Explanation Result used: (1) Ratio test : Let ∑ k = 1 ∞ a k be an infinite series and let r = lim k → ∞ | a k + 1 a k | . If r < 1 , then the series ∑ k = 1 ∞ a k is converges. (2) The Raabe’s test: If there exist an N such that for all n ≥ N , a n ≥ 0 and lim n → ∞ ( n ( a n a n + 1 − 1 ) ) = L then the aeries ∑ a n converges if L > 1 and diverges if L > 1 . Given: The function is f ( x ) = 1 + x 2 . Calculation: The first four nonzero terms of the Maclaurin series is computed as follows, Substitute x 2 for x in the Maclaurin series 1 + x = 1 + x 2 − x 2 8 + x 3 16 − ⋯ , 1 + x 2 = 1 + x 2 2 − ( x 2 ) 2 8 + ( x 2 ) 3 16 − ⋯ = 1 + x 2 2 − x 4 8 + x 6 16 − ⋯ Therefore, the first four nonzero terms of the Maclaurin series centered at 0 is 1 + x 2 2 − x 4 8 + x 6 16 − ⋯ . The summation notation of the series is ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! 4 k ( k ! ) 2 ( 1 − 2 k ) x 2 k . Let a k = ( − 1 ) k ( 2 k ) ! 4 k ( k ! ) 2 ( 1 − 2 k ) x 2 k . Then, a k + 1 = ( − 1 ) ( k + 1 ) ( 2 ( k + 1 ) ) ! 4 k + 1 ( ( k + 1 ) ! ) 2 ( 1 − 2 ( k + 1 ) ) x 2 ( k + 1 ) . Use the Ratio test to compute the value | a k + 1 a k | . | a k + 1 a k | = | ( − 1 ) ( k + 1 ) ( 2 ( k + 1 ) ) ! 4 k + 1 ( ( k + 1 ) ! ) 2 ( 1 − 2 ( k + 1 ) ) x 2 ( k + 1 ) ( − 1 ) k ( 2 k ) ! 4 k ( k ! ) 2 ( 1 − 2 k ) x 2 k | Take lim k → ∞ on both the sides, lim k → ∞ | a k + 1 a k | = lim k → ∞ | ( − 1 ) ( k + 1 ) ( 2 ( k + 1 ) ) ! 4 k + 1 ( ( k + 1 ) ! ) 2 ( 1 − 2 ( k + 1 ) ) x 2 ( k + 1 ) ( − 1 ) k ( 2 k ) ! 4 k ( k ! ) 2 ( 1 − 2 k ) x 2 k | r = lim k → ∞ | ( − 1 ) ( k + 1 ) ( 2 k + 2 ) ! 4 k + 1 ( ( k + 1 ) ! ) 2 ( 1 − 2 k + 2 ) x 2 k + 2 ⋅ 4 k ( k ! ) 2 ( 1 − 2 k ) ( − 1 ) k ( 2 k ) ! x 2 k | = lim k → ∞ | ( 2 k + 1 ) ( 1 − 2 k ) 2 ( k + 1 ) ( 3 − 2 k ) x 2 | = lim k → ∞ | x 2 2 | ( ( 2 + 1 k ) ( 1 k − 2 ) ( 1 + 1 k ) ( 3 k − 2 ) x ) That is, r = | x 2 2 ⋅ ( 2 + 1 ∞ ) ( 1 ∞ − 2 ) ( 1 + 1 ∞ ) ( 3 ∞ − 2 ) | = | x 2 2 ⋅ ( 2 ) ( − 2 ) ( 1 ) ( − 2 ) | = | x 2 | From the above result, the series is converges as | x 2 | < 1

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 9.3, Problem 45E

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To find: The first four nonzero terms of the Maclaurian series and interval of convergence by using the Maclaurian series 1+x=1+x2−x28+x316−⋯, for −1≤x≤1.

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Chapter 9.3, Problem 44E

Chapter 9.3, Problem 46E

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Chapter 9 Solutions

Calculus: Early Transcendentals (2nd Edition)

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 Ch. 9.1 - How is the remainder Rn(x) in a Taylor polynomial...Ch. 9.1 - Explain how to estimate the remainder in an...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...

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The first four nonzero terms of the Maclaurian series and interval of convergence by using the Maclaurian series 1 + x = 1 + x 2 − x 2 8 + x 3 16 − ⋯ , for − 1 ≤ x ≤ 1 .

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To find: The first four nonzero terms of the Maclaurian series and interval of convergence by using the Maclaurian series 1+x=1+x2−x28+x316−⋯, for −1≤x≤1.

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Chapter 9 Solutions