{"id":924,"date":"2025-06-20T17:22:59","date_gmt":"2025-06-20T17:22:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=924"},"modified":"2025-07-23T16:38:17","modified_gmt":"2025-07-23T16:38:17","slug":"introduction-to-power-series-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-power-series-learn-it-4\/","title":{"raw":"Introduction to Power Series: Learn It 4","rendered":"Introduction to Power Series: Learn It 4"},"content":{"raw":"<h2>Representing Functions as Power Series<\/h2>\r\n<p class=\"whitespace-normal break-words\">Why should we care about representing functions as power series? The answer lies in the incredible usefulness of polynomials.<\/p>\r\n<p class=\"whitespace-normal break-words\">Polynomials are the simplest functions to work with \u2014 they only involve basic arithmetic operations like addition, subtraction, multiplication, and division. When we represent a complicated function as an \"infinite polynomial\" (a power series), we gain several powerful advantages:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Easy differentiation and integration:<\/strong> We can differentiate or integrate term by term<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Function approximation:<\/strong> We can use partial sums to approximate function values<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Simplified analysis:<\/strong> Complex functions become more manageable<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The key question becomes: when can we actually represent a function using a power series?<\/p>\r\nLet's revisit the geometric series we've seen before:\r\n<p style=\"text-align: center;\">[latex]1 + x + x^2 + x^3 + \\cdots = \\displaystyle\\sum_{n=0}^{\\infty} x^n[\/latex]<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Geometric Series Formula<\/strong>\r\n\r\nThe geometric series [latex]a + ar + ar^2 + ar^3 + \\cdots[\/latex] converges if and only if [latex]|r| &lt; 1[\/latex]. When it converges, the sum equals [latex]\\frac{a}{1-r}[\/latex].\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">For our series with [latex]a = 1[\/latex] and [latex]r = x[\/latex], we get convergence when [latex]|x| &lt; 1[\/latex], and the sum is [latex]\\frac{1}{1-x}[\/latex]. Therefore:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]1 + x + x^2 + x^3 + \\cdots = \\frac{1}{1-x} \\text{ for } |x| &lt; 1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">We've successfully represented the function [latex]f(x) = \\frac{1}{1-x}[\/latex] as a power series. This representation is valid whenever [latex]|x| &lt; 1[\/latex].<\/p>\r\nWe can see how well this power series represents the original function by comparing the graph of [latex]f(x) = \\frac{1}{1-x}[\/latex] with the graphs of several partial sums of this infinite series.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\r\n<p id=\"fs-id1170572553449\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and the graphs of the corresponding partial sums [latex]{S}_{N}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{N}{x}^{n}[\/latex] for [latex]N=2,4,6[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex]. Comment on the approximation [latex]{S}_{N}[\/latex] as <em data-effect=\"italics\">N<\/em> increases.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1170571769944\" data-type=\"solution\">\r\n<p id=\"fs-id1170571769947\">From the graph in Figure 2 you see that as <em data-effect=\"italics\">N<\/em> increases, [latex]{S}_{N}[\/latex] becomes a better approximation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] for <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_10_01_002\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234431\/CNX_Calc_Figure_10_01_002.jpg\" alt=\"This figure is the graph of y = 1\/(1-x), which is an increasing curve with vertical asymptote at 1. Also on this graph are three partial sums of the function, S sub 6, S sub 4, and S sub 2. These curves, in order, gradually become flatter.\" width=\"487\" height=\"312\" data-media-type=\"image\/jpeg\" \/> Figure 2. The graph shows a function and three approximations of it by partial sums of a power series.[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572320906\" data-type=\"problem\">\r\n<p id=\"fs-id1170572320911\">Use a power series to represent each of the following functions [latex]f[\/latex]. Find the interval of convergence.