Pre-Calc Homework Solutions 395

Pre-Calc Homework Solutions 395 - Chapter 9 Review 4x 1 x 1...

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Unformatted text preview: Chapter 9 Review 4x 1 x 1 1 x 395 26. 4x 33. Use the Maclaurin series for e x given at the end of x x2 4x 3 4x(1 4x ... ... xn 4x n ...) 1 Section 9.2. ... xe x2 4x 2 x1 x x3 ( x 2) x 2! 5 ( x 2)2 2! ... x n! ... ( 1)n 2n 1 ( x 2 )n n! ... ... 1 27. Replace x by x in the Maclaurin series for sin x given at the end of Section 9.2. sin x x ( x)3 3! ( x)5 5! 34. Replace x by 3x in the Maclaurin series for tan ... ( 1)n ( x)2n 1 (2n 1)! x given at ... the end of Section 9.2. tan 1 3x 3x 28. Replace x by 2x in the Maclaurin series for sin x given at 3 (3x)3 3 (3x)5 5 ... ( 1)n (3x)2n 1 2n 1 ... x) 35. Replace x by 2x in the Maclaurin series for ln (1 the end of Section 9.2 2x sin 3 ( 2x 3 2x 3 3 2x 5 3 3! 5! ... ( 2x 2n+1 1)n 3 (2n 1)! ) given at the end of Section 9.2. ln (1 2x) 2x ( 2x)2 ( 2x)3 ... 2 3 n ( 2x) ... ( 1)n 1 n n 8x 3 ... (2x) 2x 2x 2 3 n 2x 3 4x 3 81 4x 5 3645 ... ( 1)n (2n 1 2x 2n 1 3 .... 1)! 29. x sin x x x3 x5 x7 ... 3! 5! 7! x 2n 1 ... ( 1)n (2n 1)! 2n 1 x3 x5 x7 ... ( 1)n x 3! 5! 7! (2n 1)! 36. Use the Maclaurin series for ln (1 Section 9.2. x ln (1 ... x x2 x x) x ln [1 2 x) given at the end of x ( x)] 3 30. ex 2 e x 1 1 2 x 1 1 2 x2 2! x2 2! ... x2 2! xn n! ... ( 1)n ... xn n! x3 2 ( x) 2 x4 3 1 ( x) 3 ... xn n 1 ( 1)n ... n 1 ( x) n ... ... 1 1 x x4 4! ... x 2n (2n)! ... 37. f (2) f (2) f (2) f (2) f (n)(2) (3 (3 2(3 6(3 x) x) x 2 2 x 2 3 x 2 4 x 2 n 1 1 ... x) x) 2, so 31. Replace x by 5x in the Maclaurin series for cos x given at f (2) 2! f (2) 6, so 3! 1 1 1 ... n!(3 1 (x (x x) the end of Section 9.2. cos 5x 1 ( 5x)2 2! 5x)4 ... 4! ( 5x)2n ... ( 1)n (2n)! n 5x (5x)2 ... ( 1)n (5x) 2! 4! (2n)! ( 3 x 2 f (n)(2) n!, so n! 2 1 x 2) 2)n 2x 2 4x) x 4) x 1 (x ... 5) x 1 1 2) (x 2) 3 1 ... 38. f ( 1) f ( 1) f ( 1) f ( 1) f (n)( 1) (x 3 (3x 2 (6x 6x 1 2 7 10, so f ( 1) 2! 32. Replace x by x in the Maclaurin series for e x given at the 2 x 2 2 x n 2 5 end of Section 9.2. e x/2 f ( 1) 6, so 3! 1 0 for n 5 2 4. 7(x 1) 5(x 1)2 (x 1)3 4 is 0. 1 1 x 2 x 2 2! 2 2 ... ... n! 1 n! 2 x n ... ... x3 2x 2 x 8 This is a finite series and the general term for n ...
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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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