Essential Mathematics for Economic Analysis 6th edition

Essential Mathematics for Economic Analysis 6th edition Knut Sydsæter Peter Hammond Andr´es Carvajal Arne Strøm Student’s Manual Contents 1 Essentials of Logic and Set Theory 1 2 Algebra 2 3 Solving Equations 9 4 Functions of One Variable 13 5 Properties of Functions 17 6 Differentiation 19 7 Derivatives in Use 24 8 Concave and Convex Functions 30 9 Optimization 30 10 Integration 34 11 Topics in Finance and Dynamics 42 12 Matrix Algebra 45 13 Determinants, inverses, and Quadratic Forms 49 14 Functions of Many Variables 61 15 Partial Derivatives in Use 63 16 Multiple Integrals 71 17 Unconstrained Optimization 72 18 Equality Constraints 80 19 Linear Programming 89 20 Nonlinear Programming 94 1 Essentials of Logic and Set Theory 1.1 Essentials of Set Theory 1.1.9 For this problem we assume that all the relevant sets are subsets of some “universal” set U . Though the answer given in the text is fairly complete, we use more words in the answer presented here. As in the answer in the book, let {Ai : i ∈ I} denote the family of sets, indexed by the set I of labels i. Their union is A∪ = S A , and their intersection is A∩ = T A . i∈I i i∈I i (a) An element x that belongs to the complement of A∪ cannot belong to any of the sets Ai, so it satisfies x ∈/ Ai, for all i in I. It follows that for every i in I, x is in the complement Ac = U Ai of T c i Ai. Thus x belongs to the intersection i∈I Ai of all the sets’ complements. T Conversely, if x belongs to the intersection i∈I Ac of all the sets’ complements, then x ∈/ Ai for all i in I. Then there is no i in I such that x ∈ Ai, which implies that x ∈/ A∪, that is, x ∈ (A∪)c. (b) If x belongs to the complement of A∩ = T A , then there must be an i in I such that x is not in i Ai. Then x belongs to the complement Ac = U Ai of Ai. It follows that x must belong to the i c union of all such complements, that is, x ∈ i∈I Ai . Conversely, if x belongs the union of all the complements Ac, then for at least one i in I, the element x belongs to Ac = U Ai, that is, x ∈/ Ai. Then x cannot belong to the intersection ∩ T T i c A = i∈I Ai, so x ∈ ( i∈I Ai) . 1.2 Essentials of Logic 1.2.5 (a) If it is false that x ≥ 0 and y ≥ 0, then either x ≥ 0 is false or y ≥ 0 is false (or both), implying that either x 0 or y 0. Conversely, if either x 0 or y 0, then either x ≥ 0 is false or y ≥ 0 is false, so it cannot be true that both x ≥ 0 and y ≥ 0. (b) If it is false that all x satisfy x ≥ a, then there exists x such that x a. Conversely, if there exists x such that x a, then it is false that all x satisfy x ≥ a. (c) If it is false that neither x nor y is less than 5, then it is false that both x ≥ 5 and y ≥ 5, so either x 5, or y 5, or both. Conversely, if either x 5, or y 5, or both, then it is false that