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 +

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

See similar textbooks

Concept explainers

Angles in Circles

Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made p…

Arcs in Circles

A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.

Videos

Question

Chapter 11.7, Problem 41E

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.7, Problem 40E

Chapter 11.7, Problem 42E

Students have asked these similar questions

i need help please

slove the initial-value problem. y'' + 7y' + 12y = 0, y(0) = 2, y'(0) = 8

Chapter 11 Solutions

Calculus - Standalone book

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

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

Videos

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

Videos

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.