Problem 13 - Essential Calculus Early Transcendentals

Question

(a) Use differentiation to find a power series representation for $$ f(x)=\frac{1}{(1+x)^{2}} $$ What is the radius of convergence? (b) Use part (a) to find a power series for $$ f(x)=\frac{1}{(1+x)^{3}} $$ (c) Use part (b) to find a power series for $$ f(x)=\frac{x^{2}}{(1+x)^{3}} $$

Step-by-Step Solution

Answer

Power series for each function: (a) \(\sum_{n=1}^{\infty} n (-1)^{n+1} x^{n-1}\), (b) \(\sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^{n-2}\), (c) \(\sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^n\). Radius of convergence is 1.

Step 1: Express the given function as a power series

Recognize that the function \( f(x) = \frac{1}{(1+x)^2} \) is the derivative of the geometric series \( \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \). Differentiating both sides with respect to \( x \) gives:\[ \frac{d}{dx} \left( \frac{1}{1+x} \right) = \sum_{n=1}^{\infty} n (-1)^n x^{n-1}. \]Thus, \( \frac{-1}{(1+x)^2} = \sum_{n=1}^{\infty} n (-1)^n x^{n-1} \), and so \( \frac{1}{(1+x)^2} = -\sum_{n=1}^{\infty} n (-1)^n x^{n-1}. \) This can also be rewritten as:\[\frac{1}{(1+x)^2} = \sum_{n=1}^{\infty} n (-1)^{n+1} x^{n-1}.\]

Step 2: Determine the radius of convergence

The series \( \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \) converges for \( |x| < 1 \). Differentiation does not change the radius of convergence, so the power series representation of \( \frac{1}{(1+x)^2} \) also has a radius of convergence of 1.

Step 3: Differentiate again to find the next power series

To find the power series for \( f(x) = \frac{1}{(1+x)^3} \), differentiate the series from Step 1:The derivative of \( \frac{1}{(1+x)^2} = \sum_{n=1}^{\infty} n (-1)^{n+1} x^{n-1} \) is:\[ \frac{d}{dx} \left( \frac{1}{(1+x)^2} \right) = \sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^{n-2} \].Thus, \( \frac{1}{(1+x)^3} = \sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^{n-2}. \)

Step 4: Adjust the series for the given function with a multiplier

For \( f(x) = \frac{x^2}{(1+x)^3} \), multiply each term of the power series from Step 3 by \( x^2 \):\[ \frac{x^2}{(1+x)^3} = x^2 \sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^{n-2}. \]This simplifies to:\[ \sum_{n=1}^{\infty} n(n-1) (-1)^{n+1} x^n. \]

Key Concepts

Differentiation

In mathematics, differentiation is a crucial concept that helps us understand the rate of change of a function. When it comes to power series representation, differentiation can be used to find new series from existing ones by calculating their derivatives. To illustrate this, let's look at the process from the original exercise. The function \( f(x) = \frac{1}{(1+x)^2} \) is noted as being the derivative of the geometric series \( \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \). By differentiating each term in the series, we generate a new power series representation.**Key Differentiation Steps:**

• Identify the base function whose derivative corresponds to the given function.
• Ddifferentiate the geometric series term by term to find the derivative series.
• Substitute into the given function to find its power series representation.

This is how a complex function can be analyzed using the straightforward method of differentiation, making power series much easier to handle.

Radius of Convergence

The radius of convergence is a fundamental concept in power series, referring to the range of values for which the series converges to a finite sum. For any power series centered at zero, like \(\sum_{n=0}^{\infty} a_n x^n,\) the radius of convergence is determined by finding the distance from the center to the nearest point where the series does not converge. In our exercise, the geometric series \( \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \) has a radius of convergence of 1, which means it converges for \(|x| < 1\). Differentiation of the series—as shown in the solution—does not change this radius.**Why does the Radius of Convergence Remain Unchanged?**

• Differentiation or integration of a power series term by term retains its convergence properties.
• This means the radius stays the same even when new series are derived from differentiated functions.

This insight is essential when working with power series, ensuring that transformations like differentiation do not alter convergence specifics.

Geometric Series

The geometric series is at the heart of many topics in calculus and analysis, serving as a foundation for constructing more complex series. A classic geometric series can be expressed as \( \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \), which is simply a sum of powers of a given base. The importance of this series lies in its simplicity and versatility, as seen in its applications to exercise problems.Here's how it works in our context:

• Start with the standard form of the geometric series.
• Use differentiation to derive other related power series, as needed in the exercise.
• Apply transformations to the series to suit different functions (e.g., multiplying by \(x^2\) for a modification).

Through these steps, the geometric series acts as a launching point, making it easier to manipulate functions into series that are more suitable for analysis and problem-solving.