{"id":2283,"date":"2021-09-21T15:11:49","date_gmt":"2021-09-21T15:11:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-infinite-series\/"},"modified":"2021-11-19T03:16:06","modified_gmt":"2021-11-19T03:16:06","slug":"skills-review-for-infinite-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-infinite-series\/","title":{"raw":"Skills Review for Infinite Series","rendered":"Skills Review for Infinite Series"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use summation notation<\/li>\r\n \t<li>Use partial fraction decomposition for linear factors<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Infinite Series section, we will begin looking at infinite sums of terms. Here we will review sigma (summation) notation and partial fraction decomposition.\r\n<h2>Expand Sigma (Summation) Notation<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Approximating Areas)<\/em><\/strong>\r\n\r\n<strong>Summation notation <\/strong>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 <strong>sigma<\/strong>\u00a0to 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 <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term of the sum. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term of the sum.\r\n\r\nIf 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.\r\n<div style=\"text-align: center;\">[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]<\/div>\r\nWe can find the sum by adding the terms:\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{5}2i=2+4+6+8+10=30[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Summation Notation<\/h3>\r\nThe sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{n}{a}_{i}[\/latex]<\/p>\r\nThis notation tells us to find the sum of [latex]{a}_{i}[\/latex] from [latex]i=1[\/latex] to [latex]i=n[\/latex].\r\n\r\n[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: EXpanding Summation Notation<\/h3>\r\nEvaluate [latex]\\displaystyle\\sum _{i=3}^{7}{i}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"14937\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14937\"]\r\n\r\nAccording 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sum _{i=3}^{7}{i}^{2}&amp; ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill &amp; =9+16+25+36+49\\hfill \\\\ \\hfill &amp; =135\\hfill \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\displaystyle\\sum _{i=2}^{5}\\left(3i - 1\\right)[\/latex].\r\n\r\n[reveal-answer q=\"812548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812548\"]\r\n\r\n38\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]222190[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use Partial Fraction Decomposition for Linear Factors<\/h2>\r\nPartial fraction <strong>decomposition<\/strong> is used to break up one fraction into two.\r\n<div style=\"text-align: center;\">[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]<\/div>\r\nWe 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/h3>\r\n<ol>\r\n \t<li>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\r\n<div style=\"text-align: center;\">[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]<\/div><\/li>\r\n \t<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\r\n \t<li>Expand the right side of the equation and collect like terms.<\/li>\r\n \t<li>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.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Decomposing a Rational Function\u00a0 with Distinct Linear Factors<\/h3>\r\nDecompose the given <strong>rational expression<\/strong> with distinct linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"828392\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"828392\"]\r\n\r\nWe will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[\/latex], or [latex]C[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{A}{\\left(x+2\\right)}+\\dfrac{B}{\\left(x - 1\\right)}[\/latex]<\/p>\r\nMultiply both sides of the equation by the common denominator to eliminate the fractions:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThe resulting equation is\r\n<p style=\"text-align: center;\">[latex]3x=A\\left(x - 1\\right)+B\\left(x+2\\right)[\/latex]<\/p>\r\nExpand the right side of the equation and collect like terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3x=Ax-A+Bx+2B\\\\ 3x=\\left(A+B\\right)x-A+2B\\end{gathered}[\/latex]<\/p>\r\nSet up a system of equations associating corresponding coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3=A+B\\\\ 0=-A+2B\\end{gathered}[\/latex]<\/p>\r\nAdd the two equations and solve for [latex]B[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3&amp;=A+B \\\\ 0&amp;=-A+2B \\\\ \\hline 3&amp;=0+3B \\\\[4mm] B&amp;=1 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]B=1[\/latex] into one of the original equations in the system.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3&amp;=A+1\\\\ 2&amp;=A\\end{align}[\/latex]<\/p>\r\nThus, the partial fraction decomposition is\r\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nAnother 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]\u00a0or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the\u00a0[latex]A-[\/latex] term becomes 0 and we can simply solve for [latex]B[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x&amp;=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(1\\right)&amp;=A\\left[\\left(1\\right)-1\\right]+B\\left[\\left(1\\right)+2\\right] \\\\ 3&amp;=0+3B\\hfill \\\\ B&amp;=1 \\end{align}[\/latex]<\/p>\r\nNext, 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x&amp;=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(-2\\right)&amp;=A\\left[\\left(-2\\right)-1\\right]+B\\left[\\left(-2\\right)+2\\right] \\\\ -6&amp;=-3A+0 \\\\ \\frac{-6}{-3}&amp;=A \\\\ A&amp;=2 \\end{align}[\/latex]<\/p>\r\nWe obtain the same values for [latex]A[\/latex] and [latex]B[\/latex] using either method, so the decompositions are the same using either method.