Pre-Calc Homework Solutions 377

# Pre-Calc Homework Solutions 377 - Section 9.3 8. cos2 x 1 2...

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Unformatted text preview: Section 9.3 8. cos2 x 1 2 1 2 1 1 2 4x 2 2 2! 1 cos (2x) 2 (2x)2 2! 16x 4 2 4! x4 3 (2x)4 4! 377 3. f(0) f (0) f (0) f (0) f (4)(0) P4(x) f (0.2) 5 sin ( x) x 5 cos x x 5 sin x x 0 0 5 sin x x 5 0, so 0 0 ... 0 ( 1)n 2 (2x)2n (2n)! ... ... f (0) 0 2! f (0) 5 5 cos x x 0 5, so 3! 6 (4) f (0) 5 sin x x 0 0, so 0 4! 5 3 5x x 6 149 P4(0.2) 0.9933 150 1 1 9. sin2 x ... ( 1)n ( 1)n 22nx 2n (2n)! x2 1 2 1 2 ... 22n 1x 2n (2n)! ... 1 cos (2x) 2 1 1 2 (2x)2 2! (2x)4 4! (2x)2n (2n)! 64x 6 2 6! 2n 2n 1 2 x (2x)6 6! ... ( 1)n 4x 2 2 2! 16x 4 2 4! ... ... ... 2n 1 2n x 1 2 4. Substituting x 2 for x in the Maclaurin series given for ln (1 ln (1 x) at the end of Section 9.2, we have x 2) x2 x2 Therefore, P4(x) 5. f (0) f (0) f (0) f (0) f (4)(0) P4(x) f (0.2) 6. xe x (1 2(1 6(1 x) 2 x 0 2 n (x 2)2 (x 2)3 ... ( 1)n 1 (x ) ... 2 3 n 2n x4 x6 ... ( 1)n 1 x ... 2 3 n x4 x2 and f (0.2) P(0.2) 0.0392. 2 ( 1)n x2 x4 3 2x 6 45 2 (2n)! ... ( 1)n (2n)! ... Note: By replacing n with n written as ( 1)n x2 1 2x 22n 1 x 2n 2 . (2n 2)! 1, the general term can be 1 2 6, so 10. x2 1 1 2x x) x) 3 x 0 4 x 2[1 x2 11. Let f (x) 5 2x 2x 3 (2x)2 4x 4 ... ... (2x)n 2n x n 2 ...] ... x3 , so we use 6 f (0) 3 2! f (0) 4 24(1 x) 5 x 0 24, so 3! (4) f (0) 120(1 x) 6 x 0 120, so 4! x 0 sin x. Then P4(x) P3(x) x the Remainder Estimation Theorem with n f (5)(x) cos x 1 for all x, we may use M 4. Since r 1, 1 2x 3x 2 1.56 x2 2! x3 2! 4x 3 5x 4 P4(0.2) x 1 x x2 x3 x 3! x3 x5 3! 5! x7 x9 7! 9! x5 giving R4(x) , so we may assure that 5! x5 4 R4(x) 5 10 by requiring 5 10 4, or 5! xn n! xn 1 n! x ... ... ... ... x 5 0.06 0.5697. Thus, the absolute error is no 10 4 greater than 5 0.56 x when 0.56 (approximately). 7. sin x x x5 5! Alternate method: Using graphing techniques, ... ... ( 1)n x 2n 1 (2n 1)! 2n 1 n x ( 1) (2n 1)! ... ... x x3 3! sin x x x3 6 5 10 4 when 0.57 x 0.57. 12. Let f (x) cos x. Then P3(x) P2(x) 1 x2 , so we may 2 Note: By replacing n with n x written as ( 1)n (2n 5)! 2n 5 2, the general term can be use the Remainder Estimation Theorem with n f (4)(x) cos x 4 3. Since r 1, 1 for all x, we may use M x . For x 4! giving R3(x) than (0.5)4 4! 0.5, the absolute error is less 0.0026 (approximately). ...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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