Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Find the partial fraction decomposition for each rational expression. $$\frac{x^{2}}{x^{4}-1}$$

Step-by-Step Solution

Answer

Decomposition: \(\frac{1/2}{x-1} + \frac{1/2}{x+1} - \frac{x}{x^2+1}\).

Step 1: Factor the Denominator

The expression given is \( \frac{x^{2}}{x^{4}-1} \). First, we need to factor the denominator, \( x^4 - 1 \). Recall that \( x^4 - 1 \) is a difference of two squares: \( (x^2)^2 - 1^2 \). This can be factored using the difference of squares formula as \((x^2 - 1)(x^2 + 1)\). Further factor \(x^2 - 1\) as \((x-1)(x+1)\), so the full factorization of the denominator is \((x-1)(x+1)(x^2+1)\).

Step 2: Set Up the Partial Fraction Decomposition

With the factorized denominator \((x-1)(x+1)(x^2+1)\), set up the partial fraction decomposition as: \( \frac{x^2}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \). The last fraction has \(Cx+D\) since \(x^2+1\) is an irreducible quadratic factor.

Step 3: Combine the Fractions

Multiply every term by the common denominator \((x-1)(x+1)(x^2+1)\) to eliminate the denominators: \[x^2 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x^2-1)\].

Step 4: Expand and Collect Like Terms

Expand each term on the right-hand side and collect like terms in powers of \(x\): - Expand \(A(x+1)(x^2+1) = A(x^3 + x^2 + x + 1)\)- Expand \(B(x-1)(x^2+1) = B(x^3 - x^2 + x - 1)\)- Expand \((Cx+D)(x^2-1) = Cx^3 - Cx + Dx^2 - D\) Combine all expansions to a single polynomial on the right.

Step 5: Solve the System of Equations

Equating coefficients from both sides of the equation \(x^2 = (A+B+C)x^3 + (A-B+D)x^2 + (A+B-C)x + (A-B-D)\), we get a system of equations:1. \(A + B + C = 0\)2. \(A - B + D = 0\)3. \(A + B - C = 1\)4. \(A - B - D = 0\)Solve these simultaneous equations to find \(A, B, C,\) and \(D\).

Step 6: Compute Values of A, B, C, and D

From the system:- From equations 2 and 4, we find \(D = 0\).- Substitute \(D = 0\) in equation 2 or 4 to get \(A = B\).- From equation 1, substitute \(A = B\) to find \(C = -2A\).- Substitute \(A = B\) and \(C = -2A\) in equation 3 to find \(A = \frac{1}{2}, B = \frac{1}{2}, C = -1\).

Step 7: Write the Partial Fraction Decomposition

Substitute \(A = \frac{1}{2}, B = \frac{1}{2}, C = -1,\) and \(D = 0\) back into the partial fractions: \[\frac{x^2}{(x-1)(x+1)(x^2+1)} = \frac{1/2}{x-1} + \frac{1/2}{x+1} - \frac{x}{x^2+1}\].

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator consist of polynomials. In the exercise, the expression \( \frac{x^{2}}{x^{4}-1} \) is an example of a rational expression. The numerator here is a simple polynomial \( x^2 \), and the denominator is \( x^4 - 1 \), also a polynomial.
To work effectively with rational expressions, it's crucial to understand how to simplify or decompose them using algebraic techniques. Partial fraction decomposition helps in expressing these complex fractions as a sum of simpler fractions. This is particularly helpful in calculus when integrating rational expressions.

Factorization

Factorization is the process of breaking down a complex expression into a product of simpler factors. In this exercise, we factored the denominator \( x^4 - 1 \).
Here's how we did it:

• Recognized \( x^4 - 1 \) as a difference of squares: \( (x^2)^2 - 1^2 \).
• Using the difference of squares formula, we split it into \( (x^2 - 1)(x^2 + 1) \).
• Further factorized \( x^2 - 1 \) into \( (x-1)(x+1) \).

After these steps, the factorization of the original denominator is complete: \( (x-1)(x+1)(x^2+1) \). Each factor plays a critical role in setting up the partial fraction decomposition.

System of Equations

A system of equations arises when we equate the original rational expression to its decomposed form. We multiplied through by the common denominator to remove all fractional terms, leading us to an expanded polynomial equation.
This system allows us to compare coefficients of corresponding powers of \( x \) from both polynomials:

• For \( x^3 \): \( A + B + C = 0 \)
• For \( x^2 \): \( A - B + D = 0 \)
• For \( x \): \( A + B - C = 1 \)
• Constant: \( A - B - D = 0 \)

This collection of linear equations is systematically solved to find the values of \( A, B, C, \) and \( D \). Each solution step unveils another piece of the puzzle towards building the partial fraction decomposition.

Irreducible Quadratic Factor

In partial fraction decomposition, an irreducible quadratic factor is a polynomial that cannot be further factored using real numbers. In this problem, \( x^2 + 1 \) is considered irreducible over the real numbers.
When such quadratic terms appear in the denominator, they require the numerator to be linear, typically represented as \( Cx + D \). This approach ensures that the decomposition is valid across all terms in the expression.
Handling irreducible quadratic factors correctly is crucial because it affects how we set up the equations and ultimately the form of the decomposition. Understanding the nature of these factors helps in accurately solving rational expressions.