Business Calc Homework w answers_Part_74

Business Calc Homework w answers_Part_74 - 366 Section 9.1...

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Unformatted text preview: 366 Section 9.1 8. (a) We graph the points n, 1 1 2n for n 2n 1, 2, 3, … . 2. (a) Note that a0 1 ,a 32 1, a1 1n . 3 an 1 , and so on. Thus 9 1, a2 1 ,a 23 (c) Note that a0 5, a1 0.5, a2 5(0.1)n 5 . 10n (b) Note that a1 ( 1)n n an 1 1 , and so on. Thus 3 . [0, 23.5] by [ 2, 2] 1 n→ 1 (b) lim an 2n 2n lim n→ 2 2 1 an 1 for n n 9. (a) We graph the points n, 2 1, 2, 3, … . 0.05, and so on. Thus 1n1 alternate 2 3. Different, since the terms of ∑ n1 between positive and negative, while the terms of 1n1 are all negative. 2 ∑ n1 4. The same, since both series can be represented as [0, 23.5] by [ 1, 3] (b) lim an 1 n lim 2 n→ n→ 10. (a) We graph the points n, 1 2 1 2 ln (n 1) for n n 1, 2, 3, … . 1 4 …. 1 8 5. The same, since both series can be represented as 1 2 1 1 4 …. 1 8 6. Different, since ∑ ∑ n [0, 23.5] by [ 1, 1] 1 ln (n 1) lim n n→ (b) lim an n→ lim n→ (n 1) 1 u4 and 1 u1 1, 1 , so 16 u1 1, u2 un n2, or * ( 1) 2n 1 1 2n 3 1 1 8 4, u3 11 , 4 u3 1 u2 9, and u4 1 , 9 9. Converges; ∑ n0 5 4 2n 3 5 4 1 5 4 3 2 3 15 4 10. Diverges, because the common ratio is 1 and the terms do not approach zero. 11. Diverges, because the terms alternate between 1 and and do not approach zero. 16. We may write n2. 12. Converges; ∑ 3( 0.1)n 1 n0 3 ( 0.1) 30 11 3 … (b) Let un represent the value of * in the nth term, starting with n 0. Then 1 u3 u3 1 u0 1, 1 , so u0 16 1, u1 16. We may write un (c) If * ( 1)3 ( 1)6 11 , 4 u2 1 u1 4, u2 (n 1 1 1 4 2 ( 1)4 n0 1)2, or * (n 1)2. 1 2 ( 1)5 1 3 3. 1 1 1 2 2( (2 2 1 2 1 1 2 n0 1 2 n 1 2 1 2 ∑ 2 n 4 9, and … , which is the same as the desired series. Thus let * 13. Converges; ∑ sin n 1 , and 9 3, the series is 2 … but 3 2 3 1 1 4 8. Diverges, because the terms do not approach zero. 1. (a) Let un represent the value of * in the n th term, starting 1. Then 1 n0 Section 9.1 Exercises with n n 7. Converges; ∑ 0 1 n1 1n1 1 1 2 2 1 1 1 …. 2 4 8 2 2 2 1)( 2 1) 2 1) 2 2 1 2 2 1 14. Diverges, because the terms do not approach zero. 1 Section 9.1 e 15. Converges; since 1, ∑ 0.865 16. Converges; ∑ n0 6n 17. Since ∑ 2 x n0 n0 1 1 function f (x) 1 2x 18. Since ∑ ( 1)n(x 1 2x ∑[ 1)n n0 (c) The partial sums alternate between positive and negative while their magnitude increases toward infinity. 1 . 2 x 1)]n, the series (x 24. Since ∑ n0 converges when (x 1) 1 1 , the series represents the 1 tan x 1 , k x k. 4 4 tan x (b) The partial sums are alternately 1 and 0. , the series represents the 1 2 , k , where k is any integer. Since the 4 23. (a) Since the terms are all positive and do not approach zero, the partial sums tend toward infinity. 11 , . Since 22 1 and the interval of convergence is the sum of the series is x function f (x) n n0 k sum of the series is 1 5 6 1 n0 1. Thus, the series converges for 4 ∑ (2x) , the series converges when nn 2x 1 tan x 1 6 15n 66 ∑ 5n ∑ (tan x)n, the series converges when n0 e 1 n1 e 22. Since ∑ tan n x 1 en ∑ en ne n0 1 and the interval of n e e , this is a geometric series with n0 e common ratio r 1.03, which is greater than one. e convergence is ( 2, 0). Since the sum of the series is 1 [ (x 1 1 1)] 1 f (x) x 2 x , 2 2 25. , the series represents the function x ∑ xn 20 n0 1 0. 1 20, x x 1 1n (x 2 19. Since ∑ n0 converges when ∑ 3)n 2 n0 3 x 2 xn 3 20 20x 19 20 1 and the interval of convergence 1 x 1 . 