Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the power series representation of \(x /\left(x^{2}-3 x+2\right)\). Hint: Use partial fractions.

Step-by-Step Solution

Answer

\( -\frac{1}{x-1} + \frac{2}{x-2} \) and expand as power series.

Step 1: Factor the Denominator

The function given is \( \frac{x}{x^2 - 3x + 2} \). Start by factoring the denominator: \( x^2 - 3x + 2 = (x-1)(x-2) \).

Step 2: Decompose into Partial Fractions

Express \( \frac{x}{(x-1)(x-2)} \) as partial fractions: \( \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \). Solve for \( A \) and \( B \) by clearing the fractions: \( x = A(x-2) + B(x-1) \).

Step 3: Solve for Coefficients

Equating coefficients, plug in suitable values for \( x \) (typically the roots of the denominator):- Let \( x = 1 \): \( 1 = A(1-2) + B(1-1) \Rightarrow A = -1 \).- Let \( x = 2 \): \( 2 = A(2-2) + B(2-1) \Rightarrow B = 2 \).Therefore, \( \frac{x}{(x-1)(x-2)} = -\frac{1}{x-1} + \frac{2}{x-2} \).

Step 4: Find the Power Series for Each Fraction

For \( -\frac{1}{x-1} \), rewrite as \( -\frac{1}{1-(x-1)} \), which is the geometric series \( -\sum_{n=0}^{\infty} (x-1)^n \).For \( \frac{2}{x-2} \), rewrite as \( \frac{2}{1-(x-2)} \), which is \( 2\sum_{n=0}^{\infty} (x-2)^n \).

Step 5: Write the Combined Power Series

Combine the series found in Step 4 as: \( -\sum_{n=0}^{\infty} (x-1)^n + 2\sum_{n=0}^{\infty} (x-2)^n \).

Step 6: Reorganize the Combined Series

Write the expression in full as a power series: \(-1 - (x-1) + (x-1)^2 - \cdots + 2 + 2(x-2) + 2(x-2)^2 + \cdots \).

Key Concepts

Partial Fractions

Partial fractions is a method used to break down complex rational expressions into simpler ones. This technique is useful for integrating functions and finding power series representations. In our problem, we want to express \( \frac{x}{(x-1)(x-2)} \) as the sum of two fractions. The idea is to represent it as \( \frac{A}{x-1} + \frac{B}{x-2} \), where \( A \) and \( B \) are constants.

• The first step is to equate \( x \) to the expression \( A(x-2) + B(x-1) \).
• Next, find suitable values of \( x \) that simplify our equation, typically the roots of the denominator.

By substitute \( x = 1 \), we determine that \( A = -1 \), and by substituting \( x = 2 \), \( B = 2 \). So, our expression becomes \(-\frac{1}{x-1} + \frac{2}{x-2} \).
This decomposition is crucial, enabling us to handle each fraction separately when we look for their power series.

Geometric Series

Geometric series are infinite series of the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio. In this exercise, each term in the partial fraction decomposition can be rewritten in a form suitable for geometric series expansion.

• For the term \(-\frac{1}{x-1}\), we can rewrite it as \(-\frac{1}{1-(x-1)}\). This is directly a geometric series with a common ratio of \( (x-1) \).
• Similarly, \( \frac{2}{x-2} \) can be seen as \( \frac{2}{1-(x-2)} \), again a geometric series where the common ratio is \( (x-2) \).

The power series for these are thus \(-\sum_{n=0}^{\infty} (x-1)^n\) and \( 2 \sum_{n=0}^{\infty} (x-2)^n\) respectively.
These series allow us to express each term as an infinite sum, which is helpful in many applications, including calculus and numerical analysis.

Factor the Denominator

Factoring the denominator is often the first step in breaking down rational expressions into simpler components. For the expression \( \frac{x}{x^2 - 3x + 2} \), the denominator is a quadratic expression. To factor this:

• Set the quadratic expression \( x^2 - 3x + 2 \) equal to zero.
• Look for two numbers that multiply to 2 (the constant term) and add up to -3 (the coefficient of the \( x \) term).

These numbers are -1 and -2. Thus, \( x^2 - 3x + 2 \) factors into \( (x-1)(x-2) \).
By factoring the polynomial, we reveal the potential for decomposition into partial fractions. This step simplifies the original problem by reducing it into terms that can be easily managed through integration and power series expansion.

Solving Coefficients

In partial fraction decomposition, solving for coefficients involves finding the constants that will satisfy the equality of the original expression and the decomposed one. Let's consider our problem where we have:

• \( A(x-2) + B(x-1) = x \)

To solve this, choose specific values for \( x \) that simplify the equation.

• When \( x = 1 \), the terms involving \( B \) vanish, leading to a straightforward equation for \( A \): \( 1 = A(-1) \), giving us \( A = -1 \).
• Similarly, when \( x = 2 \), the terms involving \( A \) cancel out, leaving us with \( 2 = B(1) \), thereby \( B = 2 \).

Finding these coefficients, \( A = -1 \) and \( B = 2 \), allows us to rewrite the original expression. This is crucial for crafting the power series as it provides the necessary pieces to reconfigure each fraction into a geometric series.