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.75 8 2 2 2 (b) 4.011025 51. Local maximum 2; local 57. Local maximum 0 when x 0; 49. (a) 4.01 minimums 1, 0 53. Local maximums 0, 1; local minimums 2, 1 55. Local maximum 0.38 when x 0.58; local minimum 0.38 when x 0.58 local minimum 13.61 when x 1.71; local minimum 73.32 when x 3.21 x 4.00 imum 0.38 when x 1.73 67. 30 times 69. 50 trees per acre 71. 600 ft by 1200 ft 73. Width 8.40 ft, height of rectangular part 4.20 ft 75. (a) 77. (a) 61. Local maximum 0.38 when x 1.73; local min- (b) 300 ft by 600 ft (b) $9.50 59. Local maximum 5.66 when 1200 2x 57,000 3000x 65. $4000, 100 units (c) $19.00 63. 25 ft SECTION 4.2 ■ page 312 1. II 3. (a), (c) 5. (a) 1 x 0 _1 2 (b) y 16 0 _2 4 (d) y 8 (b) y 27 _3 0 _9 (d) _1 _2 _4 (c) _1 _2 7. (a) _2 _8 (c) y 4 0 _8 _2 x1 9. III 11. V 13. VI Answers to Section 4.2 A23 15. 17. y 1 0 −2 y 10 1 x −2 0 2 3 x 19. 21. y 20 12 y 1 0 _1 1 3 x 0 _1 _2 2 3 2 x 3 _3 _15 23. 25. 3 y 4 0 _4 y 10 0 _1 1 3 x x 4 _30 27 29 _4 _4 _2 y 10 x 4 3 _3 _1 0 1 _10 4 x 31. P x 1 2 x2 1 x 1 2 1 x 2 2 33. P x 1 2 y 1 0 _1 _1 1 x 2 _1 _1 A24 Answers to Selected Exercises and Chapter Tests 35. P x 1 2 1 y 2x 10 0 1 2 _20 _2 _3 2 3 x x 2 1 2 2 1 x2 2x 4 2 37. P x 1
2 y 16 5 0 _2 2 x 39. P x 1 2 1 y x2 _2 _1 0 _2 1 x 2 _4 x q, y q as x q 45. y q as x q x q, y q 41. y q as 43. y q as as (b) (b) 51. 1 1, 2 1 x q 47. (a) x-intercepts 0, 4; y-intercept 0 49. (a) x-intercepts 2, 1; y-intercept 1 2, 4 2 1, 2 1, 0 2 30 local maximum 4, 16 1 2 _4 12 _50 53. _5 30 _30 local maximum local minimum 2, 25 1 2, 7 1 2 55. 30 5 _5 5 _30, 2 local minimum 3, 27 1 2 57. _3 10 _5 3 local maximum local minimum 1, 5 1 1, 1 1 2, 2 59. One local maximum, no local minimum 61. One local maximum, one local minimum 63. One local maximum, two local minima 67. One local maximum, two local minima 69. 65. No local extrema 71. _1 5 _5 c=5 c=2 c=1 c= 1 2 1 _1.5 c=2 c=1 c=0 c=_1 1.5 5 _3 Increasing the value of c stretches the graph vertically. 73. c=0 10 c=1 c=8 c=27 4 _2 Increasing the value of c moves the graph up. Increasing the value of c causes a deeper dip in the graph in the fourth quadrant and moves the positive x-intercept to the right. _40 75. (a) y 10 _3 0 3 x _10 (c) (b) Three 0, 2 1 PO1 77. (d) x x P 2 x 2 5 x PE 1 79. (a) Two local extrema 3, 8, 2 1 PE, 12, where 2 PO1 x 2 x5 6x3 2x and 10 1 −12 6 Answers to Section 4.5 A25 81. (a) 26 blenders 2 800x 4x 83. (a) V (c) Maximum volume 1539.6 cm3 3 120x (b) No; $3276.22 x 2 1 (b) 0 x 10 55. (a) 2, 2, 3
(b) 1600 0 10 SECTION 4.3 ■ page 320 2 5. 1. quotient, remainder 3. 3 2x 3 2 1 1 x 1 x2 1 11 x 3 2x2 x 1 4x 4 x2 4 9. 13. 11. x 3 1 7. 1 2x 1 2 2 3x 4 2 1 x2 3 2 1 8 x2 x 3 15 2 2x 1 57. (a) 1 2, 2 (b) 1 2 7x 11 2 59. (a) 1, 2 (b) 17. 2x 2 1, 2 23. x 4 1, 0 29. x 2 2, 3 19. x 2, 8x 1 25. x 2, 2 31. x 2 3x 1, 1 In answers 15–37 the ﬁrst polynomial given is the quotient, and the second is the remainder. 15. x 2, 16 21. 3x 1, 7x 5 27. 3x 23, 138 33. x 4 x 3 4x 2 4x 4, 2 37. x 2 3x 9, 0 47. 2159 59. x 3 3x 2 x 3 2 x3 3x2 15 63. x 2 61. x 4 8x 3 14x 2 8x 15 x 2 35. 2x 2 4x, 1 43. 7 1 16 39. 3 51. 8.279 41. 12 57. 45. 483 2 x 9 x 1 x 1 x 1 67. 65. 49 61. (a) 1, 2 (b) y 5 0 _1 1 x 2 x 2 x y 20 0 _20 y 5 0 _1 _5 y 5 0 _1 1 x _5 3, 5, 5 6, 2, 2 7. 1, 2, 4, 8, (b) 1, 1, x 1 5 2, 1; P x 2 3, 5 2, 5 1 5 6, 10, 10 3 9. 1, 7, 13. (a) 1, 3 19. x 3 2 x 2 1 2, 63. 1 positive, 2 or 0 negative; 3 or 1 real 65. 1 positive, 1 nega67. 2 or 0 positive, 0 negative; 3 or 1 real (since 0 is a tive; 2 real zero but is neither positive nor negative) 77. 2, 3 89. 2 99. (a) It began to snow again. on Saturday night
75. 3, 1 87. 2, 2, 3 97. 11.3 ft (c) Just before midnight 81. 2, 1, 3, 4 91. 1.28, 1.53 93. 1.50 101. 2.76 m 103. 88 in. (or 3.21 in.), 1 1 2, 1, 1, 4 73. 3, 2 (b) No 15 79. 1 2, 1 2 x 1 x 21. 25. 23. 17. 1 2 3, 1 1 2 x 1 a0, an, 1 11. (a) 1, 1 15., 1, 3 2 SECTION 4.4 ■ page 329 2, 1 1. 5. 1, 3 3. True b), 2 2 1, 2; P x 2 1, 2, 3; P x 1 3, 1, 1; P 1, 2; P 4, 2, 1, 1; P 2, 3 2; P 1 2, 1 3, 3; P 1, 1 2; P 3 2, 1 2, 1; P 2, 1, 3 1, 1 2, 2; P 3, 2, 1, 3; P 1; P 1 x 2 1 x 1 3, 2, 5; P 2 1 1 x 31. 39. 43. 27. 41. 35. 29. 37. 33 2x 1 2x 3 2 1 x 2 2 1 2x 3 x 3 2 1 2x 1 2x 1 2 1 2 2 1 2x 5 2 2 1 x 2 2 2x 3 2 2 1 2x 213 2x 3 2 1 3x 1 2 2 x 3 2 2 1 3x 1 2 1 2 45. 2, 1 12 49. 3, 1 15 2 51. 47. 1, 4, 1 13 2 1 2, 53. 1, 1 2, 3 110 2 2 1 x 2 1 (b) (b) SECTION 4.5 ■ page 342 1. 5; 2; 3; 1 3. n 7. (a) 0, 1 i (b) x i 2 x 2 x 2 5. (a) 0, 2i x 1 i 1 11. (a) 2, 2i x i 2 1 x 2 x 2i x 2i 2 2 1 2 1 x 1 i 13 x 1 i 13 BA 1 2 i 13, 1 2 x 1 2 i 13 2 i 13 x 1 1 1 15. (a)
x 1 2 1 (b) (b) 1 x 1 2 2 1 2 i 13 2 A x 1 2 2 i 13 2 1 A B 1 x2 (b) 1 x 1 i 2 x 2i x 2i 2 1 9. (a) i 2 13. (a) 2, 1 i 13 B 1 2 i 13 B B A A In answers 17–33 the factored form is given ﬁrst, then the zeros are listed with the multiplicity of each in parentheses. 17. x 5i B 1 x 5i x 1 x 2i 1 19. 3 21 2i 2 1 2 ; 5i, 2i 1 2 2 ; 1 i 2 4, 2i A26 Answers to Selected Exercises and Chapter Tests 23. 25. 27. 29. 31. 33. 35. 39. P 1 41, 1 59. 2 63. (a) 65. (a) (b) 1 67. (a 3i 2 1 2 1 x 3 x 3 2BA 2BA x 3i x i 2 1 x 1 2 1 16 13 2 1 2; i 2 2 1 x 2i 2 1 x i 13 2i 2 1 2 BA ;, 3i, 1 1 2 1, 3 2 1 1, 3i 2i 2 2 1 ; 1 1 2, i 13, 1 1, i 13 x3 3x2 4x 12 1 2 2 2 2 1 2, 2i x2 2x 2 37. x3 2 x2 x 2 x4 4x3 10x2 12x 5 6x4 12x3 18x2 12x 12 1 2 2 47. 43. x 1 i 13 2 53. 2, 1, 3i 55. 49. 2, 1 i 13 2 1, 2i, i 13, i 45. 2, 2i 51. 3 2, 1 i 12 57. 3 (multiplicity 2), 2i 1 1 multiplicity 2 2 x 5 x2 4 x 1 x 1 2 1 2 (b) 1 x2 9 61. 1 (multiplicity 3), 3i x 2i x 2i 3i 3i 2 x2 2x 4 1 i 13 2 1 x 2 3 x A 1 i 13 13 (b) 1 x A 3 69. (a) 4 real B 4 3 (b) 2 real, 2 imaginary (c) 4 imaginary B 4 A x2 2x 4 2 2 1
1 i 13 x B 4 3 A B 4 37. y 5 0 41. 43. 39. 2 x −3 y 2 0 −2 2 x y 5 0 _ 4 4 _5 x x-intercept 1 y-intercept 2 vertical x 2 horizontal y 4 y 5 0 _5 _10 5 x 4 x-intercept 3 4 y-intercept 7 vertical x 7 horizontal, q 3. 1, 2 5. 2, 3 q as x 2 SECTION 4.6 ■ page 356 1. 7. (a) 3, 19, 199, 1999; 5, 21, 201, 2001; 1.2500, 1.0417, 1.0204, 1.0020; 0.8333, 0.9615, 0.9804, 0.9980 q as x 2; r (b) (c) Horizontal asymptote y 1 9. (a) 22, 430, 40,300, 4,003,000; 10, 370, 39,700, 3,997,000; 0.3125, 0.0608, 0.0302, 0.0030; 0.2778, 0.0592, 0.0298, 0.0030 q as x 2 q as x 2; r (b) x 2 1 (c) Horizontal asymptote y 0 11. x-intercept 1, y-intercept 13. x-intercepts 1, 2; y-intercept 15. x-intercepts 3, 3; 1 3 no y-intercept 17. x-intercept 3, y-intercept 3, vertical x 2; horizontal y 2 19. x-intercepts 1, 1; y-intercept ; vertical 21. Vertical x 2; horizontal x 2, x 2; horizontal y 1 y 0 25. Vertical horizontal y 3 x 1 29. Vertical x 0; horizontal y 3 33. 2, x 1; horizontal y 23. Horizontal y 0 31. Vertical x 1 3, x 2; 27. Vertical x 1 35 _1 0 x 45. y y-intercept 2 vertical x 3 horizontal y 0 10 0 3 x 1 4 47. y 5 0 1 49. y 2 2 0 x-intercept 2 y-intercept 2 vertical x 1, x 4 horizontal y 0 y-intercept 1 vertical x 1, x 6 horizontal y 0 x x Answers to Section 4.6
A27 x-intercept 1 vertical x 0, x 3 horizontal y 0 slant y x 2 vertical x 2 slant y x 2 vertical x 0 slant y x 8 vertical x 3 slant y x 1 vertical x 2, x 2 63. 65. 67. 69. 71. 2 x y 1 0 y 10 _6 0 6 x _10 y 10 _6 0 6 x _10 y 30 _10 0 10 x _4 _1 _30 y 30 0 _6 6 x 51. 53. 55. 57. 59. 61 _6 x-intercept 2 3 y-intercept 4 vertical x 4, x 2 horizontal y 0 x-intercepts 2, 1 2 y-intercept 3 vertical x 1, x 3 horizontal y 1 3 x 6 x x-intercept 1 y-intercept 1 vertical x 1 horizontal y 1 _6 0 6 x _6 y 6 0 _6 y 6 x-intercepts 6, 1 y-intercept 2 vertical x 3, x 2 horizontal y 2 6 x x-intercepts 2, 3 vertical x 3, x 0 horizontal y 1 _6 0 6 x _6 y 10 _6 0 6 x y-intercept 2 vertical x 1, x 3 horizontal y 3 73. 30 vertical x 3 _10 10 _30 5. Minimum f 1 1 2 7 7. 68 feet 9. y 300 64 11. y 200 _4 0 4 x _300 _3 (_1, _32) _200 1 _30 x 13. y 2 100 31 0 _1 1 _100 x A28 Answers to Selected Exercises and Chapter Tests 75. 60 vertical x 2 vertical x 1.5 x-intercepts 0, 2.5 y-intercept 0, local maximum local minimum end behavior: y x 4 3.9, 10.4 2 0.9, 0.6 1 1 vertical x 1 x-intercept 0 y-intercept 0 local minimum end behavior: y x 2 1 1.4, 3.1 2 _10 _10 _3 _5 77. 79. 81. _30 10 _20 10 _5 100 _100 10 10 3 5 0 30 85. (a) 2.50 mg/L (b) It decreases to 0. 87. 5000 (c) 16.61 h 0 400 If the speed of the train approaches the speed of sound, then the pitch increases indeﬁnitely (a
sonic boom). CHAPTER 4 REVIEW ■ page 361 x 2 2 3 1. (a) 3. (a) x f 1 2 1 2 (b) x 4 1 2 2 17 g 1 x 2 y (b) y 2 0 2 x 15. (a) 0 (multiplicity 3), 2 (multiplicity 2) (b) y, 2 1 0 1 x vertical x 3 x-intercepts 1.6, 2.7 y-intercept 2 local maxima 2.4, 3.8 1 local minima 3.4, 54.3 2 1 end behavior y x 3, 1 1 2 0.4, 1.8 0.6, 2.3, 2 83. (a) 4000 (b) It levels off at 3000. 17. 10 x-intercepts 2.1, 0.3, 1.9 y-intercept 1 local maximum local minimum y q as y q as 1.2, 4.1 1 1.2, 2.1 1 x q x q 2 2 x-intercepts 0.1, 2.1 y-intercept 1 local minimum y q as y q as 1 x q x q 1.4, 14.5 2 3 3 _10 30 _20 S 13.8x 100 x2 1 2 (b) 0 x 10 (d) 5.8 in. _3 _2 19. 21. (a) (c) 6000 5 0 1 x 0 10 Answers to Chapter 4 Test A29 In answers 23–29 the ﬁrst polynomial given is the quotient, and the second is the remainder. 23. x 1, 3 422 9, 18 39. (a) 4, 0, 4 (b) y 27. x 3 5x 2 17x 83, 35. 8 37. (a) 1, 2, 3, 6, 41. (a) 2, 0 (multiplicity 2), 1 (b) y 25. x 2 3x 23, 94 31. 3 (b) 2 or 0 positive, 3 or 1 negative 29. 2x 3, 12 75. 30 _6 6 _30 x-intercept 2 y-intercept 4 vertical x 1, x 2 slant y x 1 local maximum local minimum 0.425, 3.599 1 4.216, 7.175 1 2 2 30 0 _4 _30 4 x 4 0 _4 _2 1 x
43. (a) 2, 1, 2, 3 (b) y 45. (a) (b) 1 2, 1 y 2, 28 77. 1 1, 26, 2 1, 2 1 2, 68, 2 1 5, 770 2 CHAPTER 4 TEST ■ page 363 1. 2 25 4 x f y x 1 2B A 1 2 1 0 1 x 20 _2 0 _10 2 x 10 _1 _10 1 x 2. 4. Minimum f 3 2 A 3 2B y 3. (a) 2500 ft (b) 1000 ft 47. 4x 3 18x 2 14x 12 49. No; since the complex conjugates of imaginary zeros will also be zeros, the polynomial would have 8 zeros, contradicting the requirement that it have degree 4. 51. 3, 1, 5 53. 1 2i, 2 (multiplicity 2) 55. 2, 59. 1, 3, 1 i 17 2 y 6 0 _6 0.25 6 x 1 (multiplicity 3) 57. 2, 61. x 0.5, 3 65. 1 i 13 63. x 0.24, 4.24 x2 2x 2 x 2 2, P x 1 2 1 2 1 2 69. 67. _5 71. _5 _3 3 10 _12 _20 x 5 4 _6 x 5 3 y 0 y 2 0 _9 73. 20 _10 10 _20 x-intercept 3 y-intercept 0.5 vertical x 3 horizontal y 0.5 no local extrema 40 19 _2 0 1 x _40 5. (a) x 3 2x 2 2, 9 6. (a) 1, 3, 3 2 (c) 1, 1 2 (d), 3 1 2, (b) (b) y x3 2x2 1 15, 2B1 10 0 1 x 2 1 x 1 x 2i 7. 3, 1 i 8. 9. x 4 2x 3 10x 2 18x 9 10. (a) 4, 2, or 0 positive; 0 negative (c) 0.17, 3.93 2 1 80 1 2 x 2i 2 _3 5 (d) Local minimum _80 2.8, 70.3 1 2 A30 Answers to Selected Exercises and Chapter Tests 11. (a) r, u (d) (b) s (c) s
(e) x 2 2x 5 60 y 6 0 _6 6 25 2 _6 _3 _10 10 6 x _60 FOCUS ON MODELING ■ page 366 1. (a) y 0.275428x 2 19.7485x 273.5523 (b) 82 (c) 35.85 lb/in2 25 48 46 3. (a) y 0.00203708x 3 0.104521x 2 1.966206x 1.45576 (d) 2.0 s (b) (c) 43 vegetables 22 0 30 5. (a) Degree 2 (b) y 16.0x 2 51.8429x 4.20714 48 0 0 3.1 (c) 0.3 s and 2.9 s (d) 46.2 ft CHAPTER 5 SECTION 5.1 ■ page 378 1. 77.880, 1.587 2.718, 0.135 11. 25, 1, 25, 15,625 5; 1 13. 3. natural; 2.71828 7. 0.885, 0.606, 0.117, 1.837 5. 2.000, 7.103, 9. 20.085, 1.259, y 1 2 0 _2 (2, 4) (_2, 9) 2 x _2 y 3 1 0 2 x (2, 22.17) y 17. y=2_x y=2x 2 x 1 −2 0 2 x y 3 0 15. _2 19. y y=7˛ y=4˛ 2 0 1 _2 2 x 21. 27. x f 1, 3x 2 q, 0 1 2 23 4B A 25. II, 29. 3, q 1 2, y 3 y 0 _1 _2 2 x (1, _3) _5 y 3 0 _5 5 x (1, _1) 31., 4, q 1 2, y 4 33., 0_1, 6) _5 y 1000 (_3, 1) 5 x _2 0 2 x 35., q, 0 1 2, y 0 37., 1, q 1 2, y 1 y 0 _1 _2 1 x (1, _2.72) y 1 (_1, 1.72) 0 _1 2 x Answers to Section 5.2 A31 39., 0, q y
1 2, y 0 41. (a) y ˝=3(2˛) 63. (a) 100 (b) 482, 999, 1168 65. (a) 11.79 billion, 11.97 billion (b) 14 (c) 1200 (c) 12 billion (2, 1) 1 0 1 x Ï=2˛ 2 _2 0 2 x 0 500 43. y g(x) = 3x (b) The graph of g is steeper than that of f. 45. (a) y 5 f(x) = x3 67. $5203.71, $5415.71, $5636.36, $5865.99, $6104.98, $6353.71 69. (a) $16,288.95 71. (a) $4,615.87 (e) $4,704.68 75. (a) $7,678.96 (b) $26,532.98 (b) $4,658.91 (b) $67,121.04 (c) $43,219.42 (g) $4,704.94 (c) $4,697.04 (f) $4,704.93 73. (i) (d) $4,703.11 SECTION 5.2 ■ page 391 1. 10x x _2 0 2 x x 103 102 101 100 101 102 103 101/2 log 200 0 2 47. (a) 20 (i) ˝=x∞ Ï=2˛ 10¶ (ii) Ï=2˛ ˝=x∞ 5 0 25 The graph of f ultimately increases much more quickly than that of g. (b) 1.2, 22.4 Ï=2˛ ˝=x∞ 50 53. (a) c=4 c=2 5 c=1 0 (iii) 10 • 0 49. c=0.5 c=0.25 3 _3 _1 The larger the value of c, the more rapidly the graph increases. 55. Local minimum q, 1.00 1 59. (a) 13 kg 61. (a) 0 (c) 100, decreasing on (b) 6.6 kg (b) 50.6 ft/s, 69.2 ft/s 0.27, 1.75 1
4 3 2 1.00, q 2 _3 3 _1 (b) The larger the value of a, the wider the graph. 57. (a) Increasing on q, 0.37 (b) 1 4 log5 125 3 (b) 52 25 3. (a) 5. Logarithmic form log8 8 1 log8 64 2 log8 4 2 log8 512 3 1 log8 2 log8 1 8 1 64 3 Exponential form 81 8 82 64 82/3 4 83 512 81 1 8 82 1 64 5 a=0.5 a=1 a=1.5 a=2 (b) 50 1 (b) e5 y 7. (a) 52 25 11. (a) ex 5 (b) log10 0.0001 4 17. (a) ln 2 x (b) 2 21. (a) 2 9. (a) 81/3 2 (b) 13. (a) log5 125 3 (b) 1 23 1 8 3 1 log2 8 (c) 2 (c) 1 (b) 0 1 2 (b) (c) 1 (b) 4 2 3 33. (a) 100 (b) 25 (c) 0.1761 (b) 1.5465 1 log8 8 19. (a) 1 15. (a) (b) ln y 3 (c) 10 23. (a) 3 27. (a) (c) 15 31. (a) 5 37. (a) 0.3010 (b) 27 (b) 3.2308 (c) 1.0051 43. y 1 25. (a) 37 (b) 8 (b) 4 (b) 4 29. (a) 32 35. (a) 2 39. (a) 1.6094 41. y 1 0 (d) 80 ft/s 1 x 0 2 x 0 100 45. y log5 x 47. y log9 x 49. I A32 Answers to Selected Exercises and Chapter Tests 51. y 5 y=4˛ y=ø› x 5 x _2 0 _2 53. 4 _1 1 55. 1 q, 0 2,, x 0 57. 01, 2) 1 75. The graph of f grows more slowly than g. 77. (a) c=4 2.6 c=3 c=2 c=1 1 x (b
) The graph of log f 2 f graph of shifted upward log c units. cx 1 x 2 1 is the 2 log x 1 2 x _10 _1 100 79. (a) 1, q (b) 2 f1 x 102x 1 f1 x 81. (a) log2 a 85. 11.5 yr, 9.9 yr, 8.7 yr 1 2 1 2 x 1 x b 87. 5.32, 4.32 (b) 0, 1 1 2 83. 2602 yr SECTION 5.3 ■ page 398 1. sum; log5 25 log5 125 2 3 x 5. 10, e; Change of Base; log7 12 times; 10 # log5 25 1.277 7. 3 2 3. log 12 log 7 59. 0, q 1 2,, x 0 61. 0, q 1, 3 2 01, 1) 2 y 1 0 x 37. 41. 43. 1 x 17. 4 19. 1 log2 x 25. log2 A 2 log2 B 31. ln a ln b 1 21 2 13. 3 x 1 15. 200 9. 2 21. 11. 3 log2 x log21 23. 10 log 6 log3 x 1 x2 1 1 2 log3 y 27. 3 log51 33. 3 log x 4 log y 6 log z 1 2 29. 2 35. log2 x log21 ln x 1 21 x2 4 1 log 2 3 1 3 ln x 1 x2 1 ln y ln z 2 log 2 x 1 2 ln 2 2 log21 1 39. 4 log x2 1 1 ln 2 3x 4 1 2 45. log3 160 47. AB/C2 49. log a 2 1 2 log2 1 x2 1 2 x2 y2 1 2 log 2 x3 7 1 2 4 2 x 1 x4 1 23 x2 1 b 2 3, q 2 63. 69. 1 65. 1 1 q, 1 1 2 2 3 20 _2 _2 71. 3 _1 _6 73. 1 _1 _3 1, q 2 0, 2 67. 2 1 1, 1 domain vertical asymptotes x 1, x 1 local maximum 0, 0 1 2 1 2 1 0, q domain vertical asymptote x 0 no maximum or minimum 2 55. 2.321928 x2 x 3 61. 3.482892 ¢
≤ 51. ln 5x2 1 1 x2 5 3 2 2 53. log 57. 2.523719 63. 59. 0.493008 _1 2 _3 4 69. (a) P c/W k 71. (a) M 2.5 log B 2.5 log B0 (b) 1866, 64 2 1 0, q domain vertical asymptote x 0 horizontal asymptote y 0 local maximum 2.72, 0.37 1 2 1 7. 0.5850 17. 43.0677 25. 2.4423 33. 1 SECTION 5.4 ■ page 408 (b) x ln 25 1. (a) ex 25 5. 0.9730 15. 1.9349 23. 2.9469 2 ln 3 0.5493 31. 41. 39. 0.01 1/ 15 0.4472 3 55. 57. 2 63. 0.57 65. 0.36 69. log 2 x log 5 73. 6.33 yr 75. 8.15 yr 43. 7 95 3 (c) 3.219 3. 1.3979 9. 1.2040 11. 0.0767 13. 0.2524 19. 2.1492 27. 14.0055 35. 0, 4 3 45. 5 47. 5 59. 2.21 21. 6.2126 29. ln 2 0.6931, 0 37. e10 22026 13 51. 4 12 61. 0.00, 1.14 49. 53. 6 67. 2 x 4 or 7 x 9 71. (a) $6435.09 (b) 8.24 yr 77. 8.30% 79. 13 days 81. (a) 7337 85. (a) t (b) 1.73 yr 1 13 5 60 I 13 ln 1 2 83. (a) P P0eh/k (b) 0.218 s (b) 56.47 kPa 69. 0.430618 73. (b) 45% (c) 1929 5. (a) 8 frogs 20,000e0.1096t 8600e0.1508t t (b) 2049 SECTION 5.5 ■ page 420 1. (a) 500 (b) 181 million (b) t n 2 1 9. (a) n 11. (a) 2029 (
