{"id":1816,"date":"2021-07-28T16:07:58","date_gmt":"2021-07-28T16:07:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1816"},"modified":"2022-03-21T23:17:16","modified_gmt":"2022-03-21T23:17:16","slug":"solving-differential-equations-and-nonelementary-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/solving-differential-equations-and-nonelementary-integrals\/","title":{"raw":"Solving Differential Equations and Nonelementary Integrals","rendered":"Solving Differential Equations and Nonelementary Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Taylor series to solve differential equations<\/li>\r\n \t<li>Use Taylor series to evaluate nonelementary integrals<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Solving Differential Equations with Power Series<\/h2>\r\n<p id=\"fs-id1167024036513\">Consider the differential equation<\/p>\r\n\r\n<div id=\"fs-id1167024036516\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }\\left(x\\right)=y[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024036543\">Recall that this is a first-order separable equation and its solution is [latex]y=C{e}^{x}[\/latex]. This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form [latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving [latex]{y}^{\\prime }=y[\/latex] to illustrate the technique.<\/p>\r\n\r\n<div id=\"fs-id1167023814845\" data-type=\"example\">\r\n<div id=\"fs-id1167023814847\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023814850\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Power Series Solution of a Differential Equation<\/h3>\r\n<div id=\"fs-id1167023814850\" data-type=\"problem\">\r\n<p id=\"fs-id1167023814855\">Use power series to solve the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1167023814858\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=y,y\\left(0\\right)=3[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1167023814897\" data-type=\"solution\">\r\n<p id=\"fs-id1167023814899\">Suppose that there exists a power series solution<\/p>\r\n\r\n<div id=\"fs-id1167023814903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024039113\">Differentiating this series term by term, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167024039116\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024039183\">If <em data-effect=\"italics\">y<\/em> satisfies the differential equation, then<\/p>\r\n\r\n<div id=\"fs-id1167024039191\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots ={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{3}{x}^{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024040266\">Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167024040274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{c}_{0}={c}_{1},\\hfill \\\\ {c}_{1}=2{c}_{2},\\hfill \\\\ {c}_{2}=3{c}_{3},\\hfill \\\\ {c}_{3}=4{c}_{4},\\hfill \\\\ \\hfill \\vdots.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023911019\">Using the initial condition [latex]y\\left(0\\right)=3[\/latex] combined with the power series representation<\/p>\r\n\r\n<div id=\"fs-id1167023911039\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y\\left(x\\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots [\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023911105\">we find that [latex]{c}_{0}=3[\/latex]. We are now ready to solve for the rest of the coefficients. Using the fact that [latex]{c}_{0}=3[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1167023911138\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{1}={c}_{0}=3=\\frac{3}{1\\text{!}},\\hfill \\\\ {c}_{2}=\\frac{{c}_{1}}{2}=\\frac{3}{2}=\\frac{3}{2\\text{!}},\\hfill \\\\ {c}_{3}=\\frac{{c}_{2}}{3}=\\frac{3}{3\\cdot 2}=\\frac{3}{3\\text{!}},\\hfill \\\\ {c}_{4}=\\frac{{c}_{3}}{4}=\\frac{3}{4\\cdot 3\\cdot 2}=\\frac{3}{4\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023806804\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167023806807\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill y&amp; =3\\left[1+\\frac{1}{1\\text{!}}x+\\frac{1}{2\\text{!}}{x}^{2}+\\frac{1}{3\\text{!}}{x}^{3}\\frac{1}{4\\text{!}}{x}^{4}+\\cdots \\right]\\hfill \\\\ &amp; =3{\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{x}^{n}}{n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023806940\">You might recognize<\/p>\r\n\r\n<div id=\"fs-id1167023779741\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023779778\">as the Taylor series for [latex]{e}^{x}[\/latex]. Therefore, the solution is [latex]y=3{e}^{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Power Series Solution of a Differential Equation.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6724949&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=EfY0D_M8Nlw&amp;video_target=tpm-plugin-nc450a0h-EfY0D_M8Nlw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.4_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.4.4\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167023779811\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023779815\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023779817\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023779817\" data-type=\"problem\">\r\n<p id=\"fs-id1167023779820\">Use power series to solve [latex]{y}^{\\prime }=2y,y\\left(0\\right)=5[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1167023779881\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023779889\">The equations for the first several coefficients [latex]{c}_{n}[\/latex] will satisfy [latex]{c}_{0}=2{c}_{1},{c}_{1}=2\\cdot 2{c}_{2},{c}_{2}=2\\cdot 3{c}_{3}[\/latex]. In general, for all [latex]n\\ge 0,{c}_{n}=2\\left(n+1\\right){C}_{n+1}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1167023779859\" data-type=\"solution\">\r\n<p id=\"fs-id1167023779861\" style=\"text-align: center;\">[latex]y=5{e}^{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167024039026\">We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation<\/p>\r\n\r\n<div id=\"fs-id1167024039030\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }-xy=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024039054\">is known as <span class=\"no-emphasis\" data-type=\"term\">Airy\u2019s equation<\/span>. It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.<\/p>\r\n\r\n<div id=\"fs-id1167024039063\" data-type=\"example\">\r\n<div id=\"fs-id1167024039065\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024039068\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Power Series Solution of Airy\u2019s Equation<\/h3>\r\n<div id=\"fs-id1167024039068\" data-type=\"problem\">\r\n<p id=\"fs-id1167023780991\">Use power series to solve<\/p>\r\n\r\n<div id=\"fs-id1167023780994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime\\prime}-xy=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023781017\">with the initial conditions [latex]y\\left(0\\right)=a[\/latex] and [latex]y^{\\prime} \\left(0\\right)=b[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1167023781058\" data-type=\"solution\">\r\n<p id=\"fs-id1167023781061\">We look for a solution of the form<\/p>\r\n\r\n<div id=\"fs-id1167023781064\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024040051\">Differentiating this function term by term, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167024040054\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {y}^{\\prime }&amp; =\\hfill &amp; {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\\cdots ,\\hfill \\\\ \\hfill y\\text{{\\'\\'} }&amp; =\\hfill &amp; 2\\cdot 1{c}_{2}+3\\cdot 2{c}_{3}x+4\\cdot 3{c}_{4}{x}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024046820\">If <em data-effect=\"italics\">y<\/em> satisfies the equation [latex]y^{\\prime\\prime}=xy[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1167024046846\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2\\cdot 1{c}_{2}+3\\cdot 2{c}_{3}x+4\\cdot 3{c}_{4}{x}^{2}+\\cdots =x\\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots \\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024046962\">Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167024046970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}2\\cdot 1{c}_{2}=0,\\hfill \\\\ 3\\cdot 2{c}_{3}={c}_{0},\\hfill \\\\ 4\\cdot 3{c}_{4}={c}_{1},\\hfill \\\\ 5\\cdot 4{c}_{5}={c}_{2},\\hfill \\\\ \\hfill \\vdots.