Problem 66 - Essential Calculus Early Transcendentals

Question

\(\begin{array}{l}{\text { (a) Expand } f(x)=\left(x+x^{2}\right) /(1-x)^{3} \text { as a power series. }} \\ {\text { (b) Use part (a) to find the sum of the series }}\end{array}\) $$\sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}}$$

Step-by-Step Solution

Answer

The sum of the series is 6.

Step 1: Recognize Inner Component and Power Series

Recognize that \( f(x) = \frac{x + x^2}{(1-x)^3} \) can be expanded as a power series. The fundamental component here is \( \frac{1}{(1-x)^3} \), which is known to be expanded as a power series through a standard formula, where \( (1-x)^{-3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \).

Step 2: Derive Inner Series

Using the binomial series expansion, \( x^n \), given by the series \( \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \), is applied to each term \( x \) and \( x^2 \). This gives separate series components for each part.

Step 3: Expand the Full Power Series

For the term \( x/(1-x)^3 \), we have: \( x \sum_{n=0}^{\infty} \binom{n+2}{2} x^n = \sum_{n=0}^{\infty} \binom{n+2}{2} x^{n+1} \). Similarly, for \( x^2/(1-x)^3 \), it expands to: \( x^2 \sum_{n=0}^{\infty} \binom{n+2}{2} x^n = \sum_{n=0}^{\infty} \binom{n+2}{2} x^{n+2} \).

Step 4: Consolidate Power Series

Combine these series to obtain: \( f(x) = \sum_{n=1}^{\infty} \binom{n+1}{2} x^n + \sum_{n=2}^{\infty} \binom{n}{2} x^n \). This requires aligning the starting indices to consolidate the coefficients at each power of \( x \).

Step 5: Simplify the Power Series

After aligning the indices, we get a combined series: \( f(x) = \sum_{n=1}^{\infty} \left( \binom{n+1}{2} + \binom{n}{2} \right) x^n \). Simplifying the coefficients, we find: \( f(x) = \sum_{n=1}^{\infty} n^2 x^n \).

Step 6: Find the Desired Series Sum

To find sum of \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \), substitute \( x = \frac{1}{2} \) into the series found in the previous step. Therefore: \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} = f\left(\frac{1}{2}\right) \).

Step 7: Calculate the Sum

Substitute \( x = \frac{1}{2} \) into \( f(x) \), so \( f\left(\frac{1}{2}\right) = \frac{\frac{1}{2} + \left(\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)^3} = \frac{\frac{1}{2} + \frac{1}{4}}{\left(\frac{1}{2}\right)^3} \). Simplify further to determine the actual value of the series sum.

Step 8: Simplify Final Expression

Calculating the above gives: \( f\left(\frac{1}{2}\right) = \frac{\frac{3}{4}}{\frac{1}{8}} = \frac{3}{4} \times 8 = 6 \).

Key Concepts

Binomial Series Expansion

The binomial series expansion helps us break down complex expressions into simpler, more manageable parts.
It is especially useful when dealing with expressions raised to a power, as it allows us to express these terms as infinite series.
For instance, in the exercise, we aim to expand \( \frac{1}{(1-x)^3} \), which is a special case of the binomial series.
Using the binomial theorem, the expression \( (1-x)^{-3} \) can be expanded into a series as follows:

• Start by recognizing the coefficient, represented by the binomial coefficient \( \binom{n+2}{2} \).
• This produces the infinite series: \( \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \).

This binomial series expansion forms the basis for further manipulation required in calculus problems and highlights the power of breaking down complex expressions.

Coefficient Simplification

Breaking down and simplifying coefficients is crucial when dealing with series to make them easier to understand and work with.
In the power series derived from the binomial expansion, each term comes with a binomial coefficient.
In the exercise, both \( \sum_{n=1}^{\infty} \binom{n+1}{2} x^n \) and \( \sum_{n=2}^{\infty} \binom{n}{2} x^n \) need simplification and alignment.

• Align the series indices for consistency, starting both series at the same point.
• Add the binomial coefficients: \( \binom{n+1}{2} + \binom{n}{2} \) for each power of \( x \).

This simplification results in a much cleaner expression: \( \sum_{n=1}^{\infty} n^2 x^n \), which is easier to substitute with different values and analyze further.

Series Summation

Summing a series, especially an infinite one, requires keen observation of its structure and behavior.
In the provided exercise, the goal is to find a numerical value for \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \).
Having already a straightforward series from the earlier simplification, substituting \( x = \frac{1}{2} \) into the power series \( \sum_{n=1}^{\infty} n^2 x^n \) efficiently provides the solution.

• Calculating the sum involves plugging \( x = \frac{1}{2} \) into the simplified series: \( f\left(\frac{1}{2}\right) \).
• This simplifies to evaluating \( \frac{\frac{3}{4}}{\frac{1}{8}} \), leading to the final answer of 6.

The process demonstrates the importance of substitution and simplifications when summing series in calculus.

Calculus Problem Solving

Solving calculus problems often involves combining various mathematical techniques.
This exercise illustrates several key steps: identifying the problem type, expanding via binomial series, simplifying coefficients, and summing series.
Here's a simplified approach to handling similar tasks:

• Identify key expressions: Recognize fundamental parts of the problem such as fractions that can be expanded.
• Expand using known series: Apply series expansions like the binomial theorem.
• Combine and simplify: Align terms and remove complexities through simplification.
• Substitute strategically: Choose suitable values for variables to find desired results.

Such strategies make solving complex calculus questions more manageable and reinforce the connection between series manipulation and problem-solving.