Guidelines on nicotine dose selection for in vivo research

Solutions to Introduction to Analysis Edward D. Gaughan March 3, 2013 Notation Symbol Ac f ∼g N LHS RHS incr. decr. cts. diff. intgbl. lg(x) ⌈x⌉ an → A lim an IVT H MVT IFT Mi mi R(x; [a, b]) U (P, f ) L(P, f ) µ(P ) S(P, P f) P an n=n0 an cgt ±ǫ a=b Meaning X \ A, where X is the universal set (assumed to be R unless otherwise stated) f tends asymptotically to g Natural numbers (symbolized as J in the book) Left-hand side Right-hand side Increasing Decreasing Continuous Differentiable Integrable log2 (x) If if a < x ≤ a + 1 where a is a positive integer, then, ⌈x⌉ = a + 1. lim an = A. n→∞ lim an n→∞ Intermediate Value Theorem l’Hospital’s Rule Mean Value Theorem Inverse Funcion Theorem Mi = supx∈(xi−1 ,xi ) f (x). (Used in tandem with a function f and partition P ). Similar to Mi , but replace sup with inf. The set of Riemann-integrable functions on the interval [a, b]. Shortened to R(x) if interval is apparent. P n If P is a partition of [a, b], then U (P, f ) = i=0 Mi (xi − xi−1 ). Similar to U (P, f ), but replace Mi with mi sup|xk − xk+1 | for partition P0 and refinements {Pn }. Called the mesh of P . k≤n Pn S(P, f ) = i=1 f (ti )(xi − xi−1 ), where ti ∈ (xi−1 , xi ). (Riemann sum) Lower bound unimportant, upper bound ∞. (Only intereseted in convergence.) Upper bound assumed to be ∞. Convergent. b−ǫ<a<b+ǫ 1 Contents 0 Prerequisites 0.1 Sets . . . . . . . . . . . . . . . . . . . 0.2 Relations and Functions . . . . . . . . 0.3 Mathematical Induction and Recursion 0.4 Equivalent and Countable Sets . . . . 0.5 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4 6 7 9 11 1 Sequences 1.1 Sequences and Convergence . . . . . . . 1.2 Cauchy Sequences . . . . . . . . . . . . 1.3 Arithmetic Operations on Sequences . . 1.4 Subsequences and Monotone Sequences Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 13 15 17 19 22 2 Limits of Functions 2.1 Definition of the Limit of a Function 2.2 Limits of Functions and Sequences . 2.3 Algebra of Limits . . . . . . . . . . . 2.4 Limits of Monotone Functions . . . . Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 23 24 26 27 28 . . . . . . Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 30 32 34 38 39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 41 44 45 48 49 . . . . . . . . . . 3 Continuity 3.1 Continuity of a Function at a Point . . . . . . . . 3.2 Algebra of Continuous Functions . . . . . . . . . 3.3 Uniform Continuity: Open, Closed, and Compact 3.4 Properties of Continuous Functions . . . . . . . . Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . 4 Differentiation 4.1 The Derivative of a Function . . . . . . . . . . . . . 4.2 The Algebra of Derivatives . . . . . . . . . . . . . . 4.3 Rolle’s Theorem and the Mean Value Theorem . . . 4.4 L’Hospital’s Rule and the Inverse-Function Theorem Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . 2 . . . . . 3 CONTENTS 5 The Riemann Integral 5.1 The Riemann Integral . . . . . . . . . . . . . . 5.2 Classes of Integrable Functions . . . . . . . . . 5.3 Riemann Sums . . . . . . . . . . . . . . . . . . 5.4 The Fundamental Theorem of Integral Calculus 5.5 Algebra of Integrable Functions . . . . . . . . . 5.6 Derivatives of Integrals . . . . . . . . . . . . . . 5.7 Mean-Value and Change-of-Variable Theorems Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 52 53 54 56 56 58 59 60 6 Infinite Series 6.1 Convergence of Infinite Series . . . . . . . . . . . 6.2 Absolute Convergence and the Comparison Test 6.3 Ratio and Root Tests . . . . . . . . . . . . . . . 6.4 Conditional Convergence . . . . . . . . . . . . . . 6.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 63 65 66 69 70 Chapter 0 Prerequisites 0.1 Sets 1. List the elements of each of the following sets: (a) N∩[0, 6) (b) Z ∩ (−6, 2] (c) {1, 2, 3, 4} ∪ {2, 3, 4, 5} (d) {1, 2, 3, 4} ∩ {2, 3, 4, 5} (a) {1, 2, 3, 4, 5} (b) {−5, −4, −3, −2, −1, 0, 1, 2} (c) {1, 2, 3, 4, 5} (d) {2, 3, 4} 2. Write each of the following in interval notation: (a) (0, 2) ∩ (1/2, 1) (b) [−1, 5] ∪ [2, 7] (a) (1/2, 1) (b) [−1, 7] 3. Prove (vi) of Theorem 0.2. (That is, prove that for sets A, B, and C, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)). (⊆) Let x ∈ A ∪ (B ∩ C). Then, x ∈ A or (x ∈ B and x ∈ C). If x ∈ A, then x ∈ (A ∪ B) ∩ (A ∪ C) since A ⊆ (A ∪ B) ∩ (A ∪ C). If x ∈ / A, then x ∈ B and x ∈ C. Thus, x ∈ A∪B and x ∈ A∪C. Thus, [A∪(B ∩C)] ⊆ [(A∪B)∩(A∪C)]. (⊇) Let x ∈ (A ∪ B) ∩ (A ∪ C). Then, (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C). Suppose x ∈ / A. Then x ∈ B and x ∈ C. Thus, x ∈ A ∪ (B ∩ C). Now, suppose x ∈ A. Then x ∈ A ∪ (B ∩ C). 4 5 CHAPTER 0. PREREQUISITES 4. Prove (ii) of Theorem 0.3. (That is, prove that for sets A, B, and C, A \ (B ∪ C) = (A \ B) ∩ (A \ C)). (⊆) Let x ∈ A \ (B ∪ C). Then, x ∈ A and x ∈ / (B ∪ C). Then, x ∈ / B and x∈ / C. Thus, x ∈ (A \ B) ∩ (A \ C). (⊇) Let x ∈ (A \ B) ∩ (A \ C). Then, (x ∈ A and x ∈ / B) and (x ∈ A and x∈ / C). Then, x is in neither B nor C, so x ∈ / (B ∪ C). Thus, x ∈ A \ (B ∪ C). 5. Prove that for all sets A, B, and C, A ∩ B ⊂ A ⊂ A ∪ C. If x ∈ A ∩ B, then x ∈ A. If x ∈ A, then x ∈ A ∪ C. 6. If A ⊂ B, prove that (C \ B) ⊂ (C \ A). Either prove the converse is true, or give a counterexample. If x ∈ C \ B, then x ∈ C and x ∈ / B. Since A ⊂ B, x ∈ / A. Thus, x ∈ C \ A. The converse is false. Let A = {1, 2, 3, 4, 5}, B = {6, 7, 8, 9, 10} and C = {6, 7}. Then, C \ A = {6, 7} but C \ B = ∅. Thus, (C \ B) ⊂ (C \ A), but A * B. 7. Under what conditions does A \ (A \ B) = B? Q4 Q3 We see A \ (A \ B) = A ∩ (A \ B)c = A ∩ (A ∩ B c )c = A ∩ (Ac ∪ B) = (A ∩ Ac ) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B. Thus, A \ (A \ B) = B if A ∩ B = B which happens when B ⊆ A. 8. Show that (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B). We see that (A ∪ B) \ (A ∩ B) = (A ∪ B) ∩ (A ∩ B)c = (A ∪ B) ∩ (Ac ∪ B c ) = [(A ∪ B) ∩ Ac ] ∪ [(A ∪ B) ∩ B c ] = [(A ∩ Ac ) ∪ (B ∩ Ac )] ∪ [(A ∩ B c ) ∪ (B ∩ B c )] = (B ∩ Ac ) ∪ (A ∩ B c ) = (A \ B) ∪ (B \ A). 9. Look up Russell’s paradox and write a brief summary of how it relates to Section 0.1. Russell’s paradox points out a flaw in naive set theory. If we examine the set A of all sets, then A ∈ A. But then, there must be another set containing A. This is the paradox. Since the theory in this section are based on naive set theory (the treatment of “set” as an undefined term), it shows that the theory is not complete. 10. Describe each of the following sets as the empty set, as R, or in interval notation, as appropriate: ∞
T (a) − n1 , n1 n=1 6 CHAPTER 0. PREREQUISITES (b) (c) (d) ∞ S n=1 ∞ T n=1 ∞ S n=1 (−n, n) − n1 , 1 + 1 n − n1 , 2 + 1 n (a) {0} (b) R (c) (0, 1] (d) (−1, 3)

11. Prove (ii) of Theorem 0.4. (That is, prove that S \(∩λ∈Λ Aλ ) = ∪λ∈Λ (S \ Aλ ).) (⊆) Let x ∈ S \ (∩λ∈Λ Aλ ). Then, x ∈ S and x ∈ / Aλ for all λ ∈ Λ. Then, x∈ / ∪λ∈Λ Aλ . Thus, x ∈ ∪λ∈Λ (S \ Aλ ). (⊇) Let x ∈ ∪λ∈Λ (S \ Aλ ) = ∪λ∈Λ (S ∩ Acλ ). Then, for each λ, x ∈ S and x∈ / Aλ . Thus, x ∈ S. Since x ∈ / Aλ for all λ, we see that x ∈ / ∩λ∈Λ Aλ . Thus, x ∈ S \ (∩λ∈Λ Sλ ). 12. Use DeMorgan’s Laws to give a different and simpler description of the following sets: ∞
T (a) R \ − n1 , n1 (b) (R \ ∞ S (R \ n=1 (a) R \ (b) n=1 0.2 n=1 ∞ S ∞ T n=1 1 + 1 + n, 2 1 n  )  )=R\

S   −∞, − n1 ∪ n1 , ∞ = R. − n1 , n1 = n, 2 1 n T1 n, 2 + 1 n  = R \ (0, 2] = (−∞, 0] ∪ (2, ∞). Relations and Functions 13. Define f : N → N by f (n) = 2n − 1 for each n ∈ N. What is im f ? Is f 1-1? Is f onto? If f has an inverse, find the domain of the inverse and give a formula for f −1 (n). We see that im f = {odd natural numbers}. We see that f is 1-1. Indeed, f (n) = f (m) ⇔ 2n−1 = 2m−1 ⇔ n = m. Since im f 6= N, f is not onto. Since f is 1-1 and im f = {odd natural numbers}, f −1 : {odd natural numbers} → N exists. To find it, set y = f −1 (n) and write n = 2y − 1 ⇔ f −1 (n) = n+1 2 . 14. What is the domain of f (x) = so, find the inverse. x x+2 ? What is im f ? Is f injective? If 7 CHAPTER 0. PREREQUISITES x+2 2 2 x = x+2−2 We see thatdom f = R \ {−2}. Since x+2 x+2 = x+2 − x+2 = 1 − x+2 , we see that im f = (1, ∞) ∪ (−∞, 1). We see that f is injective, so we find that 2 f −1 (x) = x−1 − 2. For Exercises 15-17, let A = {1, 2, 3, 4, 5}, B = {2, 3, 4, 5, 6, 7}, and C = {a, b, c, d, e}. 15. Give an example of f : A → B that is not 1-1. Let f (n) = n, if 2 ≤ n ≤ 5, and f (1) = 6 and f (1) = 7. (Note that f need not be a function.) 16. Give an example of f : A → B that has an inverse, and show the inverse. Let f (n) = n, if 2 ≤ n ≤ 5, and f (1) = 6. Then f −1 : B \ {7} → A is defined by f −1 (n) = n, if 2 ≤ n ≤ 5 and f −1 (6) = 1. 17. Give an example of f : A → B, g : B → C such that g ◦ f is 1-1, but g is not 1-1. Define f = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} and g = {(2, a), (3, b), (4, c), (5, d), (6, e), (7, e)}. Then, g ◦ f ={(1, a), (2, b), (3, c), (4, d), (5, e)} which is 1-1. But, 6 7→ e and 7 7→ e, so g is not 1-1. *18. If f : A → B is 1-1 and im f = B, prove that (f −1 ◦ f )(a) = a for all a ∈ A and (f ◦ f −1 )(b) = b for each b ∈ B. For each (x, y) ∈ f , assign (y, x) ∈ f −1 . Since f is 1-1, this assignment is also 1-1. Since im f = B, f is also onto. Now, let a ∈ A and (a, b) ∈ f . Then, (b, a) ∈ f −1 . Thus, (f −1 ◦ f )(a) = f −1 (b) = a. That (f ◦ f −1 )(b) = b follows from a similar argument. 0.3 Mathematical Induction and Recursion 19. Prove that for all n ∈ N, 1 + 2 + · · · + n = n(n+1) . 2 For n = 1, 1 = 1(2) 2 = 1. Suppose that the statement is true for all n ≤ N . +2) . Examine 1 + 2 + · · · + N + (N + 1) = N (N2+1) + (N + 1) = (N +1)(N 2 20. Prove that for all n ∈ N, 1 + 3 + 5 + · · · + (2n − 1) = n2 . CHAPTER 0. PREREQUISITES 8 For n = 1, 1 = 12 . Suppose the statement is true for all n ≤ N . Examine 1 + 3 + 5 + · · · + (2N − 1) + (2N + 1) = N 2 + 2N + 1 = (N + 1)2 . 21. Prove that n3 + 5n is divisible by 6 for each n ∈ N. For n = 1, 12 + 5(1) = 6 which is divisible by 6. Suppose the statement is true for all n ≤ N and N 3 + 5N = 6k. Then, (N + 1)3 + 5(N + 1) = N 3 + 3N 2 + 8N + 6 = (N 3 + 5N ) + (3N 2 + 3N ) + 6 = 6(k + 1) + 3N (N + 1). Thus, we need only examine the term 3N (N + 1) to prove divisibility. Suppose N is even. Then, 3N has 6 as a factor and 6 | 3N (N + 1). Now, suppose that N is odd. Then, N + 1 is even and 3(N + 1) is divisible by 6. Thus, 6 | 3N (N + 1). Thus, the statement is proven. 22. Prove that n2 < 2n for all n ∈ N, n ≥ 5. For n = 5, 25 < 32. Suppose the statement is true for all n ≤ N . Then, (N + 1)2 = N 2 + 2N + 1 < 2N + 2(2N −1 ) = 2N + 2N = 2(2N ) + 2 = 2N +1 . 23. Prove the second principle of mathematical induction. Let P (n) be a statement,P (1), P (2), ..., P (N ) are true and P (k) ⇒ P (k + 1) for all k ∈ N and k ≥ m. Suppose there is some K such that P (K) is false. Then, since P (k) ⇒ P (k + 1), by contraposition, ¬P (k + 1) ⇒ ¬P (k). By the second hypothesis, we must then have the following sequence of implications: ¬P (k) ⇒ ¬P (k − 1) ⇒ ¬P (k − 2) ⇒ · · · ⇒ ¬P (N ). Contradiction. 24. Define f : N → N by f (1) = 1, f (2) = 2, f (3) = 3 and f (n) = f (n − 1) + f (n − 2) + f (n − 3) for n ≥ 4. Prove that f (n) ≤ 2n for all n ∈ N. We see that f (4) = 6 ≤ 24 = 16. Suppose the statement is true for all n ≤ N . Then, f (N + 1) = f (N − 2) + f (N − 1) + f (N ) ≤ 2N −2 + 2N −1 + 2N = 2N +1 2N +1 2N +1 = 81 2N +1 + 14 2N +1 + 21 2N +1 = 87 2N +1 < 2N +1 . 23 + 22 + 2 p 25. Define f : N → N by f (1) = 2 and, for n ≥ 2, f (n) = 3 + f (n − 1). Prove that f (n) < 2.4 for all n ∈ N. You may want to use your calculator for this exercise. √ + 2 ≤ 2.4. Suppose the statement is true for all For n = 2, f (2) = 3p √ n ≤ N . Then, f (N + 1) = 3 + f (N ) ≤ 3 + 2.4 ≈ 2.32 < 2.4. 26. Define f : N → N by f (1) = 2, f (2) = −8, and, for n ≥ 3, f (n) = 8f (n − 1) − 15f (n − 2) + 6 · 2n . Prove that, for all n ∈ N, f (n) = −5 · 3n + 5n−1 + 2n+3 . CHAPTER 0. PREREQUISITES 9 For f (3) = 8(−8) − 15(2) + 6(8) = −5(33 ) + 52 + 26 . Suppose the statement is true for all n ≤ N . Then, f (N + 1) = 8f (N ) − 15f (N − 1) + 6 · 2N +1 = 8(−5 · 3N + 5N −1 + 2N +3 ) − 15(−5 · 3N −1 + 5N −2 + 2N +2 ) + 6 · 2N +1 = −5 · 3N +1 + 5N + 5 · 2N +3 + 3 · 23 = −5 · 3N +1 + 5N + 2N +4 . 27. Prove Theorem 0.10. (That is, suppose that P (n) is a statement and (i) for some n0 ∈ Z, P (n0 ) is true, and (ii) for each k ∈ Z, k ≥ n0 , P (k) ⇒ P (k + 1). Then, prove that P (n) is true for all n ≥ n0 ). Suppose that there is an N ≥ n0 . Then, by contraposition, P (N − 1) is false and so then is P (N − 2), etc. This eventually implies that P (n0 ) is false. Contradiction. *28. Prove the following modified version of the second principle of mathematical induction: Let P (n) be a statement for each n ∈ Z. If (a) P (n0 ), P (n0 + 1),..., P (m) is true, and (b) for k ≥ m, if P (i) is true for n0 ≤ i ≤ k, then P (k + 1) is true, then P (n) is true for n ≥ n0 , n ∈ Z. Suppose that P (N ) is false for some N . Then, there exists an i such that n0 ≤ i ≤ N − 1 such that P (i) is false. Consequently, there exists ani(1) such that n0 ≤ i(1) ≤ i − 1 such that P (i(1) ) is false. Eventually, this implies that there is an ĩ such that n0 ≤ ĩ ≤ m and P (ĩ) is false. Contradiction. 29. Define f (n) as follows for n ∈ Z, n ≥ 0, f (0) = 7, f (1) = 4, and, for n ≥ 2, f (n) = 6f (n − 2) − f (n − 1). Prove that f (n) = 5 · 2n + 2(−3)n for all n ∈ Z, n ≥ 0. For n = 2, f (2) = 6·7−4 = 38 = 5·22 +2(−3)2 . Suppose the statement is true for all n < N . Then, f (N ) = 6[5 · 2N −2 + 2(−3)N −2 ] − [5 · 2N −1 + 2(−3)N −1 ] = 5 · 2N + 2(−3)N . 0.4 Equivalent and Countable Sets 30. Prove Corollary 0.15. (That is, prove that any subset of a countable subset is countable.) Let X be a set with an uncountable subset K. Then, since K ≁ N, there cannot be a 1-1, onto function from K to N, so there consequently can’t be one from X to N. 31. Find a 1-1 function f from N onto S where S is the set of all odd integers. 10 CHAPTER 0. PREREQUISITES  n if n is odd . Then, the function is 1-1 by its −(n − 1) if n is even linear nature. It is also onto: if m ∈ S is negative, there is an even natural that is its preimage. If m ∈ S is positive, then there is an odd natural that is its preimage. Define f (n) = 32. Let Pn be the set of all polynomials of degree n with integer coefficients. Prove that Pn is countable. Consider the function f : Nn+1 → Pn defined by f (a0 , a1 , ..., an ) = an xn + an−1 xn−1 + · · · + a1 x + a0 . We have established a 1-1, onto map from Nn+1 onto Pn . Since we know that N is countable, by Theorem 0.16, Nn+1 is countable. Thus, Pn ∼ Nn+1 ∼ N. 33. Use Exercise 32 to show that the set of all polynomials with integer coefficients is a countable set. For any n, we have shown that Pn is countable. Thus, by Theorem 0.17. ∞ S Pk is countable k=0 34. Prove the following generalization of Theorem 0.17: If S is a countable set and {As }s∈S is an indexed family of countable sets, then ∪s∈S As is a countable set. Let s1 be the element of S that is in 1-1 correspondence with 1. Similarly, let sn be the element in S that is in 1-1 correspondence with n. Then, set An = Asn . Then, ∪s∈S As = ∪∞ n=1 An which is a countable union of countable sets which is countable. 35. For each p ∈ Pn , define B(p) = {x : p(x) = 0}. Prove that ∪p∈Pn B(p) is countable. We see that B(p) is just the set of roots of the polynomial p(x) which is a finite set for any p(x). We have already shown that Pn is countable, so {B(p)}p∈Pn is a countable indexed family of finite sets, which by Exercise 34 is countable. 36. An algebraic number is any number that is the root of a polynomial equation p(x) = 0 where the coefficients of p are integers. Show that the set of algebraic numbers is a countable set. Certainly A ⊆ ∪p∈Pn B(p). Any subset of a countable set is countable. 11 CHAPTER 0. PREREQUISITES 37. For a set A, let P (A) be the set of all subsets of A. Prove that A is not equivalent to P (A). Let A be finite such that |A| = n. Then, |P (A)| = 2n . (For any subset, each element can either be in the subset or not, so the total number of ways to build a subset is 2n .) Clearly no 1-1 correspondence can exist.  1 if the nth element of A is in Ak . If A is countably infinite, generate a subset Ak by ak,n = 0 otherwise Suppose that {Ak } is countable. Then we can enumerate the subsets via the sequences: A1 ↔ a1,1 a1,2 a1,3 · · · A2 ↔ a2,1 a2,2 a2,3 · · · .. .. .. .. .. .. . . . . . . . ↔ ak,1 ak,2 ak,3 · · · .. .. .. .. .. . . . . . But then the sequence {an,n } is another unique sequence that doesn’t correspond to any Ai . Thus, P (A) is uncountable. Thus, there can be no 1-1 map from A into P (A). Finally, if A is uncountable, then, any countable subset of A cannot be mapped 1-1 with the set of subsets (by the above argument), so again there can be no 1-1 map from A onto P (A). Ak .. . 38. Let a, b, c, and d be any real numbers such that a < b and c < d. Prove that [a, b] is equivalent to [c, d]. Define f : [a, b] → [c, d] via f (x) = c + x−a b−a · (d − c). This function maps x−c the position of x in [a, b] to a position in [c, d] such that x−a b−a = d−c (preserves percentage of interval. This function is linear and thus 1-1 and onto. 0.5 Real Numbers *39. If x < y, prove that x < We see that x = x 2 + x2 < x 2 x+y 2 + y2 = < y. x+y 2 . *40. If x ≥ 0 and y ≥ 0, prove that √ Similarly, y = xy ≤ y 2 + y2 > x 2 + y2 = x+y 2 . x+y 2 . Let x and y be two dimensions. Then, 2x + 2y gives the perimeter of the √ rectangle of length x and width y. We see that 4 xy gives the perimeter of a square with the same area. The perimeter of a square is always less than a √ perimeter of a proper rectangle of the same area. Thus, 4 xy ≤ 2x + 2y ⇔ √ xy ≤ x+y 2 . 12 CHAPTER 0. PREREQUISITES √ *41. If 0 < a < b, prove that 0 < a2 < b2 and 0 < a< √ b. Let b = a + δ with δ > 0. Then, b2 = a2 + δ 2 + 2aδ√> a. √ Next, suppose c√ = a and d2 √ = b =√a + δ. Then, d2 − δ = a. Thus, a = d2 − δ, while b = d. Thus, a < b. 2 42. If x, y, a, and b are greater than zero and a b. First, a b. x+a y+b = x y+b a + y+b < x y + ab . Now, since x y < ab , prove that x y < x+a y+b < < ab , this implies x y < x+a y+b < x y 43. Let A = {r : r is a rational number and r2 < 2}. Prove that A has no largest member. Let a b ∈ A. Suppose that 2 − a2 b2 = This is less than two precisely when 1+ 2 ab < c d · f e ⇔ c d + 2 ac bd < f e. c d. e2 f2 Then,  ae + 2 bf = a b + fe  e f e f 2 = + 2 ab a2 b 2 + < e2 f2 ae + 2 bf . c d. Thus, *44. If x = sup S, show that, for each ǫ > 0, there is a ∈ S such that x − ǫ < a ≤ x. Suppose that there is an ǫ0 > 0 such that there is no a between x − ǫ0 and x. Then, this implies that x − ǫ0 is an upper bound. Contradiction. *45. If y = inf S, show that, for each ǫ > 0, there is a ∈ S such that y ≤ a < y + ǫ. Suppose that there is an ǫ0 > 0 such that there is no a between y + ǫ0 and y. Then, y + ǫ0 is a lower bound. Contradiction. Chapter 1 Sequences 1.1 2 3 Sequences and Convergence 1. Show
that [0, 1] is a neighborhood of − ǫ, 23 + ǫ ⊂ [0, 1]. Choose ǫ = 31 . Then 2 3 − 31 , 23 + 1 3
= 2 3– that is, there is ǫ > 0 such that 1 3, 1
⊂ [0, 1]. *2. Let x and y be distinct real numbers. Prove there is a neighborhood P of x and a neighborhood Q of y such that P ∩ Q = ∅. Choose ǫ = P ∩ Q = ∅. |x−y| 2 . Then, let P = (x − ǫ, x + ǫ) and Q = (y − ǫ, y + ǫ). Then, *3. Suppose x is a real number and ǫ > 0. Prove that (x − ǫ, x + ǫ) is a neighborhood of each of its members; in other words, if y ∈ (x − ǫ, x + ǫ), then there is δ > 0 such that (y − δ, y + δ) ⊂ (x − ǫ, x + ǫ).   |y−(x−ǫ)| , Let δ = min |y−(x+ǫ)| . Then, (y − δ, y + δ) ⊂ (x − ǫ, x + ǫ). 2 2 4. Find upper and lower bounds for the sequence First, 3n+7 n  3n+7 n ∞ . n=1 = 3 + n7 . Thus, the lower bound is 3 and the upper bound is 10. 5. Give an example of a sequence that is bounded but not convergent. Let an = (−1)n . Then, this sequence alternates between −1 and 1, but never converges. 13 14 CHAPTER 1. SEQUENCES 6. Use the definition of convergence to prove that each of the following sequences converges:  ∞ (a) 5 + n1 n=1  2−2n ∞ (b) n n=1 −n ∞ (c) {2 } n=1 n o 3n 2n+1 (d) 2 n ∞ n=1   5 + n1 − 5 = n1 < N1 < ǫ. (a) Let ǫ > 0 be given. Let N = 1ǫ . Then,  + 2 = 2−2n+2n (b) Let ǫ > 0 be given. Let N = 2ǫ . Then, 2−2n = n n 2 < N < ǫ.