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023784631\">More generally, for [latex]n\\ge 3[\/latex], we have [latex]n\\cdot \\left(n - 1\\right){c}_{n}={c}_{n - 3}[\/latex]. In fact, all coefficients can be written in terms of [latex]{c}_{0}[\/latex] and [latex]{c}_{1}[\/latex]. To see this, first note that [latex]{c}_{2}=0[\/latex]. Then<\/p>\r\n\r\n<div id=\"fs-id1167023789230\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{3}=\\frac{{c}_{0}}{3\\cdot 2},\\hfill \\\\ {c}_{4}=\\frac{{c}_{1}}{4\\cdot 3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023789295\">For [latex]{c}_{5},{c}_{6},{c}_{7}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1167023789324\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{5}=\\frac{{c}_{2}}{5\\cdot 4}=0,\\hfill \\\\ {c}_{6}=\\frac{{c}_{3}}{6\\cdot 5}=\\frac{{c}_{0}}{6\\cdot 5\\cdot 3\\cdot 2},\\hfill \\\\ {c}_{7}=\\frac{{c}_{4}}{7\\cdot 6}=\\frac{{c}_{1}}{7\\cdot 6\\cdot 4\\cdot 3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023913967\">Therefore, the series solution of the differential equation is given by<\/p>\r\n\r\n<div id=\"fs-id1167023913971\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y={c}_{0}+{c}_{1}x+0\\cdot {x}^{2}+\\frac{{c}_{0}}{3\\cdot 2}{x}^{3}+\\frac{{c}_{1}}{4\\cdot 3}{x}^{4}+0\\cdot {x}^{5}+\\frac{{c}_{0}}{6\\cdot 5\\cdot 3\\cdot 2}{x}^{6}+\\frac{{c}_{1}}{7\\cdot 6\\cdot 4\\cdot 3}{x}^{7}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023763410\">The initial condition [latex]y\\left(0\\right)=a[\/latex] implies [latex]{c}_{0}=a[\/latex]. Differentiating this series term by term and using the fact that [latex]{y}^{\\prime }\\left(0\\right)=b[\/latex], we conclude that [latex]{c}_{1}=b[\/latex]. Therefore, the solution of this initial-value problem is<\/p>\r\n\r\n<div id=\"fs-id1167023763482\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=a\\left(1+\\frac{{x}^{3}}{3\\cdot 2}+\\frac{{x}^{6}}{6\\cdot 5\\cdot 3\\cdot 2}+\\cdots \\right)+b\\left(x+\\frac{{x}^{4}}{4\\cdot 3}+\\frac{{x}^{7}}{7\\cdot 6\\cdot 4\\cdot 3}+\\cdots \\right)[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167023801462\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023801467\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023801469\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023801469\" data-type=\"problem\">\r\n<p id=\"fs-id1167023801471\">Use power series to solve [latex]y^{\\prime\\prime}+{x}^{2}y=0[\/latex] with the initial condition [latex]y\\left(0\\right)=a[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=b[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1167023890904\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023890911\">The coefficients satisfy [latex]{c}_{0}=a,{c}_{1}=b,{c}_{2}=0,{c}_{3}=0[\/latex], and for [latex]n\\ge 4,n\\left(n - 1\\right){c}_{n}=-c_{n - 4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167023801536\" data-type=\"solution\">\r\n<p id=\"fs-id1167023801538\" style=\"text-align: center;\">[latex]y=a\\left(1-\\frac{{x}^{4}}{3\\cdot 4}+\\frac{{x}^{8}}{3\\cdot 4\\cdot 7\\cdot 8}-\\cdots \\right)+b\\left(x-\\frac{{x}^{5}}{4\\cdot 5}+\\frac{{x}^{9}}{4\\cdot 5\\cdot 8\\cdot 9}-\\cdots \\right)[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167023733638\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\r\n<p id=\"fs-id1167023733643\">Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.<\/p>\r\n<p id=\"fs-id1167023733649\">One integral that arises often in applications in probability theory is [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex]. Unfortunately, the antiderivative of the integrand [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term \"elementary function\" is not synonymous with noncomplicated function. For example, the function [latex]f\\left(x\\right)=\\sqrt{{x}^{2}-3x}+{e}^{{x}^{3}}-\\sin\\left(5x+4\\right)[\/latex] is an elementary function, although not a particularly simple-looking function. Any integral of the form [latex]\\displaystyle\\int f\\left(x\\right)dx[\/latex] where the antiderivative of [latex]f[\/latex] cannot be written as an elementary function is considered a <strong>nonelementary integral<\/strong>.<\/p>\r\n<p id=\"fs-id1167023782710\">Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167023782746\" data-type=\"example\">\r\n<div id=\"fs-id1167023782748\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023782750\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Taylor Series to Evaluate a Definite Integral<\/h3>\r\n<div id=\"fs-id1167023782750\" data-type=\"problem\">\r\n<ol id=\"fs-id1167023782755\" type=\"a\">\r\n \t<li>Express [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex] as an infinite series.<\/li>\r\n \t<li>Evaluate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex] to within an error of [latex]0.01[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1167023782839\" data-type=\"solution\">\r\n<ol id=\"fs-id1167023782841\" type=\"a\">\r\n \t<li>The Maclaurin series for [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023782866\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}^{\\text{-}{x}^{2}}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(\\text{-}{x}^{2}\\right)}^{n}}{n\\text{!}}\\hfill \\\\ &amp; =1-{x}^{2}+\\frac{{x}^{4}}{2\\text{!}}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}+\\cdots \\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023760144\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx}&amp; ={\\displaystyle\\int \\left(1-{x}^{2}+\\frac{{x}^{4}}{2\\text{!}}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}+\\cdots \\right)dx}\\hfill \\\\ &amp; =C+x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5.2\\text{!}}-\\frac{{x}^{7}}{7.3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Using the result from part a. we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023808165\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx=1-\\frac{1}{3}+\\frac{1}{10}-\\frac{1}{42}+\\frac{1}{216}-\\cdots [\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe sum of the first four terms is approximately [latex]0.74[\/latex]. By the alternating series test, this estimate is accurate to within an error of less than [latex]\\frac{1}{216}\\approx 0.0046296&lt;0.01[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using Taylor Series to Evaluate a Definite Integral.