Explanation Given: The function f ( x ) = 1 4 + x 2 , a = 0 . Result used: The Taylor series centered at a is, f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) ( x − a ) 2 2 ! + f ‴ ( a ) ( x − a ) 3 3 ! + ⋯ = ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k Calculation: Obtain the first three nonzero terms of the Taylor series f ( x ) = 1 4 + x 2 centered at a = 0 . The value of f ( x ) = 1 4 + x 2 and its derivatives at a = 0 are computed in the below table, k f k ( x ) f k ( 0 ) 0 f ( x ) = 1 4 + x 2 f ( 0 ) = 1 4 1 f ′ ( x ) = − 2 x ( 4 + x 2 ) 2 f ′ ( 0 ) = 0 2 f ″ ( x ) = − 2 ( − 3 x 2 + 4 ) ( 4 + x 2 ) 3 f ″ ( 0 ) = − 1 8 3 f ‴ ( x ) = − 24 x ( − x 2 + 4 ) ( 4 + x 2 ) 4 f ‴ ( 0 ) = 0 4 f 4 ( x ) = 24 ( 5 x 4 − 40 x 2 + 16 ) ( 4 + x 2 ) 5 f 4 ( 0 ) = 3 8 5 f 5 ( x ) = 240 x

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

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

Publisher: PEARSON

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Chapter 9 Solutions

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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 Ch. 9.1 - How is the remainder Rn(x) in a Taylor polynomial...Ch. 9.1 - Explain how to estimate the remainder in an...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...

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