both x ≥ 5 and y ≥ 5, so it is false that neither x nor y is less than 5. (d) Suppose it is false that for each ε 0, there exists a δ 0 such that B is satisfied. Then there exists an ε 0 such that there is no δ 0 such that B is satisfied. So there exists an ε 0 such that B is not satisfied for any δ 0. Conversely, suppose there exists an ε 0 such that B is not satisfied for any δ 0. Then is false that, for each ε 0, there exists a δ 0 such that B is satisfied. (e) Suppose it is false that no one can help liking cats. Then there exists someone who can resist liking cats. (Perhaps it is someone with a severe cat allergy.) Conversely, suppose someone can resist liking cats. Then it is false that no one can help liking cats. 1 (f) Suppose it is false that everyone loves somebody some of the time. Then there is someone who does not love somebody some of the time. That person never loves anyone. Conversely, if there is someone who never loves anyone, then at any time, that person fails to love anybody. So it is false that everyone loves somebody some of the time. Review exercises for Chapter 1 1.R.3 Consider the Venn diagram for three sets depicted in Fig. 1.R.3. Let nk denote the number of students in the set marked (k), for k = 1, 2, . . . , 8. Suppose the sets A, B, and C refer to those who study English, French, and Spanish, respectively. Since 10 students take all three languages, n7 = 10. There are 15 who take French and Spanish, so 15 = n2 + n7, and thus n2 = 5. Furthermore, 32 = n3 + n7, so n3 = 22. Also, 110 = n1 + n7, so n1 = 100. The rest of the information implies that 52 = n2 + n3 + n6 + n7, so n6 = 52 − 5 − 22 − 10 = 15. Moreover, 220 = n1 + n2 + n5 + n7, so n5 = 220 − 100 − 5 − 10 = 105. Finally, 780 = n1 + n3 + n4 + n7, so n4 = 780 − 100 − 22 − 10 = 648. The answers are therefore: (a) n1 = 100, (b) n3 + n4 = 648 + 22 = 670, (c) 1000 −∑7 ni = 1000 − 905 = 95. Figure 1.R.3 1.R.4 (a) ⇒ is true; ⇐ is false, because x = y = 1 also solves x + y = 2. (b) ⇒ is false, because x2 = 16 also has the solution x = −4; ⇐ true, because if x = 4, then x2 = 16. (c) ⇒ is true, because (x −3)2 ≥ 0; ⇐ false because with y −2 and x = 3, one has (x −3)2(y + 2) = 0. (d) ⇒ and ⇐ are both true, since the equation x3 = 8 has the solution x = 2 and no others.1 2 Algebra 2.3 Rules of algebra 2.3.4 (a) (2t − 1)(t2 − 2t + 1) = 2t(t2 − 2t + 1) −(t2 − 2t + 1) = 2t3 − 4t2 + 2t − t2 + 2t − 1 = 2t3 − 5t2 + 4t − 1. 1 In the terminology of Section 6.3, function f (x) = x3 is strictly increasing. See Fig. 4.3.7 and Exercise 6.3.3. 