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6724950&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZwX6LAO_TyA&amp;video_target=tpm-plugin-1zdfx8wc-ZwX6LAO_TyA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.5_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.4.5\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167023808281\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023808285\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023808287\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023808287\" data-type=\"problem\">\r\n<p id=\"fs-id1167023810006\">Express [latex]\\displaystyle\\int \\cos\\sqrt{x}dx[\/latex] as an infinite series. Evaluate [latex]{\\displaystyle\\int }_{0}^{1}\\cos\\sqrt{x}dx[\/latex] to within an error of [latex]0.01[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558809\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558809\"]\r\n<div id=\"fs-id1167023810154\" data-type=\"commentary\" data-element-type=\"hint\">\r\n\r\nUse the series found in the example: Deriving Maclaurin Series from Known Series.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558819\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1167023810066\" data-type=\"solution\">\r\n<p id=\"fs-id1167023810069\">[latex]C+\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n\\left(2n - 2\\right)\\text{!}}[\/latex] The definite integral is approximately [latex]0.514[\/latex] to within an error of [latex]0.01[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5950[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167023810172\">As mentioned above, the integral [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex] arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean [latex]\\mu [\/latex] and standard deviation [latex]\\sigma [\/latex], then the probability that a randomly chosen value lies between [latex]x=a[\/latex] and [latex]x=b[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167023806051\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{1}{\\sigma \\sqrt{2\\pi }}{\\displaystyle\\int }_{a}^{b}{e}^\\frac{{\\text{-}{\\left(x-\\mu \\right)}^{2}}{\\left(2{\\sigma }^{2}\\right)}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023806128\">(See Figure 2.)<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_10_04_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"617\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234538\/CNX_Calc_Figure_10_04_003.jpg\" alt=\"This graph is the normal distribution. It is a bell-shaped curve with the highest point above mu on the x-axis. Also, there is a shaded area under the curve above the x-axis. The shaded area is bounded by alpha on the left and b on the right.\" width=\"617\" height=\"272\" data-media-type=\"image\/jpeg\" \/> Figure 2. If data values are normally distributed with mean [latex]\\mu [\/latex] and standard deviation [latex]\\sigma [\/latex], the probability that a randomly selected data value is between [latex]a[\/latex] and [latex]b[\/latex] is the area under the curve [latex]y=\\frac{1}{\\sigma\\sqrt{2\\pi}}{e}^{\\frac{-\\left(x-\\mu\\right)^{2}}{\\left(2{\\sigma}^{2}\\right)}}[\/latex] between [latex]x=a[\/latex] and [latex]x=b[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1167023782195\">To simplify this integral, we typically let [latex]z=\\frac{x-\\mu }{\\sigma }[\/latex]. This quantity [latex]z[\/latex] is known as the [latex]z[\/latex] score of a data value. With this simplification, integral our previous equation becomes<\/p>\r\n\r\n<div id=\"fs-id1167023782231\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\dfrac{1}{\\sqrt{2\\pi }} {\\displaystyle\\int}_{\\frac{\\left(a-\\mu \\right)}{\\sigma}}^{\\frac{\\left(b-\\mu\\right)}{\\sigma}}{e}^{-z^2\\text{\/}2}dz[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023782313\">In the next example, we show how we can use this integral in calculating probabilities.<\/p>\r\n\r\n<div id=\"fs-id1167023782321\" data-type=\"example\">\r\n<div id=\"fs-id1167023782323\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023782325\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Maclaurin Series to Approximate a Probability<\/h3>\r\n<div id=\"fs-id1167023782325\" data-type=\"problem\">\r\n<p id=\"fs-id1167023782327\">Suppose a set of standardized test scores are normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =50[\/latex]. Use the equation before this example and the first six terms in the Maclaurin series for [latex]{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to approximate the probability that a randomly selected test score is between [latex]x=100[\/latex] and [latex]x=200[\/latex]. Use the alternating series test to determine how accurate your approximation is.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44548899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44548899\"]\r\n<div id=\"fs-id1167023757999\" data-type=\"solution\">\r\n<p id=\"fs-id1167023758001\">Since [latex]\\mu =100,\\sigma =50[\/latex], and we are trying to determine the area under the curve from [latex]a=100[\/latex] to [latex]b=200[\/latex], the integral becomes<\/p>\r\n\r\n<div id=\"fs-id1167023758050\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{0}^{2}{e}^{\\frac{\\text{-}{z}^{2}}{2}}dz[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023758104\">The Maclaurin series for [latex]{e}^{\\text{-}\\frac{{x}^{2}}{2}}[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167023758126\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}^{\\text{-}{x}^{2}\\text{\/}2}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-\\frac{{x}^{2}}{2}\\right)}^{n}}{n\\text{!}}\\hfill \\\\ &amp; =1-\\frac{{x}^{2}}{{2}^{1}\\cdot 1\\text{!}}+\\frac{{x}^{4}}{{2}^{2}\\cdot 2\\text{!}}-\\frac{{x}^{6}}{{2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{{2}^{n}\\cdot n\\text{!}}+\\cdots \\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2}{}^{n}}{{2}^{n}\\cdot n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024047711\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167024047714\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int {e}^{\\text{-}{z}^{2}\\text{\/}2}dz}&amp; =\\hfill &amp; \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int \\left(1-\\frac{{z}^{2}}{{2}^{1}\\cdot 1\\text{!}}+\\frac{{z}^{4}}{{2}^{2}\\cdot 2\\text{!}}-\\frac{{z}^{6}}{{2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{z}^{2n}}{{2}^{n}\\cdot n\\text{!}}+\\cdots \\right)dz}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{1}{\\sqrt{2\\pi }}\\left(C+z-\\frac{{z}^{3}}{3\\cdot {2}^{1}\\cdot 1\\text{!}}+\\frac{{z}^{5}}{5\\cdot {2}^{2}\\cdot 2\\text{!}}-\\frac{{z}^{7}}{7\\cdot {2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{z}^{2n+1}}{\\left(2n+1\\right){2}^{n}\\cdot n\\text{!}}+\\cdots \\right)\\hfill \\\\ \\hfill \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int _{0}^{2}{e}^{\\text{-}{z}^{2}\\text{\/}2}dz}&amp; =\\hfill &amp; \\frac{1}{\\sqrt{2\\pi }}\\left(2-\\frac{8}{6}+\\frac{32}{40}-\\frac{128}{336}+\\frac{512}{3456}-\\frac{{2}^{11}}{11\\cdot {2}^{5}\\cdot 5\\text{!}}+\\cdots \\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023790566\">Using the first five terms, we estimate that the probability is approximately [latex]0.4922[\/latex]. By the alternating series test, we see that this estimate is accurate to within<\/p>\r\n\r\n<div id=\"fs-id1167023790577\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{2\\pi }}\\frac{{2}^{13}}{13\\cdot {2}^{6}\\cdot 6\\text{!