The snowflake has infinite length.
b.) Let A_0 be the area of the original equilateral triangle. Show that (3)4^(n-1) new triangles are added at the nth stage, each with the area A_0/9^n (for ≥ n). Show that the toal area of the Koch Snowflake is 8/5A_0.
a.) Show that the perimeter Pn of the polygon at the nth stage satisfies
Pn = 4/3P_(n-1). Prove that lim(n to infinity) Pn = infinity. The snowflake has infinite length.
b.) Let A_0 be the area of the original equilateral triangle. Show that (3)4^(n-1) new triangles are added at the nth stage, each with the area A_0/9^n (for ≥ n). Show that the toal area of the Koch Snowflake is 8/5A_0.
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a) Without loss of generality suppose that the initial equilateral triangle has sides of length 1.
Inductively:
After the first stage, each side is trisected, now yielding 4 edges with lengths 1/3 apiece.
So, P1 = 3 * (4/3) = (4/3) P0.
Given stage k, for stage k+1, we replace each edge with 4 edges one third of the length.
Hence,
P(k+1) = (4/3) * Pk
...........= (4/3) * (4/3) P(k-1) = (4/3)^2 P(k-1)
...........= (4/3)^2 * (4/3) P(k-2) = (4/3)^3 P(k-2)
...
...........= (4/3)^k P0.
Letting k go to infinity, we see that P(k+1) also goes to infinity, because 4/3 > 1.
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b) This is done similarly.
Each time, we add a new triangle onto a preexisting edge in its middle third.
Stage 0: 1 triangle, with 3 = 3 * 4^0 edges
Stage 1: 3 new triangles, now the shape has 3 * 4 edges
Stage 2: 3 * 4 new triangles, now the shape has 3 * 4^2 edges
Stage 3: 3 * 4^2 new triangles, now the shape has 3 * 4^3 edges
...
==> Stage k: 3 * 4^(k-1) new triangles
Note that these new triangles are replicas of the previous triangles but at 1/3 scale;
so their base and heights are 1/3 of what they were before.
Hence, the extra factor of 1/3 * 1/3 = 1/9 in the area formula.
Keeping track of the new triangles from each stage, the area formula for the snowflake becomes
A0 + 3 A0/9 + 3 * 4A0/9^2 + 3 * 4^2 A0/9^3 + ... + 3 * 4^(k-1) A0/9^k + ...
= (3 A0/9) * {3 + [1 + 4/9 + (4/9)^2 + ...]}
= (A0/3) * {3 + 1/(1 - 4/9)}, by geometric series
= (8/5) A0.
I hope this helps!
Inductively:
After the first stage, each side is trisected, now yielding 4 edges with lengths 1/3 apiece.
So, P1 = 3 * (4/3) = (4/3) P0.
Given stage k, for stage k+1, we replace each edge with 4 edges one third of the length.
Hence,
P(k+1) = (4/3) * Pk
...........= (4/3) * (4/3) P(k-1) = (4/3)^2 P(k-1)
...........= (4/3)^2 * (4/3) P(k-2) = (4/3)^3 P(k-2)
...
...........= (4/3)^k P0.
Letting k go to infinity, we see that P(k+1) also goes to infinity, because 4/3 > 1.
---------------
b) This is done similarly.
Each time, we add a new triangle onto a preexisting edge in its middle third.
Stage 0: 1 triangle, with 3 = 3 * 4^0 edges
Stage 1: 3 new triangles, now the shape has 3 * 4 edges
Stage 2: 3 * 4 new triangles, now the shape has 3 * 4^2 edges
Stage 3: 3 * 4^2 new triangles, now the shape has 3 * 4^3 edges
...
==> Stage k: 3 * 4^(k-1) new triangles
Note that these new triangles are replicas of the previous triangles but at 1/3 scale;
so their base and heights are 1/3 of what they were before.
Hence, the extra factor of 1/3 * 1/3 = 1/9 in the area formula.
Keeping track of the new triangles from each stage, the area formula for the snowflake becomes
A0 + 3 A0/9 + 3 * 4A0/9^2 + 3 * 4^2 A0/9^3 + ... + 3 * 4^(k-1) A0/9^k + ...
= (3 A0/9) * {3 + [1 + 4/9 + (4/9)^2 + ...]}
= (A0/3) * {3 + 1/(1 - 4/9)}, by geometric series
= (8/5) A0.
I hope this helps!