{"id":1815,"date":"2021-07-28T16:18:51","date_gmt":"2021-07-28T16:18:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1815"},"modified":"2022-03-21T23:12:49","modified_gmt":"2022-03-21T23:12:49","slug":"finding-taylor-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/finding-taylor-polynomials\/","title":{"raw":"Finding Taylor Polynomials","rendered":"Finding Taylor Polynomials"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the procedure for finding a Taylor polynomial of a given order for a function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167024907497\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin Series<\/h2>\r\n<p id=\"fs-id1167025238767\">Consider a function [latex]f[\/latex] that has a power series representation at [latex]x=a[\/latex]. Then the series has the form<\/p>\r\n\r\n<div id=\"fs-id1167025118355\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025239871\">What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series in the above equation is a representation for [latex]f[\/latex] at [latex]x=a[\/latex], we certainly want the series to equal [latex]f\\left(a\\right)[\/latex] at [latex]x=a[\/latex]. Evaluating the series at [latex]x=a[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1167024866801\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}&amp; ={c}_{0}+{c}_{1}\\left(a-a\\right)+{c}_{2}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; ={c}_{0}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024942802\">Thus, the series equals [latex]f\\left(a\\right)[\/latex] if the coefficient [latex]{c}_{0}=f\\left(a\\right)[\/latex]. In addition, we would like the first derivative of the power series to equal [latex]{f}^{\\prime }\\left(a\\right)[\/latex] at [latex]x=a[\/latex]. Differentiating our initial equation term-by-term, we see that<\/p>\r\n\r\n<div id=\"fs-id1167024875132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025092208\">Therefore, at [latex]x=a[\/latex], the derivative is<\/p>\r\n\r\n<div id=\"fs-id1167024999505\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ \\hfill \\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; ={c}_{1}+2{c}_{2}\\left(a-a\\right)+3{c}_{3}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; ={c}_{1}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025227426\">Therefore, the derivative of the series equals [latex]{f}^{\\prime }\\left(a\\right)[\/latex] if the coefficient [latex]{c}_{1}={f}^{\\prime }\\left(a\\right)[\/latex]. Continuing in this way, we look for coefficients <em data-effect=\"italics\">c<sub>n<\/sub><\/em> such that all the derivatives of the power series will agree with all the corresponding derivatives of [latex]f[\/latex] at [latex]x=a[\/latex]. The second and third derivatives of our initial equation are given by<\/p>\r\n\r\n<div id=\"fs-id1167024955379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+4\\cdot 3{c}_{4}{\\left(x-a\\right)}^{2}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024962562\">and<\/p>\r\n\r\n<div id=\"fs-id1167025089047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(x-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024934396\">Therefore, at [latex]x=a[\/latex], the second and third derivatives<\/p>\r\n\r\n<div id=\"fs-id1167025091472\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; =2{c}_{2}+3\\cdot 2{c}_{3}\\left(a-a\\right)+4\\cdot 3{c}_{4}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; =2{c}_{2}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025018088\">and<\/p>\r\n\r\n<div id=\"fs-id1167025099708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; =3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(a-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; =3\\cdot 2{c}_{3}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025239876\">equal [latex]f^{\\prime\\prime} \\left(a\\right)[\/latex] and [latex]f^{\\prime\\prime\\prime}\\left(a\\right)[\/latex], respectively, if [latex]{c}_{2}=\\frac{f^{\\prime\\prime}\\left(a\\right)}{2}[\/latex] and [latex]{c}_{3}=\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3}\\cdot 2[\/latex]. More generally, we see that if [latex]f[\/latex] has a power series representation at [latex]x=a[\/latex], then the coefficients should be given by [latex]{c}_{n}=\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}[\/latex]. That is, the series should be<\/p>\r\n\r\n<div id=\"fs-id1167024875578\" 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}=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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025239902\">This power series for [latex]f[\/latex] is known as the Taylor series for [latex]f[\/latex] at [latex]a[\/latex]. If [latex]a=0[\/latex], then this series is known as the Maclaurin series for [latex]f[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167024973208\" 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\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167025101458\">If [latex]f[\/latex] has derivatives of all orders at [latex]x=a[\/latex], then the <strong>Taylor series<\/strong> for the function [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1167025228678\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=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}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024875404\">The Taylor series for [latex]f[\/latex] at 0 is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167025017712\">Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from Uniqueness of Power Series that power series representations are unique. Therefore, if a function [latex]f[\/latex] has a power series at [latex]a[\/latex], then it must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167025088556\" 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: Uniqueness of Taylor Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167025070215\">If a function [latex]f[\/latex] has a power series at <em data-effect=\"italics\">a<\/em> that converges to [latex]f[\/latex] on some open interval containing <em data-effect=\"italics\">a<\/em>, then that power series is the Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167024992988\">The proof follows directly from Uniqueness of Power Series.