  (c) Let ǫ > 0 be given. Let N = lg 1ǫ . Then, |2−n | = 21n < 21N < ǫ. 3 3n = − 23 = 6n−6n−3 (d) Let ǫ > 0 be given. Let N = 4ǫ Then, 2n+1 4n+2 −3 4n+2 < 3 4n < 3 4N < ǫ. ∞ *7. Show that {an }∞ n=1 converges to A iff {an − A}n=1 converges to 0. We see that |(an − A) − 0| = |an − A| < ǫ. ∞ 8. Suppose {an }∞ n=1 converges to A, and define a new sequence {bn }n=1 by an +an+1 ∞ bn = for all n. Prove that {bn }n=1 converges to A. 2 Let ǫ > 0 be given. We see that |an −A| 2 + |an+1 −A| ∃N ǫ < 2 s.t. 2 + ǫ 2 an +an+1 2 −A = an +an+1 −2A 2 = an −A+an+1 −A 2 = ǫ. Thus, bn → A. ∞ ∞ ∞ *9. Suppose {an }∞ n=1 , {bn }n=1 , and {cn }n=1 such that {an }n=1 converges ∞ to A, {bn }n=1 converges to A, and an ≤ cn ≤ bn for all n. Prove that {cn }∞ n=1 converges to A. Since an ≤ cn ≤ bn , we must have an −A ≤ cn −A ≤ bn −A. By convergence and definition of absolute value, −ǫ < an−1 − A ≤ cn − A ≤ bn − A < ǫ. Hence, |cn − A| < ǫ. Thus, cn → A. (We will call this result the Squeeze Theorem.) ∞ *10. Prove that, if {an }∞ n=1 converges to A, then {|an |}n=1 converges to |A|. Is the converse true? Justify your conclusion. TI We see that ||an | − |A|| < ||an − A|| = |an − A| < ǫ. The converse is not true. For instance, |(−1)n | → 1, but {(−1)n } diverges. *11. Let {an }∞ n=1 be a sequence such that there exist numbers α and N such that, for n ≥ N , an = α. Prove that {an }∞ n=1 converges to α. We see that |an − α| ≤ |aN − α| = 0 < ǫ for all ǫ > 0. ≤ 15 CHAPTER 1. SEQUENCES 12. Give an alternate proof of Theorem 1.1 along the following lines. Choose ǫ > 0. There is N1 such that for n ≥ N1 , |an − A| < 2ǫ , and there is N2 such that for n ≥ N2 , |an − B| < 2ǫ . Use the triangle inequality to show that this implies that |A − B| < ǫ. Let N = max(N1 , N2 ). Then, ǫ > |an − A| + |an − B| = |an − A| + |B − an | > |an − A + B − an | = |B − A|. Thus, |A − B| < ǫ. 13. Let x be any positive real number, and define a sequence {an }∞ n=1 by an = [x] + [2x] + · · · + [nx] n2 where [x] is the largest integer less than or equal to x. Prove that {an }∞ n=1 converges to x/2. Let ǫ > 0 be given and set N = x(1+···+n) n2 1.2 − x 2 = xn(n+1) 2n2 − x 2 = x 2ǫ . xn2 2n2 Then, an − + xn 2n2 − x 2 = x 2 ≤ x 2n x+2x+···+nx n2 x < 2N < ǫ. − x 2 = Cauchy Sequences 14. Prove that every Cauchy sequence is bounded. (Theorem 1.4) Suppose that {an } is not bounded. Then, for any k, there is an nk such that |ank | > k. Then, {ank }is an unbounded sequence. Then, for any N , there exist ank and anℓ such that |ank − anℓ | > |ank | − |anℓ | = k − ℓ where k − ℓ > N . Thus, {an } is not Cauchy. ∞ 15. Prove directly (do not use Theorem 1.8) that, if {an }∞ n=1 and {bn }n=1 ∞ are Cauchy, so is {an + bn }n=1 . Since {an } and {bn } are Cauchy, then for all ǫ > 0, there exist N1 and N2 such that |an − am | < 2ǫ for all m, n > N1 and |bn − bm | < 2ǫ for all m, n > N2 . Choose N = max(N1 , N2 ). Then, |an + bn − (am + bm )| = |an − am + bn − bm | < |an − am | + |bn − bm | < 2ǫ + 2ǫ = ǫ for all m, n > N . ∞ 16. Prove directly (do not use Theorem 1.9) that, if {an }∞ n=1 and {bn }n=1 ∞ are Cauchy, so is {an bn }n=1 . You will want to use Theorem 1.4. Since {bn } is Cauchy, then it is bounded (by Exercise 14). Thus, |bn | < M for ǫ some M . Since {an } is Cauchy, then for all ǫ > 0, there exist N |an − am | < M for all m, n > N and |bn − bm | < 2ǫ for all m, n > N2 . Let ǫ > 0 be given. Then, |an bn − am bm | < |an M − am M | = |M (an − am )| < ǫ. 16 CHAPTER 1. SEQUENCES 17. Prove that the sequence  2n+1 n ∞ n=1 is Cauchy. Let ǫ > 0 be given. Choose N = 1ǫ . Then, n m − mn = mn < NN2 = N1 < ǫ. m−n nm 2n+1 n − 2m+1 m = 2mn+m−2mn−n nm = 18. Give an example of a sequence with exactly two accumulation points.  1/n , if n is even . Then, an has accumulation points at Let an = 1 + 1/n , if n is odd 0 and 1. 19. Give an example of a set with a countably infinite set of accumulation points. The set Q has the propertythat every element is an accumulation point, since for any ab ∈ Q, the sequence ab + n1 converges to ab . Since Q is countable, we have found the desired set. 20. Give an example of a set that contains each of its accumulation points. The set [0, 1] contains all of its accumulation points.  21. Determine the accumulation points of the set 2n + 1 k : n and k are positive integers . The set {2n : n ∈ Z+ }∪{∞} is the set of accumulation points since 2n + k1 → 2 as k → ∞ and 2n + k1 → ∞ as n → ∞. n 22. Let S be a nonempty set of real numbers that is bounded from above (below) and let x = sup S (inf S). Prove that either x belongs to S or x is an accumulation point of S. It is clear that x ∈ S is a possibility. Suppose x ∈ / S. Then, by Exercise 0.44, for any ǫ > 0, there is an a ∈ S such that x − ǫ < a < x. Thus, for all n, there exists an an ∈ S such that x − n1 < an < x. Since x − n1 → x, we have an → x. Thus, x is an accumulation point of S. n−2 for each 23. Let a0 and a1 be distinct real numbers. Define an = an−1 +a 2 ∞ positive integer n ≥ 2. Show that {an }n=1 is a Cauchy sequence. You may want to use induction to show that  n 1 an+1 − an = − (a1 − a0 ) 2 and then use the result from Example 0.9 of Chapter 0. 17 CHAPTER 1. SEQUENCES
n The statement an+1 − an = − 12 (a1 − a0 ) is obviously true for n = 1. that it is itrue for n < N . Then, aN +2 − aN +1 = aN +12+aN − h Suppose

N +1 N (a1 − a0 ) + aN = aN +1 +a2N −2aN + − 12 (a1 − a0 ) = aN +12−aN + − 21 h
N +1
N
N +1
N
N +1 i (a1 −a0 ) = − 21 (a1 −a0 )+ − 12 (a1 −a0 ) = − 21 + − 12 (a1 − − 12 h i h i




N N N − 21 a0 ) = + − 12 − 21 (a1 − a0 ) = − 21 + 1 − 21 (a1 − a0 ) =
1 N +1 −2 (a1 − a0 ). Thus, the statement is proven by induction. l  m Now, let ǫ > 0 and n, m ∈ N be given. Choose N = lg |a1 −a0ǫ|(n−m) .
n ǫ for n ≥ N. Thus, we see that |an − am | = Then, − 21 (a1 − a0 ) < n−m |an − an−1 + an−1 − an−2 + · · · + am+1 − am | < |an − an−1 |+· · ·+|am+1 − am | < ǫ ǫ ǫ n−m + n−m + · · · + n−m = ǫ. 24. Suppose {an }∞ n=1 converges to A and {an : n ∈ N} is an infinite set. Show that A is an accumulation point of {an : n ∈ N}.
Let Nk = A − k1 , a + k1 . By convergence, there is an N such that |an − A| < 1 k for all n ≥ N . Thus, there is an element of {an : n ∈ N} in Nk for every k. Now, every neighborhood of A has Nk as a subset for some k and since there are an infinity of Nk ’s, we have an infinity of members of {an } in any neighborhood. 1.3 Arithmetic Operations on Sequences ∞ ∞ 25. Suppose {an }∞ n=1 and {bn }n=1 are sequences such that {an }n=1 and ∞ ∞ {an + bn }n=1 converge. Prove that {bn }n=1 converges. Suppose an → A and an + bn → C. Then, {bn } = {an + bn − an }. Thus, by Theorem 1.8, bn → C − A. ∞ 26. Give an example in which {an }∞ n=1 and {bn }n=1 do not converge but ∞ {an + bn }n=1 converges. Let an = (−1)n and bn = (−1)n+1 . We know that an and bn don’t converge, but an + bn → 0. ∞ ∞ 27. Suppose {an }∞ n=1 and {bn }n=1 are sequences such that {an }n=1 con∞ ∞ verges to A 6= 0 and {an bn }n=1 . Prove that {bn }n=1 converges. n o Suppose an bn → C. Then, {bn } = aannbn , so by Theorem 1.9, bn → C A. 28. If {an }∞ converges to a with an ≥ 0 for all n, show √ n=1 converges to a. √ an ∞ n=1 18 CHAPTER 1. SEQUENCES √ Let ǫ > 0 be given. Then, there is an N such that |an − a| < an − a = √ √an −a an + a = |an − a| 29. Prove that √ 1 √ an + a < |an − a| √1 a √ǫ . a Then, < ǫ for all n ≥ N .  ∞  n+k       k  (n + k)k      n=1 converges to 1 k! , We see that where  n (n+k)! n!k!(n+k)k given. Choose N = n+1 k!(n+k) − 1 k! =  1−k  ǫ o n+k k = . Then, n+1−n−k k!(n+k) = n  = (n + k)! . n!k! (n+k−1)(n+k−2)···(n+1) k!(n+k)k−1 (n+k−1)···(n+1) k!(n+k)k−1 1−k k!(n+k) < 1−k n − 1 k! o < . Now, let ǫ > 0 be (n+k)k−2 (n+1) k!(n+k)k−1 − 1 k! = < ǫ for all n ≥ N . 30. Prove the following variation on Lemma 1.10. If {bn }∞ n=1 converges to B 6= 0 and bn 6= 0 for all n, then there is M > 0 such that |bn | ≥ M for all n. Choose M = |bn | /2. Then, the statement holds. 31. Consider a sequence {an }∞ n=1 and, for each n, define αn = a1 + a2 + · · · + an . n ∞ Prove that if {an }∞ n=1 converges to A, then {αn }n=1 converges to A. Give an ∞ ∞ example in which {αn }n=1 converges, but {an }n=1 does not. Let ǫ > 0 be given. There is an N1 such that |an − A| < 2ǫ . for all n > N . Let M = |a1 − A| + |a2 − A| + · · · + |aN1 − A|. Then, there is an N2 such that a1 +a2 +···+an ǫ M −A = n < 2 for all n > N2 . Let N = max(N1 , N2 ). Then, n |aN +1 −A|+···+|an −A| |a1 −A|+···+|aN −A| |aN +1 −A|+···+|an −A| a1 +···+aN +···+an −nA M + = n+ < < n n n n ǫ ǫ 2 + 2 = ǫ. (A quicker way to show this would be to observe that by the definition of convergence, |an | < 2ǫ for all n > N for some N . Then, since n → Let an = (−1)n . Then, as we have seen before, {an } diverges, but a1 +a2 +···+a n 0. 32. Find the limit of the sequences with general term as given: 2 (a) nn2+4n −5 (b) cosn n 2 √n (c) sin n 19 CHAPTER 1. SEQUENCES (d) n 2 n −3 q (e) 4− (f) 1 n − √ n (−1)n n+7 (a) 1 (b) 0 (c) 0 (d) q 0 (e) −n 4n . 4− 1 n  2 n q  −2 n=n 4− Thus, limit is (f) 0 − 14 . 1 n −2n = √ √ 4n2 − n− 4n2 = 2 2 √ √4n −n−4n 4n2 −n+ 4n2 ∼ 33. Find the limit of the sequence in Exercise 23 when a0 = 0 and a1 = 3. You might want to look at Example 0.10. Example 0.10 says that must be 2. 1.4 an−1 +an−2 2
= − 21n 3 + 2. Since − 21n → 0, the limit Subsequences and Monotone Sequences 34. Find a convergent subsequence of the sequence ∞   1 (−1)n 1 − n n=1  Let nk = 2n. Then, the subsequence is 1 − 1 2n which converges to 0. 35. Suppose x is an accumulation point of {an : n ∈ N}. Show that there is a subsequence of {an }∞ n=1 that converges to x. Since x is an accumulation point, every neighborhood about x contains an
infinity of {an }. Thus, let ank be a member of {an : n ∈ N} ∩ x − k1 , x + k1 . Then, for any ǫ > 0, there is a K such that k1 < ǫ for all k > K. Thus, ank → x. ∞ 36. Let {an }∞ n=1 be a bounded sequence of real numbers. Prove that {an }n=1 has a convergent subsequence. Either{an } has a finite number of values or {an } has an infinite number of values. For the former, there must be some value x for which there are infinitely many k such that ank = x. Thus, ank → x. For the latter, the sequence is a 20 CHAPTER 1. SEQUENCES bounded infinite set of real numbers, so by the Bolzano-Weierstrass Theorem, an has a convergent subsequence. ∞ *37. Prove that if {an }∞ n=1 is decreasing and bounded, then {an }n=1 converges. Assume that {an } attains an infinite number of values. Suppose that inf an = intervals that sequence values M . Let ǫ > 0 be given. Then, there are a1 −M ǫ may fall. Since this is a finite number and there are an infinite number of values, at least one region must contain an infinite number of function values. Since the sequence is decreasing, the last region must contain an infinity of values; that is, an ∈ (M, M + ǫ) for all n > N for some N . Since ǫ was arbitrarily chosen, the proof is complete. The case for when {an } has only finitely many values is easy. √ 38. Prove that if c > 1, then { n c}∞ n=1 converges to 1. √ √ √ It is clear that n c > n+1 c. Thus,√{ n c} is a monotone decreasing sequence. √ Also, n c > 1, so by Theorem 1.16, n c → 1. ∞ *39. Suppose {xn }∞ n=1 converges to x0 and {yn }n=1 converge to x0 . Define ∞ a sequence {zn }n=1 as follows: z2n = xn and z2n−1 = yn . Prove that {zn }∞ n=1 converges to x0 . Both subsequences of {zn } converge to x0 . Thus, by Theorem 1.14, zn → x0 . 40. Show that the sequence defined by a1 = 6 and an = is convergent and find its limit. √ 6 + an−1 for n > 1 √ To find the limit L, set L = 6 + L ⇔ L2 = 6 + L ⇔ L2 − L − 6 = 0. The solutions are −2 and 3. The only solution √ is 3.√ Thus, an → 3. We √ that works prove that {an } is decreasing. Since 6 + an−1 < 6 + an−1 and we know √ that an−1 is decreasing, we see that the whole sequence is decreasing. Also, square roots must be greater than 0, so the sequence is bounded. Thus, the sequence is bounded below and decreasing and is thus convergent. 41. Let {xn }∞ n=1 be a bounded sequence and let E be the set of subsequential limits of that sequence. By Exercise 36, E is nonempty. Prove that E is bounded and contains both sup E and inf E. Since {xn } is bounded (by M ), its limit points must be such that they are within ǫ distance of some sequence values. Thus, limit points must be within the same bounds as {xn } or within ǫ distance of the boundary for any ǫ. Thus, E is bounded (by, say M + 1). We must ensure that members of E do not form a sequence themselves that converges to a non-limit point. So, suppose there is a sequence {en } of limit points. Then, for every ǫ, there is an N such that 21 CHAPTER 1. SEQUENCES |en − xn | < ǫ for all n > N . Thus, xn → en . Thus, all sequences of E converge in E (since they are estimated by subsequences of {xn }. Thus, sup E, inf E ∈ E. 42. Let {xn }∞ n=1 be any sequence and T : N → N be any 1-1 function. Prove that if {xn }∞ converges to x, then {xT (n) }∞ n=1 n=1 also converges to x. Explain how this relates to subsequences. Define what one might call a “rearrangement” of a sequence. What does the result imply about rearrangements of sequences? We see that {xT (n) } = {xn1 , xn2 , ...} and is a subsequence of {xn }. Since all subsequences converge, we must have xT (n) → x. Let T : N → N be any 1-1 function and let {xn } be a sequence. Then, {xT (n) } is called a rearrangement. The result implies that if {xn } converges, then so does {xT (n) } for any T . 43. Assume 0 ≤ a ≤ b. Does the sequence {(an + bn )1/n }∞ n=1 converge or diverge? If the sequence converges, find the limit. The sequence does not converge in general. For instance, if a = 1 and b = −1, then the sequence becomes {[1n + (−1)n ]1/n }. Taking even indexes, the limit is 1, and taking odd indexes, the limit is 0. Thus, not all subsequences converge to the same limit point, so the sequence is not convergent. 44. Does the sequence ( k  X n=1 1 √ 2 k +n ) ∞ k=1 diverge or converge? If the sequence converges, find the limit. Pk P∞ 1 P∞ k √ 1 √ 1 We see that n=1 k2 +n < n=1 k = k = 1. Also, n=1 k2 +n > P∞ √ 1 √ k → 1. Thus, the sequence must converge to 1 by the n=1 k2 +k = k2 +k Squeeze Theorem (established in Exercise 9). *45. Show that if x is any real number, there is a sequence of rational numbers converging to x. Let a0 .a1 a2 · · · be the decimal expansion for x. Then, define xn = a0 .a1 a2 · · · an . Then, xn ∈ Q for all n and xn → x (for there exists an N for which xn can be within 10k distance for any integer k and n > N ). *46. Show that if x is any real number, there is a sequence of irrational numbers converging to x. If x is already irrational, define xn ≡ x. Clearly, xn → x. If x is rational, define xn = x + nπ . Then, xn ∈ R \ Q for all n and xn → x. CHAPTER 1. SEQUENCES 22 47. Suppose that {an }∞ n=1 converges to A and that B is an accumulation point of {an : n ∈ N}. Prove that A = B.
