{"id":962,"date":"2025-06-20T17:25:12","date_gmt":"2025-06-20T17:25:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=962"},"modified":"2025-09-10T13:26:58","modified_gmt":"2025-09-10T13:26:58","slug":"applications-of-series-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-series-fresh-take\/","title":{"raw":"Applications of Series: Fresh Take","rendered":"Applications of Series: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write out the terms in a binomial series<\/li>\r\n \t<li>Find the Taylor series for different functions<\/li>\r\n \t<li>Use Taylor series to solve differential equations<\/li>\r\n \t<li>Use Taylor series to evaluate integrals that don't have elementary antiderivatives<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">The Binomial Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The binomial series generalizes the familiar binomial theorem to handle any real exponent [latex]r[\/latex], not just positive integers. This powerful tool lets you find power series for functions like [latex]\\sqrt{1+x}[\/latex], [latex]\\frac{1}{1+x}[\/latex], and [latex](1+x)^{-2}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The generalized binomial coefficient:<\/strong> For any real number [latex]r[\/latex] and non-negative integer [latex]n[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex](\\begin{array}{c}r \\ n\\end{array}) = \\frac{r(r-1)(r-2)\\cdots(r-n+1)}{n!}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The binomial series formula:<\/strong> [latex](1+x)^r = \\sum_{n=0}^{\\infty} (\\begin{array}{c}r \\ n\\end{array}) x^n = 1 + rx + \\frac{r(r-1)}{2!}x^2 + \\frac{r(r-1)(r-2)}{3!}x^3 + \\cdots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">When [latex]r[\/latex] is a non-negative integer, the series terminates naturally (since higher-order derivatives become zero), giving you the familiar finite binomial expansion. When [latex]r[\/latex] is any other real number, you get an infinite series.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Interval of convergence:<\/strong> The binomial series converges for [latex]|x| &lt; 1[\/latex]. Endpoint behavior depends on [latex]r[\/latex]:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]r \\geq 0[\/latex]: converges at both [latex]x = \\pm 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-1 &lt; r &lt; 0[\/latex]: converges at [latex]x = 1[\/latex], diverges at [latex]x = -1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]r &lt; -1[\/latex]: diverges at both endpoints<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Common applications:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{1+x} = (1+x)^{1\/2}[\/latex] with [latex]r = \\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\sqrt{1+x}} = (1+x)^{-1\/2}[\/latex] with [latex]r = -\\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{1+x} = (1+x)^{-1}[\/latex] with [latex]r = -1[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The binomial series is particularly valuable for approximating roots and reciprocals when [latex]x[\/latex] is small. You can use just the first few terms to get good approximations, with Taylor's theorem helping you bound the error.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\r\n<p id=\"fs-id1167024040781\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{2}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167024044103\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024044109\">Use the definition of binomial series for [latex]r=-2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167024044048\" data-type=\"solution\">\r\n<p id=\"fs-id1167024044050\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Common Functions Expressed as Taylor Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Instead of deriving Taylor series from scratch every time, you can build a powerful toolkit of standard series and use them to find series for more complex functions. This reference approach makes Taylor series much more practical and efficient.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Your essential reference table:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{1-x} = \\sum_{n=0}^{\\infty} x^n[\/latex] for [latex]|x| &lt; 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}[\/latex] for all [latex]x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sin x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n+1}}{(2n+1)!}[\/latex] for all [latex]x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n}}{(2n)!}[\/latex] for all [latex]x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\ln(1+x) = \\sum_{n=1}^{\\infty} (-1)^{n+1} \\frac{x^n}{n}[\/latex] for [latex]-1 &lt; x \\leq 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\tan^{-1} x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n+1}}{2n+1}[\/latex] for [latex]|x| \\leq 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex](1+x)^r = \\sum_{n=0}^{\\infty} \\binom{r}{n} x^n[\/latex] for [latex]|x| &lt; 1[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Building new series from old ones:<\/strong> Use substitution, differentiation, integration, and algebraic manipulation:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">For [latex]\\cos \\sqrt{x}[\/latex]: substitute [latex]\\sqrt{x}[\/latex] into the cosine series<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For [latex]\\sinh x = \\frac{e^x - e^{-x}}{2}[\/latex]: combine exponential series<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For [latex]\\frac{1}{\\sqrt{1+x}}[\/latex]: differentiate the series for [latex]\\sqrt{1+x}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Key patterns to recognize:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Exponential-related functions<\/strong>: Use [latex]e^x[\/latex] series with appropriate substitutions<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Trigonometric variants<\/strong>: Modify [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] series<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Rational functions<\/strong>: Often relate to geometric series [latex]\\frac{1}{1-x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Root and power functions<\/strong>: Use binomial series <a class=\"underline\" href=\"1+x\">latex<\/a>^r[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Instead of computing derivatives repeatedly, ask \"Which standard series can I modify to get this function?