b) 1.6 g 23. (a) 210F (b) 153 F (c) 28 min (b) 116 min 27. (a) 2.3 (b) 3.2 107 M 31. 4.8 pH 6.4 35. Twice as intense 37. 8.2 17. 18 yr (c) 70 yr (b) 3.5 2 1 (c) About 48,000 (b) About 11,600 13. 22.85 h 19. 149 h (c) 8.3 (d) 6.66 h (b) 816 frogs 7. (a) 20,000 3. (a) 233 million 15. (a) (d) 2017 (c) 4.6 h t 2 21. 3560 yr 1 25. (a) 137F n 10e0.0231t 29. (a) 103 M 33. log 20 1.3 39. 73 dB 41. (b) 106 dB Answers to Chapter 5 Test A33 71. 2.303600 10 _1 vertical asymptote x 2 horizontal asymptote y 2.72 no maximum or minimum 20 _20 75. 1.5 _1.5 2.5 _1.5 vertical asymptotes x 1, x 0, x 1 local maximum 0.58, 0.41 1 2 CHAPTER 5 REVIEW ■ page 425 1. 0.089, 9.739, 55.902 5., y 0,, 7. 3. 11.954, 2.989, 2.518 0, q y 1 2, y 3 3, q y 2 1 4 2 0 −3 x3 −3 0 3 x 1, q 2 9. 1 y,, x 1 11. 0, q 2 1 y,, x 0 77. 2.42 81. Increasing on 79. 0.16 x 3.15 q, 0 and 85. 0.579352 1 4 91. 1.83 yr 83. 1.953445 89. (a) $16,081.15 (d) $16,198.31 95. (a) n 1 (b) 1.39 105 yr 99. (a) (c) 2520 yr t n 1 103. 7.9, basic 105. 8.0 30e0.15t 101. (a) t 2 2 3, decreasing on 1.10, q 2 87. log4
258 0, 1.10 3 4 (b) $16,178.18 (c) $16,197.64 93. 4.341% (c) 19 yr 150e0.0004359t (b) 55 t n 1500e0.1515t (b) 7940 1 2 97. (a) 9.97 mg (b) 97.0 mg CHAPTER 5 TEST ■ page 428 1. (a),, y 4 4, q 1 y 2 (b) 3 _1 5 x (1, 2 13., 1, q 1, y 1 2 y 15. 0. (a) (d) 1 n t 2 1 y 2. (a) log6 25 2x 2 (e) (f) 2 (d) 3 3 x23 x4 x2 1 2 b ln 5. a 2 (b) e3 A 3. (a) 36 (b) 3 (c) 4 log x log 4. x 2 log 3 2 x2. (a) 4.32 (b) 0.77 (c) 5.39 (d) 2 1000e2.07944t (b) 22,627 (c) 1.3 h 21. 210 1024 29. 7 31. 45 10,000 1 2 1 2 A 1 2 37. 19. 39. 2 2, q 35. 3 q, 1 2B q, 2 17. 23. 10 y x 25. log2 64 6 27. log 74 x 2 43. 33. 6 3 45. log A 2 log B 3 log C 47. x2 1 2 log5 x 3 2 log51 x y 2 1 x2 y2 2 b 61. 1.15 41. 92 1 ln 2 3 1 x3 x 2 log51 x2 4 2x2 4 b 65. 15 log2 a 59. 2.60 a 63. 4, 2 1 5x 1 log 49. 55. 53. 3/2 2 2 1 2 2 ln x2 1 1 51. log 96 57. 5 67. 3 0 1 2 x 2 4 8. (a) A t 1 2 12,000 a 12t 1 0.056 12 b 3e0.069t t (b) $14,195.06 (b) 0.048 g (c) 9.249 yr (c) after 3.6 min 9. (a) A 1 2
10. 1995 times more intense A34 Answers to Selected Exercises and Chapter Tests CUMULATIVE REVIEW TEST FOR CHAPTERS 3, 4, AND 5 ■ page 429 1 q, q 1. (a) 2 (d) x 2 4, 1x 6 (f) f g x 4 g 10 4, q (b) 3, 4 h2 1x 4 x (c) 12, 0, 0, 2, (e, x 0 12 f f 1 8, 2 0 1 2 13, undeﬁned 12 g 1 2 2 0, 1 1 2 2 2. (a) 4, 4, 4, 0, 1 (g) g1 1 (b) 2 y 1 0 1 x 5 3. (a) f (c 10 0 1 x 2 13 (b) Maximum 13 q, 2 (d) Increasing on 1 2, q decreasing on 3 (e) Shift upward 5 units (f) Shift to the left 3 units 2 4 4. f, D; g, C; r, A; s, F; h, B; k, E 5. (a) 1, 2, 4, 8, (b) 2, 4, (cdb) y ab t, where a 1.180609 1015, b 1.0204139, and y is the population in millions in the year t (d) 207.8 million 3. (a) Yes (b) Yes, the scatter plot appears linear. (c) 515.9 million (e) No 7.5 −1 4 32 (c) ln E 4.551436 0.092383t, where t is years since 1970 and E is expenditure in billions of dollars where a 0.0923827621 5. (a) I0 (b) 14 22.7586444, k 0.1062398 (d) E 94.76838139e at, (e) 3478.5 billion dollars (c) 47.3 ft ; 0 45 7. (a) y ab t, where a 301.813054, b 0.819745, and t is the number of years since 1970 where a 0.002430, b 0.135159, c 2.014322, d 4.055294, e 199.092227, and t is the number of years since (c) From the graphs we see that the fourth
-degree 1970 polynomial is a better model. (b) y at 4 bt 3 ct 2 dt e, x 300 y 5 0 _5 1 2 1 2 Q 1 x 6. (a) 1 (b) x Q ; 1, 1 i, 1 i multiplicity x2 2x 2 multiplicity c) 7. x-intercepts 0, 2; y-intercept 0; horizontal asymptote y 3; vertical asymptotes x 2 and x 1 y 2 2 8 2x 3 1 2 (b) 9. (a) 4 10. (a) 4 (b) After 6.23 years 12. (a) t P 1 2 5 log x 1 log x 1 2 log 11. (a) $29,396.15 1 2 (b) ln 2, ln 4 120e0.0565t (c) 12.837 years (b) 917 (c) After 49.8 months FOCUS ON MODELING ■ page 437 1. (a) 290 1780 0 2020 0 25 (d) 202.8, 27.8; 184.0, 43.5 9. (a) 1.2 0 (b).5 0 17.5 18 0 3 −3 −3 (c) Exponential function (d) y ab x where a 0.057697 and b 1.200236 11. (a) y c 1 aebx, where a 49.10976596, b 0.4981144989, and c 500.855793 (b) 10.58 days CHAPTER 6 SECTION 6.1 ■ page 448 1. system 3. substitution, elimination, graphical 3, 9 7. 2, 2 11.,, 2, 2 5. 3, 2 1 2 13. 19. 23. 27. 33. 39. 41. 47, 1 1 3, 4 4, 16 1 2 25, 5 2 15. 1 2, 1 1 21. 2 1 1, 12 29. 1 6, 2 9. 3, 1 2 25, 5 3, 4 1 1, 12, A 2 1 2 5 2, 7 0, 0 2, 4, 4B A 2 2, 2 35., 2 2 1 15, 2 15, 2 15, 2, 1 1 3, 1 3, 1 5, 1 1 43. 2B 2B 3B 8.00,
0 49. 2.00, 20.00, 1 0.35, 4.21, 1 2 2, 27, 2B A 1, 1, 1 2 2, 6, 1.23, 3.87, 2 2 2, 1 17. 2 2, 27 2B 1 2, 4 31., 1 37. No solution 2 25. 2 2 2, 1 2 2, 3 2 4, 0 1 2 2 0.33, 5.33 15, 2 1 45. 4.51, 2.17 1 53. 1, 2 1 2.30, 0.70 15, 20, 2 2 4.91, 0.97 2 1 51. 0.48, 1.19 1 59. 2 400.50, 200.25 1 2 1 55. 12 cm by 15 cm 57. 1 2, 447.77 m 61. 2 12, 8 1 2 SECTION 6.2 ■ page 455 1. no, inﬁnitely many 2, 2 3. 1 2 y 5. No solution y 5 0 _5 5 x x-y=4 1 0 1 (2, _2) x 2x+y=2 _5 7. Inﬁnitely many solutions y 2 0 2 x Answers to Section 6.5 A35 4, 0, 3 9. 1 2 11. 5, 2, 1 2B A 13. x 2y z 4 y 4z 4 2x y z 0 15. 2x y 3z 2 x 2y z 4 3y 7z 14 2, 1, 3 c 17. 1 19. 1, 2, 1 1 2 2 23. 2 21. 5, 0, 1 1 c 29. No solution 2 2t, 2 3 33. 1 31. 4 1 3 t, t 0, 1, 2 25. 3 t, 3 2t, t 1 2 1 3t, 2t, t 2 1, 1, 1, 2 35. 27. No solution 2 2 B A 1 37. $30,000 in short-term bonds, $30,000 in intermediate-term bonds, $40,000 in long-term bonds wheat, 450 acres soybeans Mango, 60 Tropical Torrent, 30 Pineapple Power of A, 1200 shares of B, 1000 shares of C 41. Impossible 43. 50 Midnight 39. 250 acres corn, 500 acres 45. 1500 shares 5. (iii) 3. SECTION 6.4 ■
page 473 A x 1 Bx C x 2 4 B x 2 7. 9. Ax B x 2 1 A x 3 A 11. x Ex F B 2x 5 1 C 2 x 5 2 2 Gx H x 2 2 x 5 1 2 Cx 2x 15. 21 17. 23. 3 1 2 x 2 1 2x 3 4 x 2 x 2 3 2x 3 1 4 x 1 2 2 x 1 27. 2 x 1 1 x 31 4x 3 2 x 3 1 2 2x 1 1 x 2 3 x 2 2x 5 x2 x2 1 x 1 x 2 1 41. 45. 2 2 1 37 13. 19. 25. 29. 33. 35. 39. 43. 1 1 A 11. 9. 2, 2 2 10, 9 19. 2 x, 3 3 27. 2 x B 35. No solution 1 41., a a 1 3, 1 1 21. 2 2, 1 1 29. 37. 1 a 1 b 1 1 13. 2, 1 15. 1 2 1 23. No solution 31. 2 3, 7 3.87, 2.74 2 A 39. 2 1 a b, 43. a A 3, 5 17. 2 x, 1 25. x, 5 5 33. 6 x B 61.00, 20.00 1 a b b 1 1, 3 2 1 3 x 5 3B 5, 10 1 2 45. 22, 12 47. 5 dimes, 9 quarters 49. 200 gallons of regular gas, 80 gallons of premium gas 51. Plane’s speed 120 mi/h, wind speed 30 mi/h 53. 200 g of A, 40 g of B 55. 25%, 10% 57. $14,000 at 5%, $6,000 at 8% 59. John h, Mary h 61. 25 21 2 21 4 SECTION 6.3 ■ page 464 1. x 3z 1 3. Linear 5. Nonlinear 7. 1, 3, 2 1 2 2 SECTION 6.5 ■ page 479 1. equation; y x 1; test 3. y 1 0 y=x+ A36 Answers to Selected Exercises and Chapter Tests 5. 9. 13. 29. 31 = 2x + 2 11. y 1 x 2x − y = 8 4x + 5y = 20 1 0 1 x x bounded 33. x2 + y2 = 4 (0, 9) y=
9-x2 y 1 0 1 (3, 0) x y = 9 − x2 (−3, 0) bounded 35. y 1 ( 2, 2) x − y = 0 0 1 x (−2, 4) y y = x + 3 (2, 5) 1 0 1 x y 5 2x2 + y = 12 x2 − y = 0 (2, 4) 0 1 x 15. y = x2 + x2 + y2 = 25 (4, 3) 3 x y = 2x − 5 y 1 0 3x + 2y = 9 ( 5 3 ), 2 3x + 5y = 15 1 x y = 1 4 x + 2 not bounded 27. 17. 21. y 1 2x 1 19. x 2 y 2 4 23. y 3 y = x (2, 2) 0 3 x x + y = 4 not bounded 25. 1 y=_ x+5 2 y (0, 5) (2, 4) 1 0 1 (4, 0) x y=_2x+8 bounded bounded (− 2, − 2) bounded 37. y 3x − y = 0 x + 2y = 14 2 2 (−1, −3) x − y = 2 (6, 4) x not bounded 41. x + 2y = 12 1, 13 2 ) x + 1 = 0 10 3 13 3, ) 0 3 x y 2 0 (0, 3) x2 + y2 = 9 2 x x + y = 0 bounded 45 ) bounded bounded 39 bounded 43. x = 5 (5, 2) x y 1 0 x2 + y2 = 8 x = 2 (2, 2) 1 x (2 2, 0) bounded 47. (−1, 8) −5 10 −4 (11, 8) 13 49. −4 10 −6 (0.6, 3.4) (6.4, −2.4) 10 51. x number of ﬁction books y number of nonﬁction y Answers to Chapter 6 Test A37 19. 1, 1, 2 1 2 21. No solution 17. 23. 2 12.07, 1.44 11.94, 1.39, 1 1 x 4t 1, y t 1, z t x 6 5t, y 1, z t 21 7 3t 2 25. 2 Kieran is 13 years old 29. 12 nickels,
30 dimes, 8 quarters 4 x 2 x 1 4 1 x 1 33. 31. 2 27. Siobhan is 9 years old 35. 41. 37. 3 x2 2 x x2 2 1 43. books x y 100 20 y, x y x 0, y 0 c 53. x number of standard packages y number of deluxe packages 3 1 8y 80 4x 5 4x 3 8y 90 x 0, y 0 50 (20, 20) (50, 50) (80, 20) 0 50 x y 50 0 (0, 128) (70, 100) 50 (120, 0) x y 1 0 3x + y = 6 1 x 45. 47. y 1 c 0 1 x 1 2 39 x2 + y2 = 9 1 x y 1 0 1 x CHAPTER 6 REVIEW ■ page 483 2, 7 4B 2, 2 2, 1 1 1. 3. A, 2 1 1 2 5. 2, 1 1 2 y 1 0 7. x any number y 7 x 4 2 49. 51. 1 x _5 y 5 0 _5 ( − x2 + y2 =, 16 3 ) y = x + 4 x + 2y = 12 x 9. No solution y 1 0 1 x 11. 3, 3 1 2, 8, 2 1 2 13. 16 7, 14 3 B A 15. 1 21.41, 15.93 2 bounded bounded 53 55. 2, 3 CHAPTER 6 TEST ■ page 485 2. (a) nonlinear 0.43, 0.29,, B A (b) 1 2, 1 2.12, 0.56 2, 1, 1 2 2 2 5. (a) 1 1 3, 1 1 2, 3. (b) 2 1 0.55, 0.78 1. (a) linear 4, 6 1 4. Wind 60 km/h, airplane 300 km/h (b) Neither x 1 7. (a) 7 1 8. (a) 10, 0, 1 donut $0.75 9t 2 2 2 1 1 2 (b) Inconsistent 6. (a) No solution, b) neither 9. Coffee $1.50, juice $1.75, (b) Dependent y y = 2x + 5 x2 + y = 5 45. 49. 9, 2, 0 1 2 41. No solution 43. x 5
t, y 3 5t, z t 1, 0, 0, 1 3s 2 3t 4t, 7 7 55. 2 VitaMax, 1 Vitron, 2 VitaPlus 4 53. B 57. 5-mile run, 2-mile swim, 30-mile cycle 59. Impossible 3t, z s, „ t x 1 3 0, 3, 0, 3 47. y 1 1 3s 1 51. 4t, 9 2 3 4t, t 7 A, 1 2 4 4 A38 Answers to Selected Exercises and Chapter Tests 10. y x − y = −2 11. x + 2y = 4 (2, 4) 2x + y = 8 (_2, 1) 1 0 1 x 1 0 1 x 12 13. 1 x x 2 x2 3 FOCUS ON MODELING ■ page 490 1. 198, 195 3. y maximum 161 minimum 135 2x + y = 10 3 0 3 2x + 4y = 28 x 5. 3 tables, 34 chairs 7. 30 grapefruit crates, 30 orange crates 9. 15 Pasadena to Santa Monica, 3 Pasadena to El Toro, 0 Long 11. 90 stanBeach to Santa Monica, 16 Long Beach to El Toro dard, 40 deluxe 13. $7500 in municipal bonds, $2500 in bank certiﬁcates, $2000 in high-risk bonds 0 utility 15. 4 games, 32 educational, CHAPTER 7 SECTION 7.1 ■ page 505 1. dependent, inconsistent (c) x 3 t, y 5 2t, z t 3. (a) x and y 5. 3 2 (b) dependent 7. 2 1 9. 1 3 11. (a) Yes (b) Yes 13. (a) Yes (b) No (c) 15. (a) No (b) No (c) 17. (a) Yes (b) Yes (c) e (c) x 3 y 5 x 2y 8z 0 y 3z 2 0 0 x 5z 0 c 0 0 y 5z 1 x 3y „ 0 z 2 21. 19. 1, 1, 2 10, 3, 2 27. 2 33. No solution x 1 37. 23. d 1, 0, 1 2 1 1 29. No solution 35. 1, 0, 1 31. 2t 5, t 2, t 39. 1 2s t 6, y s
, z t 1 2 25. 1, 5, 0 2 1 2 3t, 3 5t, t 2 2, 1, 3 1 2 SECTION 7.2 ■ page 515 1. dimension 3. (i), (ii) 5. No 7. 11. Impossible 13. 17. No solution 19. 5 7 B 0 1 2 10 7 5 25 20 R 10 10 10 25 S 35 0 (b) Impossible 23. (a) C 25. (a) Impossible (b) 8 6 4 17 29. (a) 31. (a) B 4 45 R 49 0 (b) Impossible 35. (b) 14 14 335 C 343 0 x 2, y 1 (b) 1 1 3 5 9. 3 6 12 3 0 3 15 21. (ab) Impossible B R 27. (a) 10 0 2 4 7 14 7 1 8 1 (b) B R 33. (a) S 13 7 C x 1, y 2 S 37 41. 39 43. Only ACB is deﬁned. B B R B C ACB 3 21 2 14 B R x1 x2 x3 x4 27 6 T D S 18 4 0 5 4 C S 4,690 45. (a) Monica, Long Beach, and Anaheim, respectively. 13,210 1,690 4 3 (b) Total revenue in Santa B R 105,000 47. (a) (in ounces) of tomato sauce produced, and the second entry is the total amount (in ounces) of tomato paste produced. (b) The ﬁrst entry is the total amount 58,000 4 3 49. (a) 2 (cb. 3 5 2 3 11. 5 13 5 2 4 4 15. B 1 1 2 2 2 3 17 13. No inverse B R 3. (a) 3 4 (b) Yes (c) Yes (d) 19. No inverse 5. (a) 3 4 (b) No (c) No (d) (de) The letter E SECTION 7.3 ■ page 528 1. (a) identity (b) A, A (c) inverse 7. 1 2 7 2 3 2 21 23 27. x 126, y 50 S 1 1 0 0 1 C 25. x 8, y 12 29. x 38, y 9, z 47 33. x 3, y 2, z 1 D 31. x 20, y 10, z 16 T 35
. x 3, y 2, z 2 37. x 8, y 1, z 0, „ 3 39. 7 10 2 3 3 5 1 1 x 2 x2 41. 1 2a 1 1 1 1 43. 45. R 1 2 1 x ex 1 D ex e2x for x 0 47. (a) (b) 1 oz A, 1 oz B, 2 oz C S (c) 2 oz A, 0 oz B, 1 oz C (d) No 49. (a) C S x y 2z 675 2x y z 600 x 2y z 625 c (b She earns $125 on a standard set, $150 on a deluxe set, and $200 on a leather-bound set. S S S 675 600 625 A1 x y z (c) C C C S C SECTION 7.4 ■ page 539 1. True 13. 20, 20 3. True 5. 6 15. 12, 12 7. 4 9. Does not exist 11. 1 8 17. 0, 0 19. 4, has an inverse 21. 5000, has an inverse 25. 4, has an inverse 23. 0, does not have an inverse 31. (a) 2 29. 120 27. 18 Answers to Chapter 7 Test A39 (c) Yes 33. 2, 5 35. 0.6, 0.4 (b) 2 39. 47. 1 A 4, 2, 1 2 4, 1 1 2, 1 4, 1 B 57. 63 2 61. (a) 0, 1, 1 1 1 2 2 1 41. 43. 1, 3, 2 1 49. abcde 51. 0, 1, 2 1100a 10b c 25 1225a 15b c 333 4 1600a 40b c 40 2 45. A 53. 1, 1 2 37. 189 4, 1 2 1 29, 108 29, 88 29B 55. 21 (b) y 0.05x 2 3x c CHAPTER 7 REVIEW ■ page 544 1. (a) 2 3 (b) Yes (c) No (d) e x 2y 5 x 2y 3 x 8z 0 y 5z 1 0 0 y 3z 4 x y 3z 7 c x 2y z 2 15. 1 A 7. 2 3t 4 0, 1, 2 4 9. No solution 3t 2 3, 5 13. 3, t 19. 17
. No solution 23. Impossible 4 4 2 18 0 2 25. 27. 10 B 1 3 1 1, 0, 1, 2 11. s 1, 2s t 1, s, t 2 21. Not equal c 2 1 1, t 1, t, 0 2 0 5 29. 4 7 10 2 1 9 2 d c 31. C 30 9 22 2 S 1 4 33. 1 2 15 4 1 2 11 2 3 2 1 37. 1 3 5 2 d 1 3 c 49. 24 51. 1 65, 154 53. A 2 ¥ 1 12, 1 12, 1 12B 55. (a) Matrix A describes the number of pounds of each vegetable sold on each day; matrix B lists the price per pound of each vegetable. (b) AB 68.5 41.0 ; $68.50 was the total made on Saturday, and $41.00 was the total made on Sunday. 61. 11 63. $2,500 in bank A, $40,000 in bank B, $17,500 in bank C 57. 1 5, 9 5B A 59. 87 26, 21 26, 3 2B A B R CHAPTER 7 TEST ■ page 546 1. Row-echelon form 2. Neither 4. Reduced row-echelon form 5. 1 2 5 t, 1 5 7. 3 5 A 5t, t B 9. Incompatible dimensions A B 2, 0 3. Reduced row-echelon form 2, 5 5 6. No solution 8. Incompatible dimensions 6 10 3 2 9 36 58 0 3 28 18 3 10. 11. C S C S B ; inverse does not exist R 39. 7 2 C c 0 2 8 d 41 43. 1, 9 4 1 d 2 c 4 T ; inverse exists for all x 0 0 1 45. 0, no inverse 47. 1 § 13. B is not square 14. B is not square 19. F 0, 3 2B A ; y 3 2; 6 21. F 5 12, 0 A B A40 Answers to Selected Exercises and Chapter Tests 12. 2 3 2 1 d 1 c 15. 3 16. (a) 4 3 3 2 x y 17. A 0 0 0, B 0 0 2, B1 B B R R 19. 1.2 lb almonds, 1.8 lb walnuts C 10 30 b) 70, 90 1 2 18. 1 5, 5
, 4 2 0 0 1 S FOCUS ON MODELING ■ page 549 3. (a) Shear to the right (b) T 1 1 1.5 1 0 (c) Shear to the left (d) We get back the original square. 1 4 1 0 0 1 0.75.75 5 0.75 7 4.5 7 R 4.5 8 0 8 0 0 5. (a) D 0 0 0.75 0 B 0.75 0 (b) T TD 0 0 (c) T B B 1 0 0.25 1 TD.25 5 2.25 5 2.75 7 7.75 7 8 8 2 8 0 0 23. _3 27. _2 29. 37. 45. R y 0 _1 _3 3 x 25. _3 1 _0.5 4 _4 3 1 x2 8y y2 4x y2 3x 31. 39. y2 32x x2 20y x y2 47. 2 4py, p 1 2 33. y2 8x x2 8y 43. 2 4 12 y x 41. 49. R 51. (a) x, 1, 4, and 8 B CHAPTER 8 SECTION 8.1 ■ page 559 1. focus, directrix 3. 1 p, 0 7. II 9. VI F 2, x p, F, x 3 5. III 3, 0 1 2 (b) The closer the directrix to the vertex, the steeper the parabola. _3 p=8 p=4 ; x 5 3 12; 5 y 2 _1 0 x _2 1 _1 1 x2 40y 35. y2 16x 0 3 Order of answers: focus; directrix; focal diameter 11. F A ; y 0, 9 4B y 9 4; 9 13. F 1, 0 1 2 ; x 1; 4 y 2 1 0 _2 2 x 0 _2 1 x 1 20; 1 5 17. F 1 32, 0 B A ; x 1 32; 1 8 y 1 15. F 0, 1 20B A ; y y 4 0 _1 1 x 0 x _10 _1 _5 53. (a) y 2 12x (b) 8 115 31 cm 55. x 2 600y p=1 _1 p=1 2 SECTION 8.2 ■ page 569 1. sum; foci 0, a 3.