2 (b) (a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = (a2 + 2a + 1) + (a2 − 2a + 1) − 2(a2 − 1) = 4.2 (c) (x + y + z)2 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) = (x2 + xy + xz) + (yx + y2 + yz) + (zx + zy + z2) = x2 + y2 + z2 + 2xy + 2xz + 2yz. (d) Put a = x + y + z and b = x − y − z. Then (x + y + z)2 −(x − y − z)2 = a2 − b2 = (a + b)(a − b) = 2x(2y + 2z) = 4x(y + z) 2.4 Fractions 2.4.5 (a) 1 − 1 = x + 2 − x − 2 = x + 2 − x + 2 = 4 . x − 2 x + 2 (x − 2)(x + 2) (x + 2)(x − 2) (x − 2)(x + 2) x2 − 4 (b) Since 4x + 2 = 2(2x + 1) and 4x2 − 1 = (2x + 1)(2x − 1), the lowest common denominator, LCD, is 2(2x + 1)(2x − 1). Then, 6x + 25 6x2 + x − 2 (6x + 25)(2x − 1) − 2(6x2 + x − 2) 42x − 21 21 4x + 2 − 4x2 − 1 = = = . 2(2x + 1)(2x − 1) 2(2x + 1)(2x − 1) 2(2x + 1) 18b2 a 18b2 − a(a − 3b) + 2(a2 − 9b2) a(a + 3b) a (c) a2 − 9b2 − a + 3b + 2 = = = . (a + 3b)(a − 3b) (a + 3b)(a − 3b) a − 3b (d) 1 − 1 = (a + 2) − a = 2 = 1 . 2t − t2 5t 2t t(2 − t) 3t −t(t − 2) 3t −3t (e) • − t + 2 t − 2 t − 2 = t + 2 • t − 2 = t + 2 • t − 2 = t + 2 . a (f) Note that 1 1 2a 0 25 1 4 1 2 a 1 − 2a 0.25 = 2 −(4a − 2) = 4 − 4a = 4(1 − a). 2 1 2(x + 1) + x − 3x(x + 1) 2 − 3x2 2.4.6 (a) x + x + 1 − 3 = = x(x + 1) x(x + 1) t t (b) − = t(2t − 1) − t(2t + 1) = −2t 2t + 1 2t − 1 (2t + 1)(2t − 1) 4t2 − 1 3x 4x 2x − 1 3x(x − 2) + 4x(x + 2) −(2x − 1) 7x2 + 1 (c) x + 2 − 2 − x − (x − 2)(x + 2) = 1 1 1 1 + + = (x − 2)(x + 2) xy x2 − 4 (d) The expression equals x y = x y 1 1 y + x = = x + y. 1 xy xy • xy 1 1 1 1 2 2 x2 − y2 x2 − y2 • x y y2 − x2 (e) The expression equals 1 = 1 1 1 = y2 + x2 . x2 + y2 x2 + y2 • x2y2 2 Alternatively, apply the quadratic identity x2 + y2 − 2xy = (x − y)2 with x = a + 1 and y = a − 1 to obtain (a + 1)2 +(a − 1)2 − 2(a + 1)(a − 1) = [(a + 1) −(a − 1)]2 = 22 = 4. 3 (f) To clear the fractions within both the numerator and denominator, multiply both by xy to get a(y − x) = y − x a(y + x) y + x 2.4.8 (a) 1 1 4 − 5 5 4 = 20 − 20 = 1 , so 20 1 1 −2 4 − 5 1 −2 = 20 = 202 = 400. n n • n n2 n(n − 1) − n2 n (b) n − 1 − 1 = n − 1 = n − n − 1 = n − 1 = − n − 1 . n 1 − n • n (c) Let u = xp−q. Then 1 1 + xp−q 1 + 1 + xq−p 1 = 1 + u 1 + 1 + 1/u 1 = 1 + u u + 1 + u = 1. (d) Using x2 − 1 = (x + 1)(x − 1), one has 1 1 + (x2 − 1) x − 1 x2 − 1 (x + 1) + 1 x + 2 x − which reduces to 2 x + 1 (x2 − 1) = x(x2 − 1) − 2(x − 1) = (x − 1)[x(x + 1) − 2] , x + 2 x + 2 1 (e) Since (x − 1)(x2 + x − 2) = (x − 1)[(x + 2)(x − 1)] = (x − 1)2 . 1 1 x2 −(x + h)2 −2xh − h it follows that (x + h)2 − x2 = x2(x + h)2 = x2(x + h)2 , 1 1 (x + h)2 − x2 −2x − h h = x2(x + h)2 . 2 10x2 (f) Multiplying both numerator and denominator by x 2x — 1 = (x + 1)(x − 1) yields 5x(x − 1) , which reduces to . x − 1 2.5 Fractional Powers 2.5.5 The answers for each respective part that are given in the book emerge after multiplying both numer- ator and denominator by the following terms: (a) √7 − √5 (b) √5 − √3 (c) √3 + 2 (d) x√y − y√x (e) √x + h + √x (f) 1 − √x + 1 2 2.5.12 (a) (2 ) = 2 , which equals 2 if and only if 2x = x , or if and only if x = 0 or x = 2. (b) Correct because ap−q = ap/aq. (c) Correct because a−p = 1/ap. (d) 51/x = 1/5x = 5−x if and only if 1/x = −x or −x2 = 1, so there is no real x that satisfies the equation. 