{"id":926,"date":"2025-06-20T17:23:05","date_gmt":"2025-06-20T17:23:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=926"},"modified":"2025-09-29T16:31:15","modified_gmt":"2025-09-29T16:31:15","slug":"introduction-to-power-series-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-power-series-fresh-take\/","title":{"raw":"Introduction to Power Series: Fresh Take","rendered":"Introduction to Power Series: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Recognize power series and when they converge<\/li>\r\n \t<li>Find where a power series converges and where it doesn't<\/li>\r\n \t<li>Use power series to write functions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">What is a Power Series?<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">A power series is essentially a polynomial that never stops growing. Instead of having a finite number of terms like [latex]3 + 2x + x^2[\/latex], a power series continues with infinitely many powers of [latex]x[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The basic form:<\/strong> [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \\cdots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Here, [latex]c_n[\/latex] represents the coefficients (just numbers), and [latex]x[\/latex] is your variable. This is called a power series <strong>centered at zero<\/strong> because the powers are of [latex]x[\/latex] itself.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Centered at any point:<\/strong> You can also have [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex], which is <strong>centered at<\/strong> [latex]x = a[\/latex]. This shifts the \"center\" of your series from zero to the point [latex]a[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Unlike regular series with just numbers, power series involve a variable [latex]x[\/latex]. This means convergence depends on what value you plug in for [latex]x[\/latex]. The series might converge for some values of [latex]x[\/latex] and diverge for others.<\/p>\r\nPower series bridge the gap between polynomials (which you can easily work with) and more complex functions. They let you represent functions like [latex]e^x[\/latex], [latex]\\sin x[\/latex], and [latex]\\cos x[\/latex] as infinite sums, opening up powerful computational and analytical techniques.\r\n\r\nThe big question with any power series is: \"For which values of [latex]x[\/latex] does this infinite sum actually converge to a finite value?\"\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">The geometric series [latex]1 + x + x^2 + x^3 + \\cdots[\/latex] is actually a power series where all coefficients [latex]c_n = 1[\/latex]. It converges when [latex]|x| &lt; 1[\/latex] and diverges when [latex]|x| \\geq 1[\/latex].<\/section>\r\n<h2 data-type=\"title\">Convergence of a Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The critical question for any power series is: \"For which values of [latex]x[\/latex] does this infinite sum actually give you a finite answer?\" Unlike regular series with constant terms, power series convergence depends entirely on what you plug in for [latex]x[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>One guaranteed convergence point:<\/strong> Every power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex] always converges at its center [latex]x = a[\/latex]. When you substitute [latex]x = a[\/latex], all terms except the first become zero, leaving just [latex]c_0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Three possible behaviors:<\/strong> Every power series falls into exactly one of these categories:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Case 1: Converges only at the center<\/strong> - The series works at [latex]x = a[\/latex] but diverges everywhere else. This happens with series that grow very rapidly, like [latex]\\sum n! x^n[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Case 2: Converges everywhere<\/strong> - The series converges for all real numbers [latex]x[\/latex]. Examples include [latex]\\sum \\frac{x^n}{n!}[\/latex] (which represents [latex]e^x[\/latex]).<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Case 3: Converges in an interval<\/strong> - This is the most common case. There exists a positive number [latex]R[\/latex] (called the radius of convergence) such that:\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\">The series converges when [latex]|x-a| &lt; R[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The series diverges when [latex]|x-a| &gt; R[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">At the boundary points [latex]|x-a| = R[\/latex], you need to check individually<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nThe endpoints [latex]x = a-R[\/latex] and [latex]x = a+R[\/latex] require separate investigation. The series might converge at one endpoint, both endpoints, or neither - there's no universal rule.\r\n\r\n<\/div>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Interval and Radius of Convergence<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Once you know a power series converges in an interval, you need to find exactly where that interval is and how wide it extends. The <strong>interval of convergence<\/strong> tells you all the [latex]x[\/latex]-values where your series actually works, while the <strong>radius of convergence<\/strong> [latex]R[\/latex] measures how far you can go from the center before the series breaks down.