Problem 95 - Precalculus

Question

In calculus, when integrating rational functions, we decompose the function into partial fractions. This technique involves the solution of systems of equations. For example, suppose $$\begin{aligned} \frac{1}{x^{2}+x-2} &=\frac{1}{(x-1)(x+2)} \\ &=\frac{A}{x-1}+\frac{B}{x+2} \\ &=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)} \end{aligned}$$ and we want to find \(A\) and \(B\) such that \(1=A(x+2)+B(x-1)\) which is equivalent to \(1=(A+B) x+(2 A-B) .\) From this equation, we obtain the system of equations $$\begin{aligned} A+B &=0 \\ 2 A-B &=1 \end{aligned}$$ which solution is \(\left(\frac{1}{3},-\frac{1}{3}\right)\) Find the values of \(A\) and \(B\) that make each equation true. $$x+1=A(x+2)+B(x-3)$$

Step-by-Step Solution

Answer

\( A = \frac{4}{5}, B = \frac{1}{5} \)

Step 1: Expand the Expression

We start by expanding the right-hand side of the equation: \(A(x+2) + B(x-3)\). This results in \(Ax + 2A + Bx - 3B\).

Step 2: Combine Like Terms

Combine the terms involving \(x\) and the constant terms. This yields: \((A + B)x + (2A - 3B)\).

Step 3: Set the Expression Equal

Now, set the expanded expression equal to the left-hand side of the original equation: \((A + B)x + (2A - 3B) = x + 1\).

Step 4: Equate the Coefficients

To find the values of \(A\) and \(B\), equate the coefficients of the corresponding terms: \(A + B = 1\) and \(2A - 3B = 1\).

Step 5: Solve the System of Equations

Solve the system of equations: 1. \(A + B = 1\)2. \(2A - 3B = 1\)From the first equation, express \(A = 1 - B\), and substitute it into the second equation. Then:\[2(1-B) - 3B = 1\]\[2 - 2B - 3B = 1\]\[2 - 5B = 1\]\[-5B = -1\]\[B = \frac{1}{5}\]Now substitute \(B = \frac{1}{5}\) back into \(A = 1 - B\):\[A = 1 - \frac{1}{5} = \frac{4}{5}\].

Key Concepts

Partial Fractions

Partial fraction decomposition is a powerful technique used in calculus to simplify complex rational expressions, making them easier to integrate or manage. When integrating rational functions, we often encounter fractions that have polynomial numerators and denominators. To handle these, we can break them down into simpler, more manageable parts, called partial fractions.

Here's how the process generally works:

• First, factor the denominator of the rational function into irreducible factors if possible. This helps in setting up the partial fractions.
• Next, express the rational function as the sum of simpler fractions whose denominators are these factors. In the case of linear factors, each fraction will take the form \(\frac{A}{x-k}\), where \(A\) is a constant to be determined.
• The goal is to express the original rational function as a sum of these basic fractions, thereby simplifying the integral or the function itself.

By employing partial fractions, a complicated integral such as \(\frac{1}{(x-1)(x+2)}\) becomes much simpler, allowing the use of basic anti-differentiation techniques. This decomposition process involves setting up an equation and solving for the constants, such as \(A\) and \(B\). This sets the stage for the next fundamental step, which is solving these systems of equations.

Systems of Equations

The process of finding constants like \(A\) and \(B\), needed for partial fraction decomposition, involves setting up and solving a system of equations. A system of equations is essentially a collection of two or more equations that share common variables. The main goal is to find values for these variables that satisfy all the equations involved.

In the context of partial fractions, once you have expressed your function in terms of unknown constants, you should have an equation like \((A + B)x + (2A - 3B) = x + 1\). Here, equating the coefficients of like terms on both sides results in two separate equations:

• Equation 1: \(A + B = 1\)
• Equation 2: \(2A - 3B = 1\)

These equations can be solved simultaneously using algebraic methods, such as:

• Substitution: Solve one equation for one variable and substitute that variable in the other equation.
• Elimination: Add or subtract the equations to eliminate one variable, facilitating a solution for the other.

Solving such systems is crucial for accurately determining the values of constants in partial fraction decomposition. It's what ultimately allows us to simplify and integrate complex rational functions.

Decomposition of Fractions

The decomposition of fractions into partial fractions involves breaking down a complex fraction into a sum of simpler fractions. This simplifies the process of integration or helps in better understanding and manipulating the fraction itself. Decomposition is very much foundational in calculus and algebra.

To begin, you should have the fraction you are working with, such as \(\frac{1}{x^2+x-2}\). Once the denominator is factored, the fraction can potentially be expressed as a sum of different terms like \(\frac{A}{x-1} + \frac{B}{x+2}\). This process hinges on algebraic skills and sometimes requires polynomial long division if the numerator's degree is higher than the denominator's.

The need to decompose arises often when integrating because the integral of the original function might be challenging or impossible with elementary techniques. By breaking it down into partial fractions:

• Integration becomes straightforward, utilizing simple logarithmic or power functions that are easy to manage.
• The expressions involved are simplified, yielding solutions that are easy to interpret and compute, often necessary in applied mathematics.

Decomposition of fractions through partial fractions is an essential tool in a mathematician's toolkit, greatly enhancing our ability to work with complex rational expressions in a more intuitive and manageable way.