Problem 46 - Precalculus Mathematics for Calculus

Question

Determine \(A, B, C,\) and \(D\) in terms of \(a\) and \(b\) $$ \frac{a x^{3}+b x^{2}}{\left(x^{2}+1\right)^{2}}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{\left(x^{2}+1\right)^{2}} $$

Step-by-Step Solution

Answer

A = a, B = b, C = -a, D = -b.

Step 1: Understand the Problem

The expression given is a rational function decomposed into partial fractions. We need to find the coefficients \(A, B, C,\) and \(D\) such that the decomposition on the right-hand side is equivalent to the original fraction on the left.

Step 2: Set Up the Equation

Equate both sides of the equation:\[ \frac{a x^3 + b x^2}{(x^2+1)^2} = \frac{A x + B}{x^2+1} + \frac{C x + D}{(x^2+1)^2}.\]Multiply both sides by \((x^2+1)^2\) to eliminate the denominators, giving:\[a x^3 + b x^2 = (A x + B)(x^2 + 1) + C x + D.\]

Step 3: Expand the Right Hand Side

Expand the terms:\[(A x + B)(x^2 + 1) = A x^3 + A x + B x^2 + B.\]So, the equation becomes:\[a x^3 + b x^2 = A x^3 + B x^2 + A x + B + C x + D.\]

Step 4: Combine Like Terms

Group the terms by their powers of \(x\):- \(x^3\) term: \(A x^3\)- \(x^2\) term: \(B x^2\)- \(x\) term: \(A x + C x\)- Constant term: \(B + D\) The equation becomes:\[a x^3 + b x^2 = A x^3 + B x^2 + (A + C)x + B + D.\]

Step 5: Compare Coefficients

By comparing the coefficients on both sides:- Coefficient of \(x^3\): \(a = A\)- Coefficient of \(x^2\): \(b = B\)- Coefficient of \(x\): \(0 = A + C\)- Constant term: \(0 = B + D\)

Step 6: Solve for the Coefficients

From the system of equations, solve for \(A, B, C,\) and \(D\):- \(A = a\)- \(B = b\)- \(A + C = 0 \Rightarrow C = -a\)- \(B + D = 0 \Rightarrow D = -b\) Thus, \(A = a\), \(B = b\), \(C = -a\), and \(D = -b\).

Key Concepts

Rational Functions

Rational functions are mathematical expressions that represent the ratio of two polynomials. These functions are defined as having the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not equal to zero. This is crucial because a division by zero is undefined, making the function invalid at those points where \( Q(x) \) equals zero. Rational functions are used extensively in calculus and algebra, playing a vital role in understanding the behavior of curves and surfaces. They are typically characterized by their asymptotes, which are lines that the graph of the function approaches but never touches. Understanding rational functions helps in simplifying complex fractional expressions and analyzing the points where these expressions may potentially become undefined or infinite. This simplification is essential, particularly when performing operations like partial fraction decomposition, where we break down complex fractions into simpler ones for easier integration or other calculus operations.

Coefficient Comparison

Coefficient comparison is a technique used in algebra to determine the unknowns in polynomial equations. This involves aligning the coefficients of terms with the same degree on both sides of the equation and setting them equal to each other. This method is pivotal in problems like partial fraction decomposition, where we resolve equations by comparing polynomial degrees. For example, in the given exercise, we compared:

• The coefficients of the term \( x^3 \) to establish the equation \( a = A \).
• The coefficients of the term \( x^2 \) to get \( b = B \).
• The coefficients of the \( x \) term, resulting in \( 0 = A + C \).
• The constant terms, leading to \( 0 = B + D \).

The power of this technique lies in its ability to break complex equations into manageable systems that can then be solved for the unknowns easily.

Systems of Equations

A system of equations is a collection of two or more equations with a common set of variables. Solving a system of equations involves finding a set of values for the variables that satisfy each equation in the system. In the context of partial fraction decomposition, systems of equations arise from the method of coefficient comparison. For the given problem, after setting the coefficients equal, we derive a system of equations as follows:

• \( A = a \)
• \( B = b \)
• \( A + C = 0 \)
• \( B + D = 0 \)

Solving these equations allows us to find the values of \( A, B, C, \) and \( D \), which define the decomposed form of the original rational function. Systems of equations can be solved using various methods such as substitution, elimination, and matrix operations, depending on the number and complexity of the equations. In this exercise, simple substitution is sufficient, showcasing how interconnected solving systems are to understanding algebraic expressions in rational functions.