Problem 24 - Calculus Single Variable

Question

Use Heaviside's method to calculate the partial fraction decomposition of the given rational function. $$ \frac{2 x^{2}+8 x+13}{(x+1)(x+3)(x+5)} $$

Step-by-Step Solution

Answer

\(\frac{1/2}{x+1} - \frac{5/2}{x+3} + \frac{4}{x+5}\)

Step 1: Identify the Denominator Roots

The denominator of the rational function is \((x+1)(x+3)(x+5)\). Each factor corresponds to a root: \(-1\), \(-3\), and \(-5\). These roots will help us find the constants in the partial fraction decomposition.

Step 2: Set Up Partial Fraction Decomposition

Write the expression as the sum of simpler fractions, each with one of the identified roots: \[ \frac{2x^2 + 8x + 13}{(x+1)(x+3)(x+5)} = \frac{A}{x+1} + \frac{B}{x+3} + \frac{C}{x+5} \]where \(A\), \(B\), and \(C\) are constants we need to determine.

Step 3: Clear the Denominator

Multiply every term by \((x+1)(x+3)(x+5)\) to eliminate the denominators:\[ 2x^2 + 8x + 13 = A(x+3)(x+5) + B(x+1)(x+5) + C(x+1)(x+3) \]

Step 4: Expand and Simplify

Expand the terms on the right-hand side:- Expand \(A(x+3)(x+5)\) to get \(A(x^2 + 8x + 15)\)- Expand \(B(x+1)(x+5)\) to get \(B(x^2 + 6x + 5)\)- Expand \(C(x+1)(x+3)\) to get \(C(x^2 + 4x + 3)\)Combine these to get:\[ Ax^2 + 8Ax + 15A + Bx^2 + 6Bx + 5B + Cx^2 + 4Cx + 3C \]

Step 5: Combine Like Terms

Group terms with similar powers of x:\((A + B + C)x^2 + (8A + 6B + 4C)x + (15A + 5B + 3C)\)Set this equal to the left side of the equation:\[ 2x^2 + 8x + 13 \]

Step 6: Solve the System of Equations

Match coefficients to solve for \(A\), \(B\), and \(C\):- Equate the coefficients of \(x^2\): \(A + B + C = 2\)- Equate the coefficients of \(x\): \(8A + 6B + 4C = 8\)- Equate the constant terms: \(15A + 5B + 3C = 13\)Solve this system to find:\[ A = \frac{1}{2}, B = -\frac{5}{2}, C = 4 \]

Step 7: Write the Partial Fraction Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into the original partial fraction equation:\[ \frac{2x^2 + 8x + 13}{(x+1)(x+3)(x+5)} = \frac{1/2}{x+1} - \frac{5/2}{x+3} + \frac{4}{x+5} \]

Key Concepts

Heaviside's Method

Heaviside's Method is a straightforward approach used in partial fraction decomposition. It is particularly useful for breaking down rational functions into simpler forms, making them easier to integrate or work with in calculus.

• This technique focuses on identifying the roots of the denominator of a rational function.
• Once these roots are known, the rational function can be expressed as a sum of simpler fractions.

Heaviside's Method helps in systematically forming an equation where the original complex fraction is split into a number of simpler fractions with undetermined coefficients. These coefficients are then calculated through methods such as systems of equations, by comparing both sides of the equation. This method greatly simplifies otherwise daunting algebraic expressions.

Rational Functions

Rational functions are expressions that involve the division of two polynomials. An example of a rational function is \(\frac{2x^2 + 8x + 13}{(x+1)(x+3)(x+5)}\).

• The numerator is a polynomial: \(2x^2 + 8x + 13\).
• The denominator is a polynomial multiplied factor-wise: \((x+1)(x+3)(x+5)\).
• Rational functions can pose challenges, especially when you need to simplify or integrate them.

Partial fraction decomposition turns these functions into simpler fractions, easing the process of solving or integrating them. Mastery of rational functions allows students to engage deeply with mathematical analysis and calculus problems.

Roots of Denominator

The roots of the denominator play an essential role in partial fraction decomposition. These are the values of \(x\) that make the denominator equal to zero.

• For example, the denominator \((x+1)(x+3)(x+5)\) has roots: \(-1\), \(-3\), and \(-5\).
• Finding these roots helps identify the terms in the decomposition.
• Each root contributes to the formation of a simple fraction in the decomposition.

These values lead us to set up the equation in the partial fraction decomposition format as distinct fractions with unknown coefficients. Effectively identifying and using these roots allows the expression to be simplified and makes it easier to solve.

System of Equations

To determine the coefficients in the partial fraction decomposition, we need to solve a system of equations.

• After breaking down the initial rational function into simpler forms, each term in the equations is expanded and simplified.
• By comparing coefficients of corresponding powers of \(x\), we create a system of equations.
• For instance, if \((A + B + C)x^2 + (8A + 6B + 4C)x + (15A + 5B + 3C) = 2x^2 + 8x + 13\), we derive three distinct equations.

Solving this system reveals the values of unknown coefficients \(A\), \(B\), and \(C\). This methodical approach ensures that each coefficient accurately transforms the original rational function into the desired partial fractions, achieving a correct decomposition.