Explanation Given: The function is f ( x ) = x 5 + 2 x 3 + x . Result used: If f has a power series expansion at a , f ( x ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n , f ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + f ‴ ( a ) 3 ! ( x − a ) 3 + ⋅ ⋅ ⋅ Calculation: Obtain the first four nonzero terms of the series as follows, Since the function f ( x ) = x 5 + 2 x 3 + x and obtain f ( 2 ) = 50 . f ( 2 ) = ( 2 ) 5 + 2 ( 2 ) 3 + ( 2 ) = 32 + 16 + 2 = 50 Find the first derivative of f ( x ) at a = 2 . f ′ ( x ) = d d x ( x 5 + 2 x 3 + x ) = d d x ( x 5 ) + 2 d d x ( x 3 ) + d d x ( x ) = 5 x 4 + 2 ( 3 x 2 ) + 1 = 5 x 4 + 6 x 2 + 1 f ′ ( x ) = 5 x 4 + 6 x 2 + 1 (1) Substitute 2 for x in f ′ ( x ) , f ′ ( 2 ) = 5 ( 2 ) 4 + 6 ( 2 ) 2 + 1 = 5 ( 16 ) + 24 + 1 = 80 + 25 = 105 That is, f ′ ( 2 ) = 105 . Find the second derivative of f ( x ) at a = 2 . f ( 2 ) ( x ) = d 2 d x 2 ( f ( x ) ) = d d x ( f ′ ( x ) ) = d d x ( 5 x 4 + 6 x 2 + 1 ) ( by equation ( 1 ) ) = 5 d d x ( x 4 ) + 6 d d x ( x 2 ) + d d x ( 1 ) Simplify further and obtain f ( 2 ) ( 2 ) as shown below. f ′ ( x ) = 5 ( 4 x 3 ) + 6 ( 2 x ) + 0 = 20 x 3 + 12 x f ( 2 ) ( x ) = 20 x 3 + 12 x (2) Substitute 2 for x in f ( 2 ) ( x ) , f ( 2 ) ( 2 ) = 20 ( 2 ) 3 + 12 ( 2 ) = 160 + 24 = 184 That is, f ( 2 ) ( 2 ) = 184 Since the third derivative of f ( x ) at a = 2 . f ( 3 ) ( x ) = d 3 d x 3 ( f ( x ) ) = d d x ( f ( 2 ) ( x ) ) = d d x ( 20 x 3 + 12 x ) ( by equation ( 2 ) ) = 20 d d x ( x 3 ) + 12 d d x ( x ) Simplify further and obtain f ( 3 ) ( 2 ) , f ( 3 ) ( x ) = 20 ( 3 x 2 ) + 12 ( 1 ) = 60 x 2 + 12 f ( 3 ) ( x ) = 60 x 2 + 12 (3) Substitute 2 for x in f ( 3 ) ( x ) , f ( 3 ) ( 2 ) = 60 ( 2 ) 2 + 12 = 240 + 12 = 252 That is, f ( 3 ) ( 2 ) = 252 Find the fourth derivative of f ( x ) at a = 2 . f ( 4 ) ( x ) = d 4 d x 4 ( x 5 + 2 x 3 + x ) = d d x ( f ( 3 ) ( x ) ) = d d x ( 60 x 2 + 12 ) ( by equation(3) ) = 60 d d x ( x 2 ) + d d x ( 12 ) Simplify further and obtain f ( 4 ) ( 2 ) as shown below

Student Solutions Manual, Chapters 10-17 for Stewart's Multivariable Calculus, 8th (James Stewart Calculus)

8th Edition

ISBN: 9781305271821

Author: James Stewart

Publisher: Cengage Learning

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Chapter 11.10, Problem 19E

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To find: The Taylor series for f(x) centered at 2 and radius of convergence.

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Chapter 11.10, Problem 18E

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Chapter 11 Solutions

Student Solutions Manual, Chapters 10-17 for Stewart's Multivariable Calculus, 8th (James Stewart Calculus)

Ch. 11.1 - (a) What is a sequence? (b) What does it mean to...Ch. 11.1 - (a) What is a convergent sequence? Give two...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - List the first five terms of the sequence. 5....Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - List the first five terms of the sequence. 8....Ch. 11.1 - Prob. 9E Ch. 11.1 - List the first five terms of the sequence. 10. a1...

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The Taylor series for f ( x ) centered at 2 and radius of convergence.

Solution Summary: The author explains how to find the Taylor series for f(x) centered at 2 and radius of convergence.

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To find: The Taylor series for f(x) centered at 2 and radius of convergence.

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Chapter 11 Solutions