}}\\approx 0.00546[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167023790628\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1167023790633\">If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately [latex]95\\%[\/latex]. Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around [latex]47.5\\text{%}[\/latex]. The estimate, combined with the bound on the accuracy, falls within this range.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167023790662\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023790666\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023790668\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023790668\" data-type=\"problem\">\r\n<p id=\"fs-id1167023790670\">Use the first five terms of the Maclaurin series for [latex]{e}^{\\text{-}\\frac{{x}^{2}}{2}}[\/latex] to estimate the probability that a randomly selected test score is between [latex]100[\/latex] and [latex]150[\/latex]. Use the alternating series test to determine the accuracy of this estimate.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558699\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1167023790727\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023790735\">Evaluate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}\\frac{{z}^{2}}{2}}dz[\/latex] using the first five terms of the Maclaurin series for [latex]{e}^{\\text{-}\\frac{{z}^{2}}{2}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1167023790706\" data-type=\"solution\">\r\n<p id=\"fs-id1167023790708\">The estimate is approximately [latex]0.3414[\/latex]. This estimate is accurate to within [latex]0.0000094[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167024040461\">Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is<\/p>\r\n\r\n<div id=\"fs-id1167024040466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta }{\\sqrt{1-{k}^{2}{\\sin}^{2}\\theta }}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024040523\">An integral of this form is known as an <span class=\"no-emphasis\" data-type=\"term\">elliptic integral<\/span> of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.<\/p>\r\n\r\n<div id=\"fs-id1167024040533\" data-type=\"example\">\r\n<div id=\"fs-id1167024040535\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024040537\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Period of a Pendulum<\/h3>\r\n<div id=\"fs-id1167024040537\" data-type=\"problem\">\r\n<p id=\"fs-id1167024040542\">The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length [latex]L[\/latex] that makes a maximum angle [latex]{\\theta }_{\\text{max}}[\/latex] with the vertical, its period [latex]T[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167024040565\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta }{\\sqrt{1-{k}^{2}{\\sin}^{2}\\theta }}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023766290\">where [latex]g[\/latex] is the acceleration due to gravity and [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex] (see Figure 3). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and [latex]\\sin\\theta [\/latex] is approximated by [latex]\\theta[\/latex].) Use the binomial series<\/p>\r\n\r\n<div id=\"fs-id1167023766351\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{1+x}}=1+\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023766456\">to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if<\/p>\r\n\r\n<ol id=\"fs-id1167023766461\" type=\"a\">\r\n \t<li>you use only the first term in the binomial series, and<\/li>\r\n \t<li>you use the first two terms in the binomial series.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_04_003\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234541\/CNX_Calc_Figure_10_04_002.jpg\" alt=\"This figure is a pendulum. There are three positions of the pendulum shown. When the pendulum is to the far left, it is labeled negative theta max. When the pendulum is in the middle and vertical, it is labeled equilibrium position. When the pendulum is to the far right it is labeled theta max. Also, theta is the angle from equilibrium to the far right position. The length of the pendulum is labeled L.\" width=\"487\" height=\"435\" data-media-type=\"image\/jpeg\" \/> Figure 3. This pendulum has length [latex]L[\/latex] and makes a maximum angle [latex]{\\theta }_{\\text{max}}[\/latex] with the vertical.[\/caption]<\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44553899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44553899\"]\r\n<div id=\"fs-id1167023871808\" data-type=\"solution\">\r\n<p id=\"fs-id1167023871810\">We use the binomial series, replacing [latex]x[\/latex] with [latex]\\text{-}{k}^{2}{\\sin}^{2}\\theta [\/latex]. Then we can write the period as<\/p>\r\n\r\n<div id=\"fs-id1167023871840\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}{\\sin}^{4}\\theta +\\cdots \\right)d\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<ol id=\"fs-id1167023871957\" type=\"a\">\r\n \t<li>Using just the first term in the integrand, the first-order estimate is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023871967\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\approx 4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}d\\theta =2\\pi \\sqrt{\\frac{L}{g}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIf [latex]{\\theta }_{\\text{max}}[\/latex] is small, then [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex] is small. We claim that when [latex]k[\/latex] is small, this is a good estimate. To justify this claim, consider<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023794977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}{\\sin}^{4}\\theta +\\cdots \\right)d\\theta [\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]|\\sin{x}|\\le 1[\/latex], this integral is bounded by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023795106\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(\\frac{1}{2}{k}^{2}+\\frac{1.3}{2\\text{!}{2}^{2}}{k}^{4}+\\cdots \\right)d\\theta &lt;\\frac{\\pi }{2}\\left(\\frac{1}{2}{k}^{2}+\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}+\\cdots \\right)[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFurthermore, it can be shown that each coefficient on the right-hand side is less than [latex]1[\/latex] and, therefore, that this expression is bounded by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023835770\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\pi {k}^{2}}{2}\\left(1+{k}^{2}+{k}^{4}+\\cdots \\right)=\\frac{\\pi {k}^{2}}{2}\\cdot \\frac{1}{1-{k}^{2}}[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhich is small for [latex]k[\/latex] small.<\/li>\r\n \t<li>For larger values of [latex]{\\theta }_{\\text{max}}[\/latex], we can approximate [latex]T[\/latex] by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023785853\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill T&amp; \\approx 4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta \\right)d\\theta \\hfill \\\\ &amp; =2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167023785982\">The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.