a . Explanation Given: The function f ( x ) = cos x , a = π 6 . Result used: The n th-order Taylor polynomial centered at a is shown below, p n ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f n ( a ) n ! ( x − a ) n Calculation: The n th-order Taylor polynomial centered at π 6 is shown below, p n ( x ) = f ( π 6 ) + f ′ ( π 6 ) ( x − π 6 ) + f ″ ( π 6 ) 2 ! ( x − π 6 ) 2 + ⋯ + f n ( π 6 ) n ! ( x − π 6 ) n (1) Substitute π 6 for x in f ( x ) = cos x , f ( x ) = cos ( π 6 ) = 3 2 That is, f ( π 6 ) = 3 2 . The first derivative of f ( x ) = cos x is computed as follows, f ′ ( x ) = − sin x Substitute π 6 for x , f ′ ( π 6 ) = − sin ( π 6 ) = − ( 1 2 ) That is, f ′ ( π 6 ) = − 1 2 . The second derivative of f ( x ) = cos x is computed as follows, f ″ ( x ) = − cos x Substitute π 6 for x , f ″ ( π 6 ) = − cos ( π 6 ) = − 3 2 That is, f ″ ( π 6 ) = − 3 2 . The Taylor polynomial centered at π 6 for n = 0 , 1 and 2 is obtained as follows, Substitute 0 for n in equation (1) and obtain p 0 ( x ) , p 0 ( x ) = f ( a ) = f ( π 6 ) = 3 2 Thus, p 0 ( x ) = 3 2 b. To determine To sketch: The Taylor polynomial and the function.

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 9.1, Problem 32E

To determine

To find: The nth-order Taylor polynomial centered at the point a for n=0,1,and 2.

To determine

To sketch: The Taylor polynomial and the function.

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Chapter 9.1, Problem 31E

Chapter 9.1, Problem 33E

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Chapter 9 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The nth -order Taylor polynomial centered at the point a for n = 0 , 1 , and 2 .

Solution Summary: The nth-order Taylor polynomial centered at the point a=pi 6 and

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To find: The nth-order Taylor polynomial centered at the point a for n=0,1,and 2.

To sketch: The Taylor polynomial and the function.

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Chapter 9 Solutions