If B is an accumulation point, then B − n1 , B + n1 contains a member of {an } for all n. Thus, one can construct a subsequence of these members that converges to B, and since {an } is convergent, we must have A = B. Miscellaneous ∞ 48. Suppose that {an }∞ n=1 and {bn }n=1 are two sequences of positive real numbers. We say that an is O(bn ) (read as “big oh” of bn ) if there is an integer N and a real number M such that for n ≥ N , an ≤ M · bn . Prove that if {an /bn }∞ n=1 converges to L 6= 0, then an is O(bn ) and bn is O(an ). What can you say if L = 0? Illustrate with examples. Since an /bn → L, we guess that there is an N such that an ≤ (L + 1) · bn for all n ≥ N . We now prove this assertion. First, for any ǫ > 0, there is an N for which an /bn ∈ (L − ǫ, L + ǫ) for all n ≥ N . Thus, for the same N , an ∈ (bn [L − ǫ], bn [L + ǫ]). Thus, an ≤ bn (L + 1). Thus, an is O(bn ) if N is chosen to correspond to ǫ = 1. Similarly, bn is O(an ). If L = 0, then either an → 0 and {bn } is bounded or bn → ∞ and {an } is 2 1/n = 0. (This result is called the Limit bounded. For instance, nn3 → 0 and 2+1/n Comparison Test.) Chapter 2 Limits of Functions 2.1 Definition of the Limit of a Function 1. Define f : (−2, 0) → R by f (x) = and find it. x2 −4 x+2 . Let ǫ > 0 be given. Choose δ = ǫ. Then δ = ǫ. Thus, lim f (x) = −4. Prove that f has a limit at −2 x2 −4 x+2 + 4 = |x − 2 + 4| = |x + 2| < x→−2 2. Define f : (−2, 0) → R by f (x) = −2 and find it. 2x2 +3x−2 . x+2 Let ǫ > 0 be given. Choose δ = 2ǫ . Then Prove that f has a limit at 2x2 +3x−2 x+2 +5 = (x+2)(2x−1) x+2 +5 = |2x − 1 + 5| = |2x + 4| = 2 |x + 2| < 2δ = ǫ. 3. Give an example of a function f : (0, 1) → R that has a limit at every point except 12 . Use the definition of limit of a function to justify the example.  0, 0 < x < 12 . Then, the limit is clearly 0 for every input Let f (x) = 1, 21 ≤ x < 1

x ∈ 0, 12 and 1 for every input x ∈
12 , 1 . Let ǫ0 = 1. Let δ > 0 be given. Then, in the neighborhood 12 − δ, 12 + δ , there is an input to the left-half of the interval so that f (x) = 0 and there is an input in the right-half of the interval such that f (x) = 1. Now, the only possibilities for the limit would be either 0 or 1. However, the maximum difference between function values on the interval and either limit is 1 which is equal to ǫ0 . Thus, the limit does not exist. 4. Give an example of a function f : R → R that is bounded and has a limit at every point except −2. Use the definition to justify the example. 23 24 CHAPTER 2. LIMITS OF FUNCTIONS Change the parameters of the solution given to Exercise 3 to obtain a similar proof. *5. Suppose f : D → R with x0 an accumulation point of D. Assume L1 and L2 are limits of f at x0 . Prove L1 = L2 . Let ǫ > 0 be given. Then, there are δ1 and δ2 such that |f (x) − L1 | < 2ǫ and |f (x) − L2 | < 2ǫ . Then, |L1 − L2 | = |L1 − f (x) + f (x) − L2 | < |L1 − f (x)| + |f (x) − L2 | < 2ǫ + 2ǫ = ǫ. Thus, L1 = L2 . 6. Define f : (0, 1) → R by f (x) = cos x1 . Does f have a limit at 0? Justify. For any δ > 0, any function value between 0 and 1 can be obtained from inputs x ∈ (0, δ). Thus, cos x1 − L can be made greater than some ǫ. Thus, no limit exists at x = 0. 7. Define f : (0, 1) → R by f (x) = x cos Justify. 1 x
. Does f have a limit at x = 0? We guess that lim f (x) = 0. Let ǫ > 0 be given. Choose δ = ǫ. Then, x→0
x cos x1 < |x| < δ = ǫ. 8. Define f : (0, 1) → R by f (x) = 1. We see lim x x→1 3 −x2 +x−1 x−1 = lim x 2 x→1 2. Let ǫ > 0 be given. Choose δ = 2 x3 −x2 +x−1 . x−1 (x−1)+(x−1) x−1 √ ǫ. Then, 2 x − 1 = |x + 1| |x − 1| < δ = ǫ. 9. Define f : (−1, 1) → R by f (x) = Justify. Prove that f has a limit at = lim (x 2 +1)(x−1) x−1 x→1 x3 −x2 +x−1 −2 x−1 x+1 x2 −1 . = lim x2 + 1 = x→1 2 = x +1−2 = Does f have a limit at 1? x+1 1 We see lim xx+1 From what we know from 2 −1 = lim (x+1)(x−1) = lim x−1 . x→1 x→1 x→1 calculus, we know that the limit does not exist. To be more precise, inputs in (1, 1 + ǫ) yield outputs which become arbitrarily large as ǫ → 0 and inputs in (1 − ǫ, 1) yield outputs which become arbitrarily negative as ǫ → 0. 2.2 Limits of Functions and Sequences 10. Consider f : (0, 2) → R defined by f (x) = xx . Assume that f has a limit at 0 and find that limit. 25 CHAPTER 2. LIMITS OF FUNCTIONS 1 n
. Clearly, xn → 0. Since we assumed that f had
2 a limit, it must be the case that f (xn ) → lim f (x). Then, f (xn ) = log 1 + n1 Examine xn = log 1 + x→0 which clearly has milt 0. *11. Suppose f , g, and h : D → R where x0 is an accumulation point of D, f (x) ≤ g(x) ≤ h(x) for all x ∈ D, and f and h have limits at x0 with limx→x0 f (x) = limx→x0 h(x). Prove that g has a limit at x0 and lim f (x) = lim g(x) = lim h(x). x→x0 x→x0 x→x0 Since f and h have limits, lim f (xn ) = lim g(xn ) = lim h(xn ) for any n→∞ n→∞ n→∞ xn → x0 by the Squeeze Theorem. Thus, lim f (x) = lim g(x) = lim h(x) by x→x0 x→x0 x→x0 Theorem 2.1. *12. Suppose f : D → R has a limit at x0 . Prove that |f | : D → R has a limit at x0 and that limx→x0 |f (x)| = |limx→x0 f (x)|. Let ǫ > 0 be given. Then by definition of limit, there is a δ > 0 such that for all x ∈ (x0 − δ, x0 + δ), |f (x) − x0 | < ǫ. There are three cases to consider. If x0 > 0, then |f (x)| ≡ f (x) on the interval of interest and so the limit would be that of f . If x0 < 0, then, |f (x)| ≡ −f (x) on the interval of interest and |−f (x) + x0 | = |−1| |f (x) − x0 | < ǫ for any x ∈ (x0 − δ, x0 + δ). Finally, if x0 = 0, then ||f (x)|| = |f (x)| < ǫ for any x ∈ (x0 − δ, x0 + δ), so the limit exists. 13. Define f : R → R by f (x) = x − [x]. Determine the points at which f has a limit and justify your conclusions. We see that x − [x] leaves us with the decimal part of x. Thus, the function is continuous for (k, k + 1) for every integer k. It is not have a limit at the endpoints as there will be a jump of 1 unit. 14. Define f : R → R as follows: f (x) = 8x if x is a rational number f (x) = 2x2 + 8 if x is an irrational number. Use subsequences to guess at which points f has a limit, then use ǫ’s and δ’s to justify your conclusions. We can only hope to find a limit at an accumulation point and since there can always be found a rational number between two irrationals (and vice versa), we see that the only possibility is at x = 2. We suppose that the limit at this point is 16. Now, let ǫ > 0 be given. Choose δ1 = 2ǫ Then |8x − 16| = 8 |x − 2| < 8δ1 = ǫ. 2 Then, let δ2 = 2ǫ . Then, 2x2 + 8 − 16 = 2x2 − 8 = 2 x2 − 4 < 2 |x − 4| = 26 CHAPTER 2. LIMITS OF FUNCTIONS 2 · 2ǫ = ǫ. Choose δ = min(δ1 , δ2 ) to see that the function has a limit of 16 at x = 2. 15. Let f : D → R with x0 as an accumulation point of D. Prove that f has a limit at x0 if for each ǫ > 0, there is a neighborhood Q of x0 such that, for any x, y ∈ Q ∩ D, x 6= x0 , y 6= x0 , we have |f (x) − f (y)| < ǫ. Let ǫ > 0 be given. Since f has a limit L at x = x0 , there is a δ such that for any x ∈ (x0 − δ, x0 + δ), |f (x) − L| < 2ǫ . Let Q = (x0 − δ, x0 + δ). We see that |f (x) − f (y)| = |f (x) − L + L − f (y)| < |f (x) − L| + |f (y) − L| = 2ǫ + 2ǫ = ǫ. 2.3 Algebra of Limits 16. Define f : (0, 1) → R by f (x) = 0 and find that limit. We see that f (x) = x(x2 +6x+1) x(x−6) . we certainly see is equal to − 16 . x3 +6x2 +x x2 −6x . Prove that f has a limit at lim (x2 +6x+1) Thus, the limit is lim xx · x→0lim(x−6) x→0 which x→ 17. Define f : R → R as follows: f (x) = x − [x] if [x] is even. f (x) = x − [x + 1] if [x] is odd. Determine those points where f has a limit and justify your conclusions. As discussed in Exercise 13, x − [x] gives us the decimal part of x. It was determined that the function had no limits for any x ∈ Z. We now observe that x − [x + 1] gives us the negative value of the decimal part of 1 − x. We still have limits in the intervals (k, k + 1), but approaching the left endpoint of an odd numbered interval from the left, the limit is 1, while the limit from the right is 1. Thus, the limit is defined at odd left-endpoints. Similarly, all the endpoints have limits. Thus, f has a limit at every point in R. 18. Define g : (0, 1) → R by g(x) = and find it. √ 1+x−1 . x Prove that g has a limit at 0 √ √ · 1+x+1 = 1+x−1 = xx . The limit of We see that, for x 6= 0, g(x) = 1+x−1 x x x this function is clearly 1. (This is a proof because of the multiplication of limits is the limit of multiplications.) 19. Define f : (0, 1) → R by f (x) = and find it. √ 9−x−3 . x Prove that f has a limit at 0 27 CHAPTER 2. LIMITS OF FUNCTIONS We see that for x 6= 0, f (x) = is clearly 1. √ 9−x−3 x · √ 9−x+3 x = 9−x−9 x = − xx . The limit 20. Prove Theorem 2.5. [That is, suppose f : D → R and g : D → R, x0 is an accumulation point of D, and f and g have limits at x0 . Prove that if f (x) ≤ g(x) for all x ∈ D, then lim f (x) ≤ lim g(x).] x→x0 x→x0 Suppose lim f (x) > lim g(x). Then, we must have lim f (x) − lim g(x) = x→x0 x→x0 x→x0 x→x0 lim (f (x) − g(x)). We see that the left-hand side must be positive, but the x→x0 right-hand side is at most 0. Contradiction. 21. Suppose g : D → R with x0 an accumulation point of D and g(x) 6= 0 for all x ∈ D. Further assume that g has a limit at x0 and limx→x0 g(x) 6= 0. State and prove a lemma similar to Lemma 1.10 for such a function. [Lemma 1.10 states: “If {bn }∞ n=1 converges to B and B 6= 0, then there is a positive real number M and a positive integer N such that, if n ≥ N , then |bn | ≥ M .”] Lemma. If g(x) has a limit at x0 and x0 6= 0, then there is a positive real number M and a δ > 0 such that, if x ∈ (x0 − δ, x0 + δ), then |g(x)| ≥ M . Proof. Suppose that lim g(x) = L. Then, there is a δ > 0 such that for x→x0 x ∈ (x0 − δ, x0 + δ), we have g(x) ∈ (x0 − δ, x0 + δ). Thus, setting M = x0 − δ, we see that |g(x)| ≥ M . 22. Show by example that, even though f and g fail to have limits at x0 , it is possible for f + g to have a limit at x0 . Give similar examples for f g and fg . We have seen that the function f (x) = x1 does not have a limit at 0. Thus, g(x) = −f (x) will also not have a limit. Nonetheless, lim (f + g)(x) = 0. Also, x→0 if f is the function defined in Exercise 14, then f has no limit any where except at x = 2. If g ≡ f1 , then lim (f g)(x) = 1. For the same f , choosing g ≡ f gives lim fg = 1. x→0 x→1 2.4 Limits of Monotone Functions 23. State and prove a lemma similar to Lemma 2.7 for decreasing functions. [Lemma 2.7 states: “Let f : [α, β] → R be increasing. Let U (x) = inf{f (y) : x < y} and L(x) = sup{f (y) : y < x} for x ∈ (α, β). Then f has a limit at x0 ∈ (α, β) iff U (x0 ) = L(x0 ), and in this case lim f (x) = f (x0 ) = U (x0 ) = x→x0 L(x0 ).”] CHAPTER 2. LIMITS OF FUNCTIONS 28 Lemma. Let f : [α, β] → R be decreasing. Let L(x) = sup{f (y) : y < x} and U (x) = inf{f (y) : y > x} for x ∈ (α, β). Then f has a limit at x0 ∈ (α, β) iff U (x0 ) = L(x0 ), and in this case lim f (x) = f (x0 ) = U (x0 ) = L(x0 ). x→x0 Proof. (⇒) Suppose that lim f (x) = L. Since f is decreasing, if x ≤ y, x→x0 we have f (x) ≤ f (y). Thus, for every δ > 0, f (x0 ) ≥ L(x0 ) ≥ f (x0 + δ). Similarly,
f (x0 ) ≤ U (x
0 ) ≤ f (x0 + δ) . We also note that since the limit exists, f x0 + n1 , f x0 − n1 → L = f (x0 ). Thus, L(x0 ) = L and U (x0 ) = L. (⇐) Suppose that U (x0 ) = L(x0 ). Then, inf{f (y) : y > x0 } = sup{f (y) : y < x0 }. This implies that for every ǫ > 0, there exists a t1 ∈ (x0 , β] such that U (x0 ) ≤ f (t1 ) < U (x0 ) + ǫ. Similarly, there exists a t2 ∈ [α, x0 ) such that L(x0 ) ≥ f (t2 ) > L(x0 ) − ǫ. Since we have U (x0 ) = L(x0 ) = M , we have M + ǫ ≥ f (t1 ) ≥ M ≥ f (t2 ) > M − ǫ. Let δ = |t1 − t2 |. Then for any x ∈ (x0 − δ, x0 + δ), we have |f (x) − f (x0 )| < |f (t1 ) − f (t2 )| < ǫ. Thus, lim f (x) = f (x0 ). x→x0 b. 24. Let f : [a, b] → R be monotone. Prove that f has a limit both at a and Without loss of generality, suppose that f is increasing. Then, the only way that f might not have a limit at a is if U (a) > a (L(a) = a since there are no function values defined for inputs less
than a). Suppose that U (a) = f (a+δ0 ) = 0 < U (a) by the fact that f is increasing. f (a) + ǫ0 . But then, f (a) < f a+δ 2 Contradiction. Similar for the limit at b. (Use L(b) and argue similarly.) 25. Suppose f : [a, b] → R and define g : [a, b] → R as follows: g(x) = sup{f (t) : a ≤ t ≤ x} Prove that g has a limit at x0 if f has a limit at x0 and limt→x0 f (t) = f (x0 ). We note that g(x) is increasing. Let x⋆ be such that f (x⋆ ) is the largest function value on [a, x0 ]. There are two cases to consider: (1) f (x0 ) = f (x⋆ ) or (2) f (x0 ) < f (x⋆ ). For (1), either g ≡ f on (x0 − δ, x + δ) for some δ or g ≡ f on (x0 − δ, x0 ) and g ≡ f (x⋆ ) on (x0 , x0 + δ). Either way, since f has a limit at x0 and constant functions have limits at any point, g(x) has a limit at x0 . For (2), g ≡ f (x⋆ ) on (x0 − δ, x0 + δ) for some δ. Constant functions have limits at all points, so g(x) has a limit at x0 . Miscellaneous 26. Assume that f : R → R is such that f (x + y) = f (x)g(x) for all x, y ∈ R. If f has a limit at zero, prove that f has a limit at every point and either lim f (x) = 1 or f (x) = 0 for all x ∈ R. x→0 29 CHAPTER 2. LIMITS OF FUNCTIONS If f has a limit at 0, then for any an → 0, f (an ) → lim f (x). Then, since x→0 any input y can be written as 0 + y, we see that f (y + an ) = f (y)f (an ) → lim f (x) · f (y), showing that a limit exists for all inputs y. x→0 If we have a limit for any input, then we should have lim f (x + y) = x→x0 lim [f (x)f (y)] = lim f (x) · lim f (y). In particular, lim f (x + 0) = lim f (x) · x→x0 x→x0 x→x0 x→x0 x→0 lim f (x). Thus, lim f (x) = 1 or f ≡ 0. x→x0 x→0 27. Suppose f : D → R, g : E → R, x0 is an accumulation point of D ∩ E, and there is ǫ > 0 such that D ∩ [x0 − ǫ, x0 + ǫ] = E ∩ [x0 − ǫ, x0 + ǫ]. If f (x) = g(x) for all x ∈ D ∩ E ∩ [x0 − ǫ, x0 + ǫ], prove that f has a limit at x0 iff g has a limit at x0 . Let ǫ > 0 be given. Then, lim f (x) = L exists ⇔ |f (x) − L| < 2ǫ . At the x→x0 same time, |f (x) − g(x)| < 2ǫ . If g(x) has a limit at x0 , it must be equal to L. To see this, suppose |g(x) − L| ≥ ǫ0 for some x ∈ D∩E ∩[x0 −δ, x0 +δ]. We then have |g(x) − f (x)| = |g(x) − L + L − f (x)| > |g(x) − L| − |f (x) − L| > ǫ0 − ǫ. Contradiction. The argument is similar if we assume that g(x) has a limit at x0 . Chapter 3 Continuity 3.1 2. Continuity of a Function at a Point 1. Define f : R → R by f (x) = 3x2 − 2x + 1. Show that f is continuous at √ Let ǫ > 0 be given. Choose δ = 3x2 − 2x − 8 = x + 4 3 |x − 2| < x + 2. Define f : [−4, 0] → R by f (x) = Show that f is continuous at −3. Let ǫ > 0 be given. Choose δ = ǫ. We see that 3x2 − 2x + 1 − 9 = 4 2 3 ǫ 2. < δ 2 = ǫ. 2x2 −18 x+3 Then, for x 6= −3 and f (−3) = 12. 2x2 −18 x+3 |2(x − 3) + 3| = |2x − 3| < 2 |x − 3| < 2δ = ǫ. n√ o∞ n n+1 e 3. Use Theorem 3.1 to prove that n=1 x +3 = 2(x2 −9) x+3 +3 = is convergent and find the limit. You may assume that the function f (x) = e is continuous on R. n o n n+1 o 1 = e · e n . By Theorem 3.1, First, our sequence is equivalent to e n 1 e n → e0 = 1, so the limit of the sequence is e. 4. If x0 ∈ E, x0 is not an accumulation point of E, and f : E → R, prove that, for every sequence {xn }∞ n=1 converging to x0 with xn ∈ E for all n, converges to f (x ). {f (xn )}∞ 0 n=1 Suppose that f (xn ) 9 f (x0 ). Then, for any N , there is an ǫN such that |f (xn ) − f (x0 )| ≥ ǫN for some n > N . Thus, there always exists an xn within 30 31 CHAPTER 3. CONTINUITY δN distance of x0 such that |xn − x0 | ≥ δN . Thus, xn 9 x0 . q 5. Define f : (0, 1) → R by f (x) = √1x − x+1 x . Can one define f (0) to make f continuous at 0? x6=0 = − xx = 1. Thus, extending f so that f (0) = 1 We see that f (x) = 1−x−1 x will allow the function value to be equal to the limit, ensuring continuity. 6. Prove that f (x) = √ x is continuous for all x ≥ 0. Let xn → x0 . By Exercise 28 of Chapter 1, continuous for all x ≥ 0. √ xn → √ x0 . Thus, f (x) is 7. Suppose f√: R → R is continuous and f (r) = r2 for each rational number r. Determine f ( 2) and justify your conclusion. √ Since converging to 2 converges √ √ f is continuous, any sequence of numbers to f ( 2). By Exercise 45√from Chapter 1, 2 has a sequence of rationals that converges to it. Thus, f ( 2) = 2. 