<\/p>\r\n\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use Taylor series to solve differential equations<\/li>\n<li>Use Taylor series to evaluate nonelementary integrals<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Solving Differential Equations with Power Series<\/h2>\n<p id=\"fs-id1167024036513\">Consider the differential equation<\/p>\n<div id=\"fs-id1167024036516\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }\\left(x\\right)=y[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024036543\">Recall that this is a first-order separable equation and its solution is [latex]y=C{e}^{x}[\/latex]. This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form [latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving [latex]{y}^{\\prime }=y[\/latex] to illustrate the technique.<\/p>\n<div id=\"fs-id1167023814845\" data-type=\"example\">\n<div id=\"fs-id1167023814847\" data-type=\"exercise\">\n<div id=\"fs-id1167023814850\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Power Series Solution of a Differential Equation<\/h3>\n<div id=\"fs-id1167023814850\" data-type=\"problem\">\n<p id=\"fs-id1167023814855\">Use power series to solve the initial-value problem<\/p>\n<div id=\"fs-id1167023814858\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=y,y\\left(0\\right)=3[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023814897\" data-type=\"solution\">\n<p id=\"fs-id1167023814899\">Suppose that there exists a power series solution<\/p>\n<div id=\"fs-id1167023814903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024039113\">Differentiating this series term by term, we obtain<\/p>\n<div id=\"fs-id1167024039116\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024039183\">If <em data-effect=\"italics\">y<\/em> satisfies the differential equation, then<\/p>\n<div id=\"fs-id1167024039191\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots ={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{3}{x}^{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024040266\">Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,<\/p>\n<div id=\"fs-id1167024040274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{c}_{0}={c}_{1},\\hfill \\\\ {c}_{1}=2{c}_{2},\\hfill \\\\ {c}_{2}=3{c}_{3},\\hfill \\\\ {c}_{3}=4{c}_{4},\\hfill \\\\ \\hfill \\vdots.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023911019\">Using the initial condition [latex]y\\left(0\\right)=3[\/latex] combined with the power series representation<\/p>\n<div id=\"fs-id1167023911039\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y\\left(x\\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023911105\">we find that [latex]{c}_{0}=3[\/latex]. We are now ready to solve for the rest of the coefficients. Using the fact that [latex]{c}_{0}=3[\/latex], we have<\/p>\n<div id=\"fs-id1167023911138\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{1}={c}_{0}=3=\\frac{3}{1\\text{!}},\\hfill \\\\ {c}_{2}=\\frac{{c}_{1}}{2}=\\frac{3}{2}=\\frac{3}{2\\text{!}},\\hfill \\\\ {c}_{3}=\\frac{{c}_{2}}{3}=\\frac{3}{3\\cdot 2}=\\frac{3}{3\\text{!}},\\hfill \\\\ {c}_{4}=\\frac{{c}_{3}}{4}=\\frac{3}{4\\cdot 3\\cdot 2}=\\frac{3}{4\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023806804\">Therefore,<\/p>\n<div id=\"fs-id1167023806807\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill y& =3\\left[1+\\frac{1}{1\\text{!}}x+\\frac{1}{2\\text{!}}{x}^{2}+\\frac{1}{3\\text{!}}{x}^{3}\\frac{1}{4\\text{!}}{x}^{4}+\\cdots \\right]\\hfill \\\\ & =3{\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{x}^{n}}{n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023806940\">You might recognize<\/p>\n<div id=\"fs-id1167023779741\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023779778\">as the Taylor series for [latex]{e}^{x}[\/latex]. Therefore, the solution is [latex]y=3{e}^{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Power Series Solution of a Differential Equation.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6724949&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=EfY0D_M8Nlw&amp;video_target=tpm-plugin-nc450a0h-EfY0D_M8Nlw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.4_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.4.4&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167023779811\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023779815\" data-type=\"exercise\">\n<div id=\"fs-id1167023779817\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023779817\" data-type=\"problem\">\n<p id=\"fs-id1167023779820\">Use power series to solve [latex]{y}^{\\prime }=2y,y\\left(0\\right)=5[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Hint<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023779881\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023779889\">The equations for the first several coefficients [latex]{c}_{n}[\/latex] will satisfy [latex]{c}_{0}=2{c}_{1},{c}_{1}=2\\cdot 2{c}_{2},{c}_{2}=2\\cdot 3{c}_{3}[\/latex]. In general, for all [latex]n\\ge 0,{c}_{n}=2\\left(n+1\\right){C}_{n+1}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023779859\" data-type=\"solution\">\n<p id=\"fs-id1167023779861\" style=\"text-align: center;\">[latex]y=5{e}^{2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167024039026\">We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation<\/p>\n<div id=\"fs-id1167024039030\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }-xy=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024039054\">is known as <span class=\"no-emphasis\" data-type=\"term\">Airy\u2019s equation<\/span>. It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.<\/p>\n<div id=\"fs-id1167024039063\" data-type=\"example\">\n<div id=\"fs-id1167024039065\" data-type=\"exercise\">\n<div id=\"fs-id1167024039068\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Power Series Solution of Airy\u2019s Equation<\/h3>\n<div id=\"fs-id1167024039068\" data-type=\"problem\">\n<p id=\"fs-id1167023780991\">Use power series to solve<\/p>\n<div id=\"fs-id1167023780994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime\\prime}-xy=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023781017\">with the initial conditions [latex]y\\left(0\\right)=a[\/latex] and [latex]y^{\\prime} \\left(0\\right)=b[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Show Solution<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023781058\" data-type=\"solution\">\n<p id=\"fs-id1167023781061\">We look for a solution of the form<\/p>\n<div id=\"fs-id1167023781064\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024040051\">Differentiating this function term by term, we obtain<\/p>\n<div id=\"fs-id1167024040054\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {y}^{\\prime }& =\\hfill & {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\\cdots ,\\hfill \\\\ \\hfill y\\text{{\\'\\'} }& =\\hfill & 2\\cdot 1{c}_{2}+3\\cdot 2{c}_{3}x+4\\cdot 3{c}_{4}{x}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024046820\">If <em data-effect=\"italics\">y<\/em> satisfies the equation [latex]y^{\\prime\\prime}=xy[\/latex], then<\/p>\n<div id=\"fs-id1167024046846\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2\\cdot 1{c}_{2}+3\\cdot 2{c}_{3}x+4\\cdot 3{c}_{4}{x}^{2}+\\cdots =x\\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots \\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024046962\">Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,<\/p>\n<div id=\"fs-id1167024046970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}2\\cdot 1{c}_{2}=0,\\hfill \\\\ 3\\cdot 2{c}_{3}={c}_{0},\\hfill \\\\ 4\\cdot 3{c}_{4}={c}_{1},\\hfill \\\\ 5\\cdot 4{c}_{5}={c}_{2},\\hfill \\\\ \\hfill \\vdots.