SalasSV_11_07_ex_ans

# SalasSV_11_07_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES 2...

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES 2 A-79 Now, (d) 0 f (k ) ( x ) c e−1/x is a sum of terms of the form , where n is a positive integer and c is a constant. By part (b), f (k +1) (0) = 0. Therefore f (n) (0) = 0 x xn for all n. (e) x = 0 ∞ 65. P2 (x) = x − 1 x2 , P3 (x) = x − 1 x2 + 1 x3 , P4 (x) = x − 1 x2 + 1 x3 − 1 x4 , P5 (x) = x − 1 x2 + 1 x3 − 1 x4 + 1 x5 2 2 3 2 3 4 2 3 4 5 67. k =0 (ln 3)k k x ; R=∞ k! SECTION 11.6 1. P3 (x) = 2 + 1 (x − 4) − 4 R3 ( x ) = 5. P3 (x) = 1 (x 64 − 4)2 + 1 (x 512 − 4)3 3. P4 (x) = √ √ 2 2 π + x− 2 2 4 cos c π x− 120 4 5 √ − 2 π x− 4 4 2 √ − 2 π x− 12 4 3 √ + 2 π x− 48 4 4 −5 (x − 4)4 , 128c7/2 |c − 4| < |x − 4| R4 ( x ) = R3 (x) = , c− π π < x− 4 4 π 1 1 1 + (x − 1) − (x − 1)2 + (x − 1)3 4 2 4 12 (−∞, ∞) 13. ∞ c(1 − c2 ) (x − 1)4 , (1 + c2 )4 |c − 1| < |x − 1| (−∞, ∞) 7. 6 + 9(x − 1) + 7(x − 1)2 + 3(x − 1)3 , ∞ 9. −3 + 5(x + 1) − 19(x + 1)2 + 20(x + 1)3 − 10(x + 1)4 + 2(x + 1)5 , ∞ k =0 11. k =0 ∞ ( − 1)k 1 2 k +1 (x − 1)k , ( − 1, 3) 1 5 2 5 π 2 k (x + 2)k , 2k 91 −, 22 ( − ∞, ∞) ∞ 15. k =0 (−1)k +1 ( x − π ) 2 k +1 , (2k + 1)! ∞ (−∞, ∞) 2 3 k 17. k =0 (−1)k +1 (x − π )2k , (2k )! (−∞, ∞) ∞ 19. k =0 (−1)k (2k )! (x − 1)2k , ∞ 21. ln 3 + k =1 ∞ (−1)k +1 k 15 (x − 1)k , − , 22 23. 2 ln 2 + (1 + ln 2)(x − 2) + k =2 ∞ ( − 1)k (x − 2)k k (k − 1)2k −1 ∞ 25. k =0 ( − 1)k 2k +2 x (2k + 1)! 27. k =0 (k + 2)(k + 1) 2 k −1 (x + 2)k 5k +3 29. 1 + k =1 ( − 1)k 22k −1 (x − π )2k (2k )! ∞ k =0 31. k =0 n! (x − 1)k (n − k )!k ! ∞ 33. (a) ex = e x −a = ea ∞ k =0 (x − a)k , k! k ex =ea k =0 (x − a)k k! (b) ea+(x−a) = e x = e a ∞ k =0 (x − a)k , k! e x 1 +x 2 =e x1 ∞ k =0 k x2 = e x1 e x2 k! (c) e−a ( − 1)k (x − a) k! 35. (a) P3 (x) = 39. P6 (x) = √ 1 π 1 3 π + (x − ) − x− 2 2 6 4 6 2 √ − 3 π x− 12 6 3 (b) 0.5736 37. P2 (x) = 6 + 1 (x 12 − 36) − 1 (x 1728 − 36)2 ; √ 38 ∼ 6. 164 = π 1 1 1 1 1 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)5 + (x − 1)6 4 2 4 12 40 48 SECTION 11.7 1. (a) absolutely convergent 11. {0} 13. [− 1 , 1 ) 22 (b) absolutely convergent 15. (−1, 1) 27. (−4, 0) 39. (c) ? (d) ? 3. (−1, 1) 5. (−∞, ∞) 7. {0} 9. [−2, 2) 17. (−10, 10) 29. (−∞, ∞) 19. (−∞, ∞) 31. (−1, 1) 21. (−∞, ∞) 33. (0, 4) 23. (−3/2, 3/2) 25. converges only at x = 1 35. (− 5 , 1 ) 22 43. (a) (b) If 37. (−2, 2) |ak (−r )k | 1 1 −√ , √ 3 3 41. (a) absolutely convergent 45. (a) (b) absolutely convergent (c) ? ak x k = (a0 + a1 x + a2 x2 )x3k , a geometric series with | x3 | < 1 implies |x| < 1 a0 + a 1 x + a 2 x 1 − x3 2 |ak r k | = |ak (−r )k | converges, then ak (−r )k converges. a = a0 + a1 x + a2 x2 and r = x3 ; (b) ak xk = 47. Examine the convergence of 49. (a) R = ∞ (b) R = 1 2 |ak xk |; for (a) use the root test and for (b) use the ratio test. (c) R = 2 51. (c) ( − 1, 1) ...
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## This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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