Explanation Given: The quantity is 1 17 4 . Result used: (1) The Taylor series for x − 1 4 centered at 16 is 1 2 − 1 128 ( x − 16 ) + 5 16385 ( x − 16 ) 2 + ⋯ . (2) The remainder in the Taylor series centered at a is R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 for some c . Calculation: Obtain the value of 1 17 4 . Use the calculator the value of 1 17 4 is 0.4924790605. Obtain the approximation of 1 17 4 . The first derivatives of 1 x 4 computed as follows, y ′ = d d x ( − x 1 4 ) = 1 4 ( − x 1 4 − 1 ) = 1 4 ( − x − 3 4 ) The second derivatives of 1 x 4 computed as follows, y ″ = d d x ⋅ 1 4 ( − x 3 4 ) = 1 4 ( − 3 4 ) ( − x − 3 4 − 1 ) = − 3 16 ( x − 7 3 ) The third derivatives of 1 x 4 computed as follows, y ‴ = d d x ( − 3 16 ( x − 7 3 ) ) = ( − 3 16 ) ( − 7 3 ) ( x − 7 3 − 1 ) = 21 64 ( x − 11 3 ) In general the ( n + 1 ) th derivative of 1 x 4 is 1 ⋅ 5 ⋅ ⋯ ( 4 n + 1 ) 4 n + 1 x ( 4 n + 5 ) 4 That is | f ( n + 1 ) ( c ) | ≤ 1 ⋅ 5 ⋅ ⋯ ( 4 n + 1 ) 4 n + 1 ( 16 ) ( 4 n + 5 ) 4

Single Variable Calculus: Early Transcendentals Plus MyLab Math with Pearson eText -- Access Card Package (2nd Edition) (Briggs/Cochran/Gillett Calculus 2e)

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ISBN: 9780321965172

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

Publisher: PEARSON

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Chapter 9.3, Problem 86E

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To approximate: The given quantity with an error of 10−4 or less.

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Chapter 9.3, Problem 85E

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Chapter 9 Solutions

Single Variable Calculus: Early Transcendentals Plus MyLab Math with Pearson eText -- Access Card Package (2nd Edition) (Briggs/Cochran/Gillett Calculus 2e)

Ch. 9.1 - Prob. 1QC Ch. 9.1 - Prob. 2QC Ch. 9.1 - Prob. 3QC Ch. 9.1 - Prob. 4QC Ch. 9.1 - Prob. 5QC Ch. 9.1 - In Example 7, find an approximate upper bound for...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

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To approximate: The given quantity with an error of 10 − 4 or less.

Solution Summary: The author calculates the value of 1sqrt[4]17 using the calculator.

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Chapter 9 Solutions