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Unformatted text preview: 374 Section 9.2
(x)(e x) d ex 1 x dx xe x e x 1 x2 e e 1 1 1 (e x x2 1)(1) 16. continued (b) Since the Taylor series of f (x) can be obtained by differentiating the terms of the Taylor series of f (x), the (c) g (x) g (1) second order Taylor polynomial of f (x) is given by 5 8x 3x 2. Evaluating at x 3.52 4 4 f (1.2) ( 1)(x (x 1) 1)
3 (x 2 3 (x 2! 0.2, From the series, g (x)
2 (x 3! x2 d x 1 3! dx 2! 2x 3x 2 1 3! 4! 2! nxn 1 1)! n 1 (n x3 4! f (0.2) 17. (a) P3(x) ... xn (n 1)! ... ... 1)2
1 (x 3 1)3 nx n 1 (n 1)! ... 1)2 1)3 P3(1.2) 3.863 Therefore, g (1) (b) Since the Taylor series of f (x) can be obtained by differentiating the terms of the Taylor series of f (x), the second order Taylor polynomial of f (x) is given by 1 f (1.2) 3(x 1) 0.36 (x 1)2. Evaluating at x 1.2, n 1 n 1 n (n 1)! , which means n (n 1)! 1. 20. (a) Factor out 2 and substitute t 2 for x in the Maclaurin series for f (t)
1 at the end of Section 9.2. 1 x 2 1 t2 1 2 1 t2 2[1
1 x 1 18. (a) Since f (0)x , f (0) . 2 2! 2! f (10)(0) 10 10! x 10 (10) Since x , f (0) 11! 10 ! 11! t2 2t 2 (t 2)2 2t 4 (t 2)3 2t 6 ... ... (t 2)n ... ...] 2
1 . 11 2t 2n (b) Since G(0) ... 0, the constant term is zero and we may (b) Multiply each term of f (x) by x. g(x) (c) g(x) x ex
x2 2! x3 3! x4 4! find G(x) by integrating the terms of the series for f (x).
xn 1 (n 1)! ... G(x) 2x 2x 3 3 2x 5 5 2x 7 7 ... 2x 2n 1 2n 1 ... 1 21. (a) f (0) (1
1 (1 2 x for x in the Maclaurin series for e x shown 2
x 2 2 x n 2 x)1/2 x) x 0 1/2 1
1 2 1 f (0) , so 4 2! 3 f (0) , so 8 3! 1 16 1 8 19. (a) Substitute f (0) f (0) x 0 at the end of Section 9.2 e x/2 1 1 (b) g(x)
1 1 x 1 x x ex x x 2 x 2 1 1 (1 4 3 (1 8 x) x) 3/2 x 0 2 x2 8 ... ... n! xn 2n n! ... f (0) P4(x) 5/2 x 0 1 x 2 x2 8 x3 16 (b) Since g(x)
x 2! x3 3! x2 3!
2 f (x 2), the first four terms are
x4 8 x6 . 16 x
x2 2! x 2! x 3! 3 ...
xn n! xn 1 n! x n! n ... 1 1 ... ... ... ... x2 2 (c) Since h(0) 5, the constant term is 5. The next three 1 terms are obtained by integrating the first three terms of the answer to part (b). The first four terms of the series .... for h(x) are 5 x
x3 6 x5 . 40 This can also be written as 1
x 2! x2 3! ... xn (n 1)! ...
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 Spring '08
 GERMAN
 Taylor Series

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