{"id":1783,"date":"2021-07-27T00:30:30","date_gmt":"2021-07-27T00:30:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1783"},"modified":"2022-03-21T23:13:27","modified_gmt":"2022-03-21T23:13:27","slug":"taylors-theorem-with-remainder","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/taylors-theorem-with-remainder\/","title":{"raw":"Taylor\u2019s Theorem with Remainder and Convergence","rendered":"Taylor\u2019s Theorem with Remainder and Convergence"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the meaning and significance of Taylor\u2019s theorem with remainder<\/li>\r\n \t<li>Estimate the remainder for a Taylor series approximation of a given function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Taylor\u2019s Theorem with Remainder<\/h2>\r\n<p id=\"fs-id1167025102169\">Recall that the [latex]n[\/latex]th Taylor polynomial for a function [latex]f[\/latex] at [latex]a[\/latex]\u00a0is the [latex]n[\/latex]th partial sum of the Taylor series for [latex]f[\/latex] at [latex]a[\/latex]. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials [latex]\\left\\{{p}_{n}\\right\\}[\/latex] converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to [latex]f[\/latex]. To answer this question, we define the remainder [latex]{R}_{n}\\left(x\\right)[\/latex] as<\/p>\r\n\r\n<div id=\"fs-id1167025162616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025162665\">For the sequence of Taylor polynomials to converge to [latex]f[\/latex], we need the remainder [latex]R_{n}[\/latex] to converge to zero. To determine if [latex]R_{n}[\/latex]\u00a0converges to zero, we introduce <strong>Taylor\u2019s theorem with remainder<\/strong>. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the [latex]n[\/latex]th Taylor polynomial approximates the function.<\/p>\r\n<p id=\"fs-id1167025168160\">Here we look for a bound on [latex]|{R}_{n}|[\/latex]. Consider the simplest case: [latex]n=0[\/latex]. Let [latex]p_{0}[\/latex]\u00a0be the 0th Taylor polynomial at [latex]a[\/latex]\u00a0for a function [latex]f[\/latex]. The remainder [latex]R_{0}[\/latex]\u00a0satisfies<\/p>\r\n\r\n<div id=\"fs-id1167025168217\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {R}_{0}\\left(x\\right)&amp; =f\\left(x\\right)-{p}_{0}\\left(x\\right)\\hfill \\\\ &amp; =f\\left(x\\right)-f\\left(a\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025168302\">If [latex]f[\/latex] is differentiable on an interval [latex]I[\/latex]\u00a0containing [latex]a[\/latex] and [latex]x[\/latex], then by the Mean Value Theorem there exists a real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]\u00a0such that [latex]f\\left(x\\right)-f\\left(a\\right)={f}^{\\prime }\\left(c\\right)\\left(x-a\\right)[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167024997799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{0}\\left(x\\right)={f}^{\\prime }\\left(c\\right)\\left(x-a\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024997849\">Using the Mean Value Theorem in a similar argument, we can show that if [latex]f[\/latex] is [latex]n[\/latex]\u00a0times differentiable on an interval [latex]I[\/latex] containing [latex]a[\/latex]\u00a0and [latex]x[\/latex], then the [latex]n[\/latex]th remainder [latex]{R}_{n}[\/latex]\u00a0satisfies<\/p>\r\n\r\n<div id=\"fs-id1167024875234\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex]<\/div>\r\n<div data-type=\"equation\" data-label=\"\"><\/div>\r\n<p id=\"fs-id1167024875321\">for some real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]. It is important to note that the value [latex]c[\/latex]\u00a0in the numerator above is not the center [latex]a[\/latex], but rather an unknown value [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]. This formula allows us to get a bound on the remainder [latex]{R}_{n}[\/latex]. If we happen to know that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|[\/latex] is bounded by some real number [latex]M[\/latex]\u00a0on this interval [latex]I[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1167025236141\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025236211\">for all [latex]x[\/latex]\u00a0in the interval [latex]I[\/latex].<\/p>\r\n<p id=\"fs-id1167025236225\">We now state Taylor\u2019s theorem, which provides the formal relationship between a function [latex]f[\/latex] and its [latex]n[\/latex]th degree Taylor polynomial [latex]{p}_{n}\\left(x\\right)[\/latex]. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for [latex]f[\/latex] converges to [latex]f[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167025239317\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Taylor\u2019s Theorem with Remainder<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167025239324\">Let [latex]f[\/latex] be a function that can be differentiated [latex]n+1[\/latex] times on an interval [latex]I[\/latex] containing the real number [latex]a[\/latex]. Let [latex]p_{n}[\/latex]\u00a0be the [latex]n[\/latex]th Taylor polynomial of [latex]f[\/latex] at [latex]a[\/latex]\u00a0and let<\/p>\r\n\r\n<div id=\"fs-id1167025239373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025239420\">be the [latex]n[\/latex]th remainder. Then for each [latex]x[\/latex]\u00a0in the interval [latex]I[\/latex], there exists a real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]\u00a0such that<\/p>\r\n\r\n<div id=\"fs-id1167025101301\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025101391\">If there exists a real number [latex]M[\/latex]\u00a0such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all [latex]x\\in I[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1167025070502\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025070572\">for all [latex]x[\/latex]\u00a0in [latex]I[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167025070586\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1167025070592\">Fix a point [latex]x\\in I[\/latex] and introduce the function <em data-effect=\"italics\">g<\/em> such that<\/p>\r\n\r\n<div id=\"fs-id1167025070610\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]g\\left(t\\right)=f\\left(x\\right)-f\\left(t\\right)-{f}^{\\prime }\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}-\\cdots -\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}-{R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n+1}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025234124\">We claim that <em data-effect=\"italics\">g<\/em> satisfies the criteria of Rolle\u2019s theorem. Since <em data-effect=\"italics\">g<\/em> is a polynomial function (in <em data-effect=\"italics\">t<\/em>), it is a differentiable function. Also, <em data-effect=\"italics\">g<\/em> is zero at [latex]t=a[\/latex] and [latex]t=x[\/latex] because<\/p>\r\n\r\n<div id=\"fs-id1167025234167\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill g\\left(a\\right)&amp; =\\hfill &amp; f\\left(x\\right)-f\\left(a\\right)-{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)-\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}-{R}_{n}\\left(x\\right)\\hfill \\\\ &amp; =\\hfill &amp; f\\left(x\\right)-{p}_{n}\\left(x\\right)-{R}_{n}\\left(x\\right)\\hfill \\\\ &amp; =\\hfill &amp; 0,\\hfill \\\\ g\\left(x\\right)\\hfill &amp; =\\hfill &amp; f\\left(x\\right)-f\\left(x\\right)-0-\\cdots -0\\hfill \\\\ &amp; =\\hfill &amp; 0.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025232390\">Therefore, <em data-effect=\"italics\">g<\/em> satisfies Rolle\u2019s theorem, and consequently, there exists <em data-effect=\"italics\">c<\/em> between <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. We now calculate [latex]{g}^{\\prime }[\/latex]. Using the product rule, we note that<\/p>\r\n\r\n<div id=\"fs-id1167025232449\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dt}\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]=\\frac{-{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}+\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025113176\">Consequently,<\/p>\r\n\r\n<div id=\"fs-id1167025113179\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{g}^{\\prime }\\left(t\\right)\\hfill &amp; = -{f}^{\\prime }\\left(t\\right)+\\left[{f}^{\\prime }\\left(t\\right)-f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)\\right]+\\left[f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}\\right]+\\cdots \\hfill \\\\ &amp; +\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025154282\">Notice that there is a telescoping effect. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167025154285\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{g}^{\\prime }\\left(t\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025228050\">By Rolle\u2019s theorem, we conclude that there exists a number <em data-effect=\"italics\">c<\/em> between <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. Since<\/p>\r\n\r\n<div id=\"fs-id1167025228092\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{g}^{\\prime }\\left(c\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025227341\">we conclude that<\/p>\r\n\r\n<div id=\"fs-id1167025227344\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025240050\">Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by [latex]n+1[\/latex], we conclude that<\/p>\r\n\r\n<div id=\"fs-id1167025240066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025234879\">as desired. From this fact, it follows that if there exists <em data-effect=\"italics\">M<\/em> such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all <em data-effect=\"italics\">x<\/em> in <em data-effect=\"italics\">I<\/em>, then<\/p>\r\n\r\n<div id=\"fs-id1167025234936\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025235008\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1167025235012\">Not only does Taylor\u2019s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] at [latex]x=8[\/latex] and determine how accurate these approximations are at estimating [latex]\\sqrt[3]{11}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167024999614\" data-type=\"example\">\r\n<div id=\"fs-id1167024999616\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024999618\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Linear and Quadratic Approximations to Estimate Function Values<\/h3>\r\n<div id=\"fs-id1167024999618\" data-type=\"problem\">\r\n<p id=\"fs-id1167024999623\">Consider the function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167024999648\" type=\"a\">\r\n \t<li>Find the first and second Taylor polynomials for [latex]f[\/latex] at [latex]x=8[\/latex]. Use a graphing utility to compare these polynomials with [latex]f[\/latex] near [latex]x=8[\/latex].