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\r\nAlthough 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 <strong>Heaviside method<\/strong>, named after Charles Heaviside, a pioneer in the study of electronics.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the partial fraction decomposition of the following expression.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{\\left(x - 3\\right)\\left(x - 2\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"800291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"800291\"]\r\n\r\n[latex]\\dfrac{3}{x - 3}-\\dfrac{2}{x - 2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174463[\/ohm_question]\r\n\r\n<\/div>\r\nSome 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational expression with repeated linear factors, decompose it.<\/h3>\r\n<ol>\r\n \t<li>Use a variable like [latex]A,B[\/latex], or [latex]C[\/latex] for the numerators and account for increasing powers of the denominators.\r\n<div style=\"text-align: center;\">[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]<\/div><\/li>\r\n \t<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\r\n \t<li>Expand the right side of the equation and collect like terms.<\/li>\r\n \t<li>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.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Decomposing with Repeated Linear Factors<\/h3>\r\nDecompose the given rational expression with repeated linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[\/latex]<\/p>\r\n[reveal-answer q=\"969334\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"969334\"]\r\n\r\nThe 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,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nNext, we multiply both sides by the common denominator.\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nOn the right side of the equation, we expand and collect like terms.\r\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=A\\left({x}^{2}-4x+4\\right)+B\\left({x}^{2}-2x\\right)+Cx[\/latex]\r\n[latex]\\begin{align}&amp;=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\\\ &amp;=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A \\end{align}[\/latex]<\/p>\r\nNext, we compare the coefficients of both sides. This will give the system of equations in three variables:\r\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rr}\\hfill A+B=-1&amp; \\hfill \\text{(1)}\\\\ \\hfill -4A - 2B+C=2&amp; \\hfill \\text{(2)}\\\\ \\hfill 4A=4&amp; \\hfill \\text{(3)}\\end{array}[\/latex]<\/p>\r\nSolving for [latex]A[\/latex] , we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}4A&amp;=4 \\\\ A&amp;=1 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]A=1[\/latex] into equation (1).\r\n<p style=\"text-align: center;\">[latex]\\begin{align}A+B=-1 \\\\ \\left(1\\right)+B=-1 \\\\ B=-2 \\end{align}[\/latex]<\/p>\r\nThen, to solve for [latex]C[\/latex], substitute the values for [latex]A[\/latex] and [latex]B[\/latex] into equation (2).\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the partial fraction decomposition of the expression with repeated linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{6x - 11}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\r\n[reveal-answer q=\"623632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"623632\"]\r\n\r\n[latex]\\dfrac{6}{x - 1}-\\dfrac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174464[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use summation notation<\/li>\n<li>Use partial fraction decomposition for linear factors<\/li>\n<\/ul>\n<\/div>\n<p>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.<\/p>\n<h2>Expand Sigma (Summation) Notation<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Approximating Areas)<\/em><\/strong><\/p>\n<p><strong>Summation notation <\/strong>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 <strong>sigma<\/strong>\u00a0to 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 <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term of the sum. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term of the sum.<\/p>\n<p>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.<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<p>We can find the sum by adding the terms:<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{5}2i=2+4+6+8+10=30[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Summation Notation<\/h3>\n<p>The sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{n}{a}_{i}[\/latex]<\/p>\n<p>This notation tells us to find the sum of [latex]{a}_{i}[\/latex] from [latex]i=1[\/latex] to [latex]i=n[\/latex].<\/p>\n<p>[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: EXpanding Summation Notation<\/h3>\n<p>Evaluate [latex]\\displaystyle\\sum _{i=3}^{7}{i}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14937\">Show Solution<\/span><\/p>\n<div id=\"q14937\" class=\"hidden-answer\" style=\"display: none\">\n<p>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.<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\displaystyle\\sum _{i=2}^{5}\\left(3i - 1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812548\">Show Solution<\/span><\/p>\n<div id=\"q812548\" class=\"hidden-answer\" style=\"display: none\">\n<p>38<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm222190\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=222190&theme=oea&iframe_resize_id=ohm222190\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use Partial Fraction Decomposition for Linear Factors<\/h2>\n<p>Partial fraction <strong>decomposition<\/strong> is used to break up one fraction into two.<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<p>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.