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 p ( x ) = 2 x 6 ln ( 1 − x ) . Calculation: Obtain the power series by using the given result. Consider the function p ( x ) = 2 x 6 ln ( 1 − x ) , p ( x ) = 2 x 6 ln ( 1 − x ) = − 2 x 6 ∑ k = 1 ∞ x k k ( ∵ ln ( 1 − x ) = − ∑ k = 1 ∞ x k k ) = − 2 ∑ k = 1 ∞ x k + 6 k Therefore, the power series representation is − 2 ∑ k = 1 ∞ x k + 6 k . Let a k = x k + 6 k . Then, a k + 1 = x k + 7 k + 1 . Use the Ratio test to compute the value | a k + 1 a k | . | a k + 1 a k | = | x k + 7 k + 1 x k + 6 k | Take lim k → ∞ on both sides. lim k → ∞ | a k + 1 a k | = lim k → ∞ | x k + 7 k + 1 x k + 6 k | = lim k → ∞ | x k + 7 k + 1 ⋅ k x k + 6 | =

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

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

Publisher: PEARSON

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Chapter 11.2, Problem 49E

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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 11.2, Problem 48E

Chapter 11.2, Problem 50E

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Chapter 11 Solutions

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

Ch. 11.1 - Verify that p3 satisfies p3(k)(a)=f(k)(a), for k =...Ch. 11.1 - Prob. 2QC Ch. 11.1 - Prob. 3QC Ch. 11.1 - Write out the next two Taylor polynomials p4 and...Ch. 11.1 - Prob. 5QC Ch. 11.1 - Prob. 6QC Ch. 11.1 - Suppose you use a second-order Taylor polynomial...Ch. 11.1 - Does the accuracy of an approximation given by a...Ch. 11.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 11.1 - Prob. 4E

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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 calculates the power series representation and interval of convergence by using the given result.

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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.

Want to see the full answer?

Chapter 11 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 calculates the power series representation and interval of convergence by using the given result.

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 11 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 .