<\/li>\r\n \t<li>Use these two polynomials to estimate [latex]\\sqrt[3]{11}[\/latex].<\/li>\r\n \t<li>Use Taylor\u2019s theorem to bound the error.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1167024999710\" data-type=\"solution\">\r\n<ol id=\"fs-id1167024999712\" type=\"a\">\r\n \t<li>For [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex], the values of the function and its first two derivatives at [latex]x=8[\/latex] are as follows:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024999755\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; \\sqrt[3]{x}\\hfill &amp; &amp; &amp; \\hfill f\\left(8\\right)&amp; =\\hfill &amp; 2\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; \\frac{1}{3{x}^{\\frac{2}{3}}}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\prime }\\left(8\\right)&amp; =\\hfill &amp; \\frac{1}{12}\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; \\frac{-2}{9{x}^{\\frac{5}{3}}}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(8\\right)&amp; =\\hfill &amp; -\\frac{1}{144}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThus, the first and second Taylor polynomials at [latex]x=8[\/latex] are given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025154132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{1}\\left(x\\right)&amp; =\\hfill &amp; f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)\\hfill \\\\ &amp; =\\hfill &amp; 2+\\frac{1}{12}\\left(x - 8\\right)\\hfill \\\\ {p}_{2}\\left(x\\right)\\hfill &amp; =\\hfill &amp; f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)+\\frac{f^{\\prime\\prime}\\left(8\\right)}{2\\text{!}}{\\left(x - 8\\right)}^{2}\\hfill \\\\ &amp; =\\hfill &amp; 2+\\frac{1}{12}\\left(x - 8\\right)-\\frac{1}{288}{\\left(x - 8\\right)}^{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe function and the Taylor polynomials are shown in Figure 5.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_03_005\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234521\/CNX_Calc_Figure_10_03_005-1.jpg\" alt=\"This graph has four curves. The first is the function f(x)=cube root of x. The second function is psub1(x). The third is psub2(x). The curves are very close around x=8.\" width=\"487\" height=\"387\" data-media-type=\"image\/jpeg\" \/> Figure 5. The graphs of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] and the linear and quadratic approximations [latex]{p}_{1}\\left(x\\right)[\/latex] and [latex]{p}_{2}\\left(x\\right)[\/latex].[\/caption]<\/figure>\r\n<\/li>\r\n \t<li>Using the first Taylor polynomial at [latex]x=8[\/latex], we can estimate<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025149114\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt[3]{11}\\approx {p}_{1}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)=2.25[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUsing the second Taylor polynomial at [latex]x=8[\/latex], we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025035671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt[3]{11}\\approx {p}_{2}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)-\\frac{1}{288}{\\left(11 - 8\\right)}^{2}=2.21875[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>By the Uniqueness of Taylor Series, there exists a <em data-effect=\"italics\">c<\/em> in the interval [latex]\\left(8,11\\right)[\/latex] such that the remainder when approximating [latex]\\sqrt[3]{11}[\/latex] by the first Taylor polynomial satisfies<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025035798\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{1}\\left(11\\right)=\\frac{f^{\\prime\\prime}\\left(c\\right)}{2\\text{!}}{\\left(11 - 8\\right)}^{2}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nWe do not know the exact value of <em data-effect=\"italics\">c<\/em>, so we find an upper bound on [latex]{R}_{1}\\left(11\\right)[\/latex] by determining the maximum value of [latex]f^{\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex]. Since [latex]f^{\\prime\\prime}\\left(x\\right)=-\\frac{2}{9{x}^{\\frac{5}{3}}}[\/latex], the largest value for [latex]|f^{\\prime\\prime}\\left(x\\right)|[\/latex] on that interval occurs at [latex]x=8[\/latex]. Using the fact that [latex]f^{\\prime\\prime}\\left(8\\right)=-\\frac{1}{144}[\/latex], we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025234774\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{1}\\left(11\\right)|\\le \\frac{1}{144\\cdot 2\\text{!}}{\\left(11 - 8\\right)}^{2}=0.03125[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSimilarly, to estimate [latex]{R}_{2}\\left(11\\right)[\/latex], we use the fact that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025149246\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{2}\\left(11\\right)=\\frac{f^{\\prime\\prime\\prime}\\left(c\\right)}{3\\text{!}}{\\left(11 - 8\\right)}^{3}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]f^{\\prime\\prime\\prime}\\left(x\\right)=\\frac{10}{27{x}^{\\frac{8}{3}}}[\/latex], the maximum value of [latex]f^{\\prime\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex] is [latex]f^{\\prime\\prime\\prime}\\left(8\\right)\\approx 0.0014468[\/latex]. Therefore, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025149396\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{2}\\left(11\\right)|\\le \\frac{0.0011468}{3\\text{!}}{\\left(11 - 8\\right)}^{3}\\approx 0.0065104[\/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\nWatch the following video to see the worked solution to Example: Using Linear and Quadratic Approximations to Estimate Function Values.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1223&amp;end=1687&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1223to1687_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.3 Taylor and Maclaurin Series\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167024969435\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167024969439\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\r\n<p id=\"fs-id1167024969443\">Find the first and second Taylor polynomials for [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex]. Use these polynomials to estimate [latex]\\sqrt{6}[\/latex]. Use Taylor\u2019s theorem to bound the error.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558849\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167025098843\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025098850\">Evaluate [latex]f\\left(4\\right),{f}^{\\prime }\\left(4\\right)[\/latex], and [latex]f^{\\prime\\prime} \\left(4\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167024969487\" data-type=\"solution\">\r\n<p id=\"fs-id1167024969489\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)-\\frac{1}{64}{\\left(x - 4\\right)}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(6\\right)=2.5[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(6\\right)=2.4375[\/latex];<\/p>\r\n<p id=\"fs-id1167025098788\" style=\"text-align: center;\">[latex]|{R}_{1}\\left(6\\right)|\\le 0.0625;|{R}_{2}\\left(6\\right)|\\le 0.015625[\/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 id=\"fs-id1167025098904\" data-type=\"example\">\r\n<div id=\"fs-id1167025098906\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025098908\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Approximating sinx using maclaurin polynomials<\/h3>\r\n<div id=\"fs-id1167025098908\" data-type=\"problem\">\r\n\r\nFrom the <em>Example: Finding Maclaurin Polynomials<\/em>, the Maclaurin polynomials for [latex]\\sin{x}[\/latex] are given by\r\n<div id=\"fs-id1167025098942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {p}_{2m+1}\\left(x\\right)&amp; ={p}_{2m+2}\\left(x\\right)\\hfill \\\\ &amp; =x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}-\\frac{{x}^{7}}{7\\text{!}}+\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m+1}}{\\left(2m+1\\right)\\text{!}}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025169190\">for [latex]m=0,1,2,\\dots[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167025169217\" type=\"a\">\r\n \t<li>Use the fifth Maclaurin polynomial for [latex]\\sin{x}[\/latex] to approximate [latex]\\sin\\left(\\frac{\\pi }{18}\\right)[\/latex] and bound the error.<\/li>\r\n \t<li>For what values of\u00a0[latex]x[\/latex]\u00a0does the fifth Maclaurin polynomial approximate [latex]\\sin{x}[\/latex] to within 0.0001?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1167025089128\" data-type=\"solution\">\r\n<ol id=\"fs-id1167025089130\" type=\"a\">\r\n \t<li>The fifth Maclaurin polynomial is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025089141\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{5}\\left(x\\right)=x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUsing this polynomial, we can estimate as follows:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025089200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sin\\left(\\frac{\\pi }{18}\\right)&amp; \\approx {p}_{5}\\left(\\frac{\\pi }{18}\\right)\\hfill \\\\ &amp; =\\frac{\\pi }{18}-\\frac{1}{3\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{3}+\\frac{1}{5\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{5}\\hfill \\\\ &amp; \\approx 0.173648.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTo estimate the error, use the fact that the sixth Maclaurin polynomial is [latex]{p}_{6}\\left(x\\right)={p}_{5}\\left(x\\right)[\/latex] and calculate a bound on [latex]{R}_{6}\\left(\\frac{\\pi }{18}\\right)[\/latex]. By the Uniqueness of Taylor Series, the remainder is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024967486\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{6}\\left(\\frac{\\pi }{18}\\right)=\\frac{{f}^{\\left(7\\right)}\\left(c\\right)}{7\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{7}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor some <em data-effect=\"italics\">c<\/em> between 0 and [latex]\\frac{\\pi }{18}[\/latex]. Using the fact that [latex]|{f}^{\\left(7\\right)}\\left(x\\right)|\\le 1[\/latex] for all <em data-effect=\"italics\">x<\/em>, we find that the magnitude of the error is at most<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024967616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{7\\text{!