Final Exam Solutions for Elementary Analysis | MATH 315, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Final Exam Solutions for Elementary Analysis | MATH 315 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Final Exam — Solutions Math 315 March 16, 2005 1. [60 points] (a) Determine the following limit and indicate how you determined it: lim n→∞ (5n2)1/n = (lim 51/n)(lim n1/n)2 = 1 · 12 = 1. (b) Determine whether or not the following series converges and justify your conclusion: ∞∑ n=1 ( 2 cos n n )n α = lim 2 cos n n = 0, so by the Root Test it converges. (c) Of course, no subsequential limit of a sequence (sn) is less than lim inf sn. In particular, explain that if lim inf sn = 2, then 0 can not be a subsequential limit of (sn). Since 1 < 2 = lim inf sn, there exists some tail with each term greater than 1, so the are at most finitely many sn < 1. Thus, 0 can not be an ssl. (d) Find the interval of convergence of the following power series and justify your answer: ∞∑ n=1 2n√ n xn Here β = lim 2n+1√ n + 1 √ n 2n = 2 lim √ n n + 1 = 2. So R = 1/2. At x = −1/2, ∑ (−1)n/ √ n converges by the Alternating Series Test, but at x = 1/2, the series ∑ 1/ √ n diverges (by the p-Test), so the I of C is [−1/2, 1/2). (e) Determine whether or not the series ∑ xn/5n converges uniformly on [−2, 2]. Explain. The geometric series ∑ (3/5)n converges since 0 < 3/5 < 1. Then since |xn/5n| < (3/5)n for all x ∈ [−2, 2] and all n ∈ lN, the power series ∑ xn/5n converges uniformly on [−2, 2] by the M - test. 2. [30 points] Let (sn) be a sequence of real numbers. (a) Define what it means for (sn) to converge to some real number s. For each ε > 0, ∃ N such that n > N ⇒ |sn − s| < ε. (b) Assume now that (sn) is bounded and monotone increasing. Using your answer to part (a) prove that (sn) does converge to some real number s. By Completeness s = sup{sn : n ∈ lN} is in IR. Claim that lim sn = s. Let ε > 0. Then there is some N ∈ lN with s− ε < sN ≤ s. So n > N ⇒ s− ε < sN ≤ sn ≤ s < s + ε, so |sn − s| < ε. 3. [30 points] (a) Define what it means for a function f on IR to be continuous at x = 5. For every sequence (xn) in IR if lim xn = 5, then lim f(xn) = f(5). (b) If f is continuous at x = 5, then using induction that fn is continuous at x = 5 for all n ∈ lN. [Recall that fn is defined recursively by f1 = f and fn+1 = f1fn for all n ≥ 1.] (B) f1 = f is continuous at x = 5 by hypothesis. (IS) Let n ≥ 1 and assume that fn is continuous at x = 5. Then for every sequence (xk) in IR if lim xk = 5, then lim k→∞ fn+1(xk) = lim k→∞ f1(xk) lim k→∞ fn(xk) (IH) = f1(5)fn(5) = fn+1(5), so fn+1 is continuous at x = 5.