{"id":957,"date":"2025-06-20T17:24:59","date_gmt":"2025-06-20T17:24:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=957"},"modified":"2025-09-12T14:32:26","modified_gmt":"2025-09-12T14:32:26","slug":"applications-of-series-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-series-learn-it-1\/","title":{"raw":"Applications of Series: Learn It 1","rendered":"Applications of Series: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write out the terms in a binomial series<\/li>\r\n \t<li>Find the Taylor series for different functions<\/li>\r\n \t<li>Use Taylor series to solve differential equations<\/li>\r\n \t<li>Use Taylor series to evaluate integrals that don't have elementary antiderivatives<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">The Binomial Series<\/h2>\r\nIn the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions.\r\n\r\nOur first goal 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 <strong>binomial series<\/strong>.\r\n<p id=\"fs-id1167023795770\">We begin by considering the simplest case: [latex]r[\/latex] is a nonnegative integer. Let's look at what happens when we expand [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] for the first few values of [latex]r[\/latex]:<\/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<p id=\"fs-id1167023755079\">The expressions on the right-hand side are known as <strong>binomial expansions<\/strong> and the coefficients are known as <strong>binomial coefficients<\/strong>.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>binomial coefficients<\/h3>\r\nFor 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:\r\n<p style=\"text-align: center;\">[latex]\\left(\\begin{array}{c}r\\ n\\end{array}\\right)=\\frac{r!}{n!\\left(r-n\\right)!}[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Using this formula, we can write the general binomial expansion as:<\/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<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>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].\r\n\r\nSpecifically, 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]:\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<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]\\dfrac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}=\\dfrac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">When [latex]r[\/latex] is a nonnegative integer, the series automatically terminates (since higher derivatives become zero) and our generalized binomial coefficient formula reduces to the familiar [latex]\\left(\\begin{array}{c}r\\ n\\end{array}\\right)=\\frac{r!}{n!\\left(r-n\\right)!}[\/latex]. This confirms our approach works for both finite and infinite cases.<\/section>\r\n<p id=\"fs-id1167023863068\">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<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<p class=\"whitespace-normal break-words\">We now determine the interval of convergence for the binomial series using the ratio test:<\/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 \\dfrac{|{a}_{n+1}|}{|{a}_{n}|}&amp; =\\dfrac{{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n\\right)|x||}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\dfrac{n\\text{!}}{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)|{|x|}^{n}}\\hfill \\\\ &amp; =\\dfrac{|r-n||x|}{|n+1|}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024047434\">Since [latex]\\underset{n\\to\\infty}{\\text{lim}}\\frac{|{a}{n+1}|}{|{a}{n}|}=|x|[\/latex], we have convergence when [latex]|x|&lt;1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">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]:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">For [latex]r\\geq 0[\/latex]: the series converges at both endpoints<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For [latex]-1 &lt; r &lt; 0[\/latex]: the series converges at [latex]x=1[\/latex] and diverges at [latex]x=-1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For [latex]r&lt;-1[\/latex]: the series diverges at both endpoints<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">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].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>the binomial series<\/h3>\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:<\/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<p id=\"fs-id1167023870789\">for [latex]|x|&lt;1[\/latex].<\/p>\r\n\r\n<\/section>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].\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\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[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol 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><\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\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<ol type=\"a\">\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[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.4.1\" here (opens in new window)<\/a>.<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311966[\/ohm_question]<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write out the terms in a binomial series<\/li>\n<li>Find the Taylor series for different functions<\/li>\n<li>Use Taylor series to solve differential equations<\/li>\n<li>Use Taylor series to evaluate integrals that don&#8217;t have elementary antiderivatives<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">The Binomial Series<\/h2>\n<p>In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions.<\/p>\n<p>Our first goal 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 <strong>binomial series<\/strong>.<\/p>\n<p id=\"fs-id1167023795770\">We begin by considering the simplest case: [latex]r[\/latex] is a nonnegative integer. Let&#8217;s look at what happens when we expand [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex] for the first few values of [latex]r[\/latex]:<\/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 id=\"fs-id1167023755079\">The expressions on the right-hand side are known as <strong>binomial expansions<\/strong> and the coefficients are known as <strong>binomial coefficients<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>binomial coefficients<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]\\left(\\begin{array}{c}r\\ n\\end{array}\\right)=\\frac{r!}{n!\\left(r-n\\right)!}[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Using this formula, we can write the general binomial expansion as:<\/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<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\n<p>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].<\/p>\n<p>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 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]\\dfrac{{f}^{\\left(n\\right)}\\left(0\\right)}{n\\text{!}}=\\dfrac{r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)}{n\\text{!}}[\/latex].<\/div>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">When [latex]r[\/latex] is a nonnegative integer, the series automatically terminates (since higher derivatives become zero) and our generalized binomial coefficient formula reduces to the familiar [latex]\\left(\\begin{array}{c}r\\ n\\end{array}\\right)=\\frac{r!}{n!\\left(r-n\\right)!}[\/latex]. This confirms our approach works for both finite and infinite cases.<\/section>\n<p id=\"fs-id1167023863068\">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 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 class=\"whitespace-normal break-words\">We now determine the interval of convergence for the binomial series using the ratio test:<\/p>\n<div id=\"fs-id1167023862667\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\dfrac{|{a}_{n+1}|}{|{a}_{n}|}& =\\dfrac{{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n\\right)|x||}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\dfrac{n\\text{!}}{|r\\left(r - 1\\right)\\left(r - 2\\right)\\cdots \\left(r-n+1\\right)|{|x|}^{n}}\\hfill \\\\ & =\\dfrac{|r-n||x|}{|n+1|}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024047434\">Since [latex]\\underset{n\\to\\infty}{\\text{lim}}\\frac{|{a}{n+1}|}{|{a}{n}|}=|x|[\/latex], we have convergence when [latex]|x|<1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">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]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">For [latex]r\\geq 0[\/latex]: the series converges at both endpoints<\/li>\n<li class=\"whitespace-normal break-words\">For [latex]-1 < r < 0[\/latex]: the series converges at [latex]x=1[\/latex] and diverges at [latex]x=-1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">For [latex]r<-1[\/latex]: the series diverges at both endpoints<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">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].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>the binomial series<\/h3>\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:<\/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 id=\"fs-id1167023870789\">for [latex]|x|<1[\/latex].<\/p>\n<\/section>\n<p>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<section class=\"textbox example\" aria-label=\"Example\">\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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol 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><\/li>\n<\/ol>\n<\/li>\n<\/ol>\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<ol type=\"a\">\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<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.4.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.4.1&#8221; here (opens in new window)<\/a>.<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311966\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311966&theme=lumen&iframe_resize_id=ohm311966&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":23,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/957"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":13,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/957\/revisions"}],"predecessor-version":[{"id":2331,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/957\/revisions\/2331"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/957\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=957"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=957"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=957"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=957"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}