{"id":1814,"date":"2021-07-28T16:10:26","date_gmt":"2021-07-28T16:10:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1814"},"modified":"2022-03-21T23:14:05","modified_gmt":"2022-03-21T23:14:05","slug":"functions-as-maclaurin-and-taylor-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/functions-as-maclaurin-and-taylor-series\/","title":{"raw":"Functions as Maclaurin and Taylor Series","rendered":"Functions as Maclaurin and Taylor Series"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the terms of the binomial series<\/li>\r\n \t<li>Recognize the Taylor series expansions of common functions<\/li>\r\n \t<li>Recognize and apply techniques to find the Taylor series for a function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">The Binomial Series<\/h2>\r\n<p id=\"fs-id1167023795770\"><span style=\"font-size: 1rem; text-align: initial;\">Our first goal in this section is to determine the Maclaurin series for the function [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] for all real numbers [latex]r[\/latex]. The Maclaurin series for this function is known as the <\/span><strong style=\"font-size: 1rem; text-align: initial;\">binomial series<\/strong><span style=\"font-size: 1rem; text-align: initial;\">. We begin by considering the simplest case: [latex]r[\/latex] is a nonnegative integer. We recall that, for [latex]r=0,1,2,3,4,f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] can be written as<\/span><\/p>\r\n\r\n<section id=\"fs-id1167023802432\" data-depth=\"1\">\r\n<div id=\"fs-id1167023810404\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ f\\left(x\\right)={\\left(1+x\\right)}^{0}=1,\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{1}=1+x,\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{2}=1+2x+{x}^{2},\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{3}=1+3x+3{x}^{2}+{x}^{3},\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{4}=1+4x+6{x}^{2}+4{x}^{3}+{x}^{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023755079\">The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer [latex]r[\/latex], the binomial coefficient of [latex]{x}^{n}[\/latex] in the binomial expansion of [latex]{\\left(1+x\\right)}^{r}[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167023717125\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\text{!}}{n\\text{!}\\left(r-n\\right)\\text{!}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023773726\">and<\/p>\r\n\r\n<div id=\"fs-id1167023919069\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)&amp; ={\\left(1+x\\right)}^{r}\\hfill \\\\ &amp; =\\left(\\begin{array}{c}r\\hfill \\\\ 0\\hfill \\end{array}\\right)1+\\left(\\begin{array}{c}r\\hfill \\\\ 1\\hfill \\end{array}\\right)x+\\left(\\begin{array}{c}r\\hfill \\\\ 2\\hfill \\end{array}\\right){x}^{2}+\\left(\\begin{array}{c}r\\hfill \\\\ 3\\hfill \\end{array}\\right){x}^{3}+\\cdots +\\left(\\begin{array}{c}r\\hfill \\\\ r - 1\\hfill \\end{array}\\right){x}^{r - 1}+\\left(\\begin{array}{c}r\\hfill \\\\ r\\hfill \\end{array}\\right){x}^{r}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{r}}\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023808491\">For example, using this formula for [latex]r=5[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1167023731125\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)&amp; ={\\left(1+x\\right)}^{5}\\hfill \\\\ &amp; =\\left(\\begin{array}{c}5\\hfill \\\\ 0\\hfill \\end{array}\\right)1+\\left(\\begin{array}{c}5\\hfill \\\\ 1\\hfill \\end{array}\\right)x+\\left(\\begin{array}{c}5\\hfill \\\\ 2\\hfill \\end{array}\\right){x}^{2}+\\left(\\begin{array}{c}5\\hfill \\\\ 3\\hfill \\end{array}\\right){x}^{3}+\\left(\\begin{array}{c}5\\hfill \\\\ 4\\hfill \\end{array}\\right){x}^{4}+\\left(\\begin{array}{c}5\\hfill \\\\ 5\\hfill \\end{array}\\right){x}^{5}\\hfill \\\\ &amp; =\\frac{5\\text{!}}{0\\text{!}5\\text{!}}1+\\frac{5\\text{!}}{1\\text{!}4\\text{!}}x+\\frac{5\\text{!}}{2\\text{!}3\\text{!}}{x}^{2}+\\frac{5\\text{!}}{3\\text{!}2\\text{!}}{x}^{3}+\\frac{5\\text{!}}{4\\text{!}1\\text{!}}{x}^{4}+\\frac{5\\text{!}}{5\\text{!}0\\text{!}}{x}^{5}\\hfill \\\\ &amp; =1+5x+10{x}^{2}+10{x}^{3}+5{x}^{4}+{x}^{5}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023772884\">We now consider the case when the exponent [latex]r[\/latex] is any real number, not necessarily a nonnegative integer. If [latex]r[\/latex] is not a nonnegative integer, then [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] cannot be written as a finite polynomial. However, we can find a power series for [latex]f[\/latex]. Specifically, we look for the Maclaurin series for [latex]f[\/latex]. To do this, we find the derivatives of [latex]f[\/latex] and evaluate them at [latex]x=0[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167023788120\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)&amp; =\\hfill &amp; {\\left(1+x\\right)}^{r}\\hfill &amp; &amp; &amp; \\hfill f\\left(0\\right)&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)&amp; =\\hfill &amp; r{\\left(1+x\\right)}^{r - 1}\\hfill &amp; &amp; &amp; \\hfill f\\prime \\left(0\\right)&amp; =\\hfill &amp; r\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right){\\left(1+x\\right)}^{r - 2}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right)\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right)\\left(r - 2\\right){\\left(1+x\\right)}^{r - 3}\\hfill &amp; &amp; &amp; \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right)\\left(r - 2\\right)\\hfill \\\\ \\hfill {f}^{\\left(n\\right)}\\left(x\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right){\\left(1+x\\right)}^{r-n}\\hfill &amp; &amp; &amp; \\hfill {f}^{\\left(n\\right)}\\left(0\\right)&amp; =\\hfill &amp; r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023896718\">We conclude that the coefficients in the binomial series are given by<\/p>\r\n\r\n<div id=\"fs-id1167023896721\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}=\\frac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023863068\">We note that if [latex]r[\/latex] is a nonnegative integer, then the [latex]\\left(r+1\\right)\\text{st}[\/latex] derivative [latex]{f}^{\\left(r+1\\right)}[\/latex] is the zero function, and the series terminates. In addition, if [latex]r[\/latex] is a nonnegative integer, then the above equation for the coefficients agrees with [latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\text{!