Explanation Result used: Taylor series for ln ( 1 + x ) , ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ... = ∑ k = 1 ∞ ( − 1 ) k + 1 x k k . Given: The function is f ( x ) = ln ( 1 + x ) . Calculation: Use the above mentioned result it can be conclude that, f ( x ) = ( − 1 ) k + 1 x k k . Substitute f ( − 1 2 ) in the equation f ( x ) = ln ( 1 + x ) , ln ( 1 − 1 2 ) = ∑ k = 1 ∞

Calculus: Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

3rd Edition

ISBN: 9780134995991

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

Publisher: PEARSON

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Chapter 11.4, Problem 54E

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To evaluate: The function at f(−12) to find the value of ∑k=1∞1k⋅2k.

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Chapter 11.4, Problem 53E

Chapter 11.4, Problem 55E

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Chapter 11 Solutions

Calculus: Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

Ch. 11.1 - Verify that p3 satisfies p3(k)(a)=f(k)(a), for k =...Ch. 11.1 - Verify the following properties for f(x) = sin x...Ch. 11.1 - Why do the Taylor polynomials for sin x centered...Ch. 11.1 - Write out the next two Taylor polynomials p4 and...Ch. 11.1 - At what point would you center the Taylor...Ch. 11.1 - In Example 7, find an approximate upper bound for...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 - Suppose f(0) = 1, f(0) = 2, and f(0) = 1. Find the...

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The function at f ( − 1 2 ) to find the value of ∑ k = 1 ∞ 1 k ⋅ 2 k .

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To evaluate: The function at f(−12) to find the value of ∑k=1∞1k⋅2k.

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