0, a ; c 2a2 b2;, 1 1 5. II 7. I 2 2 0, 5, 1 2 1 0, 5 0, 3, 1 2, 1 2 0, 3 2 Order of answers: vertices; foci; eccentricity; major axis and minor axis 5, 0 ; F V 9. 4 5; 10, 6 V 11. 1 15/3; 6, 4 4, 0 0, 15 ; B y 3 0 _3 y 2 0 _2 2 x x 5 _2 4, 0 V 13. 1 13/2; 8, 4 2 ; F A 2 13, 0 ; B V 0, 13 15. B 1/ 12; 2 13, 16 A ; F y 2 0 _2 _4 x4 _2 y 2 0 _2 1, 0 17. V 1 13/2; 2, 1 2 ; F A 13/2, 0 ; B V 0, 12 19. B 13/2; 2 12, 12 A ; F y 1 0 _1 _1 y 2 1 x _ 1 0 1 _2 V 0, 1 21. 2 1 1/ 12; 2, 12 ; F A 0, 1/ 12 ; B y 1 0 _1 _1 x 1 0, 13/2 ; B A 49. (a) 6 Answers to Section 8.3 A41 (b) Common major axes and vertices; eccentricity increases as k increases. k=4 k=10 k=25 k=50 _12 12 _1 x2 0, 13/2 ; B A 51. 53. x2 2.2500 1016 2 x 1,455,642 y2 2.2491 1016 y2 1,451,610 1 1 55. 5 139/2 15.6 in. SECTION 8.3 ■ page 578 1. difference; foci 3. ; 2a2 b2; 0, a 0, a 1 5. III 7. II, 1 2 2 0, 4 0, 4, 1 2, 1 2 0, 5 0, 5, 1 2 2 1 Order of answers: vertices; foci; asymptotes 2 15, 0 V 9. 1 y 2x V 11. 1 y 2, 0 ; F ; B A 2 0, 126 ; B y 3 0 _3 _3 y 2 _2 5 x x 3 _5 1, 0 V 13. 1 y x ; F
A 2 12, 0 ; B 5 _5 23. 29. _6 33. 39. 45. 2 x 25 2 y 16 1 25. 2 x 4 2 y 8 1 27. 2 x 256 2 y 48 1 31. 7 6 _7 7 _3 y 3 _3 3 x 2 x 25 2 y 9 1 35. 2 x 2 y 4 1 37. _7 2 x 9 2 y 13 1 2 y 91 1 41. 2 x 25 2 y 5 1 43. 2 64x 225 2 64y 81 1 2 x 100 0, 2 1 _3 47. 1 _1 2 y 2 0 _2 15. V 1 0, 3 ; F 1 2 0, 134 ; y 3 5 x 2 y 5 _5 _5 x 5 C 5. Center 2, 1 ; 2 1 2 15, 1 F ; foci B 1, 1, V21 V11 vertices 5, 1 2 major axis 6, minor axis 4 A ; 2 7. Center F11 ; foci 2 2 0, 10 vertices 0, 0 major axis 10, minor axis 6 0, 5, F21, V21 2 C 1 0, 1 V11 ; 2 0, 9 ; 2 y 3 0 _1 9. Vertex focus 1 directrix 3, 1 V 1 ; 3, 1 2 y 3 F y 3 x5 ; 2 3 x _3 y 0 _5 1 11. Vertex V 2, 0 A 2, 1 1 F ; focus 16B A y 1 directrix 16 ; B y _1 0 1 x A42 Answers to Selected Exercises and Chapter Tests V 17. A y 212, 0 1 2 x B ; F A 110, 0 ; B 19. V A y 0, 1 2 x 1 2B ; F A 0, 15/2 ; B y 5 _5 _5 y 2 _2 3 x 5 x _3 21. 27. _8 31. 37. 2 x 4 y2 12 1 23. y2 16 2 x 16 1 25. 2 x 9 4y2 9 1 29. 8 _8 8 _8 8 _8 1 2 x 9 5y2 64 y2 16 5x 256 2 33. 2 y2 x 3 1 35. 2 x y2 25 1 1 39. 2 x 16 y2 16 1 41. 2 x 9 y2 16 1 43. (b) x 2 y 2 c 2/2 47. (b) 10
0 k=12 k=8 k=4 k=1 5 _5 49. x2 y2 2.3 1019 SECTION 8.4 ■ page 587 1. (a) right; left (b) upward; downward 3. Vertex (_5, 0) Focus (_3, 0) y 1 0 8 3 _3 0 _2 x _2 1, 3 13. Center C 1 4, 3 V11 vertices x 1 y 4 31 2 2, V21 2 3 ; foci 2, 3 6, 3, F21 F11 ; asymptotes 2 2 4, 3 ; 2 y As k increases, the asymptotes get steeper. 1 0 1 x ; foci F A 1, 15 ; B 15. Center vertices asymptotes V 1 1, 0 C 2 1 1, 1 ; 2 y 1 21 y x 1 2 2 x y 2 Vertex (5, 0) Vertex (_2, 1) Vertex (8, 1) Focus (3, 0) Focus (0, 1) Focus (6, 1) 1 x 1 0 1 x 17. 21. x2 y 1 y 4 1 41 2 x 2 2 1 19. 1 2 2 1 x 5 25 2 y2 16 1 Answers to Chapter 8 Review A43 ; 43. 2 x 150 1 2 18,062,500 y2 18,040,000 1 CHAPTER 8 REVIEW ■ page 590 1. V 0, 0 1. V 0, 0 ; F 1 2 1 0, 2 2 ; y 2 y 2 0 _2 1 x y 2 _2 2 0 _2 x 2 V 5. 1 x 2, 2 ; B 2, 3 V 7. 1 y 4 ; F 1 2 2, 2 ; 2 y 2 0 _2 _2 2 x y 2 0 _1 _2 x 9. ; V C 0, 0 2 1 e 4 5; axes 10, 6 1 0, 5 ; F 1 2 0, 4 ; 2 11. C 0, 0 2 1 2 13, 0 F A B ; V 4, 0 1 ; e ; 2 13 2 ; axes 8, 4 y 2 0 _2 _2 y 1 0 2 x _4 13. C 3, 0 ; V 2 1 3, 17 1 ; e 3, 4 17 4 ; 2 ; axes 8, 6 15. C 0, 2 ; V 1 15, 2 15 3 B
6 F A y 4 _4 1 x _3 y 1 0 4 x ; 2 ; axes 6, 4 3 x 23. Parabola; 4, 4 V 3 25. Hyperbola; 3 7, F2A ; F1A C 2, 2 2, 2 1, 2 B 2 1 15, 2 V ; asymptotes B y x 1 2 1 21, 5 ; B 8, 5 ; 2 1 C 27. Ellipse; 3 121, 5 F A 2, 5, V11 V11 major axis 10, minor axis 4 y 2 0 3 x _5 31. Degenerate conic (pair of lines), x 4 y 1 21 y 2 4 0 35. _2 4 x 4 3 _9 C 29. Hyperbola; 3, 0 1 3, 4 3 asymptotes 31 33. Point 1, 3 1 2 y 1 0 37. _2 (1, 3 ) 1 8 _12 x 39. (a) F 17 (b) F 17 (c) F 17 (c) The parabolas become narrower. 41. (a) _6 6 p= 1 2 p=1 p= 3 2 p=2 6 p=-2 3 p=2 p=-1 _6 1 p=2 A44 Answers to Selected Exercises and Chapter Tests ; V 0, 4 ; 2 1 ; asymptotes 2 0, 0 17. C 1 0, 5 F 2 y 4 3 x 1 19. F C 0, 0 2 1 2 16, 0 ; V 4, 0 ; asymptotes ; 1 2 A y B x 1 12 39. Ellipse; F A V11 3, 3 1/ 12 3, 4, V21 ; B 3 _3 0 3 x _3 3 x 0 _3 41. Has no graph 43. x2 4y 45. C 1 0, 0 4, 0 21. ; V11, 2 2 4 4 12, 0 ; F V21 asymptotes y x 4 8, 1 ; 3, 1 12 ; B 3, 1 2 15 B 3 x, 23. V F asymptotes y 1 y 1 3 x 2 ; y2 4 4 1 47. 51. 2 x 16 x 7 225 2 2 1 49. 2 1 x 1 3 2 1 x2 4 y 2 4 1 y2 25 2 2 1 2 1 2 y 2 100 2 1 53. (a
) 91,419,000 mi y 4 1 0 _4 y 2 1 x _3 0 x3 _2 (b) 94,581,000 mi 55. (a) _10 10 k=8 k=4 _10 k=1 10 k=2 25. y2 8x 27. y2 16 x2 9 1 29. 2 1 x 4 16 2 1 2 y 2 4 2 1 31. Parabola; F 0, 2 ; V 0, 1 1 2 1 y 2 33. Hyperbola; 0, 12 12 ; V F A 0, 12 1 2 B y 18 3 0 _3 _3 3 x _18 0 18 x _18 35. Ellipse; V F A 1, 4 2 15 A 1, 4 115 ; B B 37. Parabola; 255 ; _3 _60 3 x 64, 8 2 y 5 0 _5 x CHAPTER 8 TEST ■ page 592 0, 3 1. F 1, y 3 2 y 4, 0 2. V 1 ; F 2 y A 2 13, 0 ; 8, 4 B 2 0 _2 _4 4 x _4 2 0 _2 4 x 0, 3 3, 5 2 ; y 3 4 x _8 0 x 8 _8 4. y2 x 5. x2 16 1 2 y 3 9 2 1 6. x 2 2 2 1 y2 3 1 7 2B 4 1 8 _2 0 x y 3 0 _3 6 x 9 _4 _4 x 4 10. 12. y2 x2 16 9 x 3 2 9 2 1 1 11. x2 4x 8y 20 0 y2 25 1 13. 3 4 in. CUMULATIVE REVIEW TEST FOR CHAPTERS 6, 7, AND 8 ■ page 593 2, 2 (b) 1. (a) Nonlinear 2 (c) Circle, parabola y, 0, 0 2 (d), (e) 2, 2, 1 1 1 2 2 0 _2 2 x (b) x t 1, y t 2, z t 1 2. (a) 3, 0, 1 Yolanda 10, Zachary 6 4. (a) A B impossible; C D 2 3. Xavier 4 ; AB 9 2 4 1 2 5 0 ; CB impossible; Answers to Cumulative Review Test for Chapters 6, 7, and 8 A45 7. (a) F 0b) F
1 8. (a) Hyperbola; V 3, 0 1, F 1 2 110, 0 ; y 1 3x 2 y 1 2 x (b) Ellipse; V 3, 0 1, F 1 2 212, 0 2 y 2 0 1 x (c) Hyperbola; V 0, 1 1, F 1 2 0, 110 ; BD C ; det C1 B B 2 impossible; det R 1 C 2 2; det 1 D 2 0 9. (a) 5. (ac 10 S 15 B R (d) x 10, y 15 R R B B B R _2 0 2 x (b) (b) 6. 0, 3 15 0, 3 15 F11 F21 ellipse, ; 2 2 y 3 A46 Answers to Selected Exercises and Chapter Tests (b) 3 312, 4 3 312, 4 F11 F21 hyperbola, ; 2 2 y 4 0 4 x 57. 14 15 16 17 18 19 110 59. x 3 x 4... x 100 61. 100 a k1 k 63. 10 a k1 k2 65. 999 a k1 1 k 1 2 k 1 67. 100 a k0 xk 69. 21 2n1 /2n 2 17. 1, 2, 3, 5, 8 7. (a) 5 2, 3 2, 1 2, 1 2, 3 2 (b) 1 (c) an 1 0 _1 1 n 10. 2 1 x 5 16 2 y2 9 1 FOCUS ON MODELING ■ page 597 5. (c) discriminant x 2 mx ma a2 m2 4ma 4a2 2 1 0, m 2a 1 2 2, m 2a CHAPTER 9 1, 1 SECTION 9.1 ■ page 608 1. the natural numbers 3. 2, 3, 4, 5; 101 4, 1 7. 10,000 13. 3, 2, 0, 4, 12 19. (a) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43 (b) 9. 0, 2, 0, 2; 2 15. 1, 3, 7, 15, 31 9, 1 16; 45 1 1 5. 2, 1 4, 1 11. 1, 4, 27, 256; 100100 5; 1 3, 1 101 11 12, 6, 4, 3, 12 5,
2, 12 7, 3 2, 4 3, 6 5 11 2, 1 2, 2, 1 2, 2, 1 2, 2, 1 2, 2, 1 2 0 21. (a) (b) 14 0 23. (a) (b) 3 0 11 1 1 n 2 1 29. 2n 1 3, 4 9, 13 1 /n2 2 81, 121 27, 40 31. 243, 364 729 27. 3n 2 25. 2n 33. 1, 4, 9, 16, 25, 36 2 1 35. 1 1 3n 1 12, 1 13, 1, 1 15; Sn 27, 80 81; Sn 9, 26 3, 8 37. 43. 55. 45. 8 11 47. 31 6 11 12 13 14 15 49. 385 39. 41. 10 53. 22 1 1n 1 51. 46,438 71. (a) 2004.00, 2008.01, 2012.02, 2016.05, 2020.08, 2024.12 73. (a) 35,700, 36,414, 37,142, 37,885, 38,643 (b) $2149.16 (b) 42,665 Sn (b) $38,000 75. (b) 6898 Sn1 2000 77. (a) SECTION 9.2 ■ page 615 1. difference 5. (a) 5, 7, 9, 11, 13 (c) 3. True (b) 2 an 15 10 5 0 1 n 9. 11. an an 3 5 1 1 5 21 2 15. Not arithmetic 48 2 n 1 2 n 1 2 17. Arithmetic,, a10, a10 an 14, an 24, an 4, an 2 3 1 4 5 1 12 4 21. 11, 18, 25, 32, 39; 7; arithmetic 25. 4, 2, 8, 14, 20; 6; 3, a5 27. 29. 5, a5 31. 4, a5 33. 1.5, a5 35. s, a5 1 37. 2 47. 1090 61. $1250 1 25 1.5 31, an 2 4s, an 2 39. 100, 98, 96 49. 20,301 63. $403,500 51. 832.3 65. 20 41. 30th 2 13. Arithmetic, 3 3 2 11 7 n 1 2 4 6 an 299, a
100 499, a100 n 1 n 1 19. Arithmetic, 1.7 9, 1 5, 1 3, 1 1 n 1 7, 1 23. 1 1 2 ; not 11, a100 384, a100 2 s, a100 173.5 2 99s 43. 100 45. 460 57. Yes 59. 50 53. 46.75 67. 78 SECTION 9.3 ■ page 622 1. ratio 3. True 5. (a) 5, 10, 20, 40, 80 (c) (b) 2 an 80 60 40 20 0 1 n ft (b) 8 4 12 P k 1 So 2 Mathematical Induction follows from P 1 P n 1. Thus, by the Principle of k 2 holds for all n. 1 2 7. (a) (c) 5 2, an 5 4, 5 8, 5 16, 5 32 (b) 1 2 1 0 _1 1 n n1, a4 5 5 2 A 16 17. Not geometric 1 2B 1 2 21. 6, 18, 54, 162, 486; geometric, 4, 1 256, 1 ; geometric, 25. 0, ln 5, 2 ln 5, 3 ln 5, 4 ln 5; 64, 1 16, 1 1024 1 2 3n1 n1 0.3 2 2 1 n1 11. 375 3 5n1, a4 an 15. Geometric, 9. an 13. Geometric, 2 19. Geometric, 1.1 6 3n1 23. common ratio 3; an n1 1 1 1 an common ratio 4; 4 A 4B 162, an 27. 3, a5 not geometric 0.00243, an 0.3, a5 29. 0.3 1 144 A 1 1 1 12, a5 144, an 12B 31 311/3, an 32/3, a5 /3 1 2 41. 11th 43. 315 39. 31. 33. 37. 2n1 25 4 2 51. 3 2 53. 3 4 55. 1 648 67. (a) 57. 1000 117 160,000 Vn 4 25, 1024 625, 5 A 5B 64 59. 7 9 0.80 2 1 73. (a) 71. 75. 2801 n 8 17 9 77. 3 m 79. (a) 2 61. n1 1 33 63. 112 999 (b) 4th year 65. 10
, 20, 40 80 A 1 3B A 69. 19 ft, 18 (b) 81. 1 n 3 4B n3 35. s2/7, a5 s8/7, an s2 1 n1 /7 2 45. 441 47. 3280 49. 6141 1024 3. $13,180.79 11. $13,007.94 SECTION 9.4 ■ page 630 1. amount 9. $572.34 17. $733.76, $264,153.60 23. (a) $859.15 25. 18.16% 27. 11.68% 5. $360,262.21 13. $2,601.59 19. $583,770.65 7. $5,591.79 15. $307.24 21. $9020.60 (b) $309,294.00 (c) $1,841,519.29 SECTION 9.5 ■ page 637 1. natural; P(1) P 3. Let denote the statement n 2 1 1 P Step 1 1 Step 2 Suppose is true, since k 2 1 is true. Then P 1 2 2 1 2 2 4... 2k 2 k 1 2 4... 2n Induction hypothesis. Thus, by the Principle of k 2 holds for all n. 1 2 1 So 2 Mathematical Induction 2 1 follows from P k 1 P 1 1 P n 5. Let P n 1 2 denote the statement 5 8... n 1 3n 2 1 2 Step 1 P 1 1 2 is true, since 5 3n Step 2 Suppose P is true. Then k 1 2 5 8... 3k Answers to Section 9.5 A47 k 1 3k 7 2 2 1 3k 5 2 3k2 13k 10 Induction hypothesis P k 1 So 2 Mathematical Induction follows from P 1 P n 1. Thus, by the Principle of k 2 holds for all n is true. Then denote the statement P 7. Let is true, since Step 1 P 1 1 Step 2 Suppose Induction hypothesis 1 n P 9. Let 2 13 23... n3 denote the statement n 1 4 n2 1 2 2. Step 1 P 1 1 2 is true, since 13 12 # 1 1 1 4 2 2. Step 2 Suppose P k is true. Then 1 2 13 23... k3 Induction hypothesis So 2 Mathemat
ical Induction follows from P 1 P n 1. Thus, by the Principle of k 2 holds for all n. 1 2 denote the statement 2 2n 1 is true, since k 3 2n2 n 1 1 23 2 # 12 is true. Then 11. Let P 23 43... n 1 2 Step 1 P 1 Step 2 Suppose 1 2 2 23 43... 1 2k2 k 1 3 2 k 1 2k Induction hypothesis 1 k 1 1 3 2 2k2 8k So 2 Mathematical Induction 2 1 follows from P P 1. Thus, by the Principle of k 2 holds for all n. P n 1 2 1 2n. 4 2 1 x 1 2 k1 kx Induction hypothesis A48 Answers to Selected Exercises and Chapter Tests 1 2 1 n P 13. Let denote the statement 1 # 2 2 # 22... n # 2n 2 3 1 # 2 2 Step 1 P is true, since 1 3 Step 2 Suppose k is true. Then 2 1 # 2 2 # 22... k # 2k 2k2k1 2 1 k 1 2k 4 2 2k1 2 1 k 1 1 k 1 1 1 k2k1 2 So 2 Mathematical Induction 1 P follows from # 2k1 k 1 2 # 2k1 2 # 2k1 2. Thus, by the Principle of k 2 holds for all n. 15. Let P n denote the statement n 2 n is divisible by 2. 1 1 2 is true, since 12 1 is divisible by 2. P Step 1 1 Step 2 Suppose 2 P k is true. Now Induction hypothesis k2 2k 1 k 1 k2 k 2 k 1 2 1 2 1 But k 2 k is divisible by 2 (by the induction hypothesis), and k 1 is clearly divisible by 2, so 2 2 k 1 P k by 2. So Mathematical Induction k 1 1 1. Thus, by the Principle of follows from P 2 holds for all n is divisible 1 2 17. Let P n denote the statement n 2 n 41 is odd. 1 1 P Step 1 1 Step 2 Suppose P k 1 2 is true, since 12 1 41 is odd. is true. Now k 1 41 k2 k 41 2k 2 But k 2 k 41 is odd (by the induction hypothesis), and 2k is k P. clearly even, so their sum is odd. So 2 holds
for Thus, by the Principle of Mathematical Induction all n. follows from 19. Let P n denote the statement 8n 3n is divisible by 5. 1 1 2 is true, since 81 31 is divisible by 5. P Step 1 1 Step 2 Suppose 2 8 # 8k k P is true. Now 2 1 8k1 3k1 8 # 8k 3 # 3k 5 # 3k 8 5 which is divisible by 5 because 8k 3k is divisible by 5 (by the induction hypothesis) and 5 3k is clearly divisible by 5. So k 1 follows from P. Thus, by the Principle of Mathematical Induction # 3k 8 # holds for all n. 8k 3k 21. Let P n denote the statement n 2n. P Step 1 1 Step 2 Suppose 2 2 is true, since 1 21. is true. Then 1 1 2 k P 1 k 1 2k 1 2k 2k 2 # 2k 2k1 k 2 holds for all n follows from P denote the statement P n 2 is true, since k 1 x 2 is true. Then So 2 Mathematical Induction 1 23. Let P P Step 1 1 Step 2 Suppose Induction hypothesis Because 1 2k. Thus, by the Principle of 1 1 1 2 x kx2 So 2 Mathematical Induction follows from P 1 P n. Thus, by the Principle of 2 k 2 holds for all n. 1 2 1 denote the statement an 5 3n1. 25. Let P n 1 1 P Step 1 1 Step 2 Suppose 2 5 30 5. is true. Then Deﬁnition of ak+1 Induction hypothesis 1 2 P ak1 2 is true, since a1 k 3 # ak 3 # 5 # 3k1 5 # 3k follows from P P n P k 1 So 2 Mathematical Induction 1. Thus, by the Principle of k 2 holds for all n. denote the statement x y is a factor of x n y n. 1 2 1 27. Let P n 1 1 2 is true, since x y is a factor of x 1 y 1. P Step 1 1 Step 2 Suppose 2 P k 2 1 k1 y x is true. Now k1 x x k1 x x y k ky x x k1 ky is clearly divisible by x y, and But is divisi1 ble by x y (by the induction hypothesis), so their sum is
divisible by x y. So. Thus, by the Principle of Mathematical Induction P holds for all n. follows from denote the statement F3n is even. 1 1 1 2 is true, since F31 2, which is even. 29. Let P n P Step 1 1 Step 2 Suppose sequence 2 P is true. Now, by the deﬁnition of the Fibonacci k 1 2 F3 1 k1 2 F3k3 F3k1 F3k F3k2 F3k 2 # F3k1 F3k1 F3k1 But F3k is even (by the induction hypothesis), and 2 F3k1 is clearly. Thus, by the even, so 2 Principle of Mathematical Induction follows from P holds for all n. is even. So k 1 F3 k1 F2 1 31. Let F2 F2 2 1 P denote the statement # Fn1 1... F2 Fn. n F1 is true, since F 2 1 k P is true. Then 1... F2 F2 k # Fk1 F2 Fk12 Fk # Fk2 Step 1 P 1 Step 2 Suppose F2 2 Fk Fk11 Fk1 k 1 So 2 Mathematical Induction follows from P P n k1 k1 P 1 1 2 1 F2 (because F1 F2 1). Induction hypothesis Deﬁnition of the Fibonacci sequence. Thus, by the Principle of k 2 holds for all n. 1 1 c n 1 0 d Fn1 Fn c Fn Fn1 d. n 1 nx for x 1. 33. Let P n 1 2 denote the statement. Step 1 P 2 1 2 is true, since F3 F2 F2 F1 d. c Answers to Chapter 9 Review A49 is true. Then k 1 Step 2 Suppose P k1, 9 8, 27 16, 81 32, 243 64 13. (a) (b) an Fk1 Fk Fk1 Fk 1 1 Fk Fk1 d c Fk Fk1 Fk Fk1 Induction hypothesis Fk2 Fk1 Fk Fk1 d Deﬁnition of the Fibonacci sequence (d) Geometric, common ratio c c c (c) 633 64 n 3 2 15. Arithmetic, 7 P k 1 So 2 Mathematical Induction
follows from P 1 P n. Thus, by the Principle of k 2 holds for all n 2. 1 2 1 denote the statement Fn n. 5 (because F5 5). is true. Now 35. Let P n P Step 1 1 Step 2 Suppose Fk1 2 1 5 2 is true, since F5 k P 2 1 Fk1 Fk k Fk1 k 1 Deﬁnition of the Fibonacci sequence Induction hypothesis Because Fk1 1 P k 1 So 2 Mathematical Induction follows from P 1 1. Thus, by the Principle of k 2 holds for all n 5. P n 1 2 SECTION 9.6 ■ page 646 1. binomial 3. n! n k ; 4! 4 3 4! 2 5. x 6 6x 5y 15x 4y 2 20x 3y 3 15x 2y 4 6xy 5 y 6 3! k!! 1 2 1 9. x 5 5x 4 10x 3 10x 2 5x 1 7. 15. 5x x5/2 x4 4x2 6 4 x2 1 x4 11. x 10y 5 5x 8y 4 10x 6y 3 10x 4y 2 5x 2y 1 13. 8x 3 36x 2y 54xy 2 27y 3 5 10 x2 x7/2 19. 4950 1 x5 17. 15 25. x 4 8x 3y 24x 2y 2 32xy 3 16y 4 1 20 6 x3 x6 x5 31. 25a 26/3 a 25/3 15 15 x4 x2 29. x 20 40x 19y 760x 18y 2 35. 300a 2b 23 37. 100y 99 39. 13,440x 4y 6 43. 10 x1/2 21. 18 1 6 x 2a b x y 23. 32 27. 45. 4 3 47. 3x 2 3xh h 2 1 2 1 2 33. 48,620x 18 41. 495a 8b 8 1 3, 9 2, 4 4, 16 CHAPTER 9 REVIEW ■ page 649 5 ; 100 0, 1 1. 7. 1, 4, 9, 16, 25, 36, 49 11. (a) 7, 9, 11, 13, 15 (b) 4, 0, 1 3. 500 11 an (c) 55 32; 1 9. 1, 3, 5
, 11, 21, 43, 85 5. 1, 3, 15, 105; 654,729,075 15 10 5 0 1 n (d) Arithmetic, common difference 2 17. Arithmetic, t 1 19. Geometric, 1 t 21. Geometric, 23. 2i 25. 5 27. 81 4 4 27 32,000 n1 1 2 An 1.05 (b) $32,000, $33,600, $35,280, 29. (a) $37,044, $38,896.20, $40,841.01, $42,883.06, $45,027.21 39. 384 31. 12,288... 350 251 37. 126 33 32 24 23 41. 02 12 22... 92 35. (a) 9 6 12 3 22 43. (b) 45. 33 a k1 3k 47. 100 a k1 k2k2 49. Geometric; 4.68559 5050 15 53. Geometric, 9831 55. 5 7 59. 13 61. 65,534 63. $2390.27 51. Arithmetic, 3 13 57. 1 2A 65. Let denote the statement 3n 1 n 1 2 3 # 1 1 2 3n 2 1 is true, since 1. 1 1 2 2 2. Step 1 P 1 1 Step 2 Suppose is true. Then 3k 3k 1 2 2 3 3k 1 4 Induction hypothesis 1 3k2 k 6k 2 2 3k So 2 Mathematical Induction follows from P 1 P n 1. Thus, by the Principle of k 2 holds for all n. 1 2 67. Let 1 1 n P denote the statement 2 1... 1 1 1 1 nB 2B A 1 1 is true, since 1B k is true. Then 1B A A Step 1 P 1 Step 2 Suppose Induction hypothesis A50 Answers to Selected Exercises and Chapter Tests P k 1 So 2 Mathematical Induction follows from P 1 P n. Thus, by the Principle of k 2 holds for all n. 1 denote the statement an 2 3n 2. 1 2 2 31 2 4. is true. Then 69. Let P n P Step 1 1 Step 2 Suppose 2 1 1 ak1 2 is true, since a1 k P 2 1 4 3ak 2 # 3k 2 3 2 2 # 3k1 2 P follows from P