<\/p>\r\n<p class=\"whitespace-normal break-words\">For a power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex], the radius [latex]R[\/latex] represents the distance from the center [latex]a[\/latex] to the edge of convergence:<\/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\">[latex]R = 0[\/latex]: Series only works at [latex]x = a[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]R = \\infty[\/latex]: Series works everywhere<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]R &gt; 0[\/latex]: Series works when [latex]|x-a| &lt; R[\/latex] and fails when [latex]|x-a| &gt; R[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding the radius with the ratio test:<\/strong> Apply the ratio test to get [latex]\\rho = \\lim_{n\\to\\infty}\\left|\\frac{a_{n+1}}{a_n}\\right|[\/latex]. For convergence, you need [latex]\\rho &lt; 1[\/latex], which typically gives you a condition like [latex]|x-a| &lt; R[\/latex].<\/p>\r\nWhen [latex]|x-a| = R[\/latex] (at [latex]x = a-R[\/latex] and [latex]x = a+R[\/latex]), the ratio test gives [latex]\\rho = 1[\/latex], making it inconclusive. You must test these boundary points separately by substituting them into the original series and checking convergence using other tests.\r\n<p class=\"whitespace-normal break-words\"><strong>Common endpoint patterns:<\/strong><\/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\">Alternating harmonic-type series often converge at one endpoint<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Regular harmonic-type series usually diverge at endpoints<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Check each endpoint individually - they can behave differently<\/li>\r\n<\/ul>\r\nThe interval of convergence might be [latex](a-R, a+R)[\/latex], [latex][a-R, a+R)[\/latex], [latex](a-R, a+R][\/latex], or [latex][a-R, a+R][\/latex], depending on endpoint behavior.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572582284\" data-type=\"problem\">\r\n<p id=\"fs-id1170572582286\">Find the interval and radius of convergence for the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{\\sqrt{n}}[\/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-id1170572203628\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170571610504\">Apply the ratio test to check for absolute convergence.<\/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-id1170572207671\" data-type=\"solution\">\r\n<p id=\"fs-id1170572207673\">The interval of convergence is [latex]\\left[-1,1\\right)[\/latex]. The radius of convergence is [latex]R=1[\/latex].<\/p>\r\n\r\n<\/div>\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 above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/f25BYE_ImSc?controls=0&amp;start=551&amp;end=674&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/6.1.2_551to674_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.1.2\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]310243[\/ohm_question]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]310246[\/ohm_question]\r\n\r\n<\/section>\r\n<h2>Representing Functions as Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Power series give you a way to rewrite complicated functions as \"infinite polynomials,\" which are much easier to work with. Once you have a power series representation, you can differentiate, integrate, and evaluate the function term by term - operations that might be difficult or impossible with the original form.<\/p>\r\n<p class=\"whitespace-normal break-words\">The key insight comes from the geometric series formula. For [latex]|r| &lt; 1[\/latex], we have [latex]\\frac{a}{1-r} = a + ar + ar^2 + ar^3 + \\cdots[\/latex]. This gives us our most important power series representation:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\frac{1}{1-x} = 1 + x + x^2 + x^3 + \\cdots[\/latex] for [latex]|x| &lt; 1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Most functions won't look exactly like [latex]\\frac{1}{1-r}[\/latex], but you can often manipulate them algebraically to fit this pattern. The key is recognizing forms that can be rewritten as geometric series.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Common algebraic tricks:<\/strong><\/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\">Rewrite [latex]\\frac{1}{1+x^3}[\/latex] as [latex]\\frac{1}{1-(-x^3)}[\/latex] to use [latex]r = -x^3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Factor constants from denominators: [latex]\\frac{x^2}{4-x^2} = \\frac{x^2}{4(1-\\frac{x^2}{4})}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Look for patterns that match [latex]\\frac{a}{1-r}[\/latex] where [latex]r[\/latex] involves your variable<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Power series representations let you:<\/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\">Approximate function values using partial sums (just add up the first few terms)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate or differentiate functions that would be difficult in their original form<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Analyze function behavior near specific points<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Your power series representation is only valid within its interval of convergence. Outside this interval, the infinite sum doesn't equal your original function, so always identify where [latex]|r| &lt; 1[\/latex] for your specific [latex]r[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Recognize power series and when they converge<\/li>\n<li>Find where a power series converges and where it doesn&#8217;t<\/li>\n<li>Use power series to write functions<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">What is a Power Series?