Please show work, thanks! 1. Consider a sequence tan -1 defined by an+1 (an -1)3
ID: 2832893 • Letter: P
Question
Please show work, thanks!
1. Consider a sequence tan -1 defined by an+1 (an -1)3+ 1 for all n (a) What is the limit of tanlae 1 if a1 Justify your answer. (b) What is the limit of fan) 1 if a1 3 Justify your answer. n (c) What is the limit of n 1 ifa1 3? Justify your answer 2. Consider a sequence Han 1 defined by an 1-+ z for all n n+1 (a) Find the first 8 terms (with a calculator). (b) Show that 2 n--1 for all n. (c) Show that a2n 1 is convergent (d) show that fa2 1 is convergent. n+1 (e) Admitting that linn-40 +1 a2n 0, deduce limn- an.Explanation / Answer
1. Consider a1 = 1 + epsilon
Then, (a1 - 1)^3 + 1 = (1+epsilon-1)^3 + 1 = epsilon^3 + 1 = 1 + epsilon^3
Then, in general, an = 1 + epsilon^(3^(n-1))
Then, for a and b epsilon = -1/2 and 1/2, and lim n-> inf epsilon^(3^(n-1)) = 0, so the limit of the ai in both cases is 1.
For a1 = 3, epsilon = 2, and lim n-> inf epsilon^(3^(n-1)) = infinity, so lim an = infinity and the sequence diverges.
2.
The sequence is
a1 = 1. an+1 = 1 + an - an^2/3 (this should have been there rather than x)
Note that, if x = 1 + x - x^2/3, x^2/3 = 1, so x = sqrt(3). Thus, if the sequence converges, it converges to this.
We can show 3/2 < x2n < sqrt(3) < x2n+1 < 2, which is an even stronger statement.
Note that the derivative of 1 + x - x^2/3 = 1 - 2/3x
For n = 1, x2n = 1 + 1 - 1^2/3 = 2 - 1/3 = 5/3
This is less than sqrt(3)
a2n = 5/3 > 3/2
Note that, if x2n = 3/2, x2n+1 = 1 + 3/2 - (3/2)^2/3 = 5/2 - 3/4 = 7/4 < 2
If x2n+1 = 2, x2n+2 = 1 + 2 - 2^2/3 = 5/3 > 3/2
The derivative at 5/3 = 1 - 2/3(5/3) = 1 - 10/9= -1/9
f(sqrt(3)) = sqrt(3)
As f(5/3) = 1 + 5/3 - (5/3)^2/3 = 47/27 and f(2) = 1 + 2 -2^2/3 = 5/3, and f'(x) < 0 for x > 3/2, 47/27 < 2, and x2 = 5/3, and we have a fixed point at sqrt(3),
5/3 <= x2n < sqrt(3) < x2n+1 <= 47/27 for all n, which is an even tighter bound than
3/2 <= x2n < x2n+1 < 2 f'(47/27) = 1 - 2/3(47/27) = -13/81
Thus, in the interval, -13/81 < f'(x) < 0
c) let x2n = sqrt(3) - epsilon
Then, sqrt(3) > x2n+2 > sqrt(3) - epsilon*(13/81)^2 = sqrt(3) - 169/6561 epsilon.
Then, a2n converges monotonely to sqrt(3)
Similarly, for x2n+1 = sqrt(3) + epsilon, we have
sqrt(3) + 169/6561 epsilon > x2n+3 > sqrt(3)
Thus, x2n+1 converges monotonely to sqrt(3)
1.666666667 1.740740741 1.730681299 1.732262046 1.732018114 1.732055865 1.732050025 1.732050929