n 1 1 P k 1 So 2 Mathematical Induction 1 4 Induction hypothesis. Thus, by the Principle of k 2 holds for all n. 1 2 (c) 1.0001An1, A0 An 100, A0 1.0025An1 275,027.50, A2 275,110.02, A5 275,192.56 275,000 275,055.00, A3 275,137.53, A6 1.0001n FOCUS ON MODELING ■ page 654 1. (a) An A1 A4 A7 3. (a) An 200.25, A2 A1 100 (c) An 5. (b) A0 A4 (d) 3427.8 tons, 3428.6 tons (e) 1 100 401.50, A4 /0.0025 300.75, A3 1.0025n1 1 3 1 2400, A1 (c) An 2 3120, A2 3428.6 4 3336, A3 1 0.3n1 3420.2 3600 1 2 275,000, (b) A0 275,082.51, 275,165.04, 275,000 100, 2 (b) A0 502.51 (d) $6580.83 3400.8, 73. 32 A3 3A2B 3AB2 B3 71. 100 77. 1 6x 2 15x 4 20x 6 15x 8 6x 10 x 12 79. 1540a 3b 19 81. 17,010A 6B 4 75. 0 20 (c) 104 58 1 12,500 2 1 CHAPTER 9 TEST ■ page 651 1. 1, 6, 15, 28, 45, 66; 161 (b) 2 n 1 3 an 2. 2, 5, 13, 36, 104, 307 an 4. (a) (b) 1 4 1 2 3. (a) 3 12 1 4B A n1 (c) 3/48 5. (a) 1 5, 1 25 (b) 6. (a) 8 9, 78 (b) 60 8. (a) 1 1 52 2 1 1 (b) 1 12 2 50 321 1 9. (a) 2 58,025 59,049 1 422 2 12 2 1 (b) 1 22 1 1 32 1 2 1 42 2 1 1 2 523 1
1 2 624 10 10. Let P n 2 1 12 22... n2 denote the statement n n 1 1 is true, since 12 Step 1 P 1 1 Step 2 Suppose 2 P k 1 2 is true. Then 12 22... k2 k 1 2 2 1 2n 2k 1 2 k 1 2 2 1 Induction hypothesis 2k 1 6 2k 1 6 2k2 7k. (b) In the 35th year 9. (a) R1 R6 104, R2 127 124, R7 108, R3 (b) It approaches 200. 112, R4 116, R5 120, 250 0 50 CHAPTER 10 SECTION 10.1 ■ page 660 m n; 2 3 6 13. 120 11. 60 23. 158,184,000 33. 1050 1. 9. 120 21. No 31. 24,360 (d) 2401 41. 483,840 (e) 4802 43. 6 3. 12 5. (a) 64 15. 32 25. 1024 27. 8192 17. 216 (b) 24 7. 1024 19. 480 29. No 35. (a) 16,807 37. 936 39. (a) 1152 (b) 2520 (c) 2401 (b) 1152 5. 336 19. 330 31. 362,880 3. False 17. 56 29. 24 39. 277,200 SECTION 10.2 ■ page 669 1. False 15. 1260 27. 120 37. 15 47. 2300 57. 1,560,780 (b) 792 67. 104,781,600 (b) 8640 59. (a) 56 65. 69. 6600 75. 17,813,250 49. 2,598,960 (c) 6160 41. 2,522,520 51. 120 (b) 256 20 # 19 # C 77. 182 7. 7920 9. 100 11. 60 21. 100 23. 2730 33. 997,002,000 13. 60 25. 151,200 35. 24 43. 168 45. 20 53. 495 55. 2,035,800 61. 1024 63. (a) 15,504 18, 4 2 1 71. 161,280 1,162,800 73. (a) 20,160 79. 62,986 k 1 1 P So 2 Mathematical Induction follows from P 1 2 1 11. 32
x 5 80x 4y 2 80x 3y 4 40x 2y 6 10xy 8 y 10 k 2 holds for all n.. Thus, by the Principle of P n 12. 10 3 b 1 a 3x 13. (a) an 2 1 2 3 1 2 7 414,720x 3 0.85 1.24 n 2 2 1 (b) 3.09 lb (c) Geometric SECTION 10.3 ■ page 679 1. sample space; event 3. mutually exclusive; mutually exclusive 5. (a) 1, 2, 3, 4, 5, 6 F (b) (c) S S 5 (c) 5 1 6 15. (a) 7. (a) (b) 1 2 (c) 0 6 6 (b) E 1 4 (b) 5 2, 4, 6 3 4 (c) 6 (d) (c) 3 16 10 13 (b) 13. (a) 3 8 (c) 1 2 5 8 5 8 HH, HT, TH, TT 11. (a) 1 3 52, 5 (b) 3 13 1 13 5 17. (a) 17 0.000495 19. C 13, 5 1 2 /C 1 2 5 9. (a) 5, 6 1 6 6 (b) 7 8 21. C 1 23. (a) 12, 5 C /C 2 3, 2 52, 5 2 0.11 1 2 0.000305 C 1 C 8, 2 2 39, 5 C 52, 5 1 1 1 0.778 2 2 25. 1 (b) 5, 2 8, 2 C C 1 1 2 2 0.36 5 GGGG, GGGB, GGBG, GGBB, GBGG, GBGB, GBBG, 27. (a) GBBB, BGGG, BGGB, BGBG, BGBB, BBGG, BBGB, 9 BBBG, BBBB (c) 19 29. (d) (e) 11 16 1 8 1 3 (b) 16 8 7.15 108 2 8.18 1011 6 49, 6 1 1/486 31. 1/C 35. (a) 33. (a) (b) 1 1024 1/4818 (b) 15 128 5.47 1031 1 41. (a) Yes 4 (b) Not mutually 4/11! 1.00 107 37. (b)
No 43. (a) Mutually exclusive; 1 2 exclusive; 3 39. (a) 45. (a) Not mutually exclusive; (b) 3 4 11 26 1 (b) Mutually exclusive; 2 47. (a) 3 4 (b) 1 2 (c) 1 49. 21 38 51. 31 1001 53. (a) 3 8 (b) 1 2 (c) 11 16 (d) 13 16 55. (a) Yes (b) 1 36 57. (a) 1 16 (b) 1 32 59. 1 12 61. 1 1444 63. 1 363 2.14 105 65. (i) 67. 1 69. 600/P 40, 3 1 2 1 P 365, 8 3658 5/494 2 0.07434 71. 0.1 1 2 1 7. C 5, 1 0.7 2 2 1 1 SECTION 10.4 ■ page 687 1. two; success, failure 3. 5. 0.7 5, 0 0..3 1 2 0.3 0.7 2 0.7 1 5 2 0.3 2 2 1 5, 0 1 0.7 2 1 2 1/6 2 5/6 1 2 0.4 2 1 10, 5 2 1 10, 0 2 1 4 2 0.6 2 1 0.45 2 0.45 5, 5 5, 2 5 0.00243 4 0.02835 1 C 1 0 C 2 1 5, 4 1 5 C 0.3 4 0.20094 6 0.25082 2 5 0.55.55 2 8 0.90044 2 0 0.75 0.25 2 9. 11. 13. C 5, 4 2 1 1 C 5, 5 1 1 C 6, 2 2 1 10, 4 15. C 1 17. C 1 19. (a) (b) C 1 10, 2 21. (a) C 1 1 C 1 0.45 2 2 1 1 C 2 1 0.55 1 4, 0 2 (b) C 1 (c) 4, 2 C 1 4, 4 2 1 4, 4 C 2 1 0.75 1 23. (a) (c) C 1 25. (a) 2 1 0.52 1 10, 5 2 1 0.005 1 1 C 2 8, 0 27. 1 0.75 29. (a) 2 0.75 1 C 6, 3 2 1 1
1 C 6, 6 (c) (d) 31. (a) 1 1 1 0.25 2 1 4, 1 1 2 1 10, 8 (b) C 4, 3 1 33. (a) C 35. (a) C C 10, 10 1 1 1 0.75 4 2 1 0.75 1 2 0.25 4 2 2 1 1 2 C 0.25 1 0 0.94922 2 0.25 2 0 0.31641 2 1 1 2 2 10 1.4456 103 2 0.52 0.48 5 5 0.24413 (b) 3 1 2 2 1 0.96 2 3 1.25 107 0 0.04 2 2 1 6 0.17798 0.25 0 8 C (b) 3 0.13184 2 6 C 0.75 4 0.68359 1 C 0.75 1 2 1 0.25 2 1 0.75 3 1 1 2 2 2 1 4, 4 2 1 1 3 0.4116 2 2 C 0.6 1 0 0.0123 2 2 1 0.3 2 0.7 1 8 1 2 0.6 2 2 0.4 10 1 2 1 0.4 2 2 1 0.7 0.3 2 2 1 2 2 1 3 0.1323 0.7 5 0.3 2 0.7 1 4 2 0.3 2 2 1 5, 1 1 0.7 2 1 2 1 0 0.52822 1 0.52822 4 0.96922 0.3 2 2 1 5 0.23403 10 C 10, 1 1 0.45 0.55 1 2 1 2 2 1 9 4 0.99609 0.75 2 4, 3 3 0.25 2 1 1 2 (b) 10 6.4925 104 0.48 1 2 3 0.014925 1 0.995 1 1 0.04 2 0.96 7 1 2 1 6 2.4414 104 2 2 0.038147 8, 1 1 0.25 2 6, 5 0.25 0.75 1 2 1 2 2 1 5 0.46606 0.25 4 1 2 (b) 2 0.75 1 9 0 0.05078 0.7 1 4 0.7599 2 10, 9 0.4 2 2 1 (b) Yes 1 1 0.6 2 Answers to Chapter 10 Focus on Modeling A51 SECTION 10.5 ■
page 691 1. $10, $100 13. $0.0526 $2.10 per stock. 3. $1.50 15. $0.50 19. $0.93 5. $0.94 21. $1 7. $0.92 9. 0 11. $0.30 17. No, she should expect to lose CHAPTER 10 REVIEW ■ page 694 1. 624 3. (a) 10 (b) 20 5. 120 7. 45 9. 17,576 11. 120 13. 5 15. 14 17. (a) 240 (b) 3360 (c) 1680 19. 40,320 (b) 11,760 21. (a) 31,824 (f) 6,683,040 25. (a) S {HHH, HHT, HTH, HT T, THH, THT, T TH, T T T} (b) (c) 19,448 2 15 (d) 2808 4 (d) 5 29. (a) 23. (a) 27. (d) (b) (b) (c) (c) (c) 8 15 2 3 (e) 2808 3 52 1 48 1 78 1 2 3.69 106 (b) 1 8 C 1 2 4, 4 1 2 52, 4 1 26, 4 31. (a) C 2 # C C (c) 2 1 52, 4 37. (a) 105 (b) 55 (d) (c) 84 2 1 (c) (d) C 1 C 5, 2 2 1 1 C 1 5, 5 2 1 5, 3 2 1 0.3 2 2 0.3 0 2 0.7 1 1 2 7 8 0.3 2 0.11044 1 (c) 41. (a) 2 1 32 0.3 3 0.3087 2 3 C 0.7 1 5 0.83692 5, 4 0.7 1 2 2 2 (b) 30, 4 2. 72 P 30, 4 3. (a) 456,976,000 657,720 C 1 7. (a) 4! 24 CHAPTER 10 TEST ■ page 696 1. 81 4. (a) 2 1 6. 4 214 65,536 30 # 29 # 28 # C 10. (a) 13. (c) 12. 2 1 5 (b) # 10
1 1 # 11 13 10, 6 0.55 1,966,582,800 9 6 (c) 13 13 0.427 4 0.23837 14. (a) 27, 5 1 26 1 6 # 9 11. 8. C 12 12 12 2 6 1 10, 0 2 1 0.55 (b) C C 1 10, 2 1 2 1 2 1 0.55 2 2 1 0 1 2 0.45 2 0.45 2 2 1 10 C 0.45 1 8 0.02739 2 1 624 C C 13, 4 52, 4 2 2 0.00264 1 1 33. 1 24 35. (a) 3 (b) 0.51 39. (a) 144 (d) 75 5 0.00243 (b) 0.7 1 (b) 126 5 0.16807 2 4 0.3 1 0.7 2 1 2 2 1 43. $0.83 (b) 258,336,000 27,405 (b) 6!/3! 120 5. 12 9. (a) 1 2 /C 1 (b) 15, 3 2 1 13 0.022 C 5, 3 1 2 10, 1 0.55 2 15. $0.65 2 1 1 9 0.45 1 2 1. (a) CUMULATIVE REVIEW TEST FOR CHAPTERS 9 AND 10 ■ page 697 15, 20 340, 801 99 (b) 6 # 0.01 0.64, 1 (c) 5115/512 2, 115 5 (c) (d) 6 B 19 # 0.01 5242.88 2 (d) 9 3. $2658.15 1 (b) 88,572 2 2 (e) 12 7984 A 41 37 2 7 2 6, 12 19 5 6 B A 2. (a) 41.4 1 n2 2n 1 2 5 8x 1 32 2 n 1 2 495 16 x 1 (b) 4 1 n 1 4. Hint: Induction step is 2 an an1 2 5 40x4 20x3 5x 32x 5. (a) 263 # 104 175,760,000 6. (a) # P P (b) 10, 4 26, 3 # C 1 C (c) 26, 3 10, 4 1 8. 0.56 dollar 2 (b) 3 2 insect has at least one spot; 3 0
.16 1 3 2 10 2 2 2 2 1 1 1 1 2 1023 3072 78,624,000 546,000 (b) 9. (a) Getting 3 heads and 2 tails 7. (a) 1 36 1 8 (c) 5 108 10. (a) The event that a randomly selected (b) 2049 3072 FOCUS ON MODELING ■ page 700 1. (b) 9 10 3. (b) 7 8 7. (b) 1 2 This page intentionally left blank INDEX Abel, Niels Henrik, 332 Absolute value, 16–17, 121–22 properties of, 17 Absolute value equations, 122 Absolute value function, graph of, 218, 221 Absolute value inequalities, 122–24 properties of, 123 Addition common errors, 51–52 commutative property of, 9 of complex numbers, 99 of fractions, 11 graphical, 256 of inequalities, 113 of matrices, 508–10 of polynomials, 33–34 of rational expressions, 48 Additive identity, 10 Adleman, Leonard, 351 Algebraic expression, 33 domain, 45–46 factoring, 39–45 recognizing form of, 40–41 Algebraic representation of function, 209–10 Amortization schedule, 632 Amount (Af) of annuity, 626–27 Anagram, 694 Analogy, 59 Ancillary circle of ellipse, 570 Annual percentage yield (APY), 408 Annuity amount of, 626–27 in perpetuity, 631 present value of, 628–29 Aphelion, 570 Apolune, 570 Applied maximum and minimum problems, 296–97 Approval voting, 674 Archimedes, 85, 557 Area problems about, 76–78 of ring, 63 Arithmetic expression, evaluating, 9 Arithmetic mean, 616 Arithmetic sequence, 610–16 deﬁnition of, 611 ﬁnding terms of, 611–12 partial sums of, 613–14 Arrow, Kenneth, 674 Arrow diagram for composition of function, 257 for function, 205–6 Asset division, modeling fair, 132–33 Associative properties, 9, 512 Astronomical unit, 432 Asymptote(s), 344–45, 573, 574 horizontal, 345, 347–54 of hyperbola, 575, 577–78 oblique, 353–54 of rational functions, 347–
49 slant, 353–54 vertical, 345, 347–54 Augmented matrix, 494, 495–96 Average rate of change, 236–39 Axis of ellipse, 564 of hyperbola, 572, 573 of parabola, 553 of symmetry, 553 Bacterial growth, 393, 414–15 Base of exponential function, 371 exponential notation, 20 of logarithm, 384, 397–98 Beer-Lambert Law, 406, 438 Best ﬁt, 467–68 Bhaskara, 63, 86 Binomial, 33, 638 Binomial coefﬁcient, 640–42 key property, 642 Binomial expansion general term of, 644 using Binomial Theorem, 643–45 using Pascal’s triangle, 639–40 Binomial experiment, 685 Binomial probability, 685–89 Binomial Theorem, 642–46 proof, 645–46 Bird ﬂight, energy expended in, 108–9 Birthday paradox, 678–79 Boone, Steven, 602 Bounded region, 478 Branches of hyperbola, 572 Burton, David, 700 CAD (computer-aided design), 306 Calculator, graphing. See Graphing calculator Calculator notation for scientiﬁc notation, 24 Canceling common factors, 11, 46–47 Carbon dating, 393 Cardano, Gerolamo, 332, 340 Car purchase equal ownership cost, 131–32 modeling, 130–31 Carrying capacity, 436 Cartesian coordinate system, 245 Catenary, 373 CAT scans, 554 Cayley, Arthur, 522 Center of ellipse, 564 Central box of hyperbola, 573, 574 Certain event, 674 Chang Ch’iu-Chien, 86 Change of base formula, 397–98 Chaos, 264 Chevalier, Auguste, 323 Chevalier de Méré, 672 Chui-chang suan-shu, 85 Circle as conic section, 552 constructing, 596 Closed interval, 15 C(n, r), 665–66 Codes, unbreakable, 351 Coefﬁcient, binomial, 640–42 Coefﬁcient matrix, 525 Coefﬁcient of polynomial, 300 constant, 300 leading, 303 Cofactor (Aij) of matrix element, 531
Coin toss, Monte Carlo model of, 699–700 Columns (matrix), 495 Column transformation of determinant, 534 Combination, 665–67 C(n, r), 665–66 used in problem solving, 667–69 Comet, path of, 576 Common difference of arithmetic sequence, 611 Common factors, 11, 39–40 Common logarithm, 388–89 Common ratio of geometric sequence, 617 Commutative properties, 9 Complement of event, 675–76 Complete Factorization Theorem, 335, 336, 337 Completing the square, 88–89 to graph quadratic function, 292, 293 Complex conjugate, 99, 102 I1 I2 Index Complex number arithmetic operations on, 98–100 dividing, 99–100 imaginary part, 98 pure imaginary, 98 real part, 98 Complex numbers, 97–103 deﬁnition, 98 Complex roots/solutions, 101–2 Complex zeros, 335 in conjugate pairs, 340–41 factoring polynomial with, 337–38 linear and quadratic factors, 341–42 Composite function. See Composition of functions Composition of functions, 256–59 Compound fraction, 49–50 Compound interest, 377–78, 407–8 continuously compounded, 378, 393 formula, 377 Computer-aided design (CAD), 306 Computer graphics, 514–15, 547–50 Computers, 246 Concavity of graph, 242–43 Concentrations, problems involving, 78–79 Confocal conics, 588–89 Confocal hyperbolas, 580 Conic section(s), 551–98. See also Ellipse; Hyperbola; Parabola in architecture, 595–98 confocal, 588–89 constructing, 595–96 degenerate conic, 585–86 general equation of shifted conic, 585–86 shifted, 581–89 Conjecture, 632–33 Conjugate, complex, 99, 102 Conjugate hyperbolas, 579 Conjugate radical, 50–51 Conjugate Zeros Theorem, 340–41 Constant, 66 growth, 263–64 Constant coefﬁcient (constant term), 300 Constant function, 215 Constant rate of change, 239–40 Constraints, 487 Contestant’s dilemma, 699, 702 Continuously compounded interest, 378, 39
3 Cooling, Newton’s Law of, 416–18 Cooper, Curtis, 602 Coordinate line, 13–14 Coordinate plane, 245 Cost function for long-distance phone calls, 218–19 Counterexample, 54 Counting, calculating probability by, 675–76 Counting principles, 658–62 Counting problems, guidelines for solving, 667 Cramer’s Rule, 535–37 Cube, factoring differences and sums of, 41 Cubic formula, 332 Cumulative voting, 674 D’Alembert, Jean Le Rond, 700 Data. See also Modeling demographic, represented by matrices, 513–14 ﬁtting exponential and power curves to, 431–40 linearizing, 433–34 Data matrix, 548 Decibel scale, 420 Decimals changing to scientiﬁc notation, 23–24 repeating, 8 Decision-making, modeling for, 130–35 Decomposition of partial fraction, 469–73 distinct linear factors, 469–70 distinct quadratic factors, 471–72 repeated irreducible quadratic factor, 472–73 repeated linear factors, 471 Decreasing function, 228–30 Degenerate conic, 585–86 Degree of polynomial, 33, 300, 301, 303 Demand function, 272 Denominator, 11 irreducible quadratic factors, 471–72 least common, 48 product of distinct linear factors, 469–70 product of repeated linear factors, 471 rationalizing, 29–30, 50–51 repeated irreducible quadratic factor, 472–73 Dependent system, 451, 460, 500–504 Dependent variable, 205 Depressed cubic, 332 Descartes, René, 245 Descartes’ Rule of Signs, 325, 327 Determinant of matrix, 522, 530 area of triangle, determining, 538 expanding, 532 row and column transformations, 534 of square matrix, 530–33 Diaconis, Persi, 659 Die, rolling, 686 Difference of cubes, factoring, 41 of functions, 255 of matrices, 508–9 of squares, factoring, 41 Difference quotient, 207, 238 Difference table, 625 Dimension of matrix, 495 Diophantus, 49, 86 Directrix of parabola, 552, 553, 554 Discounted value,
628–29 Discriminant, 90–91 Distance between two points on line, 16–17 Distance-rate-time problems, 80–81, 92–93, 453–54 Distinguishable permutation, 664–65 Distributive property, 9, 10, 33, 512 Dividend, 315, 316 Division of complex numbers, 99–100 of exponents, 21 of fractions, 11 of polynomials, 315–21 of rational expressions, 47–48 of real numbers, 11–12 Division Algorithm, 315 Divisor, 11, 315, 316 Domain of algebraic expression, 45–46 of function, 205, 208–9, 227–28 of logarithmic function, 391 of rational function, 344 of relation, 225 Doppler effect, 358 Drawing diagram in problem solving, 74 Drug dose, recursive sequence modeling, 653–54 Drug effectiveness, testing, 686–87 Earthquakes, magnitude and intensity of, 419–20 Ebbinghaus, Hermann, 396–97 Ebbinghaus’ Law of Forgetting, 397, 438 Eccentricity of ellipse, 567–68 of orbit of planet, 568 Ecology, mathematical, 527 Economics, mathematical, 628 Einstein, Albert, 62 Electrical resistance, 355 Elementary row operations, 496–97 Element of set, 14 Elevation-temperature model, 3–4 Elimination method, 444–46 Ellipse, 563–71 axis of, 564 with center at origin, 565 center of, 564 as conic section, 552 constructing, 596 eccentricity of, 567–68 foci of, 565, 566, 568 geometric deﬁnition, 563–64 major axis of, 565 minor axis of, 565 reﬂection property of, 568 shifted, 581–82 sketching, 565–66 vertices, 564, 565 Empty set, 14 End behavior of polynomial function, 302–4 of rational function, 353–55 Equality of complex numbers, 98 of matrices, 507 properties of, 66 Equation(s), 66–74 of conic, shifting graphs of, 581 of ellipse, 564–67 equivalent, 66 exponential, 400–403 false, 460 family of, 73–74 linear, systems of, 450–66, 535–37 logarithmic, 403
–6 matrix, 524–28 modeling, 74–85 of parabola, 553–58 power, 69–71 quadratic, 87–97 quadratic type, 105–7 root of, 66, 328 solution of, 66 system of, 441, 442–50 that deﬁne functions, 220–21 variables of, 453 Equivalent equations, 66 Equivalent inequalities, rules for generating, 112–13 Equivalent system, 459 Eratosthenes, 603 Error-correcting codes, 106 Establishing subgoals for problem solving, 60 e (the number), 375 Euclid, 69 Euler, Leonhard, 100, 375 Even function, 249–50 Event(s), 673 certain, 674 complement of, 675–76 ﬁnding probability of, 675 impossible, 674 independent, 678–79 intersection of, 678–79 mutually exclusive, 677–78 union of, 676–78 Exact ﬁt, 467–68 Existence theorem, 333 Expanding determinant, 532 by ﬁrst row, 532–33 Expectation, 690–92 Expected value, 690–92 Experiment, 672–73 binomial, 685 Exponent(s), 20, 385 fractional, 22, 28, 42–43, 70–71, 107 integer, 19–26 Laws of, 21, 22, 28, 29 rational, 28–29 Exponential equation, 400–403 steps in solving, 401 Exponential form, 384–85 Exponential function, 370–83 calculating compound interest and, 377–78 deﬁned, 371 graphs of, 371–75 linearizing exponential data, 434 modeling with, 369, 370, 383, 411–15, 431–32, 435 natural, 375–76 Exponential growth formula, 383 Exponential growth model, 412 Exponential notation, 19–20 Expression algebraic, 33 factoring completely, 42–43 form, recognizing, 40–41 fractional, 45 rational, 45 Extraneous solutions, 68, 105 Factor(s) canceling common, 46–47 common, 11, 39–40 linear, 341–42, 469–70, 471 of polynomial, 34, 322 quadratic, 341–42, 471–73 repeated, solving inequality with, 116–17 Factorial, 641 Fact
