Explanation 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 function is f ( x ) = sin x . The n th-order Taylor polynomial centered at 0 is shown below. p n ( x ) = f ( 0 ) + f ′ ( 0 ) ( x − 0 ) + f ″ ( 0 ) 2 ! ( x − 0 ) 2 + ⋯ + f n ( 0 ) n ! ( x − 0 ) n (1) Substitute 0 for x in f ( x ) = sin x . f ( 0 ) = sin ( 0 ) = 0 The first derivative of f ( x ) = sin x is f ′ ( x ) = cos x . Substitute 0 for x in f ′ ( x ) = cos x as, f ′ ( 0 ) = cos ( 0 ) = 1 The second derivative of f ( x ) = sin x is f ″ ( x ) = − sin x . Substitute 0 for x in f ″ ( x ) = − sin x as, f ″ ( 0 ) = − sin ( 0 ) = 0 The third derivative of f ( x ) = sin x is f ‴ ( x ) = − cos x Substitute 0 for x in f ‴ ( x ) = − cos x as, f ‴ ( 0 ) = − cos ( 0 ) = − 1 The Taylor polynomial centered at 0 for n = 3 is obtained as follows

Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book

2nd Edition

ISBN: 9780321954237

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

Publisher: PEARSON

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Chapter 9.1, Problem 3QC

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Chapter 9.1, Problem 2QC

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Chapter 9 Solutions

Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book

Ch. 9.1 - Prob. 1QC Ch. 9.1 - Prob. 2QC Ch. 9.1 - Prob. 3QC Ch. 9.1 - Prob. 4QC Ch. 9.1 - Prob. 5QC Ch. 9.1 - In Example 7, find an approximate upper bound for...Ch. 9.1 - Suppose you use a second-order Taylor polynomial...Ch. 9.1 - Does the accuracy of an approximation given by a...Ch. 9.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 9.1 - Prob. 4E

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The reason why the Taylor polynomials for sin x centered at 0 consist only of odd powers of x .

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To explain: The reason why the Taylor polynomials for sin x centered at 0 consist only of odd powers of x.

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Chapter 9 Solutions