Problem 9 - Essential Calculus Early Transcendentals

Question

Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{1+x}{1-x} $$

Step-by-Step Solution

Answer

The power series is \( 1 + \sum_{n=1}^{\infty} 2x^n \), with interval \((-1, 1)\).

Step 1: Express the function in a familiar form

Recognize that the function can be divided into two separate terms: \( \frac{1}{1-x} + x \frac{1}{1-x} \). The first term \( \frac{1}{1-x} \) is a standard geometric series \( \sum_{n=0}^{\infty} x^n \). Thus, we can separate the function as \( f(x) = \sum_{n=0}^{\infty} x^n + x \sum_{n=0}^{\infty} x^n \).

Step 2: Find the power series for each term

Apply the basic geometric series formula \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) for the first term. For the second term, multiply this series by \( x \), which results in \( x \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} x^{n+1} \). Hence, we now have two series: \( \sum_{n=0}^{\infty} x^n \) and \( \sum_{n=0}^{\infty} x^{n+1} \).

Step 3: Combine the two series

Sum the power series found in Step 2 into a single series. Start from \( n=0 \) for \( \sum_{n=0}^{\infty} x^n \) and shift the index for the second series, \( \sum_{n=0}^{\infty} x^{n+1} = \sum_{m=1}^{\infty} x^m \). Thus, we combine them to form \( \sum_{n=0}^{\infty} x^n + \sum_{m=1}^{\infty} x^m = \sum_{n=0}^{\infty} x^n + x^0 = \sum_{n=0}^{\infty} 2x^n = \sum_{n=1}^{\infty} 2x^n \).

Step 4: Simplify and finalize the power series expression

After rearranging the terms, you should get the final expression: \( 1 + 2x + 2x^2 + 2x^3 + \ldots = 1 + \sum_{n=1}^{\infty} 2x^n \). This is the power series representation of the function.

Step 5: Determine the interval of convergence

For a geometric series \( \sum_{n=0}^{\infty} ar^n \) with \( -1 < r < 1 \), it converges if the absolute value of \( r \) is less than 1. In our case, both series \( \sum_{n=0}^{\infty} x^n \) and \( \sum_{n=1}^{\infty} 2x^n \) share the same convergence criteria: \( |x| < 1 \). Therefore, the interval of convergence is \( (-1, 1) \).

Key Concepts

Geometric Series

A geometric series is a fundamental concept in mathematics that deals with the sum of an infinite number of terms. Each term in the sequence is a constant multiple, called the 'common ratio', of the previous term. In mathematical notation, a geometric series can be expressed as:

• For the series: \( a, ar, ar^2, ar^3, \ldots \)
• We write it as: \( S = a + ar + ar^2 + ar^3 + \ldots \)

The sum of this series, where \( |r| < 1 \), is given by:\[S = \frac{a}{1 - r}\]In the context of the problem, the function \( \frac{1}{1-x} \) has been identified as a geometric series with \( a = 1 \) and \( r = x \). This means each term of the series is simply another increment of \( x \). This is why it can be expanded into the series \( \sum_{n=0}^{\infty} x^n \). Recognizing this, and adjusting for the term \( x \frac{1}{1-x} \), allows the function \( f(x) \) to be expressed as the combination of two geometric series.

Interval of Convergence

The interval of convergence is a crucial aspect of a power series. It determines where a series will actually sum to a finite value, which means it's all about figuring out the values for which the series holds true.
For a geometric series like \( \sum_{n=0}^{\infty} x^n \), the series converges if the absolute value of the common ratio is less than one, i.e., \( |x| < 1 \). This means the series remains valid and does not diverge into infinity within this range.
In our problem, since both series terms, \( \sum_{n=0}^{\infty} x^n \) and \( \sum_{n=1}^{\infty} 2x^n \), share the same common ratio \( x \), the interval of convergence for \( f(x) \) is \( -1 < x < 1 \). This range ensures the power series expression of the function will provide meaningful, finite results and is vital to ensuring the practical use of the series in calculations.

Series Representation

Series representation involves expressing a function as an infinite sum of terms. Each term is generated in a systematic manner relating to powers of the variable. In the context of power series, this involves representing a function using a series of powers.
In our example, the function \( f(x) = \frac{1+x}{1-x} \) was represented as two geometric series combined:

• \( \sum_{n=0}^{\infty} x^n \)
• \( \sum_{n=0}^{\infty} x^{n+1} = \sum_{n=1}^{\infty} x^n \)

These series were combined to yield the complete power series representation:\[1 + \sum_{n=1}^{\infty} 2x^n\]This approach allows complex functions to be simplified into an infinite sum, making it easier to perform calculations and analyze their behavior within their interval of convergence. By breaking down an entire function into a series, it opens up avenues for approximations and deeper analysis, especially within calculus and mathematical analysis.