orial notation, 660 Factoring, 39–45 out common factors, 39–40 formulas for, 41 by grouping, 43, 104 polynomial equation, 104 polynomial with complex zeros, 337–38 quadratic equation, 87–88 to solve nonlinear inequality, 114–18 trinomials, 40–41 Factoring formulas, 41 Factor Theorem, 319, 321, 335 Fahrenheit and Celsius scales, relationship between, 119 conversion function, representations of, 210–11 Fair asset division, modeling, 132–33, 612 Falling object, average speed of, 237–38 False equation, 460 Family of equations, 73–74 Family of functions, 216–17, 372, 386 Family of parabolas, 557–58 Family of polynomials, 311 Feasible region, 478–79, 487, 488, 489 Fechner, Gustav, 389 Fermat, Pierre de, 159, 672, 701 Ferrari, 332 Fibonacci, Leonardo, 604 Fibonacci numbers, 607 Fibonacci sequence, 603–4 Finance annuity, amount of, 626–27 annuity, present value of, 628–29 annuity in perpetuity, 631 installment buying, 629–30 mathematics of, 626–32 First difference sequence, 625 First term of sequence, 600, 611, 617 Fitts’s Law, 394 Focal diameter of parabola, 556–57 Focus (foci) of ellipse, 563, 565, 566, 568 of hyperbola, 572, 573 of parabola, 552, 553, 554 FOIL, multiplying binomials using, 34 Food price index, model for, 231–32 Forgetting, Ebbinghaus’ Law of, 397, 438 Formula(s) distance, rate (speed), and time, 80, 92 factoring, 41 geometric, 76–78 path of projectile, 93 simple interest, 75 visualizing, 38 Fractal image compression, 621 Fractals, 621 Fraction(s), 11–12 adding and subtracting, 48 compound, 49–50 dividing, 47–48 multiplying, 47 partial, 469–74 properties of, 11 writing repeated decimal as, 622 Fractional exponent, 22, 28 equation involving, 70–71, 107 factoring expression with, 42–43 Fractional
expression, 45 equation involving, 104–5 Fuel tank, determining size of, 328–29 Function(s), 203, 204–36 algebra of, 255 arrow diagram, 205–6 average rate of change of, 236–39 combining, 254–62 composition of, 256–59 constant, 215 decreasing, 228–30 deﬁnition of, 205–6 dependent variable in, 205 Index I3 difference of, 255 domain of, 205, 208–9, 227–28 equations that deﬁne, 220–21 evaluating, 206–8 even, 249–50 examples of, 204 exponential, 370–83, 411–15, 431–32, 435 family of, 216–17, 372, 386 four ways to represent, 209–11 graphical addition of, 256 graphs of, 204–5, 214–24. See also Graph(s) greatest integer, 218 horizontal shifting, 244–46 increasing, 228–30 independent variable in, 205 inverse, 266–69 iterates of, 263–64 linear, 215 linear, constant rate of change of, 239–40 local maximum and minimum values of, 230–32 logarithmic, 384–94 machine diagram, 205 maximum value of, 293–95 minimum value of, 293–95 modeling with, 280–89 objective, 487, 488, 489 odd, 249–50 one-to-one, 265–66 piecewise-deﬁned, 207 polynomial. See Polynomial function(s) power, 216–17, 432–33, 435 product of, 255 quadratic, 292–300 quotient of, 255 range of, 205, 227–28 rational, 343–59 as relation, 225–26 sum of, 254–55 value of, 205, 227–28 vertical shifting, 243–44 Fundamental Counting Principle, 658–60, 667 Fundamental Theorem of Algebra, 335–36 Galilei, Galileo, 293, 562 Galois, Evariste, 323, 332 Game theory, 628 Garden, model for fencing, 282–83 Gas mileage maximum, 296 model for, 4–5 Gateway Arch, 373 Gauss, Carl Friedrich, 335, 338, 498, 613 Gaussian elimination, 459, 497–99 Gauss-Jordan elimination, 499–500 General equation of shifted conic, 585–86 Geometric sequence, 616–26 deﬁn
ition of, 617 ﬁnding terms of, 617–18 partial sums, 619 Geometric series inﬁnite, 621–22 sum of, 621 Global positioning system (GPS), 441, 450, 462 Golden ratio, 607 Googol, 394 Googolplex, 394 Graham, Ronald, 666 I4 Index Graph(s) concavity of, 242–43 of ellipse, 564–67 of exponential functions, 371–75 ﬁnding domain and range of function from, 227–28 ﬁnding local maxima and minima from, 231 ﬁnding values of function from, 227 of function, 204–5, 214–24 of function, by type, 221 horizontal shift of, 244–46 of hyperbola, 573–78 of increasing and decreasing functions, 228, 229 of inequalities, 14, 474–76 of interval, 15 of inverse of function, 269–70 of logarithmic function, 385–88 of monomial, 301 of parabola, 554–58 of polynomial, 301–2, 327–29 of polynomial, shape near zero of multiplicity m, 309–10 of polynomial, using zeros in, 304–8 of power function, 216–17 of quadratic function, using standard form, 292–93 of rational function, 354–55 reﬂecting, 246–47 shifted, 581 stretching and shrinking, horizontal, 248–49 stretching and shrinking, vertical, 247–48 transformations of functions and, 243–54 vertical shift of, 243–44, 246 Graphical addition of functions, 256 Graphical method, 446–47 Graphing calculator ellipse, 565–66 exponential functions, 371 family of polynomials, 311 graphing functions with, 216–17 hyperbola, 574–75 local extrema of polynomial, 311 polynomial of best ﬁt, 364–68 rational function, 349–53 reduced row-echelon command, 500 “row-echelon form” command, 499 shifted hyperbola, 584–85 solving system of equations, 447 solving system of linear inequalities, 477–78 table of values, 308 Gravitation, Newton’s Law of, 71, 235 Greater than, 14 Greatest integer function,
218, 221 Grouping, factoring by, 43, 104 Growth constant, 263–64 Half-life, 415 Harmonic mean, 615 Harmonic sequence, 615 Hilbert, David, 531 Horizontal asymptote, 345, 347–54 Horizontal Line Test, 265 Horizontal shift of graph, 244–46 Horizontal stretching and shrinking of graphs, 248–49 Hyperbola, 572–80 branches of, 572 with center at the origin, 573 confocal, 580 as conic section, 552 conjugate, 579 degenerate, 586 equation, 572, 573–78 foci of, 574, 575, 576–77 geometric deﬁnition of, 572–73 reﬂective property of, 577 shifted, 583–85 sketching, 574 vertices of, 572, 573, 574, 575, 576–78 Hyperbolic cosine and sine functions, 379 Identity function, 272 Identity matrix, 520–24 Image of x under f, 205 Imaginary number, 98 Imaginary part of complex number, 98 Impossible event, 674 Inconsistent system, 451, 460, 500–504 Increasing function, 228–30 Independent events, 678–79 Independent variable, 205 Index of summation, 606 Indirect reasoning, 60 Induction, mathematical, 633–38 principle of, 634 Induction hypothesis, 634 Induction step, 633 Inequality(ies), 112–21 equivalent, 112–13 graphing, 14, 474–76 involving absolute value, 122–24 linear, 113–14, 477 modeling with, 118–19 nonlinear, 114–18 proving, by mathematical induction, 636–37 reversing direction of, 113 rules for, 113 simultaneous, solving pair of, 114 solving, 112–18 system of, 476–81 Inﬁnite geometric series, 621–22 Inﬁnite series, 620–22 Inﬁnity (symbol), 15 Inner product, 510 Input in function, 205 Installment buying, 629–30 Instantaneous rates of change, 238 Integer, 7 Integer exponents, 19–26 Intensity levels of sounds, 420 Interest calculating rate of, from size of monthly payments, 630 compound, 377–78, 407–8 continuously compounded, 378, 393 model, constructing, 75–76
Linear inequality, 113–14, 477 systems of, 477–79 Linearizing data, 433–34 Linear programming, 486–92 guidelines for, 488 Linear system modeling with, 452–55, 463, 504 number of solutions, 460–62 in row-echelon form, solutions of, 501–4 in several variables, 457–66 in three variables, 504 in two variables, 450–57 in two variables, Cramer’s rule for, 535–37 Lithotripsy, 568 Loan amount, 629 Local extrema of polynomial, 310–11 Local maximum point, 310–11 Local maximum value, 230–32 Local minimum point, 310–11 Local minimum value, 230–32 Logarithm(s) base of, 388 change of base, 397–98 common, 388–89 exponential form, 384–85 expressions, expanding and combining, 395–97 laws of, 394–400 natural, 390–91 properties of, 385 Logarithmic equation, 403–6 steps in solving, 404 Logarithmic form, 384–85 Logarithmic function, 384–94 deﬁnition of, 384 graph of, 385–88 Logarithmic model, 440 Logarithmic scales, 418–20 Logistic function, 263, 436 Logistic growth model, 381, 414 modeling with, 436 Long division of polynomials, 315–17 LORAN system, 578 Lottery jackpot, dividing, 108 Lower bound for root, 325–27 Machine diagram for composition of function, 256 for function, 205 Main diagonal of matrix, 520 Major axis of ellipse, 564, 565 Majority voting, 674 Mandelbrot, Benoit, 621 Mathematical induction, 633–38 principle of, 634 proof by, 634–36 used in problem solving, 60 Mathematics in modern world, 23 automotive design, 306 CAT scans and MRIs, 554 computers, 246 error-correcting codes, 106 fair division of assets, 612 fair voting methods, 674 fractals, 621 global positioning system (GPS), 462 law enforcement and, 387 mathematical ecology, 527 mathematical economics, 628 splines, 302, 306 unbreakable codes, 351 weather prediction, 447 Matrix equation, 524–28 modeling with, 526–28 steps in solving, 525 Matrix (matrices),
493–550 addition of, 508–10 augmented, of linear system, 494, 495–96 coefﬁcient, 525 computer graphics, 514–15, 547–50 data, 548–49 deﬁnition, 495 determinant of, 522, 530–33, 534, 538 dimension of, 510 elementary row operations, 496–97 entries, 495 equality of, 507 Gaussian elimination, 497–99 Gauss-Jordan elimination, 499–500 identity, 520–24 inverse of, 520–24 main diagonal of, 523 moving images in plane, 548–49 moving points in plane, 547–48 multiplication of, 510–14 product of, 510–14 reduced row-echelon form, 497–500 scalar product of, 508–10 square root, 518 stochastic, 514 subtraction of, 508–9 transition, 519 zero, 529 Maximum and minimum problems, 296–97 Maximum proﬁt, manufacturing for, 486–87 Maximum value of function, local, 230–32 Maximum value of quadratic, 293–95 Mean arithmetic, 616 harmonic, 615 Mendel, Gregor, 672 Mersenne numbers, 602 Minimum surface area of can, modeling, 284–85 Minimum value of function, local, 230–32 Minimum value of quadratic, 293–95 Minor axis of ellipse, 564, 565 Minor of element (Mij), 531 Mixture problems, 78–79, 454–55 Model, 2 Modeling, 2–7 best ﬁt versus exact ﬁt, 467–68 for best (or optimal) decisions, 130–35 computer graphics, 547–50 conics in architecture, 595–98 deﬁned, 2 equations, 74–85 with exponential functions, 369, 370, 383, 411–15, 431–32, 435 with functions, 280–89 with inequalities, 118–19 linear programming, 486–92 with linear systems, 452–55, 463, 504 with logistic functions, 436 making algebra models, 4–5 with matrix equations, 526–28 Index I5 Monte Carlo method, 699–702 polynomial curves, ﬁtting to data, 364–68 with power functions, 432–33, 435 with
quadratic equations, 91–94 with quadratic functions, 296–97 real-life problems, 107–9 with recursive sequences, 652–56 using algebra models, 2–4 Monomial, 33, 301 graph of, 301 transformations of, 301–2 Monte Carlo method, 699–702 Mortgage payment amortization schedule, 632 calculating monthly, 629–30 Multiplication of algebraic expressions, 34 of complex numbers, 99 of exponents, 21 of fractions, 11 of inequalities, 113 of matrices, 510–14 of polynomials, 34 of powers, 20 properties of, 51 of rational expressions, 47 Multiplicative identity, 11 Multiplicity of zero, 309–10, 337–39 Mutually exclusive events, 677–78 Napier, John, 388 Nash, John, 628 Natural exponential function, 375–76 Natural logarithm, 390–91 properties of, 390 Natural number, 7 Negative exponent, 20 Negative number, square root of, 100–101 Negative of real number, 10–11 Newton, Sir Isaac, 644 Newton’s Law of Cooling, 416–18 Newton’s Law of Gravitation, 71, 235 n! (n factorial), 641 Noether, Emmy, 534 Nonlinear inequality, solving, 114–18 nth partial sum, 605, 613, 619 nth power, 20 nth root(s), 27 properties of, 27 solving equations with, 69–71 nth term of sequence, 600 of arithmetic sequence, 611 of geometric sequence, 617 Number(s) complex, 97–103 imaginary, 98 integer, 7 irrational, 7, 8 natural, 7 polygonal, 625 pure imaginary, 98 rational, 7 real, 7–13 rectangular array, 494. See also Matrix (matrices) Numerator, 11 rationalizing, 51 Numerical representation of function, 209–10 I6 Index Objective function, 487, 488, 489 Oblique asymptote, 353–54 Odd function, 249–50 One-to-one function, 265–66 ﬁnding inverse of, 268–69 Open interval, 15, 16 Order on real line, 14 Outcome of experiment, 672 Output of function, 205 Palindrome, 662, 695 Parabola, 292–93, 551. See also Quadratic function(s) applications for, 5
58–59 axis of symmetry, 553 as conic section, 552 constructing, 596 directrix of, 552, 554 equations and graphs of, 553–58 family of, 557–58 focal diameter of, 556–57 focal length of, 561 focus of, 552, 554 geometric deﬁnition, 552–53 with horizontal axis, 555–56 latus rectum of, 556 motion of falling ball modeled by, 562 prime focus of, 561 shifted, 582–83 vertex of, 292, 553, 554 with vertical axis, 554–55 Parabolic reﬂector, 558–59, 560, 561 Parameter, 74 of system of solutions, 462 Pareto, Vilfredo, 399 Pareto’s Principle, 399 Partial fraction decomposition, 469–73 distinct linear factors, 469–70 distinct quadratic factors, 471–72 repeated irreducible quadratic factor, 472–73 repeated linear factors, 471 Partial fractions, 469–74 long division to prepare for, 473 Partial sums of arithmetic sequences, 613–14 of geometric sequences, 619 of sequence, 605–6 Partitions, ﬁnding number of, 665 Pascal, Blaise, 636, 672, 701 Pascal’s triangle, 639–40, 642 Patterns ﬁnding and recognizing, 59 problem solving, 59 Pay, model for, 2–3 Perfect cube, 63 Perfect square, 42 Perihelion, 570 Perilune, 570 Periodic rent, 627 Period of planet, 432 Piecewise-deﬁned function, 207 graphing, 217–19 Piston tolerances, 123–24 Planet aphelion, 570 eccentricity of orbit, 567 perihelion, 570 Plotting points, graphing functions by, 215–16 Plurality voting, 674 P(n, r), 663 Points, moving in the plane, 547–48 Poll results, 684 Pollutant outputs, restricting, 478–79 Polya, George, 59 Polygonal number, 625 Polynomial(s), 33 addition of, 33–34 of best ﬁt, 364–68 coefﬁcient(s) of, 300 degree of, 33, 300, 301, 303
division of, 315–21 factors of, 34, 322 family of, 311 Intermediate Value Theorem, 305, 333 multiplication of, 34 subtraction of, 33–34 Polynomial equation, 103–5 Polynomial function(s), 291, 300–315 end behavior, 302–4 graph of, 301–2, 327–29 graph of, guidelines for, 305 graph of, shape near zero of multiplicity m, 309–10 local extrema of, 310–11 as models, 364–68 terms of, 300 zeros of, 304–8, 321–34 Population growth carrying capacity and, 436 comparing different rates of, 412–13 exponential functions for modeling, 369, 370, 383, 411–15 ﬁnding initial population, 413–14 predicting size of population, 412 recursive sequence modeling, 653 world population projections, 413 Positive square root, 27 Power, 20 Power equations, 69–71 Power function, 216–17, 221 comparing exponential and, 374–75 linearizing power data, 434 modeling with, 432–33, 435 Powers, sums of, 636 Preference voting, 674 Present value of annuity, 628–29 of sum, 382 Prime focus, 561 Prime number(s), 633 power model for planetary periods, 432–33 Permutation(s), 663–65 distinguishable, 664–65 ﬁnding number of, 664 P(n, r), 663 used in problem solving, 667–69 pH scale, 418–19 large, 602 Principal, 377 Principal nth root, 27 Principal square root of negative number, 100 Principle of Substitution, 35 Principles of problem solving, 59–60 Probability, 657, 672–83 binomial, 685–89 calculating by counting, 675–76 of complement of event, 676 deﬁnition of, 674 expected value, 690–92 of intersection of independent events, 678–79 of union of mutually exclusive events, 677–78 of union of two events, 676–77 Problem solving with combinations, 667–69 general principles, 59–60 with permutations, 667–69 with quadratic equations, 91–94 Product of functions, 255 inner, 510 of matrices, 510–14 of polynomials, 34 scalar, 508–10 sign of, 114 Product formulas, 35 Projectile, path of
, 93–94 Proof, 633 of Binomial Theorem, 645–46 by contradiction, 60 by mathematical induction, 634–36 Pure imaginary number, 98 Pyramid, volume of, 63 Pythagoras, 284 Pythagorean Theorem, 63, 284 Quadratic equation, 87–97 complex roots/solutions, 101–2 deﬁnition, 87 modeling with, 91–94 solving, by completing the square, 88–89 solving, by factoring, 87–88 Quadratic factors, 341–42 distinct, 471–72 repeated irreducible, 472–73 Quadratic formula, 89–90 using Rational Zeros Theorem and, 324–25 Quadratic function(s), 292–300 deﬁned, 292 graph of, 292–93 maximum value of, 293–95 minimum value of, 293–95 modeling with, 296–97 standard form of, 292–93 Quadratic polynomial, factoring, 87 Quadratic type equations, 105–7 Quotient, 11 difference, 207, 238 of functions, 255 of polynomials, 315, 316 raising to power, 21 sign of, 114 solving inequality involving, 117–18 Radical, 27–28, 29, 69 equations involving, 105 Radioactive decay, 415–16 Radioactive element, 415 Radioactive waste, 416 Radiocarbon dating, 402 Rainfall and crop yield, modeling, 364–65 Ramanujan, Srinivasa, 618 Range of function, 205, 227–28 of relation, 225 Rational exponent, 28–29 Rational expression(s), 45 adding, 48 dividing, 47–48 multiplying, 47 simplifying, 46–47 subtracting, 48 Rational function, 343–59 applications, 355 asymptotes of, 347–49 domain of, 344 graphing with graphing calculator, 349–53 steps in graphing, 349 transformations to graph, using, 346 Rationalizing denominator/numerator, 29–30, 50–51 Rational number, 7 Rational zero of polynomial, 322–25, 336, 339 Rational Zeros Theorem, 322–23, 324–25 Real line, 13–14 order on, 14 Real number, 7–13 operations on, 8–9 properties of, 9–12 types of, 7–8 Real part of complex number, 98 Reciprocal function, 221 Reciprocal of inequalities, taking, 113 Rec