}}{n\\text{!}\\left(r-n\\right)\\text{!}}[\/latex] for the coefficients, and the formula for the binomial series agrees with the equation that follows for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number [latex]r[\/latex], we define<\/p>\r\n\r\n<div id=\"fs-id1167024038814\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024046788\">With this notation, we can write the binomial series for [latex]{\\left(1+x\\right)}^{r}[\/latex] as<\/p>\r\n\r\n<div id=\"fs-id1167024036610\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}=1+rx+\\frac{r\\left(r - 1\\right)}{2\\text{!}}{x}^{2}+\\cdots +\\frac{r\\left(r - 1\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}{x}^{n}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023862661\">We now need to determine the interval of convergence for the binomial series of the above equation. We apply the ratio test. Consequently, we consider<\/p>\r\n\r\n<div id=\"fs-id1167023862667\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{|{a}_{n+1}|}{|{a}_{n}|}&amp; =\\frac{{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n\\right)|x||}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)|{|x|}^{n}}\\hfill \\\\ &amp; =\\frac{|r-n||x|}{|n+1|}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024047434\">Since<\/p>\r\n\r\n<div id=\"fs-id1167024047437\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{|{a}_{n+1}|}{|{a}_{n}|}=|x|&lt;1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023811485\">if and only if [latex]|x|&lt;1[\/latex], we conclude that the interval of convergence for the binomial series is [latex]\\left(-1,1\\right)[\/latex]. The behavior at the endpoints depends on [latex]r[\/latex]. It can be shown that for [latex]r\\ge 0[\/latex] the series converges at both endpoints; for [latex]-1&lt;r&lt;0[\/latex], the series converges at [latex]x=1[\/latex] and diverges at [latex]x=-1[\/latex]; and for [latex]r&lt;-1[\/latex], the series diverges at both endpoints. The binomial series does converge to [latex]{\\left(1+x\\right)}^{r}[\/latex] in [latex]\\left(-1,1\\right)[\/latex] for all real numbers [latex]r[\/latex], but proving this fact by showing that the remainder [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex] is difficult.<\/p>\r\n\r\n<div id=\"fs-id1167023778748\" 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-id1167023778752\">For any real number [latex]r[\/latex], the Maclaurin series for [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] is the binomial series. It converges to [latex]f[\/latex] for [latex]|x|&lt;1[\/latex], and we write<\/p>\r\n\r\n<div id=\"fs-id1167023823038\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\left(1+x\\right)}^{r}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}\\hfill \\\\ &amp; =1+rx+\\frac{r\\left(r - 1\\right)}{2\\text{!}}{x}^{2}+\\cdots +\\frac{r\\left(r - 1\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}{x}^{n}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023870789\">for [latex]|x|&lt;1[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167023870809\">We can use this definition to find the binomial series for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex] and use the series to approximate [latex]\\sqrt{1.5}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167023918533\" data-type=\"example\">\r\n<div id=\"fs-id1167023918535\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023918537\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Binomial Series<\/h3>\r\n<div id=\"fs-id1167023918537\" data-type=\"problem\">\r\n<ol id=\"fs-id1167023918542\" type=\"a\">\r\n \t<li>Find the binomial series for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex].<\/li>\r\n \t<li>Use the third-order Maclaurin polynomial [latex]{p}_{3}\\left(x\\right)[\/latex] to estimate [latex]\\sqrt{1.5}[\/latex]. Use Taylor\u2019s theorem to bound the error. Use a graphing utility to compare the graphs of [latex]f[\/latex] and [latex]{p}_{3}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167023809878\" data-type=\"solution\">\r\n<ol id=\"fs-id1167023809880\" type=\"a\">\r\n \t<li>Here [latex]r=\\frac{1}{2}[\/latex]. Using the definition for the binomial series, we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023809903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sqrt{1+x}&amp; =1+\\frac{1}{2}x+\\frac{\\left(1\\text{\/}2\\right)\\left(\\text{-}1\\text{\/}2\\right)}{2\\text{!}}{x}^{2}+\\frac{\\left(1\\text{\/}2\\right)\\left(\\text{-}1\\text{\/}2\\right)\\left(\\text{-}3\\text{\/}2\\right)}{3\\text{!}}{x}^{3}+\\cdots \\hfill \\\\ &amp; =1+\\frac{1}{2}x-\\frac{1}{2\\text{!}}\\frac{1}{{2}^{2}}{x}^{2}+\\frac{1}{3\\text{!}}\\frac{1\\cdot 3}{{2}^{3}}{x}^{3}-\\cdots +\\frac{{\\left(-1\\right)}^{n+1}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 3\\right)}{{2}^{n}}{x}^{n}+\\cdots \\hfill \\\\ &amp; =1+{\\displaystyle\\sum _{n=1}^{\\infty}}\\frac{{\\left(-1\\right)}^{n+1}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 3\\right)}{{2}^{n}}{x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>From the result in part a. the third-order Maclaurin polynomial is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023809917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{3}\\left(x\\right)=1+\\frac{1}{2}x-\\frac{1}{8}{x}^{2}+\\frac{1}{16}{x}^{3}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023809982\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sqrt{1.5}&amp; =\\sqrt{1+0.5}\\hfill \\\\ &amp; \\approx 1+\\frac{1}{2}\\left(0.5\\right)-\\frac{1}{8}{\\left(0.5\\right)}^{2}+\\frac{1}{16}{\\left(0.5\\right)}^{3}\\hfill \\\\ &amp; \\approx 1.2266.