Business Calc Homework w answers_Part_77

Business Calc Homework w answers_Part_77 - Section 9.4 30....

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Unformatted text preview: Section 9.4 30. (a) sin2 x 1 (1 cos 2x) 2 1 1 (2x)2 1 2 2 2! 381 (b) e ax cos bx dx e(a a a2 a a2 a2 e a2 ax i e ax sin bx dx (2x)4 (2x)6 ... 4! 6! (2x)2n ... ( 1)n (2n)! 256x 8 4x 2 16x 4 64x 6 2 8! 2 2! 2 4! 2 6! 1024x10 ... 2 10! x4 2x 6 x8 2x10 ... x2 3 45 315 14,175 bi)x dx eax bi (a bi)x e b2 bi ax e (cos bx b2 b2 i sin bx) b sin bx ai sin bx) bi cos bx (a cos bx b2 [(a cos bx b sin bx) b cos bx)] (b) derivative (c) part (b) 2x 2x 4x 3 (2x)3 3! 3 4x 15 (2x)5 5! 5 8x ... 315 (2x)7 ... 7! 7 i(a sin bx sin 2x Separating the real and imaginary parts gives e ax cos bx dx e ax sin bx dx e ax (a cos bx a2 b2 ax e (a sin bx a2 b2 31. (a) It works. For example, let n 2. Then P 3.14 and P sin P 3.141592653, which is accurate to more than 6 decimal places. (b) Let P x where x is the error in the original b sin bx) and b cos bx) estimate. Then P sin P ( x) sin ( x) sin x x sin x x3 . 6 s Section 9.4 Radius of Convergence (pp. 487496) Exploration 1 Test 1. For 1 :L n 1 n But by the Remainder Theorem, x Finishing the Proof of the Ratio 1 Therefore, the difference between the new estimate P 32. (a) ei 2 cos 2 cos 2 i sin 2 sin P and e i x3 is less than . 6 i sin ) (cos ( ) i sin ( 2 cos i sin )) lim n n 1 n 1 lim n n n 1 n2 (n 1. 1 (cos For 2 : L 2. (a) (b) 1 dx x 1 dx x2 lim n (n 1 n2 1)2 lim n 1)2 1. cos i sin ) i sin ) 2i (cos ( 2i (cos ) i sin ( )) 1 lim ln x k k 1 1k lim ln k k . 1 k (b) ei e 2i i (cos (cos 2i sin 2i 1 lim k x 1 1 n lim k 1 1. 1 dx. Since x 1 dx. x2 i sin ) 3. Figure 9.14a shows that is greater than the integral diverges, so must the series. Figure 9.14b shows that 1 is less than 1 n2 1 sin d 33. [eax(cos bx dx ax 1 i sin bx)] bi cos bx) bi cos bx) i sin bx) (aeax)(cos bx a(cos bx a(cos bx i sin bx) Since the integral converges, so must the series. i sin bx)] i sin bx)] 4. These two examples prove that L 1 can be true for either a divergent series or a convergent series. The Ratio Test itself is therefore inconclusive when L 1. (e )( b sin bx (eax)[(bi2 sin bx (eax)[bi(cos bx (a (a ax bi)(e )(cos bx bi)e (a bi)x i sin bx) Exploration 2 1. L lim n Revisiting a Maclaurin Series 1 n xn x n n 1 lim n n n 1 x x . The series 34. (a) The derivative of the right-hand side is a a2 bi (a bi)e(a bi)x b2 a 2 (bi)2 (a bi)x e a2 b2 a 2 b 2 (a bi)x e e(a bi)x, a2 b2 converges absolutely when x convergence is 1. 2. When x 1 1 2 1 3 1, so the radius of 1, the series becomes ... 1 .... n 1 n Each term in this series is the negative of the corresponding term in the divergent series of Figure 9.14a. Just as diverges to , this series diverges to . which confirms the antiderivative formula. 