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

The power series representation is \(\sum_{n=0}^{\infty} (x-1)^n - \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n\)."}

Step 1: Break down the expression into partial fractions

The given expression is \(\frac{x}{x^2 - 3x + 2}\). First, we factor the denominator: \(x^2 - 3x + 2 = (x-1)(x-2)\). Thus, the partial fraction decomposition of \(\frac{x}{(x-1)(x-2)}\) is \(\frac{A}{x-1} + \frac{B}{x-2}\).

Step 2: Solve for the constants in the partial fraction

Set \(\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\). Multiply through by the common denominator \((x-1)(x-2)\) to get \(x = A(x-2) + B(x-1)\). Equating coefficients or using inspection, solve for \(A\) and \(B\). Setting \(x = 1\) gives \(A = 1\), and setting \(x = 2\) gives \(B = -1\). Thus, \(\frac{x}{x^2 - 3x + 2} = \frac{1}{x-1} - \frac{1}{x-2}\).

Step 3: Represent each fraction as a geometric series

We express \(\frac{1}{x-1}\) and \(\frac{1}{x-2}\) as geometric series. For \(\frac{1}{x-1}\), write it as \(\frac{1}{1-(x-1)}\) and for \(\frac{1}{x-2}\), as \(-\frac{1}{2-(x)}\). Both expressions can be expanded using the formula for geometric series: \(\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}\).

Step 4: Expand each series

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

Step 5: Combine both series into one power series

Sum the two series obtained: \(\sum_{n=0}^{\infty} (x-1)^n - \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n\). This is the power series representation of \(\frac{x}{x^2 - 3x + 2}\).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to rewrite a complicated fraction into a sum of simpler fractions. This is particularly useful when dealing with rational functions, similar to the one in our exercise: \( \frac{x}{x^2 - 3x + 2} \).
The idea is to express this fraction as a sum of fractions with simpler denominators. To start, we factor the denominator: \( x^2 - 3x + 2 = (x-1)(x-2) \).
Now, the original fraction can be written as the sum of two fractions: \( \frac{A}{x-1} + \frac{B}{x-2} \). By doing this, we've decomposed the original fraction into simpler parts.

• First, identify the polynomial factorization in the denominator.
• Second, express the original fraction as a sum of fractions, each having one of the factors as a denominator.
• Then, solve for the unknown constants in front of the simpler fractions.

This process simplifies integration or further manipulation of the expression and is a powerful tool used across calculus.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It is a common way to express certain functions as infinite series for ease of manipulation and summation.
The formula for a geometric series is \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \) when \(|r| < 1\). Here, \(a\) is the first term and \(r\) is the common ratio.
In our exercise, after partial fraction decomposition, we rewrite each simpler fraction into a geometric series.

• For \( \frac{1}{x-1} \), it becomes \( \frac{1}{1-(x-1)} \), which can be expanded to the series \( \sum_{n=0}^{\infty} (x-1)^n \).
• For \( \frac{-1}{x-2} \), by rewriting it as \( \frac{1}{2-(x)} \), it turns into the series \( -\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n \).

Understanding the expansion of rational expressions into geometric series is crucial for simplifying complex expressions and solving them analytically.

Series Expansion

Series expansion allows us to represent functions as infinite sums of terms. This is beneficial in calculus for approximations, solving differential equations, and finding power series.
In our problem, we need to express \( \frac{x}{x^2 - 3x + 2} \) as a power series. After employing partial fraction decomposition, each term can be approximated using a geometric series expansion.

• Convert each of the simplified fractions into a series representation using geometric series formula.
• The infinite series for \( \frac{1}{x-1} \) is \( \sum_{n=0}^{\infty} (x-1)^n \).
• The infinite series for \( -\frac{1}{x-2} \) is \( -\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n \).

By summing these series, we achieve a combined power series representation of the original function. Recognizing and utilizing series expansion is a fundamental step in advanced calculus and mathematical analysis.

Rational Functions

Rational functions are quotients of polynomials and are central to calculus due to their versatility in modeling real-world problems.
The function \( \frac{x}{x^2 - 3x + 2} \) is a rational function where the numerator is a linear polynomial and the denominator is a quadratic polynomial.

• These functions can often be simplified using algebraic techniques such as factoring and partial fraction decomposition.
• They may be integral or differentiable, providing versatility in calculus operations.
• When analyzing behavior such as identifying limits or asymptotes, rational functions require careful examination of both numerator and denominator.

Rational functions can represent a wide range of phenomena, making their study crucial for students of mathematics and applied fields alike.