8. Suppose f : (a, b) → R is continuous and f (r) = 0 for each rational number r ∈ (a, b). Prove that f (x) = 0 for all x ∈ (a, b). By the same argument as above, since any irrational number has a sequence of rationals converging to it, f (j) = 0 for all irrational numbers in (a, b). 9. Define f : (0, 1) → R by f (x) = x sin x1 . Can one define f (0) to make f continuous at 0? Explain. Yes. Define f (0) = 0. Let ǫ > 0 be given and choose δ = ǫ. We then see that x sin x1 = |x| sin x1 < |x| < δ = ǫ for any x ∈ (−δ, δ). *10. Suppose f : E → R is continuous at x0 and x0 ∈ F ⊂ E. Define g : F → R by g(x) = f (x) for all x ∈ F . Prove that g(x) is continuous at x0 . Show by example that the continuity of g at x0 need not imply the continuity of f at x0 . Let ǫ > 0 be given. Then, since f is continuous at x0 , there is a δ > 0 such that for all x ∈ (x0 − δ, x0 + δ) we have |f (x) − f (x0 )| < ǫ. Then, choose δ ′ = |sup[(x0 − δ, x0 + δ) ∩ F ] − x0 |. Then, (x0 − δ ′ , x0 + δ ′ ) ⊆ (x0 − δ, x0 + δ), so we must have |g(x) − g(x0 )| < ǫ. If g(x) is continuous at x0 then f (x) is not necessarily continuous at x0 .  x if x ∈ R \ {2} Suppose for instance we have the function f (x) = . Let 1 if x = 2 32 CHAPTER 3. CONTINUITY E = R and F = {2}. Then, g(x) is trivially continuous at 2, but f (x) is not continuous at 2 since its left- and right-hand limits are not equal. 11. Define f : R → R by f (x) = 8x if x is rational and f (x) = 2x2 +8 if x is irrational. Prove from the definition that f is continuous at 2 and discontinuous at 1. Let ǫ > 0 be given. Choose δ1 = 8ǫ . Then, |8x − 16| = 8 |x − 2| ≤ 8δ1 = ǫ. Choose δ2 = min(2, 8ǫ ). Then, 2x2 + 8 − 16 = 2x2 − 8 = 2 x2 − 4 = 2 |x + 2| |x − 2| < 8 |x − 2| < 8δ = ǫ. Thus, choosing δ = min(δ1 , δ2 ) gives |f (x) − 16| < ǫ. Let ǫ0 = 2. Let δ > 0 be given. Then, there is always an irrational number 2 x ∈ (1 − δ, 1 + δ) so that |f (x) − 8| = 2x2 + 8 − 8 = 2x2 = 2 |x| . Since 1 ∈ (x − δ, x + δ) for all δ > 0, we see that there is always an x ∈ (1 − δ, 1 + δ) 2 such that |f (x) − 8| = 2 |x| = 2 = ǫ0 . 3.2 Algebra of Continuous Functions *12. Let p and q be polynomials and x0 be a zero of q of multiplicity m. Prove that p/q can be assigned a value at x0 such that the function thus defined will be continuous iff x0 is a zero of p of multiplicity greater than or equal to m. p(x) (x−x0 )m+ℓ (x−x0 )m+ℓ p̃(x) (x−x0 )ℓ p̃(x) then have p(x) . Define q(x) = (x−x0 )m q̃(x) = q̃(x) p̃(x) . Choose > 0 be given. Let M = max x∈(x −1,x +1) q̃(x) (⇐) If x0 is a zero of p of multiplicity n = m + ℓ, then set p̃(x) = and q̃(x) = q(x) (x−x0 )m . We p(x0 )/q(x0 ) = 0. Let ǫ 0 0 ǫ δ=M . Then, |f (x)| < |M (x − x0 )| < δM = ǫ. Thus, f (x) is continuous at x0 . (⇒, by way of contrapositive) Suppose that x0 is a zero of p of multiplicity n = m − ℓ. Then, as above, f (x) = (x−xp̃(x) . Let ǫ0 = 1. Let δ > 0 be ℓ 0 ) q̃(x) given and m = |f (x) − y| = min p̃(x) . Then suppose f (x0 ) = y. We then have x∈(x0 −δ,x0 +δ) q̃(x) p̃(x) m (x−x0 )q̃(x) − y > (x−x0 ) − y . Since 1 x−x0 gets arbitrarily close to zero, there will always be an x ∈ (x − δ, x + δ) such that |f (x) − y| > continuous. m(y+1) m 1 x−x0 > y+1 m . Thus, − y = 1 = ǫ0 . Thus, f (x0 ) cannot be defined so as to be 13. Let f : D → R be continuous at x0 ∈ D. Prove that there is M > 0 and a neighborhood Q of x0 such that |f (x)| ≤ M for all x ∈ Q ∩ D. Since f is continuous at x0 , we have that for every ǫ > 0, there is a δǫ > 0, such that if x ∈ (x0 − δǫ , x0 + δǫ ), we have −ǫ < f (x) − f (x0 ) < ǫ ⇔ f (x0 ) − ǫ < 33 CHAPTER 3. CONTINUITY f (x) < f (x0 ) + ǫ. Thus, let M = max{f (x0 ) ± ǫ} and Q = (x0 − δǫ , x0 + δǫ ), ǫ>0 where ǫ is the value chosen in the definition of M . Then, by construction, we have that |f (x)| ≤ M for every x ∈ Q ∩ D. 14. If f : D → R is continuous at x0 ∈ D, prove that the function |f | : D → R such that |f | (x) = |f (x)| is continuous at x0 . Let ǫ > 0 be given. Since f (x) is continuous at x0 , there exists a δ > 0 such that |f (x) − f (x0 )| < ǫ for any x ∈ (x0 − δ, x0 + δ). Now we see that ||f (x)| − f (x0 )| < |f (x) − f (x0 )| < ǫ for any x ∈ (x0 − δ, x0 + δ). 15. Suppose f, g : D → R are both continuous on D. Define h : D → R by h(x) = max{f (x), g(x)}. Show that h is continuous on D. Let x0 ∈ D and without loss of generality, suppose h(x0 ) = f (x0 ). Suppose that there is an ǫ0 > 0 such that for every δ > 0, we have an x ∈ (x0 − δ, x0 + δ) such that |h(x) − f (x0 )| ≥ ǫ0 . Since f (x) is continuous at x0 , there is a δ0 1 such that |f (x) − f (x0 )| < ǫ0 for all x ∈ (x0 − δ0 , x0 + δ0 ). Take δn = k+n 1 where k < δ0 . Construct {xn } by selecting xn ∈ (x0 − δn , x0 + δn ) such that |h(xn ) − f (x0 )| ≥ ǫ0 . By the construction of the sequence, we have h(xn ) = g(xn ) for all n. Then, xn → x0 , and since h(xn ) = g(xn ) and g(x) is continuous, we have h(xn ) → h(x0 ) = g(x0 ) 6= f (x0 ). Contradiction. 16. Assume the continuity of f (x) = ex and g(x) = ln(x). Define h(x) = xx by xx = xx ln x . Show that h is continuous for x > 0. I’m pretty sure that this question contains a typo as xx 6= xx ln x (just plug in x = 2 to see that there is no equality). I believe that the author intended to write xx = ex ln x . In this case, we know that the function k(x) = x is continuous, so h(x) = f (x) ◦ (k(x) · g(x)), and so is continuous. 17. Suppose that f√: D → R with f (x) ≥ 0 for all x ∈ D. Show that, if f is continuous at x0 , then f is continuous at x0 . Let ǫ > 0 be given. Then, there is a δ > 0 such that for any x ∈ (x0 −δ, x0 +δ), p p p p p p we have |f (x) − f (x0 )| < ǫ. Now, f (x) − f (x0 ) < f (x) − f (x0 ) f (x) + f (x0 ) = |f (x) − f (x0 )| < ǫ for all x ∈ (x0 − δ, x0 + δ). 18. Define f : R → R as follows: f (x) = x − [x] if [x] is even. f (x) = x − [x + 1] if [x] is odd. Determine where f is continuous. Justify. 34 CHAPTER 3. CONTINUITY We have already seen in Exercise 2.17 that this function has limits everywhere, and the function values at those endpoints in question are equal to the limit. Thus, the function is continuous everywhere. 3.3 Uniform Continuity: Open, Closed, and Compact Sets 19. Let f, g : D → R be uniformly continuous. Prove that f + g : D → R is uniformly continuous. What can be said about f g? Justify. Let ǫ > 0 be given. Then, there is some δ > 0 such that |f (x) − f (y)| < and |g(x) − g(y)| < 2ǫ if |x − y| < δ. Now, |f (x) + g(x) − f (y) − g(y)| ≤ |f (x) − f (y)| + |g(x) − g(y)| < 2ǫ + 2ǫ = ǫ. Now, f g is not necessarily uniformly continuous. For example, let f (x) = g(x) = x. Let ǫ0 = 1. Assume without loss of generality that y > x. Then, for any δ > 0, x2 − y 2 = |x + y| |x − y| > |x + y| |y + δ − y + δ| = |x + y| · 2δ > 1 1 , we get x2 − y 2 > 2 · 4δ · 2δ = 1 = ǫ0 . Thus, f (x)g(x) is not 2y · 2δ. If y = 4δ uniformly continuous. ǫ 2 20. Let f : A → B and g : B → C be uniformly continuous. What can be said about g ◦ f : A → C? Justify. The function g ◦ f is
not necessarily uniformly continuous if A 6= Ā. For instance g : (1, 2) → 1, 12 defined by g(x) = x1 is uniformly continuous on (1, 2). 1 Also, f (x) = x−1 is uniformly continuous everywhere. However, g(f (x)) = x−1 is not uniformly continuous on (1, 2) (for the same reason x1 isn’t uniformly continuous on (0, 1)). 2 . Show that f is uniformly 21. Define f : [3.4, 5] → R by f (x) = x−3 continuous on [3.4, 5] without using Theorem 3.8– that is, use the methods of Example 3.2. Let ǫ > 0 be given. Choose δ = 2 y−x (x−3)(y−3) ≤ 2 0.16 |x − y| ≤ 2 0.16 δ 0.16 2 ǫ. Then, 2 x−3 − 2 y−3 =2 y−3−x+3 (x−3)(y−3) = = ǫ. 22. Define f : (2, 7) → R by f (x) = x3 − x + 1. Show that f is uniformly continuous on (2, 7) without using Theorem 3.8– that is, use the methods of Example 3.3. ǫ . Then, x3 − x + 1 − y 3 + y − 1 = Let ǫ > 0 be given. Choose δ = 148 3 2 2 x − y + y − x = (x − y)(x + xy + y ) + x − y = (x − y)(x2 + xy + y 2 + 1) < 148 |x − y| < 148δ = ǫ. 3 CHAPTER 3. CONTINUITY 35 23. A function f : R → R is periodic iff there is a real number h 6= 0 such that f (x + h) = f (x) for all x ∈ R. Prove that if f : R → R is periodic and continuous, then f is uniformly continuous. Define fn : R → R by fn (x) = f (x − nh) if x ∈ [(n − 1)h, nh] and fn (x) = 0 otherwise. Then, fn is continuous on P [(n − 1)h, nh] which is compact, making it uniformly continuous. Then, f (x) = fn (x) is a sum of uniformly continuous functions which is uniformly continuous by Exercise 19. 24. Suppose A is bounded and not compact. Prove that there is a function that is continuous on A but not uniformly continuous. Give an example of a set that is not compact, but every function continuous on that set is uniformly continuous. If A is bounded and not compact, then it is not closed (Heine-Borel The1 will orem). Thus, suppose that x0 ∈ Ā \ A. Then, the function f (x) = x−x 0 not be uniformly continuous. (Since it is an accumulation point, inputs can get 1 arbitrarily large. Then, the function fails arbitrarily close to x0 making x−x 0 to be uniformly continuous for the same reason that x1 fails to be uniformly continuous.) It is nonetheless continuous (we’ve seen why before). A set that is not compact would be N. Every function that is continuous on N (which is to say any function defined on N) is also uniformly continuous on N. (Since δ < 1 implies that x = y and |f (x) − f (x)| = 0 < ǫ.) 25. Give an example of sets A and B and a continuous function f : A∪B → R such that f is uniformly continuous on A and uniformly continuous on B, but not uniformly continuous on A ∪ B. Let A = (−∞, 0] and B = (0, ∞). Define f (x) = x if x ∈ A and f (x) = x + 1 if x ∈ B. Then, f is not continuous at 0, let alone uniformly continuous. Nevertheless, f is uniformly continuous on both A and B. *26. Let E ⊂ R. Prove that E is closed if, for every x0 such that there is a sequence {xn }∞ n=1 of points of E converging to x0 , it is true that x0 ∈ E. In other words, prove E is closed if it contains all limits of sequences of members of E. Suppose that for every xn → x0 , we have x0 ∈ E. Then, by the properties of convergence and the fact that xn ∈ E for all xn , for any ǫ > 0, we have a xk ∈ {xn } ⊆ E such that xk ∈ (x0 − ǫ, x0 + ǫ) ∩ E. Thus, x0 is a point of closure. Thus, E is closed. *27. Prove that every set of the form {x : a < x < b} is open and every set of the form {x : a ≤ x ≤ b} is closed. CHAPTER 3. CONTINUITY 36 Let x0 ∈ {x : a < x < b} = (a, b). Then, choose ǫ0 = min(x0 − a, b − x0 ). Then by construction, we have (x0 − ǫ0 , x0 + ǫ0 ) ⊂ (a, b). Thus, (a, b) is open. Let x0 ∈ {x : a ≤ x ≤ b} = [a, b]. Let ǫ > 0 be given. Then, x0 ∈ (x0 − ǫ, x0 + ǫ) ∩ [a, b], so x0 is a point of closure. Thus, [a, b] is closed. 28. Let D ⊂ R, and let D′ be the set of accumulation points of D. Prove that D̄ = D ∪ D′ is closed and if F is any closed set that contains D, then D̄ ⊂ F . D̄ is called the closure of D. Since D̄ contains all the accumulation points of D (since D̄ = D ∪ D′ ), D̄ is closed by Exercise 27. Now, since F is closed, F also contains all of its accumulation points. Since D ⊂ F , we know that F must contain all of the accumulation points of D. Thus, D̄ ⊂ F . 29. If D ⊂ R, prove that D̄ is bounded. Suppose that D̄ is unbounded. Let M = sup D and x0 ∈ D̄ \ D with |x0 − M | > 100. Then, choosing ǫ0 = 50, we see that (x0 − ǫ0 , x0 + ǫ0 ) ∩ D = ∅. This contradicts that D̄ is closed. Thus, D̄ is bounded. 30. Suppose f : R → R is continuous and let r0 ∈ R. Prove that {x : f (x) 6= r0 } is an open set. Define the preimage of y under f to be f ← (y) = {x ∈ R : f (x) = y}. Let x0 ∈ {x : f (x) 6= r0 }. Choose x̄ ∈ f ← (r0 ) such that |x̄ − x0 | is minimal. Choose ǫ0 = inf{|x − y| : x, y ∈ f ← (r0 )}. Then, by the continuity of f , there is a δ0 > 0 such that |f (x) − r0 | < ǫ0 . Let δ = min(|x − x̄| , δ0 ). Then, by construction, (x0 − δ, x0 + δ) ⊆ {x : f (x) 6= r0 }. 31. Suppose f : [a, b] → R and g : [a, b] → R are both continuous. Let T = {x : f (x) = g(x)}. Prove that T is closed. Let x0 ∈ T and let ǫ > 0 be given. Then, x0 ∈ (x0 − ǫ, x0 + ǫ). Thus, T is closed. (This problem is quite trivial.) 32. If D ⊂ R, then x ∈ D is said to be an interior point of D iff there is a neighborhood Q of x such that Q ⊂ D. Define D◦ to be the set of interior points of D. Prove that D◦ is open and that if S is any open set contained in D, then S ⊂ D◦ ⊂ D. D◦ is called the interior of D. Since for any point x0 ∈ D◦ , there is a ǫ > 0 such that (x0 − ǫ, x0 + ǫ) ⊂ D◦ , we know that D◦ is open. Let S ⊂ D be open. Then, for any x0 ∈ S, there must be ǫ > 0 such that (x0 − ǫ, x0 + ǫ) ⊂ S. Since S ⊂ D, we have (x0 − ǫ, x0 + ǫ) ⊂ S ⊂ D◦ . Thus, S ⊂ D◦ ⊂ D. 37 CHAPTER 3. CONTINUITY 33. Find an open cover of {x : x > 0} with no finite subcover. Let Gn = (0, n]. Then, subcover. S∞ n=1 Gn is a cover of (0, ∞) that has no finite 34. Find an open cover of (1, 2) with no finite subcover.   S∞ Let Gn = 1 − n1 , 2 − n1 . Then, n=1 Gn = (1, 2) that has no finite subcover. *35. Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E. Since E is compact, it is closed and bounded. Now, sup E and inf E are both accumulation points of E, so since E must be bounded, inf E, sup E ∈ E. 36. If E1 , ..., En are compact sets, prove that E = ∪ni=1 Ei is compact. Since each Ei is bounded, we clearly have that ∪Ei is bounded. Next, if x0 ∈ ∪Ei , then x0 ∈ Ei for some i. Since each Ei is closed, x0 is a point of closure. Thus, ∪Ei is closed. 37. Let f : [a, b] → R have a limit at each x ∈ [a, b]. Prove that f is bounded. Suppose that for any n, there is an xn ∈ [a, b] such that f (xn ) > n. Then, f (xn ) → ∞, but xn 9 ∞ since [a, b] contains all of its accumulation points. This contradicts the fact that f has a limit for every x ∈ [a, b]. Thus, f is bounded. 38. Suppose f : D → R is continuous with D compact. Prove that {x : 0 ≤ f (x) ≤ 1} is compact. Since D is compact, it is closed and bounded. Thus, by Exercise 37, {x : 0 ≤ f (x) ≤ 1} is bounded. Let x0 ∈ {x : 0 ≤ f (x) ≤ 1}. Now, since f is continuous, for ǫ0 = min (f (x0 ), 1 − f (x0 )), there is a δ for which x ∈ (x0 − δ, x0 + δ) ∩ D implies that |f (x) − f (x0 )| < ǫ. Since this is true of all x in this interval, define δn = δ/n. We can then construct {xn } such that xn ∈ (x0 − δn , x0 + δn ) for all n. Thus, the sequence xn → x0 (since δn → 0) is a sequence of members of D. Thus, x0 is a point of closure. Thus, {x : 0 ≤ f (x) ≤ 1} is closed. 39. Suppose f : R → R is continuous and has the property that for each ǫ > 0, there is M > 0 such that if |x| > M , then |f (x)| < ǫ. Show that f is uniformly continuous. Let ǫ > 0 be given. Then, since |f (x)| < ǫ for all x ∈ (−∞, −M ) ∪ (M, ∞), |f (x) − f (y)| < |f (x)| < ǫ for all x, y ∈ (−∞, −M ) ∪ (M, ∞). Thus, f is CHAPTER 3. CONTINUITY 38 uniformly continuous on (−∞, −M ) ∪ (M, ∞). Then, by Theorem 3.8, f is uniformly continuous on [−M, M ] (since it is compact). Thus, f is uniformly continuous. 40. Give an example of a function f : R → R that is continuous and bounded but not uniformly continuous. Let f (x) = sin(x2 ). Then, f is clearly bounded and continuous. However, the further away from the origin we go, we see that the maxima become closer and closer to each other. Therefore, for any delta we choose, we can find x values far enough away from the origin so that sin(x2 ) − sin(y 2 ) = 1. 3.4 Properties of Continuous Functions 41. Find an interval of length 1 that contains a root of xex = 1. We see that 0 · e0 − 1 = −1 and 1 · e1 − 1 = e − 1 > 0. Thus, by Bolzano’s Theorem, [0,1] contains a root of xex = 1. 42. Find an interval of length 1 that contains a root of the equation x3 − 6x2 + 2.826 = 0. We see that 03 − 6 · 02 + 2.826 = 2.826 while 13 − 6 · 12 + 2.826 < 0. Thus, by Bolzano’s Theorem, [0, 1] contains a root of x3 − 6x + 2.826 = 0. 