Problem 32 - Essential Calculus

Question

Use the power series $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},|x|<1.$$ Find the series representation of the function and determine its interval of convergence. $$ f(x)=\frac{x}{(1-x)^{2}} $$

Step-by-Step Solution

Answer

The series representation of the function \( f(x) = \frac{x}{{(1-x)}^2} \) is \( f(x) = \sum_{n=1}^{\infty} n x^n \), and the interval of convergence is \( -1 < x < 1 \).

Step 1: Use Given Power Series

We use the given power series \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, |x|<1 \). This serves as the base which we will modify to get the required function \( f(x) \).

Step 2: Differentiate the Power Series

To get from \( \frac{1}{1-x} \) to \( f(x) \), we differentiate the given power series \( \frac{1}{1-x} \). Using the power rule \( (x^n)' = n x^{n-1} \) for all \( n \), the derivative of the power series is \( \sum_{n=1}^{\infty} n x^{n-1}, |x|<1 \). Notice that the lower limit is 1 because the term with \( n=0 \) vanishes after differentiation.

Step 3: Multiply the Result by x

We now have \( \frac{1}{{(1-x)}^2} = \sum_{n=1}^{\infty} n x^{n-1} \) which is close to the desired function. To reach this, we multiply the whole equation by \( x \) to get \( f(x) = \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n \).

Step 4: Find the Interval of Convergence

We now need to find the interval of convergence. We know from the given power series that the radius of convergence is 1. Because \( f(x) \) is a simple multiplication of the original series, the radius of convergence does not change. Hence, convergence holds for \( |x|<1 \), but does not include the endpoints \( x=1 \) or \( x=-1 \). Therefore the interval of convergence is \( -1 < x < 1 \).

Key Concepts

Series Representation

When we talk about the series representation of a function, we're essentially describing a way to write the function as an infinite sum of terms. This can be particularly useful for complex functions, as it breaks them down into more manageable parts. One common example of a series representation is the power series, which expresses a function as a sum of powers of a variable.

In the context of our problem, we take the well-known power series for the function \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n}, |x|<1 \). This series is only valid when the absolute value of the variable \(x\) is less than 1. In mathematics, this condition is known as the interval of convergence, which refers to the range of values for which the series converges or adds up to a finite number.

Interval of Convergence

The interval of convergence is crucial for using series to represent functions since it tells us where the series accurately describes the function. For the power series \( \sum_{n=0}^{\infty} x^{n} \), the interval of convergence is \( |x|<1 \), which means for all real numbers \( x \) where \( x \) is greater than \( -1 \) and less than \( 1 \) the series converges.

To determine the interval of convergence for a modified power series, like in our exercise, we need to examine any operations we performed on the original series. For the function \( f(x)=\frac{x}{(1-x)^{2}} \), we find that the interval of convergence remains \( -1 < x < 1 \) even after differentiation and multiplication by \( x \) because these operations do not affect the series' convergence properties. Thus, our solution remains accurate and represents our function within this interval.

Differentiation of Power Series

Differentiating a power series term-by-term is a powerful technique in calculus. This approach applies the power rule of differentiation \( (x^n)' = n x^{n-1} \) to each term of the series individually. For the series \( \sum_{n=0}^{\infty} x^{n} \) that represents \( \frac{1}{1-x} \), we differentiate each term to find the series representation for the function we're seeking, namely, \( \frac{x}{{(1-x)}^2} \).

As a result, we yield the new series \( \sum_{n=1}^{\infty} n x^{n-1} \) after applying differentiation, and with a slight transformation - multiplying by \(x\) - we achieve the desired series \( \sum_{n=1}^{\infty} n x^n \). This process shows that differentiation of power series is a tool that enables us to obtain new functions and their series representations from known ones. It is essential to keep in mind the interval of convergence throughout this process, as the differentiated series will only be valid within the same convergence interval as the original series.