<\/p>\r\n<p id=\"fs-id1167025087959\">To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as <span data-type=\"term\">Taylor polynomials<\/span>.<\/p>\r\n\r\n<div id=\"fs-id1167025087961\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\nVisit the MacTutor History of Mathematics archive to <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Taylor\/\" target=\"_blank\" rel=\"noopener\">read a biography of Brook Taylor<\/a> and a <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Maclaurin\/\" target=\"_blank\" rel=\"noopener\">biography of\u00a0Colin Maclaurin<\/a> and how they developed the concepts named after them.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1167025234017\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Taylor Polynomials<\/h2>\r\n<p id=\"fs-id1167025239889\">The <em data-effect=\"italics\">n<\/em>th partial sum of the Taylor series for a function [latex]f[\/latex] at [latex]a[\/latex] is known as the <em data-effect=\"italics\">n<\/em>th Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by<\/p>\r\n\r\n<div id=\"fs-id1167024875102\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=f\\left(a\\right),\\hfill \\\\ {p}_{1}\\left(x\\right)=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right),\\hfill \\\\ {p}_{2}\\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},\\hfill \\\\ {p}_{3}\\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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3},\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025018662\">respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of [latex]f[\/latex] at [latex]a[\/latex], respectively. If [latex]a=0[\/latex], then these polynomials are known as <span data-type=\"term\">Maclaurin polynomials<\/span> for [latex]f[\/latex]. We now provide a formal definition of Taylor and Maclaurin polynomials for a function [latex]f[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167025242377\" 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\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167024983571\">If [latex]f[\/latex] has <em data-effect=\"italics\">n<\/em> derivatives at [latex]x=a[\/latex], then the <em data-effect=\"italics\">n<\/em>th Taylor polynomial for [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1167025091916\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{n}\\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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots +\\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-id1167025134804\">The [latex]n[\/latex]th Taylor polynomial for [latex]f[\/latex] at [latex]0[\/latex] is known as the [latex]n[\/latex]th\u00a0Maclaurin polynomial for [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167024858718\">We now show how to use this definition to find several Taylor polynomials for [latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]x=1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167025228652\" data-type=\"example\">\r\n<div id=\"fs-id1167025228654\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025228656\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Taylor Polynomials<\/h3>\r\n<div id=\"fs-id1167025228656\" data-type=\"problem\">\r\n<p id=\"fs-id1167024978555\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]x=1[\/latex]. Use a graphing utility to compare the graph of [latex]f[\/latex] with the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167025040285\" data-type=\"solution\">\r\n<p id=\"fs-id1167025154312\">To find these Taylor polynomials, we need to evaluate [latex]f[\/latex] and its first three derivatives at [latex]x=1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167025154330\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; \\text{ln}x\\hfill &amp; &amp; &amp; \\hfill f\\left(1\\right)&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; \\frac{1}{x}\\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{1}{{x}^{2}}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(1\\right)&amp; =\\hfill &amp; -1\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; \\frac{2}{{x}^{3}}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)&amp; =\\hfill &amp; 2\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167025040302\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167025040305\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{0}\\left(x\\right)&amp; =\\hfill &amp; f\\left(1\\right)=0,\\hfill \\\\ \\hfill {p}_{1}\\left(x\\right)&amp; =\\hfill &amp; f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)=x - 1,\\hfill \\\\ \\hfill {p}_{2}\\left(x\\right)&amp; =\\hfill &amp; f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}=\\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2},\\hfill \\\\ \\hfill {p}_{3}\\left(x\\right)&amp; =\\hfill &amp; f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(1\\right)}{3\\text{!}}{\\left(x - 1\\right)}^{3}\\hfill \\\\ &amp; =\\hfill &amp; \\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2}+\\frac{1}{3}{\\left(x - 1\\right)}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024971031\">The graphs of [latex]y=f\\left(x\\right)[\/latex] and the first three Taylor polynomials are shown in Figure 1.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_10_03_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234510\/CNX_Calc_Figure_10_03_001.jpg\" alt=\"This graph has four curves. The first is the function f(x)=ln(x). The second function is psub1(x)=x-1. The third is psub2(x)=(x-1)-1\/2(x-1)^2. The fourth is psub3(x)=(x-1)-1\/2(x-1)^2 +1\/3(x-1)^3. The curves are very close around x = 1.\" width=\"487\" height=\"364\" data-media-type=\"image\/jpeg\" \/> Figure 1. The function [latex]y=\\text{ln}x[\/latex] and the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] at [latex]x=1[\/latex] are plotted on this graph.[\/caption]<\/figure>\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 Taylor Polynomials.