{"id":960,"date":"2025-06-20T17:25:06","date_gmt":"2025-06-20T17:25:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=960"},"modified":"2025-09-12T14:38:02","modified_gmt":"2025-09-12T14:38:02","slug":"applications-of-series-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-series-learn-it-4\/","title":{"raw":"Applications of Series: Learn It 4","rendered":"Applications of Series: Learn It 4"},"content":{"raw":"<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\r\n<p id=\"fs-id1167023733643\">Power series offer a powerful solution when you encounter integrals that can't be solved with standard techniques. You've already seen how power series help solve differential equations\u2014now let's explore their second major application: evaluating integrals involving functions whose antiderivatives don't exist in elementary form.<\/p>\r\nConsider the integral [latex]\\displaystyle\\int e^{-x^2}dx[\/latex], which appears frequently in probability theory and statistics. You might try various integration techniques, but you'll quickly discover that none work. The reason? The antiderivative of [latex]e^{-x^2}[\/latex] cannot be expressed as an elementary function - a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The term \"elementary\" doesn't mean \"simple\"\u2014for instance, [latex]f(x) = \\sqrt{x^2-3x} + e^{x^3} - \\sin(5x+4)[\/latex] is elementary despite its complexity.<\/section>When the antiderivative of a function isn't elementary, we classify the integral as <strong>nonelementary<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>nonelementary integrals<\/h3>\r\nAny integral [latex]\\displaystyle\\int f(x)dx[\/latex] where the antiderivative of [latex]f[\/latex] cannot be written as an elementary function. These integrals require special techniques beyond basic integration methods.\r\n\r\n<\/section>When faced with nonelementary integrals, you can use this strategy: express the integrand as a power series, then integrate term by term. This approach transforms an impossible integral into a manageable infinite series.\r\n<p class=\"whitespace-normal break-words\">We'll demonstrate this technique using [latex]\\displaystyle\\int e^{-x^2}dx[\/latex] as our key example.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<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\" data-mce-fragment=\"1\"><\/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>.<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311968[\/ohm_question]<\/section>\r\n<p class=\"whitespace-normal break-words\">The integral [latex]\\displaystyle\\int e^{-x^2}dx[\/latex] isn't just a mathematical curiosity\u2014it plays a crucial role in analyzing real-world data. You'll encounter this integral when working with normally distributed data sets, where values follow the familiar bell-shaped curve pattern.<\/p>\r\n<p class=\"whitespace-normal break-words\">When data is normally distributed with mean [latex]\\mu[\/latex] and standard deviation [latex]\\sigma[\/latex], you can find the probability that a randomly chosen value falls between [latex]x = a[\/latex] and [latex]x = b[\/latex] using:<\/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<p id=\"fs-id1167023806128\">This represents the area under the normal curve between the two boundary values. (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 calculations, statisticians use the substitution [latex]z = \\frac{x-\\mu}{\\sigma}[\/latex]. This quantity [latex]z[\/latex] is called the z-score of a data value\u2014it measures how many standard deviations away from the mean a particular value lies.<\/p>\r\n<p class=\"whitespace-normal break-words\">With this substitution, our probability integral becomes the more manageable form:<\/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<p id=\"fs-id1167023782313\">This standardized version allows us to work with universal probability tables and makes calculations more efficient. In the next example, we'll show how to use this integral to calculate actual probabilities.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\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&nbsp;\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[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<p id=\"fs-id1167024040461\">Nonelementary integrals appear in many areas of physics and engineering. One important example comes from analyzing pendulum motion, where the period involves the integral:<\/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<div data-type=\"equation\" data-label=\"\">This integral cannot be evaluated using standard techniques. An integral of this form is known as an elliptic integral of the first kind. These integrals originally arose when mathematicians attempted to calculate the arc length of an ellipse, which explains their name. Elliptic integrals appear throughout physics and engineering\u2014from pendulum periods to planetary motion calculations. Since we can't solve them analytically, power series provide an essential tool for obtaining numerical approximations.<\/div>\r\n<p class=\"whitespace-normal break-words\">We'll now demonstrate how to use power series techniques to approximate this integral and make it useful for practical calculations.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>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.","rendered":"<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\n<p id=\"fs-id1167023733643\">Power series offer a powerful solution when you encounter integrals that can&#8217;t be solved with standard techniques. You&#8217;ve already seen how power series help solve differential equations\u2014now let&#8217;s explore their second major application: evaluating integrals involving functions whose antiderivatives don&#8217;t exist in elementary form.