Download Midterm Exam with Solution on Analysis | MATH 131A and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Math 131a Midterm 1 Lecture 2 Spring 2009 Name: Instructions: • There are 4 problems. Make sure you are not missing any pages. • Unless stated otherwise (or unless it trivializes the problem), you may use without proof anything proven in the sections of the book covered by this test (excluding the exercises). • Give complete, convincing, and clear answers (or points will be deducted). • No calculators, books, or notes are allowed. • Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. Question Points Score 1 10 2 10 3 10 4 10 Total: 40 1. (10 points) Suppose that f : C → D is an injective function, and that A,B ⊂ C. Prove that f(A ∩ B) = f(A) ∩ f(B). (Here f(A) = {f(x) : x ∈ A} and similarly for B and A ∩B.) Solution: First, we show f(A∩B) ⊂ f(A)∩ f(B). Suppose that y ∈ f(A∩B). Then by definition, there is an x ∈ A ∩ B with f(x) = y. Since x ∈ A, we then have y ∈ f(A). Since x ∈ B, we have y ∈ f(B). Thus y ∈ f(A) ∩ f(B). We finish by showing that f(A) ∩ f(B) ⊂ f(A ∩ B). Suppose that y ∈ f(A) ∩ f(B). Since y ∈ f(A), there is an x1 ∈ A with f(x1) = y. Since y ∈ f(B) there is an x2 ∈ B with f(x2) = y. Since f is injective and f(x1) = f(x2), we have x1 = x2 and hence x1 ∈ A ∩B. It follows that y = f(x1) ∈ f(A ∩B). 4. (10 points) Let (bn) ∞ n=1 be a real-valued sequence, and b ∈ R with limn→∞ bn = b. Sup- pose that b > 0 and that for every n, bn > 0. Show that there is a real number m > 0 such that for every n ∈ N, bn ≥ m. (hint: you can prove this using limit theorems, or by using the definition of a limit.) Solution 1: Since limn→∞ bn = b, and b 6= 0, and for every n, bn 6= 0, we may apply a limit theorem to see that limn→∞ 1 bn = 1 b . Another limit theorem tells us that convergent sequences are bounded, and so there is a real number M > 0 such that 1 bn ≤M for every n ∈ N. Since bn > 0 for every n, we conclude that 0 < 1 M ≤ bn for every n. Solution 2: Since b/2 > 0 and limn→∞ bn = b, there is an N ∈ N such that |bn − b| < b/2 for every n ≥ N . Then for every n ≥ N , we have bn > b2 (since b = |b| ≤ |b− bn|+ |bn| < b/2+ bn). Since, for every n we have bn 6= 0, it follows that m = min(b1, . . . , bN , b2) > 0 and bn ≥ m for every n. Extra Scratch Paper:
