Solution for Final Exam - Elementary Real Analysis | MATH 444, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Solution for Final Exam - Elementary Real Analysis | MATH 444 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! MATH 444: SOLUTIONS FOR FINAL 1 (10 points): Prove that, for any positive integer n, ∑n k=1(k − 1)/k! = 1 − 1/n!. Denote the statement ∑n k=1(k − 1)/k! = 1 − 1/n! by P (n). We show by induction that P (n) is true for any positive integer n. The basic step consists of verifying P (1), which is clearly true: (1 − 1)/1! = 1 − 1/1!. To perform the inductive step, we have to show that P (n) implies P (n + 1): in other words, if ∑n k=1(k − 1)/k! = 1 − 1/n!, then ∑n+1 k=1(k − 1)/k! = 1 − 1/(n + 1)!. By the induction hypothesis, n+1 ∑ k=1 k − 1 k! = n ∑ k=1 k − 1 k! + (n + 1) − 1 (n + 1)! = 1 − 1 n! + n + 1 (n + 1)! − 1 (n + 1)! = 1 − 1 n! + 1 n! − 1 (n + 1)! = 1 − 1 (n + 1)! , as desired. 2 (10 points): For x ∈ R, let F (x) = ∫ x3 1 sin(t2) dt. Compute F ′(2). Let G(u) = ∫ u 1 sin(t2) dt, and φ(x) = x3. Then F = G ◦ φ. Moreover, φ′(x) = 3x2, and, by Fundamental Theorem of Calculus, G′(u) = sin u2. By the Chain Rule, F ′(x) = G′(φ(x))φ′(x) = 3x2 sin x6. In particular, F ′(2) = 12 sin 64. 3 (14 points): Compute the following limits. (a) lim x→1 x1/3 − 1√ x − 1 . Hint. a3 − b3 = (a − b)(a2 + ab + b2). Multiplying and dividing by the “conjugate,” we obtain: x1/3 − 1√ x − 1 = x1/3 − 1√ x − 1 · x2/3 + x1/3 + 1 x2/3 + x1/3 + 1 · √ x + 1√ x + 1 = (x1/3 − 1)(x2/3 + x1/3 + 1) ( √ x − 1)(√x + 1) · √ x + 1 x2/3 + x1/3 + 1 = x − 1 x − 1 · √ x + 1 x2/3 + x1/3 + 1 , hence lim x→1 x1/3 − 1√ x − 1 = limx→1 √ x + 1 x2/3 + x1/3 + 1 = 2 3 . (b) lim n!/nn. lim n!/nn = 0. To see this, let k = ⌊n/2⌋ (here, ⌊x⌋ denotes the integer part of x, that is, the largest integer not exceeding x). Then 0 < n! nn = k! nk · (k + 1) . . . n nn−k ≤ k k nk ≤ (1 2 )k < (1 2 )n/2−1 = 2 √ 1 2 n . 1 2 MATH 444: SOLUTIONS FOR FINAL However, lim an = 0 for any a ∈ (0, 1). Thus, by Squeeze Theorem, lim n!/nn = 0. Alternative solution. Let xn = n!/n n. Then xn > 0, and xn+1 xn = (n + 1)! n! · n n (n + 1)n+1 = ( n n + 1 )n . We know (see Section 3.3 of the textbook) that lim(1 + 1/n)n = e ∈ (5/2, 3), hence lim xn+1/xn < 1. Thus, by Theorem 3.2.11, lim xn = 0. 4 (14 points): For x ∈ R, define f(x) = { x3 sin(1/x3) x 6= 0 0 x = 0 . (a) Is f differentiable at 0? If yes, compute the derivative. f is differentiable at 0, and f ′(0) = 0. Indeed, we have to show that limx→0 f(x)/x exists, and equals 0. But | sin θ| ≤ 1 for every θ ∈ R, hence |f(x)| ≤ |x|3 for any x 6= 0. Therefore, |f(x)/x| ≤ x2 for x 6= 0. As limx→0 x2 = 0, Squeeze Theorem implies limx→0f(x)/x = 0. (b) Is f Lipschitz on [−1, 1]? Recall that a function g is called Lipschitz on a set S if there exists a constant K s.t. |g(x) − g(y)| ≤ K|x − y| for any x, y ∈ S. f is not Lipschitz on [−1, 1]. Indeed, if |f(x)−f(y)| ≤ K|x−y| for any x, y ∈ (a, b), and f is differentiable at c ∈ (a, b), then |f ′(c)| ≤ K. To see this, recall that f ′(x) = limx→c(f(x) − f(c))/(x − c), and |f(x) − f(c)|/|x − c| ≤ K (actually, we mentioned this in class). However, for x 6= 0, f ′(x) = 3x2 sin(1/x3)−3 cos(1/x3)/x. In particular, |f ′((nπ)−1/3)| = 3(nπ)−1/3 for any n ∈ N. As there is no K ∈ R s.t. 3(nπ)−1/3 ≤ K for each n, the function f cannot be Lipschitz. 5 (10 points): Prove that x − y ≤ tanx − tan y ≤ 4(x − y) whenever −π/3 ≤ y < x ≤ π/3. Let f(t) = tan t. Then f ′(t) = sec2 t = 1/ cos2 t. By the Mean Value Theorem, for x and y as above there exists c ∈ (y, x) ⊂ (−π/3, π/3) s.t. f(x)− f(y) = f ′(c)(x− y). Recall that cos t > 0 for t ∈ (−π/2, π/2), and cos′ t = − sin t. Therefore, cos increases on [−π/3, 0], and decreases on [0, π/3]. Thus, for any c ∈ (−π/3, π/3), cos c ≥ min{cos(−π/3), cos π/3} = 1/2, hence sec2 c ≤ 4. Furthermore, | cos c| ≤ 1, hence sec2 c ≥ 1. Thus, 1 ≤ f ′(c) < 4. We complete the proof by combining this inequality with f(x) − f(y) = f ′(c)(x − y). 6 (10 points): Prove that, for any x > 0, 1 + x 3 − x 2 9 ≤ (1 + x)1/3 ≤ 1 + x 3 − x 2 9(1 + x)5/3 .