<\/p>\n<p>Consider the integral [latex]\\displaystyle\\int e^{-x^2}dx[\/latex], which appears frequently in probability theory and statistics. You might try various integration techniques, but you&#8217;ll quickly discover that none work. The reason? The antiderivative of [latex]e^{-x^2}[\/latex] cannot be expressed as an elementary function &#8211; a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The term &#8220;elementary&#8221; doesn&#8217;t mean &#8220;simple&#8221;\u2014for instance, [latex]f(x) = \\sqrt{x^2-3x} + e^{x^3} - \\sin(5x+4)[\/latex] is elementary despite its complexity.<\/section>\n<p>When the antiderivative of a function isn&#8217;t elementary, we classify the integral as <strong>nonelementary<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>nonelementary integrals<\/h3>\n<p>Any integral [latex]\\displaystyle\\int f(x)dx[\/latex] where the antiderivative of [latex]f[\/latex] cannot be written as an elementary function. These integrals require special techniques beyond basic integration methods.<\/p>\n<\/section>\n<p>When faced with nonelementary integrals, you can use this strategy: express the integrand as a power series, then integrate term by term. This approach transforms an impossible integral into a manageable infinite series.<\/p>\n<p class=\"whitespace-normal break-words\">We&#8217;ll demonstrate this technique using [latex]\\displaystyle\\int e^{-x^2}dx[\/latex] as our key example.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558829\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<\/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\" data-mce-fragment=\"1\"><\/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>.<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311968\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311968&theme=lumen&iframe_resize_id=ohm311968&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p class=\"whitespace-normal break-words\">The integral [latex]\\displaystyle\\int e^{-x^2}dx[\/latex] isn&#8217;t just a mathematical curiosity\u2014it plays a crucial role in analyzing real-world data. You&#8217;ll encounter this integral when working with normally distributed data sets, where values follow the familiar bell-shaped curve pattern.<\/p>\n<p class=\"whitespace-normal break-words\">When data is normally distributed with mean [latex]\\mu[\/latex] and standard deviation [latex]\\sigma[\/latex], you can find the probability that a randomly chosen value falls between [latex]x = a[\/latex] and [latex]x = b[\/latex] using:<\/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 id=\"fs-id1167023806128\">This represents the area under the normal curve between the two boundary values. (See Figure 2.)<\/p>\n<figure id=\"CNX_Calc_Figure_10_04_002\"><figcaption><\/figcaption><figure style=\"width: 617px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1167023782195\">To simplify calculations, statisticians use the substitution [latex]z = \\frac{x-\\mu}{\\sigma}[\/latex]. This quantity [latex]z[\/latex] is called the z-score of a data value\u2014it measures how many standard deviations away from the mean a particular value lies.<\/p>\n<p class=\"whitespace-normal break-words\">With this substitution, our probability integral becomes the more manageable form:<\/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 id=\"fs-id1167023782313\">This standardized version allows us to work with universal probability tables and makes calculations more efficient. In the next example, we&#8217;ll show how to use this integral to calculate actual probabilities.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44548899\">Show Solution<\/button><\/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<p>&nbsp;<\/p>\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<\/section>\n<p id=\"fs-id1167024040461\">Nonelementary integrals appear in many areas of physics and engineering. One important example comes from analyzing pendulum motion, where the period involves the integral:<\/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<div data-type=\"equation\" data-label=\"\">This integral cannot be evaluated using standard techniques. An integral of this form is known as an elliptic integral of the first kind. These integrals originally arose when mathematicians attempted to calculate the arc length of an ellipse, which explains their name. Elliptic integrals appear throughout physics and engineering\u2014from pendulum periods to planetary motion calculations. Since we can&#8217;t solve them analytically, power series provide an essential tool for obtaining numerical approximations.<\/div>\n<p class=\"whitespace-normal break-words\">We&#8217;ll now demonstrate how to use power series techniques to approximate this integral and make it useful for practical calculations.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\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<figure style=\"width: 487px\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44553899\">Show Solution<\/button><\/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<\/section>\n<p>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","protected":false},"author":15,"menu_order":26,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/960"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/960\/revisions"}],"predecessor-version":[{"id":2333,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/960\/revisions\/2333"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/960\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=960"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=960"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=960"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=960"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}