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFrom Taylor\u2019s theorem, the error satisfies<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023828674\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{3}\\left(0.5\\right)=\\frac{{f}^{\\left(4\\right)}\\left(c\\right)}{4\\text{!}}{\\left(0.5\\right)}^{4}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor some [latex]c[\/latex] between [latex]0[\/latex] and [latex]0.5[\/latex]. Since [latex]{f}^{\\left(4\\right)}\\left(x\\right)=-\\frac{15}{{2}^{4}{\\left(1+x\\right)}^{\\frac{7}{2}}}[\/latex], and the maximum value of [latex]|{f}^{\\left(4\\right)}\\left(x\\right)|[\/latex] on the interval [latex]\\left(0,0.5\\right)[\/latex] occurs at [latex]x=0[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024042978\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{3}\\left(0.5\\right)|\\le \\frac{15}{4\\text{!}{2}^{4}}{\\left(0.5\\right)}^{4}\\approx 0.00244[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe function and the Maclaurin polynomial [latex]{p}_{3}[\/latex] are graphed in Figure 1.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_10_04_001\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234536\/CNX_Calc_Figure_10_04_001.jpg\" alt=\"This graph has two curves. The first one is f(x)= the square root of (1+x) and the second is psub3(x). The curves are very close at y = 1.\" width=\"487\" height=\"208\" data-media-type=\"image\/jpeg\" \/> Figure 1. The third-order Maclaurin polynomial [latex]{p}_{3}\\left(x\\right)[\/latex] provides a good approximation for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex] for [latex]x[\/latex] near zero.[\/caption]<\/figure>\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: Finding Binomial Series.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6724921&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nWgiXcz2aL0&amp;video_target=tpm-plugin-lxhj90gz-nWgiXcz2aL0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.4.1\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167024040774\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167024040777\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\r\n<p id=\"fs-id1167024040781\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{2}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167024044103\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024044109\">Use the definition of binomial series for [latex]r=-2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167024044048\" data-type=\"solution\">\r\n<p id=\"fs-id1167024044050\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/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]169468[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1167024044127\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Common Functions Expressed as Taylor Series<\/h2>\r\n<p id=\"fs-id1167024044133\">At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]. In the table below, we summarize the results of these series. We remark that the convergence of the Maclaurin series for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] at the endpoint [latex]x=1[\/latex] and the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] at the endpoints [latex]x=1[\/latex] and [latex]x=-1[\/latex] relies on a more advanced theorem than we present here. (Refer to Abel\u2019s theorem for a discussion of this more technical point.)<\/p>\r\n\r\n<table id=\"fs-id1167024047165\" summary=\"This table has seven rows and three columns. The header row labels the columns \"><caption><span data-type=\"title\">Maclaurin Series for Common Functions<\/span><\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th data-align=\"left\">Function<\/th>\r\n<th data-align=\"left\">Maclaurin Series<\/th>\r\n<th data-align=\"left\">Interval of Convergence<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]-1&lt;x&lt;1[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)={e}^{x}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\text{-}\\infty &lt;x&lt;\\infty [\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\sin{x}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\text{-}\\infty &lt;x&lt;\\infty [\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\cos{x}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\text{-}\\infty &lt;x&lt;\\infty [\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]-1&lt;x\\le 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{2n+1}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]-1&lt;x\\le 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]-1&lt;x&lt;1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1167024038494\">Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in <a class=\"autogenerated-content\" href=\"#fs-id1167024047165\">[link]<\/a>, to create Maclaurin series for other functions.<\/p>\r\n\r\n<div id=\"fs-id1167024038502\" data-type=\"example\">\r\n<div id=\"fs-id1167024038504\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024038506\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Deriving Maclaurin Series from Known Series<\/h3>\r\n<div id=\"fs-id1167024038506\" data-type=\"problem\">\r\n<p id=\"fs-id1167024038511\">Find the Maclaurin series of each of the following functions by using one of the series listed in the table.