<\/p>\r\n\r\n<ol id=\"fs-id1170572320919\" type=\"a\">\r\n \t<li>[latex]f\\left(x\\right)=\\frac{1}{1+{x}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{4-{x}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"970045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"970045\"]\r\n<ol id=\"fs-id1170571546704\" type=\"a\">\r\n \t<li>You should recognize this function <em data-effect=\"italics\">f<\/em> as the sum of a geometric series, because[latex]\\frac{1}{1+{x}^{3}}=\\frac{1}{1-\\left(\\text{-}{x}^{3}\\right)}[\/latex]\r\nUsing the fact that, for [latex]|r|&lt;1,\\frac{a}{1-r}[\/latex] is the sum of the geometric series\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }a{r}^{n}=a+ar+a{r}^{2}+\\cdots [\/latex],<\/p>\r\nwe see that, for [latex]|\\text{-}{x}^{3}|&lt;1[\/latex],\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+{x}^{3}}&amp; =\\frac{1}{1-\\left(\\text{-}{x}^{3}\\right)}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-{x}^{3}\\right)}^{n}\\hfill \\\\ &amp; =1-{x}^{3}+{x}^{6}-{x}^{9}+\\cdots .\\hfill \\end{array}[\/latex]<\/p>\r\nSince this series converges if and only if [latex]|\\text{-}{x}^{3}|&lt;1[\/latex], the interval of convergence is [latex]\\left(-1,1\\right)[\/latex], and we have\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\\cdots \\text{for}|x|&lt;1[\/latex].<\/p>\r\n<\/li>\r\n \t<li>This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate <em data-effect=\"italics\">f<\/em> to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}&amp; =\\frac{{x}^{2}}{4\\left(\\frac{1-{x}^{2}}{4}\\right)}\\hfill \\\\ &amp; =\\frac{{x}^{2}}{4\\left(1-{\\left(\\frac{x}{2}\\right)}^{2}\\right)}.\\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}&amp; =\\frac{{x}^{2}}{4\\left(1-{\\left(\\frac{x}{2}\\right)}^{2}\\right)}\\hfill \\\\ &amp; =\\frac{\\frac{{x}^{2}}{4}}{1-{\\left(\\frac{x}{2}\\right)}^{2}}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{{x}^{2}}{4}{\\left(\\frac{x}{2}\\right)}^{2n}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe series converges as long as [latex]|{\\left(\\frac{x}{2}\\right)}^{2}|&lt;1[\/latex] (note that when [latex]|{\\left(\\frac{x}{2}\\right)}^{2}|=1[\/latex] the series does not converge). Solving this inequality, we conclude that the interval of convergence is [latex]\\left(-2,2\\right)[\/latex] and\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{{x}^{2n+2}}{{4}^{n+1}}\\hfill \\\\ &amp; =\\frac{{x}^{2}}{4}+\\frac{{x}^{4}}{{4}^{2}}+\\frac{{x}^{6}}{{4}^{3}}+\\cdots \\hfill \\end{array}[\/latex]<\/p>\r\nfor [latex]|x|&lt;2[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2>Representing Functions as Power Series<\/h2>\n<p class=\"whitespace-normal break-words\">Why should we care about representing functions as power series? The answer lies in the incredible usefulness of polynomials.<\/p>\n<p class=\"whitespace-normal break-words\">Polynomials are the simplest functions to work with \u2014 they only involve basic arithmetic operations like addition, subtraction, multiplication, and division. When we represent a complicated function as an &#8220;infinite polynomial&#8221; (a power series), we gain several powerful advantages:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Easy differentiation and integration:<\/strong> We can differentiate or integrate term by term<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Function approximation:<\/strong> We can use partial sums to approximate function values<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Simplified analysis:<\/strong> Complex functions become more manageable<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The key question becomes: when can we actually represent a function using a power series?<\/p>\n<p>Let&#8217;s revisit the geometric series we&#8217;ve seen before:<\/p>\n<p style=\"text-align: center;\">[latex]1 + x + x^2 + x^3 + \\cdots = \\displaystyle\\sum_{n=0}^{\\infty} x^n[\/latex]<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Geometric Series Formula<\/strong><\/p>\n<p>The geometric series [latex]a + ar + ar^2 + ar^3 + \\cdots[\/latex] converges if and only if [latex]|r| < 1[\/latex]. When it converges, the sum equals [latex]\\frac{a}{1-r}[\/latex].\n\n<\/section>\n<p class=\"whitespace-normal break-words\">For our series with [latex]a = 1[\/latex] and [latex]r = x[\/latex], we get convergence when [latex]|x| < 1[\/latex], and the sum is [latex]\\frac{1}{1-x}[\/latex]. Therefore:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]1 + x + x^2 + x^3 + \\cdots = \\frac{1}{1-x} \\text{ for } |x| < 1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">We&#8217;ve successfully represented the function [latex]f(x) = \\frac{1}{1-x}[\/latex] as a power series. This representation is valid whenever [latex]|x| < 1[\/latex].<\/p>\n<p>We can see how well this power series represents the original function by comparing the graph of [latex]f(x) = \\frac{1}{1-x}[\/latex] with the graphs of several partial sums of this infinite series.