4 (e) Put u = ax and v = ay, which reduces the equation to uv = u + v, or 0 = uv − u − v = (u − 1)(v − 1) − 1. This is true only for special values of u and v and so for special values of x and y. In particular, the equation is false when x = y = 1. (f) Putting u = √x and v = √y reduces the equation to 2u • 2v = 2uv, which holds if and only if uv = u + v, as in (e) above. 2.6 Inequalities 2.6.4 (a) It is easy to see by means of a sign diagram that x(x + 3) 0 precisely when x lies in the open interval (−3, 0). Therefore we have ⇒, but not ⇐: for example, if x = 10, then x(x + 3) = 130. (b) x2 9 ⇐⇒ −3 x 3, so x2 9 ⇒ x 3. If x = −5, for instance, we have x 3 but x2 9, hence we cannot have ⇐ here. (c) If x 0, then x2 0, but x2 0 also when x 0. So, we have ⇐ but not ⇒. (d) As y2 ≥ 0 for any real number y, we have that x 0 whenever x y2. But x = 1 0 does not imply that x y2 for y ≥ 1. That is, we have ⇒ but not ⇐ . 2.6.8 Note that for any x ≥ 0 and y ≥ 0, x − 2√xy + y = (√x − √y)2 ≥ 0, which implies that 1 (x + y) ≥ √xy. (∗) Note also that the inequality is strict unless x = y. To show that mA ≥ mG, simply let x = a and y = b, so that 1 mA = 2 (a + b) ≥ with strict inequality whenever a /= b. ab = mG, To show that mG ≥ mH, use x = 1/a and y = 1/b, so that, from (∗), r 1 1 1 1 1 Rearranging, we get that + 2 a b a • b ≥ 0. −1 r !−1 1 1 1 1 1 mH = 2 a + b ≤ a • b = mG, where, again, the inequality is strict unless a = b. 2.8 Sign Diagrams 2.8.1 (a) This inequality has the same solutions as 3x + 1 − 2 0, or 3x + 1 − 2(2x + 4) 0, or −x − 7 0. A sign diagram reveals that the inequality is satisfied for −7 x −2. A serious error is to multiply the inequality by 2x + 4, without checking the sign of 2x + 4. If 2x + 4 0, multiplying by this number will reverse the inequality sign.3 3 It could be instructive to test the inequality for some values of x. For example, for x = 0 it is not true. What about x = −5? 5 (b) The inequality is equivalent to 120/n ≤ 0.75 = 3/4, or (480 − 3n)/4n ≤ 0. A sign diagram reveals that the inequality is satisfied for n 0 and for n ≥ 160.4 (c) This is easy: g(g − 2) ≤ 0 and so 0 ≤ g ≤ 2. (d) Note that p2 − 4p + 4 = (p − 2)2, so the inequality reduces to (p − 2) + 3 = p + 1 ≥ 0. (p − 2)2 The fraction makes no sense if p = 2. The conclusion that p ≥ −1 and p (e) The inequality is equivalent to (p − 2)2 2 follows. −n − 2 − 2 0 ⇐⇒ −n − 2 − 2n − 8 0 ⇐⇒ −3n − 10 0 ⇐⇒ −4 n − 10 (f) x4 − x2 = x2(x2 − 1) 0 ⇐⇒ x =/ 0 and x2 1 ⇐⇒ −1 x 0 or 0 x 1 2.9 Summation Notation 2.9.3 (a)–(d): In each case, look at the last term in the sum and replace n by k to get an expression for the kth term. Call it sk. Then in (a), (b), and (d) the sum is ∑n sk, and