<\/p>\r\n\r\n<ol id=\"fs-id1167024038518\" type=\"a\">\r\n \t<li>[latex]f\\left(x\\right)=\\cos\\sqrt{x}[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)=\\text{sinh}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<div id=\"fs-id1167024038568\" data-type=\"solution\">\r\n<ol id=\"fs-id1167024038570\" type=\"a\">\r\n \t<li>Using the Maclaurin series for [latex]\\cos{x}[\/latex] we find that the Maclaurin series for [latex]\\cos\\sqrt{x}[\/latex] is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024044163\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{\\left(\\sqrt{x}\\right)}^{2n}}{\\left(2n\\right)\\text{!}}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{x}^{n}}{\\left(2n\\right)\\text{!}}\\hfill \\\\ &amp; =1-\\frac{x}{2\\text{!}}+\\frac{{x}^{2}}{4\\text{!}}-\\frac{{x}^{3}}{6\\text{!}}+\\frac{{x}^{4}}{8\\text{!}}-\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThis series converges to [latex]\\cos\\sqrt{x}[\/latex] for all [latex]x[\/latex] in the domain of [latex]\\cos\\sqrt{x}[\/latex]; that is, for all [latex]x\\ge 0[\/latex].<\/li>\r\n \t<li>To find the Maclaurin series for [latex]\\text{sinh}x[\/latex], we use the fact that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023775252\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{sinh}x=\\frac{{e}^{x}-{e}^{\\text{-}x}}{2}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUsing the Maclaurin series for [latex]{e}^{x}[\/latex], we see that the [latex]n\\text{th}[\/latex] term in the Maclaurin series for [latex]\\text{sinh}x[\/latex] is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024038392\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{n}}{n\\text{!}}-\\frac{{\\left(\\text{-}x\\right)}^{n}}{n\\text{!}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFor [latex]n[\/latex] even, this term is zero. For [latex]n[\/latex] odd, this term is [latex]\\frac{2{x}^{n}}{n\\text{!}}[\/latex]. Therefore, the Maclaurin series for [latex]\\text{sinh}x[\/latex] has only odd-order terms and is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167024047295\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}=x+\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}+\\cdots [\/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: Deriving Maclaurin Series from Known Series.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6724922&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=bpMzrl8U7Lw&amp;video_target=tpm-plugin-iro0wr30-bpMzrl8U7Lw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.3_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.4.3\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167024047395\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167024047399\" data-type=\"exercise\">\r\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\r\n<p id=\"fs-id1167024047404\">Find the Maclaurin series for [latex]\\sin\\left({x}^{2}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167024041386\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024041393\">Use the Maclaurin series for [latex]\\sin{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167024041315\" data-type=\"solution\">\r\n<p id=\"fs-id1167024041317\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/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]169473[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167024041411\">We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In the next example, we differentiate the binomial series for [latex]\\sqrt{1+x}[\/latex] term by term to find the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex]. Note that we could construct the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex] directly from the definition, but differentiating the binomial series for [latex]\\sqrt{1+x}[\/latex] is an easier calculation.<\/p>\r\n\r\n<div id=\"fs-id1167023802476\" data-type=\"example\">\r\n<div id=\"fs-id1167023802478\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023802480\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Differentiating a Series to Find a New Series<\/h3>\r\n<div id=\"fs-id1167023802480\" data-type=\"problem\">\r\n<p id=\"fs-id1167023802485\">Use the binomial series for [latex]\\sqrt{1+x}[\/latex] to find the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1167023802518\" data-type=\"solution\">\r\n<p id=\"fs-id1167023802520\">The two functions are related by<\/p>\r\n\r\n<div id=\"fs-id1167023802524\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\sqrt{1+x}=\\frac{1}{2\\sqrt{1+x}}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023802566\">so the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167023777262\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{\\sqrt{1+x}}&amp; =2\\frac{d}{dx}\\sqrt{1+x}\\hfill \\\\ &amp; =1+{\\displaystyle\\sum _{n=1}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167023914385\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023914389\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\r\n<p id=\"fs-id1167023914394\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{\\frac{3}{2}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1167024036470\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167024036478\">Differentiate the series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1167023914438\" data-type=\"solution\">\r\n<p id=\"fs-id1167023914440\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167024036501\">In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.