43. Suppose f : [a, b] → R is continuous and f (b) ≤ y ≤ f (a). Prove that there is c ∈ [a, b] such that f (c) = y. If y ∈ (f (b), f (a)), then by the Intermediate Value Theorem, there is c ∈ (a, b) such that f (c) = y. Otherwise, f (a) = y or f (b) = y. (This is a trivial problem.) 44. Suppose f : [a, b] → [a, b] is continuous. Prove that there is at least one fixed point in [a, b]– that is x such that f (x) = x. Define F (x) = f (x) − x. We see that F (x) is also continuous. Now, sup f (x) ≤ b, so F (x) ≤ 0 for some x = c1 . Also, inf f (x) ≥ a, so F (x) ≥ 0 for some c2 . Then, by Bolzano’s Theorem, there is c ∈ [c1 , c2 ] such that F (c) = 0. Thus, f (x) has a fixed point. 45. If f : [a, b] → R is 1-1 and has the intermediate value property– that is, if y is between f (u) and f (v), there is x between u and v such that f (x) = y– show that f is continuous. (Hint: First show that f is monotone.) 39 CHAPTER 3. CONTINUITY First, f is monotone. To see this, suppose not. Then, there is some point x0 where f (x) < f (x0 ) on one side and f (x) > f (x0 ) on the other for values x ∈ (x0 − δ, x0 + δ) for some δ > 0. Without loss of generality, let us assume that it is increasing to the left of x0 and increasing to the right of x0 . By the intermediate value property, f attains all values in [f (x0 − δ), f (x0 )] and [f (x0 ), f (x0 + δ)]. Let ǫ = min (|f (x0 ) − f (x0 − δ)| , |f (x0 ) − f (x0 + δ)|). By the intermediate value property, every function value in (f (x0 ) − ǫ, f (x0 )) must be attained in both intervals, showing that at least one (many more than one, actually) function value must have two elements in its preimage. Contradiction. Without loss of generality, suppose that f is increasing. Let x0 ∈ [a, b] and let ǫ > 0 be given. Since ǫ > 0 is only of interest when it is small, suppose (f (x0 ) − ǫ, f (x0 ) + ǫ) ⊂ (f (a), f (b)). Then, by the intermediate value property, this implies that there are x1 , x2 ∈ [a, b] such that f (x1 ) = f (x0 )−ǫ and f (x2 ) = f (x0 )+ǫ. Take δ = min (|x0 − x1 | , |x0 − x2 |). Then, for every x ∈ (x0 −δ, x0 +δ) we have |f (x0 ) − f (x)| < ǫ. Thus, f is continuous at x0 . 46. Prove that there is no continuous function f : R → R such that, for each c ∈ R, the equation f (x) = c has exactly two solutions. Since f is continuous, it has the intermediate value property. From this, we can narrow down the form that f can have. We see that f must be increasing, then decreasing or vice versa. If f changes its increasing/decreasing status any more, some function value will have 3 elements in its preimage. Also, if f is monotone, then f is 1-1 (and so won’t fit the criteria outlined in the exercise). Without loss of generality, assume f is increasing on some interval (−∞, x0 ) then decreasing on (x0 , ∞). (Note that f cannot be constant at any interval since this would result in f (x) = const. for infinitely many x.) We examine f (x0 ). Now, we know that to the left of x0 , f (x) < f (x0 ) and to the right, f (x0 ) > f (x) (again, we do not have equality because of the above discussion). Thus, f (x0 ) can have only one element in its preimage. Miscellaneous 47. Let f : R → R be additive. (See Project 2.1 at the end of Chapter 2.) That is, f (x + y) = f (x) + f (y) for all x, y ∈ R. In addition, assume there are M > 0 and a > 0 such that if x ∈ [−a, a], then |f (x)| ≤ M . Prove that f is uniformly continuous. In particular, prove that there is a real number m such that f (x) = mx for all x ∈ R. From Project 2.1, we know that an additive function that fits the above criteria has a limit at each point and lim f (x) = f (x0 ). Now, notice that x→x0 |f (x) − f (y)| = |f (x − y)|. So, we actually only need to show that for any CHAPTER 3. CONTINUITY 40 ǫ > 0, there is a δ > 0 such that |f (x)| < ǫ for any x ∈ (−δ, δ). Now, since f is additive, we have f (x) = f (0 + x) = f (0) + f (x). Thus, f (x) = 0. Finally, since f is continuous at each point, it is continuous at x = 0. Thus, there is δ > 0 such that |f (x) − 0| = |f (x)| < ǫ. Thus, f is uniformly continuous. By the axioms for the real numbers, if f (x) = y there is an m such that y = mx. Now, suppose for x1 and x2 , there are m1 and m2 for which f (x1 ) = m1 x1 and f (x2 ) = m2 x2 . Then, since f is additive, we must have f (x1 +x2 ) = f (x1 )+ f (x2 ) = m1 x1 + m2 x2 . By the distributive law, this implies that m1 = m2 = m so that f (x1 + x2 ) = m(x + y) = mx1 + mx2 . 48. Let f : [a, b] → R be continuous, and define g : [a, b] → R by g(t) = sup{f (x) : a ≤ x ≤ t} Prove that g is continuous. We see that g is either identical to f or constant. Now, let ǫ > 0 and x0 ∈ [a, b] be given. By the continuity of f , there is a δ > 0 for which |f (x) − f (x0 )| < ǫ. Now, if g ≡ f on (x0 −δ, x0 +δ), then g is clearly continuous. If g(x) 6= f (x) for some x ∈ (x0 − δ, x0 + δ), then g(x) = const. and we can find x̄ = inf{x ∈ (x0 − δ, x0 + δ) : g(x) 6= f (x)}. Choose δ ′ = min(δ, |x̄ − x0 |). Then, by construction, |g(x0 ) − g(x)| < ǫ for all x ∈ (x0 − δ ′ , x0 + δ ′ ). 49. Suppose that g : D → R is continuous at x0 and that x0 is also an accumulation point of D. Define D0 = {x : g(x) = 0}. If g(x) 6= 0, prove that x0 is an accumulation point of D0 . Choose x̄ to be such that g(x̄) = 0 and |x̄ − x0 | is minimum. Then, by the Intermediate Value Theorem, g attains all values between g(x) and g(x0 ) for any |x − x0 | < |x̄ − x0 | (of which there are infinitely many). Thus, any neighborhood of x0 contains an infinity of points of D0 . Thus, x0 ∈ D0′ . 50. Suppose f : D → R , g : E → R, and x0 ∈ D ∩ E. Suppose further that there is ǫ > 0 such that D ∩ [x0 − ǫ, x0 + ǫ] = E ∩ [x0 − ǫ, x0 + ǫ] and f (x) = g(x) for all x ∈ D ∩ E ∩ [x0 − ǫ, x0 + ǫ]. Prove that f is continuous at x0 iff g is continuous at x0 . If f is continuous at x0 , then for all ǫ′ > 0, there is a δ > 0 such that for all x ∈ (x0 − δ, x0 + δ), we have |f (x) − f (x0 )| < ǫ′ . If δ < ǫ, then we have g ≡ f on (x0 − δ, x0 + δ) since D ∩ [x0 − δ, x0 + δ] = E ∩ [x0 − δ, x0 + δ]. If δ ≥ ǫ, then for all x ∈ (x0 − ǫ, x0 + ǫ), we also have |f (x) − f (x0 )| < ǫ′ . (Since (x0 − ǫ, x0 + ǫ) ⊆ (x0 − δ, x0 + δ).) Thus, f (x) and f (x0 ) can be replaced by g(x) and g(x0 ) respectively. The argument is similar if we assume that g(x) is continuous at x0 . Chapter 4 Differentiation 4.1 The Derivative of a Function 1. Let (x0 , y0 ) be an arbitrary point on the graph of the function f (x) = x2 . For x0 6= 0, find the equation of the line tangent to f at that point by finding a line that intersects the curve in exactly one point. Do not use the derivative to find this line. We need x20 = mx0 + b ⇔ x20 − mx0 − b = 0. Since we want only one solution, 2 we must have the discriminant (−m)2 − 4(−b) = m2 + 4b = 0. Thus, b = −m 4 2 and the equation is x20 − mx0 + m4 = 0. This implies that m = 2x0 . Thus, the equation of the tangent line is y − x20 = 2x0 (x − x0 ). 2. Prove that the definition of the derivative and the alternate definition of the derivative are equivalent. Set x + h = x0 . Then, since as x → x0 , h → 0, we have lim x→x0 f (x)−f (x0 ) x−x0 = (x) lim f (x+h)−f . h h→0 √ 3. Use the definition to find the derivative of f (x) = x, for x > 0. Is f differentiable at zero? √ √ √ √  x √x+h+ x 1 √ 1 √ √ = lim √x+h+ = 2√ = lim h(√x+h−x . The lim x+h− h x x+h+ x x+h+ x) x h→0 h→0 h→0 limit does not exist at x = 0, so f is not differentiable there. 4. Use the definition to find the derivative of g(x) = x2 . 41 42 CHAPTER 4. DIFFERENTIATION 2 lim (x+h)h h→0 −x2 = lim x 2 h→0 +h2 +2xh−x2 h 3 = lim h(h+2x) = lim (h + 2x) = 2x. h h→0 h→0 1 x 5. Define f (x) = x sin for x 6= 0 and h(0) = 0. Show that h is differentiable everywhere and that h′ is continuous everywhere but fails to have a derivative at one point. You may use the rules for differentiating products, sums and quotients of elementary functions that you learned in calculus.
3 = We see that f ′ (x) = 3x2 sin x1 + cos x1 · − x12 · x3 (⋆). Now, lim x sin(x+h) h lim x h→0 ′ 3 h→0 sin x sin h−cos x cos h h = lim x3 sin x · h→0 sin h h − cos x · cos h h ′ = 0 at x = 0. Thus, f (0) = 0 and is defined everywhere. Now, from (⋆), f is continuous at everywhere except possibly at x = 0. Using elementary calculus, we see that the limit of f ′ as x → 0 is −∞. Thus, f ′ is not continuous at x = 0. 6. Suppose f : (a, b) → R is differentiable at x ∈ (a, b). Prove that lim h→0 f (x + h) − f (x − h) 2h exists and equals f ′ (x). Give an example where this limit exists, but the function is not differentiable. (x) exists, then by the same token, lim f (x)−fh(x−h) exists. If lim f (x+h)−f h h→0 h→0 h i (x) f (x+h)+f (x−h) f (x)−f (x−h) Thus, lim f (x+h)−f = lim + exists by the limit h h h h→0 h→0 (x−h) must also exist. (It’s just the last limit multiplied laws, so lim f (x+h)−f 2h h→0 (x) by 1/2.) Then, since lim f (x+h)−f = lim f (x)−fh(x−h) = f ′ (x), we see that h (x−h) lim f (x+h)+f = 2h h→0 h→0 2f ′ (x) 2 h→0 = f ′ (x). (x−h) = Let f (x) = |x|. Then, lim f (x+h)+f 2h h→0 2h 2h = 1 near x = 0. Thus, the limit exists, but from what we know of calculus, f is not differentiable at x = 0. 7. A function f : (a, b) → R satisfies a Lipschitz condition at x ∈ (a, b) iff there is M > 0 and ǫ > 0 such that |x − y| < ǫ and y ∈ (a, b) imply that |f (x) − f (y)| < M |x − y|. Give an example of a function that fails to satisfy a Lipschitz condition at a point of continuity. If f is differentiable at x, prove that f satisfies a Lipschitz condition at x. (y)| We first make the observation that |f (x) − f (y)| ≤ |x − y| M ⇔ |f (x)−f ≤ |x−y| M (if x 6= y), so essentially the criterion is that the average rate of change be p 2 if x ∈ [0, 2) p1 − (x − 1) . tween x and y is bounded. Define f (x) = 1 − (x + 1)2 if x ∈ (−2, 0] The following is the graph of the function. 43 CHAPTER 4. DIFFERENTIATION We see that at x = 0, the graph has a vertical tangent line, so as x → 0, the average rate of change from 0 to x will go to ∞. Thus, f is not Lipschitz at x = 0. (y) = M . Thus, for Now, if f is differentiable at x, we have lim f (x)−f x−y y→x any ǫ > 0, there is a δ > 0 such that for any y ∈ (x − δ, x + δ), we have f (x)−f (y) x−y (y) (y) (y) − M ≤ f (x)−f < − M < ǫ. Then, f (x)−f − M , so f (x)−f x−y x−y x−y ǫ + M and |f (x) − f (y)| < |x − y| (ǫ + M ). Thus f is Lipschitz at x. 8. A function f : (a, b) → R is said to be uniformly differentiable iff f is differentiable on (a, b) and for each ǫ > 0, there is δ > 0 such that 0 < |x − y| < δ and x, y ∈ (a, b) imply that f (x) − f (y) − f ′ (x) < ǫ. x−y Prove that if f is uniformly differentiable on (a, b), then f ′ is continuous on (a, b). Since f is uniformly continuous, for every ǫ > 0, there is a δ for which if |x − (y) (y) y| < δ we have −ǫ/2 < f (x)−f − f ′ (x) < ǫ/2. Thus, for the same δ, f (x)−f − x−y x−y ǫ/2 < f ′ (x) < f (x)−f (y) x−y ′ f (x)−f (y) x−y ′ f (x)−f (y) − ǫ/2 < f ′ (y) < x−y   ǫ/2 − f (x)−f (y) + ǫ/2 = ǫ. x−y + ǫ/2. By the same token, + ǫ/2. Thus, |f (x) − f ′ (y)| < Thus, f (x) is continuous on (a, b). f (x)−f (y) x−y − 9. Suppose f : (a, b) → R is continuous on (a, b) and differentiable at x0 ∈ (a, b). Define g(x) = f (x) − f (x0 ) for x ∈ (a, b) \ {x0 }, g(x0 ) = f ′ (x0 ). x − x0 Prove that g is continuous on (a, b). 44 CHAPTER 4. DIFFERENTIATION We see that g(x) represents the average rate of change from the variable point x 6= x0 to the set point x0 . Since f itself is continuous, g is also continuous here. At x = x0 , g is the derivative, which is defined to be the limit of the average rate of change as x → x0 . Thus, limit = function value and g is continuous at x0 . 10. Suppose f, g, and h are defined on (a, b) and a < x0 < b. Assume f and h are differentiable at x0 , f (x0 ) = h(x0 ), and f (x) ≤ g(x) ≤ h(x) for all x ∈ (a, b). Prove that g is differentiable at x0 and f ′ (x0 ) = g ′ (x0 ) = h′ (x0 ). f (xn )−f (x0 ) → f ′ (x0 ) = h′ (x0 ) ← xn −x0 g(xn )−g(x0 ) (x0 ) 0) 0) as n → ∞. Thus, since f (xxnn)−f ≤ g(xxnn)−g(x ≤ h(xxnn)−h(x xn −x0 −x0 −x0 −x0 ′ ′ for all n, we see that the middle sequence must converge to f (x0 ) = h (x0 ). Define xn = x0 + 1/n, repeat the same argument (except that the inequalities 0) in the middle). By proving the convergence will be reversed with g(xxnn)−g(x −x0 of this sequence, there is a δ = 1/N (where N is chosen to prove that the 0) sequences converge) such that g(x)−g(x < ǫ. Thus, g ′ (x0 ) exists and is equal x−x0 to f ′ (x0 ) = h′ (x0 ). Define xn = x0 − 1/n. Then, we see that 4.2 The Algebra of Derivatives 11. Prove f : (0, 1) → R defined by f (x) = on (0, 1) and compute the derivative. √ 2x2 − 3x + 6 is differentiable √ We know that x is differentiable on (0, ∞). We know that 2x2 − 3x + 6 √ 2 is differentiable on R. Thus, 2x − 3x + 6 is differentiable for all x such that 2x2 − 3x + 6 > 0. This is always the case, so the function is differentiable and the derivative is 1/2(2x2 − 3x + 6)−1/2 (4x − 3). 12. Suppose f : [a, b] → [c, d], g : [c, d] → [p, q], and h : [p, q] → R, with f differentiable at x0 ∈ [a, b], g differentiable at f (x0 ), and h differentiable at g(f (x0 )). Prove that h ◦ (g ◦ f ) is differentiable at x0 and find the derivative. The chain rule implies that [h(g(f (x0 )))]′ = h′ (g(f (x0 )) · g ′ (f (x0 )) · f ′ (x0 ). Done. 13. Suppose f : [a, b] → [c, d] and g : [c, d] → R are differentiable on [a, b] and [c, d], respectively. Suppose f ′ : [a, b] → R and g ′ : [c, d] → R are also differentiable on [a, b] and [c, d], respectively. Show that (g ◦f )′ : [a, b] → R is differentiable and find the derivative. CHAPTER 4. DIFFERENTIATION 45 By the chain rule, [g(f (x))]′ = g ′ (f (x)) · f ′ (x). Since this is defined for all possible inputs, this must indeed be the derivative. 14. Suppose f : R → R is differentiable and define g(x) = x2 f (x3 ).Show that g is differentiable and compute g ′ . Chain rule + product rule: g ′ (x) = 2x · f (x3 ) + x2 · f ′ (x3 ) · 3x2 . q p √ x + x + x for x ≥ 0. Determine where f is 15. Define f (x) = differentiable and compute the derivative. p √ √ Chain rule. f ′ (x) = 1/2(x+ x + x)−1/2 ·(1+ 1/2(x+ x)−1/2 ·(1+ 1/2x−1/2 )). Now, x cannot be 0 since there will be a factor in the denominator. 4.3 Rolle’s Theorem and the Mean Value Theorem √ 16. Define f : [0, 2] → R by f (x) = 2x − x2 . Show that f satisfies the conditions of Rolle’s theorem and find c such that f ′ (c) = 0. The function is clearly continuous on the interval, and the chain rule will show that the function is differentiable on (0, 2). Also, f (0) = f (2) = 0. Next, f ′ (x) = 1/2(2x − x2 )(2 − 2x). This equals 0 at x = 1/2. 17. Define f : R → R by f (x) = 1/(1 + x2 ). Prove that f has a maximum value and find the point at which the maximum occurs. 2x f ′ (x) = − (1+x 2 )2 , so f has its only extremum at x = 0. Since the slope of the tangent line is positive to the left of x = 0 and negative to the right, it follows that f attains its maximum value at x = 0. 18. Prove that the equation x3 − 3x + b = 0 has at most one root in the interval [−1, 1]. The function f (x) = x3 − 3x + b has derivative f ′ (x) = 3x2 − 3. Thus, the extrema occur at ±1. This means that the slope of the tangent line switches from negative to positive only once on this interval. Thus, only one root (at the most) exists.   19. Show that cos x = x3 + x2 + 4x has exactly one root in 0, π2 . 46 CHAPTER 4. DIFFERENTIATION Since cos x is decreasing on the interval and x3 + x2 + 4x is increasing on the interval, their graphs can intersect but once. 20. Suppose f : [0, 2] → R is differentiable, f (0) = 0, f (1) = 2, and f (2) = 2. Prove that 1. there is c1 such that f ′ (c1 ) = 0, 2. there is c2 such that f ′ (c2 ) = 2, and 3. there is c3 such that f ′ (c3 ) = 23 . (1) Since f (1) = f (2) = 2, Rolle’s Theorem implies that there is c1 such that f ′ (c1 ) = 0. (0) = 2−0 (2) By MVT, there is c2 such that f ′ (c) = f (1)−f 1 1 = 2. ′ (3) Since f is continuous, IVT implies that there is c3 such that f ′ (c3 ) = 3/2. 