}}\\cdot {\\left(\\frac{\\pi }{18}\\right)}^{7}\\le 9.8\\times {10}^{-10}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>We need to find the values of <em data-effect=\"italics\">x<\/em> such that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025131308\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{7\\text{!}}{|x|}^{7}\\le 0.0001[\/latex].<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\"><span data-type=\"newline\">\r\n<\/span>\r\nSolving this inequality for [latex]x[\/latex], we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as [latex]|x|&lt;0.907[\/latex].<\/p>\r\n<\/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: Approximating sin x Using Maclaurin Polynomials.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1693&amp;end=1936&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1693to1936_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.3 Taylor and Maclaurin Series\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167025131370\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167025131374\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025131376\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167025131376\" data-type=\"problem\">\r\n<p id=\"fs-id1167025131379\">Use the fourth Maclaurin polynomial for [latex]\\cos{x}[\/latex] to approximate [latex]\\cos\\left(\\frac{\\pi }{12}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558819\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1167025131419\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025131426\">The fourth Maclaurin polynomial is [latex]{p}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1167025131413\" data-type=\"solution\">\r\n<p id=\"fs-id1167025131415\">0.96593<\/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-id1167025131484\">Now that we are able to bound the remainder [latex]{R}_{n}\\left(x\\right)[\/latex], we can use this bound to prove that a Taylor series for [latex]f[\/latex] at [latex]a[\/latex] converges to [latex]f[\/latex].<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5936[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section>\r\n<h2 data-type=\"title\">Representing Functions with Taylor and Maclaurin Series<\/h2>\r\n<p id=\"fs-id1167025009061\">We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.<\/p>\r\n\r\n<div id=\"fs-id1167025009066\" data-type=\"example\">\r\n<div id=\"fs-id1167025009068\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025009070\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Taylor Series<\/h3>\r\n<div id=\"fs-id1167025009070\" data-type=\"problem\">\r\n<p id=\"fs-id1167025009075\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] at [latex]x=1[\/latex]. Determine the interval of convergence.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558999\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558999\"]\r\n<div id=\"fs-id1167025009112\" data-type=\"solution\">\r\n<p id=\"fs-id1167025009114\">For [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex], the values of the function and its first four derivatives at [latex]x=1[\/latex] are<\/p>\r\n\r\n<div id=\"fs-id1167025009149\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; \\frac{1}{x}\\hfill &amp; &amp; &amp; \\hfill f\\left(1\\right)&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; -\\frac{1}{{x}^{2}}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\prime }\\left(1\\right)&amp; =\\hfill &amp; -1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; \\frac{2}{{x}^{3}}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(1\\right)&amp; =\\hfill &amp; 2\\text{!}\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; -\\frac{3\\cdot 2}{{x}^{4}}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)&amp; =\\hfill &amp; -3\\text{!}\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)&amp; =\\hfill &amp; \\frac{4\\cdot 3\\cdot 2}{{x}^{5}}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\left(4\\right)}\\left(1\\right)&amp; =\\hfill &amp; 4\\text{!.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025097493\">That is, we have [latex]{f}^{\\left(n\\right)}\\left(1\\right)={\\left(-1\\right)}^{n}n\\text{!}[\/latex] for all [latex]n\\ge 0[\/latex]. Therefore, the Taylor series for [latex]f[\/latex] at [latex]x=1[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167025097564\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(1\\right)}{n\\text{!}}{\\left(x - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025097683\">To find the interval of convergence, we use the ratio test. We find that<\/p>\r\n\r\n<div id=\"fs-id1167025097686\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{|{\\left(-1\\right)}^{n+1}{\\left(x - 1\\right)}^{n+1}|}{|{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}|}=|x - 1|[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025095544\">Thus, the series converges if [latex]|x - 1|&lt;1[\/latex]. That is, the series converges for [latex]0&lt;x&lt;2[\/latex]. Next, we need to check the endpoints. At [latex]x=2[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1167025095598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(2 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024969122\">diverges by the divergence test. Similarly, at [latex]x=0[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1167024969136\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(0 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024969245\">diverges. Therefore, the interval of convergence is [latex]\\left(0,2\\right)[\/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: Finding a Taylor Series.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1946&amp;end=2189&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1946to2189_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.3 Taylor and Maclaurin Series\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167024969269\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167024969273\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\r\n<p id=\"fs-id1167024969277\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{2x}[\/latex] at [latex]x=2[\/latex] and determine its interval of convergence.<\/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-id1167025232134\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025232142\" style=\"text-align: center;\">[latex]{f}^{\\left(n\\right)}\\left(2\\right)=\\frac{{\\left(-1\\right)}^{n}n\\text{!}}{{2}^{n+1}}[\/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-id1167024969311\" data-type=\"solution\">\r\n<p id=\"fs-id1167024969313\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2-x}{{2}^{n+2}}\\right)}^{n}[\/latex]. The interval of convergence is [latex]\\left(0,4\\right)[\/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-id1167025232201\">We know that the Taylor series found in this example converges on the interval [latex]\\left(0,2\\right)[\/latex], but how do we know it actually converges to [latex]f?[\/latex] We consider this question in more generality in a moment, but for this example, we can answer this question by writing<\/p>\r\n\r\n<div id=\"fs-id1167025232231\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{x}=\\frac{1}{1-\\left(1-x\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025232281\">That is, [latex]f[\/latex] can be represented by the geometric series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex]. Since this is a geometric series, it converges to [latex]\\frac{1}{x}[\/latex] as long as [latex]|1-x|&lt;1[\/latex]. Therefore, the Taylor series found in the previous example does converge to [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] on [latex]\\left(0,2\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1167025149892\">We now consider the more general question: if a Taylor series for a function [latex]f[\/latex] converges on some interval, how can we determine if it actually converges to [latex]f?[\/latex] To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em>, the <em data-effect=\"italics\">n<\/em>th partial sum is given by the <em data-effect=\"italics\">n<\/em>th Taylor polynomial <em data-effect=\"italics\">p<sub>n<\/sub><\/em>. Therefore, to determine if the Taylor series converges to [latex]f[\/latex], we need to determine whether<\/p>\r\n\r\n<div id=\"fs-id1167025149944\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{p}_{n}\\left(x\\right)=f\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025149991\">Since the remainder [latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex], the Taylor series converges to [latex]f[\/latex] if and only if<\/p>\r\n\r\n<div id=\"fs-id1167025150045\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025166973\">We now state this theorem formally.<\/p>\r\n\r\n<div id=\"fs-id1167025166976\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Theorem: Convergence of Taylor Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167025166984\">Suppose that [latex]f[\/latex] has derivatives of all orders on an interval [latex]I[\/latex]\u00a0containing [latex]a[\/latex]. Then the Taylor series<\/p>\r\n\r\n<div id=\"fs-id1167025167001\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025167069\">converges to [latex]f\\left(x\\right)[\/latex] for all [latex]x[\/latex]\u00a0in [latex]I[\/latex]\u00a0if and only if<\/p>\r\n\r\n<div id=\"fs-id1167025167096\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to\\infty}\\lim {R}_{n}\\left(x\\right)=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025167134\">for all [latex]x[\/latex] in [latex]I[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167025167149\">With this theorem, we can prove that a Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em> converges to [latex]f[\/latex] if we can prove that the remainder [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex]. To prove that [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex], we typically use the bound<\/p>\r\n\r\n<div id=\"fs-id1167025117660\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025117731\">from Taylor\u2019s theorem with remainder.<\/p>\r\n<p id=\"fs-id1167025117735\">In the next example, we find the Maclaurin series for <em data-effect=\"italics\">e<sup>x<\/sup><\/em> and [latex]\\sin{x}[\/latex] and show that these series converge to the corresponding functions for all real numbers by proving that the remainders [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex] for all real numbers <em data-effect=\"italics\">x<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1167025117785\" data-type=\"example\">\r\n<div id=\"fs-id1167025117787\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025117789\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Maclaurin Series<\/h3>\r\n<div id=\"fs-id1167025117789\" data-type=\"problem\">\r\n<p id=\"fs-id1167025117794\">For each of the following functions, find the Maclaurin series and its interval of convergence. Use Taylor's Theorem with Remainder to prove that the Maclaurin series for [latex]f[\/latex] converges to [latex]f[\/latex] on that interval.