{"id":4108,"date":"2022-04-14T18:15:55","date_gmt":"2022-04-14T18:15:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-series-solutions-of-differential-equations\/"},"modified":"2022-11-09T16:51:37","modified_gmt":"2022-11-09T16:51:37","slug":"skills-review-for-series-solutions-of-differential-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-series-solutions-of-differential-equations\/","title":{"raw":"Skills Review for Series Solutions of Differential Equations","rendered":"Skills Review for Series Solutions of Differential Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply factorial notation<\/li>\r\n \t<li>Use summation notation<\/li>\r\n \t<li>Write the terms of a sequence defined by a recursive formula<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Series Solutions of Differential Equations section, we will use infinite series to solve and express solutions of differential equations. Here we will review how to use factorial notation, expand sigma (summation) notation, and evaluate a recursive formula.\r\n<h2>Apply Factorial Notation<\/h2>\r\nRecall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&amp;=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&amp;=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nAn example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nThe factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]\r\n<div class=\"textbox\">\r\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\r\n<strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\r\nThe special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&amp;amp;theme=oea&amp;amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex](n+3)![\/latex].\r\n[reveal-answer q=\"953701\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953701\"]\r\n\r\n[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]\r\n\r\nNotice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Expand Sigma (Summation) Notation<\/h2>\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>Evaluate a Recursive Formula<\/h2>\r\nA <strong>recursive formula<\/strong> is\u00a0a formula that defines its value at a particular input using the result of the previous input(s).\r\n\r\nA recursive formula always has two parts: the value of an initial input and an equation defining each term in terms of preceding terms. For example, suppose we know the following:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{x}_{1}=3 \\\\ &amp;{x}_{n}=2{x}_{n - 1}-1, \\text{ for } n\\ge 2 \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: left;\">We can find the subsequent terms of the recursive formula using the first term.<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{x}_{1}=3\\\\ &amp;{x}_{2}=2{x}_{1}-1=2\\left(3\\right)-1=5\\\\ &amp;{x}_{3}=2{x}_{2}-1=2\\left(5\\right)-1=9\\\\ &amp;{x}_{4}=2{x}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/div>\r\nSo, the first four terms are [latex]3,5,9,\\text{ and},17[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Recursive Formula<\/h3>\r\nWrite the first five terms defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{x}_{1}=9\\\\ &amp;{x}_{n}=3{x}_{n - 1}-20\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"638659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"638659\"]\r\n\r\nThe first term is given in the formula. For each subsequent term, we replace [latex]{x}_{n - 1}[\/latex] with the value of the preceding term.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1&amp;&amp; {x}_{1}=9 \\\\ &amp;n=2&amp;&amp; {x}_{2}=3{x}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &amp;n=3&amp;&amp; {x}_{3}=3{x}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &amp;n=4&amp;&amp; {x}_{4}=3{x}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &amp;n=5&amp;&amp; {x}_{5}=3{x}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71\\end{align}[\/latex]<\/p>\r\nThe first five terms are [latex]9, 7, 1, -17, \\text{ and}, -71[\/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]221978[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply factorial notation<\/li>\n<li>Use summation notation<\/li>\n<li>Write the terms of a sequence defined by a recursive formula<\/li>\n<\/ul>\n<\/div>\n<p>In the Series Solutions of Differential Equations section, we will use infinite series to solve and express solutions of differential equations. Here we will review how to use factorial notation, expand sigma (summation) notation, and evaluate a recursive formula.<\/p>\n<h2>Apply Factorial Notation<\/h2>\n<p>Recall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>An example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>The factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\n<p><strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\n<p>The special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm61071\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&#38;theme=oea&#38;iframe_resize_id=ohm61071&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex](n+3)![\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953701\">Show Solution<\/span><\/p>\n<div id=\"q953701\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]<\/p>\n<p>Notice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Expand Sigma (Summation) Notation<\/h2>\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>Evaluate a Recursive Formula<\/h2>\n<p>A <strong>recursive formula<\/strong> is\u00a0a formula that defines its value at a particular input using the result of the previous input(s).<\/p>\n<p>A recursive formula always has two parts: the value of an initial input and an equation defining each term in terms of preceding terms. For example, suppose we know the following:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{x}_{1}=3 \\\\ &{x}_{n}=2{x}_{n - 1}-1, \\text{ for } n\\ge 2 \\end{align}[\/latex]<\/div>\n<div><\/div>\n<div style=\"text-align: left;\">We can find the subsequent terms of the recursive formula using the first term.<\/div>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{x}_{1}=3\\\\ &{x}_{2}=2{x}_{1}-1=2\\left(3\\right)-1=5\\\\ &{x}_{3}=2{x}_{2}-1=2\\left(5\\right)-1=9\\\\ &{x}_{4}=2{x}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/div>\n<p>So, the first four terms are [latex]3,5,9,\\text{ and},17[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Recursive Formula<\/h3>\n<p>Write the first five terms defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{x}_{1}=9\\\\ &{x}_{n}=3{x}_{n - 1}-20\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q638659\">Show Solution<\/span><\/p>\n<div id=\"q638659\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first term is given in the formula. For each subsequent term, we replace [latex]{x}_{n - 1}[\/latex] with the value of the preceding term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&n=1&& {x}_{1}=9 \\\\ &n=2&& {x}_{2}=3{x}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &n=3&& {x}_{3}=3{x}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &n=4&& {x}_{4}=3{x}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &n=5&& {x}_{5}=3{x}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71\\end{align}[\/latex]<\/p>\n<p>The first five terms are [latex]9, 7, 1, -17, \\text{ and}, -71[\/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=\"ohm221978\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=221978&theme=oea&iframe_resize_id=ohm221978\" 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-4108\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/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>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/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":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4108","chapter","type-chapter","status-publish","hentry"],"part":4150,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4108","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4108\/revisions"}],"predecessor-version":[{"id":6497,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4108\/revisions\/6497"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4150"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4108\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4108"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4108"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4108"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4108"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}