<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A power series is essentially a polynomial that never stops growing. Instead of having a finite number of terms like [latex]3 + 2x + x^2[\/latex], a power series continues with infinitely many powers of [latex]x[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The basic form:<\/strong> [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \\cdots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Here, [latex]c_n[\/latex] represents the coefficients (just numbers), and [latex]x[\/latex] is your variable. This is called a power series <strong>centered at zero<\/strong> because the powers are of [latex]x[\/latex] itself.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Centered at any point:<\/strong> You can also have [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex], which is <strong>centered at<\/strong> [latex]x = a[\/latex]. This shifts the &#8220;center&#8221; of your series from zero to the point [latex]a[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Unlike regular series with just numbers, power series involve a variable [latex]x[\/latex]. This means convergence depends on what value you plug in for [latex]x[\/latex]. The series might converge for some values of [latex]x[\/latex] and diverge for others.<\/p>\n<p>Power series bridge the gap between polynomials (which you can easily work with) and more complex functions. They let you represent functions like [latex]e^x[\/latex], [latex]\\sin x[\/latex], and [latex]\\cos x[\/latex] as infinite sums, opening up powerful computational and analytical techniques.<\/p>\n<p>The big question with any power series is: &#8220;For which values of [latex]x[\/latex] does this infinite sum actually converge to a finite value?&#8221;<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">The geometric series [latex]1 + x + x^2 + x^3 + \\cdots[\/latex] is actually a power series where all coefficients [latex]c_n = 1[\/latex]. It converges when [latex]|x| < 1[\/latex] and diverges when [latex]|x| \\geq 1[\/latex].<\/section>\n<h2 data-type=\"title\">Convergence of a Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The critical question for any power series is: &#8220;For which values of [latex]x[\/latex] does this infinite sum actually give you a finite answer?&#8221; Unlike regular series with constant terms, power series convergence depends entirely on what you plug in for [latex]x[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>One guaranteed convergence point:<\/strong> Every power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex] always converges at its center [latex]x = a[\/latex]. When you substitute [latex]x = a[\/latex], all terms except the first become zero, leaving just [latex]c_0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Three possible behaviors:<\/strong> Every power series falls into exactly one of these categories:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Case 1: Converges only at the center<\/strong> &#8211; The series works at [latex]x = a[\/latex] but diverges everywhere else. This happens with series that grow very rapidly, like [latex]\\sum n! x^n[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Case 2: Converges everywhere<\/strong> &#8211; The series converges for all real numbers [latex]x[\/latex]. Examples include [latex]\\sum \\frac{x^n}{n!}[\/latex] (which represents [latex]e^x[\/latex]).<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Case 3: Converges in an interval<\/strong> &#8211; This is the most common case. There exists a positive number [latex]R[\/latex] (called the radius of convergence) such that:\n<ul>\n<li class=\"whitespace-normal break-words\">The series converges when [latex]|x-a| < R[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The series diverges when [latex]|x-a| > R[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">At the boundary points [latex]|x-a| = R[\/latex], you need to check individually<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>The endpoints [latex]x = a-R[\/latex] and [latex]x = a+R[\/latex] require separate investigation. The series might converge at one endpoint, both endpoints, or neither &#8211; there&#8217;s no universal rule.<\/p>\n<\/div>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Interval and Radius of Convergence<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you know a power series converges in an interval, you need to find exactly where that interval is and how wide it extends. The <strong>interval of convergence<\/strong> tells you all the [latex]x[\/latex]-values where your series actually works, while the <strong>radius of convergence<\/strong> [latex]R[\/latex] measures how far you can go from the center before the series breaks down.