\" This approach is faster, less error-prone, and builds your intuition about how different functions relate to each other.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\r\n<p id=\"fs-id1167024047404\">Find the Maclaurin series for [latex]\\sin\\left({x}^{2}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167024041386\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024041393\">Use the Maclaurin series for [latex]\\sin{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167024041315\" data-type=\"solution\">\r\n<p id=\"fs-id1167024041317\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\r\n<p id=\"fs-id1167023914394\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{\\frac{3}{2}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1167024036470\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024036478\">Differentiate the series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1167023914438\" data-type=\"solution\">\r\n<p id=\"fs-id1167023914440\" style=\"text-align: center;\">[latex]\\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]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Solving Differential Equations with Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Power series offer a systematic approach to solving differential equations that can't be handled by separation of variables, integrating factors, or other elementary techniques. The method transforms a differential equation into a problem about finding coefficients in a power series.<\/p>\r\n<p class=\"whitespace-normal break-words\">Assume your solution has the form [latex]y = \\sum_{n=0}^{\\infty} c_n x^n[\/latex], then use the differential equation to determine what the coefficients [latex]c_n[\/latex] must be.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Assume a power series solution:<\/strong> [latex]y = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \\cdots[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Differentiate term-by-term:<\/strong> [latex]y' = c_1 + 2c_2 x + 3c_3 x^2 + \\cdots[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Substitute into the differential equation:<\/strong> Replace [latex]y[\/latex], [latex]y'[\/latex], etc. with their series<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Match coefficients:<\/strong> Use the uniqueness of power series to equate coefficients of like powers<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply initial conditions:<\/strong> Use [latex]y(0)[\/latex], [latex]y'(0)[\/latex], etc. to find specific coefficient values<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">The uniqueness of power series representations means that if two power series are equal, their corresponding coefficients must be equal. This gives you a system of equations to solve for the coefficients.<\/p>\r\n<p class=\"whitespace-normal break-words\">Often you'll find that [latex]c_{n+k} = f(c_n)[\/latex] for some function [latex]f[\/latex]. This creates patterns where coefficients can be expressed in terms of just a few initial coefficients.<\/p>\r\n<p class=\"whitespace-normal break-words\">Power series solutions are especially valuable for:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Linear differential equations with polynomial coefficients<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Equations where elementary methods fail (like Airy's equation [latex]y'' - xy = 0[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Problems requiring solutions near specific points<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Sometimes your coefficient pattern will match a known Taylor series (like [latex]e^x[\/latex]), giving you a closed-form solution. Other times, the power series itself is the most useful form of the solution.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Some integrals simply can't be solved using the integration techniques you've learned. Functions like [latex]e^{-x^2}[\/latex], [latex]\\frac{\\sin x}{x}[\/latex], and [latex]\\frac{1}{\\sqrt{1-k^2\\sin^2\\theta}}[\/latex] have antiderivatives that aren't expressible as elementary functions. Power series provide the key to unlocking these \"impossible\" integrals.<\/p>\r\n<p class=\"whitespace-normal break-words\">When you can't integrate directly, represent the integrand as a power series, then integrate term by term. This transforms an unsolvable integral into a manageable infinite series.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Express the integrand as a power series<\/strong> using known series or substitutions<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate term by term<\/strong> - power series can be integrated just like polynomials<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate definite integrals<\/strong> by substituting limits into your series<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Estimate accuracy<\/strong> using alternating series test or other convergence tools<\/li>\r\n<\/ol>\r\nPower series don't just solve differential equations - they extend your integration capabilities to a vast class of functions that would otherwise be inaccessible. When traditional integration fails, power series succeed by transforming the problem into polynomial arithmetic.