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=331&amp;end=545&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\/6.3TaylorAndMaclaurinSeries331to545_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-id1167025019196\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167025019199\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\r\n<p id=\"fs-id1167025019203\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] at [latex]x=1[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167025112903\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025118432\">Find the first three derivatives of [latex]f[\/latex] and evaluate them at [latex]x=1[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167024971078\" data-type=\"solution\">\r\n<p id=\"fs-id1167024971080\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1 - 2\\left(x - 1\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}-4{\\left(x - 1\\right)}^{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]25579[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167025118454\">We now show how to find Maclaurin polynomials for <em data-effect=\"italics\">e<sup>x<\/sup><\/em>, [latex]\\sin{x}[\/latex], and [latex]\\cos{x}[\/latex]. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.<\/p>\r\n\r\n<div id=\"fs-id1167025118487\" data-type=\"example\">\r\n<div id=\"fs-id1167025118489\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025111740\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Maclaurin Polynomials<\/h3>\r\n<div id=\"fs-id1167025111740\" data-type=\"problem\">\r\n<p id=\"fs-id1167025111745\">For each of the following functions, find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] with [latex]f[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167025238039\" type=\"a\">\r\n \t<li>[latex]f\\left(x\\right)={e}^{x}[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)=\\sin{x}[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)=\\cos{x}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<ol id=\"fs-id1167025118800\" type=\"a\">\r\n \t<li>Since [latex]f\\left(x\\right)={e}^{x}[\/latex], we know that [latex]f\\left(x\\right)={f}^{\\prime }\\left(x\\right)=f^{\\prime\\prime}\\left(x\\right)=\\cdots ={f}^{\\left(n\\right)}\\left(x\\right)={e}^{x}[\/latex] for all positive integers <em data-effect=\"italics\">n<\/em>. Therefore,<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167025069158\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(0\\right)={f}^{\\prime }\\left(0\\right)=f^{\\prime\\prime}\\left(0\\right)=\\cdots ={f}^{\\left(n\\right)}\\left(0\\right)=1[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor all positive integers <em data-effect=\"italics\">n<\/em>. Therefore, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024875791\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{0}\\left(x\\right)&amp; =\\hfill &amp; f\\left(0\\right)=1,\\hfill \\\\ \\hfill {p}_{1}\\left(x\\right)&amp; =\\hfill &amp; f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x=1+x,\\hfill \\\\ \\hfill {p}_{2}\\left(x\\right)&amp; =\\hfill &amp; f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2\\text{!}}{x}^{2}=1+x+\\frac{1}{2}{x}^{2},\\hfill \\\\ \\hfill {p}_{3}\\left(x\\right)&amp; =\\hfill &amp; f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2}{x}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(0\\right)}{3\\text{!}}{x}^{3}\\hfill \\\\ &amp; =\\hfill &amp; 1+x+\\frac{1}{2}{x}^{2}+\\frac{1}{3\\text{!}}{x}^{3},\\hfill \\\\ \\hfill {p}_{n}\\left(x\\right)&amp; =\\hfill &amp; f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2}{x}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(0\\right)}{3\\text{!}}{x}^{3}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}{x}^{n}\\hfill \\\\ &amp; =\\hfill &amp; 1+x+\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{3}}{3\\text{!}}+\\cdots +\\frac{{x}^{n}}{n\\text{!}}\\hfill \\\\ &amp; =\\hfill &amp; {\\displaystyle\\sum _{k=0}^{n}}\\frac{{x}^{k}}{k\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe function and the first three Maclaurin polynomials are shown in Figure 2.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_03_002\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234513\/CNX_Calc_Figure_10_03_002.jpg\" alt=\"This graph has four curves. The first is the function f(x)=e^x. The second function is psub0(x)=1. The third is psub1(x) which is an increasing line passing through y=1. The fourth function is psub3(x) which is a curve passing through y=1. The curves are very close around y= 1.\" width=\"487\" height=\"321\" data-media-type=\"image\/jpeg\" \/> Figure 2. The graph shows the function [latex]y={e}^{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex].[\/caption]<\/figure>\r\n<ul>\r\n \t<li>For [latex]f\\left(x\\right)=\\sin{x}[\/latex], the values of the function and its first four derivatives at [latex]x=0[\/latex] are given as follows:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024874943\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; \\sin{x}\\hfill &amp; &amp; &amp; \\hfill f\\left(0\\right)&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; \\cos{x}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\prime }\\left(0\\right)&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; -\\sin{x}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; -\\cos{x}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; -1\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)&amp; =\\hfill &amp; \\sin{x}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\left(4\\right)}\\left(0\\right)&amp; =\\hfill &amp; 0.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the fourth derivative is [latex]\\sin{x}[\/latex], the pattern repeats. That is, [latex]{f}^{\\left(2m\\right)}\\left(0\\right)=0[\/latex] and [latex]{f}^{\\left(2m+1\\right)}\\left(0\\right)={\\left(-1\\right)}^{m}[\/latex] for [latex]m\\ge 0[\/latex]. Thus, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025229347\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=0,\\hfill \\\\ {p}_{1}\\left(x\\right)=0+x=x,\\hfill \\\\ {p}_{2}\\left(x\\right)=0+x+0=x,\\hfill \\\\ {p}_{3}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}=x-\\frac{{x}^{3}}{3\\text{!