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023784631\">More generally, for [latex]n\\ge 3[\/latex], we have [latex]n\\cdot \\left(n - 1\\right){c}_{n}={c}_{n - 3}[\/latex]. In fact, all coefficients can be written in terms of [latex]{c}_{0}[\/latex] and [latex]{c}_{1}[\/latex]. To see this, first note that [latex]{c}_{2}=0[\/latex]. Then<\/p>\n<div id=\"fs-id1167023789230\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{3}=\\frac{{c}_{0}}{3\\cdot 2},\\hfill \\\\ {c}_{4}=\\frac{{c}_{1}}{4\\cdot 3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023789295\">For [latex]{c}_{5},{c}_{6},{c}_{7}[\/latex], we see that<\/p>\n<div id=\"fs-id1167023789324\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {c}_{5}=\\frac{{c}_{2}}{5\\cdot 4}=0,\\hfill \\\\ {c}_{6}=\\frac{{c}_{3}}{6\\cdot 5}=\\frac{{c}_{0}}{6\\cdot 5\\cdot 3\\cdot 2},\\hfill \\\\ {c}_{7}=\\frac{{c}_{4}}{7\\cdot 6}=\\frac{{c}_{1}}{7\\cdot 6\\cdot 4\\cdot 3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023913967\">Therefore, the series solution of the differential equation is given by<\/p>\n<div id=\"fs-id1167023913971\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y={c}_{0}+{c}_{1}x+0\\cdot {x}^{2}+\\frac{{c}_{0}}{3\\cdot 2}{x}^{3}+\\frac{{c}_{1}}{4\\cdot 3}{x}^{4}+0\\cdot {x}^{5}+\\frac{{c}_{0}}{6\\cdot 5\\cdot 3\\cdot 2}{x}^{6}+\\frac{{c}_{1}}{7\\cdot 6\\cdot 4\\cdot 3}{x}^{7}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023763410\">The initial condition [latex]y\\left(0\\right)=a[\/latex] implies [latex]{c}_{0}=a[\/latex]. Differentiating this series term by term and using the fact that [latex]{y}^{\\prime }\\left(0\\right)=b[\/latex], we conclude that [latex]{c}_{1}=b[\/latex]. Therefore, the solution of this initial-value problem is<\/p>\n<div id=\"fs-id1167023763482\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=a\\left(1+\\frac{{x}^{3}}{3\\cdot 2}+\\frac{{x}^{6}}{6\\cdot 5\\cdot 3\\cdot 2}+\\cdots \\right)+b\\left(x+\\frac{{x}^{4}}{4\\cdot 3}+\\frac{{x}^{7}}{7\\cdot 6\\cdot 4\\cdot 3}+\\cdots \\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023801462\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023801467\" data-type=\"exercise\">\n<div id=\"fs-id1167023801469\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023801469\" data-type=\"problem\">\n<p id=\"fs-id1167023801471\">Use power series to solve [latex]y^{\\prime\\prime}+{x}^{2}y=0[\/latex] with the initial condition [latex]y\\left(0\\right)=a[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=b[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558839\">Hint<\/span><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023890904\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023890911\">The coefficients satisfy [latex]{c}_{0}=a,{c}_{1}=b,{c}_{2}=0,{c}_{3}=0[\/latex], and for [latex]n\\ge 4,n\\left(n - 1\\right){c}_{n}=-c_{n - 4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558849\">Show Solution<\/span><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023801536\" data-type=\"solution\">\n<p id=\"fs-id1167023801538\" style=\"text-align: center;\">[latex]y=a\\left(1-\\frac{{x}^{4}}{3\\cdot 4}+\\frac{{x}^{8}}{3\\cdot 4\\cdot 7\\cdot 8}-\\cdots \\right)+b\\left(x-\\frac{{x}^{5}}{4\\cdot 5}+\\frac{{x}^{9}}{4\\cdot 5\\cdot 8\\cdot 9}-\\cdots \\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167023733638\" data-depth=\"1\">\n<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\n<p id=\"fs-id1167023733643\">Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.<\/p>\n<p id=\"fs-id1167023733649\">One integral that arises often in applications in probability theory is [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex]. Unfortunately, the antiderivative of the integrand [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term &#8220;elementary function&#8221; is not synonymous with noncomplicated function. For example, the function [latex]f\\left(x\\right)=\\sqrt{{x}^{2}-3x}+{e}^{{x}^{3}}-\\sin\\left(5x+4\\right)[\/latex] is an elementary function, although not a particularly simple-looking function. Any integral of the form [latex]\\displaystyle\\int f\\left(x\\right)dx[\/latex] where the antiderivative of [latex]f[\/latex] cannot be written as an elementary function is considered a <strong>nonelementary integral<\/strong>.<\/p>\n<p id=\"fs-id1167023782710\">Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex].<\/p>\n<div id=\"fs-id1167023782746\" data-type=\"example\">\n<div id=\"fs-id1167023782748\" data-type=\"exercise\">\n<div id=\"fs-id1167023782750\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using Taylor Series to Evaluate a Definite Integral<\/h3>\n<div id=\"fs-id1167023782750\" data-type=\"problem\">\n<ol id=\"fs-id1167023782755\" type=\"a\">\n<li>Express [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex] as an infinite series.<\/li>\n<li>Evaluate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex] to within an error of [latex]0.01[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558829\">Show Solution<\/span><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023782839\" data-type=\"solution\">\n<ol id=\"fs-id1167023782841\" type=\"a\">\n<li>The Maclaurin series for [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023782866\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}^{\\text{-}{x}^{2}}& ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(\\text{-}{x}^{2}\\right)}^{n}}{n\\text{!}}\\hfill \\\\ & =1-{x}^{2}+\\frac{{x}^{4}}{2\\text{!}}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023760144\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx}& ={\\displaystyle\\int \\left(1-{x}^{2}+\\frac{{x}^{4}}{2\\text{!}}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{n\\text{!}}+\\cdots \\right)dx}\\hfill \\\\ & =C+x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5.2\\text{!}}-\\frac{{x}^{7}}{7.3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>Using the result from part a. we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023808165\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx=1-\\frac{1}{3}+\\frac{1}{10}-\\frac{1}{42}+\\frac{1}{216}-\\cdots[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe sum of the first four terms is approximately [latex]0.74[\/latex]. By the alternating series test, this estimate is accurate to within an error of less than [latex]\\frac{1}{216}\\approx 0.0046296<0.01[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using Taylor Series to Evaluate a Definite Integral.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6724950&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZwX6LAO_TyA&amp;video_target=tpm-plugin-1zdfx8wc-ZwX6LAO_TyA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.5_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.4.5&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167023808281\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023808285\" data-type=\"exercise\">\n<div id=\"fs-id1167023808287\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023808287\" data-type=\"problem\">\n<p id=\"fs-id1167023810006\">Express [latex]\\displaystyle\\int \\cos\\sqrt{x}dx[\/latex] as an infinite series. Evaluate [latex]{\\displaystyle\\int }_{0}^{1}\\cos\\sqrt{x}dx[\/latex] to within an error of [latex]0.01[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558809\">Hint<\/span><\/p>\n<div id=\"q44558809\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023810154\" data-type=\"commentary\" data-element-type=\"hint\">\n<p>Use the series found in the example: Deriving Maclaurin Series from Known Series.