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\n<p id=\"fs-id1170572553449\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and the graphs of the corresponding partial sums [latex]{S}_{N}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{N}{x}^{n}[\/latex] for [latex]N=2,4,6[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex]. Comment on the approximation [latex]{S}_{N}[\/latex] as <em data-effect=\"italics\">N<\/em> increases.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571769944\" data-type=\"solution\">\n<p id=\"fs-id1170571769947\">From the graph in Figure 2 you see that as <em data-effect=\"italics\">N<\/em> increases, [latex]{S}_{N}[\/latex] becomes a better approximation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] for <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_10_01_002\"><figcaption><\/figcaption><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\/11234431\/CNX_Calc_Figure_10_01_002.jpg\" alt=\"This figure is the graph of y = 1\/(1-x), which is an increasing curve with vertical asymptote at 1. Also on this graph are three partial sums of the function, S sub 6, S sub 4, and S sub 2. These curves, in order, gradually become flatter.\" width=\"487\" height=\"312\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 2. The graph shows a function and three approximations of it by partial sums of a power series.<\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572320906\" data-type=\"problem\">\n<p id=\"fs-id1170572320911\">Use a power series to represent each of the following functions [latex]f[\/latex]. Find the interval of convergence.<\/p>\n<ol id=\"fs-id1170572320919\" type=\"a\">\n<li>[latex]f\\left(x\\right)=\\frac{1}{1+{x}^{3}}[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{4-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q970045\">Show Solution<\/button><\/p>\n<div id=\"q970045\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170571546704\" type=\"a\">\n<li>You should recognize this function <em data-effect=\"italics\">f<\/em> as the sum of a geometric series, because[latex]\\frac{1}{1+{x}^{3}}=\\frac{1}{1-\\left(\\text{-}{x}^{3}\\right)}[\/latex]<br \/>\nUsing the fact that, for [latex]|r|<1,\\frac{a}{1-r}[\/latex] is the sum of the geometric series\n\n\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }a{r}^{n}=a+ar+a{r}^{2}+\\cdots[\/latex],<\/p>\n<p>we see that, for [latex]|\\text{-}{x}^{3}|<1[\/latex],\n\n\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+{x}^{3}}& =\\frac{1}{1-\\left(\\text{-}{x}^{3}\\right)}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-{x}^{3}\\right)}^{n}\\hfill \\\\ & =1-{x}^{3}+{x}^{6}-{x}^{9}+\\cdots .\\hfill \\end{array}[\/latex]<\/p>\n<p>Since this series converges if and only if [latex]|\\text{-}{x}^{3}|<1[\/latex], the interval of convergence is [latex]\\left(-1,1\\right)[\/latex], and we have\n\n\n<p style=\"text-align: center;\">[latex]\\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\\cdots \\text{for}|x|<1[\/latex].<\/p>\n<\/li>\n<li>This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate <em data-effect=\"italics\">f<\/em> to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}& =\\frac{{x}^{2}}{4\\left(\\frac{1-{x}^{2}}{4}\\right)}\\hfill \\\\ & =\\frac{{x}^{2}}{4\\left(1-{\\left(\\frac{x}{2}\\right)}^{2}\\right)}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}& =\\frac{{x}^{2}}{4\\left(1-{\\left(\\frac{x}{2}\\right)}^{2}\\right)}\\hfill \\\\ & =\\frac{\\frac{{x}^{2}}{4}}{1-{\\left(\\frac{x}{2}\\right)}^{2}}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{{x}^{2}}{4}{\\left(\\frac{x}{2}\\right)}^{2n}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The series converges as long as [latex]|{\\left(\\frac{x}{2}\\right)}^{2}|<1[\/latex] (note that when [latex]|{\\left(\\frac{x}{2}\\right)}^{2}|=1[\/latex] the series does not converge). Solving this inequality, we conclude that the interval of convergence is [latex]\\left(-2,2\\right)[\/latex] and\n\n\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{{x}^{2}}{4-{x}^{2}}& ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{{x}^{2n+2}}{{4}^{n+1}}\\hfill \\\\ & =\\frac{{x}^{2}}{4}+\\frac{{x}^{4}}{{4}^{2}}+\\frac{{x}^{6}}{{4}^{3}}+\\cdots \\hfill \\end{array}[\/latex]<\/p>\n<p>for [latex]|x|<2[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":8,"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\/924"}],"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":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/924\/revisions"}],"predecessor-version":[{"id":1458,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/924\/revisions\/1458"}],"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\/924\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=924"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=924"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=924"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=924"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}