Could someone be so kind to help me out with this project! it\'s has mutiple par
ID: 3122350 • Letter: C
Question
Could someone be so kind to help me out with this project! it's has mutiple parts bu i dont really understand how to complete it. i have tried and it doesnt seem right so i would love to see another persons take on it.
Construct a sequence of figures, To, Th,T2, Tn,... (see below) in the following way. To is an equilateral triangle of side length one; T is obtained from To by replacing the middle third of each edge of To by an outward equilateral triangle whose side length is of the side length of the triangle To. Now, T is obtained from Th by replacing the middle third of each edge of Ti by an outward facing triangle who side length is i of the side length of the triangles in T1 and so on. (The limit of this sequence of figures is called the "Koch Snowflake")Explanation / Answer
a. Length of one side of T_0 =L_0 1.
Length of one side of T_1 = L_1 (1/3)*1 = 1/3
Length of one side of T_2 = L_2 = (1/3)*L_1 = 1/9
Length of one side of T_3 = L_3 = (1/3)*L_2 =(1/3)(1/9) = 1/27
Similarly L_4= 1/81 and and L_5 = 1/243.
b. Clearly as a function of 'n' L_n = 1/3^n. As a recursive definition L_0 = 1; and L_n = L_{n-1}*(1/3), for n>0.
c. Notice that when we replace the middle third of side with an outward equilateral triangle we essentially are introducing three more sides, the two sides of the outward triangle and one because of the existing side getting divided into two. Thus the contribution of one side in T_n towards the number of sides in T_{n+1} is 2+1+1=4. Thus we see that
S_0 = 3, S_1 = 12, S_2 = 48, S_3 = 192, S_4 = 768.
d. From above we see that recursively S_0 = 3; S_{n} = 4*S_{n-1} for n>0.
As a function of n we have S_n = 3*4^n.
e. P_n = S_n*L_n. Thus P_0 = 3; P_1 = 12*1/3 = 4; P_2 = 48*(1/9) = 16/3; P_3 = 192*(1/27) = 64/9; P_4 = 768*(1/81) = 256/27.
f. P_n = S_n*L_n = 3*4^n*(1/3^n) = 4^n/3^{n-1}. As recursive definition P_0 = 3 and P_n = P_{n-1}*(4/3), n>0.
g. Let us denote the limit of P_n as P_n tends to infinity as Lim P_n then, Lim P_n = Lim 4^n/3^{n-1} = Lim 4 (4/3)^{n-1} = 4 Lim (4/3)^{n-1} = infinity. Thus Koch snowflake has no finite perimeter.
h. To each side a new triangle is added at each stage of iteration. And the area of each such triangle is (1/9) times of area of triangles added the previous time (because side is 1/3 times of previous step). Let us denote area of T_n by A_n and for convenience we let A_0 = A_0.
Then the area of each new triangle added at 1st stage is A_0/9. At 2nd stage this is A_1/9 = (A_0/9)/9= A_0/81. Continuing this process we see that the area added to the figure due to addition of a single triangle is A_0/9^n. But there are S_{n-1} such triangles added so total area added is S_{n-1}*A_0/9^n = A_0(3*4^{n-1}/9^n) = A_0 (3/4)(4/9)^n.
So A_1 = A_0+A_0 (3/4)(4/9)^n = A_0(1+ (3/4)(4/9)); A_2 = A_1 + A_0 (3/4)(4/9)^2 = A_0 (1+(3/4)(4/9)+(3/4)(4/9)^2)
Notice that {A_n} is a monotonically increasing sequence. Also notice that 1+(3/4)(4/9)+(3/4)(4/9)^2 + ...+ (3/4)(4/9)^n+... converges (because it is a geometric series with common difference less than 1 but greater than -1). So the sequence {A_n} is bounded above. Bounded monotone sequences converge, so Koch Snowflake as finite area.
i. The area of Koch Snowflake is Lim A_n = A_0 (1+(3/4)(4/9)+(3/4)(4/9)^2 + ...+ (3/4)(4/9)^n+... ) =A_0 (1+(3/4)(4/9+(4/9)^2 + ...+ (4/9)^n+... ) = A_0( 1 + (3/4) ((4/9) (1/(1-4/9))) = A_0( 1 + (3/4) ((4/9) (9/5)) = A_0( 1 + (3/4) (4/5) ) = A_0 (1+3/5) = A_0(8/5).
Thus area of Koch snowflake is (8/5)A_0. Now A_0 is (root(3)/4) (1^2) = (root(3)/4). Thus area of Koch snowflake is 2*root(3)/5.