Chapter 8 Answers. Lesson a. This a bipartite graph. b. 1, 4, 9 c. Number of Couples

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1 Chapter 8 Answers Lesson a. 2. a. This a bipartite graph. b. 1, 4, 9 c. Number of Couples d. 2n 1 e. H n = H n 1 + 2n 1 Number of Handshakes Recurrence Relation H 2 = H H 3 = H H 4 = H H 5 = H Couples b. 2n 2. c. H n = H n 1 + 2n 2 3. a. i. H n = H n Handshakes ii. H n = (H n 1 ) 2 iii. H n = H n 1 + n iv. H n = (H n 1 ) n b. 16, 64, 36, 5040 Chapter 8 Answers 1

2 4. a. A given number of the sequence is the sum of the previous two numbers. The last number should be = 13. b. t n = t n 1 + t n 2 5. a. 4 b. 1 c. 0 d. None 6. a. 6, 10 b. A n = A n 1 + n 1 7. a. 8. a. c. A n = d. 45 n(n! 1) 2 Term Number Number of First Differences Second Differences Handshakes The values in the second difference column are constant. b. Second degree, 2 Term Number Value First Differences Second Differences Differences , , b. 3 c. They are the same. Chapter 8 Answers 2

3 9. a. Levels Vertices b. V n = V n n 10. a. Term Number Number of Bees 0 5, , , , , b. B n = 1.12B n 1 c. After 27 years, in 2014 Chapter 8 Answers 3

4 11. a. Year Amount It takes 15 years for Susie's money to double. b. M n = (M n 1 ) a. It is one-third the length of a segment in the previous figure. b. There are four times as many. c. L n = 4 3 L n!1 Chapter 8 Answers 4

5 Lesson a. 2 b. 1 c. 2 d. This sequence was not generated by a polynomial; it was generated by an exponential function (note that each term is 3 times the previous term). 2. a. H n = 2n 2 5n b. H n = (n 1) c. H n = n 2 5n + 4 d. H n = 3 n a. T n = T n 1 + n 1 b. 0 n(n! 1) c. T n = 2 4. a. 1, 2, 11, 38, 119, 362 b a. D n = D n 1 + n 2 b. D n = n2 2! 3n 2 6. a. Row Number Number of Seats Total Seats b. S n = S n c. S n = n 2 d e. See the table in the answer to part a. f. T n = T n n 2 g. T n = n n Chapter 8 Answers 5

6 7. a. Year Value , , , , , , , , , , , , , , a. b. V n = (V n 1 ) 1.05 c. The differences never become constant. Year b. In 2058 Deer c. T n = 1.04T n 1 d. The differences never become constant. 9. A recurrence relation is S n = S n 1 + n. A closed-form formula is S n = n2 + n For a third-degree polynomial, the third differences are constant. Each third difference is 6 times the leading coefficient.. Chapter 8 Answers 6

7 Lesson a. i. Arithmetic 2. 0 ii. Geometric iii. Geometric iv. Neither v. Geometric vi. Neither b. i. H n = H n ii. H n = H n!1 2 iii. H n = (H n 1 ) 1.2 iv. H n = H n 1 + H n 2 v. H n = (H n 1 ) 0.1 vi. H n = H n 1 + n 1 c. i. H n = 2 + 3(n 1) ii. H n = 64(0.5 n 1 ) iii. H n = 10(1.2 n 1 ) v. H n = 0.3(0.1 n 1 ) 3. There is no fixed point. 4. a. C n = C n b. C n = (n 1) c. $76,500, $61, a. M n = (M n 1 ) b. M n = 5000(1.048 n ) c. $5, d. 0.4% e. M n = (M n 1 ) f. M n = 5000(1.004 n ) g. $5, h. Yearly = $5, Monthly = $5, Monthly compounding produces $17.65 more interest than yearly compounding. Chapter 8 Answers 7

8 6. 7. a. 74; 585 Year 4.8% monthly 5% yearly 0 $5, $5, $5, $5, $5, $5, $5, $5, b ; c. 3272, d. " 5n! 1 = 585 ;! 1.25n = ; "!27n = 51, After 19 years 9. a. C n = C n b. Arithmetic, 14.95, $117, Approximately 6.5% 12. a. 4 b. 67 c a , or approximately 1.4 feet b , or approximately 47.5 feet , a ( ), or about $71,300 b. A little over $22, a b. 2,318 c. After 33 hours 17. a meters b meters Chapter 8 Answers 8

9 18. a. S n = 4 3 S n 1 b. Assuming that "now" means that sales are currently $8 million, predicted sales in 5 years are 8 4 5! $ " # 3% &, or about $33.7 million. In 10 years, about $142 million. c. No is a reasonable answer because predictions are unlikely to remain accurate for 10 years. 19. a b. 200( ) 4.7 billion Chapter 8 Answers 9