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558819\">Show Solution<\/span><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023810066\" data-type=\"solution\">\n<p id=\"fs-id1167023810069\">[latex]C+\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n\\left(2n - 2\\right)\\text{!}}[\/latex] The definite integral is approximately [latex]0.514[\/latex] to within an error of [latex]0.01[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5950\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5950&theme=oea&iframe_resize_id=ohm5950&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167023810172\">As mentioned above, the integral [latex]\\displaystyle\\int {e}^{\\text{-}{x}^{2}}dx[\/latex] arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean [latex]\\mu[\/latex] and standard deviation [latex]\\sigma[\/latex], then the probability that a randomly chosen value lies between [latex]x=a[\/latex] and [latex]x=b[\/latex] is given by<\/p>\n<div id=\"fs-id1167023806051\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{1}{\\sigma \\sqrt{2\\pi }}{\\displaystyle\\int }_{a}^{b}{e}^\\frac{{\\text{-}{\\left(x-\\mu \\right)}^{2}}{\\left(2{\\sigma }^{2}\\right)}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023806128\">(See Figure 2.)<\/p>\n<figure id=\"CNX_Calc_Figure_10_04_002\"><figcaption><\/figcaption><div style=\"width: 627px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234538\/CNX_Calc_Figure_10_04_003.jpg\" alt=\"This graph is the normal distribution. It is a bell-shaped curve with the highest point above mu on the x-axis. Also, there is a shaded area under the curve above the x-axis. The shaded area is bounded by alpha on the left and b on the right.\" width=\"617\" height=\"272\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. If data values are normally distributed with mean [latex]\\mu [\/latex] and standard deviation [latex]\\sigma [\/latex], the probability that a randomly selected data value is between [latex]a[\/latex] and [latex]b[\/latex] is the area under the curve [latex]y=\\frac{1}{\\sigma\\sqrt{2\\pi}}{e}^{\\frac{-\\left(x-\\mu\\right)^{2}}{\\left(2{\\sigma}^{2}\\right)}}[\/latex] between [latex]x=a[\/latex] and [latex]x=b[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167023782195\">To simplify this integral, we typically let [latex]z=\\frac{x-\\mu }{\\sigma }[\/latex]. This quantity [latex]z[\/latex] is known as the [latex]z[\/latex] score of a data value. With this simplification, integral our previous equation becomes<\/p>\n<div id=\"fs-id1167023782231\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\dfrac{1}{\\sqrt{2\\pi }} {\\displaystyle\\int}_{\\frac{\\left(a-\\mu \\right)}{\\sigma}}^{\\frac{\\left(b-\\mu\\right)}{\\sigma}}{e}^{-z^2\\text{\/}2}dz[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023782313\">In the next example, we show how we can use this integral in calculating probabilities.<\/p>\n<div id=\"fs-id1167023782321\" data-type=\"example\">\n<div id=\"fs-id1167023782323\" data-type=\"exercise\">\n<div id=\"fs-id1167023782325\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using Maclaurin Series to Approximate a Probability<\/h3>\n<div id=\"fs-id1167023782325\" data-type=\"problem\">\n<p id=\"fs-id1167023782327\">Suppose a set of standardized test scores are normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =50[\/latex]. Use the equation before this example and the first six terms in the Maclaurin series for [latex]{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to approximate the probability that a randomly selected test score is between [latex]x=100[\/latex] and [latex]x=200[\/latex]. Use the alternating series test to determine how accurate your approximation is.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44548899\">Show Solution<\/span><\/p>\n<div id=\"q44548899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023757999\" data-type=\"solution\">\n<p id=\"fs-id1167023758001\">Since [latex]\\mu =100,\\sigma =50[\/latex], and we are trying to determine the area under the curve from [latex]a=100[\/latex] to [latex]b=200[\/latex], the integral becomes<\/p>\n<div id=\"fs-id1167023758050\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{0}^{2}{e}^{\\frac{\\text{-}{z}^{2}}{2}}dz[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023758104\">The Maclaurin series for [latex]{e}^{\\text{-}\\frac{{x}^{2}}{2}}[\/latex] is given by<\/p>\n<div id=\"fs-id1167023758126\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}^{\\text{-}{x}^{2}\\text{\/}2}& ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-\\frac{{x}^{2}}{2}\\right)}^{n}}{n\\text{!}}\\hfill \\\\ & =1-\\frac{{x}^{2}}{{2}^{1}\\cdot 1\\text{!}}+\\frac{{x}^{4}}{{2}^{2}\\cdot 2\\text{!}}-\\frac{{x}^{6}}{{2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{{2}^{n}\\cdot n\\text{!}}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2}{}^{n}}{{2}^{n}\\cdot n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024047711\">Therefore,<\/p>\n<div id=\"fs-id1167024047714\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int {e}^{\\text{-}{z}^{2}\\text{\/}2}dz}& =\\hfill & \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int \\left(1-\\frac{{z}^{2}}{{2}^{1}\\cdot 1\\text{!}}+\\frac{{z}^{4}}{{2}^{2}\\cdot 2\\text{!}}-\\frac{{z}^{6}}{{2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{z}^{2n}}{{2}^{n}\\cdot n\\text{!}}+\\cdots \\right)dz}\\hfill \\\\ & =\\hfill & \\frac{1}{\\sqrt{2\\pi }}\\left(C+z-\\frac{{z}^{3}}{3\\cdot {2}^{1}\\cdot 1\\text{!}}+\\frac{{z}^{5}}{5\\cdot {2}^{2}\\cdot 2\\text{!}}-\\frac{{z}^{7}}{7\\cdot {2}^{3}\\cdot 3\\text{!}}+\\cdots +{\\left(-1\\right)}^{n}\\frac{{z}^{2n+1}}{\\left(2n+1\\right){2}^{n}\\cdot n\\text{!}}+\\cdots \\right)\\hfill \\\\ \\hfill \\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int _{0}^{2}{e}^{\\text{-}{z}^{2}\\text{\/}2}dz}& =\\hfill & \\frac{1}{\\sqrt{2\\pi }}\\left(2-\\frac{8}{6}+\\frac{32}{40}-\\frac{128}{336}+\\frac{512}{3456}-\\frac{{2}^{11}}{11\\cdot {2}^{5}\\cdot 5\\text{!}}+\\cdots \\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023790566\">Using the first five terms, we estimate that the probability is approximately [latex]0.4922[\/latex]. By the alternating series test, we see that this estimate is accurate to within<\/p>\n<div id=\"fs-id1167023790577\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{2\\pi }}\\frac{{2}^{13}}{13\\cdot {2}^{6}\\cdot 6\\text{!}}\\approx 0.00546[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023790628\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1167023790633\">If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately [latex]95\\%[\/latex]. Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around [latex]47.5\\text{%}[\/latex]. The estimate, combined with the bound on the accuracy, falls within this range.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023790662\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023790666\" data-type=\"exercise\">\n<div id=\"fs-id1167023790668\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023790668\" data-type=\"problem\">\n<p id=\"fs-id1167023790670\">Use the first five terms of the Maclaurin series for [latex]{e}^{\\text{-}\\frac{{x}^{2}}{2}}[\/latex] to estimate the probability that a randomly selected test score is between [latex]100[\/latex] and [latex]150[\/latex]. Use the alternating series test to determine the accuracy of this estimate.