ursive sequence, 603–4 as models, 652–56 Reduced row-echelon form, 497–500 Reﬂecting graphs, 246–47 Reﬂection property, 558 of ellipse, 568 of hyperbola, 577 of parabola, 558–59 Region bounded, 478 feasible, 478–79, 487, 488, 489 unbounded, 478 Regression line, 457 Relation, 225 functions and, 225–26 Relativity, Theory of, 213 Remainder, 315, 316 Remainder Theorem, 318 Repeating decimal, 8 as fraction, writing, 622 Rhind papyrus, 86, 470 Richter, Charles, 419 Richter scale, 419 Rivest, Ron, 351 Roberval, Gilles Personne de, 700, 701 Robinson, Julia, 508 Root of equation, 66 nth, 27 of polynomial equation, 328 square. See Square root(s) upper and lower bounds for, 325–27 Root function, 221 Row, 495 expanding the determinant by the ﬁrst, 532–33 transformations, 534 Row-echelon form, 497–99 RSA code, 351 Sample space of experiment, 672, 673 Sampling, 684 Scalar product, 508–10 Scientiﬁc notation, 23–24 Seating capacity of amphitheater, ﬁnding, 614 Second difference sequence, 625 Sequence, 600–619 arithmetic, 610–16 arithmetic mean of, 616 deﬁnition of, 600 difference, 625 geometric, 616–26 harmonic, 615 nth term of, ﬁnding, 602–3 of partial sums, 605–6 of polygonal numbers, ﬁnding patterns for, 625 recursive, 603–4, 652–56 terms of, 600, 601–3 Series, inﬁnite geometric, 621–22 Set(s), 14 element of, 14 empty, 14 intersection of, 14–15 union of, 14–15 Set-builder notation, 14 Shamir, Adi, 351 Shifted conics, 581–89 Shifting graph, 243–46, 248 Shipping problem, 488–89 Shrinking graph, 247–49 Sigma notation, 606–8 Sign of product or quotient, 114 Similar Tri
angles, 77–78 Simple interest, 75 Simplifying rational expressions, 46–47 Slant asymptote, 353–54 Solar power plants, 551 Solution(s) absolute value equations, 122 of equation, 66 extraneous, 68, 105 of linear system in two variables, number of, 451–53 of quadratic, 87 of system of equations, 442, 498, 499–500, 501 Solving the equation, 66 linear equations, 66–69 for one variable in terms of others, 71–72 power equations, 69–71 Sørensen, Søren Peter Lauritz, 418 Sound common logarithms, 389 intensity levels of, 420 inverse square law for, 423 Special product formulas, 35 Splines, 302, 306 Square matrix, 530–33 Square root(s), 27 of matrices, 518 of negative number, 100–101 principal, 100 elementary operations, 496–97 Standard form of quadratic function, 292–93 Index I7 Stefan Boltzmann Law, 235 Step function, 218, 224 Stochastic matrix, 514 Stretching graph, 247–49 Substitution method, 443–44 Subtraction, 10 of complex numbers, 99 of inequalities, 113 of matrices, 508–9 of polynomials, 33–34 of rational expressions, 48 Sum of cubes, factoring, 41 of functions, 254–55 of inﬁnite geometric series, 621–22 of matrices, 508–10 of powers, 636 properties of, 608 writing in sigma notation, 607 Summation variable, 606 Synthetic division, 317–18 System of equations, 441, 442–50 deﬁned, 442 dependent, 500–504 elimination method of solving, 444–46 graphical method of solving, 446–47 inconsistent, 500–504 leading variable of, 500 linear, Cramer’s rule for, 535–37 linear, in several variables, 457–66 linear, in two variables, 450–57 solutions of, 442 steps in solving, 501 substitution method of solving, 443–44 System of inequalities, 476–81 linear inequalities, 477–79 Tangent line, 596 Taussky-Todd, Olga, 514 Temperature average rate of change, 239 elevation-temperature model, 3–4 Fahrenheit and Celsius scales, relationship between, 119, 210–11 Term(s
) of polynomial, 300 of sequence, 600, 601–3 Test point, 475 Test value, 115 Third difference sequence, 625 Ticket price plans, carnival, 118 Ticket sales revenue, maximizing, 296–97 Time needed to do job, problems about, 80 Torricelli’s Law, 213, 271 Transformation of determinant of matrix, 534 of matrix, 547–48 of monomial, 301–2 Transformations of functions, effect on graphs, 243–54 even and odd functions, 249–50 horizontal shifting, 244–46 horizontal shrinking and stretching, 248–49 rational function, 346 reﬂecting graphs, 246–47 vertical shifting, 243–44, 246 vertical stretching and shrinking, 247–48 I8 Index Transition matrix, 519 Transverse axis of hyperbola, 572, 573 horizontal, 574–75 vertical, 575–76 Tree diagram, 658 Trial and error, factoring by, 40 Triangle(s) area of, 538 Pascal’s, 639–40, 642 similar, 77–78 Triangular form of system, 458 Trinomial, 33 factoring, 40–41 Turing, Alan, 160 Unbounded region, 478 Union of events, 676–78 probability formula for, 676 Union of mutually exclusive events, 677–78 probability formula for, 678 Union of sets, 14–15 Upper and Lower Bounds Theorem, 326–27 Upper bound for root, 325–27 Value of function, 205, 227–28 Variable, 32, 66 dependent, 205 domain of, 45–46 identifying, in word problems, 74 independent, 205 modeling with systems of equations and, 453 solving for one, in terms of others, 71–72 Variation in sign, 325 Verbal representation of function, 209–10 Vertex (vertices) of ellipse, 564, 565 of feasible region, 487 of hyperbola, 572, 573, 574, 575, 576–78 of parabola, 292, 553, 554 of system of inequalities, 477 Vertical asymptote, 345, 347–54 Vertical Line Test, 219 Vertical shift of graph, 243–44 combining horizontal shift and, 246 Vertical stretching and shrinking of graphs, 247–48 Viète, François, 89 Virus, exponential model for spread of, 376 Visualizing formula, 38 Visual representation of function, 209
–11 Volume of box, modeling, 280–81 of pyramid, 63 Von Neumann, John, 246 Voting methods, 674 Weber-Fechner Law, 420 Weight of astronaut, 208 Whispering gallery, 568 Word problems examples of, 75–81 guidelines for solving, 74 Working backward, problem solving by, 60 World population, 413 exponential model for, 431–32 x-intercept, 350, 351, 352, 353, 354 y-intercept, 350, 351, 352, 353, 354 Zero complex. See Complex zeros double, 338 Intermediate Value Theorem, 305, 333 irreducible, 341 multiplicity of, 309–10, 337–39 numerical method for ﬁnding, 333–34 of polynomial, 304–8, 321–34 rational, 322–25, 336, 339 speciﬁed, ﬁnding polynomials with, 319, 339, 341 Zero exponent, 20 Zero matrix, 529 Zero Product Property, 87 Zeros Theorem, 337 PHOTO CREDITS This page constitutes an extension of the copyright page. We have made every effort to trace the ownership of all copyrighted material and to secure permission from copyright holders. In the event of any question arising as to the use of any material, we will be pleased to make the necessary corrections in future printings. Thanks are due to the following authors, publishers, and agents for permission to use the material indicated. CHAPTER P 1 © Daimler AG 59 Stanford University News Service 62 © Bettmann/CORBIS CHAPTER 1 65 © Creatas/Photolibrary 69 © Mary Evans Picture Library/ Alamy 86 The British Museum 89 Library of Congress 100 Library of Congress 106 NASA CHAPTER 2 137 © Pal Hermansen/Getty Images 159 Bettmann/CORBIS 160 National Portrait Gallery 194 © AP/Wide World Photos 196 © Eric & David Hosking/ CORBIS CHAPTER 3 CHAPTER 6 203 Courtesy Blue Skies Skydiving Adventures, Inc., www.blueskiesskydiving.com 441 © imagebroker/Alamy 447 © USA Today, reprinted with permission 220 Stanford University News 462 NASA Service 245 Library of Congress 249 The Granger Collection, New York CHAPTER 4 291 © Tom Mackie/Alamy 293 The Granger Collection, New York 306 Courtesy of Ford Motor Company 323 Library of Congress 338 © CORBIS 340 Copyright © North Wind/North Wind Picture Archives. All rights reserved
. 365 © Ted Wood/The Image Bank/Getty Images CHAPTER 5 369 © David Iliff 373 © Garry McMichael/Photo Researchers, Inc. 387 © Bettmann/CORBIS 387 © Hulton-Deutsch Collection/ CORBIS 388 Library of Congress 416 © Joel W. Rogers/CORBIS 420 © Roger Ressmeyer/CORBIS 432 © Chabruken/The Image Bank/ Getty Images CHAPTER 7 493 © 2006 Mirage Studios, Inc. Game © 2006 Ubisoft Entertainment 508 The Archives of the National Academy of Sciences 514 Courtesy of CalTech 514 © E.O. Hoppé/CORBIS 519 © Art Wolfe/Stone/Getty Images 522 The Granger Collection, New York 527 © Volvox/Index Stock 531 © CORBIS 534 The Granger Collection, New York CHAPTER 8 551 SCE/Sandia National Laboratory 554 © Roger Ressmeyer/CORBIS 585 © North Wind/North Wind 595 Picture Archives—All rights reserved. (top left) © Nick Wheeler/ CORBIS; (top center) Courtesy of The Architect of the Capitol; (top right) © Stone/Getty Images; (bottom left) © Richard T. Nowitz/CORBIS; (bottom center) Courtesy of Chamber of P1 Commerce, St. Louis, MO; (bottom right) © O. Alamany & E. Vincens/CORBIS 598 © skyscrapers.com 598 © Bob Krist/CORBIS CHAPTER 9 599 © Zia Soleil/Getty Images 604 The Granger Collection, New York 607 © CORBIS 618 © Michael Ng 621 © Bill Ross/CORBIS 636 © Archivo Iconograﬁco, S. A./ CORBIS 644 © Bill Sanderson/Science Photo Library/Photo Researchers, Inc. CHAPTER 10 657 © Øystein Wika 659 Courtesy of Stanford University/ Department of Mathematics 666 © Ken Regan/Camera 5 P2 SEQUENCES AND SERIES COUNTING Arithmetic Fundamental counting principle a, a d, a 2d, a 3d, a 4d,... an Sn a n 1d n a k1 ak n 2 3 2a n 1 1 d 4 2 n a an 2 a b Geometric a, ar, ar 2, ar 3, ar 4,... an Sn
ar n1 n a k1 ak a 1 r n 1 r If r 0 0 1, then the sum of an inﬁnite geometric series is S a 1 r THE BINOMIAL THEOREM a b n 2 1 n 0 b a an n 1 b a an1 b p n n 1 b a abn1 n n b a bn FINANCE Compound interest A P nt 1 r n b a where A is the amount after t years, P is the principal, r is the interest rate, and the interest is compounded n times per year. Amount of an annuity Af R 1 n 1 1 i 2 i where Af is the ﬁnal amount, i is the interest rate per time period, and there are n payments of size R. Present value of an annuity Ap R 1 n 1 1 i i 2 where Ap is the present value, i is the interest rate per time period, and there are n payments of size R. Installment buying R iAp 1 i n 2 1 1 where R is the size of each payment, i is the interest rate per time period, Ap is the amount of the loan, and n is the number of payments. Suppose that two events occur in order. If the ﬁrst can occur in m ways and the second can occur in n ways (after the ﬁrst has occurred), then the two events can occur in order in m n ways. Permutations and combinations The number of permutations of n objects taken r at a time is n, r P 1 2 n! n r 1! 2 The number of combinations of n objects taken r at a time is n, r C 1 2 n! n r! 2 r! 1 The number of subsets of a set with n elements is 2n. The number of distinguishable permutations of n elements, with ni elements of the ith kind (where n1 n k n), is n2 n! n1!n2! p nk! PROBABILITY Probability of an event: If S is a sample space consisting of equally likely outcomes, and E is an event in S, then the probability of E is number of elements in E number of elements in S 2 Complement of an event: 1 E′ P 1 2 1 P E 1 2 Union of two events: P E F P E P F 2 Intersection of two independent events If a game gives payoffs of a1
, a2,..., an with probabilities p1, p2,..., pn, respectively, then the expected value is E a1 p1 a2 p2 a n pn Binomial Probability: If an experiment has the outcomes “success” and “failure” with probabilities p and q 1 p respectively, then P(r successes in n trials) C prqnr n, r 1 2 GEOMETRIC FORMULAS CONIC SECTIONS Formulas for area A, perimeter P, circumference C, volume V: Rectangle A l„ P 2l 2„ Box V l„ h Circles „ Triangle A 1 2 bh l l Pyramid V 1 3 ha 2 h „ Parabolas x 2 4py y p p>0 x p<0 y 0 r C(h, k) x y 2 4px y p<0 p>0 p x Focus 0, p 2 1, directrix y p Focus p, 0 2 1, directrix x p h b h a a Circle A r 2 C 2r Sphere V 4 3 pr 3 A 4r 2 y 0 (h, k) x r r y a 1 a 0, x h 2 k, 2 h 0, (h, k) x 2 k, x h 2 h 0, k 0 Ellipses >>b y a c _a _c c a x _b b x _b _c _a Foci c, 0 2 1, c2 a2 b2 Foci 1 0, c 2, c2 a2 b2 Hyperbolas _b _c _a _b b x _a _c Foci c, 0 2 1, c2 a2 b2 Foci 1 0, c 2, c2 a2 b2 Cylinder V r 2h Cone V 1 3 pr 2h r h PYTHAGOREAN THEOREM In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. a2 b2 c2 a SIMILAR TRIANGLES h r c b Two triangles are similar if corresponding angles are equal If ABC is similar to ABC, then ratios of corresponding sides are equal: a a¿ b b¿ c c¿e sets. Example. With n = 3, the singletons {1}, {2}, {3
} form a 1-uniform cover, and so does {1}, {2, 3}. Also, {1, 2}, {1, 3} and {2, 3} form a uniform 2-cover. However, {1, 2} and {2, 3} do not form a uniform cover of [3]. Note that we allow repetitions. Example. {1}, {1}, {2, 3}, {2}, {3} is a 2-uniform cover of [3]. Theorem (Uniform cover inequality). If A1,..., Ar is a uniform k-cover of [n], then \|K\|k = \|KA\|. r i=1 25 6 Projections III Combinatorics Proof. Let A be a k-uniform cover of [k]. Note that A is a multiset. Write A− = {A ∈ A : n ∈ A} A+ = {A \ {n} ∈ A : n ∈ A} We have \|A+\| = k, and A+ ∪ A− forms a k-uniform cover of [n − 1]. Now note that if K = Rn and n ∈ A, then for all x. Also, if n ∈ A, then \|KA\| ≥ \|K(x)A\| \|KA\| = \|K(x)A\{n}\| dx. (1) (2) In the previous proof, we used Cauchy–Schwarz. What we need here is H¨older’s inequality f g dx ≤ f p dx 1/p 1/q, gq dx where 1 p + 1 q = 1. Iterating this, we get f1 · · · fk dx ≤ k 1/k. f k i dx i=1 Now to perform the proof, we induct on n. We are done if n = 1. Otherwise, given K ⊆ Rn and n ≥ 2, by induction, \|K\| = \|K(x)\| dx \|K(x)A\|1/k \|K(x)A\|1/k dx A∈A− A∈A+ (by induction) \|KA\|1/k A∈A− A∈A+ \|K(x)A\|1/k dx (by (1)) \|KA\|1/k 1/k \|K(x)A
\| A≤\|A− A∈A+ \|KA\|1/k \|KA∪{n}\|1/k. (by H¨older) (by (2)) ≤ ≤ ≤ = A∈A A∈A+ This theorem is great, but we can do better. In fact, Theorem (Box Theorem (Bollob´as, Thomason)). Given a body K ⊆ Rn, i.e. a non-empty bounded open set, there exists a box L such that \|L\| = \|K\| and \|LA\| ≤ \|KA\| for all A ⊆ [n]. Of course, this trivially implies the uniform cover theorem. Perhaps more surprisingly, we can deduce this from the uniform cover inequality. To prove this, we ﬁrst need a lemma. 26 6 Projections III Combinatorics Deﬁnition (Irreducible cover). A uniform k-cover is reducible if it is the disjoint union of two uniform covers. Otherwise, it is irreducible. Lemma. There are only ﬁnitely many irreducible covers of [n]. Proof. Let A and B be covers. We say A < B if A is a “subset” of B, i.e. for each A ⊆ [n], the multiplicity of A in A is less than the multiplicity in B. Then note that the set of irreducible uniform k-covers form an anti-chain, and observe that there cannot be an inﬁnite anti-chain. Proof of box theorem. For A an irreducible cover, we have Also, \|K\|k ≤ A∈A \|KA\|. \|KA\| ≤ i∈A \|K{i}\|. Let {xA : A ⊆ [n]} be a minimal array with xA ≤ \|KA\| such that for each irreducible k-cover A, we have and moreover \|K\|k ≤ xA A∈A xA ≤ i∈A x{i} (1) (2) for all A ⊆ [n]. We know this exists since there are only ﬁnitely many inequalities to be satisﬁed, and we can just decrease the xA’s one by one. Now again by
ﬁniteness, for each xA, there must be at least one inequality involving xA on the right-hand side that is in fact an equality. Claim. For each i ∈ [n], there exists a uniform ki-cover Ci containing {i} with equality \|K\|ki = xA. A∈Ci Indeed if xi occurs on the right of (1), then we are done. Otherwise, it occurs on the right of (2), and then there is some A such that (2) holds with equality. Now there is some cover A containing A such that (1) holds with equality. Then replace A in A with {{j} : j ∈ A}, and we are done. Now let C = n i=1 Ci, C = C \ {{1}, {2},..., {n}}, k = n i=1 ki. Then \|K\|k = A∈C xA = A∈C1 xA ≥ \|K\|k−1 n i=1 xi. 27 6 Projections So we have \|K\| ≥ n i=1 xi. III Combinatorics But we of course also have the reverse inequality. So it must be the case that they are equal. Finally, for each A, consider A = {A} ∪ {{i} : i ∈ A}. Then dividing (1) by i∈A xi gives us xi ≤ xA. By (2), we have the inverse equality. So we have i∈A xA = i∈A xi for all i. So we are done by taking L to be the box with side length xi. Corollary. If K is a union of translates of the unit cube, then for any (not necessarily uniform) k-cover A, we have \|K\|k ≤ A∈A \|KA\|. Here a k-cover is a cover where every element is covered at least k times. Proof. Observe that if B ⊆ A, then \|KB\| ≤ \|KA\|. So we can reduce A to a uniform k-cover. 28 7 Alon’s combinatorial Nullstellensatz III Combinatorics 7 Alon’s combinatorial Nullstellensatz Alon’s combinatorial Nullstellensatz is a seemingly unexciting result that has surprisingly many useful consequences.