10 Lesson a. $93, b. $48,000 c. $45, d. $202,107.52, $93, e. $118,789.44, $93, f. $139,605.33, $93, a , or approximately b. Approximately years. 3. a. You must show that a one-disk puzzle can be completed in = 1 moves, which is obvious. 4. a. 5. a. b. Assume that a k-disk puzzle can be completed in 2 k 1 moves and prove that a (k+1)-disk puzzle can be completed in 2 k+1 1 moves. c. The recurrence relation states that the number of moves in a (k+1)-disk puzzle is 2(2 k 1) + 1 = 2 2 k = 2 k+1 1. N (in months) T n 0 2, , , , b. M n = (M n 1 ) c. 2,443.31; 2, T(in months) T n 0 12, , , , b. A n = (A n 1 ) c. It takes 52 months. d. 40,000; 40,000; 40,000 the fixed point Chapter 8 Answers 10

11 6. a. t n t n 1 = 0.08(t n 1 70) b. t n = 0.92 (t n 1 ) c d. 70 degrees. It is the temperature of the room. 7. a. M n = (M n 1 ) b. $9,352; $8, c. 10% d. $20,000. The interest would equal the payments, and the loan amount would never decrease (1.05) = $14, a. T n = (T n 1 ) b. 1076; a. The values of k and P are 0.4 and 1125, respectively. T n T n 1 = 0.4(1125 T n 1 ), or T n T n 1 = T n 1, or T n = T n a. b It is the total population. Term Term number b. You can see that the terms are approaching 2. Chapter 8 Answers 11

12 Lesson a. 3; T n = 4(2 n 1 ) 3; b. 3.5, T n = 1.5(3 n 1 ) + 3.5, c. 5 3 ; T n = 1 3 (4 n 1 ) + 5 ; d. No fixed point; T n = 2n 6; 194! 2. a. B n = (B n 1 ) $ " # 12 % & b. c. 22,500 Month Balance ! d. 22, $ " # 12 % & n'1 e. $225, f. 472 months g. Approximately $ a. B n = (B n 1 ) b.! 22,500, or 22, $ " # 12 % & n 22,500 n B n 0 12, , , , , c d. B n = 14,140(1.01 n 1 ) + 26,000, or B n = 14,000(1.01 n )+ 26,000 e. $ f. 62 months Chapter 8 Answers 12

13 g = $16,120; $4,120 h. $ i. If Jilian borrowed this amount ($26,000), the interest would equal the payments, and the loan balance would never decrease. 4. a. (t 1 p)(a 1 1 ) + p = t 1 p + p = t 1 b. t n = (t 1 p)(a n 1 ) + p; t n+1 = (t 1 p)(a n ) + p c. t n+1 = t n (a) + b d. t n+1 = a((t 1 p)(a n 1 ) + p) + b = a(t 1 p)(a n 1 ) + ap + b = at 1 a n 1 pa a n 1 + ap + b = t 1 a n pa n + ap + b = (t 1 p)a n + a b 1! a + b ab + b! ab = (t 1 p)a n + 1! a = (t 1 p)a n + p 5. a. T n = 0.92(T n 1 ) + 5.6; The fixed point is 70. b. T n = 100(0.92 n ) + 70 c degrees d. Minutes Temperature e. No, because solving the equation requires taking the log of a negative number. 6. a. P n = 0.5(P n 1 ) b. 400 c. P n = 320(0.5 n 1 ) d. 399 people e. As with the previous disease problem, the fixed point of the recurrence relation equals the population. Chapter 8 Answers 13

14 7. a. The following table shows the growth of the account. Age Amount Chapter 8 Answers 14

15 The amounts are generated with the mixed recurrence relation A n = 1.07(A n 1 ) The final balance is approximately what is claimed in the article. Chapter 8 Answers 15

16 b. The following table tracks the account for 20 years Age Amount Chapter 8 Answers 16

17 The amounts are generated with the mixed recurrence relation A n = 1.07(A n 1 ) + 1,000. The 20-year balance is about what is claimed in the article.! c. 214, $ " # 12 % &, or about $1,427. Chapter 8 Answers 17

18 8. The following table shows the growth in Plan A. Age Amount 21 $2, $4, $6, $9, $11, $13, $14, $15, $16, $18, $20, $21, $23, $25, $28, $30, $33, $36, $39, $43, $47, $51, $56, $61, $67, $73, $79, $86, $94, $103, $112, $122, Chapter 8 Answers 18

19 53 $133, $145, $158, $173, $188, $205, $224, $244, $266, $290, $316, $344, $375, The first five years are generated with the mixed recurrence relation A n = 1.09(A n 1 ) + 2,000; the remaining years with the geometric recurrence relation A n = 1.09(A n 1 ). Chapter 8 Answers 19

20 The following table tracks the account in Plan B. Age Amount 45 $2, $4, $6, $9, $11, $15, $18, $22, $26, $30, $35, $40, $45, $52, $58, $66, $73, $82, $92, $102, $113, The amounts are generated with the mixed recurrence relation A n = 1.09(A n 1 ) + 2,000. Sasha is wrong. Plan A has over three times the amount that Plan B has. 9. If a = 1, the recurrence relation is arithmetic. An arithmetic sequence does not have a fixed point (unless the constant difference is 0). The closed form for the recurrence relation can be found by applying previous results for arithmetic sequences. Chapter 8 Answers 20