21. Let f : [0, 1] → R and g : [0, 1] → R be differentiable with f (0) = g(0) and f ′ (x) > g ′ (x) for all x ∈ [0, 1]. Prove that f (x) > g(x) for all x ∈ (0, 1]. Define D(x) = f (x) − g(x) for all x ∈ [0, 1]. We already have D(0) = 0. Suppose that g(x0 ) ≥ f (x0 ) for some x0 . Then, there is c for which f (c) = g(c) by IVT. Thus, D(c) = 0. Then, by Rolle’s Theorem, there is ξ ∈ (0, c) for which D′ (ξ) = f ′ (ξ) − g ′ (ξ) = 0. This implies that f ′ (ξ) = g ′ (ξ). Contradiction. 22. Use the Mean-Value Theorem to prove that ny n−1 (x − y) ≤ xn − y n ≤ nxn−1 (x − y) if n ≥ 1 and 0 ≤ y ≤ x. (y) ⇔ Define f (z) = z n . By MVT, there is c ∈ (x, y) for which f ′ (c) = f (x)−f x−y ncn−1 (x − y) = xn − y n . The only critical point of f is at z = 0. Thus, this equality holds for all x, y. Since 0 ≤ y ≤ x, this means that the lower bound for xn − y n is ny n−1 (x − y), while the upper bound is nxn−1 (x − y). Hence, ny n−1 (x − y) ≤ xn − y n ≤ nxn−1 (x − y). 23. Use the Mean-Value Theorem to prove that √ 1 + h < 1 + 1/2h for h > 0. ? h>0 Since LHS, RHS > 1, we square both sides and obtain 1+h < 1+ 1/4h+h ⇔ X 0 < 1/4h. I don’t know how or why MVT would be used to solve this. CHAPTER 4. DIFFERENTIATION 47 24. Generalize Exercise 23 as follows: If 0 < p < 1 and h > 0, then show that (1 + h)p < 1 + ph. You may assume the usual rules about differentiating powers. (0) . Thus, By MVT, there is always a c ∈ (0, h) such that f ′ (c) = f (h)−f h ′ if f (c) < 0, we have f (h) < f (0). Define f (x) = (1 + px) − (1 + x)p . We see f (0) = 0 and f ′ (x) = p − p(1 + x)p−1 = p[1 − (1 + x)p−1 ]. For c > 0 (the only possible ones), we clearly have f ′ (c) > 0, so f (h) > f (0). Thus, (1 + ph) − (1 + h)p > 0 ⇔ (1 + h)p < 1 + px. 25. Suppose f : (a, b) → R is differentiable and |f ′ (x)| ≤ M for all x ∈ (a, b). Prove that f is uniformly continuous on (a, b). Give an example of a function f : (0, 1) → R that is differentiable and uniformly continuous on (0, 1), but such that f ′ is unbounded. (y) for all x, y ∈ (a, b). By MVT, there is c ∈ (x, y) for which f ′ (c) = f (x)−f x−y ǫ Let ǫ > 0 be given and choose δ = /M Now, |f (x) − f (y)| ≤ f ′ (c)|x − y| ≤ M |x − y| < M δ = ǫ. Thus, √ f is uniformly continuous. The function f (x) = 1 − x2 is differentiable on (0, 1). It is also uniformly continuous on (0, 1) (we have seen why). But, we have also shown (Q7) that the tangent lines to the graph tend to a vertical line as x → 1− and x → 0− , so f ′ (x) → ∞ as x → 0+ , 1− and is unbounded. 26. Suppose f is differentiable on (a, b) except possibly at x0 ∈ (a, b), and is continuous on [a, b]; assume limx→x0 f ′ (x) exists. Prove that f is differentiable at x0 and f ′ is continuous at x0 . f (x)−f (x0 ) for x−x0 f (x)−f (x0 ) f (x)−f (x0 ) ′ lim f (cδ ) = lim x−x0 = lim = x−x0 x→x0 δ→0 δ→0 at x0 and since f ′ (x0 ) = lim f ′ (x), f ′ is also x→x0 Let δ be given. Then, by MVT, there is cδ such that f ′ (cδ ) = x ∈ (x0 − δn , x0 + δn ). Then, f ′ (x). Thus f is differentiable continuous. 27. Define f (x) = x + 2x2 sin x1 for x 6= 0 and f (0) = 0. Prove that f is differentiable everywhere. Show that there exists a number a such that f ′ (a) > 0, but there does not exist a neighborhood of a in which f is increasing. First, f is continuous since lim f (x) = 0 = f (0). Thus, by Q26, f ′ (0) exists. x→0 Thus, f is differentiable everywhere. (0) By MVT, there is c ∈ (0, a) such that f ′ (c) = f (a)−f = f (a) a−0 a . Thus. ′ f (c) < 0 precisely when f (a) < 0. There are an infinity of such points. Now, 2x2 cos 1 x near 0, f ′ (x) = 1 + 4x sin x1 − = 1 + 4x sin x1 − 2 cos x1 . We thus have x2 horizontal tangent lines at 2/kπ for each k. CHAPTER 4. DIFFERENTIATION 48 28. Prove that the function f (x) = 2x3 + 3x2 − 36x + 5 is 1-1 on the interval [−1, 1]. Is f increasing or decreasing. f ′ (x) = 6x2 + 6x − 36 = (x + 3)(x − 2). This function has no roots in [−1, 1], so f is 1-1. f ′ (0) = −6 < 0, so f is decreasing. 29. Show that the function f (x) = x3 − 3x2 + 17 is not 1-1 on the interval [−1, 1]. f ′ (x) = 3x2 − 6x = 3x(x − 2) which has a root at x = 0 ∈ [−1, 1] and so f is not 1-1. 30. Give an example of a function f : R → R that is differentiable and 1-1, but f ′ (x) = 0 for some x ∈ R. Let f (x) = x3 . Then, f ′ (x) = 3x2 which has a root at x = 0. As we know, x is a 1-1 function. 3 31. If f : [a, b] → R is differentiable at c, a < c < b and f ′ (c) > 0, prove that there is x, c < x < b, such that f (x) > f (c). If f ′ (c) > 0, then f is increasing on some interval about c and so there is x such that f (x) > f (c). 4.4 L’Hospital’s Rule and the Inverse-Function Theorem 32. Assume the rules for differentiating the elementary functions, and L’Hospital’s Rule and find the following limits: ln x a. lim x−1 x→1 b. lim exx−1 x→0 c. lim sinx x x→0 H 1/x = 1. x→1 1 H lim xx = lim e1x = 1. x→0 x→0 e −1 ln x (a) lim x−1 = lim x→1 (b) H (c) lim sinx x = lim cos1 x = 1. x→0 x→0 49 CHAPTER 4. DIFFERENTIATION 33. Use L’Hospital’s Rule to find the limit: x2 sin x . x→0 sin x − x cos x lim H 2 2 H 2 2 H x+(2−x ) sin x ) cos x−6x sin x sin x 2x sin x+x cos x lim sinxx−x = lim 4x cos = lim (6−x = cos x = lim x sin x x cos x+sin x 2 cos x−x sin x 3. x→0 x→0 x→0 x→0 34. Prove the following variant of Theorem 4.14. Suppose f : [a, b] → R is 1-1. If f is differentiable at c, f ′ (c) 6= 0, and f −1 is continuous at d = f (c), then f −1 is differentiable at d and (f −1 )′ (d) = 1 . f ′ (c) Suppose that the tangent line to f at c is y − d = f ′ (c)(x − c). The inverse of this tangent line is given by the equation x − d = f ′ (c)(y − c) ⇔ y − c = 1 1 f ′ (c) (x − d). Thus, the slope of the tangent line passing through (d, c) is f ′ (c) . Done. 35. Find the equation for the line tangent to the graph of f −1 at the point (3, 1) if f (x) = x3 + 2x2 − x + 1. f ′ (x) = 3x2 + 4x − 1. f ′ (1) = 3 + 4 − 1 = 6. Thus, (f −1 (3))−1 = 1/6. 36. Use the Inverse-Function Theoremto derive  the formula for the derivative of the inverse of sin x on the interval − π2 , π2 . You may assume the usual facts about the function f (x) = sin x. By IFT, (sin−1 )′ (sin x) = 1 (sin x)′ = 1 cos(x) = sec(x). 37. Suppose both f and f −1 are twice-differentiable functions. Derive a formula for (f −1 )′′ .   1 By IFT, (f −1 )′ (f (x)) = f ′1(x) . Thus, (f −1 )′′ (f (x)) = f ′1(x) = − f ′ (x) 2 · ′′ (x) f ′′ (x) = − (ff′ (x)) 2. Miscellaneous 50 CHAPTER 4. DIFFERENTIATION 38. Suppose f : (a, b) → Ris differentiable at x0 ∈ (a, b) with {αn }∞ n=1 and two sequences in (a, b) \ {x }converging to x such that the sequence {βn }∞ 0 0 n=1 ∞  βn − x0 βn − αn n=1 is bounded. Prove that converges to f ′ (x0 ).  f (βn ) − f (αn ) βn − αn ∞ n=1 For each n, we have cn between αn and βn such that f ′ (c) = Since αn , βn → x0 and since f ′ is continuous, f (cn ) → f (x0 ). f (βn )−f (αn ) . βn −αn 39. Suppose f : R → R is such that f (x + y) = f (x)f (y), f is differentiable at zero, and f is not identically zero. Prove that f is differentiable everywhere and that f ′ (x) = f (x)f ′ (0). Assuming the properties of the exponential function, prove that f (x) = ecx where c = f ′ (0). (x)f (h) (h) = f ′ (0). Then, lim f (x)−fh(x+h) = lim f (0)f (x)−f = We know that lim f (0)−f h h h→0 h→0 h→0 (h)) = f (x)f ′ (0) for any x. Thus, f is differentiable everywhere. lim f (x)(f (0)−f h h→0 ′ ′ Next, if f (x) = ef (0)x , then f ′ (x) = f ′ (0)ef (0)x = f ′ (0)f (x). Thus, f (x) = ef (0)x + K, where K ∈ R. But, f (x) = f (x + 0) = f (0)f (x) for all x, so ′ ′ f (0) = 1. Thus, 1 = ef (0)(0) + K = e0 + K. Thus, K = 0. Thus, f (x) = ef (0)x . ′ 40. Suppose f : [a, b] → R is differentiable and f ′′ exists at t ∈ (a, b). Prove that f (t + h) − 2f (t) + f (t − h) f ′′ (t) = lim . h→0 h2 Give an example where this limit exists, but f ′ is not differentiable at t. H (t−h) = lim f First, lim f (t+h)−2fh(t)+f 2 2f ′′ (t) 2 h→0 h→0 ′′ = f (t).  1/2x2 , Let f (x) = −1/2x2 , 0. Also, lim+ h→0 1/2h2 −1/2h2 h2 ′ ′′ ′′ (t+h)−f ′ (t−h) H (t−h) = lim f (t+h)+f = 2h 2 h→0 2 1 1 if x ≥ 0 /2h2 = . Then at x = 0, we see that lim − /2hh+ 2 − if x < 0 h→0 = 0. Thus, the limit at 0 is 0. But f ′ (x) = |x| which we know is not differentiable at x = 0. 41. Suppose f : D → R, g : E → R, x0 ∈ D ∩ E, x0 an accumulation point of D ∩ E. Suppose further that there is ǫ > 0 such that D ∩ [x0 − ǫ, x0 + ǫ] = E ∩ [x0 − ǫ, x0 + ǫ] CHAPTER 4. DIFFERENTIATION 51 with f (x) = g(x) for all x ∈ D ∩ E ∩ [x0 − ǫ, x0 + ǫ]. Prove that f is differentiable at x0 iff g is differentiable at x0 . |h|<ǫ 0 +h) f ′ (x0 ) = lim f (x0 )−fh(x0 +h) = lim g(x0 )−g(x = g ′ (x0 ). h h→0 h→0 Chapter 5 The Riemann Integral 5.1 The Riemann Integral 1. Use Theorem 5.2 to prove directly that the function f (x) = x3 is integrable on [0, 1].
Pn k+1 3 1 Let P = {0, 1/n, 2/n, ..., 1}. Then, U (P, f ) = k=0 f) = n · n and L(P, 
Pn Pn (n+1)3 3k2 +3k+1 k 3 1 = n4 → 0 as · n . Thus, U (P, f ) − L(P, f ) = k=0 k=0 n n4 n → ∞. Thus, f (x) is integrable. 2. Use Theorem 5.2. to prove directly that f (x) = x is integrable on [0, 1]. Find the integral of f by finding a number A such that L(P, f ) ≤ A ≤ U (P, f ) for all partitions of [0, 1]. Pn 1 Let P = {0, 1/n, 2/n, ..., 1}. Then, U (P, f ) = k=0 k+1 n · n and L(P, f ) = Pn k 1 Pn n+1 1 k=0 n · n . Thus, U (P, f ) − L(P, f ) = k=0 n2 = n2 → 0 as n → ∞. Thus, f (x) is integrable. ´1 1 → 21 ← n+1 Now, U (P, f ) = (n+1)(n+2) 2n2 2n = L(P, f ), so 0 f (x) dx = 2 . 3. Define f (x) = x if x is rational and f (x) = 0 if x is irrational. Compute ´1 f (x) dx and f (x) dx. Is f integrable on [0, 1]? You may wish to look at the 0 0 results of Exercise 2. ´1 From Q2, ´1 0 f (x) dx = 1 2. But, ´1 0 f (x) dx = lim ||P ||→0 P mi (xi+1 − xi ) = 0. (mi = 0 since there is always an irrational number between two rational ones. To see the truth of this statement, consider two terminating or repeating decimal expansions and see that one can easily find a non-terminating, non-repeating decimal expansion between them.) Thus, f (x) is not integrable on [0, 1]. 52 CHAPTER 5. THE RIEMANN INTEGRAL 53 4. A set A ⊂ [0, 1] is dense in [0, 1] iff every open interval that intersects [0, 1] contains a point of A. Suppose f : [0, 1] → R is integrable and f (x) = 0 ´1 for all x ∈ A with A dense in [0, 1]. Show that 0 f (x) dx = 0. ´ ´ ´ ´ If f is integrable, then = . If ≥ 0, then = 0, because mi = 0 for any possible partition. (There will always be a point of A in-between any possible ´1 partition points of [0, 1] since A is dense.) Thus, 0 f (x) dx = 0. Similarly, if ´ ´ ≤ 0, then = 0 because Mi = 0 for similar reasons. 5. Define f : [0, 2] → R by f (x) = 1 for 0 ≤ x ≤ 1 and f (x) = 2 for 1 < x ≤ 2. Show that f ∈ R(x) on [0, 2] and compute the integral. 2 4 PnLet P = {0, /n, /n, ..., 1, ..., 2} where n = 2k. Then, U (P, f ) − L(P, f ) = 2 k=0 (Mk − mk ) · /n. By how we have insisted that our partition include a point at x = 1, we know that mk = Mk for all k. Thus, this sum is 0. Hence, f ∈ R(x). The integral is 3. 5.2 Classes of Integrable Functions 6. If f : [a, b] → Ris decreasing, prove f ∈ R(x) on [a, b]. 2(b−a) , ..., xn = b} be a partition. Let P = {x0 = a, x1 = a + b−a n , x2 = a + n Then, Mk = f (xk ) and mi = f (xk+1 ), since f is decreasing. Thus, U (P, f ) − n P b−a [f (xk ) − f (xk+1 )] · b−a L(P, f )= lim n = lim n ·[f (a) − f (x1 )+f (x1 ) − n→∞k=0 n→∞ f (x2 )+· · · − f (xn−1 ) + f (xn−1 ) − f (b)]. As we see, all the middle terms cancel, showing U (P, f ) − L(P, f ) = lim b−a n · [f (a) − f (b)] = 0. Thus, f ∈ R(x) on n→∞ [a, b]. *7. Suppose g : [a, b] → R is continuous except at x0 ∈ (a, b) and bounded. Prove that g(x) ∈ R(x) on [a, b]. See Exercises 24 and 25 for generalizations of this result. First, even though g is not continuous at x0 , g is bounded at x0 . Thus, g(x) will still be defined for all i (since inf Mi = sup g(x) and mi = xi ≤x≤xi+1 xi ≤x≤xi+1 there will be a right- and left-hand limit for g at x0 ). Thus, g ∈ R(x). 8. Find the integral of f (x) = x on [1, 3] using the techniques of Example 5.5. 54 CHAPTER 5. THE RIEMANN INTEGRAL 2 Consider the function g(x) = x2 and let P be any partition of [1, 3]. Then, P MVT P f (ti )(xi − xi−1 ) = [g(xi ) − g(xi−1 )] = g(3) − g(1) = 9/2 − 1/2 = 4. 9. Assume f : [a, b] → R is continuous and f (x) ≥ 0 for all x ∈ [a, b]. Prove ´b that if a f dx = 0, then f (x) = 0 for all x ∈ [a, b]. Let P0 be a partition of [a, b] and let {Pn } be a sequence of refinements. n ´b P sup f (x)(xi − xi−1 ) = 0. Now, since Since a f (x) dx = 0, we have lim n→∞i=0x∈(x i−1 ,xi ) f (x) ≥ 0 and (xi − xi−1 ) > 0 for all i, the only way that this sum can be 0 is if f (x) = 0 on each interval. But, since this f (x) is the supremum of the interval, this implies that f ≡ 0. 5.3 Riemann Sums 10. Prove Theorem 5.7. [Theorem 5.7 states “Suppose f : [a, b] → R is bounded. Then, f (x) ∈ R(x) on [a, b] iff, for each sequence {Pn }∞ n=1 of marked ∞ partitions with {µ(Pn )}∞ converging to zero, the sequence {S(P n , f )}n=1 is n=1 convergent. If the condition is satisfied, then each of the sequences {S(Pn , f )}∞ n=1 will ´b converge to a f dx.”] (⇒) Let P be a partition for [a, b] for which U = L. Then, since µ(Pn ) → 0, we know that there is N such that PN is a refinement of P (since partition points of Pn are required to get closer together, it follows that the intervals of Pn will eventually have to fit inside the established intervals of the fixed partition P ) and so U (Pk , f ) = L(Pk , f ) for k ≥ N . Since L(Pk , f ) ≤ S(Pk , f ) ≤ U (Pk , f ) ´b ´b and L(Pk , f ), U (Pk , f ) → a f (x) dx, it follows that S(Pk , f ) → a f (x) dx by the squeeze theorem. (⇐) Suppose that S(Pn , f ) → S. Then, there is N such that |S(Pn , f ) − S)| < ǫ for all n > N . Thus, there exists P = PN such that |S(P, f ) − S| < ǫ. Now, by definition of S and the fact that f is bounded, we can choose ti ∈ (xi−1 , xi ) such that S = U or S = L. (What we mean to say is that the convergence of S implies the convergence of U and L.) Thus, we get that ´b U (P, f ) = L(P, f ) as a natural consequence. Hence, S(P, f ) = a f (x) dx. 11. Show that, for a > 1 and b > 1, the function f (x) = x1 is ´integrable a on [1, a] and on [b, ab]. Use the results of Section 5.3 to show that 1 x1 dx = ´ ab 1 dx. b x ´ ab b The function f (x) is continuous and bounded on the intervals, so dx x exist. ´a 1 dx x and CHAPTER 5. THE RIEMANN INTEGRAL 55 Choose {P ´ na } of partitions of [1, a] such that µ(Pn ) → 0. We know that S(Pn , f ) → 1 dx x . Now, choose Qn = b · Pn . (By this we mean to set Qn to be the partition obtained by multiplying every point of Pn by b.) Clearly, {Qn } is a sequence of partitions of P [b, ab] and µ(Qn ) → 0. P Then, S(Pn , f ) = P n n n 1 1 1 (x −x ) while S(Q , f ) = ·b(x −x ) = i i−1 n i i−1 i=0 ti i=0 bti i=0 ti (xi −xi−1 ). ´ ab ´ a dx Thus, S(Qn , f ) = S(Pn , f ) and since S(Pn , f ) → 1 x and S(Qn , f ) → b dx x , ´ a dx ´ ab dx we have that 1 x = b x . 12. Suppose f ∈ R(x) on [0, 1]. Define n an = 1X f n k=1 for all n. Prove {an }∞ n=1 converges to ´1 0   k n f dt. This is the integral of f over sequence of
partitionsPPn = Pn [0, 1] using the P n n {0, 1/n, 2/n, ..., 1} and ti = i/n: i=0 ti (xi −xi−1 ) = k=0 f nk ·1/n = 1/n k=0 f 13. Suppose f : [a, b] → R is bounded and for each ǫ > 0 there is a partition P such that for any refinements Q1 and Q2 of P , regardless of how marked, |S(Q1 , f ) − S(Q2 , f )| < ǫ. Prove that f is integrable on [a, b]. k n