<\/p>\r\n\r\n<\/section><section id=\"fs-id1167024036508\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the terms of the binomial series<\/li>\n<li>Recognize the Taylor series expansions of common functions<\/li>\n<li>Recognize and apply techniques to find the Taylor series for a function<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">The Binomial Series<\/h2>\n<p id=\"fs-id1167023795770\"><span style=\"font-size: 1rem; text-align: initial;\">Our first goal in this section is to determine the Maclaurin series for the function [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] for all real numbers [latex]r[\/latex]. The Maclaurin series for this function is known as the <\/span><strong style=\"font-size: 1rem; text-align: initial;\">binomial series<\/strong><span style=\"font-size: 1rem; text-align: initial;\">. We begin by considering the simplest case: [latex]r[\/latex] is a nonnegative integer. We recall that, for [latex]r=0,1,2,3,4,f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] can be written as<\/span><\/p>\n<section id=\"fs-id1167023802432\" data-depth=\"1\">\n<div id=\"fs-id1167023810404\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ f\\left(x\\right)={\\left(1+x\\right)}^{0}=1,\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{1}=1+x,\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{2}=1+2x+{x}^{2},\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{3}=1+3x+3{x}^{2}+{x}^{3},\\hfill \\\\ f\\left(x\\right)={\\left(1+x\\right)}^{4}=1+4x+6{x}^{2}+4{x}^{3}+{x}^{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023755079\">The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer [latex]r[\/latex], the binomial coefficient of [latex]{x}^{n}[\/latex] in the binomial expansion of [latex]{\\left(1+x\\right)}^{r}[\/latex] is given by<\/p>\n<div id=\"fs-id1167023717125\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\text{!}}{n\\text{!}\\left(r-n\\right)\\text{!}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023773726\">and<\/p>\n<div id=\"fs-id1167023919069\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={\\left(1+x\\right)}^{r}\\hfill \\\\ & =\\left(\\begin{array}{c}r\\hfill \\\\ 0\\hfill \\end{array}\\right)1+\\left(\\begin{array}{c}r\\hfill \\\\ 1\\hfill \\end{array}\\right)x+\\left(\\begin{array}{c}r\\hfill \\\\ 2\\hfill \\end{array}\\right){x}^{2}+\\left(\\begin{array}{c}r\\hfill \\\\ 3\\hfill \\end{array}\\right){x}^{3}+\\cdots +\\left(\\begin{array}{c}r\\hfill \\\\ r - 1\\hfill \\end{array}\\right){x}^{r - 1}+\\left(\\begin{array}{c}r\\hfill \\\\ r\\hfill \\end{array}\\right){x}^{r}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{r}}\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023808491\">For example, using this formula for [latex]r=5[\/latex], we see that<\/p>\n<div id=\"fs-id1167023731125\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={\\left(1+x\\right)}^{5}\\hfill \\\\ & =\\left(\\begin{array}{c}5\\hfill \\\\ 0\\hfill \\end{array}\\right)1+\\left(\\begin{array}{c}5\\hfill \\\\ 1\\hfill \\end{array}\\right)x+\\left(\\begin{array}{c}5\\hfill \\\\ 2\\hfill \\end{array}\\right){x}^{2}+\\left(\\begin{array}{c}5\\hfill \\\\ 3\\hfill \\end{array}\\right){x}^{3}+\\left(\\begin{array}{c}5\\hfill \\\\ 4\\hfill \\end{array}\\right){x}^{4}+\\left(\\begin{array}{c}5\\hfill \\\\ 5\\hfill \\end{array}\\right){x}^{5}\\hfill \\\\ & =\\frac{5\\text{!}}{0\\text{!}5\\text{!}}1+\\frac{5\\text{!}}{1\\text{!}4\\text{!}}x+\\frac{5\\text{!}}{2\\text{!}3\\text{!}}{x}^{2}+\\frac{5\\text{!}}{3\\text{!}2\\text{!}}{x}^{3}+\\frac{5\\text{!}}{4\\text{!}1\\text{!}}{x}^{4}+\\frac{5\\text{!}}{5\\text{!}0\\text{!}}{x}^{5}\\hfill \\\\ & =1+5x+10{x}^{2}+10{x}^{3}+5{x}^{4}+{x}^{5}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023772884\">We now consider the case when the exponent [latex]r[\/latex] is any real number, not necessarily a nonnegative integer. If [latex]r[\/latex] is not a nonnegative integer, then [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] cannot be written as a finite polynomial. However, we can find a power series for [latex]f[\/latex]. Specifically, we look for the Maclaurin series for [latex]f[\/latex]. To do this, we find the derivatives of [latex]f[\/latex] and evaluate them at [latex]x=0[\/latex].<\/p>\n<div id=\"fs-id1167023788120\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & {\\left(1+x\\right)}^{r}\\hfill & & & \\hfill f\\left(0\\right)& =\\hfill & 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & r{\\left(1+x\\right)}^{r - 1}\\hfill & & & \\hfill f\\prime \\left(0\\right)& =\\hfill & r\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & r\\left(r - 1\\right){\\left(1+x\\right)}^{r - 2}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(0\\right)& =\\hfill & r\\left(r - 1\\right)\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & r\\left(r - 1\\right)\\left(r - 2\\right){\\left(1+x\\right)}^{r - 3}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(0\\right)& =\\hfill & r\\left(r - 1\\right)\\left(r - 2\\right)\\hfill \\\\ \\hfill {f}^{\\left(n\\right)}\\left(x\\right)& =\\hfill & r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right){\\left(1+x\\right)}^{r-n}\\hfill & & & \\hfill {f}^{\\left(n\\right)}\\left(0\\right)& =\\hfill & r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023896718\">We conclude that the coefficients in the binomial series are given by<\/p>\n<div id=\"fs-id1167023896721\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}=\\frac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023863068\">We note that if [latex]r[\/latex] is a nonnegative integer, then the [latex]\\left(r+1\\right)\\text{st}[\/latex] derivative [latex]{f}^{\\left(r+1\\right)}[\/latex] is the zero function, and the series terminates. In addition, if [latex]r[\/latex] is a nonnegative integer, then the above equation for the coefficients agrees with [latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\text{!}}{n\\text{!}\\left(r-n\\right)\\text{!}}[\/latex] for the coefficients, and the formula for the binomial series agrees with the equation that follows for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number [latex]r[\/latex], we define<\/p>\n<div id=\"fs-id1167024038814\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right)=\\frac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024046788\">With this notation, we can write the binomial series for [latex]{\\left(1+x\\right)}^{r}[\/latex] as<\/p>\n<div id=\"fs-id1167024036610\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}=1+rx+\\frac{r\\left(r - 1\\right)}{2\\text{!