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# ch10 - 10.1 SOLUTIONS 689 CHAPTER TEN Solutions for Section...

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10.1 SOLUTIONS 689 CHAPTER TEN Solutions for Section 10.1 Exercises 1. Let f ( x ) = 1 1 - x = (1 - x ) - 1 . Then f (0) = 1 . f 0 ( x ) = 1!(1 - x ) - 2 f 0 (0) = 1! , f 00 ( x ) = 2!(1 - x ) - 3 f 00 (0) = 2! , f 000 ( x ) = 3!(1 - x ) - 4 f 000 (0) = 3! , f (4) ( x ) = 4!(1 - x ) - 5 f (4) (0) = 4! , f (5) ( x ) = 5!(1 - x ) - 6 f (5) (0) = 5! , f (6) ( x ) = 6!(1 - x ) - 7 f (6) (0) = 6! , f (7) ( x ) = 7!(1 - x ) - 8 f (7) (0) = 7! . P 3 ( x ) = 1 + x + x 2 + x 3 , P 5 ( x ) = 1 + x + x 2 + x 3 + x 4 + x 5 , P 7 ( x ) = 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 . 2. Let 1 1 + x = (1 + x ) - 1 . Then f (0) = 1 . f 0 ( x ) = - 1!(1 + x ) - 2 f 0 (0) = - 1 , f 00 ( x ) = 2!(1 + x ) - 3 f 00 (0) = 2! , f 000 ( x ) = - 3!(1 + x ) - 4 f 000 (0) = - 3! , f (4) ( x ) = 4!(1 + x ) - 5 f (4) (0) = 4! , f (5) ( x ) = - 5!(1 + x ) - 6 f (5) (0) = - 5! , f (6) ( x ) = 6!(1 + x ) - 7 f (6) (0) = 6! , f (7) ( x ) = - 7!(1 + x ) - 8 f (7) (0) = - 7! , f (8) ( x ) = 8!(1 + x ) - 9 f (8) (0) = 8! . P 4 ( x ) = 1 - x + x 2 - x 3 + x 4 , P 6 ( x ) = 1 - x + x 2 - x 3 + x 4 - x 5 + x 6 , P 8 ( x ) = 1 - x + x 2 - x 3 + x 4 - x 5 + x 6 - x 7 + x 8 . 3. Let f ( x ) = 1 + x = (1 + x ) 1 / 2 . Then f (0) = 1 , and f 0 ( x ) = 1 2 (1 + x ) - 1 / 2 f 0 (0) = 1 2 , f 00 ( x ) = - 1 4 (1 + x ) - 3 / 2 f 00 (0) = - 1 4 , f 000 ( x ) = 3 8 (1 + x ) - 5 / 2 f 000 (0) = 3 8 , f (4) ( x ) = - 15 16 (1 + x ) - 7 / 2 f (4) (0) = - 15 16 . Thus, P 2 ( x ) = 1 + 1 2 x - 1 8 x 2 , P 3 ( x ) = 1 + 1 2 x - 1 8 x 2 + 1 16 x 3 , P 4 ( x ) = 1 + 1 2 x - 1 8 x 2 + 1 16 x 3 - 5 128 x 4 .

