Learning Outcomes
- Use summation notation
- Use partial fraction decomposition for linear factors
In the Infinite Series section, we will begin looking at infinite sums of terms. Here we will review sigma (summation) notation and partial fraction decomposition.
Expand Sigma (Summation) Notation
(also in Module 1, Skills Review for Approximating Areas)
Summation notation is used to represent long sums of values in a compact form. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms of the sum. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term of the sum. The number above the sigma, called the upper limit of summation, is the number used to generate the last term of the sum.
If we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{i}=2i[/latex] for [latex]i=1[/latex] through [latex]i=5[/latex]. We can begin by substituting the terms for [latex]i[/latex] and listing out the terms.
[latex]\begin{array}{l} {a}_{1}=2\left(1\right)=2 \\ {a}_{2}=2\left(2\right)=4\hfill \\ {a}_{3}=2\left(3\right)=6\hfill \\ {a}_{4}=2\left(4\right)=8\hfill \\ {a}_{5}=2\left(5\right)=10\hfill \end{array}[/latex]
We can find the sum by adding the terms:
[latex]\displaystyle\sum _{i=1}^{5}2i=2+4+6+8+10=30[/latex]
A General Note: Summation Notation
The sum of the first [latex]n[/latex] terms of a series can be expressed in summation notation as follows:
[latex]\displaystyle\sum _{i=1}^{n}{a}_{i}[/latex]
This notation tells us to find the sum of [latex]{a}_{i}[/latex] from [latex]i=1[/latex] to [latex]i=n[/latex].
[latex]k[/latex] is called the index of summation, 1 is the lower limit of summation, and [latex]n[/latex] is the upper limit of summation.
Example: EXpanding Summation Notation
Evaluate [latex]\displaystyle\sum _{i=3}^{7}{i}^{2}[/latex].
Show Solution
According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{i}^{2}[/latex] from [latex]i=3[/latex] to [latex]i=7[/latex]. We find the terms of the series by substituting [latex]i=3\text{,}4\text{,}5\text{,}6[/latex], and [latex]7[/latex] into the function [latex]{i}^{2}[/latex]. We add the terms to find the sum.
[latex]\begin{align}\sum _{i=3}^{7}{i}^{2}& ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\hfill \\ \hfill & =9+16+25+36+49\hfill \\ \hfill & =135\hfill \end{align}[/latex]
Try It
Evaluate [latex]\displaystyle\sum _{i=2}^{5}\left(3i - 1\right)[/latex].
Use Partial Fraction Decomposition for Linear Factors
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
[latex]\dfrac{3}{x - 3}-\dfrac{2}{x - 2}[/latex]
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
[latex]\dfrac{6}{x - 1}-\dfrac{5}{{\left(x - 1\right)}^{2}}[/latex]
