Problem 14 - Thomas Calculus

Question

Find the Maclaurin series for the functions \(\frac{2+x}{1-x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \(\frac{2+x}{1-x}\) is: \(2 + 3x + 3x^2 + 3x^3 + \ldots\).

Step 1: Recognize the Known Series

First, recognize that the function can be broken down using the geometric series formula. The formula for a geometric series is given by \[ \sum_{n=0}^{\infty} ar^n = \frac{1}{1-r} \]if \(|r| < 1\). Here, we identify the function corresponding to this structure.

Step 2: Adjust to the Series Form

Given the function \(\frac{2+x}{1-x}\), express it as \(2 \cdot \frac{1}{1-x} + x \cdot \frac{1}{1-x} \). This is essentially splitting the function into two parts that are easier to expand into series.

Step 3: Expand Each Part into Series

For each part of the function:1. \(2 \cdot \frac{1}{1-x}\) can be expanded using \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), therefore: \( 2 \sum_{n=0}^{\infty} x^n = 2 + 2x + 2x^2 + 2x^3 + \ldots \)2. \(x \cdot \frac{1}{1-x} = x \sum_{n=0}^{\infty} x^n = \sum_{n=1}^{\infty} x^n\), thus: \( x + x^2 + x^3 + \ldots \)

Step 4: Combine Series Components

Add the two series from the previous step to form the Maclaurin series:\[ \left(2 + 2x + 2x^2 + 2x^3 + \ldots \right) + \left(x + x^2 + x^3 + \ldots \right) \]Combine like terms to get: \( 2 + 3x + 3x^2 + 3x^3 + \ldots \)

Step 5: Summarize the Maclaurin Series

The final Maclaurin series of the function \( \frac{2+x}{1-x} \) is:\[ 2 + 3x + 3x^2 + 3x^3 + \cdots = \sum_{n=0}^{\infty} (2 + n)x^n \]

Key Concepts

geometric series

A geometric series is a series of terms where each successive term is a constant multiple of the previous term. This constant is known as the 'common ratio'. Understanding this is crucial since many functions can be expressed or approximated as geometric series, thus simplifying complex expressions. In mathematical terms, a geometric series can be written as:\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \ldots \]where:

• \(a\) is the first term in the series,
• \(r\) is the common ratio,
• The series converges if the absolute value of \(r\) is less than 1 \(|r| < 1\).

In the context of the Maclaurin series, we utilize this concept by manipulating functions into a form that closely resembles the structure of a geometric series. For example, in the problem \( \frac{2+x}{1-x} \), we recognize \( \frac{1}{1-x} \) as the foundation for a geometric series expansion.

power series expansion

Power series expansion involves expressing a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is an essential tool for representing functions in mathematics because it provides an approximation that becomes more accurate with more terms included.A power series is generally of the form:\[ \sum_{n=0}^{\infty} a_n (x - c)^n \]where:

• \(a_n\) represents the coefficients determined by the function values or derivatives,
• \(x\) is a variable,
• \(c\) is the center of the series (for Maclaurin series, \(c = 0\)).

In the exercise, the power series expansion is critical because it transforms the function \( \frac{2+x}{1-x} \) into a format that we can easily manipulate and use to form the Maclaurin series. For the function \( \frac{1}{1-x} \), we recognize the power series expansion as an infinite series: \(1 + x + x^2 + x^3 + \ldots \).

series expansion

Series expansion is a method of writing a complex function as an infinite sum of simpler terms. This technique, especially in calculus, is vital for understanding the behavior of functions and performing computations based on approximations.There are different types of series expansions, such as Taylor and Maclaurin series. They are expressed as:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \]For the Maclaurin series, \(c\) is always zero, hence simplifying it to:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]In our exercise, series expansion is applied to the function \( \frac{2+x}{1-x} \). We decomposed it into simpler parts: \(2\cdot \frac{1}{1-x}\) and \(x \cdot \frac{1}{1-x}\), then expanded these into series. The final series expansion combines these results to yield a full Maclaurin series: \(2 + 3x + 3x^2 + 3x^3 + \ldots \). Series expansion helps to understand how the function behaves near the point of expansion, which is especially useful in analysis and approximation.