Problem 65 - Essential Calculus Early Transcendentals

Question

\(\begin{array}{l}{\text { (a) Expand } f(x)=x /(1-x)^{2} \text { as a power series. }} \\ {\text { (b) Use part (a) to find the sum of the series }}\end{array}\) $$\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$

Step-by-Step Solution

Answer

The sum of the series is 2.

Step 1: Recognize the function structure

Recognize that we need to expand the function \( f(x) = \frac{x}{(1-x)^2} \) as a power series. This function resembles the derivative of a geometric series.

Step 2: Recall geometric series expansion

Recall that the geometric series \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \) when \(|x| < 1\). Differentiating this series with respect to \(x\) gives \( \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \ldots\).

Step 3: Multiply by x

To find the expansion of \(\frac{x}{(1-x)^2}\), multiply the series obtained in Step 2 by \(x\): \[f(x) = x (1 + 2x + 3x^2 + 4x^3 + \ldots) = x + 2x^2 + 3x^3 + 4x^4 + \ldots\]

Step 4: Identify the form to relate it to the summation

Observe that the power series in Step 3 is \(x + 2x^2 + 3x^3 + \ldots\). Let's express this as \(\sum_{n=1}^{\infty} n x^n\).

Step 5: Substitute to find the sum

Now, to find \(\sum_{n=1}^{\infty} \frac{n}{2^n}\), substitute \(x = \frac{1}{2}\) into our power series from Step 4: \[f\left(\frac{1}{2}\right) = \sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^n\]This simplifies to \(\frac{\frac{1}{2}}{(1-\frac{1}{2})^2} = \frac{1/2}{(1/2)^2} = 2\).

Step 6: Conclusion: Sum of the series

Based on the expansion and substitution, \( \sum_{n=1}^{\infty} \frac{n}{2^n} = 2 \).

Key Concepts

Geometric Series

A geometric series is a series of numbers where each term is a constant multiple of the previous term. It has a basic structure formed by terms that build upon a single initial term using a common ratio. The standard geometric series can be written as:

• First term: 1
• Common ratio: x
• General form: 1 + x + x^2 + x^3 + \ldots

The significance of geometric series becomes apparent when discussing convergence; the series converges when the absolute value of the common ratio is less than one, \(|x| < 1\). This convergence implies that we can accurately find the sum at the given constraints, using the formula: \[\frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots\]This formula helps us in problems that require understanding the behavior of functions similar to the classic geometric series, like the one shown in the exercise.

Series Expansion

Series expansion is a way to express a function as a sum of terms calculated from its derivatives at a single point. For instance, when you begin with a function similar to the classic geometric series, understanding its series expansion can aid in solving complex problems.
Series expansion often helps to approximate functions in simpler forms over specific intervals. Consider the function \( f(x) = \frac{x}{(1-x)^2} \) from our exercise. First, we identify its relationship with the geometric series by recognizing it as a resulting form after a differentiation.
By understanding its relation, we can express it through a series in simpler terms, such as:

• Initial form: \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \)
• Derivative form: \( \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \ldots \)

These series expansions provide a useful tool for representing functions, making it easier to work with them in various mathematical contexts, such as translating them into known series to solve for specific answers.

Differentiation of Series

Differentiation of series involves taking the derivative of each term in the series individually. This concept is crucial for transforming a given function into a power series that we can use to solve exercises, much like the one we encountered here.
Let's take a moment to recall how the derivative applies to our geometric series:

• Original series: \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \)
• Derivative: \( \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \ldots \)

This process helps us see not just how the original function works, but also how it behaves under various transformations. The differentiation allows us to apply this understanding to determine additional properties of the series and offer solutions to related problems.