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 harmonic series is diverges. (3) The alternating harmonic series converges. Given: The function is f ( − 4 x ) = ln ( 1 + 4 x ) . Calculation: Obtain the power series by using the given result. Substitute -4 x for x in f ( − 4 x ) = ln ( 1 + 4 x ) , f ( − 4 x ) = ln ( 1 + 4 x ) = ln ( 1 − ( − 4 x ) ) = − ∑ k = 1 ∞ ( − 4 x ) k k ( ∵ ln ( 1 − x ) = − ∑ k = 1 ∞ x k k ) Therefore, the power series representation is − ∑ k = 1 ∞ ( − 4 x ) k k . Let a k = ( − 4 x ) k k . Then, a k + 1 = ( − 4 x ) k + 1 k + 1 . Use the Ratio test to compute the value | a k + 1 a k | . | a k + 1 a k | = | ( − 4 x ) k + 1 k + 1 ( − 4 x ) k k | Take lim k → ∞ on both sides, lim k → ∞ | a k + 1 a k | = lim k → ∞ | ( − 4 x ) k + 1 k + 1 ( − 4 x ) k k | = lim k → ∞ | ( − 4 x ) k + 1 k + 1 ⋅ k ( − 4 x )

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 9.2, Problem 40E

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To find: The power series representation and interval of convergence by using the power series representation f(x)=ln(1−x)=−∑k=1∞xkk, for −1≤x<1.

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Chapter 9.2, Problem 39E

Chapter 9.2, Problem 41E

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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 power series representation and interval of convergence by using the power series representation f ( x ) = ln ( 1 − x ) = − ∑ k = 1 ∞ x k k , for − 1 ≤ x < 1 .

Solution Summary: The author explains how to find the power series representation and interval of convergence.

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To find: The power series representation and interval of convergence by using the power series representation f(x)=ln(1−x)=−∑k=1∞xkk, for −1≤x<1.

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Chapter 9 Solutions

The power series representation and interval of convergence by using the power series representation f ( x ) = ln ( 1 − x ) = − ∑ k = 1 ∞ x k k , for − 1 ≤ x &lt; 1 .

Solution Summary: The author explains how to find the power series representation and interval of convergence.

Videos

To find: The power series representation and interval of convergence by using the power series representation f(x)=ln(1−x)=−∑k=1∞xkk, for −1≤x<1.

Want to see the full answer?

Chapter 9 Solutions

The power series representation and interval of convergence by using the power series representation f ( x ) = ln ( 1 − x ) = − ∑ k = 1 ∞ x k k , for − 1 ≤ x < 1 .