}},\\hfill \\\\ {p}_{4}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}+0=x-\\frac{{x}^{3}}{3\\text{!}},\\hfill \\\\ {p}_{5}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}+0+\\frac{1}{5\\text{!}}{x}^{5}=x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}},\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nand for [latex]m\\ge 0[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025241412\" 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{!}}-\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m+1}}{\\left(2m+1\\right)\\text{!}}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{k=0}^{m}}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{\\left(2k+1\\right)\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nGraphs of the function and its Maclaurin polynomials are shown in Figure 3.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_03_003\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234515\/CNX_Calc_Figure_10_03_003.jpg\" alt=\"This graph has four curves. The first is the function f(x)=sin(x). The second function is psub1(x). The third is psub3(x). The fourth function is psub5(x). The curves are very close around x=0.\" width=\"487\" height=\"350\" data-media-type=\"image\/jpeg\" \/> Figure 3. The graph shows the function [latex]y=\\sin{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{1},{p}_{3}[\/latex] and [latex]{p}_{5}[\/latex].[\/caption]<\/figure>\r\n<\/li>\r\n \t<li>For [latex]f\\left(x\\right)=\\cos{x}[\/latex], the values of the function and its first four derivatives at [latex]x=0[\/latex] are given as follows:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025069824\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; \\cos{x}\\hfill &amp; &amp; &amp; \\hfill f\\left(0\\right)&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; -\\sin{x}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\prime }\\left(0\\right)&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; -\\cos{x}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; -1\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; \\sin{x}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)&amp; =\\hfill &amp; \\cos{x}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\left(4\\right)}\\left(0\\right)&amp; =\\hfill &amp; 1.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the fourth derivative is [latex]\\sin{x}[\/latex], the pattern repeats. In other words, [latex]{f}^{\\left(2m\\right)}\\left(0\\right)={\\left(-1\\right)}^{m}[\/latex] and [latex]{f}^{\\left(2m+1\\right)}=0[\/latex] for [latex]m\\ge 0[\/latex]. Therefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025087805\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=1,\\hfill \\\\ {p}_{1}\\left(x\\right)=1+0=1,\\hfill \\\\ {p}_{2}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}=1-\\frac{{x}^{2}}{2\\text{!}},\\hfill \\\\ {p}_{3}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0=1-\\frac{{x}^{2}}{2\\text{!}},\\hfill \\\\ {p}_{4}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0+\\frac{1}{4\\text{!}}{x}^{4}=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}},\\hfill \\\\ {p}_{5}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0+\\frac{1}{4\\text{!}}{x}^{4}+0=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}},\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nand for [latex]n\\ge 0[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167025230186\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {p}_{2m}\\left(x\\right)&amp; ={p}_{2m+1}\\left(x\\right)\\hfill \\\\ &amp; =1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}-\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m}}{\\left(2m\\right)\\text{!}}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{k=0}^{m}}{\\left(-1\\right)}^{k}\\frac{{x}^{2k}}{\\left(2k\\right)\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nGraphs of the function and the Maclaurin polynomials appear in Figure 4.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_03_004\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234518\/CNX_Calc_Figure_10_03_004.jpg\" alt=\"This graph has four curves. The first is the function f(x)=cos(x). The second function is psub0(x). The third is psub2(x). The fourth function is psub4(x). The curves are very close around y=1\" width=\"487\" height=\"312\" data-media-type=\"image\/jpeg\" \/> Figure 4. The function [latex]y=\\cos{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{2}[\/latex] and [latex]{p}_{4}[\/latex] are plotted on this graph.[\/caption]<\/figure>\r\n<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding Maclaurin Polynomials.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oQkN46wsyJs?controls=0&amp;start=546&amp;end=1086&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\/6.3TaylorAndMaclaurinSeries546to1086_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-id1167025235544\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167025235548\" data-type=\"exercise\">\r\n<div id=\"fs-id1167025235550\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167025235550\" data-type=\"problem\">\r\n<p id=\"fs-id1167025235553\">Find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{1+x}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial. Write your anwer using sigma notation.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167025102143\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025102151\">Evaluate the first four derivatives of [latex]f[\/latex] and look for a pattern.