Problem 11 - Essential Calculus Early Transcendentals

Question

Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence. $$ f(x)=\frac{3}{x^{2}-x-2} $$

Step-by-Step Solution

Answer

The power series is \( \sum_{n=0}^{\infty} \left( -\frac{1}{2} \cdot \left(\frac{x}{2}\right)^n + (-1)^n x^n \right) \) with interval of convergence \(|x|<1\).

Step 1: Factor the Denominator

Start by factoring the denominator of the function. The denominator \(x^2 - x - 2\) can be factored as \((x-2)(x+1)\). This gives us the fraction: \(\frac{3}{(x-2)(x+1)}\).

Step 2: Set Up Partial Fractions

Write the expression \( \frac{3}{(x-2)(x+1)} \) as a sum of simpler fractions: \( \frac{A}{x-2} + \frac{B}{x+1} \).

Step 3: Solve for A and B

Multiply through by the common denominator \((x-2)(x+1)\) to get \[ 3 = A(x+1) + B(x-2) \]Expand and rearrange terms to find A and B by plugging in suitable values of \(x\). By substituting \(x = 2\), \(3 = A(2+1) + B(0) \rightarrow 3 = 3A \rightarrow A = 1\).By substituting \(x = -1\), \(3 = A(0) + B(-3) \rightarrow 3 = -3B \rightarrow B = -1\).Thus, \(A = 1\) and \(B = -1\).

Step 4: Write as a Sum of Partial Fractions

Now substitute \(A\) and \(B\) back into the partial fractions:\[ \frac{3}{(x-2)(x+1)} = \frac{1}{x-2} - \frac{1}{x+1} \]

Step 5: Expand Each Fraction into a Power Series

You can write \(\frac{1}{x-2}\) and \(\frac{1}{x+1}\) as power series:\( \frac{1}{x-2} = \frac{-1}{2} \cdot \frac{1}{1-\frac{x}{2}} \), which is the geometric series \[ -\frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n \text{, valid for } |x|<2 \]\( \frac{1}{x+1} = \frac{1}{1-(-x)} \), which is the geometric series \[ \sum_{n=0}^{\infty} (-x)^n \text{, valid for } |x|<1 \]

Step 6: Combine the Power Series

Combine the two power series obtained in Step 5:\[ -\frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n - \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} \left( -\frac{1}{2} \cdot \left(\frac{x}{2}\right)^n + (-1)^n x^n \right)\]The combined series converges for overlapped interval of both series, valid for \( |x|<1 \).

Step 7: Identify the Interval of Convergence

The interval of convergence is found where overlapping conditions hold. The series for \(\frac{1}{x+1}\) converges for \(|x| < 1\) and for \(\frac{-1}{2} \frac{1}{x/2}\), it converges for \(|x| < 2\). Thus, the function representing \(f(x)\) as a sum of a power series has an interval of convergence of \(|x| < 1\), since both converge within this range.

Key Concepts

Partial Fractions

Partial fractions are used to simplify complex rational expressions, especially when integrating or finding power series. The method expresses a fraction as the sum of two or more simpler fractions. This is useful when the denominator can be factored into linear terms or irreducible quadratic factors.

In the example given, the function \( f(x) = \frac{3}{x^2 - x - 2} \) is first factored in the denominator to form \( (x-2)(x+1) \). This allows it to be rewritten as a sum of partial fractions: \( \frac{A}{x-2} + \frac{B}{x+1} \). By solving for \(A\) and \(B\), the terms become more manageable for further calculations.

The process involves:

• Factoring the denominator
• Setting up an equation for the separate fractions
• Solving for unknown coefficients by substituting suitable values

Once these components are determined, it becomes easier to synthesize power series.

Interval of Convergence

The interval of convergence is the set of values for which a power series converges to a finite sum. It is crucial for determining the valid range of \(x\) values for an expression in series form.

When a function is expressed in partial fractions and expanded as a power series, we require each series to converge independently. Only then can we determine the combined interval of convergence.

For \( \frac{1}{1-u} \), a geometric series, convergence is valid if \( |u|<1 \). In the presented example, each partial fraction generates its series:

• \( \frac{1}{x-2} = \frac{-1}{2} \cdot \frac{1}{1-\frac{x}{2}} \), converging for \(|x| < 2\)
• \( \frac{1}{x+1} = \frac{1}{1-(-x)} \), converging for \(|x| < 1\)

The overall interval of convergence is the more restrictive \(|x| < 1\), as both series must satisfy this condition.

Geometric Series

A geometric series is a type of power series that is structured based on repeated multiplication by a constant ratio. It takes the form \( \sum_{n=0}^{\infty} a r^n \), where \(a\) is the initial term and \(r\) is the common ratio.

In math, the geometric series is a powerful tool for expressing functions as infinite sums. The series converges when \(|r| < 1\).

Using geometric series to rewrite rational expressions involves transforming functions like \( \frac{1}{1-u} \) into an infinite series, thus making the calculus operations easier. In the given exercise:

• The term \( \frac{1}{x+1} = \frac{1}{1-(-x)} \) translates to a series valid for \(|x| < 1\).
• Meanwhile, \( \frac{1}{x-2} \) is rewritten as \( \frac{-1}{2} \cdot \frac{1}{1-\frac{x}{2}} \), which converges for \(|x| < 2\).

By interpreting the series in their geometric form, it becomes feasible to approximate and compute values of functions otherwise difficult to handle analytically.