Skills Review for Properties of Power Series

Learning Outcomes

Use partial fraction decomposition for linear factors
Simplify expressions using the power property of pxponents

In the Properties of Power Series section, we will look at how to combine, differentiate, and integrate power series. Here we will review how to use partial fraction decomposition for linear factors and how to use the power property of exponents.

Use Partial Fraction Decomposition for Linear Factors

(also in Module 5, Skills Review for Infinite Series)

Partial fraction decomposition is used to break up one fraction into two.

[latex]\begin{align}\underset{\text{ }\\ \text{Simplified sum}}{\frac{x+7}{{x}^{2}-x - 6}}=\underset{\text{ }\\ \text{Partial fraction decomposition}}{\frac{2}{x - 3}+\frac{-1}{x+2}}\\ \text{ }\end{align}[/latex]

We will investigate rational expressions with linear factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.

How To: Given a rational expression with distinct linear factors in the denominator, decompose it.

Use a variable for the original numerators, usually [latex]A,B,[/latex] or [latex]C[/latex], depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use [latex]{A}_{n}[/latex] for each numerator
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}[/latex]
Multiply both sides of the equation by the common denominator to eliminate fractions.
Expand the right side of the equation and collect like terms.
Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

Example: Decomposing a Rational Function with Distinct Linear Factors

Decompose the given rational expression with distinct linear factors.

[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}[/latex]

Solution

We will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[/latex], or [latex]C[/latex].

[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{A}{\left(x+2\right)}+\dfrac{B}{\left(x - 1\right)}[/latex]

Multiply both sides of the equation by the common denominator to eliminate the fractions:

[latex]\cancel{\left(x+2\right)}\cancel{\left(x - 1\right)}\left[\dfrac{3x}{\cancel{\left(x+2\right)}\cancel{\left(x - 1\right)}}\right]=\cancel{\left(x+2\right)}\left(x - 1\right)\left[\dfrac{A}{\cancel{\left(x+2\right)}}\right]+\left(x+2\right)\cancel{\left(x - 1\right)}\left[\dfrac{B}{\cancel{\left(x - 1\right)}}\right][/latex]

The resulting equation is

[latex]3x=A\left(x - 1\right)+B\left(x+2\right)[/latex]

Expand the right side of the equation and collect like terms.

[latex]\begin{gathered}3x=Ax-A+Bx+2B\\ 3x=\left(A+B\right)x-A+2B\end{gathered}[/latex]

Set up a system of equations associating corresponding coefficients.

[latex]\begin{gathered}3=A+B\\ 0=-A+2B\end{gathered}[/latex]

Add the two equations and solve for [latex]B[/latex].

[latex]\begin{align}3&=A+B \\ 0&=-A+2B \\ \hline 3&=0+3B \\[4mm] B&=1 \end{align}[/latex]

Substitute [latex]B=1[/latex] into one of the original equations in the system.

[latex]\begin{align}3&=A+1\\ 2&=A\end{align}[/latex]

Thus, the partial fraction decomposition is

[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{2}{\left(x+2\right)}+\dfrac{1}{\left(x - 1\right)}[/latex]

Another method to use to solve for [latex]A[/latex] or [latex]B[/latex] is by considering the equation that resulted from eliminating the fractions and substituting a value for [latex]x[/latex] that will make either the [latex]A-[/latex] or [latex]B-[/latex]term equal 0. If we let [latex]x=1[/latex], the [latex]A-[/latex] term becomes 0 and we can simply solve for [latex]B[/latex].

[latex]\begin{align}3x&=A\left(x - 1\right)+B\left(x+2\right) \\ 3\left(1\right)&=A\left[\left(1\right)-1\right]+B\left[\left(1\right)+2\right] \\ 3&=0+3B\hfill \\ B&=1 \end{align}[/latex]

Next, either substitute [latex]B=1[/latex] into the equation and solve for [latex]A[/latex], or make the [latex]B-[/latex]term 0 by substituting [latex]x=-2[/latex] into the equation.

[latex]\begin{align}3x&=A\left(x - 1\right)+B\left(x+2\right) \\ 3\left(-2\right)&=A\left[\left(-2\right)-1\right]+B\left[\left(-2\right)+2\right] \\ -6&=-3A+0 \\ \frac{-6}{-3}&=A \\ A&=2 \end{align}[/latex]

We obtain the same values for [latex]A[/latex] and [latex]B[/latex] using either method, so the decompositions are the same using either method.

