Explanation Result used: The Taylor series for ( 1 + x ) − 1 2 , ( 1 + x ) − 1 2 = 1 − x 2 + 1 ⋅ 3 x 2 2 ⋅ 4 − 1 ⋅ 3 ⋅ 5 x 3 2 ⋅ 4 ⋅ 6 + ... The Taylor series for ln ( 1 + t ) , ln ( 1 + t ) = t − t 2 2 + t 3 3 − t 4 4 + t 5 5 − t 6 6 + t 7 7 − ... (1) Given: The inverse hyperbolic sine function sinh − 1 x = ln ( x + x 2 + 1 ) = ∫ 0 x d t 1 + t 2 . Calculation: The Taylor series for x + x 2 + 1 − 1 is x + x 2 + 1 − 1 = x + ( 1 + x 2 2 − x 4 8 + x 6 16 − ... ) − 1 x + x 2 + 1 − 1 = x + x 2 2 − x 4 8 + x 6 16 − ... (2) Substitute t = x + x 2 + 1 − 1 in the equation (1), ln ( 1 + x + x 2 + 1 − 1 ) = ln ( x + x 2 + 1 ) = x − x 3 6 + 3 x 5 40 − 5 x 7 112 +

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ISBN: 9780135182543

Author: Briggs

Publisher: PEARSON

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Chapter 11.4, Problem 70E

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To find: The first four terms of the Taylor series of sinh−1x.

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Chapter 11.4, Problem 69E

Chapter 11.4, Problem 71E

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Chapter 11 Solutions

CALCULUS:EARLY TRANSCENDENTALS-PACKAGE

Ch. 11.1 - Verify that p3 satisfies p3(k)(a)=f(k)(a), for k =...Ch. 11.1 - Verify the following properties for f(x) = sin x...Ch. 11.1 - Why do the Taylor polynomials for sin x centered...Ch. 11.1 - Write out the next two Taylor polynomials p4 and...Ch. 11.1 - At what point would you center the Taylor...Ch. 11.1 - In Example 7, find an approximate upper bound for...Ch. 11.1 - Suppose you use a second-order Taylor polynomial...Ch. 11.1 - Does the accuracy of an approximation given by a...Ch. 11.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 11.1 - Suppose f(0) = 1, f(0) = 2, and f(0) = 1. Find the...

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