}}{x}^{2}+\\cdots +\\frac{r\\left(r - 1\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}{x}^{n}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023862661\">We now need to determine the interval of convergence for the binomial series of the above equation. We apply the ratio test. Consequently, we consider<\/p>\n<div id=\"fs-id1167023862667\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{|{a}_{n+1}|}{|{a}_{n}|}& =\\frac{{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n\\right)|x||}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)|{|x|}^{n}}\\hfill \\\\ & =\\frac{|r-n||x|}{|n+1|}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024047434\">Since<\/p>\n<div id=\"fs-id1167024047437\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{|{a}_{n+1}|}{|{a}_{n}|}=|x|<1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023811485\">if and only if [latex]|x|<1[\/latex], we conclude that the interval of convergence for the binomial series is [latex]\\left(-1,1\\right)[\/latex]. The behavior at the endpoints depends on [latex]r[\/latex]. It can be shown that for [latex]r\\ge 0[\/latex] the series converges at both endpoints; for [latex]-1<r<0[\/latex], the series converges at [latex]x=1[\/latex] and diverges at [latex]x=-1[\/latex]; and for [latex]r<-1[\/latex], the series diverges at both endpoints. The binomial series does converge to [latex]{\\left(1+x\\right)}^{r}[\/latex] in [latex]\\left(-1,1\\right)[\/latex] for all real numbers [latex]r[\/latex], but proving this fact by showing that the remainder [latex]{R}_{n}\\left(x\\right)\\to 0[\/latex] is difficult.<\/p>\n<div id=\"fs-id1167023778748\" 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-id1167023778752\">For any real number [latex]r[\/latex], the Maclaurin series for [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] is the binomial series. It converges to [latex]f[\/latex] for [latex]|x|<1[\/latex], and we write<\/p>\n<div id=\"fs-id1167023823038\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\left(1+x\\right)}^{r}& ={\\displaystyle\\sum _{n=0}^{\\infty}}\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}\\hfill \\\\ & =1+rx+\\frac{r\\left(r - 1\\right)}{2\\text{!}}{x}^{2}+\\cdots +\\frac{r\\left(r - 1\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}{x}^{n}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023870789\">for [latex]|x|<1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023870809\">We can use this definition to find the binomial series for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex] and use the series to approximate [latex]\\sqrt{1.5}[\/latex].<\/p>\n<div id=\"fs-id1167023918533\" data-type=\"example\">\n<div id=\"fs-id1167023918535\" data-type=\"exercise\">\n<div id=\"fs-id1167023918537\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding Binomial Series<\/h3>\n<div id=\"fs-id1167023918537\" data-type=\"problem\">\n<ol id=\"fs-id1167023918542\" type=\"a\">\n<li>Find the binomial series for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex].<\/li>\n<li>Use the third-order Maclaurin polynomial [latex]{p}_{3}\\left(x\\right)[\/latex] to estimate [latex]\\sqrt{1.5}[\/latex]. Use Taylor\u2019s theorem to bound the error. Use a graphing utility to compare the graphs of [latex]f[\/latex] and [latex]{p}_{3}[\/latex].<\/li>\n<\/ol>\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-id1167023809878\" data-type=\"solution\">\n<ol id=\"fs-id1167023809880\" type=\"a\">\n<li>Here [latex]r=\\frac{1}{2}[\/latex]. Using the definition for the binomial series, we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023809903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sqrt{1+x}& =1+\\frac{1}{2}x+\\frac{\\left(1\\text{\/}2\\right)\\left(\\text{-}1\\text{\/}2\\right)}{2\\text{!}}{x}^{2}+\\frac{\\left(1\\text{\/}2\\right)\\left(\\text{-}1\\text{\/}2\\right)\\left(\\text{-}3\\text{\/}2\\right)}{3\\text{!}}{x}^{3}+\\cdots \\hfill \\\\ & =1+\\frac{1}{2}x-\\frac{1}{2\\text{!}}\\frac{1}{{2}^{2}}{x}^{2}+\\frac{1}{3\\text{!}}\\frac{1\\cdot 3}{{2}^{3}}{x}^{3}-\\cdots +\\frac{{\\left(-1\\right)}^{n+1}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 3\\right)}{{2}^{n}}{x}^{n}+\\cdots \\hfill \\\\ & =1+{\\displaystyle\\sum _{n=1}^{\\infty}}\\frac{{\\left(-1\\right)}^{n+1}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 3\\right)}{{2}^{n}}{x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>From the result in part a. the third-order Maclaurin polynomial is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023809917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{p}_{3}\\left(x\\right)=1+\\frac{1}{2}x-\\frac{1}{8}{x}^{2}+\\frac{1}{16}{x}^{3}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023809982\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\sqrt{1.5}& =\\sqrt{1+0.5}\\hfill \\\\ & \\approx 1+\\frac{1}{2}\\left(0.5\\right)-\\frac{1}{8}{\\left(0.5\\right)}^{2}+\\frac{1}{16}{\\left(0.5\\right)}^{3}\\hfill \\\\ & \\approx 1.2266.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFrom Taylor\u2019s theorem, the error satisfies<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023828674\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{3}\\left(0.5\\right)=\\frac{{f}^{\\left(4\\right)}\\left(c\\right)}{4\\text{!}}{\\left(0.5\\right)}^{4}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor some [latex]c[\/latex] between [latex]0[\/latex] and [latex]0.5[\/latex]. Since [latex]{f}^{\\left(4\\right)}\\left(x\\right)=-\\frac{15}{{2}^{4}{\\left(1+x\\right)}^{\\frac{7}{2}}}[\/latex], and the maximum value of [latex]|{f}^{\\left(4\\right)}\\left(x\\right)|[\/latex] on the interval [latex]\\left(0,0.5\\right)[\/latex] occurs at [latex]x=0[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024042978\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{3}\\left(0.5\\right)|\\le \\frac{15}{4\\text{!