in (c) we have ∑n sk. (e) The coefficients are the powers 3n for n = 1, 2, 3, 4, 5, so the general term is 3nxn. (f) –(g) See the answers in the Solutions section of the book. (h) This is trickier: one has to see that each term is 198 larger than the previous term.5 2.9.7 (a) Valid: ∑n ck2 = c • 12 + c • 22 + • • • + c • n2 = c(12 + 22 + • • • + n2) = c ∑n k2. (b) Wrong, even for n = 2: the left-hand side is (a1 + a2)2 = a2 + 2a1a2 + a2, but the right-hand side 1 2 is a2 + a2, which is different unless a1a2 = 0. 1 2 (c) Valid: both sides equal b1 + b2 + • • • + bN. (d) Valid: both sides equal 51 + 52 + 53 + 54 + 55. (e) Valid: both sides equal a2 + • • • + a2 . (f) Wrong, even for n = 2: the left-hand side is a1 + a2/2, but the right-hand side is (1/k)(a1 + a2). 2.12 Double Sums 2.12.1 (a) See the solution in the book. rs 2 2 4 rs (b) Note first that r + s = 0 when s = 0. So ∑ ∑ s=0 r=2 r + s reduces to 2 ∑ s=1 " 2 2s + 2 + s 2 3s + 3 + s 4s 4 + s 2# = 2 2 + 3 3 2 + 4 4 2 + 5 4 2 + 4 6 2 + 5 8 2 6 4 9 which equals + + 16 + 1 + 36 + 16 = (4 + 16)• 400 + 9 • 225 + (16 + 36)• 144 + 1 9 16 25 25 9 3600 8000 + 2025 + 7488 or 3600 + 1 = 17513 3600 + 1 = 5 + 3113 . 3600 4 Note that for n = 0 the inequality makes no sense. For n = 160, we have equality. 5 This problem is related to the story about Gauss in Section 2.9. 6 (c) Because ∑m n j=1 (i + j2) = ∑n m i=1 m i=1 n j=1 j2, we can use formulae (2.9.4) and (2.9.5) along with the equality ∑p a = pa to write the sum as n 1 m 1 1 1 ∑ 2 m(m + 1) + ∑ 6 n(n + 1)(2n + 1) = n 2 m(m + 1) + m 6 n(n + 1)(2n + 1) j=1 i=1 = 1 mn[3(m + 1) + (n + 1)(2n + 1)] = 1 mn(2n2 + 3n + 3m + 4) 6 6 (d) Using formulae (2.9.4) and (2.9.5) again, we get m 2 m m m ∑ ∑ i j = ∑(i + i2) = ∑ i + ∑ i2 = 1 m(m + 1) + 1 m(m + 1)(2m + 1) i=1 j=1 i=1 i=1 2 6 i=1 = 1 m(m + 1)[3 + (2m + 1)] = 1 m(m + 1)(2m + 4) = 1 m(m + 1)(m + 2). 6 6 3 2.12.4 a¯ is the mean of the column means a¯ j, because ! 1 n 1 n 1 m 1 m n n ∑ a¯ j = n ∑ m ∑ ar j = mn ∑ ∑ ar j = a¯. j=1 j=1 r=1 r=1 j=1 To prove (∗), note that because ar j − a¯ is independent of the summation index s, it is a common factor when we sum over s, so ∑m (ar j − a¯)(as j − a¯) = (ar j − a¯) ∑m (as j − a¯) for each r. Next, summing over r gives m m " #" # m m ∑ ∑(ar j a¯)(as j a¯) = r=1 s=1 ∑(ar j a¯) ∑(as j a¯) r=1 s=1 , (∗∗) because ∑m (ar j − a¯) is a common factor when we sum over r. Using the properties of sums and the definition of a¯ j, we have m m m ∑(ar j − a¯) = ∑ ar j − ∑ a¯ = ma¯ j − ma¯ = m(a¯ j − a¯). r=1 r=1 r=1 Similarly, replacing r with s as the index of summation, one also has ∑m (as j − a¯) = m(a¯ j − a¯). Substituting these values into (∗∗) then confirms (∗). Review exercises for Chapter 2 2.R.5 (a) (2x)4 = 24x4 = 16x4. (b) 2−1 − 4−1 = 1 − 1 = 1 , so (2−1 − 4−1)−1 = 4. (c) Cancel the common factor 4x2yz2. (d) Here −(−ab3)−3 = −(−1)−3a−3b−9 = a−3b−9, so [−(−ab3)−3(a6b6)2]3 = [a−3b−9a12b12]3 = [a9b3]3 = a27b9. (e) a5 • a3 • a−2 a−3 • a6 a6 = a3 = a3. (f) x 3 2 • 8 x−2 −3 x3 = 8 • 8x −3 = (x5)−3 = x−15. 