Download Exam 1 Questions with Solution - Analysis 1 | MATH 3001 and more Exams Advanced Calculus in PDF only on Docsity! Math 3001 Exam 1 Solutions 1. (20 points) Prove that if lim n→∞ sn = 3, then lim n→∞ s2n = 9. Use only the definition of limit. Solution: We have |s2n − 9| = |(sn + 3)(sn − 3)| = |sn + 3||sn − 3|. Since limn→∞ sn = 3, there is an N1 such that n > N1 implies |sn − 3| < 1, i.e., that 2 < sn < 4. Hence for n > N1 we have |s2n − 9| < 7|sn − 3|. Now let ε > 0 be arbitrary. Choose N2 so that n > N2 implies |sn − 3| < ε/7. Set N = max{N1, N2}. Then n > N implies |s2n − 9| < 7|sn − 3| < ε. Since we can find such an N for any ε > 0, we have proved that limn→∞ s 2 n = 9. 2. (20 points) Let A and B be two subsets of R which are bounded above. (a) (5 points) How do you know that L = supA and M = supB exist in R? Solution: We know that supA and supB exist since these sets are bounded above and R is complete. (b) (10 points) Let C = {x + y |x ∈ A, y ∈ B}. Prove that supC exists and that supC ≤ L + M . Solution: It is sufficient to show that supA+supB is an upper bound for C. This is obvious, because if x+y ∈ C where x ∈ A and y ∈ B, then x+y ≤ supA+supB because supA and supB are upper bounds. (c) (10 points) Prove that supC ≥ L + M . Solution: Assume not: then supC < supA + supB. Let ε = supA + supB − supC > 0. Since supA is the supremum of A, we know supA − ε/2 is not an upper bound of A, so there is an x ∈ A such that x > supA − ε/2. Similarly there is a y ∈ B such that y > supB − ε/2. Thus x + y > supA + supB − ε = supC, a contradiction since x + y ∈ C. 3. Consider the field F of rational functions anx n + · · ·+ a0 bkxk + · · ·+ b0 , where n and k are nonneg- ative integers, and all the coefficients ai and bj are real numbers with bk 6= 0. Here x is not a number but just a placeholding parameter. Recall that this is an ordered field under the usual rules for addition and multiplication of functions, and with the order defined by anx n + · · ·+ a0 bkxk + · · ·+ b0 > 0⇐⇒ anbk > 0. 1