}{2}^{4}}{\\left(0.5\\right)}^{4}\\approx 0.00244[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe function and the Maclaurin polynomial [latex]{p}_{3}[\/latex] are graphed in Figure 1.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_10_04_001\">\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\/11234536\/CNX_Calc_Figure_10_04_001.jpg\" alt=\"This graph has two curves. The first one is f(x)= the square root of (1+x) and the second is psub3(x). The curves are very close at y = 1.\" width=\"487\" height=\"208\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The third-order Maclaurin polynomial [latex]{p}_{3}\\left(x\\right)[\/latex] provides a good approximation for [latex]f\\left(x\\right)=\\sqrt{1+x}[\/latex] for [latex]x[\/latex] near zero.<\/p>\n<\/div>\n<\/figure>\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: Finding Binomial Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6724921&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=nWgiXcz2aL0&amp;video_target=tpm-plugin-lxhj90gz-nWgiXcz2aL0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.4.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167024040774\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167024040777\" data-type=\"exercise\">\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167024040779\" data-type=\"problem\">\n<p id=\"fs-id1167024040781\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{2}}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><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-id1167024044103\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024044109\">Use the definition of binomial series for [latex]r=-2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><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-id1167024044048\" data-type=\"solution\">\n<p id=\"fs-id1167024044050\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169468\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169468&theme=oea&iframe_resize_id=ohm169468&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1167024044127\" data-depth=\"1\">\n<h2 data-type=\"title\">Common Functions Expressed as Taylor Series<\/h2>\n<p id=\"fs-id1167024044133\">At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]. In the table below, we summarize the results of these series. We remark that the convergence of the Maclaurin series for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] at the endpoint [latex]x=1[\/latex] and the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] at the endpoints [latex]x=1[\/latex] and [latex]x=-1[\/latex] relies on a more advanced theorem than we present here. (Refer to Abel\u2019s theorem for a discussion of this more technical point.)<\/p>\n<table id=\"fs-id1167024047165\" summary=\"This table has seven rows and three columns. The header row labels the columns\">\n<caption><span data-type=\"title\">Maclaurin Series for Common Functions<\/span><\/caption>\n<thead>\n<tr valign=\"top\">\n<th data-align=\"left\">Function<\/th>\n<th data-align=\"left\">Maclaurin Series<\/th>\n<th data-align=\"left\">Interval of Convergence<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex]<\/td>\n<td data-align=\"left\">[latex]-1<x<1[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)={e}^{x}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\text{-}\\infty <x<\\infty[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\sin{x}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\text{-}\\infty <x<\\infty[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\cos{x}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\text{-}\\infty <x<\\infty[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}[\/latex]<\/td>\n<td data-align=\"left\">[latex]-1<x\\le 1[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{2n+1}[\/latex]<\/td>\n<td data-align=\"left\">[latex]-1<x\\le 1[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/td>\n<td data-align=\"left\">[latex]-1<x<1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1167024038494\">Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in <a class=\"autogenerated-content\" href=\"#fs-id1167024047165\">[link]<\/a>, to create Maclaurin series for other functions.<\/p>\n<div id=\"fs-id1167024038502\" data-type=\"example\">\n<div id=\"fs-id1167024038504\" data-type=\"exercise\">\n<div id=\"fs-id1167024038506\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Deriving Maclaurin Series from Known Series<\/h3>\n<div id=\"fs-id1167024038506\" data-type=\"problem\">\n<p id=\"fs-id1167024038511\">Find the Maclaurin series of each of the following functions by using one of the series listed in the table.<\/p>\n<ol id=\"fs-id1167024038518\" type=\"a\">\n<li>[latex]f\\left(x\\right)=\\cos\\sqrt{x}[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)=\\text{sinh}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<div id=\"fs-id1167024038568\" data-type=\"solution\">\n<ol id=\"fs-id1167024038570\" type=\"a\">\n<li>Using the Maclaurin series for [latex]\\cos{x}[\/latex] we find that the Maclaurin series for [latex]\\cos\\sqrt{x}[\/latex] is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024044163\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{\\left(\\sqrt{x}\\right)}^{2n}}{\\left(2n\\right)\\text{!}}& ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{x}^{n}}{\\left(2n\\right)\\text{!}}\\hfill \\\\ & =1-\\frac{x}{2\\text{!}}+\\frac{{x}^{2}}{4\\text{!}}-\\frac{{x}^{3}}{6\\text{!}}+\\frac{{x}^{4}}{8\\text{!}}-\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis series converges to [latex]\\cos\\sqrt{x}[\/latex] for all [latex]x[\/latex] in the domain of [latex]\\cos\\sqrt{x}[\/latex]; that is, for all [latex]x\\ge 0[\/latex].