X3 8 x 2 x 2 x 2 1 2 1 2 xn x2 x 4 3 a r n 1x 02

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Unformatted text preview: os x n n 4x n! 0 2 1 2 2 2n [Note: 1 nn x x 1 4 nn 1, 2! x n! 1, 2 4 12 0 12 n 4 1, 1, 1, 4 x 4 3! n . 1, 487 1, 1, . . .] 3 x 4 4! 4 ... 488 Chapter 9 5. For c Infinite Series 1, we have, fx ln x f1 0 fx 1 x f1 1 1 x2 fx 2 x3 x f f 4 5 f 6 x4 x f f1 x 1 f 4 5 2 1 f 24 x5 1 1 6 24 and so on. Therefore, we have: n f ln x n 1x n! 0 0 x x 1 7. For c n x 1 1 2 2x 1 x 2 n n 1 6x x 1 3 1 4 24 x 4! x 3 1 0 3 3! 2 n 1 2! x 1 n 1 1 4 x 4 1 5 ... 5! 1 5 ... 5 1 . 0, we have: fx sin 2x f0 0 fx 2 cos 2x f0 2 0 fx 4 sin 2x f0 fx 8 cos 2x f0 f 4 f 5 f 6 f 7 x x f x 128 cos 2x 0 32 6 0 0 f 64 sin 2x 0 5 f 32 cos 2x 0 f 16 sin 2x x 8 4 7 0 128 and so on. Therefore, we have: f sin 2x n 0 0 xn n! 8x3 3! 2x 9. For c n 0 2x 32x5 5! 0x 2 2! 128x7 7! 8x3 3! 0x 4 4! 32x5 5! 128x7 7! 1 n 2x 2n 1 . 2n 1 ! ... n 0x6 6! 0 0, we have: fx f0 1 fx sec x tan x f0 0 fx sec3 x f0 1 fx f sec x 5 sec3 x tan x f0 0 4 x 5 sec5 f sec x n 0 n sec x tan2 x x 18 sec x tan3 x sec3 n 0x n! tan2 x 2 1 x 2! x sec x 4 5x 4! . . .. tan4 x f 4 0 5 ... S ection 9.10 11. The Maclaurin series for f x Because f 1 x n n→ n f 0 ≤ Rn x Since lim 0 ± sin x or ± cos x, we have f x 1 n n 1 zn x 1! x n ≤ 1 n 13. The Maclaurin series for f x f 1 1 ≤ 1 for all z. Hence by Taylor’s Theorem, z 1 1! . . Hence, the Maclaurin series for cos x converges to cos x for all x. x 2n 1 . 2n 1 ! sinh x is n n n 0, it follows that Rn x → 0 as n → 1! 0 sinh x or cosh x . For fixed x, x n f 0 ≤ Rn x 1 n zn x 1! n The argument is the same if f 15. Since 1 k x 1 1 2 x 1 sinh z n x n 1! 1 1 x 1 kk 2 3 x2 2! 2 x3 1k 3! 2! 2x → 0 as n → . cosh x . Hence, the Maclaurin series for sinh x converges to sinh x for all x. 1 x2 kk kx 2 3 4 x3 3! . . ., we have 2 3 4 5 x4 4! ... 1 3x 2 2x 1 n 1 17. 4 1 2 x2 1 4 x 12 x2 we have 1 xn 0 n! ex 21. n 2 ex 2 n 23. 0 x2 2 n! n n n 0 0 1 n 3x 2n 2n 1 ! sin 3x n n n x2 2! x 2n n 0 2 n! 1 n 3 2 x2 2 1 1 1 1 x 5 . . . 2n 2n n! 3 1 x 2 12 12 x n x3 3! 1 1 1 x2 5 . . . 2n 2n n! 1 3 x 2n 2n n! 2 x4 4! x2 2 x5 5! x4 222! ... n 1 xn 3 , we have 5 . . . 2n 1n! 1 x2n 23n 1 . 3 xn 5 . . . 2n 3 1 n1 1 2 5x 4 1 xn. n 5 . . . 2n 2n n! 3 1 2n n 4x3 0 1 n1 1 n 1 nx2n 1 2n 1 ! sin x 12 and since 1 n 1 1 2 1 n1 1 1 2 x2 19. Since 1 x 2 1 489 1 n x2n . 2n ! cos x is n n Taylor and Maclaurin Series . ... x6 233! x8 244! 25. ... 1 n x 2n 2n ! cos x n 0 0 x3 2n ! 0 1 n x3n 2n ! 1 cos x3 1 2 n n 1 x3 2! n 1 x6 4! 2 2n ... x2 2! x4 4! ... 490 Chapter 9 Infinite Series ex 1 x x2 2! x3 3! x4 4! x5 5! ... e x 1 x x2 2! x3 3! x4 4! x5 5! ... e x 2x 27. ex 2x3 3! 1x e 2 sinh x e 2x5 5! x5 5! 