<\/p>\r\n\r\n<ol id=\"fs-id1167025117811\" type=\"a\">\r\n \t<li>[latex]e^{x}[\/latex]<\/li>\r\n \t<li>[latex]\\sin{x}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558599\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1167025117839\" data-type=\"solution\">\r\n<ol id=\"fs-id1167025117842\" type=\"a\">\r\n \t<li>Using the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial for [latex]e^{x}[\/latex]\u00a0found in the example: Finding Maclaurin Polynomials (a), we find that the Maclaurin series for [latex]e^{x}[\/latex]\u00a0is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025117876\" 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<span data-type=\"newline\">\r\n<\/span>\r\nTo determine the interval of convergence, we use the ratio test. Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025146566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{|x|}^{n}}=\\frac{|x|}{n+1}[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwe have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025146689\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{|x|}{n+1}=0[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor all <em data-effect=\"italics\">x<\/em>. Therefore, the series converges absolutely for all <em data-effect=\"italics\">x<\/em>, and thus, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. To show that the series converges to [latex]e^{x}[\/latex]\u00a0for all [latex]x[\/latex], we use the fact that [latex]{f}^{\\left(n\\right)}\\left(x\\right)={e}^{x}[\/latex] for all [latex]n\\ge 0[\/latex] and [latex]e^{x}[\/latex]\u00a0is an increasing function on [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Therefore, for any real number [latex]b[\/latex], the maximum value of [latex]e^{x}[\/latex]\u00a0for all [latex]|x|\\le b[\/latex] is [latex]e^{b}[\/latex]. Thus,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025241662\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{e}^{b}}{\\left(n+1\\right)\\text{!}}{|x|}^{n+1}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince we just showed that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025241737\" 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<span data-type=\"newline\">\r\n<\/span>\r\nconverges for all <em data-effect=\"italics\">x<\/em>, by the divergence test, we know that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025241787\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}=0[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor any real number [latex]x[\/latex]. By combining this fact with the squeeze theorem, the result is [latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex].<\/li>\r\n \t<li>Using the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial for [latex]\\sin{x}[\/latex] found in the example: Finding Maclaurin Polynomials (b), we find that the Maclaurin series for [latex]\\sin{x}[\/latex] is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024984713\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIn order to apply the ratio test, consider<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024984788\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{{|x|}^{2n+3}}{\\left(2n+3\\right)\\text{!}}\\cdot \\frac{\\left(2n+1\\right)\\text{!}}{{|x|}^{2n+1}}=\\frac{{|x|}^{2}}{\\left(2n+3\\right)\\left(2n+2\\right)}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025008815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{|x|}^{2}}{\\left(2n+3\\right)\\left(2n+2\\right)}=0[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor all [latex]x[\/latex], we obtain the interval of convergence as [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. To show that the Maclaurin series converges to [latex]\\sin{x}[\/latex], look at [latex]{R}_{n}\\left(x\\right)[\/latex]. For each [latex]x[\/latex]\u00a0there exists a real number [latex]c[\/latex]\u00a0between 0 and [latex]x[\/latex] such that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025008960\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{x}^{n+1}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]|{f}^{\\left(n+1\\right)}\\left(c\\right)|\\le 1[\/latex] for all integers [latex]n[\/latex] and all real numbers [latex]c[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024999263\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor all real numbers [latex]x[\/latex]. Using the same idea as in part a., the result is [latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex] for all [latex]x[\/latex], and therefore, the Maclaurin series for [latex]\\sin{x}[\/latex] converges to [latex]\\sin{x}[\/latex] for all real [latex]x[\/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: Finding Maclaurin Series.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=2255&amp;end=2670&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries2255to2670_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.3 Taylor and Maclaurin Series\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167024999404\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167024999408\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024999410\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167024999410\" data-type=\"problem\">\r\n<p id=\"fs-id1167024999413\">Find the Maclaurin series for [latex]f\\left(x\\right)=\\cos{x}[\/latex]. Use the ratio test to show that the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Show that the Maclaurin series converges to [latex]\\cos{x}[\/latex] for all real numbers [latex]x[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558399\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558399\"]\r\n<div id=\"fs-id1167025241321\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025241328\">Use the Maclaurin polynomials for [latex]\\cos{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558499\"]\r\n<div id=\"fs-id1167025241158\" data-type=\"solution\">\r\n<p id=\"fs-id1167025241160\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\r\n<p id=\"fs-id1167025241218\">By the ratio test, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Since [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex], the series converges to [latex]\\cos{x}[\/latex] for all real <em data-effect=\"italics\">x<\/em>.<\/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 id=\"fs-id1167025241347\" class=\"project\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3 data-type=\"title\">Activity: Proving that [latex]e[\/latex] is Irrational<\/h3>\r\n<p id=\"fs-id1167025241359\">In this project, we use the Maclaurin polynomials for [latex]e^{x}[\/latex]\u00a0to prove that <em data-effect=\"italics\">e<\/em> is irrational. The proof relies on supposing that <em data-effect=\"italics\">e<\/em> is rational and arriving at a contradiction. Therefore, in the following steps, we suppose [latex]e=\\frac{r}{s}[\/latex] for some integers [latex]r[\/latex] and [latex]s[\/latex]\u00a0where [latex]s\\ne 0[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167024996743\" type=\"1\">\r\n \t<li>Write the Maclaurin polynomials [latex]{p}_{0}\\left(x\\right),{p}_{1}\\left(x\\right),{p}_{2}\\left(x\\right),{p}_{3}\\left(x\\right),{p}_{4}\\left(x\\right)[\/latex] for <em data-effect=\"italics\">e<sup>x<\/sup><\/em>. Evaluate [latex]{p}_{0}\\left(1\\right),{p}_{1}\\left(1\\right),{p}_{2}\\left(1\\right),{p}_{3}\\left(1\\right),{p}_{4}\\left(1\\right)[\/latex] to estimate <em data-effect=\"italics\">e<\/em>.<\/li>\r\n \t<li>Let [latex]{R}_{n}\\left(x\\right)[\/latex] denote the remainder when using [latex]{p}_{n}\\left(x\\right)[\/latex] to estimate [latex]e^{x}[\/latex]. Therefore, [latex]{R}_{n}\\left(x\\right)={e}^{x}-{p}_{n}\\left(x\\right)[\/latex], and [latex]{R}_{n}\\left(1\\right)=e-{p}_{n}\\left(1\\right)[\/latex]. Assuming that [latex]e=\\frac{r}{s}[\/latex] for integers <em data-effect=\"italics\">r<\/em> and <em data-effect=\"italics\">s<\/em>, evaluate [latex]{R}_{0}\\left(1\\right),{R}_{1}\\left(1\\right),{R}_{2}\\left(1\\right),{R}_{3}\\left(1\\right),{R}_{4}\\left(1\\right)[\/latex].<\/li>\r\n \t<li>Using the results from part 2, show that for each remainder [latex]{R}_{0}\\left(1\\right),{R}_{1}\\left(1\\right),{R}_{2}\\left(1\\right),{R}_{3}\\left(1\\right),{R}_{4}\\left(1\\right)[\/latex], we can find an integer [latex]n[\/latex]\u00a0such that [latex]k{R}_{n}\\left(1\\right)[\/latex] is an integer for [latex]n=0,1,2,3,4[\/latex].<\/li>\r\n \t<li>Write down the formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial [latex]{p}_{n}\\left(x\\right)[\/latex] for [latex]e^{x}[\/latex]\u00a0and the corresponding remainder [latex]{R}_{n}\\left(x\\right)[\/latex]. Show that [latex]sn\\text{!}{R}_{n}\\left(1\\right)[\/latex] is an integer.<\/li>\r\n \t<li>Use Taylor\u2019s theorem to write down an explicit formula for [latex]{R}_{n}\\left(1\\right)[\/latex]. Conclude that [latex]{R}_{n}\\left(1\\right)\\ne 0[\/latex], and therefore, [latex]sn\\text{!}{R}_{n}\\left(1\\right)\\ne 0[\/latex].<\/li>\r\n \t<li>Use Taylor\u2019s theorem to find an estimate on [latex]{R}_{n}\\left(1\\right)[\/latex]. Use this estimate combined with the result from part 5 to show that [latex]|sn\\text{!}{R}_{n}\\left(1\\right)|&lt;\\frac{se}{n+1}[\/latex]. Conclude that if [latex]n[\/latex]\u00a0is large enough, then [latex]|sn\\text{!}{R}_{n}\\left(1\\right)|&lt;1[\/latex]. Therefore, [latex]sn\\text{!}{R}_{n}\\left(1\\right)[\/latex] is an integer with magnitude less than 1. Thus, [latex]sn\\text{!}{R}_{n}\\left(1\\right)=0[\/latex]. But from part 5, we know that [latex]sn\\text{!}{R}_{n}\\left(1\\right)\\ne 0[\/latex]. We have arrived at a contradiction, and consequently, the original supposition that <em data-effect=\"italics\">e<\/em> is rational must be false.