Theorem (Alon’s combinatorial Nullstellensatz). Let F be a ﬁeld, and let S1,..., Sn be non-empty ﬁnite subsets of F with \|Si\| = di + 1. Let f ∈ F[X1,..., Xn] have degree d = n n be non-zero. Then f is not identically zero on S = S1 × · · · × Sn. i=1 di, and let the coeﬃcient of X d1 1 · · · X dn Its proof follows from generalizing facts we know about polynomials in one variable. Here R will always be a ring; F always a ﬁeld, and Fq the unique ﬁeld of order q = pn. Recall the following result: Proposition (Division algorithm). Let f, g ∈ R[X] with g monic. Then we can write where deg h ≤ deg f − deg g and deg r < deg g. f = hg + r, Our convention is that deg 0 = −∞. Let X = (X1,..., Xn) be a sequence of variables, and write R[X] = R[X1,..., Xn]. Lemma. Let f ∈ R[X], and for i = 1,..., n, let gi(Xi) ∈ R[Xi] ⊆ R[X] be monic of degree deg gi = degXi gi = di. Then there exists polynomials h1,..., hn, r ∈ R[X] such that f = figi + r, where for all i, j. deg hi ≤ deg f − deg di degXi hi ≤ degXi f − di degXj hi ≤ degXj f degXi r ≤ di − 1 degXi r ≤ degXi f deg r ≤ deg f Proof. Consider f as a polynomial with coeﬃcients in R[X2,..., Xn], then divide by g1 using the division algorithm. So we write Then we have f = h1g1 + r1. degX1 h1 ≤ degX1 f − d1 deg h1 ≤ deg f degXj h1 ≤ degXj
f degX1 r1 ≤ d1 − 1 degXj r1 ≤ degXj f deg r ≤ deg f. Then repeat this with f replaced by r1, g1 by g2, and X1 by X2. We also know that a polynomial of one variable of degree n ≥ 1 over a ﬁeld has at most n zeroes. 29 7 Alon’s combinatorial Nullstellensatz III Combinatorics Lemma. Let S1,..., Sn be non-empty ﬁnite subsets of a ﬁeld F, and let h ∈ F[X] be such that degXi h < \|Si\| for i = 1,..., n. Suppose h is identically 0 on S = S1 × · · · × Sn ⊆ Fn. Then h is the zero polynomial. Proof. Let di = \|Si\| − 1. We induct on n. If n = 1, then we are done. For n ≥ 2, consider h as a one-variable polynomial in F [X1,..., Xn−1] in Xn. Then we can write dn h = gi(X1,..., Xn−1)X i m. i=0 Fix (x1,..., xn−1) ∈ S1 × · · · Sn−1, and set ci = gi(x1,..., xn−1) ∈ F. Then dn n vanishes on Sn. So ci = gi(x1,..., xn−1) = 0 for all (x1,..., xn−1) ∈ S1 × · · · × Sn−1. So by induction, gi = 0. So h = 0. i=0 ciX i Another fact we know about polynomials in one variables is that if f ∈ F[X] vanishes at z1,..., zn, then f is a multiple of n Lemma. For i = 1,..., n, let Si be a non-empty ﬁnite subset of F, and let i=1(X − zi). gi(Xi) = s∈Si (Xi − s)
∈ F[Xi] ⊆ F [X]. Then if f ∈ F[X] is identically zero on S = S1 × · · · × Sn, then there exists hi ∈ F[X], deg hi ≤ deg f − \|Si\| and f = n i=1 higi. Proof. By the division algorithm, we can write f = n i=1 higi + r, where r satisﬁes degXi r < deg gi. But then r vanishes on S1 × · · · × Sn, as both f and gi do. So r = 0. We ﬁnally get to Alon’s combinatorial Nullstellensatz. Theorem (Alon’s combinatorial Nullstellensatz). Let S1,..., Sn be non-empty ﬁnite subsets of F with \|Si\| = di + 1. Let f ∈ F[X] have degree d = n i=1 di, and let the coeﬃcient of X d1 n be non-zero. Then f is not identically zero on S = S1 × · · · × Sn. 1 · · · X dn Proof. Suppose for contradiction that f is identically zero on S. Deﬁne gi(Xi) and hi as before such that f = higi. Since the coeﬃcient of X d1 But that’s impossible, since 1 · · · X dn n is non-zero in f, it is non-zero in some hjgj. deg hj ≤ di − deg gj = n i=1 i=j di − 1, and so hj cannot contain a X d1 1 · · · ˆXj dj · · · X dn n term. 30 7 Alon’s combinatorial Nullstellensatz III Combinatorics Let’s look at some applications. Here p is a prime, q = pk, and Fq is the unique ﬁeld of order q. Theorem (Chevalley, 1935). Let f1,..., fm ∈ Fq[X1,..., Xn] be such that m i=1 deg fi < n. Then the fi cannot have exactly one common zero. Proof. Suppose
not. We may assume that the common zero is 0 = (0,..., 0). Deﬁne m f = (1 − fi(X)q−1) − γ n (Xi − s), i=1 i=1 s∈F× q where γ is chosen so that F (0) = 0, namely the inverse of s∈F× q m (−s). Now observe that for any non-zero x, the value of fi(x)q−1 = 1, so f (x) = 0. Thus, we can set Si = Fq, and they satisfy the hypothesis of the theorem. In is γ = 0. However, f vanishes on Fn q. · · · X q−1 n particular, the coeﬃcient of X q−1 This is a contradiction. 1 It is possible to prove similar results without using the combinatorial Nullstellensatz. These results are often collectively refered to as Chevalley–Warning theorems. Theorem (Warning). Let f (X) = f (X1,..., Xn) ∈ Fq[X] have degree < n. Then N (f ), the number of zeroes of f is a multiple of p. One nice trick in doing these things is that ﬁnite ﬁelds naturally come with an “indicator function”. Since the multiplicative group has order q − 1, we know that if x ∈ Fq, then xq−. Proof. We have Thus, we know 1 − f (x)q−1 = 1 0 f (x) = 0 otherwise. N (f ) = x∈Fn q (1 − f (x)q−1) = − f (x)q−1 ∈ Fq. x∈Fn q Further, we know that if k ≥ 0, then xk = x∈Fn q −1 k = q − 1 otherwise 0. So let’s write f (x)q−1 as a linear combination of monomials. Each monomial has degree < n(q − 1). So there is at least one k such that the power of Xk in that monomial is < q − 1. Then the sum over Xk vanishes for this monomial. So each monomial contributes 0 to
the sum. 31 7 Alon’s combinatorial Nullstellensatz III Combinatorics We can use Alon’s combinatorial Nullstellensatz to eﬀortlessly prove some of our previous theorems. Theorem (Cauchy–Davenport theorem). Let p be a prime and A, B ⊆ Zp be non-empty subsets with \|A\| + \|B\| ≤ p + 1. Then \|A + B\| ≥ \|A\| + \|B\| − 1. Proof. Suppose for contradiction that A + B ⊆ C ⊆ Zp, and \|C\| = \|A\| + \|B\| − 2. Let’s come up with a polynomial that encodes the fact that C contains the sum A + B. We let f (X, Y ) = (X + Y − c). c∈C Then f vanishes on A × B, and deg f = \|C\|., To apply the theorem, we check that the coeﬃcient of X \|A\|−1Y \|B\|−1 is \|C\| \|A\|−1 which is non-zero in Zp, since C < p. This contradicts Alon’s combinatorial Nullstellensatz. We can also use this to prove Erd¨os–Ginzburg–Ziv again. Theorem (Erd¨os–Ginzburg–Ziv). Let p be a prime and a1,..., a2p+1 ∈ Zp. Then there exists I ∈ [2p − 1](p) such that Proof. Deﬁne ai = 0 ∈ Zp. i∈I f1(X1,..., X2p−1) = f2(X1,..., X2p−1) = 2p−1 i=1 2p−1 i=1 X p−1 i. aiX p−1 i. Then by Chevalley’s theorem, we know there cannot be exactly one common zero. But 0 is one common zero. So there must be another. Take this solution, and let I = {i : xi = 0}. Then f1(X) = 0 is the same as saying \|I\| = p, and f2(
X) = 0 is the same as saying i∈I ai = 0. We can also consider restricted sums. We set · + B = {a + b : a ∈ A, b ∈ B, a = b}. A Example. If n = m, then · + [m] = {3, 4,..., m + n} [n] · + [n] = {3, 4,..., 2n − 1} [n] From this example, we show that if \|A\| ≥ 2, then \|A · + A\| can be as small as 2\|A\| − 3. In 1964, Erd¨os and Heilbronn Conjecture (Erd¨os–Heilbronn, 1964). If 2\|A\| ≤ p + 3, then \|A · + A\| ≥ 2\|A\| − 3. 32 7 Alon’s combinatorial Nullstellensatz III Combinatorics This remained open for 30 years, and was proved by Dias da Silva and Hamidoune. A much much simpler proof was given by Alon, Nathanson and Ruzsa in 1996. Theorem. Let A, B ⊆ Zp be such that 2 ≤ \|A\| < \|B\| and \|A\| + \|B\| ≤ p + 2. Then A · + B ≥ \|A\| + \|B\| − 2. The above example shows we cannot do better. Proof. Suppose not. Deﬁne f (X, Y ) = (X − Y ) (X + Y − c), c∈C where A · + B ⊆ C ⊆ Zp and \|C\| = \|A\| + \|B\| − 3. Then deg g = \|A\| + \|B\| − 2, and the coeﬃcient of X \|A\|−1Y \|B\|−1 is \|A\| + \|B\| − 3 \|A\| − 2 − \|A\| + \|B\| − 3 \|A\| − 1 = 0. Hence by Alon’s combinatorial Nullstellensatz, f (x, y) is not identically zero on A × B. A contradiction. Corollary (Erd¨os–Heilbronn conjecture). If A, B ⊆ Zp, non-empty and \|A
\| + · + B\| ≥ \|A\| + \|B\| − 3. \|B\| ≤ p + 3, and p is a prime, then \|A Proof. We may assume 2 ≤ \|A\| ≤ \|B\|. Pick a ∈ A, and set A = A \ {a}. Then · + B\| ≥ \|A · + B\| ≥ \|A\| + \|B\| − 2 = \|A\| + \|B\| − 3. \|A Now consider the following problem: suppose we have a circular table Z2n+1. Suppose the host invites n couples, and the host, being a terrible person, wants the ith couple to be a disatnce di apart for some 1 ≤ di ≤ n. Can this be done? Theorem. If 2n + 1 is a prime, then this can be done. Proof. We may wlog assume the host is at 0. We want to partition Zp \ {0} = Z× p into n pairs {xi, xi + di}. Consider the polynomial ring Zp[X1,..., Xn] = Zp[X]. We deﬁne f (x) = i Xi(Xi+di) i<j (Xi−Xj)(Xi+di−Xj)(Xi−Xj −dj)(Xi+di−Xj −dj). We want to show this is not identically zero on Zn p First of all, we have deg f = 4 n 2 + 2n = 2n2. So we are good. The coeﬃcient of X 2n 1 · · · X 2n n is the same as that in X 2 i i<j (Xi − Xj)4 = X 2 i (Xi − Xj)2 = i=j 33 X 2n i 1 − i=j 2. Xi Xj 7 Alon’s combinatorial Nullstellensatz III Combinatorics This, we are looking for the constant term in 1 − i=j 2. Xi Xj By a question on the example sheet, this is 2n 2, 2,..., 2 = 0 in Zp. Our ﬁnal example is as follows: suppose we are in Zp, and a1,..., ap and c1,..., cp are enumerations of the elements, and bi = ci
− ai. Then clearly we have bi = 0. Is the converse true? The answer is yes! Theorem. If b1,..., bp ∈ Zp are such that bi = 0, then there exists numerations a1,..., ap and b1,..., bp of the elements of Zp such that for each i, we have ai + bi = ci. Proof. It suﬃces to show that for all (bi), there are distinct a1, · · ·, ap−1 such that ai + bi = aj + bj for all i = j. Consider the polynomial The degree is (Xi − Xj)(Xi + bi − Xj − bj). i<j p − 1 2 2 = (p − 1)(p − 2). We then inspect the coeﬃcient of X p−2 non-zero is the same as above. 1 · · · X p−2 p−1, and checking that this is 34 Index Index (k, )-regular, 3 X (≤r), 5 X (r), 5 X ≥r, 5 Γ(S), 3 P(X), 6 · +, 32 n-sections, 24 Alon’s combinatorial Nullstellensatz, 29, 30 anti-chain, 6 ball Hamming, 17 binary order, 20 bipartite graph, 3 biregular graph, 3 boundary, 16 box theorem, 26 Cauchy–Davenport theorem, 22, 32 chain, 6 Chevalley–Warning theorems, 31 co-dimension 1 compression, 18 colex order, 12 combinatorial Nullstellensatz, 29, 30 complete matching, 3 compression operator, 11 cover, 25 discrete cube, 16 distinct representatives, 4 division algorithm, 29 III Combinatorics downward-expanding poset, 7 edge boundary, 20 expanding, 7 graded poset, 6 Hall’s condition, 3 Hall’s theorem, 3 Hamming ball, 17 irreducible cover, 27 isoperimetric inequality, 16 lex order, 12 LYM inequality, 7 neighbourhood, 16 poset regular, 8 power set, 6 reducible cover, 27 regular poset, 8 restricted sum, 32 shadow, 7 shadow function, 13 simplicial ordering, 17 symmetric chain, 9 uniform cover, 25 uniform cover inequality
, 25 upward-expanding, 7 weight, 7 35lly, we get to the integration part. Suppose we picked all our γw to be the ﬁxed straight line segment from a0. Then for antiderivative to be diﬀerentiable, we needed f (z) dz = f (z) dz. γw∗δh γw+h In other words, we needed to the integral along the path γw ∗ δh ∗ (−γw+h) to vanish. This is a rather simple kind of paths. It is just (the boundary of) a triangle, consisting of three line segments. Deﬁnition (Triangle). A triangle in a domain U is what it ought to be — the Euclidean convex hull of 3 points in U, lying wholly in U. We write its boundary as ∂T, which we view as an oriented piecewise C 1 path, i.e. a contour. good bad very bad Our earlier result on constructing antiderivative then shows: Proposition. If U is a star domain, and f : U → C is continuous, and if ∂T f (z) dz = 0 for all triangles T ⊆ U, then f has an antiderivative on U. Proof. As before, taking γw = [a0, w] ⊆ U if U is star-shaped about a0. 24 2 Contour integration IB Complex Analysis This is in some sense a weaker proposition — while our hypothesis only requires the integral to vanish over triangles, and not arbitrary closed loops, we are restricted to star domains only. But well, this is technically a weakening, but how is it useful? Surely if we can somehow prove that the integral of a particular function vanishes over all triangles, then we can easily modify the proof such that it works for all possible curves. Turns out, it so happens that for triangles, we can ﬁddle around with some geometry to prove the following result: Theorem (Cauchy’s theorem for a triangle). Let U be a domain, and let f : U → C be holomorphic. If T ⊆ U is a triangle, then ∂T f (z) dz = 0. So for holomorphic functions, the hypothesis of the previous theorem auto- matically holds. We immediately get the following coroll
ary, which is what we will end up using most of the time. Corollary (Convex Cauchy). If U is a convex or star-shaped domain, and f : U → C is holomorphic, then for any closed piecewise C 1 paths γ ∈ U, we must have f (z) dz = 0. γ Proof of corollary. If f is holomorphic, then Cauchy’s theorem says the integral over any triangle vanishes. If U is star shaped, our proposition says f has an antiderivative. Then the fundamental theorem of calculus tells us the integral around any closed path vanishes. Hence, all we need to do is to prove that fact about triangles. Proof of Cauchy’s theorem for a triangle. Fix a triangle T. Let η = ∂T f (z) dz, = length(∂T ). The idea is to show to bound η by ε, for every ε > 0, and hence we must have η = 0. To do so, we subdivide our triangles. Before we start, it helps to motivate the idea of subdividing a bit. By subdividing the triangle further and further, we are focusing on a smaller and smaller region of the complex plane. This allows us to study how the integral behaves locally. This is helpful since we are given that f is holomorphic, and holomorphicity is a local property. We start with T = T 0 : We then add more lines to get T 0 which). it doesn’t really matter which is 25 2 Contour integration IB Complex Analysis We orient the middle triangle by the anti-clockwise direction. Then we have ∂T 0 f (z) dz = a,b,c,d ∂T 0· f (z) dz, since each internal edge occurs twice, with opposite orientation. For this to be possible, if η = ∂T 0 f (z) dz, then there must be some subscript in {a, b, c, d} such that ∂T 0· f (z) dz ≥ η 4. We call this T 0· = T 1. Then we notice ∂T 1 has length length(∂T 1) = 2. Iterating this, we obtain triangles T 0 ⊇ T 1 ⊇ T 2 ⊇ · ·
· such that η 4i, Now we are given a nested sequence of closed sets. By IB Metric and Topological Spaces (or IB Analysis II), there is some z0 ∈ length(∂T i) = f (z) dz 2i. ∂T i ≥ i≥0 T i. Now ﬁx an ε > 0. Since f is holomorphic at z0, we can ﬁnd a δ > 0 such that \|f (w) − f (z0) − (w − z0)f (z0)\| ≤ ε\|w − z0\| whenever \|w − z0\| < δ. Since the diameters of the triangles are shrinking each time, we can pick an n such that T n ⊆ B(z0, ε). We’re almost there. We just need to do one last thing that is slightly sneaky. Note that ∂T n 1 dz = 0 = ∂T n z dz, since these functions certainly do have anti-derivatives on T n. Therefore, noting that f (z0) and f (z0) are just constants, we have (f (z) − f (z0) − (z − z0)f (z0)) dz f (z) dz = ∂T n ∂T n ≤ \|f (z) − f (z0) − (z − z0)f (z0)\| dz ∂T n ≤ length(∂T n)ε sup z∈∂T n \|z − z0\| ≤ ε length(∂T n)2, 26 2 Contour integration IB Complex Analysis where the last line comes from the fact that z0 ∈ T n, and the distance between any two points in the triangle cannot be greater than the perimeter of the triangle. Substituting our formulas for these in, we have So η 4n ≤ 1 4n 2ε. η ≤ 2ε. Since is ﬁxed and ε was arbitrary, it follows that we must have η = 0. Is this the best we can do? Can we formulate this for an arbitrary domain, and not just star-shaped ones? It is obviously not true if the domain is not z deﬁned on C \ {0}. However, it turns out simply connected, e
.g. for f (z) = 1 Cauchy’s theorem holds as long as the domain is simply connected, as we will show in a later part of the course. However, this is not surprising given the Riemann mapping theorem, since any simply connected domain is conformally equivalent to the unit disk, which is star-shaped (and in fact convex). We can generalize our result when f : U → C is continuous on the whole of U, and holomorphic except on ﬁnitely many points. In this case, the same conclusion holds — γ f (z) dz = 0 for all piecewise smooth closed γ. Why is this? In the proof, it was suﬃcient to focus on showing ∂T f (z) dz = 0 for a triangle T ⊆ U. Consider the simple case where we only have a single point of non-holomorphicity a ∈ T. The idea is again to subdivide. a We call the center triangle T. Along all other triangles in our subdivision, we get f (z) dz = 0, as these triangles lie in a region where f is holomorphic. So ∂T f (z) dz = ∂T f (z) dz. Note now that we can make T as small as we like. But ∂T f (z) dz ≤ length(∂T ) sup z∈∂T \|f (z)\|. Since f is continuous, it is bounded. As we take smaller and smaller subdivisions, length(∂T ) → 0. So we must have ∂T f (z) dz = 0. From here, it’s straightforward to conclude the general case with many points of non-holomorphicity — we can divide the triangle in a way such that each small triangle contains one bad point. 27 2 Contour integration IB Complex Analysis 2.3 The Cauchy integral formula Our next amazing result will be Cauchy’s integral formula. This formula allows us to ﬁnd the value of f inside a ball B(z0, r) just given the values of f on the boundary ∂B(z0, r). Theorem (Cauchy integral formula). Let U be a domain, and f : U → C be holomorphic. Suppose there is some B(z0; r) ⊆ U for
some z0 and r > 0. Then for all z ∈ B(z0; r), we have f (z) = 1 2πi ∂B(z0;r) f (w) w − z dw. Recall that we previously computed 1 z dz = 2πi. This is indeed a special case of the Cauchy integral formula. We will provide two proofs. The ﬁrst proof relies on the above generalization of Cauchy’s theorem. ∂B(0,1) Proof. Since U is open, there is some δ > 0 such that B(z0; r + δ) ⊆ U. We deﬁne g : B(z0; r + δ) → C by g(w) = f (w)−f (z) w−z f (z) w = z w = z, where we have ﬁxed z ∈ B(z0; r) as in the statement of the theorem. Now note that g is holomorphic as a function of w ∈ B(z0, r + δ), except perhaps at w = z. But since f is holomorphic, by deﬁnition g is continuous everywhere on B(z0, r + δ). So the previous result says ∂B(z0;r) g(w) dw = 0. This is exactly saying that ∂B(z0;r) f (w) w − z dw = ∂B(z0;r) f (z) w − z dw. We now rewrite 1 w − z = 1 w − z0 · 1 z−z0 w−z0 = ∞ n=0 1 − (z − z0)n (w − z0)n+1. Note that this sum converges uniformly on ∂B(z0; r) since z − z0 w − z0 < 1 for w on this circle. By uniform convergence, we can exchange summation and integration. So ∂B(z0;r) f (w) w − z ∞ dw = n=0 ∂B(z0,r) f (z) (z − z0)n (w − z0)n+1 dw. 28 2 Contour integration IB Complex Analysis We note that f (z)(z − z0
)n is just a constant, and that we have previously proven ∂B(z0;r) (w − z0)k dw = 2πi k = −1 k = −1 0. So the right hand side is just 2πif (z). So done. Corollary (Local maximum principle). Let f : B(z, r) → C be holomorphic. Suppose \|f (w)\| ≤ \|f (z)\| for all w ∈ B(z; r). Then f is constant. In other words, a non-constant function cannot achieve an interior local maximum. Proof. Let 0 < ρ < r. Applying the Cauchy integral formula, we get \|f (z)\| = 1 2πi ∂B(z;ρ) f (w) w − z dw Setting w = z + ρe2πiθ, we get = 1 0 ≤ sup f (z + ρe2πiθ) dθ \|f (w)\| \|z−w\|=ρ ≤ f (z). So we must have equality throughout. When we proved the supremum bound for the integral, we showed equality can happen only if the integrand is constant. So \|f (w)\| is constant on the circle \|z − w\| = ρ, and is equal to f (z). Since this is true for all ρ ∈ (0, r), it follows that \|f \| is constant on B(z; r). Then the Cauchy–Riemann equations then entail that f must be constant, as you have shown in example sheet 1. Going back to the Cauchy integral formula, recall that we had B(z0; r) ⊆ U, f : U → C holomorphic, and we want to show f (z) = 1 2πi ∂B(z0;r) f (w) w − z dw. When we proved it last time, we remember we know how to integrate things of the form (w−z0)n, and manipulated the formula such that we get the integral is made of things like this. 1 The second strategy is to change the contour of integration instead of changing the integrand. If we can change it so that the integral is performed over a circle around z instead of z0, then we know what to do. z z0 29 2 Contour integration IB Complex