. If |S(Q1 , f )−S(Q2 , f )| < ǫ for any refinements of P , then there is a sequence {Qn } of refinements of P whose mesh is 0 for which |S(Qn , f ) − S(Qm , f )| < ǫ ´b for all n, m. Thus {Qn } is Cauchy and so S(Qn , f ) → a f (x) dx by Theorem 5.7. 14. Suppose f is integrable on [−b, b] and f is an odd function– that is, ´b f (−t) = −f (t) for all t ∈ [−b, b]. Prove that −b f dx = 0. If f is even– that is, ´b ´b f (−t) = f (t) for all t ∈ [−b, b]– prove that −b f dx = 2 0 f dx. Suppose f is odd. Let P = {x0 = 0, x1 , ..., xn = b} be a partition of ´b ´0 Pn ⋆ [0, b]. Let 0 f (x)dx = lim i=0 f (xi )(xi+1 − xi ). Then, −b f (x) dx = µ(P )→0 ´b Pn Pn odd ⋆ ⋆ lim i=0 f (−xi )(−xi −(−xi+1 )) = lim i=0 −f (xi )(xi+1 −xi ) =− 0 f (x) dx = µ(P )→0 µ(P )→0 ´b −I. Hence, −b f (x) = −I + I = 0. ´0 Pn even ⋆ Suppose f is even. Then, −b f (x) dx = lim i=0 f (−xi )(−xi −(−xi+1 )) = µ(P )→0 ´b ´b ´b Pn lim i=0 f (x⋆i )(xi+1 − xi )= 0 f (x) dx. Thus, −b f (x) dx = 2I = 2 0 f (x) dx. µ(P ) 15. Suppose that f : R → R is periodic and integrable ´ p on every ´ a+pclosed interval. If p is the period of f , prove that for any a ∈ R, 0 f dx = a f dx. 56 CHAPTER 5. THE RIEMANN INTEGRAL ´ a+p ´p Write a = kp + r where k ∈ Z and 0 ≤ r < p. Now, a f (x)dx = 0 f (x + ´p ´ p+r ´p ´ p+r per. ´ p a)dx= 0 f (x+kp+r)dx = 0 f (x+r)dx= r f (x)dx= r f (x)dx+ p f (x)dx= ´p ´r ´r ´p per. ´ p f (x)dx + 0 f (x + p)dx = r f (x)dx + 0 f (x)dx= 0 f (x)dx. (If the first r equality hasn’t been established yet, it is easily proven using the definition of integral and periodicity.) 5.4 The Fundamental Theorem of Integral Calculus 16. Use the Fundamental Theorem of Integral Calculus to compute the following: ´3 a. 0 (x2 − x) dx ´4 b. −2 (1 − x3 − x2 ) dx ´ π/2 c. 0 x sin x2 dx 3 2 (a) F (x) = x3 − x2 ; F (3) − F (0) = 29 . 3 4 (b) F (x) = x − x4 − x3 ; F (4) − f (−2) = −78. (c) F (x) = − cos(x2 ) ; 2 F (π/2) − F (0) = 1 2 − cos(π 2 /4) . 2 17. Define f : [0, 2] → R by f (x) = 2x−x2 for 0 ≤ x ≤ 1 and f (x) = (x−2)2 for 1 < x ≤ 2. Prove that f is integrable on [0, 2] and find the integral of f over [0, 2]. Do not use Theorem 5.10, but rather find the integral by methods similar to those used in the proof of Theorem 5.8. [Theorem 5.8 is FTC and Theorem ´b ´c ´b 5.10 is a = a + c .] ´2 0 Let P = {x0 = 0, x1 , ..., 1, ..., xn = 2} be apartition of [0, 2]. Then,  Pn Pk Pn ⋆ ⋆ ⋆ lim i=0 f (xi )(xi+1 −xi )= lim i=0 f (xi )(xi+1 − xi ) + i=k f (xi )(xi+1 − xi ) f (x) dx = µ(P )→0 µ(P )→0 where xk = 1. ( x2 , 0≤x≤1 . We see that f (x) ≡ 1<x≤2 ⋆ F (xi+1 )−F (xi ). f (x⋆i )(xi+1 −xi ) =  F ′ (x). By MVT, P we can select xi such that P Pk k n ⋆ ⋆ Thus, lim i=0 f (xi )(xi+1 − xi ) + i=k f (xi )(xi+1 − xi ) = lim i=0 [F (xi+1 )− µ(P )→0 µ(P )→0 Pn tel. − + 2 F (xi )] + lim i=k [F (xi+1 ) − F (xi )] = F (1 ) − F (0) + F (2) − F (1 ) = /3. Define the function F (x) = µ(P )→0 5.5 (x−2)3 , 3 ser. Algebra of Integrable Functions 57 CHAPTER 5. THE RIEMANN INTEGRAL *18. Suppose f and g are differentiable on [a, b] and f ′ and g ′ are integrable ´b on [a, b]. Prove that f ′ g and g ′ f are integrable on [a, b] and that a f ′ g dx = ´b f (b)g(b) − f (a)g(a) − a g ′ f dx. d By the product rule, dx (f (x)g(x)) = f ′ (x)g(x) + g ′ (x)f (x). Integrating ´ ´b b both sides, f (x)g(x)|ba = a f ′ (x)g(x) dx + a g ′ (x)f (x) dx. 1 f 19. Suppose f ∈ R(x) on [a, b] and ∈ R(x) on [a, b]. Since f ∈ R(x) and 1 x ∈ R(x), 1 f (x) = 1 x 1 f is bounded on [a, b]. Prove that ◦ f (x) ∈ R(x). √ 20. Suppose f ∈ R(x) on [a, b] and f (x) ≥ 0 for all x ∈ [a, b]. Prove that f ∈ R(x) on [a, b]. Since f ∈ R(x) and √ x ∈ R(x) for x ≥ 0, we have √ f= √ x ◦ f ∈ R(x). ´ π/4 21. Use Exercise 18 to calculate 0 x cos x dx. You may assume what you need about the derivatives of the trigonometric functions. By Q18, ´ π/4 0 x cos x dx = x2 2 π/4 cos x|0 − ´ π/4 0 sin x dx = √ π 2 8 + √ 2 2 − 1. 22. Suppose on [0, 1]. Define gn = f (xn ) for n = 1, 2, .... o∞ n´ f is continuous 1 converges to f (0). Prove that 0 gn (x) dx n=1 ´1 For each n, we have min |f (x) ≤ 0 f (xn ) dx ≤ max |f (xn )|. Thus, lim min |f (x)| ≤ ´1 lim f (xn ) dx ≤ lim max |f (xn )| and since f is continuous, min |f (lim xn ) | ≤ ´1 0 n ´1 ´1 f (x ) dx ≤ max |f (lim xn ) |. Thus, f (0) ≤ 0 f (x) dx ≤ f (0) so 0 f (x) dx = 0 f (0). 23. Define γn = 1 + 21 + 31 + · · · + n1 − ´n 1 1 t dt. Prove that {γn }∞ n=1 converges. i P h 1 ´ k+1 dt dt/t as n → ∞. t → k − k ´n Notice that 1 + 1/2 + 1/3 + · · · + 1/n − 1 P1 By visualizing of areas of rectangles of base 1 and height k, we k as the sum i h P 1 ´ k+1 dt/t is just the difference between the aforementioned find that k − k ´ k+1 dt decr. 1 1 1 rectangles and the curve 1/t. Next, k1 − k = k(k+1) . Thus, < k − k+1 t P h 1 ´ k+1 dt i P 1 calc < k − k t k(k+1) < ∞. Thus, the sequence converges by the comparison test (proven in ch. 1). *24. Suppose g : [a, b] → R is bounded and continuous except at x1 , ..., xn ∈ [a, b]. Prove that g ∈ R(x) on [a, b]. 58 CHAPTER 5. THE RIEMANN INTEGRAL We have continuity on the intervals [a, x1 ), (x1 , x2 ), ..., (xn , b], so clearly ´x ´b g(x) dx+ x12 g(x) dx+· · ·+ xn g(x) dx (⋆) exists. Now, let P = {a, x1 , x2 , ..., xn , b} a be a partition of [a, b]. We can easily find refinements Pn of P such that µ(Pn ) → 0 as n → ∞. But these refinements must be unions of refinements of ´b {a, x1 }, {x1 , x2 }, ..., {xn , b} by how P was defined. Thus, a g(x) dx exists and equals (⋆). ´ x1 *25. Suppose {an }∞ n=1 is a sequence of members of [a, b] converging to x0 in [a, b]. If f is bounded on [a, b] and continuous on [a, b] except at x0 and the points of the sequence {an }∞ n=1 , prove that f is integrable on [a, b]. We repeat a very similar argument to Q24. Let Pk = {an }kn=1 ∪ {a, a + (b−a)/n, ..., b}. Clearly, µ(P ) → 0 as k → ∞. Since we have continuity at all k points between partition points, the integral exists. 26. Prove that 1 √ 3 2 ≤ ´1 0 2 √x dx 1+x2 ≤ 13 . On the interval (0, 1), we find that x2 √ 2 ≤ function over (0, 1), we establish the inequality 2 √x 1+x2 1 √ ≤ 3 2 ≤ x2 . Integrating each ´ 1 x2 √ dx ≤ 31 . 0 1+x2 27. Suppose f and g are integrable on [a, b]. Define h(x) = max{f (x), g(x)}. Prove that h is integrable on [a, b]. ´b ´ ´ h(x) dx = G g(x) dx + F f (x) dx where F = {x|f (x) ≥ g(x)} and G = {x|f (x) < g(x)}(= F c ). Since the only places at which h is discontinuous is at the points where f (x) = g(x), and f (x− ) 6= g(x− ), we can invoke Q25 to substantiate our claim since this can only happen countably many times. a 5.6 Derivatives of Integrals 28. Suppose f : [0, 1] → R is continuous and x ∈ [0, 1]. Prove that f (x) = 0 for all x ∈ [0, 1]. ´x 0 f (t) dt = ´1 x f (t) dt for all ´x ´1 ´x d d d f (t) dt = f (x). Also, dx f (t) dt = − dx f (t) dt = −f (x). First, dx 0 x 1 Thus, f (x) = −f (x) which implies that f ≡ 0. 29. Suppose f and g are continuous on [a, b] and Prove that there is c ∈ [a, b] such that f (c) = g(c). ´b a f (x) dx = ´b a g(x) dx. ´x ´x Let F (x) = 0 f (t) dt and G(x) = 0 g(t) dt. By Cauchy’s MVT, there is a c ∈ (a, b) such that (F (b) − F (a))G′ (c) = (G(b) − G(a))F ′ (c). Since F (b) − ´b ´b F (a) = a f (x) dx = a g(x) dx = G(b) − G(a) and G′ (x) = g(x) and F ′ (x) = f (x), the result follows. 59 CHAPTER 5. THE RIEMANN INTEGRAL 30. Find f´′ where f is defined on [0, 1] as indicated: x√ a. f (x) = 0 t2 + 1 dt ´1 1 b. f (x) = x cos t+1 dt ´ 2x c. f (x) = x2 sin t2 dt ´ √x 1 d. f (x) = x 1+t 3 dt √ (a) f ′ (x) = ´x2 + 1 x dt 1 1 (b) −f (x) = 1 cos t+1 ⇒−f ′ (x) = cos x+1 ⇒f ′ (x) = − cos x+1 . 2 ´ ´ 2x ´0 ´ chain x 2x d d sin t2 dt+ dx sin t2 dt ⇒f ′ (x) = (c) f (x) = x2 sin t2 dt+ 0 sin t2 dt ⇒f ′ (x) = − dx 0 0 − sin((x2 )2 ) · (2x) + sin(2x2 ) √· 2. ´ x dt ´ x dt 1 ′ (d) f (x) = − 0 1+t 3 + 0 1+t3 ⇒f (x) = − 1+x3 · 1 + rule 1 √ 1+( x)3 31. Let f : R → R be continuous and δ > 0. Define g(t) = all t ∈ R. Prove that g is differentiable and compute g ′ . ´ t+δ t−δ · 1/2x−1/2 . f (x) dx for ´ t+δ ´0 ´ t+δ ´t We see that g(t) = t−δ f (x) dx = t−δ f (x) dx + 0 f (x) dx= − δ f (x − ´t δ) dx + −δ f (x + δ) dx. The last equality is a valid integral since f (t + const.) is continuous, and thus integrable. The resulting function is thus differentiable (as all integrals of continuous functions are). Thus, g ′ (x) = f (t + δ) − f (t − δ). *32. Suppose f : [a, b] → R is continuous and g : [c, d] → [a, b] is differen´ g(x) tiable. Define F (x) = a f (t) dt for x ∈ [c, d]. Prove that F is differentiable and compute F ′ . Since we have continuity of f , we can use FTC and since g is differentiable, we can use the chain rule. Thus, F ′ (x) = f (g(x))g ′ (x). 5.7 Mean-Value and Change-of-Variable Theorems 33. Suppose f : R → R is continuous and has period p so that f (x + p) = ´ x+p f (x) for all x ∈ R. Show that x f (t) dt is independent of x in that, for all ´ x+p ´ y+p ´p x, y, x f (t) dt = y f (t) dt. Show, then, that 0 [f (x + a) − f (x)]dx = 0 for any real number a. Conclude that for any real number a, there is x such that f (x + a) = f (x). ´ p ´ a+p ´ x+p ´ y+p ´ p By Q15, 0 = a for any a ∈ R. Thus, for all x, y, x = y = 0. ´p ´p ´p ´ a+p Then, 0 [f (x + a) − f (x)]dx = 0 f (x + a) dx − 0 f (x) dx= a f (x) dx − 60 CHAPTER 5. THE RIEMANN INTEGRAL ´ a+p f (x) = 0. The first term of the last equality comes from the change-ofa variable theorem and the second comes from Q15. *34. Prove the following variation on Theorem 5.18. Assume that φ : [a, b] → R is differentiable, 1-1, and increasing with φ(a) = c and φ(b) = d. If f : [c, d] → R is integrable on [c, d] and (f ◦ φ)φ′ is also integrable on [a, b], then ´d ´b f (x) dx = a f (φ(t))φ′ (t) dt. c ´b ´b It is easily seen that f (φ(b))φ′ (b)−f (φ(a))φ′ (a) = a f (φ(t))φ′ (t) dt= 0 f (φ(t))φ′ (t) dt− ´a ´d f (φ(t))φ′ (t) dt= c f (x) dx. The last equality comes from FTC and Q32. 0 35. Use Theorem 5.18 to evaluate the integrals [Theorem 5.18 is the changeof-variable theorem]: ´3 √ a. 0 3 1 + x2 x dx ´4 √ 3 √ dx b. 1 ( x+2) x ´ √3 √x2 −9 c. 1 dx ´ 1 x2x √ d. 0 1−x2 dx ´3 (a) (b) 2 1/3 (1 + x ) ´04 (√x+2)3 √ ´ 10 1/3 · x dx = u du = ´4 3 1 dx = 2 3 u du = 1/2u4 |43 = 1/2 3 4u 175 2 . 1/2 4/3 10 1 =   1 3 10·10 /3 −1 8 . 1 x √ (c) Clearly, something is wrong. Both 1 and 3 have squares less than or equal to 9, so this result is complex. It isn’t even a nice complex integral either. The answer is about ei(1.57) · 1.47. d 1 , let u = sin−1 (x), then, x2 = sin2 (u) (sin−1 (x)) = √1−x (d) Recognizing dx 2 ´ π/2 2 and the integral becomes 0 sin (u) du = π/4. 36. Use Theorem 5.18 to establish the result in Exercise 11. 1 b ´a 1 dx x = ´a dx 1 bx = ´ ab b du u . 37. Use Theorem 5.18 to establish the result in Example 5.6. [Example 5.6 ´ 1/a dx ´1 establishes the identity a x2dx +1 = 1 x2 +1 .] ´1 dx a x2 +1 = ´ 1/a 1 a·du (au)2 +1 =a· ´ 1/a 1 du (au)2 +1 =a· ´ 1/a 1 dx a[(x)2 +1] = ´ 1/a 1 dx x2 +1 . Miscellaneous 38. Assume f : [a, b] → R is continuous, x ∈ [a, b], and h f (x) ≥ 0ifor all 1/n ∞ ´b M = sup{f (x) : x ∈ [a, b]}. Show that converges to [f (x)]n dx a n=1 61 CHAPTER 5. THE RIEMANN INTEGRAL M. Let Iδ = {x ∈ [a, b] : |f (x)| ≥ M − δ}. (The set exists by MVT.) Now, ´ 1/n ´ 1/n b M (b − a)1/n ≥ a [f (x)]n dx ≥ Iδ [f (x)]n dx ≥ (M − δ)ℓ(Iδ )1/n where ℓ(Iδ ) is the sum of lengths of intervals composing Iδ . (The first inequality is the upper bound of the integral over (a, b), and the third is a similar lower bound for the integral over Iδ .) Since any set real number raised to 1/n tends to 1 as ´ 1/n b n → ∞, we see that M ≥ lim a [f (x)]n dx ≥ (M − δ). Taking δ → 0, we 1/n ´ b = M. see that lim a [f (x)]n dx h

2
2 i 2 39. For each positive integer n, define an = n1 n1 + n2 + · · · + nn . Find the limit of the sequence {an }∞ n=1 .
2 Pn an = n1 k=1 nk = (n+1)(2n+1) → 31 . (The second equality was proven 6n2 somewhere in Chapter 0. I don’t know what this has to do with integration.) 40. For each positive integer n, define an = Find the limit of the sequence {an }∞ n=1 . 1 n   nπ sin nπ + sin 2π n + · · · + sin n .
Pn which is easily seen to be a left-handed Riemann an = k=1 1/n · sin kπ ´1 n sum. Thus, lim an = 0 sin(πx) dx = π2 . 41. If f and g are integrable on [a, b], show that ´b t= 2 (f (x) + tg(x)) dx = a´ b (f (x))2 dx a ´b f (x)g(x)dx a ´b a (f (x))2 dx + 2 ´b a h´ b a f g dx i2 f (x)g(x)dx + t2 ≤ ´b ´b (g(x))2 dx. Set a a f 2 dx ´b a g 2 dx. (unless f g ≡ 0, in which case the inequality is apparent). Then, by performing straightforward, carefulalgebra using this value of t,  we ´b ´b   2 2 ´b ´b  (f (x)) dx a (g(x)) dx  find that a [f (x)+tg(x)]dx = a (f (x))2 dx  a ´ − 1. 2 b f (x)g(x)dx a 2 ´ ´ ´b ´b b b Since LHS≥0 and a (f (x))2 dx ≥ 0, it follows that a f (x)g(x)dx ≤ a (f (x))2 dx a (g(x))2 dx. 42. Use Exercise 41 to prove the Minkowski inequality– that is, if f and h´ i1/2 h´ i1/2 b b g are integrable on [a, b], then a [f (x) + g(x)]2 dx ≤ a [f (x)]2 dx + i1/2 h´ b . [g(x)]2 dx a ´b ´b ´b ´b ´b (f (x)+g(x))2 dx = a (f (x))2 dx+2 a f (x)g(x)dx+ a (g(x))2 dx ≤ a (f (x))2 dx+  1/2 2 1/2 ´ 1/2 ´ 1/2 ´ ´ ´b b b b b 2 2 2 2 . + (g(x)) dx = (g(x)) dx (g(x)) dx + (f (x)) dx 2 a (f (x))2 dx a a a a a CHAPTER 5. THE RIEMANN INTEGRAL Take the square root of both sides to see the inequality. 62 Chapter 6 Infinite Series 6.1 Convergence of Infinite Series P∞ 1. Let {an }∞ numbers. Prove that n=1 (an − an+1 ) n=1 be a sequence of real P ∞ converges iff {an }∞ n=1 converges. If n=1 (an − an+1 ) converges, what is the sum? P (⇒) Suppose (an − an+1 ) < ∞. Then, we must have (an − an+1 ) → 0 (test for divergence). Let m > n. Then, for any ǫ > 0, there is N such that ǫ |an − an+1 | < m−n for all n > N . Thus, for m > n > N , |an − am | = |an − an+1 + an+1 − an+2 + · · · + am−1 − am | ≤ |an − an+1 | + · · · + |am−1 − am | < ǫ = ǫ. Thus, {an } is Cauchy and thus convergent. (m − n) (m−n) (⇐) Suppose an → L. Then, {an } is Cauchy, so an − an+1 → 0. Then, P (an − an+1 ) = a1 − a2 + a2 − a3 + · · · = lim(a1 − an ) = a1 − L. P∞ 2. Let n=1 an converge. Let {nk }∞ k=1 be a subsequence of the sequence of positive integers. For each k,P define bk = ank−1 +1 + · · · + P ank where nP 0 = 0. ∞ ∞ ∞ Prove that k=1 bk converges and that k=1 bk = n=1 an . P This is the identical sum. All that is happening in bk is that we choose terms of an and add the terms between them (including the an terms themselves). Thus, there is really no analysis to perform here. 3. Prove that P∞ n=1 2n rn converges if |r| < 1/2 and find the sum. PN For a finite series, n=0 (2r)n = 1 + 2r + 4r2 + · · · . Multiplying both sides PN by (1 − 2r), we see that (1 − 2r) n=0 (2r)n = (1 − 2r)(1 + 2r + 4r2 + · · · )= N +1 PN 1 − 2r + 2r − 4r2 + 4r2 − · · · = 1 − (2r)N +1 . Thus, n=0 (2r)n = 1−(2r) 1−(2r) . 63 64 CHAPTER 6. INFINITE SERIES (We will use this for Q4.) If we instead set the lower bound as n = 1, we find N +1 2r<1 PN P∞ 2r−(2r)N +1 2r 2 . Thus, n=1 (2r)n = lim 2r−(2r) . = 1−2r n=1 (2r) = 1−2r 1−2r 4. Prove that P∞ n=0 3−n converges and find the limit. Q3 shows that the sum is finite. The sum is 5. Determine whether the series Justify your conclusion. P∞ n=1 ( √ 1 1−1/3 n+1− = √ 1 2/3 = 32 . n)converges or diverges. √ √ √ √ √ √ √ PN √ √ ( n + 1− n) = 2− 1+ 3− 2+· · ·+ N − N − 1+ N + 1− n=1 √ √ √ N = N + 1 − 1. Taking N → ∞, we see that the sum is ∞. √ 6. Use induction to show that 1 + √12 + · · · + √1n ≥ n for n ≥ 1. Can you P∞ use this fact to determine whether the series n=1 √1n converges or diverges? For n = 1, the inequality is √ true. Suppose it is true for n = N . Then, 1 + √12 + · · · + √1N + √N1+1 ≥ N + √N1+1 . This inequality holds if and only   √ √ if N + 1 1 + √12 + · · · + √1N + 1 ≥ N + 1 holds. By subtracting 1 from   √ √ each side, we obtain N + 1 1 + √12 + · · · + √1N ≥ N which is immediately √ apparent since 1 + √12 + · · · + √1N ≥ N by the inductive hypothesis. Thus, the inequalityPhas been proven by induction. √ P 1 N √ = ∞. Now, n=0 √1n ≥ N , so as N → ∞, we see that n
1 1 1 1 7. Use induction P∞ to1show that 2 1 + 8 + 27 + · · · + n3 < 3 − n2 for n ≥ 2. Does the series n=0 n3 converge? Justify your conclusions. One can easily find  that the statement is true forn = 2. Suppose it is true 1 1 2 < 3 − N12 + (N +1) + · · · + N13 + (N +1) for n = N . Then, 2 1 + 81 + 27 3 3 which
(N +1)3 1 1 1 3 3 is equivalent to (N +1) ·2 1 + 8 + 27 + · · · + N 3 +2 < 3(N +1) − N 2 +2. As expected, we subtract 2 from both
sides and find that  this is equivalent to 1 (N + 1)3 · 2 1 + 18 + 27 + · · · + N13 < (N + 1)3 3 − N12 which is immediately apparent from the inductive hypothesis. Thus, the inequality has been proven by induction.
