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

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ISBN: 9781323178522

Author: Briggs

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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