[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{2}{\left(x+2\right)}+\dfrac{1}{\left(x - 1\right)}[/latex]

Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics.

Try It

Find the partial fraction decomposition of the following expression.

[latex]\dfrac{x}{\left(x - 3\right)\left(x - 2\right)}[/latex]

Show Solution

Try It

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

How To: Given a rational expression with repeated linear factors, decompose it.

Use a variable like [latex]A,B[/latex], or [latex]C[/latex] for the numerators and account for increasing powers of the denominators.
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\dfrac{{A}_{n}}{{\left(ax+b\right)}^{n}}[/latex]
Multiply both sides of the equation by the common denominator to eliminate fractions.
Expand the right side of the equation and collect like terms.
Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

Example: Decomposing with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.

[latex]\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[/latex]

Show Solution

The denominator factors are [latex]x{\left(x - 2\right)}^{2}[/latex]. To allow for the repeated factor of [latex]\left(x - 2\right)[/latex], the decomposition will include three denominators: [latex]x,\left(x - 2\right)[/latex], and [latex]{\left(x - 2\right)}^{2}[/latex]. Thus,

[latex]\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\dfrac{A}{x}+\dfrac{B}{\left(x - 2\right)}+\dfrac{C}{{\left(x - 2\right)}^{2}}[/latex]

Next, we multiply both sides by the common denominator.

[latex]\begin{gathered}x{\left(x - 2\right)}^{2}\left[\dfrac{-{x}^{2}+2x+4}{x{\left(x - 2\right)}^{2}}\right]=\left[\dfrac{A}{x}+\dfrac{B}{\left(x - 2\right)}+\dfrac{C}{{\left(x - 2\right)}^{2}}\right]x{\left(x - 2\right)}^{2} \\[2mm] -{x}^{2}+2x+4=A{\left(x - 2\right)}^{2}+Bx\left(x - 2\right)+Cx \end{gathered}[/latex]

On the right side of the equation, we expand and collect like terms.

[latex]-{x}^{2}+2x+4=A\left({x}^{2}-4x+4\right)+B\left({x}^{2}-2x\right)+Cx[/latex]
[latex]\begin{align}&=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\ &=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A \end{align}[/latex]

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

[latex]-{x}^{2}+2x+4=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A[/latex]

[latex]\begin{array}{rr}\hfill A+B=-1& \hfill \text{(1)}\\ \hfill -4A - 2B+C=2& \hfill \text{(2)}\\ \hfill 4A=4& \hfill \text{(3)}\end{array}[/latex]

Solving for [latex]A[/latex] , we have

[latex]\begin{align}4A&=4 \\ A&=1 \end{align}[/latex]

Substitute [latex]A=1[/latex] into equation (1).

[latex]\begin{align}A+B=-1 \\ \left(1\right)+B=-1 \\ B=-2 \end{align}[/latex]

Then, to solve for [latex]C[/latex], substitute the values for [latex]A[/latex] and [latex]B[/latex] into equation (2).

[latex]\begin{align}-4A - 2B+C=2\\ -4\left(1\right)-2\left(-2\right)+C=2\\ -4+4+C=2\\ C=2\end{align}[/latex]

Thus,

[latex]\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\dfrac{1}{x}-\dfrac{2}{\left(x - 2\right)}+\dfrac{2}{{\left(x - 2\right)}^{2}}[/latex]

Try It

Find the partial fraction decomposition of the expression with repeated linear factors.

[latex]\dfrac{6x - 11}{{\left(x - 1\right)}^{2}}[/latex]

Show Solution

Try It

Simplify Expressions Using the Power Property for Exponents

A General Note: The Power Rule of Exponents

For any real number [latex]a[/latex] and positive integers [latex]m[/latex] and [latex]n[/latex], the power rule of exponents states that

[latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex]

Example: USING the Power Rule of Exponents

Write each of the following products with a single base. Do not simplify further.

[latex]{\left({x}^{2}\right)}^{7}[/latex]
[latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex]
[latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex]

Show Solution

Power Series: Prerequisite Material