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558699\">Hint<\/span><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023790727\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023790735\">Evaluate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}\\frac{{z}^{2}}{2}}dz[\/latex] using the first five terms of the Maclaurin series for [latex]{e}^{\\text{-}\\frac{{z}^{2}}{2}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023790706\" data-type=\"solution\">\n<p id=\"fs-id1167023790708\">The estimate is approximately [latex]0.3414[\/latex]. This estimate is accurate to within [latex]0.0000094[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167024040461\">Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is<\/p>\n<div id=\"fs-id1167024040466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta }{\\sqrt{1-{k}^{2}{\\sin}^{2}\\theta }}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024040523\">An integral of this form is known as an <span class=\"no-emphasis\" data-type=\"term\">elliptic integral<\/span> of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.<\/p>\n<div id=\"fs-id1167024040533\" data-type=\"example\">\n<div id=\"fs-id1167024040535\" data-type=\"exercise\">\n<div id=\"fs-id1167024040537\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Period of a Pendulum<\/h3>\n<div id=\"fs-id1167024040537\" data-type=\"problem\">\n<p id=\"fs-id1167024040542\">The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length [latex]L[\/latex] that makes a maximum angle [latex]{\\theta }_{\\text{max}}[\/latex] with the vertical, its period [latex]T[\/latex] is given by<\/p>\n<div id=\"fs-id1167024040565\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta }{\\sqrt{1-{k}^{2}{\\sin}^{2}\\theta }}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023766290\">where [latex]g[\/latex] is the acceleration due to gravity and [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex] (see Figure 3). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and [latex]\\sin\\theta[\/latex] is approximated by [latex]\\theta[\/latex].) Use the binomial series<\/p>\n<div id=\"fs-id1167023766351\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\sqrt{1+x}}=1+\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023766456\">to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if<\/p>\n<ol id=\"fs-id1167023766461\" type=\"a\">\n<li>you use only the first term in the binomial series, and<\/li>\n<li>you use the first two terms in the binomial series.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_04_003\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234541\/CNX_Calc_Figure_10_04_002.jpg\" alt=\"This figure is a pendulum. There are three positions of the pendulum shown. When the pendulum is to the far left, it is labeled negative theta max. When the pendulum is in the middle and vertical, it is labeled equilibrium position. When the pendulum is to the far right it is labeled theta max. Also, theta is the angle from equilibrium to the far right position. The length of the pendulum is labeled L.\" width=\"487\" height=\"435\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. This pendulum has length [latex]L[\/latex] and makes a maximum angle [latex]{\\theta }_{\\text{max}}[\/latex] with the vertical.<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44553899\">Show Solution<\/span><\/p>\n<div id=\"q44553899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023871808\" data-type=\"solution\">\n<p id=\"fs-id1167023871810\">We use the binomial series, replacing [latex]x[\/latex] with [latex]\\text{-}{k}^{2}{\\sin}^{2}\\theta[\/latex]. Then we can write the period as<\/p>\n<div id=\"fs-id1167023871840\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}{\\sin}^{4}\\theta +\\cdots \\right)d\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<ol id=\"fs-id1167023871957\" type=\"a\">\n<li>Using just the first term in the integrand, the first-order estimate is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023871967\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\approx 4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}d\\theta =2\\pi \\sqrt{\\frac{L}{g}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIf [latex]{\\theta }_{\\text{max}}[\/latex] is small, then [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex] is small. We claim that when [latex]k[\/latex] is small, this is a good estimate. To justify this claim, consider<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023794977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}{\\sin}^{4}\\theta +\\cdots \\right)d\\theta[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]|\\sin{x}|\\le 1[\/latex], this integral is bounded by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023795106\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(\\frac{1}{2}{k}^{2}+\\frac{1.3}{2\\text{!}{2}^{2}}{k}^{4}+\\cdots \\right)d\\theta <\\frac{\\pi }{2}\\left(\\frac{1}{2}{k}^{2}+\\frac{1\\cdot 3}{2\\text{!}{2}^{2}}{k}^{4}+\\cdots \\right)[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFurthermore, it can be shown that each coefficient on the right-hand side is less than [latex]1[\/latex] and, therefore, that this expression is bounded by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023835770\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\pi {k}^{2}}{2}\\left(1+{k}^{2}+{k}^{4}+\\cdots \\right)=\\frac{\\pi {k}^{2}}{2}\\cdot \\frac{1}{1-{k}^{2}}[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhich is small for [latex]k[\/latex] small.<\/li>\n<li>For larger values of [latex]{\\theta }_{\\text{max}}[\/latex], we can approximate [latex]T[\/latex] by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023785853\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill T& \\approx 4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta \\right)d\\theta \\hfill \\\\ & =2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023785982\">The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.<\/p>\n<\/section>\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-1816\">\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>6.4.4. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>6.4.5. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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":17533,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"6.4.4\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"6.4.5\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"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-1816","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1816","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1816\/revisions"}],"predecessor-version":[{"id":2696,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1816\/revisions\/2696"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/161"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1816\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1816"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1816"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1816"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1816"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}