Pre-Calc Homework Solutions 397

Pre-Calc Homework Solutions 397 - Chapter 9 Review 53. This...

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Unformatted text preview: Chapter 9 Review 53. This is a telescoping series. (c) The fourth order Taylor polynomial for g(x) at x x 397 4 is n 3 (2n s1 s2 s3 sn S 1 3)(2n 1) n 1 2(2 3 3) 2(2 1 1 1 6 10 10 1 1 1 6 10 10 1 1 6 2(2n 1) 1 lim sn 6 n 1 1 3) 2(2n 1) 3 2(2n 1 1 1 3 1) 6 10 1 1 1 6 14 14 1 1 1 1 6 14 14 18 [7 4 3(t 3 (t 2 4) 4)2 3 (x 2 5(t 5 (t 3 4)2 4)3 5 (x 3 2(t 1 (t 2 4)3] dx 4)4 1 (x 2 x 4 7t 7(x 1 18 4) 4)2 4)3 4)4 (d) No. One would need the entire Taylor series for f (x), and it would have to converge to f (x) at x 3. 57. (a) Use the Maclaurin series for sin x given at the end of Section 9.2. 54. This is a telescoping series. 5 sin 2 1 n 2 n(n s1 s2 s3 sn S n 2 1) 2 2 n 2 2 n x 2 x 2 (x/2)3 (x/2)5 3! 5! 5x 3 x5 ... 48 768 n 5 5x 2 ... 1 1 1 lim sn 2 2 1 3 3 2 2 3 3 2 2 3 3 2 n 2 (x/2)2n 1 ... (2n 1)! 5 x 2n+1 ... ( 1)n (2n 1)! 2 ( 1)n 2 4 2 4 1 2 4 2 4 2 5 1 2 5 (b) The series converges for all real numbers, according to the Ratio Test: lim n an an 1 1 f (3) f (3)(x f (3) (x 3! n (2n lim lim 5 n (2n x 2n 3 3)! 2 x/2 2 3)(2n 2) (2n 5 1)! 2 2n+1 x 0 5 2n 55. (a) P3(x) 3) 3) 3(x 3 f (3) (x 2! 3)2 (c) Note that the absolute value of f (n)(x) is bounded by for all x and all n 1, 2, 3, .... 3)3 1 f (3.2) 4(x 3) 3)2 2(x We may use the Remainder Estimation Theorem with M So if 5 and r 2 x 1 . 2 P3(3.2) 1.936 (b) Since the Taylor series for f can be obtained by termby-term differentiation of the Taylor Series for f, the second order Taylor polynomial for f at x 4 6(x 3) 2.74. 6, which 3. 6(x 3)2. Evaluated at x 3 is 2.7, 2, the truncation error using Pn is bounded by 5 2n 1 2n 1 (n 1)! 5 (n 1)! . 4. So, two To make this less than 0.1 requires n terms (up through degree 4) are needed. f (2.7) (c) It underestimates the values, since f (3) means the graph of f is concave up near x 56. (a) Since the constant term is f (4), f (4) 2 f (4) , f (4) 3! 58. (a) Substitute 2x for x in the Maclaurin series for given at the end of Section 9.2. 1 1 2x 1 1 x 1 1 2x 2x (2x)2 4x 2 (2x)3 8x 3 ... ... (2x)n (2x)n ... ... 7. Since 12. (b) 1 1 1 , . The series for is known to converge for 2 2 1 t (b) Note that P4 (x) 3 10(x 4) 6(x 4)2 24(x 4)3. The second degree polynomial for f at x 4 is given by the first three terms of this expression, namely 3 10(x 4) 6(x 4)2. Evaluating at x 4.3, f (4.3) 0.54. 1 t 1, so by substituting t 1 2x, we find the 2x 1. resulting series converges for ...
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