Explanation Given: | x | < 1 Calculation: Step 1: Using the Maclaurin series for ln (1 + x) and ln (1 − x), we have for |x| < 1 ln ( 1 + x ) − ln ( 1 − x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n x n = ∑ n = 1 ∞ ( − 1 ) n − 1 n ( − x ) n = ∑ n = 1 ∞ ( − 1 ) n − 1 n x n + ∑ n = 1 ∞ x n n = ∑ n = 1 ∞ 1 + ( − 1 ) n − 1 n x n Step 2: Since 1 + ( − 1 ) n − 1 = 0 for even n and 1 +

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Chapter 11.8, Problem 43E

To determine

Use the Maclaurin series for ln(1+x) and ln(1-x),

$\frac{1}{2} \ln (\frac{1 + x}{1 - x}) = x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...$ .

And, conclude by comparing this result with that of Exercise 40.

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Chapter 11.8, Problem 42E

Chapter 11.8, Problem 44E

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Chapter 11 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Use the Maclaurin series for ln(1+x) and ln(1-x), 1 2 ln ( 1 + x 1 − x ) = x + x 3 3 + x 5 5 + ... . And, conclude by comparing this result with that of Exercise 40.

Solution Summary: The author concludes by comparing this result with that of Exercise 40.

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Use the Maclaurin series for ln(1+x) and ln(1-x),

$\frac{1}{2} \ln (\frac{1 + x}{1 - x}) = x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...$ .

And, conclude by comparing this result with that of Exercise 40.

Want to see the full answer?

Chapter 11 Solutions

Use the Maclaurin series for ln(1+x) and ln(1-x), 1 2 ln ( 1 + x 1 − x ) = x + x 3 3 + x 5 5 + ... . And, conclude by comparing this result with that of Exercise 40.

Solution Summary: The author concludes by comparing this result with that of Exercise 40.

Concept explainers

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Use the Maclaurin series for ln(1+x) and ln(1-x), 12ln(1+x1−x)=x+x33+x55+.... And, conclude by comparing this result with that of Exercise 40.

Want to see the full answer?

Chapter 11 Solutions

Use the Maclaurin series for ln(1+x) and ln(1-x),

$\frac{1}{2} \ln (\frac{1 + x}{1 - x}) = x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...$ .

And, conclude by comparing this result with that of Exercise 40.