sol11-1 - Chapter 11 Exercises 11.1 1. p3 ( x) = sin 0 +...

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357 Chapter 11 Exercises 11.1 1. p 3 ( x ) = sin0 + cos(0) x + 2! x 2 + cos0 3! x 3 = x 1 6 x 3 2. p 3 ( x ) = e 0/2 1 2 e 0/ 2 x + 1 4 e x 2 2! 1 8 e 0/ 2 x 3 = 1 1 2 x + 1 8 x 2 1 48 x 3 3. p 3 ( x ) = 5 e 2(0) + 10 e x + 20 e 2(0) x 2 + 40 e 2(0) x 3 = 5 + 10 x + 10 x 2 + 20 3 x 3 4. p 3 ( x ) = cos π+ (5sin π ) x (25cos π ) x 2 (125sin π ) x 3 =− 1 + 25 2 x 2 5. p 3 ( x ) = 1 + 2 x 4 x 2 + 24 x 3 = 1 + 2 x 2 x 2 + 4 x 3 6. p 3 ( x ) = 1 2 1 2 2 x + 2 2 3 x 2 6 2 4 x 3 = 1 2 1 4 x + x 2 8 x 3 16 7. f ( x ) = xe 3 x ; f ( x ) = e 3 x + 3 xe 3 x f ( x ) = 3 e 3 x + 3 e 3 x + 9 xe 3 x = 6 e 3 x + 9 xe 3 x f ( x ) = 18 e 3 x + 9 e 3 x + 27 xe 3 x = 27 e 3 x + 27 xe 3 x p 3 ( x ) = 0 + (1 + 0) x + (6 + 0) x 2 + (27 + 0) x 3 = x + 3 x 2 + 9 2 x 3 8. p 3 ( x ) = 1 1 2 x 1 4 x 2 3 8 x 3 = 1 1 2 x 1 8 x 2 1 16 x 3 9. p 4 ( x ) = e 0 + e 0 x + e 0 x 2 + e 0 x 3 + e 0 x 4 4! = 1 + x + x 2 2 + x 3 6 + x 4 24 e .01 1 + .01 + (.01) 2 2 + (.01) 3 6 + (.01) 4 24 = 1.01005 10. f ( x ) = ln(1 – x ); f ( x ) 1 x ) ; f ( x ) 1 x ) 2 f ( x ) 2 x ) 3 f (4) ( x ) 6 x ) 4 p 4 ( x ) = 0 x x 2 2 x 3 6 x 4 4! x 1 2 x 2 1 3 x 3 1 4 x 4 ln(.9) = ln(1 – .1) –.1 (.1) 2 2 (.1) 3 3 (.1) 4 4 .10536
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Chapter 11: Taylor Polynomials and Infinite Series ISM: Calculus & Its Applications, 11e 358 11. f ( x ) and p 1 ( x ) = 1 + x x y 1 f ( x ) = 1 1 – x f ( x ) and p 2 ( x ) = 1 + x + x 2 x y 1 f ( x ) = 1 1 – x (0, 1) f ( x ) and p 3 ( x ) = 1 + x + x 2 + x 3 x y 1 f ( x ) = 1 1 – x (0, 1) 12. f ( x ) and p 1 ( x ) = p 2 ( x ) = x x y f ( x ) and p 3 ( x ) = x x 3 6 x y 13. p n ( x ) = e 0 + e 0 x + e 0 x 2 2! + L + e 0 x n n ! = 1 + x + x 2 2 + x 3 6 + L + x n n ! 14. p 0 ( x ) = 0 2 + 2(0) + 1 = 1; p 1 ( x ) = 1 + (2(0) + 2) x = 1 + 2 x ; p 2 ( x ) = 1 + 2 x + 2 x 2 2! = 1 + 2 x + x 2 Since f ( x ) = 0 for all n 2, p n ( x ) = p 2 ( x ) = f ( x ) for all n 2. 15. f ( x ) = ln(1 + x 2 ); f ( x ) = 2 x 1 + x 2 f ( x ) = 2(1 + x 2 ) 4 x 2 (1 + x 2 ) 2 = 2 x 2 + 2 + x 2 ) 2 p 2 ( x ) = ln(1) + (0) x + 2 x 2 2! = x 2 ln(1 + x 2 ) dx 0 1/2 x 2 dx 0 = x 3 3 0 = 1 24 .0417 16. sin () c o s ; () ; 2c o s x fx x f x x == ( ) 21 / 2 cos 2 cos sin (cos ) 4cos xx x x x −+ ′′ = p 2 ( x ) = 1 + 0 2(1) x 1 2 x 2 2! = 1 1 4 x 2 cos xdx 1 1 1 1 4 x 2 dx 1 1 = x 1 12 x 3 1 1 = 11 6
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ISM: Calculus & Its Applications, 11e Chapter 11: Taylor Polynomials and Infinite Series 359 17. f ( x ) = 1 5 x ; f ( x ) = 1 (5 x ) 2 f ( x ) = 2 x ) 3 f ( x ) = 6 x ) 4 p 3 ( x ) = 1 5 4 + 1 4) 2 ( x 4) + 2 4) 3 ( x 2 2! + 6 4) 4 ( x 4) 3 3! = 1 + ( x 4) + ( x 4) 2 + ( x 4) 3 18. f ( x ) = ln x ; f ( x ) = 1 x f ( x ) = 1 x 2 f ( x ) = 2 x 3 ; f (4) ( x ) = 6 x 4 p 4 ( x ) = ln(1) + 1 1 ( x 1) 1 1 2 ( x 1) 2 + 2 1 3 ( x 1) 3 + 6 1 4 ( x 4 4! = ( x 1 2 ( x 1) 2 + 1 3 ( x 1) 3 1 4 ( x 4 19. f ( x ) = cos x ; f ( x ) =− sin x ; f ( x ) cos x ; f ( x ) = sin x f (4 ) ( x ) = cos x p 3 ( x ) = cos π− sin π ( x −π ) cos π ( x ) 2 2! + sin π ( x ) 3 1 + 1 2 ( x ) 2 p 4 ( x ) = p 3 ( x ) + cos π ( x ) 4 4! 1 + 1 2 ( x ) 2 1 24 ( x ) 4 20. f ( x ) = x 3 + 3 x 1; f ( x ) = 3 x 2 + 3; f ( x ) = 6 x ; f ( x ) = 6; f ( n ) ( x ) = 0 for all n > 3.
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sol11-1 - Chapter 11 Exercises 11.1 1. p3 ( x) = sin 0 +...

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