Problem 18 - Essential Calculus

Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ f(x)=\frac{2}{(x+1)^{3}}=\frac{d^{2}}{d x^{2}}\left[\frac{1}{x+1}\right] $$

Step-by-Step Solution

Answer

The power series for the function \( f(x) \) is \( \sum_{n=2}^\infty 2n(n-1)(-1)^nx^{n-2} \). Its interval of convergence is \( -1 < x < 1 \).

Step 1: Find the First Derivative of the Power Series

Differentiate \( \sum_{n=0}^{\infty}(-1)^{n} x^{n} \) term-by-term: its derivative is \( \sum_{n=1}^\infty -n(-1)^{n}x^{n-1} \).

Step 2: Find the Second Derivative of the Power Series

Take the derivative of step 1 result again term-by-term: the second derivative is \( \sum_{n=2}^\infty n(n-1)(-1)^nx^{n-2} \).

Step 3: Adjust the Power Series for the Desired Function

Multiply by 2 since \( f(x) = \frac{2}{(x+1)^3} \): the power series is now \( \sum_{n=2}^\infty 2n(n-1)(-1)^nx^{n-2} \).

Step 4: Determine the Interval of Convergence

Because no additional multiplicative functions have been applied to the original power series, the interval of convergence doesn't change: \( -1 < x < 1 \).

Key Concepts

Convergence Interval

The concept of a convergence interval is crucial in understanding power series, as it indicates the values of the variable for which the series converges to a function.

Power series, represented by a sum of terms involving powers of a variable, have a specific range where the series will produce a finite sum. This is known as the interval of convergence. For the power series used in our exercise: \[ \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n} \], the interval of convergence is the set of values for which the series converges, or in simpler terms, 'adds up' to the function it represents: \[ \frac{1}{1+x} \].

Whenever we perform operations on power series, like differentiation, we need to reassess the interval to ensure that convergence is still preserved after the operation. However, in our case, differentiation does not change the radius of convergence. Step 4 of our solution confirms that the convergence interval remains as \( -1 < x < 1 \), meaning the power series for the original function and its derivatives will converge for all values of \( x \) within this interval.

Term-by-term Differentiation

Power series can often be manipulated using term-by-term differentiation, a technique pivotal to calculus involving power series.

This technique is based on differentiating each term of the power series separately, as demonstrated in our exercise. Starting with the given power series for \( \frac{1}{1+x} \), the first derivative is determined by differentiating each term to obtain \[ \( \sum_{n=1}^\infty -n(-1)^{n}x^{n-1} \) \]. Similarly, taking the second derivative involves applying the differentiation rule to each term of the first derivative series, producing \[ \( \sum_{n=2}^\infty n(n-1)(-1)^nx^{n-2} \) \].

Term-by-term differentiation is justified due to the uniform convergence of power series within their radius of convergence. It's important to remember that this method works seamlessly within the interval of convergence of the original series.

Power Series Representation

Power series representation is the expression of a function as an infinite sum of terms involving powers of variables. It is a fundamental concept in calculus, offering a way to analyze and manipulate functions in a versatile manner.

In our exercise, we seek a power series representation for the function \( f(x) = \frac{2}{(x+1)^3} \). This is achieved by twice differentiating the power series for \( \frac{1}{1+x} \), then scaling the result by applying the constant multiple rule, which states that the derivative of a constant times a function is the constant times the derivative of the function. This is illustrated in step 3 of the solution, resulting in \[ \( \sum_{n=2}^\infty 2n(n-1)(-1)^nx^{n-2} \)\] as the required power series representation.

Such representations allow for a range of operations, including integration, differentiation, and estimation of function values, making them a powerful tool in the study and application of calculus.