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167025235824\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div style=\"text-align: left;\" data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain the meaning and significance of Taylor\u2019s theorem with remainder<\/li>\n<li>Estimate the remainder for a Taylor series approximation of a given function<\/li>\n<\/ul>\n<\/div>\n<h2>Taylor\u2019s Theorem with Remainder<\/h2>\n<p id=\"fs-id1167025102169\">Recall that the [latex]n[\/latex]th Taylor polynomial for a function [latex]f[\/latex] at [latex]a[\/latex]\u00a0is the [latex]n[\/latex]th partial sum of the Taylor series for [latex]f[\/latex] at [latex]a[\/latex]. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials [latex]\\left\\{{p}_{n}\\right\\}[\/latex] converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to [latex]f[\/latex]. To answer this question, we define the remainder [latex]{R}_{n}\\left(x\\right)[\/latex] as<\/p>\n<div id=\"fs-id1167025162616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025162665\">For the sequence of Taylor polynomials to converge to [latex]f[\/latex], we need the remainder [latex]R_{n}[\/latex] to converge to zero. To determine if [latex]R_{n}[\/latex]\u00a0converges to zero, we introduce <strong>Taylor\u2019s theorem with remainder<\/strong>. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the [latex]n[\/latex]th Taylor polynomial approximates the function.<\/p>\n<p id=\"fs-id1167025168160\">Here we look for a bound on [latex]|{R}_{n}|[\/latex]. Consider the simplest case: [latex]n=0[\/latex]. Let [latex]p_{0}[\/latex]\u00a0be the 0th Taylor polynomial at [latex]a[\/latex]\u00a0for a function [latex]f[\/latex]. The remainder [latex]R_{0}[\/latex]\u00a0satisfies<\/p>\n<div id=\"fs-id1167025168217\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {R}_{0}\\left(x\\right)& =f\\left(x\\right)-{p}_{0}\\left(x\\right)\\hfill \\\\ & =f\\left(x\\right)-f\\left(a\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025168302\">If [latex]f[\/latex] is differentiable on an interval [latex]I[\/latex]\u00a0containing [latex]a[\/latex] and [latex]x[\/latex], then by the Mean Value Theorem there exists a real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]\u00a0such that [latex]f\\left(x\\right)-f\\left(a\\right)={f}^{\\prime }\\left(c\\right)\\left(x-a\\right)[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1167024997799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{0}\\left(x\\right)={f}^{\\prime }\\left(c\\right)\\left(x-a\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024997849\">Using the Mean Value Theorem in a similar argument, we can show that if [latex]f[\/latex] is [latex]n[\/latex]\u00a0times differentiable on an interval [latex]I[\/latex] containing [latex]a[\/latex]\u00a0and [latex]x[\/latex], then the [latex]n[\/latex]th remainder [latex]{R}_{n}[\/latex]\u00a0satisfies<\/p>\n<div id=\"fs-id1167024875234\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex]<\/div>\n<div data-type=\"equation\" data-label=\"\"><\/div>\n<p id=\"fs-id1167024875321\">for some real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]. It is important to note that the value [latex]c[\/latex]\u00a0in the numerator above is not the center [latex]a[\/latex], but rather an unknown value [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]. This formula allows us to get a bound on the remainder [latex]{R}_{n}[\/latex]. If we happen to know that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|[\/latex] is bounded by some real number [latex]M[\/latex]\u00a0on this interval [latex]I[\/latex], then<\/p>\n<div id=\"fs-id1167025236141\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025236211\">for all [latex]x[\/latex]\u00a0in the interval [latex]I[\/latex].<\/p>\n<p id=\"fs-id1167025236225\">We now state Taylor\u2019s theorem, which provides the formal relationship between a function [latex]f[\/latex] and its [latex]n[\/latex]th degree Taylor polynomial [latex]{p}_{n}\\left(x\\right)[\/latex]. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for [latex]f[\/latex] converges to [latex]f[\/latex].<\/p>\n<div id=\"fs-id1167025239317\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Taylor\u2019s Theorem with Remainder<\/h3>\n<hr \/>\n<p id=\"fs-id1167025239324\">Let [latex]f[\/latex] be a function that can be differentiated [latex]n+1[\/latex] times on an interval [latex]I[\/latex] containing the real number [latex]a[\/latex]. Let [latex]p_{n}[\/latex]\u00a0be the [latex]n[\/latex]th Taylor polynomial of [latex]f[\/latex] at [latex]a[\/latex]\u00a0and let<\/p>\n<div id=\"fs-id1167025239373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025239420\">be the [latex]n[\/latex]th remainder. Then for each [latex]x[\/latex]\u00a0in the interval [latex]I[\/latex], there exists a real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]\u00a0such that<\/p>\n<div id=\"fs-id1167025101301\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025101391\">If there exists a real number [latex]M[\/latex]\u00a0such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all [latex]x\\in I[\/latex], then<\/p>\n<div id=\"fs-id1167025070502\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025070572\">for all [latex]x[\/latex]\u00a0in [latex]I[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167025070586\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1167025070592\">Fix a point [latex]x\\in I[\/latex] and introduce the function <em data-effect=\"italics\">g<\/em> such that<\/p>\n<div id=\"fs-id1167025070610\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]g\\left(t\\right)=f\\left(x\\right)-f\\left(t\\right)-{f}^{\\prime }\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}-\\cdots -\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}-{R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n+1}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025234124\">We claim that <em data-effect=\"italics\">g<\/em> satisfies the criteria of Rolle\u2019s theorem. Since <em data-effect=\"italics\">g<\/em> is a polynomial function (in <em data-effect=\"italics\">t<\/em>), it is a differentiable function. Also, <em data-effect=\"italics\">g<\/em> is zero at [latex]t=a[\/latex] and [latex]t=x[\/latex] because<\/p>\n<div id=\"fs-id1167025234167\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill g\\left(a\\right)& =\\hfill & f\\left(x\\right)-f\\left(a\\right)-{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)-\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}-{R}_{n}\\left(x\\right)\\hfill \\\\ & =\\hfill & f\\left(x\\right)-{p}_{n}\\left(x\\right)-{R}_{n}\\left(x\\right)\\hfill \\\\ & =\\hfill & 0,\\hfill \\\\ g\\left(x\\right)\\hfill & =\\hfill & f\\left(x\\right)-f\\left(x\\right)-0-\\cdots -0\\hfill \\\\ & =\\hfill & 0.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025232390\">Therefore, <em data-effect=\"italics\">g<\/em> satisfies Rolle\u2019s theorem, and consequently, there exists <em data-effect=\"italics\">c<\/em> between <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. We now calculate [latex]{g}^{\\prime }[\/latex]. Using the product rule, we note that<\/p>\n<div id=\"fs-id1167025232449\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dt}\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]=\\frac{-{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}+\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025113176\">Consequently,<\/p>\n<div id=\"fs-id1167025113179\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{g}^{\\prime }\\left(t\\right)\\hfill & = -{f}^{\\prime }\\left(t\\right)+\\left[{f}^{\\prime }\\left(t\\right)-f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)\\right]+\\left[f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}\\right]+\\cdots \\hfill \\\\ & +\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025154282\">Notice that there is a telescoping effect. Therefore,<\/p>\n<div id=\"fs-id1167025154285\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{g}^{\\prime }\\left(t\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025228050\">By Rolle\u2019s theorem, we conclude that there exists a number <em data-effect=\"italics\">c<\/em> between <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. Since<\/p>\n<div id=\"fs-id1167025228092\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{g}^{\\prime }\\left(c\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025227341\">we conclude that<\/p>\n<div id=\"fs-id1167025227344\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025240050\">Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by [latex]n+1[\/latex], we conclude that<\/p>\n<div id=\"fs-id1167025240066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025234879\">as desired. From this fact, it follows that if there exists <em data-effect=\"italics\">M<\/em> such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all <em data-effect=\"italics\">x<\/em> in <em data-effect=\"italics\">I<\/em>, then<\/p>\n<div id=\"fs-id1167025234936\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025235008\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167025235012\">Not only does Taylor\u2019s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] at [latex]x=8[\/latex] and determine how accurate these approximations are at estimating [latex]\\sqrt[3]{11}[\/latex].<\/p>\n<div id=\"fs-id1167024999614\" data-type=\"example\">\n<div id=\"fs-id1167024999616\" data-type=\"exercise\">\n<div id=\"fs-id1167024999618\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using Linear and Quadratic Approximations to Estimate Function Values<\/h3>\n<div id=\"fs-id1167024999618\" data-type=\"problem\">\n<p id=\"fs-id1167024999623\">Consider the function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex].<\/p>\n<ol id=\"fs-id1167024999648\" type=\"a\">\n<li>Find the first and second Taylor polynomials for [latex]f[\/latex] at [latex]x=8[\/latex]. Use a graphing utility to compare these polynomials with [latex]f[\/latex] near [latex]x=8[\/latex].<\/li>\n<li>Use these two polynomials to estimate [latex]\\sqrt[3]{11}[\/latex].<\/li>\n<li>Use Taylor\u2019s theorem to bound the error.