{"id":1829,"date":"2021-07-28T16:52:27","date_gmt":"2021-07-28T16:52:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1829"},"modified":"2022-03-21T23:09:55","modified_gmt":"2022-03-21T23:09:55","slug":"representing-functions-as-power-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/representing-functions-as-power-series\/","title":{"raw":"Representing Functions as Power Series","rendered":"Representing Functions as Power Series"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use a power series to represent a function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572443568\">Being able to represent a function by an \"infinite polynomial\" is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?<\/p>\r\n<p id=\"fs-id1170572351575\">Consider again the geometric series<\/p>\r\n\r\n<div id=\"fs-id1170572443573\" style=\"text-align: center;\" data-type=\"equation\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots =\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572337139\">Recall that the geometric series<\/p>\r\n\r\n<div id=\"fs-id1170571652074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n\r\nconverges if and only if [latex]|r|&lt;1[\/latex]. In that case, it converges to [latex]\\frac{a}{1-r}[\/latex]. Therefore, if [latex]|x|&lt;1[\/latex], the series in the example: Representing a Function with a Power Series converges to [latex]\\frac{1}{1-x}[\/latex] and we write\r\n<div id=\"fs-id1170572560653\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots =\\frac{1}{1-x}\\text{for}|x|&lt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572621739\">As a result, we are able to represent the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] by the power series<\/p>\r\n\r\n<div id=\"fs-id1169739062432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots \\text{when}|x|&lt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572480877\">We now show graphically how this series provides a representation for the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] by comparing the graph of <em data-effect=\"italics\">f<\/em> with the graphs of several of the partial sums of this infinite series.<\/p>\r\n\r\n<div id=\"fs-id1170572498722\" data-type=\"example\">\r\n<div id=\"fs-id1170572553441\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Function and Partial Sums of its Power Series<\/h3>\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572498722\" data-type=\"example\">\r\n<div id=\"fs-id1170572553441\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{1-{x}^{2}}[\/latex] and the corresponding partial sums [latex]{S}_{N}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{N}{x}^{2n}[\/latex] for [latex]N=2,4,6[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571673193\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170571673196\" data-type=\"exercise\">\r\n\r\n[reveal-answer q=\"44558893\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1170572460220\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572460227\">[latex]{S}_{N}\\left(x\\right)=1+{x}^{2}+\\cdots +{x}^{2N}=\\frac{1-{x}^{2\\left(N+1\\right)}}{1-{x}^{2}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1170572432260\" data-type=\"solution\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234435\/CNX_Calc_Figure_10_01_003.jpg\" alt=\"This figure is the graph of y = 1\/(1-x^2), which is a curve concave up, symmetrical about the y axis. 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 3.[\/caption]\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>\r\n<\/div>\r\n<p id=\"fs-id1170572180130\">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>\r\n\r\n<div id=\"fs-id1170572180135\" data-type=\"example\">\r\n<div id=\"fs-id1170572180137\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572320906\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Representing a Function with a Power Series<\/h3>\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572481632\" data-type=\"problem\">\r\n<p id=\"fs-id1170572481634\">Represent the function [latex]f\\left(x\\right)=\\frac{{x}^{3}}{2-x}[\/latex] using a power series and find the interval of convergence.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1170572392069\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572392076\">Rewrite <em data-effect=\"italics\">f<\/em> in the form [latex]f\\left(x\\right)=\\frac{g\\left(x\\right)}{1-h\\left(x\\right)}[\/latex] for some functions <em data-effect=\"italics\">g<\/em> and <em data-effect=\"italics\">h<\/em>.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n\r\n<span style=\"font-size: 1rem; text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n+3}}{{2}^{n+1}}[\/latex] with interval of convergence [latex]\\left(-2,2\\right)[\/latex]<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F2wZiUPsHk0?controls=0&amp;start=468&amp;end=566&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.1.3_468to566_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.1.3\" here (opens in new window)<\/a>.\r\n<p id=\"fs-id1170571546052\">In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1170571546058\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use a power series to represent a function<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572443568\">Being able to represent a function by an &#8220;infinite polynomial&#8221; is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?