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167025168862\" data-type=\"solution\">\r\n<p id=\"fs-id1167025168865\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1-x[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1-x+{x}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{n}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}+\\cdots +{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}{x}^{k}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167025102163\" data-depth=\"1\"><\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the procedure for finding a Taylor polynomial of a given order for a function<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167024907497\" data-depth=\"1\">\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin Series<\/h2>\n<p id=\"fs-id1167025238767\">Consider a function [latex]f[\/latex] that has a power series representation at [latex]x=a[\/latex]. Then the series has the form<\/p>\n<div id=\"fs-id1167025118355\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025239871\">What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series in the above equation is a representation for [latex]f[\/latex] at [latex]x=a[\/latex], we certainly want the series to equal [latex]f\\left(a\\right)[\/latex] at [latex]x=a[\/latex]. Evaluating the series at [latex]x=a[\/latex], we see that<\/p>\n<div id=\"fs-id1167024866801\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}& ={c}_{0}+{c}_{1}\\left(a-a\\right)+{c}_{2}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={c}_{0}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024942802\">Thus, the series equals [latex]f\\left(a\\right)[\/latex] if the coefficient [latex]{c}_{0}=f\\left(a\\right)[\/latex]. In addition, we would like the first derivative of the power series to equal [latex]{f}^{\\prime }\\left(a\\right)[\/latex] at [latex]x=a[\/latex]. Differentiating our initial equation term-by-term, we see that<\/p>\n<div id=\"fs-id1167024875132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025092208\">Therefore, at [latex]x=a[\/latex], the derivative is<\/p>\n<div id=\"fs-id1167024999505\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ \\hfill \\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& ={c}_{1}+2{c}_{2}\\left(a-a\\right)+3{c}_{3}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={c}_{1}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025227426\">Therefore, the derivative of the series equals [latex]{f}^{\\prime }\\left(a\\right)[\/latex] if the coefficient [latex]{c}_{1}={f}^{\\prime }\\left(a\\right)[\/latex]. Continuing in this way, we look for coefficients <em data-effect=\"italics\">c<sub>n<\/sub><\/em> such that all the derivatives of the power series will agree with all the corresponding derivatives of [latex]f[\/latex] at [latex]x=a[\/latex]. The second and third derivatives of our initial equation are given by<\/p>\n<div id=\"fs-id1167024955379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+4\\cdot 3{c}_{4}{\\left(x-a\\right)}^{2}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024962562\">and<\/p>\n<div id=\"fs-id1167025089047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(x-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024934396\">Therefore, at [latex]x=a[\/latex], the second and third derivatives<\/p>\n<div id=\"fs-id1167025091472\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& =2{c}_{2}+3\\cdot 2{c}_{3}\\left(a-a\\right)+4\\cdot 3{c}_{4}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & =2{c}_{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025018088\">and<\/p>\n<div id=\"fs-id1167025099708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& =3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(a-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & =3\\cdot 2{c}_{3}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025239876\">equal [latex]f^{\\prime\\prime} \\left(a\\right)[\/latex] and [latex]f^{\\prime\\prime\\prime}\\left(a\\right)[\/latex], respectively, if [latex]{c}_{2}=\\frac{f^{\\prime\\prime}\\left(a\\right)}{2}[\/latex] and [latex]{c}_{3}=\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3}\\cdot 2[\/latex]. More generally, we see that if [latex]f[\/latex] has a power series representation at [latex]x=a[\/latex], then the coefficients should be given by [latex]{c}_{n}=\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}[\/latex]. That is, the series should be<\/p>\n<div id=\"fs-id1167024875578\" 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}=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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025239902\">This power series for [latex]f[\/latex] is known as the Taylor series for [latex]f[\/latex] at [latex]a[\/latex]. If [latex]a=0[\/latex], then this series is known as the Maclaurin series for [latex]f[\/latex].<\/p>\n<div id=\"fs-id1167024973208\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167025101458\">If [latex]f[\/latex] has derivatives of all orders at [latex]x=a[\/latex], then the <strong>Taylor series<\/strong> for the function [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\n<div id=\"fs-id1167025228678\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=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}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024875404\">The Taylor series for [latex]f[\/latex] at 0 is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167025017712\">Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from Uniqueness of Power Series that power series representations are unique. Therefore, if a function [latex]f[\/latex] has a power series at [latex]a[\/latex], then it must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].<\/p>\n<div id=\"fs-id1167025088556\" 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: Uniqueness of Taylor Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167025070215\">If a function [latex]f[\/latex] has a power series at <em data-effect=\"italics\">a<\/em> that converges to [latex]f[\/latex] on some open interval containing <em data-effect=\"italics\">a<\/em>, then that power series is the Taylor series for [latex]f[\/latex] at <em data-effect=\"italics\">a<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167024992988\">The proof follows directly from Uniqueness of Power Series.<\/p>\n<p id=\"fs-id1167025087959\">To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as <span data-type=\"term\">Taylor polynomials<\/span>.<\/p>\n<div id=\"fs-id1167025087961\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p>Visit the MacTutor History of Mathematics archive to <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Taylor\/\" target=\"_blank\" rel=\"noopener\">read a biography of Brook Taylor<\/a> and a <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Maclaurin\/\" target=\"_blank\" rel=\"noopener\">biography of\u00a0Colin Maclaurin<\/a> and how they developed the concepts named after them.