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: Setting up the Equations

Start by multiplying both sides of the equation by the common denominator \((x^2+1)^2\). This gives us: \[ ax^3 + bx^2 = (Ax+B)(x^2+1) + (Cx+D) \] This simplifies to: \[ ax^3 + bx^2 = Ax^3 + Ax + Bx^2 + B + Cx + D \] We will now equate the coefficients of like terms from both sides.

Step 2: Equate Coefficients

Compare coefficients of each power of \(x\) from both the left and right sides of the equation: - Coefficient of \(x^3\): \( a = A \) - Coefficient of \(x^2\): \( b = B \) - Coefficient of \(x\): \( 0 = A + C \) - Constant term: \( 0 = B + D \)

Step 3: Solve for A and B

From the equations in Step 2, we have: - \( a = A \), thus \( A = a \) - \( b = B \), thus \( B = b \)

Step 4: Solve for C and D

Using the coefficients of \(x\): - Given \( 0 = A + C \), replace \(A\) with \(a\): \( 0 = a + C \) Thus, \( C = -a \) For the constant term: - Given \( 0 = B + D \), replace \(B\) with \(b\): \( 0 = b + D \) Thus, \( D = -b \)

Step 5: Final Result

The values of \(A, B, C, \) and \(D\) are: - \( A = a \) - \( B = b \) - \( C = -a \) - \( D = -b \)

Key Concepts

Polynomial Equations

Polynomial equations are a central concept in algebra that involve expressions where variables are raised to whole number powers. A polynomial equation is usually written in the form of \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\), where each \(a\) is a coefficient and \(n\) represents the degree of the polynomial.

Understanding polynomial equations is crucial because they form the basis for many calculations in algebra, calculus, and beyond. They can represent a wide range of mathematical models, from simple linear equations to more complex cubic or quadratic equations.

• Linear Equations: Polynomials of degree 1, e.g., \(ax + b = 0\)
• Quadratic Equations: Polynomials of degree 2, e.g., \(ax^2 + bx + c = 0\)
• Cubic Equations: Polynomials of degree 3, e.g., \(ax^3 + bx^2 + cx + d = 0\)

In our exercise, the polynomial involves cubic (\(x^3\)) and quadratic (\(x^2\)) terms which need to be rearranged and simplified using partial fraction decomposition. By equating and solving for the coefficients of like terms, we can reassign these values into a decomposed form, making complex equations more approachable.

Equating Coefficients

Equating coefficients is a standard method used for simplifying polynomial or rational expressions. This technique involves matching the coefficients of corresponding powers of a variable on both sides of an equation.

For instance, if we have a polynomial equation from our exercise, initially given as \( ax^3 + bx^2 = Ax^3 + Ax + Bx^2 + B + Cx + D \), we need to ensure both sides of the equation match for every power of \(x\).

• Matching Coefficient of \(x^3\): The coefficient of \(x^3\) on the left is \(a\), thus \(a = A\).
• Matching Coefficient of \(x^2\): On the left it is \(b\), leading to \(b = B\).
• Coefficient of \(x\): Equate to zero, so \(0 = A + C\).
• Constant Term: Also equate to zero, \(0 = B + D\).

By systematically equating coefficients for each degree of \(x\), we can solve for unknowns like \(A\), \(B\), \(C\), and \(D\). This process simplifies complex equations and forms the basis for partial fraction decomposition.

Rational Functions

Rational functions are fractions in which both the numerator and the denominator are polynomials. These functions are essential in various fields of mathematics because they can model complex relationships in a simple fraction form.

A typical rational function is written as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. In our exercise, we started with a rational function where the numerator and denominator are cubic and quadratic polynomials, respectively.

• Numerator: \(a x^3 + b x^2\) - represents the cubic polynomial
• Denominator: \((x^2 + 1)^2\) - represents the squared quadratic polynomial

In partial fraction decomposition, we aim to break down this complex rational function into simpler parts, which are easier to integrate or differentiate. By expressing the original rational function as a sum of simpler fractions, each with a simpler denominator, we can analyze and manipulate these functions with ease.

This decomposition helps in solving integrals and differential equations, making it a powerful tool in calculus and higher-level math.