2 2 the series represents the function f (x) x) x 1 ,1 x 5. For any real number a use a 2 a 4 1 2 use 1 20. For ∑ 3 n0 x 1n x 2 a 16 1 8 ∑ 2r n 1… . 32 1 1 3 x 5r , the series 3 x , 1 x Series: ∑ 2 3. n1 21. Since ∑ sin x n n0 sin x ∑ (sin x) , the series converges when n n0 1. Thus, the series converges for all values of x except odd integer multiples of 2 , that is, x for integers k. Since the sum of the series is series represents the function f (x) (2k 1) . 2 3 5 1 1 1 , sin x (2k 1) (b) ∑ n1 3n1 5 13 2 13 n 1 r 2 1 r 13 2 1 5 5r 5r r Series: n1 13 2 5r 5, r 2 1 , the sin x 1 3 2 6 represents the function f (x) 1) 5 r 6 3 (x 1: 5, r r 2 Since the sum of the series is x … . To get 0, a 32 1 16 2 1 n1 1 and the interval of convergence is ( 1, 3). 2 a 8 1 4 0, 27. Assuming the series begins at n , the series converges when (a) 1 20x 26. One possible answer: 2 1 (3 1 19 x , the series is (1, 5). Since the sum of the series is 367 3n1 10 3 2 3 10 368 Section 9.1 28. Let a 0.21 21 and r 100 0.21 1 , giving 100 33. 1.414 0.21(0.01)2 0.21(0.01) 1 0.414 0.414(0.001)2 0.414(0.001) ∑ 0.414(0.001)n 1 n0 … 0.21(0.01)3 ∑ 0.21(0.01)n 1 n0 157 111 0.21 1 0.01 0.21 0.99 7 33 29. Let a 0.234 34. 1.24123 1 , giving 1000 0.234(0.001) 0.234(0.001)3 0.234(0.001)2 … 124 100 41,333 33,300 0.234 1 0.001 0.234 0.999 26 111 0.7(0.1) 35. 3.142857 0.7(0.1)2 0.7(0.1)3 3 3 3 36. Total distance …) 0.142857 ∑ 0.000001n (0.142857) 1 1 0.000001 0.142857 0.999999 1 7 0.06(0.1) ∑ 0.06(.1)n 4(0.6)2 2[4(0.6) 4 2 4(0.6)3 …] 2 ∑ 2.4(0.6)n 4 …] 4 4 0.13 26 n0 ∑ 0.06 1 0.1 0.06 0.9 1 15 0.000001 22 7 d [1 0.1 0.12 10 d (0.1)n 10 n 0 d 1 10 1 0.1 d1 10 0.9 d 9 n0 0.142857(1 n0 … 0.7 1 0.1 0.7 0.9 7 9 0.06 3 3 ∑ 0.7(0.1)n 32. 0.06 … 0.00123 1 0.001 n1 0.00123 0.999 41 33,300 0.0000012 n0 31. 0.d 0.00123(0.001) ∑ 1.24 ∑ 0.234(0.001)n 0.7 0.00123 1.24 n0 30. 0.7 1.24 0.00123(0.001)2 234 and r 1000 0.234 0.414 1 0.001 46 111 1 1 2.4 0.6 16m 37. Total time 0.06(0.1)2 0.06(0.1)3 4 4.9 … 4(0.6)2 4.9 4(0.6) 4. 9 2 4(0.6)3 4.9 … 4 4.9 2 4(0.6) [1 4. 9 4 4.9 2 4(0.6) 4.9 7.113 sec 0.6 1 1 0.6 ( 0.6)2 …] 369 Section 9.1 38. The area of each square is half of the area of the preceding square, so the total of all the areas is ∑ 4 1 8 m 2. 1 2 1n 2 n1 n1 ∑ 2 ∑ 45. Comparing a a S a ar n 1r 41. Using the notation Sn ar a 1 4 1, then lim r n n→ ar 2 ar ar 3 1 r 4) , the first term is a (x 4). … 4)2 (x a ar n 1r n→ a 1 r ∑ ar n 4 1 3x a 1 r the common ratio is r 3x 9x 1 … 4)n 1, so the 1) with a 1 r , the leading term is a ( 3x) . 1 and (x 1 (x 4 1) … 1)2 x 2x with the common ratio is r 2x 2 4x 3 3x 1 2 x as (x 1 (x 1 1) a r , the first term is a 11 ,. 22 1, so the 2n 1 x n a 1 1 and the common ratio is r … r x , … 1) (x 1) Rewriting 1 2 2 (x 1) 1 n 1. 1, so the 1 1 x 21 and comparing with x 2 1 and the common ratio is r 2 the first is a 1 2 as 1 x 4 12 x 8 … a 1 r , x . 2 … 1n x 2n 1 x and 48. 1 eb e 2b e 3b … x 2 1, so the interval ∑ (e b)n n0 1 Interval: The series converges when 2x of convergence is 1 … of convergence is ( 2, 2). 2x. … x. 1)n and comparing with 1, so the 11 ,. 33 1 1 1 ( 1)n(x 4 Interval: The series converges when 1 1) interval of convergence is (0, 2). Series: … n Interval: The series converges when interval of convergence is , the first term is Alternate solution: 3x. … 2 1 (x Interval: The series converges when x 1 1, then the nth partial sum is na, which goes to with 1 (x 4 Series: 1 . a ar n has no finite limit and 1r n 1 1 41 The first term is a 1 n1 diverges. Series: x ( 1)n(x 1 and the common ratio is r 4 47. Rewriting 1, then r n has no finite limit as n → , so the expression 43. Comparing 1 ar n 1, ... a ar . 