382 Section 9.4 Section 9.4 Exercises 1. Diverges by the nth-Term Test, since lim n 3. Geometrically, we chart the progress of the partial sums as in the figure below: 2 +3 1 4 1 +5 1 1 n n 2n 1 1 1 0. . (The 2. Diverges by the nth-Term Test, since lim Ratio Test can also be used.) 1 etc. 1 1 1 2 + 3 n n 12 1 L 3. Converges by the Ratio Test, since lim n an an 1 lim n (n 2 1) 2 n 1 1 n 2 2n 1 1 1 1 1 2 + 3 4 1 2 1. 1 , 8 4. The series converges at the right-hand endpoint. As shown in the picture above, the partial sums are closing in on some limit L as they oscillate left and right by constantly decreasing amounts. 5. We know that the series does not converge absolutely at the right-hand endpoint, because of this section). 1 diverges (Exploration 1 n 4. Converges, because it is a geometric series with r so r 1. 5. Converges by the Ratio Test, since 3n 1 2 1. 3 2n n n 2 Alternately, note that n for all n. 3 3 1 n n 2 2 Since converges, converges by the Direct n 1 n 1 3 n 1 3 lim an 1 an n n (3 lim 2n 1 1 1) 2n Quick Review 9.4 1. lim n nx n 1 3 1) x lim n n n 1 Comparison Test. x n2 n 2 2. lim n2 x n n(n xn n! 6. Diverges by the nth-Term Test, since n x 3 lim n x 3 lim n sin n 1 n 1 0 3. lim n 0 7. Converges by the Ratio Test, since lim n (Note: This limit is similar to the limit which is discussed at the end of Example 3 in Section 9.3.) (n 1) 4 x 2 (2n)4 2 an an 1 lim n (n 1)2e n 2e n n 1 e 1 1. 8. Converges by the Ratio Test, since 4. lim n x lim n n4 4n 3 x 16 2 6n 2 16n 4 4n 1 lim n an an 1 x 2x 1 5. lim n 1 2x n 2 n 1 n 2 1 (n 1)10 10 n 1 n lim 10 n n10 1 10 1. 16 9. Converges by the Ratio Test, since 2x 2 1 2 1n lim n 2x 2 1 lim n an an 1 6. Since n 2 7. Since 5 n 5n for n n for n ln n for n 5 6, an 6, an 1, an n 2, bn 5 , bn n 5n, and N n and N 5 6. 6. 8. Since n N 1. n n, bn 1 for n n! ln n, and (n 4)! 3!(n 1)!3n n 4 lim 1) n 3(n 1 1. 3 lim n 1 3!n! 3n (n 3)! 10. Diverges by the nth-Term Test, since 25, lim 1 n 1 n! and hence n 9. Since 10 10 1 1 an ,b , and N 25. 10 n n n! 1 n n e 0. 2 , 3 11. Converges, because it is a geometric series with r n 3 10. Since n 2 n 2, an n 3 and hence 1 ,b n2 n 1 n2 for so r 1. n 3 , and N 2. 12. Diverges by the Ratio Test, since lim n an an 1 lim n (n 1)!e n!e n n 1 lim (n n 1)e 1 . (The nth-Term Test can also be used.) Section 9.4 13. Diverges by the Ratio Test, since lim n 383 an an 1 3n 1 lim 1)3 2n n (n 3n 3 lim (n 1)3(2) n 3 1. 2 1 n32 n 3n 22. lim n an an 1 lim n 3x 2 n n 1 2 3 1 n 3x 2n 3x 2 3 2 The series converges for 3x diverges for x 1 , and 3 1 1 , so the radius of convergence is . 3 3 2 1, or x 23. This is a geometric series which converges only for x 10 2 (The nth Term Test can also be used.) 1, or x 2 10, so the radius of convergence is 14. Converges by the Ratio Test, since lim n 10. n an an 1 lim n (n 1 lim n 2 1 1. 