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write out the terms in a binomial series<\/li>\n<li>Find the Taylor series for different functions<\/li>\n<li>Use Taylor series to solve differential equations<\/li>\n<li>Use Taylor series to evaluate integrals that don&#8217;t have elementary antiderivatives<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">The Binomial Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The binomial series generalizes the familiar binomial theorem to handle any real exponent [latex]r[\/latex], not just positive integers. This powerful tool lets you find power series for functions like [latex]\\sqrt{1+x}[\/latex], [latex]\\frac{1}{1+x}[\/latex], and [latex](1+x)^{-2}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The generalized binomial coefficient:<\/strong> For any real number [latex]r[\/latex] and non-negative integer [latex]n[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex](\\begin{array}{c}r \\ n\\end{array}) = \\frac{r(r-1)(r-2)\\cdots(r-n+1)}{n!}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The binomial series formula:<\/strong> [latex](1+x)^r = \\sum_{n=0}^{\\infty} (\\begin{array}{c}r \\ n\\end{array}) x^n = 1 + rx + \\frac{r(r-1)}{2!}x^2 + \\frac{r(r-1)(r-2)}{3!}x^3 + \\cdots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">When [latex]r[\/latex] is a non-negative integer, the series terminates naturally (since higher-order derivatives become zero), giving you the familiar finite binomial expansion. When [latex]r[\/latex] is any other real number, you get an infinite series.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Interval of convergence:<\/strong> The binomial series converges for [latex]|x| < 1[\/latex]. Endpoint behavior depends on [latex]r[\/latex]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]r \\geq 0[\/latex]: converges at both [latex]x = \\pm 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-1 < r < 0[\/latex]: converges at [latex]x = 1[\/latex], diverges at [latex]x = -1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]r < -1[\/latex]: diverges at both endpoints<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Common applications:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{1+x} = (1+x)^{1\/2}[\/latex] with [latex]r = \\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\sqrt{1+x}} = (1+x)^{-1\/2}[\/latex] with [latex]r = -\\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{1+x} = (1+x)^{-1}[\/latex] with [latex]r = -1[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The binomial series is particularly valuable for approximating roots and reciprocals when [latex]x[\/latex] is small. You can use just the first few terms to get good approximations, with Taylor&#8217;s theorem helping you bound the error.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\n<p id=\"fs-id1167024040781\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{2}}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024044103\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024044109\">Use the definition of binomial series for [latex]r=-2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024044048\" data-type=\"solution\">\n<p id=\"fs-id1167024044050\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Common Functions Expressed as Taylor Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Instead of deriving Taylor series from scratch every time, you can build a powerful toolkit of standard series and use them to find series for more complex functions. This reference approach makes Taylor series much more practical and efficient.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Your essential reference table:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{1-x} = \\sum_{n=0}^{\\infty} x^n[\/latex] for [latex]|x| < 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}[\/latex] for all [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sin x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n+1}}{(2n+1)!}[\/latex] for all [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n}}{(2n)!}[\/latex] for all [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\ln(1+x) = \\sum_{n=1}^{\\infty} (-1)^{n+1} \\frac{x^n}{n}[\/latex] for [latex]-1 < x \\leq 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\tan^{-1} x = \\sum_{n=0}^{\\infty} (-1)^n \\frac{x^{2n+1}}{2n+1}[\/latex] for [latex]|x| \\leq 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex](1+x)^r = \\sum_{n=0}^{\\infty} \\binom{r}{n} x^n[\/latex] for [latex]|x| < 1[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Building new series from old ones:<\/strong> Use substitution, differentiation, integration, and algebraic manipulation:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">For [latex]\\cos \\sqrt{x}[\/latex]: substitute [latex]\\sqrt{x}[\/latex] into the cosine series<\/li>\n<li class=\"whitespace-normal break-words\">For [latex]\\sinh x = \\frac{e^x - e^{-x}}{2}[\/latex]: combine exponential series<\/li>\n<li class=\"whitespace-normal break-words\">For [latex]\\frac{1}{\\sqrt{1+x}}[\/latex]: differentiate the series for [latex]\\sqrt{1+x}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Key patterns to recognize:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Exponential-related functions<\/strong>: Use [latex]e^x[\/latex] series with appropriate substitutions<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Trigonometric variants<\/strong>: Modify [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] series<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Rational functions<\/strong>: Often relate to geometric series [latex]\\frac{1}{1-x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Root and power functions<\/strong>: Use binomial series <a class=\"underline\" href=\"1+x\">latex<\/a>^r[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Instead of computing derivatives repeatedly, ask &#8220;Which standard series can I modify to get this function?&#8221; This approach is faster, less error-prone, and builds your intuition about how different functions relate to each other.