1 1 2 31. x sin x 1 x7 7! n 2x 2 2! x3 3! xx x5 5! x4 2x 4 4! 1 1 2 ... n 1 n 2x 2n ! 0 33. sin x x 2n x x3 3! 0 ix 2 2! e ix 1 ix ix 2! eix e ix 2ix eix e ix x 1 2ix3 3! ix 3 3! 2 ix 4 4! 3 ix 3! 2ix5 5! 2ix7 7! ... 4 ix 4! ... x3 3! x5 5! x7 7! x2 x3 x4 2 6 24 x3 2 x x2 x4 2! 4! ... n x3 6 1 x2 2! ix ... 1 ix3 3! ix x4 4! ix5 5! x6 6! x x x2 x x2 x2 2! ix3 3! x4 4! ix5 5! x6 6! x3 3 x5 30 ... n 1 n x2n 1 2n 1 ! 0 sin x ... x4 6 x x4 6 x3 x5 6 120 x5 120 x5 12 ... x5 24 P5 f ... −6 x4 2 6 −2 ... x x2 4 24 ... x x2 x3 x4 x5 2 3 4 P5 5 ... −3 9 h x x2 2 x x2 2 ... 14 cos x ln 1 1 0 1 nx2n ,x 2n 1 ! ... ex sin x 1 ... x5 5! 1 nx2n 2 2n 1 ! ix 39. h x 2x 6 6! 5! 1 37. f x 0 ... eix 2i n x2n 1 2n 1 ! ... x6 3! x2 35. ... cos 2x 1 1 2 29. cos2 x ... x x3 3! x 2x7 7! x3 3 x3 6 x3 2 x4 4 3x5 40 ... x4 4 x5 5 x5 6 x5 24 ... −4 0 S ection 9.10 sin x . Divide the series for sin x by 1 1x 41. g x x 1 x x 5x3 6 x3 6 x2 2 0x fx 0x4 x4 3! g x5 120 ... −6 0x4 5x4 6 5x4 6 5x4 6 x5 120 5x5 6 45. x 1 n t2n n! 1 dt 0 0 n 0 x 0 1 49. Since ln x n we have e2 n 1 x 1 1 n 2 1 dt n 1 x 1 2 1 3 1 4 1 x x2 2! x3 3! 2 22 2! 23 3! 2n 1 2 2 1 n 0 x 1 ... xn 0 n! n 1 n cos x n 1 cos x we have lim x→0 0 1 nx 2n 2n ! 2n 0 n! n x4 4! x2 2! x4 4! x6 6! x8 8! x 2! x3 x5 x7 ... 4! 6! x6 6! ... 8! 1 cos x x 11 n x x2 2! xe x 1 2n x 2n x8 8! ... 1 nx 2n 2 2n 2 ! 0 n 0 x 1 x 1! 3n 3 0 3 1 nx 2n 1 2n 2 ! 1 2! 0. 1 n 1 x 0 2n 4 4 0.6931. (10,001 terms) 7.3891. (12 terms) 0 n lim x→0 n 1 t2n . . ., ... x2 2! 1 n 3 53. Since cos x x1 1 dt 1 n 1t2n n 1! 0 n 1 51. Since ex n 0 we have ln 2 1 x3 2! x2 x Matches (a) x t2 y fx x sin x e 6 P4 Matches (c) 47. ... 4 x3 3! xx 5x 4 6 −4 6 x3 5x3 6 5x3 6 x2 x2 5x4 6 5x 3 6 x2 x x3 x2 y gx x2 x 43. x. Taylor and Maclaurin Series n 1x 2n 3n . . ., 3 1! 0<x≤2 491 492 Chapter 9 1 55. 0 1 sin x dx x 0 Since 1 7 1 Infinite Series 0 1 1 3 5 sin x x x→0 0.3 1 3! Note: We are using lim x3 dx 1 0.1 Since 0.37 0.17 ... 5! x3 2 1 0.1 1 56 x6 8 x3 d x 1 0.3 1! 0 n 2n 0 1! 2 1 nx 4n 2n ! x cos x dx 0 2 23 2 x9 16 5x12 128 . . . dx x4 8 x x7 56 x10 160 0.3 5x13 1664 ... 0.1 n 0.201. 0 12 dx n 0 1 nx 4n 3 2 4n 3 2n ! 2 n 1 n2x 4n 3 2 4n 3 2n ! 0 2 0 0 10! < 0.0001, we need five terms. 23 1 x cos x dx 0.14 2 2 0 using three nonzero terms 0.9461. 1 0.34 8 0.1 0.1 Since 2 2n 1n 1 2n 1 < 0.0001, we need two terms. 0.3 59. 0 1 1. 0.3 57. n 1 nx 2n 1 2n 7! < 0.0001, we need three terms: sin x dx x 0 n 1 nx 2n dx 2n 1 ! 2 0 2 3 32 27 14 2 2...
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This note was uploaded on 01/11/2013 for the course MATH 111 taught by Professor Man during the Spring '13 term at University of Washington-Tacoma Campus.

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