Practice Final Exam with Solution - Real Analysis I | MATH 323, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Practice Final Exam with Solution - Real Analysis I | MATH 323 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! December 10, 2001 Math 323 — Practice Final Exam 1. (20 points) Suppose f : [a, b]→ R. (a) Prove that, if a partition P of [a, b] is a refinement of a partition Q, then U(f, P ) ≤ U(f,Q). (For simplicity, look at one subinterval [xi−1, xi] in Q and suppose there is only one point x∗i in P with xi−1 < x ∗ i < xi.) (b) Prove that, for any partitions P,Q of [a, b], L(f, P ) ≤ U(f,Q). 2. (16 points) Let f : [a, b] → R be decreasing. Prove that f is integrable on [a, b]. (Hint: Use a partition with subintervals of equal length.) 3. (20 points) The Fundamental Theorem of Calculus says that, if f is continuous on [a, b], then for any antiderivative F of f , ∫ b a f = F (b) − F (a). This equation clearly holds for the particular antiderivative G(x) = ∫ x a f (because G(b)−G(a) = ∫ b a f − 0); but why does it hold for any antiderivative F? 4. (20 points) Define f : [0, 1]→ R by f(x) = 1 if x = 1− 1n+1 for some n ∈ N and 0 otherwise. Prove that f is integrable on [0, 1] and find the value of ∫ 1 0 f . (Hint: Given ε > 0, suppose that only for n = 1, 2, . . . , N is 1n+1 > ε 2 . For each such n, let x2n−1 = (1 − 1 n+1) − ε 4N and x2n = (1 − 1n+1) + ε 4N ; and let x0 = 0, x2N+1 = 1 − ε 2 , and x2n+2 = 1. Then why is U(f, P ) = ε?) 5. (24 points) Decide which of the following conjectures are true and supply a short proof. For those that are not true, give a counterexample. (a) If f is discontinuous at a point, then F (x) = ∫ x a f is not differentiable at that point. (b) If ∫ b a f > 0, then there is a subinterval [c, d] of [a, b] and a δ > 0 such that f(x) ≥ δ for all x ∈ [c, d]. (c) A function that is discontinuous at infinitely many points is not integrable, even if it is bounded. (d) Of the families of functions on a closed interval [a, b], the integrable functions are the largest, containing (as a subset) the continuous functions, which in turn contain the differentiable functions. Math 323 — Solutions to Practice Final Exam 1. (a) We want to show that each term sup{f(x) : x ∈ [xi−1, xi}(xi − xi−1) in the sum U(f,Q) is replaced by one or more terms in U(f, P ) that add up to no more than that term. As suggested in the problem, we suppose for simplicity that there is only one point x∗i of P that is strictly between xi−1 and xi. Then sup{f(x) : x ∈ [xi−1, x∗i ]} ≤ sup{f(x) : x ∈ [xi−1, xi]} , sup{f(x) : x ∈ [x∗i , xi]} ≤ sup{f(x) : x ∈ [xi−1, xi]} and (x∗i − xi−1) + (xi − x∗i ) = xi − xi−1, so sup{f(x) : x ∈ [xi−1, x∗i ]}(x∗i − xi−1) + sup{f(x) : x ∈ [x∗i , xi]}(xi − x∗i ) ≤ sup{f(x) : x ∈ [xi−1, xi]}(xi − xi−1) Because U(f, P ) is the sum of all terms like those on the left and U(f,Q) is the sum of terms like those on the right, we get U(f, P ) ≤ U(f,Q). (b) It is clear that, for any partition R, we have L(f,R) ≤ U(f,R), because inf{f(x) : x ∈ [xi−1, xi]} ≤ sup{f(x) : x ∈ [xi−1, xi]} for each i. Let R = P ∪Q, a common refinement of P and Q. Then by (a) and the corresponding reverse inequality on lower sums, L(f, P ) ≤ L(f,R) ≤ U(f,R) ≤ U(f,Q) , as required. 2. Let P = {xi = a + (i(b − a)/n) : i = 0, 1, 2, . . . , n}, the partition of [a, b] into n subintervals of equal length xi−xi−1 = (b− a)/n. Then because f is decreasing, it attains its inf and sup at the right and left ends respectively of each subinterval [xi−1, xi], so that U(f, P ) = n∑ i=1 f(xi−1)(xi − xi−1) = b− a n n∑ i=1 f(xi−1) and similarly L(f, P ) = b− a n n∑ i=1 f(xi) . Thus, their difference has a “telescoping sum”: U(f, P )− L(f, P ) = b− a n n∑ i=1 (f(xi−1)− f(xi)) = b− a n (f(x0)− f(xn)) = b− a n (f(a)− f(b)) . Because b, a, f(b), f(a) are constants, we see that we can make this difference as small as desired by dividing [a, b] into as large a number of (equal-length) subintervals as needed; so f is integrable.