21 Lesson a. 4, repelling b. 2.5, neither c. 6, attracting d. 12, attracting 3. Attracting when a < 1, repelling when a > 1 Chapter 8 Answers 21

22 4. a. 1 b. 3 c. n t n d. t n = 2t n e. The fixed point is 3. It can be found algebraically by solving the equation x = 2x a. The population can be modeled by the recurrence relation P n = 1.04P n 1 + b, b where b is the annual harvest. The fixed point is = 25b. Thus, the 1! 1.04 closed form is ( b)1.04 n 1 25b. Setting this equal to 10,000, assigning n the value 10, and solving for b gives approximately 669. Thus issuing 670 permits annually for 10 years would bring the population to about 10,000 in 10 years, assuming the each permit is fulfilled. 6. a. b. Approximately 400 permits a year would hold the population constant. c. The plans seems reasonable assuming the nearly all permits are fulfilled. However, the decrease from 670 to 400 permits at the end of 10 years could prove unpopular with hunters. n t n Chapter 8 Answers 22

23 b. c. The behavior is unpredictable. d. 2 is a fixed point. e. The terms diverge. f. 2, 4 g. Attracted when t 1 = 0 or 2; repelled when t 1 = 5; unpredictable when t 1 = 1 " " 7. a. t n = ! t n!1 % % # $ # $ 10 & ' & ' t n!1 b. Approaches 10,000 but never reaches it c. The population approaches 10,000. d. 0 and 10,000 are the fixed points. Chapter 8 Answers 23

24 Chapter 8 Review 1. A reasonable summary should include the following points: A description of recurrence relations, perhaps with a few examples, including recursive and closed-form representations. A discussion of finite difference techniques for finding polynomial closed forms. Arithmetic and geometric recurrence relations, including closed forms. Mixed recurrence relations, including fixed points and their role in closed forms. Cobweb diagrams, repelling and attracting fixed points. Important applications, particularly financial applications. 2. a. H n = H n 1 + 4; H n = 2 + 4(n 1); 398 b. H n = 3H n 1 1; H n = 5 2 3n ; c. H n = (H n 1 ) 2; H n = 3(2 n 1 ); d. H n = 2H n 1 + 3; H n = 2 n 3; a. Arithmetic b. Neither c. Geometric d. Neither 4. a. 0; 0, 0, 0, 0 b. 0.75; 0.75, 0.75, 0.75, 0.75 c. No fixed point d. 1; 1, 1, 1, 1 5. a. H n = 2(5 n 1 ); b. H n = 2.75(5 n 1 ) 0.75; c. H n = 2 + ( 3)(n 1); 295 d. H n = ( 2) n 1 + 1; a. Geometric b. Neither c. Arithmetic d. Neither 7. a. O n = 1.03 O n 1 b. C n = C n 1 Chapter 8 Answers 24

25 8. a. c. n 3 N S n First Differences Second Differences Third Differences n2 2 + n 6 ; Second degree; H n = n a. b. G n = G n 1 + n Day Gifts That Day Total Gifts T n = T n 1 + n2 2 + n 2 c. G n = n2 2 + n 2 T n = n3 6 + n2 2 + n a. P n = P n a. b. P n = (n 1) Month b. B n = 1.004(B n 1 ) Balance 0 $1,000 1 $1,004 2 $1, $1, Chapter 8 Answers 25

26 c. B n = 1000(1.004) n d. Solve the equation 1000(1.004) n = 2000, which gives 14 years, 6 months. 13. a. Month Balance 0 $5,000 1 $5, $5, $5, " b. B n = (B n!1 ) % # $ 12 & ' + 100! c. B n = 23, $ " # 12 % & d. $13, e. 200 months 14. a. C n = C n b. C n = n c ! n = 4 or $ a. B n = 1.008(B n 1 ) 230 ' 18,750 b. B n = 17,750(1.008) n + 28,750 c. 61 months d. $3,030 e. $ a. R n = 0.9R n 1 b. R n = 1000(0.9 n ) c. About 6.58 minutes or 6 minutes, 35 seconds 17. a. V n = 1.5V n 1 + 4,000 b. V n = 16,000(1.5) n Chapter 8 Answers 26

27 a. A n = 0.4A n b. The amount of medication in the body stabilizes at 833 mg. Chapter 8 Answers 27

28 1000 c. The cobweb would be attracted to the point (833.33, ), which is the intersection of y = x and y = 0.4x + 500, as shown in this figure d. The amount in the body reaches the stable value (833 mg, in this case) more quickly. The stable value is probably near the optimal dosage of the drug. 20. a. C n = 0.8C n b. The daily concentration sequence is 2, 2.6, 3.08, 3.464,.The concentration quickly exceeds the recommended maximum of 3 ppm and gradually approaches 5 ppm. c. Sample answer: a daily addition of 0.4 ppm. The recurrence relation C n = 0.8C n has a fixed point of 2. Daily additions of 0.4 ppm keep the concentration at a constant 2 ppm, which is the middle of the recommended range. Chapter 8 Answers 28