2 1)ln (n 1) 2 2n 1 n ln n n 1 ln (n 1) n ln n 24. lim n an an 1 lim n (n n 1) x n 3 1 n 2 n xn lim x n x 1, so The series converges for x 1 and diverges for x the radius of convergence is 1. an an xn (n x lim n 3 n 1 15. Converges by the Ratio Test, since lim n 25. lim n 1 lim n 13 n 1 an an 1 (2n 1)! 1)! n! 3)! n 1 lim 3)(2n 2) n (2n 1 lim 0 1. 3) n 2(2n (n n (2n lim 1) n x 3 n 3n xn The series converges for x 3 and diverges for x 3, so the radius of convergence is 3. 26. lim n 16. Converges by the Ratio Test, since lim n an an 1 lim n lim n lim n lim n (n 1)! n (n 1)n 1 n! (n 1)n n (n 1)(n 1)n n n n 1 1 (1 1/n)n n an an 1 x 2n 3 1)! n (n lim n! x 2n 1 lim n x2 n 1 0 The series converges for all values of x, so the radius of convergence is . an an (n x 5 1) x 5n 3 3n 1 1 27. lim n 1 lim n 5n nx 3n 1 e 1 lim n x 5 3 The series converges for x 17. One possible answer: 1 diverges (see Exploration 1 in this section) even n 1 n 1 though lim 0. n n 3 5 and diverges for x 3 an an 5, so the radius of convergence is 5. (n 1) x n [(n 1)2 n 4 x x lim 4 n 4 1 28. lim n 1 lim n 1 1] 4n(n2 1) nxn 18. One possible answer: Let an a 2 n and bn 3 n Then an and bn are convergent geometric series, but The series converges for x 4 and diverges for x 4, so bn n 3 n is a divergent geometric series. 2 the radius of convergence is 4. 1, an an n 3 1xn n 1 1 19. This is a geometric series which converges only for x so the radius of convergence is 1. 20. This is a geometric series which converges only for x 5 1, so the radius of convergence is 1. 21. This is a geometric series which converges only for (4x 1) 1, or x 1 4 1 4 1 , so the radius of 4 29. lim n 1 lim n 3n n x n lim n x 3 x 3 The series converges for x 3 and diverges for x 3, so the radius of convergence is 3. convergence is . 384 Section 9.4 an an (n 1)! x 4 n n! x 4 n 1 30. lim n 1 lim n 35. This is a geometric series with first term a common ratio r (x 4 1) 2 1 and (x 4 1)2 . It converges only when lim (n n 1) x 4) 4 1, so the interval of convergence is 3. a 1 r 1 4 x2 x2 1 (x 4 1)2 (x 1 4, so the radius of Sum x The series converges only for x convergence is 0. an an ( 2) n 1 31. lim n 1 lim n (n 2) x 1 2n (n 1) x 1 n n 1 4 (x 1)2 lim 2 x n 1 4 2x 4 2x 3 2x 1 1 3 The series converges for x x 1 an an 1 and diverges for 2 1 1 , so the radius of convergence is . 2 2 36. This is a geometric series with first term a common ratio r (x 9 1)2 (x 9 1)2 1 and . It converges only when 1, so the interval of convergence is 2. a 1 r 1 32. lim n 1 lim n 4x (n 5 1)3/2 2n 3 n 4x 3/2 5 2n 1 4 Sum x lim (4x n 5)2 (4x 5)2 5) 2 The series converges for (4x to 4x x 5 4 an an 1 1, which is equivalent 1 9 (x 9 1)2 5 1 and diverges for 4 1 1 . The radius of convergence is . 