Problem 19 Use the methods of this section ... [FREE SOLUTION]

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Chapter 1: Problem 19

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ \ln \sqrt{\frac{1+x}{1-x}}=\int_{0}^{x} \frac{d t}{1-t^{2}} $$

Short Answer

Expert verified

The first few terms of the Maclaurin series are $ x + \frac{x^3}{3} + \frac{x^5}{5} + \text{...} $.

Step by step solution

Identify the Function

The given function is $\frac{d t}{1-t^{2}}$. We need to find the Maclaurin series for it.

Expand the Function

We will expand the integral function $ \frac{1}{1-t^{2}} $ as a series. $\frac{1}{1-t^{2}} = \frac{1}{1 - t + t^2 - t^3 + \text{...}} $. This is a geometric series with the first term 1 and common ratio $t^{2} $.

Integrate Term-by-Term

Next, integrate the series term-by-term from 0 to x. $ \frac{1}{1-t} $ integrates to $ t + \frac{t^3}{3} + \frac{t^5}{5} + \text{...} $. Applying the limits, we get: $ x + \frac{x^3}{3} + \frac{x^5}{5} + \text{...} $.

Express the Maclaurin Series

The Maclaurin series for the given function $ \frac{d t}{1-t^{2}} $ is $ x + \frac{x^3}{3} + \frac{x^5}{5} + \text{...} $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

geometric series

A geometric series is a series of the form $\text{ sum } = a + ar + ar^2 + ar^3 + \text{...} $, where $a$ is the first term and $r$ is the common ratio. This type of series is important because its sum can be calculated if $ | r | < 1 $. The given problem simplifies $ \frac{1}{1-t^2}$ as a geometric series $ 1 + t^2 + t^4 + t^6 + \text{...}$. This simplification makes it easier to integrate term-by-term.

integral functions

Integral functions involve finding the area under a curve described by a function. In the problem, we need to integrate the function $ \frac{1}{1-t^2} $. Integration helps us express the problem as a sum of simpler terms.

term-by-term integration

Term-by-term integration is a key technique for handling series. Instead of integrating a complex function directly, the series allows us to split it into simpler terms and integrate them individually. For example, integrating $ 1 + t^2 + t^4 + \text{...} $ gives $ t + \frac{t^3}{3} + \frac{t^5}{5} + \text{...} $. This method can make complicated problems more manageable.

function expansion

Function expansion means rewriting a function as a series. This is useful when we want to approximate the function near a specific point. For instance, the Maclaurin series is the Taylor series expansion of a function around zero. In the problem, expanding $ \frac{1}{1-t^2} $ into a series allows us to integrate and obtain terms like $ x + \frac{x^3}{3} + \frac{x^5}{5} + \text{...}$, making it easier to analyze the function near zero.

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