a. Explanation Given: The sine integral function is S i ( x ) = ∫ 0 x sin t t . Formula used: The Taylor series for sin x is sin x = x − x 3 3 ! + x 5 5 ! − ⋯ + ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! + ⋯ = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! . Calculation: Consider, sin t = t − t 3 3 ! + t 5 5 ! − ⋯ + ( − 1 ) k t 2 k + 1 ( 2 k + 1 ) ! + ⋯ = ∑ k = 0 ∞ ( − 1 ) k t 2 k + 1 ( 2 k + 1 ) ! b. To determine To find: The Taylor series for Si. c. To determine To approximate: The sine function S i ( 0.5 ) and S i ( 1 ) .

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 9.4, Problem 78E

To determine

To expand: The integrand in a Taylor series about 0.

To determine

To find: The Taylor series for Si.

To determine

To approximate: The sine function Si(0.5) and Si(1).

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Chapter 9.4, Problem 77E

Chapter 9.4, Problem 79E

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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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To expand: The integrand in a Taylor series about 0.

Solution Summary: The author explains how to simplify the Taylor series for mathrmsint.

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To expand: The integrand in a Taylor series about 0.

To find: The Taylor series for Si.

To approximate: The sine function Si(0.5) and Si(1).

Want to see the full answer?

Chapter 9 Solutions