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1167025234017\" data-depth=\"1\">\n<h2 data-type=\"title\">Taylor Polynomials<\/h2>\n<p id=\"fs-id1167025239889\">The <em data-effect=\"italics\">n<\/em>th partial sum of the Taylor series for a function [latex]f[\/latex] at [latex]a[\/latex] is known as the <em data-effect=\"italics\">n<\/em>th Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by<\/p>\n<div id=\"fs-id1167024875102\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=f\\left(a\\right),\\hfill \\\\ {p}_{1}\\left(x\\right)=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right),\\hfill \\\\ {p}_{2}\\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},\\hfill \\\\ {p}_{3}\\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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3},\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025018662\">respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of [latex]f[\/latex] at [latex]a[\/latex], respectively. If [latex]a=0[\/latex], then these polynomials are known as <span data-type=\"term\">Maclaurin polynomials<\/span> for [latex]f[\/latex]. We now provide a formal definition of Taylor and Maclaurin polynomials for a function [latex]f[\/latex].<\/p>\n<div id=\"fs-id1167025242377\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167024983571\">If [latex]f[\/latex] has <em data-effect=\"italics\">n<\/em> derivatives at [latex]x=a[\/latex], then the <em data-effect=\"italics\">n<\/em>th Taylor polynomial for [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\n<div id=\"fs-id1167025091916\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{n}\\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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots +\\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-id1167025134804\">The [latex]n[\/latex]th Taylor polynomial for [latex]f[\/latex] at [latex]0[\/latex] is known as the [latex]n[\/latex]th\u00a0Maclaurin polynomial for [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167024858718\">We now show how to use this definition to find several Taylor polynomials for [latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]x=1[\/latex].<\/p>\n<div id=\"fs-id1167025228652\" data-type=\"example\">\n<div id=\"fs-id1167025228654\" data-type=\"exercise\">\n<div id=\"fs-id1167025228656\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding Taylor Polynomials<\/h3>\n<div id=\"fs-id1167025228656\" data-type=\"problem\">\n<p id=\"fs-id1167024978555\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]x=1[\/latex]. Use a graphing utility to compare the graph of [latex]f[\/latex] with the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025040285\" data-type=\"solution\">\n<p id=\"fs-id1167025154312\">To find these Taylor polynomials, we need to evaluate [latex]f[\/latex] and its first three derivatives at [latex]x=1[\/latex].<\/p>\n<div id=\"fs-id1167025154330\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\text{ln}x\\hfill & & & \\hfill f\\left(1\\right)& =\\hfill & 0\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & \\frac{1}{x}\\hfill & & & \\hfill {f}^{\\prime }\\left(1\\right)& =\\hfill & 1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & -\\frac{1}{{x}^{2}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(1\\right)& =\\hfill & -1\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{2}{{x}^{3}}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)& =\\hfill & 2\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167025040302\">Therefore,<\/p>\n<div id=\"fs-id1167025040305\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{0}\\left(x\\right)& =\\hfill & f\\left(1\\right)=0,\\hfill \\\\ \\hfill {p}_{1}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)=x - 1,\\hfill \\\\ \\hfill {p}_{2}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}=\\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2},\\hfill \\\\ \\hfill {p}_{3}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(1\\right)}{3\\text{!}}{\\left(x - 1\\right)}^{3}\\hfill \\\\ & =\\hfill & \\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2}+\\frac{1}{3}{\\left(x - 1\\right)}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024971031\">The graphs of [latex]y=f\\left(x\\right)[\/latex] and the first three Taylor polynomials are shown in Figure 1.<\/p>\n<figure id=\"CNX_Calc_Figure_10_03_001\"><figcaption><\/figcaption><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\/11234510\/CNX_Calc_Figure_10_03_001.jpg\" alt=\"This graph has four curves. The first is the function f(x)=ln(x). The second function is psub1(x)=x-1. The third is psub2(x)=(x-1)-1\/2(x-1)^2. The fourth is psub3(x)=(x-1)-1\/2(x-1)^2 +1\/3(x-1)^3. The curves are very close around x = 1.\" width=\"487\" height=\"364\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The function [latex]y=\\text{ln}x[\/latex] and the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] at [latex]x=1[\/latex] are plotted on this graph.<\/p>\n<\/div>\n<\/figure>\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 Taylor 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=331&amp;end=545&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\/6.3TaylorAndMaclaurinSeries331to545_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-id1167025019196\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167025019199\" data-type=\"exercise\">\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\n<p id=\"fs-id1167025019203\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] at [latex]x=1[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025112903\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025118432\">Find the first three derivatives of [latex]f[\/latex] and evaluate them at [latex]x=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=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024971078\" data-type=\"solution\">\n<p id=\"fs-id1167024971080\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1 - 2\\left(x - 1\\right)[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}-4{\\left(x - 1\\right)}^{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm25579\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25579&theme=oea&iframe_resize_id=ohm25579&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167025118454\">We now show how to find Maclaurin polynomials for <em data-effect=\"italics\">e<sup>x<\/sup><\/em>, [latex]\\sin{x}[\/latex], and [latex]\\cos{x}[\/latex]. