Explanation Result used: Taylor series for e x , e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + ... . Given: The function f ( x ) = ( e x − 1 ) x , x ≠ 0 , f ( 0 ) = 1 . Calculation: The Taylor series for the function e x − 1 x is expressed as follows, e x − 1 x = ( 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + ... ) − 1 x = x + x 2 2 ! + x 3 3 ! + x 4 4 ! + ... x = 1 + x 2 ! + x 2 3 ! + x 3 4 ! + ... On further simplification, e x − 1 x = ∑ k = 0 ∞ x k k ! − 1 x = ∑ k = 1 ∞ x k k ! x = ∑ k = 1 ∞ x k − 1 k ! = ∑ k = 0 ∞ x k ( k + 1 ) ! Differentiate the Taylor series for f ( x ) , d d x

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 9.4, Problem 52E

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To find: The value of ∑k=1∞k2k−1(k+1)! and evaluate f′(2).

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Chapter 9.4, Problem 51E

Chapter 9.4, Problem 53E

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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 value of ∑ k = 1 ∞ k 2 k − 1 ( k + 1 ) ! and evaluate f ′ ( 2 ) .

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To find: The value of ∑k=1∞k2k−1(k+1)! and evaluate f′(2).

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Chapter 9 Solutions

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