<\/li>\n<\/ol>\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-id1167024999710\" data-type=\"solution\">\n<ol id=\"fs-id1167024999712\" type=\"a\">\n<li>For [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex], the values of the function and its first two derivatives at [latex]x=8[\/latex] are as follows:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024999755\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\sqrt[3]{x}\\hfill & & & \\hfill f\\left(8\\right)& =\\hfill & 2\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & \\frac{1}{3{x}^{\\frac{2}{3}}}\\hfill & & & \\hfill {f}^{\\prime }\\left(8\\right)& =\\hfill & \\frac{1}{12}\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{-2}{9{x}^{\\frac{5}{3}}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(8\\right)& =\\hfill & -\\frac{1}{144}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThus, the first and second Taylor polynomials at [latex]x=8[\/latex] are given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025154132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{1}\\left(x\\right)& =\\hfill & f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)\\hfill \\\\ & =\\hfill & 2+\\frac{1}{12}\\left(x - 8\\right)\\hfill \\\\ {p}_{2}\\left(x\\right)\\hfill & =\\hfill & f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)+\\frac{f^{\\prime\\prime}\\left(8\\right)}{2\\text{!}}{\\left(x - 8\\right)}^{2}\\hfill \\\\ & =\\hfill & 2+\\frac{1}{12}\\left(x - 8\\right)-\\frac{1}{288}{\\left(x - 8\\right)}^{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe function and the Taylor polynomials are shown in Figure 5.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_03_005\">\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\/11234521\/CNX_Calc_Figure_10_03_005-1.jpg\" alt=\"This graph has four curves. The first is the function f(x)=cube root of x. The second function is psub1(x). The third is psub2(x). The curves are very close around x=8.\" width=\"487\" height=\"387\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The graphs of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] and the linear and quadratic approximations [latex]{p}_{1}\\left(x\\right)[\/latex] and [latex]{p}_{2}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<li>Using the first Taylor polynomial at [latex]x=8[\/latex], we can estimate<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025149114\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt[3]{11}\\approx {p}_{1}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)=2.25[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUsing the second Taylor polynomial at [latex]x=8[\/latex], we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025035671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sqrt[3]{11}\\approx {p}_{2}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)-\\frac{1}{288}{\\left(11 - 8\\right)}^{2}=2.21875[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>By the Uniqueness of Taylor Series, there exists a <em data-effect=\"italics\">c<\/em> in the interval [latex]\\left(8,11\\right)[\/latex] such that the remainder when approximating [latex]\\sqrt[3]{11}[\/latex] by the first Taylor polynomial satisfies<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025035798\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{1}\\left(11\\right)=\\frac{f^{\\prime\\prime}\\left(c\\right)}{2\\text{!}}{\\left(11 - 8\\right)}^{2}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nWe do not know the exact value of <em data-effect=\"italics\">c<\/em>, so we find an upper bound on [latex]{R}_{1}\\left(11\\right)[\/latex] by determining the maximum value of [latex]f^{\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex]. Since [latex]f^{\\prime\\prime}\\left(x\\right)=-\\frac{2}{9{x}^{\\frac{5}{3}}}[\/latex], the largest value for [latex]|f^{\\prime\\prime}\\left(x\\right)|[\/latex] on that interval occurs at [latex]x=8[\/latex]. Using the fact that [latex]f^{\\prime\\prime}\\left(8\\right)=-\\frac{1}{144}[\/latex], we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025234774\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{1}\\left(11\\right)|\\le \\frac{1}{144\\cdot 2\\text{!}}{\\left(11 - 8\\right)}^{2}=0.03125[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSimilarly, to estimate [latex]{R}_{2}\\left(11\\right)[\/latex], we use the fact that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025149246\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{2}\\left(11\\right)=\\frac{f^{\\prime\\prime\\prime}\\left(c\\right)}{3\\text{!}}{\\left(11 - 8\\right)}^{3}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]f^{\\prime\\prime\\prime}\\left(x\\right)=\\frac{10}{27{x}^{\\frac{8}{3}}}[\/latex], the maximum value of [latex]f^{\\prime\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex] is [latex]f^{\\prime\\prime\\prime}\\left(8\\right)\\approx 0.0014468[\/latex]. Therefore, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025149396\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{2}\\left(11\\right)|\\le \\frac{0.0011468}{3\\text{!}}{\\left(11 - 8\\right)}^{3}\\approx 0.0065104[\/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>Watch the following video to see the worked solution to Example: Using Linear and Quadratic Approximations to Estimate Function Values.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1223&amp;end=1687&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1223to1687_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.3 Taylor and Maclaurin Series&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167024969435\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167024969439\" data-type=\"exercise\">\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\n<p id=\"fs-id1167024969443\">Find the first and second Taylor polynomials for [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex]. Use these polynomials to estimate [latex]\\sqrt{6}[\/latex]. Use Taylor\u2019s theorem to bound the error.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558849\">Hint<\/span><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025098843\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025098850\">Evaluate [latex]f\\left(4\\right),{f}^{\\prime }\\left(4\\right)[\/latex], and [latex]f^{\\prime\\prime} \\left(4\\right)[\/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=\"q44558859\">Show Solution<\/span><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024969487\" data-type=\"solution\">\n<p id=\"fs-id1167024969489\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)-\\frac{1}{64}{\\left(x - 4\\right)}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(6\\right)=2.5[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(6\\right)=2.4375[\/latex];<\/p>\n<p id=\"fs-id1167025098788\" style=\"text-align: center;\">[latex]|{R}_{1}\\left(6\\right)|\\le 0.0625;|{R}_{2}\\left(6\\right)|\\le 0.015625[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167025098904\" data-type=\"example\">\n<div id=\"fs-id1167025098906\" data-type=\"exercise\">\n<div id=\"fs-id1167025098908\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Approximating sinx using maclaurin polynomials<\/h3>\n<div id=\"fs-id1167025098908\" data-type=\"problem\">\n<p>From the <em>Example: Finding Maclaurin Polynomials<\/em>, the Maclaurin polynomials for [latex]\\sin{x}[\/latex] are given by<\/p>\n<div id=\"fs-id1167025098942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {p}_{2m+1}\\left(x\\right)& ={p}_{2m+2}\\left(x\\right)\\hfill \\\\ & =x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}-\\frac{{x}^{7}}{7\\text{!}}+\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m+1}}{\\left(2m+1\\right)\\text{!}}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025169190\">for [latex]m=0,1,2,\\dots[\/latex].<\/p>\n<ol id=\"fs-id1167025169217\" type=\"a\">\n<li>Use the fifth Maclaurin polynomial for [latex]\\sin{x}[\/latex] to approximate [latex]\\sin\\left(\\frac{\\pi }{18}\\right)[\/latex] and bound the error.<\/li>\n<li>For what values of\u00a0[latex]x[\/latex]\u00a0does the fifth Maclaurin polynomial approximate [latex]\\sin{x}[\/latex] to within 0.0001?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558839\">Show Solution<\/span><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025089128\" data-type=\"solution\">\n<ol id=\"fs-id1167025089130\" type=\"a\">\n<li>The fifth Maclaurin polynomial is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025089141\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{5}\\left(x\\right)=x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUsing this polynomial, we can estimate as follows:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025089200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sin\\left(\\frac{\\pi }{18}\\right)& \\approx {p}_{5}\\left(\\frac{\\pi }{18}\\right)\\hfill \\\\ & =\\frac{\\pi }{18}-\\frac{1}{3\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{3}+\\frac{1}{5\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{5}\\hfill \\\\ & \\approx 0.173648.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo estimate the error, use the fact that the sixth Maclaurin polynomial is [latex]{p}_{6}\\left(x\\right)={p}_{5}\\left(x\\right)[\/latex] and calculate a bound on [latex]{R}_{6}\\left(\\frac{\\pi }{18}\\right)[\/latex]. By the Uniqueness of Taylor Series, the remainder is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024967486\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{6}\\left(\\frac{\\pi }{18}\\right)=\\frac{{f}^{\\left(7\\right)}\\left(c\\right)}{7\\text{!}}{\\left(\\frac{\\pi }{18}\\right)}^{7}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor some <em data-effect=\"italics\">c<\/em> between 0 and [latex]\\frac{\\pi }{18}[\/latex]. Using the fact that [latex]|{f}^{\\left(7\\right)}\\left(x\\right)|\\le 1[\/latex] for all <em data-effect=\"italics\">x<\/em>, we find that the magnitude of the error is at most<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024967616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{7\\text{!}}\\cdot {\\left(\\frac{\\pi }{18}\\right)}^{7}\\le 9.8\\times {10}^{-10}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>We need to find the values of <em data-effect=\"italics\">x<\/em> such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025131308\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{7\\text{!}}{|x|}^{7}\\le 0.0001[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\"><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSolving this inequality for [latex]x[\/latex], we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as [latex]|x|<0.907[\/latex].<\/p>\n<\/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: Approximating sin x Using Maclaurin Polynomials.