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Chapter 7: Problem 69

Describe two ways of solving for the constants in a partial fraction decomposition.

Short Answer

Expert verified
Two methods for finding the constants in a partial fraction decomposition are by direct substitution of roots into the decomposed equation, and by equating coefficients of terms of the same degree in the expanded version of the partial fractions.

Step by step solution

01

Method 1: Direct Substitution

Direct substitution will involve substituting specific values for the dependent variable (usually denoted as \(x\)) that will effectively 'cancel' out some of the fractions, thus allowing for easier solving. First, solve the denominator of the original algebraic fraction and get the roots. These roots will give you the decomposed fractions. Substitute these roots into the equation to solve for the constants one-by-one.
02

Method 2: Equating Coefficients

The principle behind equating coefficients is matching the polynomial on the left side of an equation to an expanded version of the partial fractions on the right side, i.e., organizing terms in the same degree and setting their coefficients equal. First, decompose the fraction into partial fractions. Then, expand the right side of the equation. Write down and solve the system of equations that appears when the coefficients of the terms of the same degree on both sides of the equation are set to be equal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Substitution
The direct substitution method is a straightforward way to determine the constants in partial fraction decomposition by picking specific values for the variable. Usually, we select values that cancel out parts of the equation. Let's walk through the process:
  • Identify Roots: Start by solving the denominator of your original algebraic fraction to find its roots. These values are essential, as they will be used for substitution.
  • Substitute Roots: Insert these root values back into the equation. By doing so, you will "cancel" parts of the fraction and isolate some constants.
  • Solve for Constants: With parts of the equation simplified, you can easily solve for one or more constants.
This method is often praised for its simplicity and effectiveness, particularly when the roots lead directly to zeroing out some terms.
Equating Coefficients
Equating coefficients is a method where you force two expressions to match by ensuring each coefficient on both sides aligns perfectly. To use this method efficiently:
  • Start with Decomposition: Decompose the original fraction into a sum of partial fractions.
  • Expand the Equation: Expand the right side, where your partial fractions are, to align with the structure of the left side.
  • Align Degrees: Make sure that terms of the same degree appear on both sides of the equation. This requires focusing on powers of the variable.
  • Form Equations: Each unique degree of the variable yields an equation when their coefficients are equated. This creates a system of equations to solve for the constants.
Equating coefficients can handle more complex problems, especially when direct substitution doesn't easily simplify the fractions involved.
Algebraic Fractions
An algebraic fraction is simply a fraction containing variables in its numerator, denominator, or both. Let's explore their utility and characteristics:
  • What They Are: These fractions involve expressions containing variables and can represent complex relationships or equations.
  • Breakdown: Algebraic fractions may seem daunting, but they can usually be decomposed into simpler parts, known as partial fractions.
  • Role in Decomposition: In partial fraction decomposition, we express a given algebraic fraction as a sum of simpler fractions, which makes integration and other operations more manageable.
  • Rationality: While working with these, it's crucial to ensure the fractions remain rational, meaning the degree of the polynomial in the numerator is less than that in the denominator.
Mastering algebraic fractions and their decomposition plays a key role in simplifying complex equations, making subsequent calculations more straightforward.

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Most popular questions from this chapter

The graphs of the two equations appear to be parallel. Yet, when you solve the system algebraically, you find that the system does have a solution. Find the solution and explain why it does not appear on the portion of the graph shown. $$\left\\{\begin{array}{c} 100 y-x=200 \\ 99 y-x=-198 \end{array}\right.$$

An investor has up to \(450,000\) to invest in two types of investments. Type \(A\) pays \(6 \%\) annually and type B pays \(10 \%\) annually. To have a well- balanced portfolio, the investor imposes the following conditions. At least one-half of the total portfolio is to be allocated to type A investments and at least one-fourth of the portfolio is to be allocated to type \(B\) investments. What is the optimal amount that should be invested in each type of investment? What is the optimal return?

Solve the system of linear equations and check any solutions algebraically. $$\left\\{\begin{array}{l} 3 x-5 y+5 z=1 \\ 5 x-2 y+3 z=0 \\ 7 x-y+3 z=0 \end{array}\right.$$

Find values of \(x, y,\) and \(\lambda\) that satisfy the system. These systems arise in certain optimization problems in calculus, and \(\lambda\) is called a Lagrange multiplier. $$\left\\{\begin{aligned} 2 x-2 x \lambda &=0 \\ -2 y+\lambda &=0 \\ y-x^{2} &=0 \end{aligned}\right.$$

Briefly explain whether it is possible for a consistent system of linear equations to have exactly two solutions.
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