<\/li>\n<li>To find the Maclaurin series for [latex]\\text{sinh}x[\/latex], we use the fact that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023775252\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{sinh}x=\\frac{{e}^{x}-{e}^{\\text{-}x}}{2}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUsing the Maclaurin series for [latex]{e}^{x}[\/latex], we see that the [latex]n\\text{th}[\/latex] term in the Maclaurin series for [latex]\\text{sinh}x[\/latex] is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024038392\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{n}}{n\\text{!}}-\\frac{{\\left(\\text{-}x\\right)}^{n}}{n\\text{!}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor [latex]n[\/latex] even, this term is zero. For [latex]n[\/latex] odd, this term is [latex]\\frac{2{x}^{n}}{n\\text{!}}[\/latex]. Therefore, the Maclaurin series for [latex]\\text{sinh}x[\/latex] has only odd-order terms and is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167024047295\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}=x+\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}+\\cdots[\/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: Deriving Maclaurin Series from Known Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6724922&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=bpMzrl8U7Lw&amp;video_target=tpm-plugin-iro0wr30-bpMzrl8U7Lw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.3_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.4.3&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167024047395\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167024047399\" data-type=\"exercise\">\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167024047401\" data-type=\"problem\">\n<p id=\"fs-id1167024047404\">Find the Maclaurin series for [latex]\\sin\\left({x}^{2}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><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-id1167024041386\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024041393\">Use the Maclaurin series for [latex]\\sin{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><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-id1167024041315\" data-type=\"solution\">\n<p id=\"fs-id1167024041317\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169473\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169473&theme=oea&iframe_resize_id=ohm169473&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167024041411\">We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In the next example, we differentiate the binomial series for [latex]\\sqrt{1+x}[\/latex] term by term to find the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex]. Note that we could construct the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex] directly from the definition, but differentiating the binomial series for [latex]\\sqrt{1+x}[\/latex] is an easier calculation.<\/p>\n<div id=\"fs-id1167023802476\" data-type=\"example\">\n<div id=\"fs-id1167023802478\" data-type=\"exercise\">\n<div id=\"fs-id1167023802480\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Differentiating a Series to Find a New Series<\/h3>\n<div id=\"fs-id1167023802480\" data-type=\"problem\">\n<p id=\"fs-id1167023802485\">Use the binomial series for [latex]\\sqrt{1+x}[\/latex] to find the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023802518\" data-type=\"solution\">\n<p id=\"fs-id1167023802520\">The two functions are related by<\/p>\n<div id=\"fs-id1167023802524\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\sqrt{1+x}=\\frac{1}{2\\sqrt{1+x}}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023802566\">so the binomial series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex] is given by<\/p>\n<div id=\"fs-id1167023777262\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{\\sqrt{1+x}}& =2\\frac{d}{dx}\\sqrt{1+x}\\hfill \\\\ & =1+{\\displaystyle\\sum _{n=1}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023914385\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023914389\" data-type=\"exercise\">\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023914391\" data-type=\"problem\">\n<p id=\"fs-id1167023914394\">Find the binomial series for [latex]f\\left(x\\right)=\\frac{1}{{\\left(1+x\\right)}^{\\frac{3}{2}}}[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Hint<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024036470\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167024036478\">Differentiate the series for [latex]\\frac{1}{\\sqrt{1+x}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023914438\" data-type=\"solution\">\n<p id=\"fs-id1167023914440\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n\\text{!}}\\frac{1\\cdot 3\\cdot 5\\cdots \\left(2n - 1\\right)}{{2}^{n}}{x}^{n}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167024036501\">In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.<\/p>\n<\/section>\n<section id=\"fs-id1167024036508\" data-depth=\"1\"><\/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-1814\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>6.4.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>6.4.3. <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":15,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"6.4.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"6.4.3\",\"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-1814","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1814","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":5,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1814\/revisions"}],"predecessor-version":[{"id":2252,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1814\/revisions\/2252"}],"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\/1814\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1814"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1814"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1814"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}