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690 Chapter Ten /SOLUTIONS 4. Let f ( x ) = 3 1 - x = (1 - x ) 1 / 3 . Then f (0) = 1 , and f 0 ( x ) = - 1 3 (1 - x ) - 2 / 3 f 0 (0) = - 1 3 , f 00 ( x ) = - 2 3 2 (1 - x ) - 5 / 3 f 00 (0) = - 2 3 2 , f 000 ( x ) = - 10 3 3 (1 - x ) - 8 / 3 f 000 (0) = - 10 3 3 , f (4) ( x ) = - 80 3 4 (1 - x ) - 11 / 3 f (4) (0) = - 80 3 4 . Then, P 2 ( x ) = 1 - 1 3 x - 1 2! 2 3 2 x 2 = 1 - 1 3 x - 1 9 x 2 , P 3 ( x ) = P 2 ( x ) - 1 3! 10 3 3 x 3 = 1 - 1 3 x - 1 9 x 2 - 5 81 x 3 , P 4 ( x ) = P 3 ( x ) - 1 4! 80 3 4 x 4 = 1 - 1 3 x - 1 9 x 2 - 5 81 x 3 - 10 243 x 4 . 5. Let f ( x ) = cos x. Then f (0) = cos(0) = 1 , and f 0 ( x ) = - sin x f 0 (0) = 0 , f 00 ( x ) = - cos x f 00 (0) = - 1 , f 000 ( x ) = sin x f 000 (0) = 0 , f (4) ( x ) = cos x f (4) (0) = 1 , f (5) ( x ) = - sin x f (5) (0) = 0 , f (6) ( x ) = - cos x f (6) (0) = - 1 . Thus, P 2 ( x ) = 1 - x 2 2! , P 4 ( x ) = 1 - x 2 2! + x 4 4! , P 6 ( x ) = 1 - x 2 2! + x 4 4! - x 6 6! . 6. Let f ( x ) = ln(1 + x ) . Then f (0) = ln 1 = 0 , and f 0 ( x ) = (1 + x ) - 1 f 0 (0) = 1 , f 00 ( x ) = ( - 1)(1 + x ) - 2 f 00 (0) = - 1 , f 000 ( x ) = 2(1 + x ) - 3 f 000 (0) = 2 , f (4) ( x ) = - 3!(1 + x ) - 4 f (4) (0) = - 3! , f (5) ( x ) = 4!(1 + x ) - 5 f (5) (0) = 4! , f (6) ( x ) = - 5!(1 + x ) - 6 f (6) (0) = - 5! , f (7) ( x ) = 6!(1 + x ) - 7 f (7) (0) = 6! , f (8) ( x ) = - 7!(1 + x ) - 8 f (8) (0) = - 7! , f (9) ( x ) = 8!(1 + x ) - 9 f (9) (0) = 8! . So, P 5 ( x ) = x - x 2 2 + x 3 3 - x 4 4 + x 5 5 , P 7 ( x ) = x - x 2 2 + x 3 3 - x 4 4 + x 5 5 - x 6 6 + x 7 7 , P 9 ( x ) = x - x 2 2 + x 3 3 - x 4 4 + x 5 5 - x 6 6 + x 7 7 - x 8 8 + x 9 9 .
10.1 SOLUTIONS 691 7. Let f ( x ) = arctan x. Then f (0) = arctan 0 = 0 , and f 0 ( x ) = 1 / (1 + x 2 ) = (1 + x 2 ) - 1 f 0 (0) = 1 , f 00 ( x ) = ( - 1)(1 + x 2 ) - 2 2 x f 00 (0) = 0 , f 000 ( x ) = 2!(1 + x 2 ) - 3 2 2 x 2 + ( - 1)(1 + x 2 ) - 2 2 f 000 (0) = - 2 , f (4) ( x ) = - 3!(1 + x 2 ) - 4 2 3 x 3 + 2!(1 + x 2 ) - 3 2 3 x + 2!(1 + x 2 ) - 3 2 2 x f (4) (0) = 0 . Therefore, P 3 ( x ) = P 4 ( x ) = x - 1 3 x 3 . 8. Let f ( x ) = tan x. So f (0) = tan 0 = 0 , and f 0 ( x ) = 1 / cos 2 x f 0 (0) = 1 , f 00 ( x ) = 2 sin x/ cos 3 x f 00 (0) = 0 , f 000 ( x ) = (2 / cos 2 x ) + (6 sin 2 x/ cos 4 x ) f 000 (0) = 2 , f (4) ( x ) = (16 sin x/ cos 3 x ) + (24 sin 3 x/ cos 5 x ) f (4) (0) = 0 . Thus, P 3 ( x ) = P 4 ( x ) = x + x 3 3 . 9. Let f ( x ) = 1 1 + x = (1 + x ) - 1 / 2 . Then f (0) = 1 . f 0 ( x ) = - 1 2 (1 + x ) - 3 / 2 f 0 (0) = - 1 2 , f 00 ( x ) = 3 2 2 (1 + x ) - 5 / 2 f 00 (0) = 3 2 2 , f 000 ( x ) = - 3 · 5 2 3 (1 + x ) - 7 / 2 f 000 (0) = - 3 · 5 2 3 , f (4) ( x ) = 3 · 5 · 7 2 4 (1 + x ) - 9 / 2 f (4) (0) = 3 · 5 · 7 2 4 Then, P 2 ( x ) = 1 - 1 2 x + 1 2! 3 2 2 x 2 = 1 - 1 2 x + 3 8 x 2 , P 3 ( x ) = P 2 ( x ) - 1 3! 3 · 5 2 3 x 3 = 1 - 1 2 x + 3 8 x 2 - 5 16 x 3 , P 4 ( x ) = P 3 ( x ) + 1 4! 3 · 5 · 7 2 4 x 4 = 1 - 1 2 x + 3 8 x 2 - 5 16 x 3 + 35 128 x 4 . 10. Let f ( x ) = (1 + x ) p . (a) Suppose that p = 0 . Then f ( x ) = 1 and f ( k ) ( x ) = 0 for any k 1 . Thus P 2 ( x ) = P 3 ( x ) = P 4 ( x ) = 1 . (b) If p = 1 then f ( x ) = 1 + x, so f (0) = 1 , f 0 ( x ) = 1 , f ( k ) ( x ) = 0 k 2 .

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ch10 - 10.1 SOLUTIONS 689 CHAPTER TEN Solutions for Section...

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