a. Explanation Given: The function is f ( x ) = sin x centered at 0 and π . Result used: The n th-order Taylor polynomial centered at a is as follows, p n ( x ) = f 0 ( a ) 0 ! ( x − a ) 0 + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f n ( a ) n ! ( x − a ) n . Calculation: The n th-order Taylor polynomial centered at a = 0 is as follows, p n ( x ) = f ( 0 ) + f ′ ( 0 ) 1 ! ( x ) + f ″ ( 0 ) 2 ! ( x ) 2 + ⋯ + f n ( 0 ) n ! ( x ) n . The value of f ( x ) = sin x and its derivatives at x = 0 as shown in the below table. n f n ( x ) f n ( 0 ) 0 f ( x ) = sin x f ( 0 ) = 0 1 f ′ ( x ) = cos x f ′ ( 0 ) = 1 2 f ″ ( x ) = − sin x f ″ ( 0 ) = 0 3 f ‴ ( x ) = − cos x f ‴ ( 0 ) = − 1 4 f 4 ( x ) = sin x f 4 ( 0 ) = 0 5 f 5 ( x ) = cos x f 5 ( 0 ) = 1 Plug the required value in p 5 ( x ) from the table, p 5 ( x ) = f ( 0 ) + f ′ ( 0 ) 1 ! ( x ) + f ″ ( 0 ) 2 ! ( x ) 2 + f ‴ ( 0 ) 3 ! ( x ) 3 + f 4 ( 0 ) 4 ! ( x ) 4 + f 5 ( 0 ) 5 ! ( x ) 5 = 0 + 1 1 ! ( x ) + 0 2 ! ( x ) 2 + ( − 1 ) 3 ! ( x ) 3 + 0 4 ! ( x ) 4 + 1 5 ! ( x ) 5 = x − x 3 3 ! + x 5 5 ! Therefore, the 5th-order Taylor polynomial centered at a = 0 is p 5 ( x ) = x − x 3 3 ! + x 5 5 ! . The n th-order Taylor polynomial centered at a = π is as follows, q n ( x ) = f 0 ( π ) 0 ! ( x − π ) 0 + f ′ ( π ) 1 ! ( x − π ) 1 + f ″ ( π ) 2 ! ( x − π ) 2 + ⋯ + f n ( π ) n ! ( x − π ) n b. To determine To sketch: The function f , the Taylor polynomials p 5 and q 5 . on the interval [ − π , 2 π ] and to find which interval p 5 better approximation to f than q 5 and which interval q 5 better approximation to f than p 5 . c. To determine To find: The error approximation of | sin x − p 5 ( x ) | and | sin x − q 5 ( x ) | . d. To determine To find: The point from the given table such that p 5 ( x ) is better approximation to f than q 5 ( x ) as well as comparable approximation between them and comments on that approximations.

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 95E

To determine

To find: The Taylor series for p5 and q5.

To determine

To sketch: The function f, the Taylor polynomials p5 and q5. on the interval [−π,2π] and to find which interval p5 better approximation to f than q5 and which interval q5 better approximation to f than p5.

To determine

To find: The error approximation of |sinx−p5(x)| and |sinx−q5(x)|.

To determine

To find: The point from the given table such that p5(x) is better approximation to f than q5(x) as well as comparable approximation between them and comments on that approximations.

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Chapter 9.1, Problem 94E

Chapter 9.1, Problem 96E

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Chapter 9 Solutions

Calculus: Early Transcendentals (2nd Edition)

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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The Taylor series for p 5 and q 5 .

Solution Summary: The author calculates the n th-order Taylor polynomial centered at a=0.

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To find: The Taylor series for p5 and q5.

To sketch: The function f, the Taylor polynomials p5 and q5. on the interval [−π,2π] and to find which interval p5 better approximation to f than q5 and which interval q5 better approximation to f than p5.

To find: The error approximation of |sinx−p5(x)| and |sinx−q5(x)|.

To find: The point from the given table such that p5(x) is better approximation to f than q5(x) as well as comparable approximation between them and comments on that approximations.

Want to see the full answer?

Chapter 9 Solutions