Explanation 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: Consider the function f ( x ) = x 6 − x 4 + 2 centered at a = − 2 . Obtain the first four nonzero terms of the series as follows, The function f ( x ) = x 6 − x 4 + 2 and obtain f ( − 2 ) . f ( 2 ) = ( − 2 ) 6 − ( − 2 ) 4 + 2 = 64 − 16 + 2 = 50 That is, f ( − 2 ) = 50 . The first derivative of f ( x ) at a = − 2 as follows, f ′ ( x ) = d d x ( x 6 − x 4 + 2 ) = d d x ( x 6 ) − d d x ( x 4 ) + d d x ( 2 ) = 6 x 5 − ( 4 x 3 ) + 0 = 6 x 5 − 4 x 3 f ′ ( x ) = 6 x 5 − 4 x 3 (1) Obtain f ′ ( − 2 ) . f ′ ( − 2 ) = 6 ( − 2 ) 5 − 4 ( − 2 ) 3 = 6 ( − 32 ) − 4 ( − 8 ) = − 192 + 32 = − 160 That is, f ′ ( − 2 ) = − 160 The second derivative of f ( x ) at a = − 2 as follows, f ( 2 ) ( x ) = d 2 d x 2 ( f ( x ) ) = d d x ( f ′ ( x ) ) = d d x ( 6 x 5 − 4 x 3 ) ( by equation(1) ) = 6 d d x ( x 5 ) − 4 d d x ( x 3 ) Simplify further and obtain f ( 2 ) ( − 2 ) , f ′ ( x ) = 6 ( 5 x 4 ) − 4 ( 3 x 2 ) = 30 x 4 − 12 x 2 f ( 2 ) ( x ) = 30 x 4 − 12 x 2 (2) Compute f ( 2 ) ( − 2 ) . f ( 2 ) ( − 2 ) = 30 ( − 2 ) 4 − 12 ( − 2 ) 2 = 30 ( 16 ) − 12 ( 4 ) = 480 − 48 = 432 That is, f ( 2 ) ( − 2 ) = 432 The third derivative of f ( x ) at a = − 2 as follows, f ( 3 ) ( x ) = d 3 d x 3 ( f ( x ) ) = d d x ( f ( 2 ) ( x ) ) = d d x ( 30 x 4 − 12 x 2 ) ( by equation(2) ) = 30 d d x ( x 4 ) − 12 d d x ( x 2 ) Obtain f ( 3 ) ( − 2 ) . f ( 3 ) ( x ) = 30 ( 4 x 3 ) − 12 ( 2 x ) = 120 x 3 − 24 x f ( 3 ) ( x ) = 120 x 3 − 24 x (3) Compute f ( 3 ) ( − 2 ) . f ( 3 ) ( − 2 ) = 120 ( − 2 ) 3 − 24 ( − 2 ) = 120 ( − 8 ) + 48 = − 960 + 48 = − 912 That is, f ( 3 ) ( − 2 ) = − 912 . The fourth derivative of f ( x ) at a = − 2 as follows, f ( 4 ) ( x ) = d 4 d x 4 ( f ( x ) ) = d d x ( f ( 3 ) ( x ) ) = d d x ( 120 x 3 − 24 x ) ( by equation(3) ) = 120 d d x ( x 3 ) − 24 d d x ( x ) f ( 4 ) ( x ) = 120 ( 3 x 2 ) − 24 ( 1 ) = 360 x 2 − 24 f ( 4 ) ( x ) = 360 x 2 − 24 (4) Compute f ( 4 ) ( − 2 )

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Author: Stewart

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Chapter 11.10, Problem 20E

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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 19E

Chapter 11.10, Problem 21E

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Chapter 11 Solutions

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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 −2 and radius of convergence.

Solution Summary: The author calculates 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