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1693&amp;end=1936&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1693to1936_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.3 Taylor and Maclaurin Series&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167025131370\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167025131374\" data-type=\"exercise\">\n<div id=\"fs-id1167025131376\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167025131376\" data-type=\"problem\">\n<p id=\"fs-id1167025131379\">Use the fourth Maclaurin polynomial for [latex]\\cos{x}[\/latex] to approximate [latex]\\cos\\left(\\frac{\\pi }{12}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558819\">Hint<\/span><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025131419\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025131426\">The fourth Maclaurin polynomial is [latex]{p}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}[\/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=\"q44558829\">Show Solution<\/span><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025131413\" data-type=\"solution\">\n<p id=\"fs-id1167025131415\">0.96593<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167025131484\">Now that we are able to bound the remainder [latex]{R}_{n}\\left(x\\right)[\/latex], we can use this bound to prove that a Taylor series for [latex]f[\/latex] at [latex]a[\/latex] converges to [latex]f[\/latex].<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5936\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5936&theme=oea&iframe_resize_id=ohm5936&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Representing Functions with Taylor and Maclaurin Series<\/h2>\n<p id=\"fs-id1167025009061\">We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.<\/p>\n<div id=\"fs-id1167025009066\" data-type=\"example\">\n<div id=\"fs-id1167025009068\" data-type=\"exercise\">\n<div id=\"fs-id1167025009070\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Taylor Series<\/h3>\n<div id=\"fs-id1167025009070\" data-type=\"problem\">\n<p id=\"fs-id1167025009075\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] at [latex]x=1[\/latex]. Determine the interval of convergence.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558999\">Show Solution<\/span><\/p>\n<div id=\"q44558999\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025009112\" data-type=\"solution\">\n<p id=\"fs-id1167025009114\">For [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex], the values of the function and its first four derivatives at [latex]x=1[\/latex] are<\/p>\n<div id=\"fs-id1167025009149\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\frac{1}{x}\\hfill & & & \\hfill f\\left(1\\right)& =\\hfill & 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & -\\frac{1}{{x}^{2}}\\hfill & & & \\hfill {f}^{\\prime }\\left(1\\right)& =\\hfill & -1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{2}{{x}^{3}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(1\\right)& =\\hfill & 2\\text{!}\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & -\\frac{3\\cdot 2}{{x}^{4}}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)& =\\hfill & -3\\text{!}\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)& =\\hfill & \\frac{4\\cdot 3\\cdot 2}{{x}^{5}}\\hfill & & & \\hfill {f}^{\\left(4\\right)}\\left(1\\right)& =\\hfill & 4\\text{!.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025097493\">That is, we have [latex]{f}^{\\left(n\\right)}\\left(1\\right)={\\left(-1\\right)}^{n}n\\text{!}[\/latex] for all [latex]n\\ge 0[\/latex]. Therefore, the Taylor series for [latex]f[\/latex] at [latex]x=1[\/latex] is given by<\/p>\n<div id=\"fs-id1167025097564\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(1\\right)}{n\\text{!}}{\\left(x - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025097683\">To find the interval of convergence, we use the ratio test. We find that<\/p>\n<div id=\"fs-id1167025097686\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{|{\\left(-1\\right)}^{n+1}{\\left(x - 1\\right)}^{n+1}|}{|{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}|}=|x - 1|[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025095544\">Thus, the series converges if [latex]|x - 1|<1[\/latex]. That is, the series converges for [latex]0<x<2[\/latex]. Next, we need to check the endpoints. At [latex]x=2[\/latex], we see that<\/p>\n<div id=\"fs-id1167025095598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(2 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024969122\">diverges by the divergence test. Similarly, at [latex]x=0[\/latex],<\/p>\n<div id=\"fs-id1167024969136\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(0 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024969245\">diverges. Therefore, the interval of convergence is [latex]\\left(0,2\\right)[\/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: Finding a Taylor Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=1946&amp;end=2189&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries1946to2189_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.3 Taylor and Maclaurin Series&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167024969269\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167024969273\" data-type=\"exercise\">\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\n<p id=\"fs-id1167024969277\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{2x}[\/latex] at [latex]x=2[\/latex] and determine its interval of convergence.<\/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-id1167025232134\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025232142\" style=\"text-align: center;\">[latex]{f}^{\\left(n\\right)}\\left(2\\right)=\\frac{{\\left(-1\\right)}^{n}n\\text{!}}{{2}^{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=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024969311\" data-type=\"solution\">\n<p id=\"fs-id1167024969313\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2-x}{{2}^{n+2}}\\right)}^{n}[\/latex]. The interval of convergence is [latex]\\left(0,4\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167025232201\">We know that the Taylor series found in this example converges on the interval [latex]\\left(0,2\\right)[\/latex], but how do we know it actually converges to [latex]f?[\/latex] We consider this question in more generality in a moment, but for this example, we can answer this question by writing<\/p>\n<div id=\"fs-id1167025232231\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{x}=\\frac{1}{1-\\left(1-x\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025232281\">That is, [latex]f[\/latex] can be represented by the geometric series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex]. Since this is a geometric series, it converges to [latex]\\frac{1}{x}[\/latex] as long as [latex]|1-x|<1[\/latex]. Therefore, the Taylor series found in the previous example does converge to [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] on [latex]\\left(0,2\\right)[\/latex].<\/p>\n<p id=\"fs-id1167025149892\">We now consider the more general question: if a Taylor series for a function [latex]f[\/latex] converges on some interval, how can we determine if it actually converges to [latex]f?[\/latex] To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em>, the <em data-effect=\"italics\">n<\/em>th partial sum is given by the <em data-effect=\"italics\">n<\/em>th Taylor polynomial <em data-effect=\"italics\">p<sub>n<\/sub><\/em>. Therefore, to determine if the Taylor series converges to [latex]f[\/latex], we need to determine whether<\/p>\n<div id=\"fs-id1167025149944\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{p}_{n}\\left(x\\right)=f\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025149991\">Since the remainder [latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex], the Taylor series converges to [latex]f[\/latex] if and only if<\/p>\n<div id=\"fs-id1167025150045\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025166973\">We now state this theorem formally.<\/p>\n<div id=\"fs-id1167025166976\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Theorem: Convergence of Taylor Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167025166984\">Suppose that [latex]f[\/latex] has derivatives of all orders on an interval [latex]I[\/latex]\u00a0containing [latex]a[\/latex]. Then the Taylor series<\/p>\n<div id=\"fs-id1167025167001\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025167069\">converges to [latex]f\\left(x\\right)[\/latex] for all [latex]x[\/latex]\u00a0in [latex]I[\/latex]\u00a0if and only if<\/p>\n<div id=\"fs-id1167025167096\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to\\infty}\\lim {R}_{n}\\left(x\\right)=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025167134\">for all [latex]x[\/latex] in [latex]I[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167025167149\">With this theorem, we can prove that a Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em> converges to [latex]f[\/latex] if we can prove that the remainder [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex]. To prove that [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex], we typically use the bound<\/p>\n<div id=\"fs-id1167025117660\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025117731\">from Taylor\u2019s theorem with remainder.<\/p>\n<p id=\"fs-id1167025117735\">In the next example, we find the Maclaurin series for <em data-effect=\"italics\">e<sup>x<\/sup><\/em> and [latex]\\sin{x}[\/latex] and show that these series converge to the corresponding functions for all real numbers by proving that the remainders [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex] for all real numbers <em data-effect=\"italics\">x<\/em>.<\/p>\n<div id=\"fs-id1167025117785\" data-type=\"example\">\n<div id=\"fs-id1167025117787\" data-type=\"exercise\">\n<div id=\"fs-id1167025117789\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding Maclaurin Series<\/h3>\n<div id=\"fs-id1167025117789\" data-type=\"problem\">\n<p id=\"fs-id1167025117794\">For each of the following functions, find the Maclaurin series and its interval of convergence. Use Taylor&#8217;s Theorem with Remainder to prove that the Maclaurin series for [latex]f[\/latex] converges to [latex]f[\/latex] on that interval.<\/p>\n<ol id=\"fs-id1167025117811\" type=\"a\">\n<li>[latex]e^{x}[\/latex]<\/li>\n<li>[latex]\\sin{x}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558599\">Show Solution<\/span><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025117839\" data-type=\"solution\">\n<ol id=\"fs-id1167025117842\" type=\"a\">\n<li>Using the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial for [latex]e^{x}[\/latex]\u00a0found in the example: Finding Maclaurin Polynomials (a), we find that the Maclaurin series for [latex]e^{x}[\/latex]\u00a0is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025117876\" 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><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo determine the interval of convergence, we use the ratio test. Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025146566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{|x|}^{n}}=\\frac{|x|}{n+1}[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwe have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025146689\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{|x|}{n+1}=0[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor all <em data-effect=\"italics\">x<\/em>. Therefore, the series converges absolutely for all <em data-effect=\"italics\">x<\/em>, and thus, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. To show that the series converges to [latex]e^{x}[\/latex]\u00a0for all [latex]x[\/latex], we use the fact that [latex]{f}^{\\left(n\\right)}\\left(x\\right)={e}^{x}[\/latex] for all [latex]n\\ge 0[\/latex] and [latex]e^{x}[\/latex]\u00a0is an increasing function on [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Therefore, for any real number [latex]b[\/latex], the maximum value of [latex]e^{x}[\/latex]\u00a0for all [latex]|x|\\le b[\/latex] is [latex]e^{b}[\/latex]. Thus,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025241662\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{e}^{b}}{\\left(n+1\\right)\\text{!}}{|x|}^{n+1}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince we just showed that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025241737\" 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><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nconverges for all <em data-effect=\"italics\">x<\/em>, by the divergence test, we know that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025241787\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}=0[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor any real number [latex]x[\/latex]. By combining this fact with the squeeze theorem, the result is [latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex].<\/li>\n<li>Using the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial for [latex]\\sin{x}[\/latex] found in the example: Finding Maclaurin Polynomials (b), we find that the Maclaurin series for [latex]\\sin{x}[\/latex] is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024984713\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn order to apply the ratio test, consider<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024984788\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{{|x|}^{2n+3}}{\\left(2n+3\\right)\\text{!}}\\cdot \\frac{\\left(2n+1\\right)\\text{!}}{{|x|}^{2n+1}}=\\frac{{|x|}^{2}}{\\left(2n+3\\right)\\left(2n+2\\right)}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025008815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{|x|}^{2}}{\\left(2n+3\\right)\\left(2n+2\\right)}=0[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor all [latex]x[\/latex], we obtain the interval of convergence as [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. To show that the Maclaurin series converges to [latex]\\sin{x}[\/latex], look at [latex]{R}_{n}\\left(x\\right)[\/latex]. For each [latex]x[\/latex]\u00a0there exists a real number [latex]c[\/latex]\u00a0between 0 and [latex]x[\/latex] such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025008960\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{x}^{n+1}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]|{f}^{\\left(n+1\\right)}\\left(c\\right)|\\le 1[\/latex] for all integers [latex]n[\/latex] and all real numbers [latex]c[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024999263\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor all real numbers [latex]x[\/latex]. Using the same idea as in part a., the result is [latex]\\underset{n\\to \\infty }{\\text{lim}}{R}_{n}\\left(x\\right)=0[\/latex] for all [latex]x[\/latex], and therefore, the Maclaurin series for [latex]\\sin{x}[\/latex] converges to [latex]\\sin{x}[\/latex] for all real [latex]x[\/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: Finding Maclaurin Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=2255&amp;end=2670&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.3TaylorAndMaclaurinSeries2255to2670_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.3 Taylor and Maclaurin Series&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167024999404\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167024999408\" data-type=\"exercise\">\n<div id=\"fs-id1167024999410\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167024999410\" data-type=\"problem\">\n<p id=\"fs-id1167024999413\">Find the Maclaurin series for [latex]f\\left(x\\right)=\\cos{x}[\/latex]. Use the ratio test to show that the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Show that the Maclaurin series converges to [latex]\\cos{x}[\/latex] for all real numbers [latex]x[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558399\">Hint<\/span><\/p>\n<div id=\"q44558399\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025241321\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025241328\">Use the Maclaurin polynomials for [latex]\\cos{x}[\/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=\"q44558499\">Show Solution<\/span><\/p>\n<div id=\"q44558499\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025241158\" data-type=\"solution\">\n<p id=\"fs-id1167025241160\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\n<p id=\"fs-id1167025241218\">By the ratio test, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Since [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex], the series converges to [latex]\\cos{x}[\/latex] for all real <em data-effect=\"italics\">x<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167025241347\" class=\"project\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3 data-type=\"title\">Activity: Proving that [latex]e[\/latex] is Irrational<\/h3>\n<p id=\"fs-id1167025241359\">In this project, we use the Maclaurin polynomials for [latex]e^{x}[\/latex]\u00a0to prove that <em data-effect=\"italics\">e<\/em> is irrational. The proof relies on supposing that <em data-effect=\"italics\">e<\/em> is rational and arriving at a contradiction. Therefore, in the following steps, we suppose [latex]e=\\frac{r}{s}[\/latex] for some integers [latex]r[\/latex] and [latex]s[\/latex]\u00a0where [latex]s\\ne 0[\/latex].<\/p>\n<ol id=\"fs-id1167024996743\" type=\"1\">\n<li>Write the Maclaurin polynomials [latex]{p}_{0}\\left(x\\right),{p}_{1}\\left(x\\right),{p}_{2}\\left(x\\right),{p}_{3}\\left(x\\right),{p}_{4}\\left(x\\right)[\/latex] for <em data-effect=\"italics\">e<sup>x<\/sup><\/em>. Evaluate [latex]{p}_{0}\\left(1\\right),{p}_{1}\\left(1\\right),{p}_{2}\\left(1\\right),{p}_{3}\\left(1\\right),{p}_{4}\\left(1\\right)[\/latex] to estimate <em data-effect=\"italics\">e<\/em>.<\/li>\n<li>Let [latex]{R}_{n}\\left(x\\right)[\/latex] denote the remainder when using [latex]{p}_{n}\\left(x\\right)[\/latex] to estimate [latex]e^{x}[\/latex]. Therefore, [latex]{R}_{n}\\left(x\\right)={e}^{x}-{p}_{n}\\left(x\\right)[\/latex], and [latex]{R}_{n}\\left(1\\right)=e-{p}_{n}\\left(1\\right)[\/latex]. Assuming that [latex]e=\\frac{r}{s}[\/latex] for integers <em data-effect=\"italics\">r<\/em> and <em data-effect=\"italics\">s<\/em>, evaluate [latex]{R}_{0}\\left(1\\right),{R}_{1}\\left(1\\right),{R}_{2}\\left(1\\right),{R}_{3}\\left(1\\right),{R}_{4}\\left(1\\right)[\/latex].<\/li>\n<li>Using the results from part 2, show that for each remainder [latex]{R}_{0}\\left(1\\right),{R}_{1}\\left(1\\right),{R}_{2}\\left(1\\right),{R}_{3}\\left(1\\right),{R}_{4}\\left(1\\right)[\/latex], we can find an integer [latex]n[\/latex]\u00a0such that [latex]k{R}_{n}\\left(1\\right)[\/latex] is an integer for [latex]n=0,1,2,3,4[\/latex].<\/li>\n<li>Write down the formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial [latex]{p}_{n}\\left(x\\right)[\/latex] for [latex]e^{x}[\/latex]\u00a0and the corresponding remainder [latex]{R}_{n}\\left(x\\right)[\/latex]. Show that [latex]sn\\text{!}{R}_{n}\\left(1\\right)[\/latex] is an integer.<\/li>\n<li>Use Taylor\u2019s theorem to write down an explicit formula for [latex]{R}_{n}\\left(1\\right)[\/latex]. Conclude that [latex]{R}_{n}\\left(1\\right)\\ne 0[\/latex], and therefore, [latex]sn\\text{!}{R}_{n}\\left(1\\right)\\ne 0[\/latex].<\/li>\n<li>Use Taylor\u2019s theorem to find an estimate on [latex]{R}_{n}\\left(1\\right)[\/latex]. Use this estimate combined with the result from part 5 to show that [latex]|sn\\text{!}{R}_{n}\\left(1\\right)|<\\frac{se}{n+1}[\/latex]. Conclude that if [latex]n[\/latex]\u00a0is large enough, then [latex]|sn\\text{!}{R}_{n}\\left(1\\right)|<1[\/latex]. Therefore, [latex]sn\\text{!}{R}_{n}\\left(1\\right)[\/latex] is an integer with magnitude less than 1. Thus, [latex]sn\\text{!}{R}_{n}\\left(1\\right)=0[\/latex]. But from part 5, we know that [latex]sn\\text{!}{R}_{n}\\left(1\\right)\\ne 0[\/latex]. We have arrived at a contradiction, and consequently, the original supposition that <em data-effect=\"italics\">e<\/em> is rational must be false.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167025235824\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div style=\"text-align: left;\" data-type=\"glossary\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1783\">\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>6.3 Taylor and Maclaurin Series. <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":12,"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\":\"6.3 Taylor and Maclaurin Series\",\"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-1783","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1783","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":10,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1783\/revisions"}],"predecessor-version":[{"id":2250,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1783\/revisions\/2250"}],"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\/1783\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1783"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1783"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1783"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}