Pre-Calc Homework Solutions 377

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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). ...
View Full Document

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.

Ask a homework question - tutors are online