7 2.R.9 Parts (a), (b), (d), (e), and (f) are straightforward; their solutions appear in the book. For the other parts: (c) − 3 3 − 6 = −3 + 3 6 = −3 + 3 3 2 = −3 + 3 2. (g) (1 + x + x2 + x3)(1 − x) = (1 + x + x2 + x3) − (1 + x + x2 + x3)x = 1 − x4. (h) (1 + x)4 = (1 + x)2(1 + x)2 = (1 + 2x + x2)(1 + 2x + x2), and so on. - 2.R.11 Parts (a) and (b) are easy, so we focus on the others: (c) ax + ay + 2x + 2y = a(x + y) + 2(x + y) = (a + 2)(x + y). (d) 2x2 − 5yz + 10xz − xy = 2x2 + 10xz − (xy + 5yz) = 2x(x + 5z) − y(x + 5z) = (2x − y)(x + 5z). (e) p2 − q2 + p − q = (p − q)(p + q) + (p − q) = (p − q)(p + q + 1). (f) u3 + v3 − u2v − v2u = u2(u − v) + v2(v − u) = (u2 − v2)(u − v) = (u + v)(u − v)(u − v), which simplifies to (u + v)(u − v)2. s s 2.R.15 (a) − = s(2s + 1) − s(2s − 1) = 2s . 2s − 1 2s + 1 (2s − 1)(2s + 1) 4s2 − 1 x (b) — 1 − x − 24 = −x(x + 3) − (1 − x)(x − 3) − 24 = −7x − 21 = 7 . 3 − x x + 3 x2 − 9 (x − 3)(x + 3) (x − 3)(x + 3) 3 − x (c) Multiplying both numerator and denominator by x2y2 yields y − x = y − x = 1 y2 − x2 (y − x)(y + x) x + y 2.R.16 (a) Simply cancel the factor 25ab. (b) Factor x2 − y2 = (x + y)(x − y), and then cancel x + y. (c) The fraction can be written as (2a 3b)2 = (2a − 3b)(2a + 3b) 2a − 3b . 2a + 3b (d) 4x x3 4 − 4x + x2 = x(4 x2) (2 − x)2 = x(2 − x)(2 + x) = (2 − x)2 x(2 + x) 2 − x 2.R.24 Let each side have length s. Then the area K of the equilateral triangle ABC shown in Fig. 2.R.24 is the sum of the areas of the three triangles ABP, BCP, and CAP, which equals 1 sh1 + 1 sh2 + 1 sh3 = K. 2 2 2 It follows that h1 + h2 + h3 = 2K/s, which is independent of where P is placed in the triangle. C A s B Figure 2.R.24 8 2.R.28 (a) Using the trick that led to Eq. (2.9.4) in the book, R = 3 + 5 + 7 + • • • + 197 + 199 + 201 R = 201 + 199 + 197 + • • • + 7 + 5 + 3. Summing vertically term by term, noting that there are 100 terms, gives 2R = 204 + 204 + 204 + • • • + 204 + 204 + 204 = 100 × 204 = 20400, and thus R = 10200. (b) Here one has S = 1001 + 2002 + 3003 + • • • + 8008 + 9009 + 10010 = 1001(1 + 2 + 3 + • • • + 8 + 9 + 10) = 1001 • 55 = 55055 3 Solving Equations 3.1 Solving Equations 3.1.3 (a) We note first that x = −3 and x = −4 both make the equation absurd. Multiplying the equation by the common denominator (x + 3)(x + 4) yields (x − 3)(x + 4) = (x + 3)(x − 4), i.e. x2 + x − 12 = x2 − x − 12, and thus x = 0. (b) Multiplying by the common denominator (x − 3)(x + 3) yields 3(x + 3) − 2(x − 3) = 9, from which we get x = −6. (c) Multiplying by the common denominator 15x, assuming that x 15x + 8x2, from which we get x = 5. 0, yields 18x2 − 75 = 10x2 −