4 4 1, or x 5 4 9 (x 1)2 9 2x x2 9 2x 8 x2 8 33. lim n lim n x n n 1 n x n 37. This is a geometric series with first term a common ratio r x 2 x 2 1 and 1 lim x n 1. It converges only when x 0 The series converges for x x an an 1 x 16. a 1 1, so the interval of convergence is 1 and diverges for Sum r 1 1 x 2 2 1 4 x 1, so the radius of convergence is 1. x 2n 2 2n 1 3 34. lim n 1 lim n 2n x 2 2n 1 38. This is a geometric series with first term a common ratio r 1 and 1, 1 (x 2)2 n 2 1 (x 2)2 2 1 The series converges for (x 2 ln x. It converges only when ln x 1 e lim so the interval of convergence is Sum 2)2 1, which is a 1 r 1 1 ln x x e. 39. This is a geometric series with first term a common ratio x2 3 1 1 and x2 3 1 equivalent to x x 2 2 2, and diverges for 2. . It converges only when 2 3 (x 2 1, 2. The radius of convergence is so the interval of convergence is Sum a 1 r 1 1 x2 3 1 x 2. 3 3 1) 4 x2 Section 9.4 40. This is a geometric series with first term a sin x sin x common ratio . Since 2 2 385 1 and 46. 1 for all x, the interval . 2 sin x n 1 (2n s1 s2 s3 sn S of convergence is Sum a 1 r 1 1 sin x 2 x 3 (3 (3 3 6 1)(2n 3 3 1) n 1 3 2n 1 2n 3 1 1) 1) 3 2n 1 1 1 2 3 5 3 5 3 3 5 3 5 3 7 3 3 7 41. Almost, but the Ratio Test won't determine whether there is convergence or divergence at the endpoints of the interval. 42. (a) For k a1 ... N, it's obvious that ak N, ak ... ... aN aN a1 ... cN 1 a1 ... aN n N 1 n lim sn 3 cn. ... For all k a1 a1 a1 ... 47. ak n 1 (2n s1 s2 s3 sn S aN ... aN ck 1 n N 1 cn (b) Since all of the an are nonnegative, the partial sums of the series form a nondecreasing sequence of real numbers. Part (a) shows that the sequence is bounded above, so it must converge to a limit. 43. (a) For k d1 ... N, it's obvious that dk N, dk ... ... dN dN d1 ... aN 1 40n 1)2(2n 1)2 5 5 9 5 5 5 9 9 5 5 5 9 9 5 5 (2n 1)2 n 1 5 (2n 1)2 (2n 5 1)2 5 25 5 25 5 5 25 5 25 5 49 5 5 49 n lim sn 2n (n 1 1)2 1 4 1 4 1 4 5 d1 ... dN n N 1 48. an. ... n 1n s1 2 n 1 1 4 1 4 1 n2 1 (n 1)2 For all k d1 d1 d1 ... 1 1 1 1 dN ... dN ak 1 dk s2 s3 sn S n N 1 1 9 1 9 1 1 9 1 9 1 16 1 1 16 an 1 (n 1)2 (b) If an converged, that would imply that dn was also convergent. 44. Answers will vary. 4 45. 3)(4n n 1 (4n 1 s1 1 5 1 s2 1 5 1 s3 1 5 1 sn 1 4n 1 n lim sn 1 1 1 1 lim sn 1 2 1 1 49. s1 1) n 1 1 9 1 9 1 4n 3 4n 1 1 s2 s3 1 2 1 2 1 1 3 1 3 1 1 3 1 3 1 4 2 1 2 1 n 1 1 4 1 5 1 5 1 1 9 1 9 1 13 sn 1 1 13 S 50. s1 s2 s3 1 n S n lim sn 1 sn S 1 1 ln 3 ln 2 1 1 1 ln 3 ln 2 ln 4 1 1 1 ln 3 ln 2 ln 4 1 1 ln 5 ln 2 1 1 ln(n 2) ln 2 1 lim sn ln 2 n 1 ln 3 1 ln 3 1 ln 4 1 ln 5 1 ln 2 1 ln 4 ...
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