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\n<p id=\"fs-id1167024047404\">Find the Maclaurin series for [latex]\\sin\\left({x}^{2}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024041386\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024041393\">Use the Maclaurin series for [latex]\\sin{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024041315\" data-type=\"solution\">\n<p id=\"fs-id1167024041317\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\n<p id=\"fs-id1167023914394\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{\\frac{3}{2}}}[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Hint<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024036470\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024036478\">Differentiate the series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023914438\" data-type=\"solution\">\n<p id=\"fs-id1167023914440\" style=\"text-align: center;\">[latex]\\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]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Solving Differential Equations with Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Power series offer a systematic approach to solving differential equations that can&#8217;t be handled by separation of variables, integrating factors, or other elementary techniques. The method transforms a differential equation into a problem about finding coefficients in a power series.<\/p>\n<p class=\"whitespace-normal break-words\">Assume your solution has the form [latex]y = \\sum_{n=0}^{\\infty} c_n x^n[\/latex], then use the differential equation to determine what the coefficients [latex]c_n[\/latex] must be.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Assume a power series solution:<\/strong> [latex]y = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \\cdots[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Differentiate term-by-term:<\/strong> [latex]y' = c_1 + 2c_2 x + 3c_3 x^2 + \\cdots[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Substitute into the differential equation:<\/strong> Replace [latex]y[\/latex], [latex]y'[\/latex], etc. with their series<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Match coefficients:<\/strong> Use the uniqueness of power series to equate coefficients of like powers<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply initial conditions:<\/strong> Use [latex]y(0)[\/latex], [latex]y'(0)[\/latex], etc. to find specific coefficient values<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">The uniqueness of power series representations means that if two power series are equal, their corresponding coefficients must be equal. This gives you a system of equations to solve for the coefficients.<\/p>\n<p class=\"whitespace-normal break-words\">Often you&#8217;ll find that [latex]c_{n+k} = f(c_n)[\/latex] for some function [latex]f[\/latex]. This creates patterns where coefficients can be expressed in terms of just a few initial coefficients.<\/p>\n<p class=\"whitespace-normal break-words\">Power series solutions are especially valuable for:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Linear differential equations with polynomial coefficients<\/li>\n<li class=\"whitespace-normal break-words\">Equations where elementary methods fail (like Airy&#8217;s equation [latex]y'' - xy = 0[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Problems requiring solutions near specific points<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Sometimes your coefficient pattern will match a known Taylor series (like [latex]e^x[\/latex]), giving you a closed-form solution. Other times, the power series itself is the most useful form of the solution.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558839\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Evaluating Nonelementary Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Some integrals simply can&#8217;t be solved using the integration techniques you&#8217;ve learned. Functions like [latex]e^{-x^2}[\/latex], [latex]\\frac{\\sin x}{x}[\/latex], and [latex]\\frac{1}{\\sqrt{1-k^2\\sin^2\\theta}}[\/latex] have antiderivatives that aren&#8217;t expressible as elementary functions. Power series provide the key to unlocking these &#8220;impossible&#8221; integrals.<\/p>\n<p class=\"whitespace-normal break-words\">When you can&#8217;t integrate directly, represent the integrand as a power series, then integrate term by term. This transforms an unsolvable integral into a manageable infinite series.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Express the integrand as a power series<\/strong> using known series or substitutions<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate term by term<\/strong> &#8211; power series can be integrated just like polynomials<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate definite integrals<\/strong> by substituting limits into your series<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Estimate accuracy<\/strong> using alternating series test or other convergence tools<\/li>\n<\/ol>\n<p>Power series don&#8217;t just solve differential equations &#8211; they extend your integration capabilities to a vast class of functions that would otherwise be inaccessible. When traditional integration fails, power series succeed by transforming the problem into polynomial arithmetic.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558809\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558819\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558699\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558799\">Show Solution<\/button><\/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<\/section>\n","protected":false},"author":15,"menu_order":28,"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\/962"}],"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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/962\/revisions"}],"predecessor-version":[{"id":2281,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/962\/revisions\/2281"}],"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\/962\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=962"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=962"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=962"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=962"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}