<\/p>\n<p id=\"fs-id1170572351575\">Consider again the geometric series<\/p>\n<div id=\"fs-id1170572443573\" style=\"text-align: center;\" data-type=\"equation\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots =\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572337139\">Recall that the geometric series<\/p>\n<div id=\"fs-id1170571652074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>converges if and only if [latex]|r|<1[\/latex]. In that case, it converges to [latex]\\frac{a}{1-r}[\/latex]. Therefore, if [latex]|x|<1[\/latex], the series in the example: Representing a Function with a Power Series converges to [latex]\\frac{1}{1-x}[\/latex] and we write\n\n\n<div id=\"fs-id1170572560653\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots =\\frac{1}{1-x}\\text{for}|x|<1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572621739\">As a result, we are able to represent the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] by the power series<\/p>\n<div id=\"fs-id1169739062432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+x+{x}^{2}+{x}^{3}+\\cdots \\text{when}|x|<1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572480877\">We now show graphically how this series provides a representation for the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] by comparing the graph of <em data-effect=\"italics\">f<\/em> with the graphs of several of the partial sums of this infinite series.<\/p>\n<div id=\"fs-id1170572498722\" data-type=\"example\">\n<div id=\"fs-id1170572553441\" data-type=\"exercise\">\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Function and Partial Sums of its Power Series<\/h3>\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\"><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-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><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\/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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The graph shows a function and three approximations of it by partial sums of a power series.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572498722\" data-type=\"example\">\n<div id=\"fs-id1170572553441\" data-type=\"exercise\">\n<div id=\"fs-id1170572553443\" data-type=\"problem\">\n<div data-type=\"title\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{1-{x}^{2}}[\/latex] and the corresponding partial sums [latex]{S}_{N}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{N}{x}^{2n}[\/latex] for [latex]N=2,4,6[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571673193\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170571673196\" data-type=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Hint<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572460220\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572460227\">[latex]{S}_{N}\\left(x\\right)=1+{x}^{2}+\\cdots +{x}^{2N}=\\frac{1-{x}^{2\\left(N+1\\right)}}{1-{x}^{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=\"q44558894\">Show Solution<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572432260\" data-type=\"solution\">\n<div style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234435\/CNX_Calc_Figure_10_01_003.jpg\" alt=\"This figure is the graph of y = 1\/(1-x^2), which is a curve concave up, symmetrical about the y axis. 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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572180130\">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<div id=\"fs-id1170572180135\" data-type=\"example\">\n<div id=\"fs-id1170572180137\" data-type=\"exercise\">\n<div id=\"fs-id1170572320906\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Representing a Function with a Power Series<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q970045\">Show Solution<\/span><\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572481632\" data-type=\"problem\">\n<p id=\"fs-id1170572481634\">Represent the function [latex]f\\left(x\\right)=\\frac{{x}^{3}}{2-x}[\/latex] using a power series and find the interval of convergence.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Hint<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572392069\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572392076\">Rewrite <em data-effect=\"italics\">f<\/em> in the form [latex]f\\left(x\\right)=\\frac{g\\left(x\\right)}{1-h\\left(x\\right)}[\/latex] for some functions <em data-effect=\"italics\">g<\/em> and <em data-effect=\"italics\">h<\/em>.<\/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=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<p><span style=\"font-size: 1rem; text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n+3}}{{2}^{n+1}}[\/latex] with interval of convergence [latex]\\left(-2,2\\right)[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F2wZiUPsHk0?controls=0&amp;start=468&amp;end=566&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/6.1.3_468to566_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.1.3&#8221; here (opens in new window)<\/a>.<\/p>\n<p id=\"fs-id1170571546052\">In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1170571546058\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1829\">\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>6.1.3. <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":4,"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\":\"6.1.3\",\"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-1829","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1829","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\/1829\/revisions"}],"predecessor-version":[{"id":2245,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1829\/revisions\/2245"}],"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\/1829\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1829"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1829"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1829"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1829"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}