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/h3>\n<ol>\n<li>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\n<div style=\"text-align: center;\">[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]<\/div>\n<\/li>\n<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n<li>Expand the right side of the equation and collect like terms.<\/li>\n<li>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.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing a Rational Function\u00a0 with Distinct Linear Factors<\/h3>\n<p>Decompose the given <strong>rational expression<\/strong> with distinct linear factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q828392\">Solution<\/span><\/p>\n<div id=\"q828392\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[\/latex], or [latex]C[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{A}{\\left(x+2\\right)}+\\dfrac{B}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>Multiply both sides of the equation by the common denominator to eliminate the fractions:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>The resulting equation is<\/p>\n<p style=\"text-align: center;\">[latex]3x=A\\left(x - 1\\right)+B\\left(x+2\\right)[\/latex]<\/p>\n<p>Expand the right side of the equation and collect like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3x=Ax-A+Bx+2B\\\\ 3x=\\left(A+B\\right)x-A+2B\\end{gathered}[\/latex]<\/p>\n<p>Set up a system of equations associating corresponding coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3=A+B\\\\ 0=-A+2B\\end{gathered}[\/latex]<\/p>\n<p>Add the two equations and solve for [latex]B[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3&=A+B \\\\ 0&=-A+2B \\\\ \\hline 3&=0+3B \\\\[4mm] B&=1 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]B=1[\/latex] into one of the original equations in the system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3&=A+1\\\\ 2&=A\\end{align}[\/latex]<\/p>\n<p>Thus, the partial fraction decomposition is<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>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]\u00a0or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the\u00a0[latex]A-[\/latex] term becomes 0 and we can simply solve for [latex]B[\/latex].<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>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.<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>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.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>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 <strong>Heaviside method<\/strong>, named after Charles Heaviside, a pioneer in the study of electronics.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the partial fraction decomposition of the following expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{\\left(x - 3\\right)\\left(x - 2\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q800291\">Show Solution<\/span><\/p>\n<div id=\"q800291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{3}{x - 3}-\\dfrac{2}{x - 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174463\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174463&theme=oea&iframe_resize_id=ohm174463\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>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.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with repeated linear factors, decompose it.<\/h3>\n<ol>\n<li>Use a variable like [latex]A,B[\/latex], or [latex]C[\/latex] for the numerators and account for increasing powers of the denominators.\n<div style=\"text-align: center;\">[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]<\/div>\n<\/li>\n<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n<li>Expand the right side of the equation and collect like terms.<\/li>\n<li>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.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing with Repeated Linear Factors<\/h3>\n<p>Decompose the given rational expression with repeated linear factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q969334\">Show Solution<\/span><\/p>\n<div id=\"q969334\" class=\"hidden-answer\" style=\"display: none\">\n<p>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,<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>Next, we multiply both sides by the common denominator.<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>On the right side of the equation, we expand and collect like terms.<\/p>\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=A\\left({x}^{2}-4x+4\\right)+B\\left({x}^{2}-2x\\right)+Cx[\/latex]<br \/>\n[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]<\/p>\n<p>Next, we compare the coefficients of both sides. This will give the system of equations in three variables:<\/p>\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A[\/latex]<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>Solving for [latex]A[\/latex] , we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}4A&=4 \\\\ A&=1 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]A=1[\/latex] into equation (1).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}A+B=-1 \\\\ \\left(1\\right)+B=-1 \\\\ B=-2 \\end{align}[\/latex]<\/p>\n<p>Then, to solve for [latex]C[\/latex], substitute the values for [latex]A[\/latex] and [latex]B[\/latex] into equation (2).<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the partial fraction decomposition of the expression with repeated linear factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{6x - 11}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q623632\">Show Solution<\/span><\/p>\n<div id=\"q623632\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{6}{x - 1}-\\dfrac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174464\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174464&theme=oea&iframe_resize_id=ohm174464\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2283\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen 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