Question

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

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 \( x + \frac{x^3}{3} + \frac{x^5}{5} + \text{...} \).

Key concepts

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.