1r 0 and so ∑ ar n lim 1 or r Series: 1 a with interval of convergence is (0, 2). the formula from Exercise 40 is Sn 42. Comparing 4) Interval: The series converges when x n If r 1, so the interval Series: n rS S (1 n→ (x 46. Comparing r: ar n r) S If r … 3x 3n interval of convergence is (3, 5). (a ar ar 2 ar 3 ... ar n 2 ar n 1) (ar ar 2 ar 3 ar 4 ... ar n 1 ar n) a ar n a lim Sn … Interval: The series converges when x (b) Just factor and divide by 1 If r 3 and x 3. 3x 6 1 (x 1 Series: 1 1 2 2 rS , the first term is a and the common ratio is r 4 40. (a) S r of convergence is ( 1, 1). 1n 42 1 a 1 Interval: The series converges when x 3 12 2n 1 2 n0 3x 3 Series: 3 ∑ 2n 39. Total area with x3 1 the common ratio is r 2 n0 4 3 44. Comparing 1n … 1, so the interval 1 eb 9 1 9 9e b 8 eb 8 9 b ln 9e b 8 9 ln 8 ln 9 370 Section 9.1 49. (a) When t n0 1 1 t. x (1 1 t t 1 t, which is true when t For t 1 . 2 (1 ∑ n0 2(1 n t 1 10 when t 1 1 t t 1 1 (1 t t) 1 t t, so 4 1 t2 with a 1 r , the first term First four terms: 4 4t n 4t (1 4t 2n 0, so the constant term of the power series for G (x) will be 0. Integrate the terms for f (x) to obtain the terms for G (x). 43 45 47 x x x 3 5 7 4 General term: ( 1)n x 2n 1 2n 1 First four terms: 4x (c) The series in part (a) converges when 2 x) C. 4x 3 … nx n x) d (1 dx 2 2x 3x 2 4x 3 … 3 x) 0 2 2 2 …) 12x 2 6x n (n 1)x n ... … 12x 2 6x 1)x n n (n (n …. 1)x n 2)(n 1, so the interval (b) No, because if you integrate it again, you would have the original series for f, but by Theorem 2, that would have to converge for 2 x 2, which contradicts the assumption that the original series converges only for 1 x 1. t2 1, so the lim an. Then by definition of convergence, for n→ there corresponds an N such that for all m and n, n, m N ⇒ am L 2 and an L 2 Now, 1, which result in the am an 4 4 3 4 5 4 7 … ( 1)n 4 2n 1 … and G ( 1) 4 4 3 respectively. 4 5 4 7 … ( 1)n 4 1 2n 1 …, whenever m am L am convergent series G (1) …. 53. (a) No, because if you differentiate it again, you would have the original series for f, but by Theorem 1, that would have to converge for 2 x 2, which contradicts the assumption that the original series converges only for 1 x 1. 54. Let L interval of convergence is ( 1, 1). (d) The two numbers are x 2 … 2 2, this may be written as 12x 2 6x 1 … 1 of convergence is ( 1, 1). 6 General term: ( 1) (4t ) (b) Note that G (0) dx Interval: The series converges when x t 2. 4 and the common ratio is r 4 3 3x 2 Replacing n by n t 2 x) 2x 1 Thus, f (x) f (x) is a 2 x) 9. 50. (a) Comparing f (t) … 4 nx n 1 , we have 2 (c) For t 2(1 d (1 dx 1 . 2 Thus, S converges for all t 1)4 (x 1 dt t 1 Using the result from Example 4, we have: t, which is always true. S 1)2 (x 1)3 2 3 ( 1)n(x 1)n n (x 1 f (x) dx 1 (x 52. To determine our starting point, we note that 0, the inequality is equivalent to 0, the inequality is equivalent to t x 1) 1)2 … 1)3 (x 1)n t), which is always false. For (x … , we may write ln x 1 1, this inequality is equivalent to t S 1 x ( 1)n(x 1, or t t 51. Since 2. 1 2 1 t (b) S converges when For t 1 1n 2 ∑ 1, S L L N and n an an L N. 2 2 = . 2 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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