Explanation Given: The function is, f ( x ) = e 2 x , a = 3 . Result used: (1) 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 + ⋅ ⋅ ⋅ (2) The Ratio Test : (i) If lim n → ∞ | a n + 1 a n | = L < 1 , then the series ∑ n = 1 ∞ a n is absolutely convergent (and therefore convergent.) (ii) If lim n → ∞ | a n + 1 a n | = L > 1 or lim n → ∞ | a n + 1 a n | = ∞ , then the series ∑ n = 1 ∞ a n is divergent. (ii) If lim n → ∞ | a n + 1 a n | = 1 , the Ratio Test inconclusive; that is, no conclusion can be drawn about the convergence or divergence of ∑ n = 1 ∞ a n . Calculation: Consider the function f ( x ) = e 2 x centered at a = 3 . Since the function f ( x ) = e 2 x at 3 is f ( 3 ) = e 6 . Find the first derivative of f ( x ) at a = 3 . f ′ ( x ) = d d x ( e 2 x ) = 2 e 2 x f ′ ( x ) = 2 e 2 x (1) Obtain f ′ ( 3 ) . f ′ ( 3 ) = 2 e 6 Find the second derivative of f ( x ) at a = 3 . f ( 2 ) ( x ) = d 2 d x 2 ( f ( x ) ) = d d x ( f ′ ( x ) ) = d d x ( 2 e 2 x ) ( by equation ( 1 ) ) = 2 ( 2 ) e 2 x That is, f ( 2 ) ( x ) = 2 2 e 2 x (2) Compute f ( 2 ) ( 3 ) . f ( 2 ) ( 3 ) = 4 e 2 ( 3 ) = 4 e 6 That is, f ( 2 ) ( 3 ) = 4 e 6 . Find the third derivative of f ( x ) at a = 3 as follows. f ( 3 ) ( x ) = d 3 d x 3 ( e 2 x ) = d d x ( f ( 2 ) ( x ) ) = d d x ( 2 2 e 2 x ) ( by equation ( 2 ) ) = 2 2 ( 2 ) e 2 x That is, f ( 3 ) ( x ) = 2 3 e 2 x (3) Compute f ( 3 ) ( 3 ) . f ( 3 ) ( 3 ) = 2 3 e 2 ( 3 ) = 8 e 6 Find the fourth derivative of f ( x ) at a = 3 . f ( 4 ) ( x ) = d 4 d x 4 ( e 2 x ) = d d x ( f ( 3 ) ( x ) ) = d d x ( 2 3 e 2 x ) ( by equation ( 3 ) ) = 2 3 ( 2 ) e 2 x That is, f ( 4 ) ( x ) = 2 4 e 2 x (4) Compute f ( 4 ) ( 3 )

Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

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Chapter 11.10, Problem 23E

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To find: The Taylor series for f(x) centered at 3 and radius of convergence.

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Chapter 11.10, Problem 22E

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Chapter 11 Solutions

Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

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 - List the first five terms of the sequence. 3....Ch. 11.1 - List the first five terms of the sequence. 4....Ch. 11.1 - List the first five terms of the sequence. 5....Ch. 11.1 - List the first five terms of the sequence. 6....Ch. 11.1 - List the first five terms of the sequence. 7....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 3 and radius of convergence.

Solution Summary: The author explains how to find the Taylor series for f(x) centered at 3 and radius of convergence.

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To find: The Taylor series for f(x) centered at 3 and radius of convergence.

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Chapter 11 Solutions