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.<\/p>\n<div id=\"fs-id1167025118487\" data-type=\"example\">\n<div id=\"fs-id1167025118489\" data-type=\"exercise\">\n<div id=\"fs-id1167025111740\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding Maclaurin Polynomials<\/h3>\n<div id=\"fs-id1167025111740\" data-type=\"problem\">\n<p id=\"fs-id1167025111745\">For each of the following functions, find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] with [latex]f[\/latex].<\/p>\n<ol id=\"fs-id1167025238039\" type=\"a\">\n<li>[latex]f\\left(x\\right)={e}^{x}[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)=\\sin{x}[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)=\\cos{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=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167025118800\" type=\"a\">\n<li>Since [latex]f\\left(x\\right)={e}^{x}[\/latex], we know that [latex]f\\left(x\\right)={f}^{\\prime }\\left(x\\right)=f^{\\prime\\prime}\\left(x\\right)=\\cdots ={f}^{\\left(n\\right)}\\left(x\\right)={e}^{x}[\/latex] for all positive integers <em data-effect=\"italics\">n<\/em>. Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<div id=\"fs-id1167025069158\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(0\\right)={f}^{\\prime }\\left(0\\right)=f^{\\prime\\prime}\\left(0\\right)=\\cdots ={f}^{\\left(n\\right)}\\left(0\\right)=1[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor all positive integers <em data-effect=\"italics\">n<\/em>. Therefore, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024875791\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {p}_{0}\\left(x\\right)& =\\hfill & f\\left(0\\right)=1,\\hfill \\\\ \\hfill {p}_{1}\\left(x\\right)& =\\hfill & f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x=1+x,\\hfill \\\\ \\hfill {p}_{2}\\left(x\\right)& =\\hfill & f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2\\text{!}}{x}^{2}=1+x+\\frac{1}{2}{x}^{2},\\hfill \\\\ \\hfill {p}_{3}\\left(x\\right)& =\\hfill & f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2}{x}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(0\\right)}{3\\text{!}}{x}^{3}\\hfill \\\\ & =\\hfill & 1+x+\\frac{1}{2}{x}^{2}+\\frac{1}{3\\text{!}}{x}^{3},\\hfill \\\\ \\hfill {p}_{n}\\left(x\\right)& =\\hfill & f\\left(0\\right)+{f}^{\\prime }\\left(0\\right)x+\\frac{f^{\\prime\\prime}\\left(0\\right)}{2}{x}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(0\\right)}{3\\text{!}}{x}^{3}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}{x}^{n}\\hfill \\\\ & =\\hfill & 1+x+\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{3}}{3\\text{!}}+\\cdots +\\frac{{x}^{n}}{n\\text{!}}\\hfill \\\\ & =\\hfill & {\\displaystyle\\sum _{k=0}^{n}}\\frac{{x}^{k}}{k\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe function and the first three Maclaurin polynomials are shown in Figure 2.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_03_002\">\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\/11234513\/CNX_Calc_Figure_10_03_002.jpg\" alt=\"This graph has four curves. The first is the function f(x)=e^x. The second function is psub0(x)=1. The third is psub1(x) which is an increasing line passing through y=1. The fourth function is psub3(x) which is a curve passing through y=1. The curves are very close around y= 1.\" width=\"487\" height=\"321\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The graph shows the function [latex]y={e}^{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex].<\/p>\n<\/div>\n<\/figure>\n<ul>\n<li>For [latex]f\\left(x\\right)=\\sin{x}[\/latex], the values of the function and its first four derivatives at [latex]x=0[\/latex] are given as follows:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024874943\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\sin{x}\\hfill & & & \\hfill f\\left(0\\right)& =\\hfill & 0\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & \\cos{x}\\hfill & & & \\hfill {f}^{\\prime }\\left(0\\right)& =\\hfill & 1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & -\\sin{x}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(0\\right)& =\\hfill & 0\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & -\\cos{x}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)& =\\hfill & -1\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)& =\\hfill & \\sin{x}\\hfill & & & \\hfill {f}^{\\left(4\\right)}\\left(0\\right)& =\\hfill & 0.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the fourth derivative is [latex]\\sin{x}[\/latex], the pattern repeats. That is, [latex]{f}^{\\left(2m\\right)}\\left(0\\right)=0[\/latex] and [latex]{f}^{\\left(2m+1\\right)}\\left(0\\right)={\\left(-1\\right)}^{m}[\/latex] for [latex]m\\ge 0[\/latex]. Thus, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025229347\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=0,\\hfill \\\\ {p}_{1}\\left(x\\right)=0+x=x,\\hfill \\\\ {p}_{2}\\left(x\\right)=0+x+0=x,\\hfill \\\\ {p}_{3}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}=x-\\frac{{x}^{3}}{3\\text{!}},\\hfill \\\\ {p}_{4}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}+0=x-\\frac{{x}^{3}}{3\\text{!}},\\hfill \\\\ {p}_{5}\\left(x\\right)=0+x+0-\\frac{1}{3\\text{!}}{x}^{3}+0+\\frac{1}{5\\text{!}}{x}^{5}=x-\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}},\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nand for [latex]m\\ge 0[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025241412\" 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{!}}-\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m+1}}{\\left(2m+1\\right)\\text{!