Analysis Proof. (of Cauchy integral formula again) Given ε > 0, we pick δ > 0 such that B(z, δ) ⊆ B(z0, r), and such that whenever \|w − z\| < δ, then \|f (w) − f (z)\| < ε. This is possible since f is uniformly continuous on the neighbourhood of z. We now cut our region apart: z z0 z z0 We know f (w) w−z is holomorphic on suﬃciently small open neighbourhoods of the half-contours indicated. The area enclosed by the contours might not be star-shaped, but we can deﬁnitely divide it once more so that it is. Hence the integral of f (w) w−z around the half-contour vanishes by Cauchy’s theorem. Adding these together, we get ∂B(z0,r) f (w) w − z dw = ∂B(z,δ) f (w) w − z dw, where the balls are both oriented anticlockwise. Now we have f (z) − 1 2πi ∂B(z0,r) f (w) w − z dw = f (z) − 1 2πi ∂B(z,δ) f (w) w − z dw. Now we once again use the fact that ∂B(z,δ) 1 w − z dz = 2πi to show this is equal to 1 2πi ∂B(z,δ) f (z) − f (w) w − z dw ≤ 1 2π · 2πδ · 1 δ · ε = ε. Taking ε → 0, we see that the Cauchy integral formula holds. Note that the subdivision we did above was something we can do in general. Deﬁnition (Elementary deformation). Given a pair of C 1-smooth (or piecewise smooth) closed paths φ, ψ : [0, 1] → U, we say ψ is an elementary deformation of φ if there exists convex open sets C1, · · ·, Cn ⊆ U and a division of the interval 0 = x0 < x1 < · · · < xn = 1 such that on [xi
−1, xi], both φ(t) and ψ(t) belong to Ci. 30 2 Contour integration IB Complex Analysis ψ φ(xi−1) φ(xi) φ φ(xi−1) φ(xi) Then there are straight lines γi : φ(xi) → ψ(xi) lying inside Ci. If f is holomorphic on U, considering the shaded square, we ﬁnd φ f (z) dz = ψ f (z) dz when φ and ψ are convex deformations. We now explore some classical consequences of the Cauchy Integral formula. The next is Liouville’s theorem, as promised. Theorem (Liouville’s theorem). Let f : C → C be an entire function (i.e. holomorphic everywhere). If f is bounded, then f is constant. This, for example, means there is no interesting holomorphic period functions like sin and cos that are bounded everywhere. Proof. Suppose \|f (z)\| ≤ M for all z ∈ C. We ﬁx z1, z2 ∈ C, and estimate \|f (z1) − f (z2)\| with the integral formula. Let R > max{2\|z1\|, 2\|z2\|}. By the integral formula, we know \|f (z1) − f (z2)\| = = ≤ = 1 2πi 1 2πi ∂B(0,R) ∂B(0,R) f (w) w − z1 − f (w) w − z2 f (w)(z1 − z2) (w − z1)(w − z2) dw dw M \|z1 − z2\| (R/2)2 · 2πR · 1 2π 4M \|z1 − z2\| R. Note that we get the bound on the denominator since \|w\| = R implies \|w−zi\| > R 2 by our choice of R. Letting R → ∞, we know we must have f (z1) = f (z2). So f is constant. Corollary (Fundamental theorem of algebra). A non-constant complex polynomial has a root in C. Proof. Let P (z) = anzn + an−1zn−1 +
· · · + a0, 31 2 Contour integration IB Complex Analysis where an = 0 and n > 0. So P is non-constant. Thus, as \|z\| → ∞, \|P (z)\| → ∞. In particular, there is some R such that for \|z\| > R, we have \|P (z)\| ≥ 1. Now suppose for contradiction that P does not have a root in C. Then consider f (z) = 1 P (z), which is then an entire function, since it is a rational function. On B(0, R), we know f is certainly continuous, and hence bounded. Outside this ball, we get \|f (z)\| ≤ 1. So f (z) is constant, by Liouville’s theorem. But P is non-constant. This is absurd. Hence the result follows. There are many many ways we can prove the fundamental theorem of algebra. However, none of them belong wholely to algebra. They all involve some analysis or topology, as you might encounter in the IID Algebraic Topology and IID Riemann Surface courses. This is not surprising since the construction of R, and hence C, is intrinsically analytic — we get from N to Z by requiring it to have additive inverses; Z to Q by requiring multiplicative inverses; R to C by requiring the root to x2 + 1 = 0. These are all algebraic. However, to get from Q to R, we are requiring something about convergence in Q. This is not algebraic. It requires a particular of metric on Q. If we pick a diﬀerent metric, then you get a diﬀerent completion, as you may have seen in IB Metric and Topological Spaces. Hence the construction of R is actually analytic, and not purely algebraic. 2.4 Taylor’s theorem When we ﬁrst met Taylor series, we were happy, since we can express anything as a power series. However, we soon realized this is just a fantasy — the Taylor series of a real function need not be equal to the function itself. For example, the function f (x) = e−x−2 has vanishing Taylor series at 0, but does not vanish in any neighbourhood of 0. What we do have is Taylor’s theorem, which gives you an expression for what the remainder is if we truncate our series, but
is otherwise completely useless. In the world of complex analysis, we are happy once again. Every holomorphic function can be given by its Taylor series. Theorem (Taylor’s theorem). Let f : B(a, r) → C be holomorphic. Then f has a convergent power series representation f (z) = ∞ n=0 cn(z − a)n on B(a, r). Moreover, cn = f (n)(a) n! = 1 2πi ∂B(a,ρ) f (z) (z − a)n+1 dz for any 0 < ρ < r. Note that the very statement of the theorem already implies any holomorphic function has to be inﬁnitely diﬀerentiable. This is a good world. 32 2 Contour integration IB Complex Analysis Proof. We’ll use Cauchy’s integral formula. If \|w − a\| < ρ < r, then f (w) = 1 2πi ∂B(a,ρ) f (z) z − w dz. Now (cf. the ﬁrst proof of the Cauchy integral formula), we note that 1 z − w = (z − a) 1 1 − w−a z−a = n n=0 (w − a)n (z − a)n+1. This series is uniformly convergent everywhere on the ρ disk, including its boundary. By uniform convergence, we can exchange integration and summation to get ∞ f (w) = 1 2πi ∂B(a,ρ) f (z) (z − a)n+1 dz (w − a)n n=0 ∞ n=0 = cn(w − a)n. Since cn does not depend on w, this is a genuine power series representation, and this is valid on any disk B(a, ρ) ⊆ B(a, r). Then the formula for cn in terms of the derivative comes for free since that’s the formula for the derivative of a power series. This tells us every holomorphic function behaves like a power series. In particular, we do not get weird things like e−x−2 on R that have a trivial Taylor series expansion, but is itself non-trivial. Similarly, we know that there are no “
bump functions” on C that are non-zero only on a compact set (since power series don’t behave like that). Of course, we already knew that from Liouville’s theorem. Corollary. If f : B(a, r) → C is holomorphic on a disc, then f is inﬁnitely diﬀerentiable on the disc. Proof. Complex power series are inﬁnitely diﬀerentiable (and f had better be inﬁnitely diﬀerentiable for us to write down the formula for cn in terms of f (n)). This justiﬁes our claim from the very beginning that Re(f ) and Im(f ) are harmonic functions if f is holomorphic. Corollary. If f : U → C is a complex-valued function, then f = u + iv is holomorphic at p ∈ U if and only if u, v satisfy the Cauchy–Riemann equations, and that ux, uy, vx, vy are continuous in a neighbourhood of p. Proof. If ux, uy, vx, vy exist and are continuous in an open neighbourhood of p, then u and v are diﬀerentiable as functions R2 → R2 at p, and then we proved that the Cauchy–Riemann equations imply diﬀerentiability at each point in the neighbourhood of p. So f is diﬀerentiable at a neighbourhood of p. On the other hand, if f is holomorphic, then it is inﬁnitely diﬀerentiable. In particular, f (z) is also holomorphic. So ux, uy, vx, vy are diﬀerentiable, hence continuous. 33 2 Contour integration IB Complex Analysis We also get the following (partial) converse to Cauchy’s theorem. Corollary (Morera’s theorem). Let U ⊆ C be a domain. Let f : U → C be continuous such that f (z) dz = 0 γ for all piecewise-C 1 closed curves γ ∈ U. Then f is holomorphic on U. Proof. We have previously shown that the condition implies that f has an antiderivative F : U → C, i.
e. F is a holomorphic function such that F = f. But F is inﬁnitely diﬀerentiable. So f must be holomorphic. Recall that Cauchy’s theorem required U to be suﬃciently nice, e.g. being star-shaped or just simply-connected. However, Morera’s theorem does not. It just requires that U is a domain. This is since holomorphicity is a local property, while vanishing on closed curves is a global result. Cauchy’s theorem gets us from a local property to a global property, and hence we need to assume more about what the “globe” looks like. On the other hand, passing from a global property to a local one does not. Hence we have this asymmetry. Corollary. Let U ⊆ C be a domain, fn; U → C be a holomorphic function. If fn → f uniformly, then f is in fact holomorphic, and f (z) = lim n f n(z). Proof. Given a piecewise C 1 path γ, uniformity of convergence says γ fn(z) dz → γ f (z) dz uniformly. Since f being holomorphic is a local condition, so we ﬁx p ∈ U and work in some small, convex disc B(p, ε) ⊆ U. Then for any curve γ inside this disk, we have fn(z) dz = 0. γ Hence we also have γ f (z) dz = 0. Since this is true for all curves, we conclude f is holomorphic inside B(p, ε) by Morera’s theorem. Since p was arbitrary, we know f is holomorphic. We know the derivative of the limit is the limit of the derivative since we can express f (a) in terms of the integral of f (z) (z−a)2, as in Taylor’s theorem. There is a lot of passing between knowledge of integrals and knowledge of holomorphicity all the time, as we can see in these few results. These few sections are in some sense the heart of the course, where we start from Cauchy’s theorem and Cauchy’s integral formula, and derive all the other amazing consequences. 2.5 Zeroes Recall that for a po
lynomial p(z), we can talk about the order of its zero at z = a by looking at the largest power of (z − a) dividing p. A priori, it is not clear how we can do this for general functions. However, given that everything is a Taylor series, we know how to do this for holomorphic functions. 34 2 Contour integration IB Complex Analysis Deﬁnition (Order of zero). Let f : B(a, r) → C be holomorphic. Then we know we can write ∞ f (z) = cn(z − a)n as a convergent power series. Then either all cn = 0, in which case f = 0 on B(a, r), or there is a least N such that cN = 0 (N is just the smallest n such that f (n)(a) = 0). n=0 If N > 0, then we say f has a zero of order N. If f has a zero of order N at a, then we can write f (z) = (z − a)N g(z) on B(a, r), where g(a) = cN = 0. Often, it is not the actual order that is too important. Instead, it is the ability to factor f in this way. One of the applications is the following: Lemma (Principle of isolated zeroes). Let f : B(a, r) → C be holomorphic and not identically zero. Then there exists some 0 < ρ < r such that f (z) = 0 in the punctured neighbourhood B(a, ρ) \ {a}. Proof. If f (a) = 0, then the result is obvious by continuity of f. The other option is not too diﬀerent. If f has a zero of order N at a, then we can write f (z) = (z − a)N g(z) with g(a) = 0. By continuity of g, g does not vanish on some small neighbourhood of a, say B(a, ρ). Then f (z) does not vanish on B(a, ρ) \ {a}. A consequence is that given two holomorphic functions on the same domain, if they agree on suﬃciently many points, then they must in fact be equal. Corollary (Identity theorem). Let U
⊆ C be a domain, and f, g : U → C be holomorphic. Let S = {z ∈ U : f (z) = g(z)}. Suppose S contains a non-isolated point, i.e. there exists some w ∈ S such that for all ε > 0, S ∩ B(w, ε) = {w}. Then f = g on U. Proof. Consider the function h(z) = f (z) − g(z). Then the hypothesis says h(z) has a non-isolated zero at w, i.e. there is no non-punctured neighbourhood of w on which h is non-zero. By the previous lemma, this means there is some ρ > 0 such that h = 0 on B(w, ρ) ⊆ U. Now we do some topological trickery. We let U0 = {a ∈ U : h = 0 on some neighbourhood B(a, ρ) of a in U }, U1 = {a ∈ U : there exists n ≥ 0 such that h(n) = 0}. Clearly, U0 ∩ U1 = ∅, and the existence of Taylor expansions shows U0 ∪ U1 = U. Moreover, U0 is open by deﬁnition, and U1 is open since h(n)(z) is continuous near any given a ∈ U1. Since U is (path) connected, such a decomposition can happen if one of U0 and U1 is empty. But w ∈ U0. So in fact U0 = U, i.e. h vanishes on the whole of U. So f = g. In particular, if two holomorphic functions agree on some small open subset of the domain, then they must in fact be identical. This is a very strong result, and is very false for real functions. Hence, to specify, say, an entire function, all we need to do is to specify it on an arbitrarily small domain we like. 35 2 Contour integration IB Complex Analysis Deﬁnition (Analytic continuiation). Let U0 ⊆ U ⊆ C be domains, and f : U0 → C be holomorphic. An analytic continuation of f is a holomorphic function h : U → C such that h\|U0 = f, i.e. h(
z) = f (z) for all z ∈ U0. By the identity theorem, we know the analytic continuation is unique if it exists. Thus, given any holomorphic function f : U → C, it is natural to ask how far we can extend the domain, i.e. what is the largest U ⊇ U such that there is an analytic continuation of f to U. There is no general method that does this for us. However, one useful trick is to try to write our function f in a diﬀerent way so that it is clear how we can extend it to elsewhere. Example. Consider the function f (z) = n≥0 deﬁned on B(0, 1). zn = 1 + z + z2 + · · · By itself, this series diverges for z outside B(0, 1). However, we know well that this function is just f (z) = 1 1 − z. This alternative representation makes sense on the whole of C except at z = 1. So we see that f has an analytic continuation to C \ {1}. There is clearly no extension to the whole of C, since it blows up near z = 1. Example. Alternatively, consider f (z) = z2n. n≥0 Then this again converges on B(0, 1). You will show in example sheet 2 that there is no analytic continuation of f to any larger domain. Example. The Riemann zeta function ζ(z) = ∞ n=1 n−z deﬁnes a holomorphic function on {z : Re(z) > 1} ⊆ C. Indeed, we have \|n−z\| = \|nRe(z)\|, and we know n−t converges for t ∈ R>1, and in fact does so uniformly on any compact domain. So the corollary of Morera’s theorem tells us that ζ(z) is holomorphic on Re(z) > 1. We know this cannot converge as z → 1, since we approach the harmonic series which diverges. However, it turns out ζ(z) has an analytic continuation to C \ {1}. We will not prove this. At least formally, using the fundamental theorem of arithmetic, we can expand n as a product of its prime factors, and write ζ(z) = (1 + p−z + p
−2z + · · · ) = primes p primes p 1 1 − p−z. If there were ﬁnitely many primes, then this would be a well-deﬁned function on all of C, since this is a ﬁnite product. Hence, the fact that this blows up at z = 1 implies that there are inﬁnitely many primes. 36 2 Contour integration IB Complex Analysis 2.6 Singularities The next thing to study is singularities of holomorphic functions. These are places where the function is not deﬁned. There are many ways a function can be ill-deﬁned. For example, if we write f (z) = 1 − z 1 − z, then on the face of it, this function is not deﬁned at z = 1. However, elsewhere, f is just the constant function 1, and we might as well deﬁne f (1) = 1. Then we get a holomorphic function. These are rather silly singularities, and are singular solely because we were not bothered to deﬁne f there. Some singularities are more interesting, in that they are genuinely singular. For example, the function f (z) = 1 1 − z is actually singular at z = 1, since f is unbounded near the point. It turns out these are the only possibilities. Proposition (Removal of singularities). Let U be a domain and z0 ∈ U. If f : U \ {z0} → C is holomorphic, and f is bounded near z0, then there exists an a such that f (z) → a as z → z0. Furthermore, if we deﬁne g(z) = f (z) a z ∈ U \ {z0} z = z0, then g is holomorphic on U. Proof. Deﬁne a new function h : U → C by h(z) = (z − z0)2f (z) 0 z = z0 z = z0. Then since f is holomorphic away from z0, we know h is also holomorphic away from z0. Also, we know f is bounded near z0. So suppose \|f (z)\| < M in some neighbourhood of z0. Then we have h(z) − h(z0) z −
z0 ≤ \|z − z0\|M. So in fact h is also diﬀerentiable at z0, and h(z0) = h(z0) = 0. So near z0, h has a Taylor series h(z) = an(z − z0)n. Since we are told that a0 = a1 = 0, we can deﬁne a g(z) by n≥0 g(z) = n≥0 an+2(z − z0)n, 37 2 Contour integration IB Complex Analysis deﬁned on some ball B(z0, ρ), where the Taylor series for h is deﬁned. By construction, on the punctured ball B(z0, ρ)\{z0}, we get g(z) = f (z). Moreover, g(z) → a2 as z → z0. So f (z) → a2 as z → z0. Since g is a power series, it is holomorphic. So the result follows. This tells us the only way for a function to fail to be holomorphic at an isolated point is that it blows up near the point. This won’t happen because f fails to be continuous in some weird ways. However, we are not yet done with our classiﬁcation. There are many ways in which things can blow up. We can further classify these into two cases — the case where \|f (z)\| → ∞ as z → z0, and the case where \|f (z)\| does not converge as z → z0. It happens that the ﬁrst case is almost just as boring as the removable ones. Proposition. Let U be a domain, z0 ∈ U and f : U \ {z0} → C be holomorphic. Suppose \|f (z)\| → ∞ as z → z0. Then there is a unique k ∈ Z≥1 and a unique holomorphic function g : U → C such that g(z0) = 0, and f (z) = g(z) (z − z0)k. Proof. We shall construct g near z0 in some small neighbourhood, and then apply analytic continuation to the whole of U. The idea is that since f (z) blows up nicely as z → z0, we know 1 f (z) behaves
sensibly near z0. We pick some δ > 0 such that \|f (z)\| ≥ 1 for all z ∈ B(z0; δ) \ {z0}. In particular, f (z) is non-zero on B(z0; δ) \ {z0}. So we can deﬁne h(z) = 1 f (z) 0 z ∈ B(z0; δ) \ {z0} z = z0. Since \| 1 f (z) \| ≤ 1 on B(z0; δ)\{z0}, by the removal of singularities, h is holomorphic on B(z0, δ). Since h vanishes at the z0, it has a unique deﬁnite order at z0, i.e. there is a unique integer k ≥ 1 such that h has a zero of order k at z0. In other words, h(z) = (z − z0)k(z), for some holomorphic : B(z0; δ) → C and (z0) = 0. Now by continuity of, there is some 0 < ε < δ such that (z) = 0 for all z ∈ B(z0, ε). Now deﬁne g : B(z0; ε) → C by g(z) = 1 (z). Then g is holomorphic on this disc. By construction, at least away from z0, we have g(z) = 1 (z) = 1 h(z) · (z − z0)k = (z − z0)kf (z). g was initially deﬁned on B(z0; ε) → C, but now this expression certainly makes sense on all of U. So g admits an analytic continuation from B(z0; ε) to U. So done. 38 2 Contour integration IB Complex Analysis We can start giving these singularities diﬀerent names. We start by formally deﬁning what it means to be a singularity. Deﬁnition (Isolated singularity). Given a domain U and z0 ∈ U, and f : U \ {z0} → C holomorphic, we say z0 is an isolated singularity of f. Deﬁnition (Removable singular

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