PN P 1 3 Now, n=0 n13 < 1/2 3 − N12 . Taking N → ∞, we find that n3 < 2 , proving convergence. 8. Write an infinite series for the repeating decimal for the rational number 5/9 and prove that it converges to 5/9. P P We claim that 59 = n=1 5(10)−n . That n=1 5(10)−n = 5/9 can be proven by the division algorithm or it can be proven by a method similar to Q9. 65 CHAPTER 6. INFINITE SERIES 9. Find the rational number that is the limit of the repeating decimal 0.15. We have x = .151515..., so 100x = 15.1515.... thus, 100x − x = 15 implying 15 99x = 15 or x = 99 . 10. Prove Theorem 6.3. [Theorem 6.3 is P (αan +βbn ) = α P an +β P bn .] We already have the result that lim(cxn + dyn ) = c lim xn + d lim yn , so since an infinite series is the limit to the sequence of partial sums, we have the desired result. 11. Prove that the series P∞ n=2 5 2n − 3 5n
converges and find the limit. P P By Q10, the sum can be written as 5 · n=2 (1/2)n − 3 · (1/5)n . The sum 1 1 converges by an argument similar to Q3. The limit is 5 · 1−/14/2 − 3 · 1−/25 1/5 . 12. Show that the sequence We have seen that by Theorem 6.3. 6.2 P 1 n P∞ n=1 2 n + 1 2n
diverges. Use Theorem 6.3. diverges, so the sequence in question must diverge Absolute Convergence and the Comparison Test P∞ ∞ 13. P∞Suppose n=1 an converges absolutely and {bn }n=1 is bounded. Prove that n=1 an bn converges absolutely. Let M = sup{bn }. Then, P |an bn | ≤ P |an M | = |M | · P |an | < ∞. P∞ 2 P∞ 14. If n=1 aP n converges absolutely, prove that n=1 an converges. Is the ∞ statement true if n=1 an converges conditionally? Let M = sup{an } (which must be finite aP conn → 0 as a necessary P since P 2 dition for convergence. We then have |a | = |a ||a | < |a ||M | < n n n n P |M | |an | < ∞. 15.PDetermine which of the following infinite series converge. ∞ 1 a. (n+1)(2n−1) Pn=1 ∞ n b. n=1 n+1 P∞ 2 c. 3k √ √ Pk=1 ∞ m+1− m d. m=1 m P∞ 3m e. P m=1 5m+1
∞ 100m+1 1 f. m=1 m2 m 66 CHAPTER 6. INFINITE SERIES a,e, and f converge; the rest diverge. 16. (Limit-comparison test.) P Prove the following generalization of Theorem P∞ ∞ 6.5. Suppose that a and n n=0 n=0 bn are series of positive terms such that n o∞ P∞ P∞ an converges to L 6= 0. Then n=0 an and n=0 bn either both diverge bn n=1 or both converge. What can be concluded if L = 0? P an <P ∞. Then, (by We have lim abnn = L, so lim an = L · lim bn . Suppose P a = definition of convergence) there is N for which 0 = n n=N n=N P L · bn = P P bn < ∞. L ·P n=N bn . Since L 6= 0, we conclude that n=N bn = 0 and so If an = ∞, then the proof is very similar. (Just change the first equality to <.) We can P see from the proof that if L = 0, than we cannot determine if the tail sum of bn goes to 0. Thus, we cannot conclude convergence or divergence. 17. Apply the limit-comparison test (Exercise 16) to the following series. √ P∞ n+17 a. 2 n −64n−112 Pn=1 ∞ 1 √ √ b. n=1 n2 +64n+ n2 +3 √ n+17 √ We see that the quotient of the sequences is (n2 −54n−112)(√ n2 +64n+ n2 +3) which clearly tends to 0. We cannot conclude anything about the series. (For the (b) doesn’t since (a) is of the order P 1good of the order, (a) converges whileP 1 with p > 1 while (b) is of the order p n np with p = 1.) 6.3 Ratio and Root Tests 18. Give an example of an infinite series for which Theorem 6.8 yields a < 1 for n > N conclusion, but Theorem 6.9 does not. [Theorem 6.8 is aan+1 n P for some N ⇒ an is absolutely convergent, and Theorem 6.9 is lim aan+1 < n P 1⇒ an is absolutely convergent.] P 1 1/(n+1)2 n2 Examine the sum = lim n2 +2n+1 = 1 and so the n2 . Now, lim 1/n2 traditional ratio test is inconclusive. However, the fraction on the RHS of the first equality isPalways less than 1 (top<bottom), so Theorem 6.8 allows us to 1 conclude that n2 is absolutely convergent. 19. Use Theorem 6.9 to determine the values of r for which converges. n+1 n+1 n+1 lim (n+1)r = lim nrnrn + rnrn = lim r · nr n r. Thus, the sum converges if r < 1. nr n nr n +r· rn nr n P∞ n=0 nrn = lim |r + r/n| = 67 CHAPTER 6. INFINITE SERIES 20. Prove that {nxn }∞ n=1 converges to zero if |x| < 1. By Q19, we have shown the sequence of partial sums of this sequence converges. A necessary condition for the convergence of the series is a convergence of the sequence to 0. 21. Prove version of the root test. If np the ofollowing ∞ series and |an | converges to L, then n=1 1. if L < 1, the series converges absolutely; and 2. if L > 1, the series diverges. P∞ n=0 an is an infinite p 6 0, the series must There must be a typo in the book, since if lim |ap n| = √ diverge. Thus, we will assume that we should replace |an | with n an in the problem p statement. P an can only If n |an | → L, then |an | → Ln . Immediately, we see that hope to converge if L < 1 (otherwise an 9 0). Next, if |an | → Ln with L < 1, P ±ǫ P n then there is N for which P n=N L < ∞. (See Q3 for the reason n=N |an | = that this converges.) Thus, an is absolutely convergent. o∞ n an+1 22. If {an }∞ n=1 is a sequence of positive real numbers such that an n=1 √ ∞ converges to L, prove that n an n=1 converges to L. ±ǫ Suppose that an+1 /an = L for all n > N for some N . This implies that ±ǫ ±ǫ ±ǫ ±ǫ an+1 = L · an . For any k, we see that aN +k = L · aN +k−1 = L2 · aN +k−2 = ±ǫ · · · = Lk aN using the previous formula. Substituting n = N + k (n > N ), we n−N ±ǫ ±ǫ see that an = Ln−N · aN which is the same as an 1/n = L n aN 1/n . Taking √ n → ∞, lim n an = L1 · a0N = L. n√ o∞ n n! 23. Prove that converges and find the limit. You might want to n n=1 look at Example 6.12. √ n+1 1/(n+1) (n+1)! (n)1/(n+1) ···(1)·n n · √ = (n+1) . We see that for all n, the n n+1 (n+1)(n)1/n (n−1)1/n ···(1) n! denominator is greater than the numerator, so the ratio is less than 1. Thus, √ n n P √ n! n! < ∞ by the Ratio Test. Thus, → 0. n n 24.PTest the following series for convergence. ∞ p n a. Pn=1 n√ p ,p>0 ∞ n b. P n=1 ( n − 1)n ∞ c. Pn=1 n−1−(1/n) ∞ 1 d. n=2 pn −q n , 0 < q < p 68 CHAPTER 6. INFINITE SERIES p n+1 p (a) Suppose that lim (n+1) = L. Then, doing some algebra, we see that np p n
p n+1 n+1 · p = L since n → 1, we find that L = p and so the sum is finite lim n for 0 < p < 1. √ (b) Suppose that lim( n n − 1)n )1/n = L. This means that lim n1/n = L + 1. Taking the natural log of both sides, lim lnnn = ln(L + 1). By l’Hospital’s rule, the LHS is equal to lim 1/n 1 = 0. Hence, 0 = ln(L + 1) and so 1 = L + 1 or L = 0 < 1. Thus, the series is absolutelyP convergent. 1 . Using the ratio test, we see (c) Using algebra, the sum becomes n·n1/n n·n1/n (n+1)(n+1)1/n 1/n 1 (d) pn −q n that < 1 for all n. Thus, the sum is absolutely convergent. ≤ 1 (pn )1/n −(q n )1/n 1 |p−q| < 1 iff |p−q| < 1. (The inequality P 1 is a simple case of Minkowski’s inequality [see Q5.42].) Thus, n=2 pn −q n < ∞ if |p − q| < 1. = 25. Determine those values of p for which P∞ 1 n=2 n(log n)p converges.  p i n(log n)p log n n Using the ratio test, examine lim (n+1)(log[n+1]) · . p = lim n+1 log(n+1) If p ≥ 0, both factors have the numerator is greater than the denominator, which would imply convergence. If p < 0, then the second factor has a quotient greater than 1. Thus, for p < 0, the sum is infinite. h 26. For each positive integer n, define γn = 1 + 21 + · · · + n1 − log n. Prove ´ x dt that {γn }∞ n=1 converges. (Use the fact that log x = 1 t for x > 0.) We could visualize 1 + 12 + · · · + n1 as the area of n rectangles, each with base 1 and height n1 . As mentioned in the hint, we can visualize log(n) as the area under the curve 1t from 1 to n. Thus, the difference 1 + 12 + · · · + n1 − log(n) can be visualized by the area between the afformentioned rectangles and the curve 1t from 1 to n. The following is an illustration of the first three rectangles described along with the curve 1t . The area that {γn } represents is equal to the sum of areas of the rectangles above 1t . CHAPTER 6. INFINITE SERIES 69 The area mentioned at the end of the previous paragraph is surely lessthan 1 the sum of areas of rectangles whose base is 1 and height is n1 − n+1 . In the figure, these rectangles are the “sub-rectangles” defined by the portion P 1 of the whole rectangles above the dashed lines. Thus, n=2 n − log n <  P 1 1 1 1 1 n=2 n − n+1 = 2 − lim n+1 = 2 < ∞. Thus, {γn } converges. 6.4 Conditional Convergence P∞ ∞ P∞27. If n=1 an converges and {bn }n=1 is monotone and bounded, prove that n=1 an bn converges. Since {bn } is monotone and bounded, bn → L for some L. Thus, for any P ±ǫ ǫ there is N1 for which bn = L for n > N1 . Also, since an < ∞, there is P ±ǫ N for which a = 0 for the same ǫ. Let N = max(N 1 , N2 ). Then, P P Pn=N2 n P2 a . We see that the a + ǫ (L + ǫ)a = L a b < n n=N n n=N n n=N n=N n n RHS becomes L(0 + ǫ) + ǫ(0 + ǫ) = ǫ(L + ǫ). Since ǫP is arbitrary, we can take ǫ → 0 to conclude that RHS → 0, or, in other words, an bn < ∞. 28. Suppose {an }∞ n=1 is a sequence of positive real numbers converging to 0 such that a ≥ an+1 for all n. Then, by the alternating series test, n P∞ n (−1) a converges; call the sum S. Let Sn be the nth partial sum of n Pn=1 ∞ n (−1) a . Prove that |Sn − S| ≤ an+1 . n n=1 Since an ≥ an+1 , we can easily find that |Sn − Sn+1 | = |Sn − (Sn + (−1)n+1 an+1 )| = |an+1 |. Similarly, |Sn −Sn+2 | = |Sn −(Sn+1 +(−1)n+2 an+2 )| = |Sn − (Sn + (−1)n+1 an+1 + (−1)n+2 an+2 )| = |an+1 − an+2 | < an+1 . In general, Pn+k TI Pn+k ℓ |Sn − Sn+k | = ℓ odd |aℓ − aℓ+1 |. Now, taking a random ℓ=n+1 (−1) aℓ < term of the series on the RHS of the inequality, say |ak − ak+1 |, we know that 70 CHAPTER 6. INFINITE SERIES the next term is |ak+2 − ak+3 | < ak+2 < ak+1 . Thus, what is being added in subsequent terms is never more than what is initially subtracted. Hence, {|Sn − Sk |}∞ k=n+1 is a decreasing sequence. Thus, |Sn − S| ≤ an+1 . P∞ 29. Let and {nk }∞ k=1 a subsequence of the n=1 an be an infinite series P ∞ sequence of positive integers. Prove that if a converges n n=1 P∞absolutely, then P∞ a converges absolutely. What can be concluded if n k n=1 an converges k=1 conditionally? P P P We see that ∞ > |an | ≥ |ank | since all terms are positive. Thus, a nk is absolutely convergent. P If an is conditionally convergent, then we cannot say anything about the convergence of subseries a priori. For series test, P example, by the alternating P P (−1)n n1 < ∞. However, we have n even (−1)n n1 = ∞ and n odd (−1)n n1 < ∞. 30. Prove that P∞ k=1 sin kx k converges. Let ak = sin kx. Then, sin kx = n+1 cos( n−1 2 )x−cos( 2 )x 2·sin( x 2) (a formula usually derived in a trigonometry class), so the partial sums are always telescoping and thus, since the trig functions always give outputs less than or equal to 1, we can 1 bound each sum. Next, P set bk = k , which we know to be decreasing toward 0. By Theorem 6.12, an bn < ∞. 31. Test each of the following series for absolute convergence, conditional convergence, or absolute convergence. P∞ (−1)n n a. n2 −5n+1 Pn=1 ∞ (−1)n (n−5) b. n3 −7n−9   √ Pn=1 √ ∞ c. ( n + 1 − n)an where an = (−1)f (n) and f (n) = n2 ; that is, n=1 f (n) isP the integer part of n2 . ∞ n n d. n=1 (−1) n+1 P (−1)n n P (−1)n n P (−1)n (a) n2 −5n+1 < n2 −5n = n−5 < ∞. Thus, the sum is conditionally convergent by the alternating series test. P P 1 P (−1)n (n−5) (−1)n n (b) n=9 n3 −8n < ∞ since the sum is of the form n3 −7n−9 < np for p >P 1. √ Thus, the √ sum is absolutely convergent. (c) ( n + 1 − n)an is a telescoping series which is convergent since 0 < an < 1.P Since all terms are positive, it is absolutely convergent. n is conditionally convergent by the ratio test. (d) (−1)n n+1 6.5 Power Series 71 CHAPTER 6. INFINITE SERIES P∞ ∞ 32. Suppose that n=0 aP n diverges and that {an }n=0 is bounded. Prove that ∞ n the radius of convergence of n=0 an x is equal to 1. Let M = max{an }. Then, convergence is 1. P an xn < M · P xn implying that the radius of P∞ 33. Suppose P∞ n=1 ann converges conditionally. Prove that the radius of convergence of n=1 an x is equal to 1. n+1 x lim an+1 an xn lim an+1 an · x n+1 x = lim aan+1 · n Since P an < ∞, we have an+1 an < lim x. Thus, the radius of convergence is 1. 34. Show that lim (n+1)!x n!x xn+1 x . n+1 P∞ n=1 ≤ 1. Hence, n!xn converges only for x = 0. = lim (n+1)x = ∞ unless x = 0. 1 35. Show that xn+1 xn n=1 n P∞ converges iff −1 ≤ x < 1. n+1 n = lim xxn (n+1) lim n+1 < lim xn xn n = x. Thus, the radius of convergence is 1. n If x = −1, it is an alternating series whose terms are decreasing, so the series would be convergent. P 1/n = ∞. If x = 1, then the sum would be 36. Determine the radius of convergence of the power series lim 2n+1 (n+1)! n+1 x (n+1)n+1 2n n! n nn x n+1 n n 2n n! n n=1 nn x . P∞ n+1 2(n+1)·n 2(n+1) (n+1)!·n = lim 22n n!(n+1) n+1 x = lim (n+1)n+1 x < lim (n+1)n+1 x < 2x. Thus, the radius of convergence is 1/2. P∞ 37. Determine the interval of convergence of the power series n=1 n1p xn for different values of p. q n x x Let lim n xnp = lim √ = L. Then, log(L) = log(x)−lim p/n log(n). n p = lim np/n n H Thus, L = exp[log(x) − lim p/n log(n)] = exp[log(x) − 0] = x. Thus, radius of convergence is 1, regardless ofPthe value of p. 1 If x = 1, then the sum is p which is convergent if p > 1. Pn (−1)n which is convergent for p > 0. If x = −1, then the sum is np P∞ P∞ 38. Show that the power series n=0 an xn and n=1 nan xn−1 either both converge for all x, both converge for x = 0, or both have the same finite nonzero radius of convergence. n+1 x Suppose that lim an+1 an xn we must have lim (n+1)an+1 xn nan xn−1 = lim aan+1 · x = L. Then, by l’Hospital’s rule, n P (n+1)an+1 = lim · x = L. Thus, n=1 nan xn−1 nan 72 CHAPTER 6. INFINITE SERIES P converges or diverges with n=0 an xn . Also, by the factorization of the limit, we see that the radii of convergence must coincide. 39. Let {an }∞ n=1 be a sequence of real numbers bounded from below, and let A = {p : there is a subsequence of {an }∞ n=1 converging to p}. Suppose A is nonvoid. Define a = inf A. Prove that a ∈ A and that, for each ǫ > 0, there is N such that for all n ≥ N, a − ǫ < an and there are infinitely many m such that am < a + ǫ. A is the set of limit points of {an }. Since a = inf A, either a is the smallest limit point of the sequence, or else we have a sequence of limit points converging to a. Suppose the latter is true. Now, each limit point has an infinity of terms of {an } in any ǫ neighborhood of itself. Thus, since a contains an infinity of limit points within any ǫ neighborhood (definition of convergence), a consequently contains an infinity of terms of {an } within any ǫ neighborhood. Thus, a is a limit point of {an }. Thus, a ∈ A. Since a is itself a limit point with ank → a, we know that for any ǫ > 0, there is N so that we have −ǫ < ank − a < ǫ for all nk > N . As a natural consequence, a − ǫ < ank < a + ǫ, and so, obviously, ank < a + ǫ for all nk > N . Thus, since {ank } ⊆ {an }, there are infinitely many m such that an < a + ǫ. For the same ǫ and N as above, we have a−ǫ < ank for all nk > N . However, since a is the smallest limit point, this must hold true for all n; that is, a−ǫ < an for all n > N . 40. State and prove theorems similar to Theorems 6.8 and 6.21 in terms of lim inf n→∞ an+1 an and lim sup n→∞ an+1 . an P (Theorem 6.8) Let an be an infinite series of nonzero terms. Then, if P an+1 L = lim sup an < 1, then |an | < ∞ and if L = lim inf aan+1 > 1, then n P an = ∞. (Proof) If L < 1, then given ǫ > 0, there is N for which aan+1 < 1 − ǫ for all n P N . Thus, |an+1 | < (1P − ǫ)|an | and so n=N |an+1 | = (1 − ǫ)|aN | + (1 − ǫ)2 |aN | + 3 (1 − ǫ) + · · · = |aN | n=N (1 − ǫ)n . Since 0 < 1 − ǫ < 1 for small ǫ, the sum is (absolutely) convergent. > 1 + ǫ for If L > 1, then for the same ǫ, there is N such that aan+1 n P all nP> N . Performing the same steps as above, we see that n=N |an+1 | > |an | n=N (1 + ǫ)n = ∞.PThus, the sum is divergent. (Theorem 6.21) Let an xn be a power series with an 6= 0 for all n. Then, P let L = lim sup aan+1 and L = lim inf aan+1 . Then, an xn < ∞if |x| < L1 and n n P an xn = ∞ if |x| > L1 . (Proof) Follows from the altered Theorem 6.8. 73 CHAPTER 6. INFINITE SERIES 41. Find the interval of convergence of the power series an is as given below. Be sure to check the endpoints. a. an = (n + 1)(n + 2) b. an = sin n√ c. an = 3−n n (n!)2 d. an = (2n)!
n2 e. an = 1 + n1 n n f. an = 3n + 2n2 P∞ n=1 an xn where (n+2)(n+3) n+3 (a) lim (n+1)(n+2) = lim n+1 = 1. Thus, interval of convergence is (−1, 1). The endpont x = −1 allows for convergence, but x = 1 causes the sum to diverge. (b) The√radius of convergence is 1 by Q32. Neither of the endpoints converge. √ n n+1·3 n+1 1 1 √ √ (c) lim 3n+1 · n = lim 3 n = L ⇒ L2 = lim 19 · n+1 n = lim /9(1 + /n) = 1/9 ⇒ L = 1/3. Thus, the radius of convergence is 3. The negative endpoint causes convergence (AST). 2 (n+1)2 (n!)2 ·(2n)! (n+1)! ·(2n)! (d) lim [(n+1)!] (2n+2)!(n!)2 = lim (2n+2)(2n+1)(2n!)(n!)2 = lim (2n+2)(2n+1) = ∞ (by l’Hospital’s rule, e.g.). Thus, the series converges only for x = 0. (n+1)(n+1) 1 n+1 (1+ n+1 ) = lim e en = e > 1. Hence, the radius of conver(e) lim 1 n·n (1+ n ) gence is 1/e. The negative endpoint converges. n+1 (n+1)+2n+1 )·n2 (f) lim (3 (n+1) = 3. Thus, the radius of convergence is 1/3. 2 ·(3n n+2n ) Neither endpoint converges.