}}\\hfill \\\\ & ={\\displaystyle\\sum _{k=0}^{m}}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{\\left(2k+1\\right)\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nGraphs of the function and its Maclaurin polynomials are shown in Figure 3.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_03_003\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234515\/CNX_Calc_Figure_10_03_003.jpg\" alt=\"This graph has four curves. The first is the function f(x)=sin(x). The second function is psub1(x). The third is psub3(x). The fourth function is psub5(x). The curves are very close around x=0.\" width=\"487\" height=\"350\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The graph shows the function [latex]y=\\sin{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{1},{p}_{3}[\/latex] and [latex]{p}_{5}[\/latex].<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<li>For [latex]f\\left(x\\right)=\\cos{x}[\/latex], the values of the function and its first four derivatives at [latex]x=0[\/latex] are given as follows:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025069824\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\cos{x}\\hfill & & & \\hfill f\\left(0\\right)& =\\hfill & 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & -\\sin{x}\\hfill & & & \\hfill {f}^{\\prime }\\left(0\\right)& =\\hfill & 0\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & -\\cos{x}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(0\\right)& =\\hfill & -1\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & \\sin{x}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)& =\\hfill & 0\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)& =\\hfill & \\cos{x}\\hfill & & & \\hfill {f}^{\\left(4\\right)}\\left(0\\right)& =\\hfill & 1.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the fourth derivative is [latex]\\sin{x}[\/latex], the pattern repeats. In other words, [latex]{f}^{\\left(2m\\right)}\\left(0\\right)={\\left(-1\\right)}^{m}[\/latex] and [latex]{f}^{\\left(2m+1\\right)}=0[\/latex] for [latex]m\\ge 0[\/latex]. Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025087805\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{p}_{0}\\left(x\\right)=1,\\hfill \\\\ {p}_{1}\\left(x\\right)=1+0=1,\\hfill \\\\ {p}_{2}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}=1-\\frac{{x}^{2}}{2\\text{!}},\\hfill \\\\ {p}_{3}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0=1-\\frac{{x}^{2}}{2\\text{!}},\\hfill \\\\ {p}_{4}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0+\\frac{1}{4\\text{!}}{x}^{4}=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}},\\hfill \\\\ {p}_{5}\\left(x\\right)=1+0-\\frac{1}{2\\text{!}}{x}^{2}+0+\\frac{1}{4\\text{!}}{x}^{4}+0=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}},\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nand for [latex]n\\ge 0[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167025230186\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {p}_{2m}\\left(x\\right)& ={p}_{2m+1}\\left(x\\right)\\hfill \\\\ & =1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}-\\cdots +{\\left(-1\\right)}^{m}\\frac{{x}^{2m}}{\\left(2m\\right)\\text{!}}\\hfill \\\\ & ={\\displaystyle\\sum _{k=0}^{m}}{\\left(-1\\right)}^{k}\\frac{{x}^{2k}}{\\left(2k\\right)\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nGraphs of the function and the Maclaurin polynomials appear in Figure 4.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_03_004\">\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\/11234518\/CNX_Calc_Figure_10_03_004.jpg\" alt=\"This graph has four curves. The first is the function f(x)=cos(x). The second function is psub0(x). The third is psub2(x). The fourth function is psub4(x). The curves are very close around y=1\" width=\"487\" height=\"312\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The function [latex]y=\\cos{x}[\/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{2}[\/latex] and [latex]{p}_{4}[\/latex] are plotted on this graph.<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding 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=546&amp;end=1086&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\/6.3TaylorAndMaclaurinSeries546to1086_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-id1167025235544\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167025235548\" data-type=\"exercise\">\n<div id=\"fs-id1167025235550\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167025235550\" data-type=\"problem\">\n<p id=\"fs-id1167025235553\">Find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{1+x}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial. Write your anwer using sigma notation.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025102143\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025102151\">Evaluate the first four derivatives of [latex]f[\/latex] and look for a pattern.<\/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=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025168862\" data-type=\"solution\">\n<p id=\"fs-id1167025168865\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1-x[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1-x+{x}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{n}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}+\\cdots +{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}{x}^{k}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167025102163\" data-depth=\"1\"><\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1815\">\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":11,"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-1815","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1815","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1815\/revisions"}],"predecessor-version":[{"id":2721,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1815\/revisions\/2721"}],"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\/1815\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1815"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1815"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1815"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}