<\/p>\n<p class=\"whitespace-normal break-words\">For a power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex], the radius [latex]R[\/latex] represents the distance from the center [latex]a[\/latex] to the edge of convergence:<\/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\">[latex]R = 0[\/latex]: Series only works at [latex]x = a[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]R = \\infty[\/latex]: Series works everywhere<\/li>\n<li class=\"whitespace-normal break-words\">[latex]R > 0[\/latex]: Series works when [latex]|x-a| < R[\/latex] and fails when [latex]|x-a| > R[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Finding the radius with the ratio test:<\/strong> Apply the ratio test to get [latex]\\rho = \\lim_{n\\to\\infty}\\left|\\frac{a_{n+1}}{a_n}\\right|[\/latex]. For convergence, you need [latex]\\rho < 1[\/latex], which typically gives you a condition like [latex]|x-a| < R[\/latex].<\/p>\n<p>When [latex]|x-a| = R[\/latex] (at [latex]x = a-R[\/latex] and [latex]x = a+R[\/latex]), the ratio test gives [latex]\\rho = 1[\/latex], making it inconclusive. You must test these boundary points separately by substituting them into the original series and checking convergence using other tests.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Common endpoint patterns:<\/strong><\/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\">Alternating harmonic-type series often converge at one endpoint<\/li>\n<li class=\"whitespace-normal break-words\">Regular harmonic-type series usually diverge at endpoints<\/li>\n<li class=\"whitespace-normal break-words\">Check each endpoint individually &#8211; they can behave differently<\/li>\n<\/ul>\n<p>The interval of convergence might be [latex](a-R, a+R)[\/latex], [latex][a-R, a+R)[\/latex], [latex](a-R, a+R][\/latex], or [latex][a-R, a+R][\/latex], depending on endpoint behavior.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572582284\" data-type=\"problem\">\n<p id=\"fs-id1170572582286\">Find the interval and radius of convergence for the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{\\sqrt{n}}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572203628\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170571610504\">Apply the ratio test to check for absolute convergence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572207671\" data-type=\"solution\">\n<p id=\"fs-id1170572207673\">The interval of convergence is [latex]\\left[-1,1\\right)[\/latex]. The radius of convergence is [latex]R=1[\/latex].<\/p>\n<\/div>\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 above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/f25BYE_ImSc?controls=0&amp;start=551&amp;end=674&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/6.1.2_551to674_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.1.2&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm310243\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=310243&theme=lumen&iframe_resize_id=ohm310243&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm310246\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=310246&theme=lumen&iframe_resize_id=ohm310246&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h2>Representing Functions as Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Power series give you a way to rewrite complicated functions as &#8220;infinite polynomials,&#8221; which are much easier to work with. Once you have a power series representation, you can differentiate, integrate, and evaluate the function term by term &#8211; operations that might be difficult or impossible with the original form.<\/p>\n<p class=\"whitespace-normal break-words\">The key insight comes from the geometric series formula. For [latex]|r| < 1[\/latex], we have [latex]\\frac{a}{1-r} = a + ar + ar^2 + ar^3 + \\cdots[\/latex]. This gives us our most important power series representation:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\frac{1}{1-x} = 1 + x + x^2 + x^3 + \\cdots[\/latex] for [latex]|x| < 1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Most functions won&#8217;t look exactly like [latex]\\frac{1}{1-r}[\/latex], but you can often manipulate them algebraically to fit this pattern. The key is recognizing forms that can be rewritten as geometric series.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Common algebraic tricks:<\/strong><\/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\">Rewrite [latex]\\frac{1}{1+x^3}[\/latex] as [latex]\\frac{1}{1-(-x^3)}[\/latex] to use [latex]r = -x^3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Factor constants from denominators: [latex]\\frac{x^2}{4-x^2} = \\frac{x^2}{4(1-\\frac{x^2}{4})}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Look for patterns that match [latex]\\frac{a}{1-r}[\/latex] where [latex]r[\/latex] involves your variable<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Power series representations let you:<\/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\">Approximate function values using partial sums (just add up the first few terms)<\/li>\n<li class=\"whitespace-normal break-words\">Integrate or differentiate functions that would be difficult in their original form<\/li>\n<li class=\"whitespace-normal break-words\">Analyze function behavior near specific points<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Your power series representation is only valid within its interval of convergence. Outside this interval, the infinite sum doesn&#8217;t equal your original function, so always identify where [latex]|r| < 1[\/latex] for your specific [latex]r[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558893\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Show Solution<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572432260\" data-type=\"solution\">\n<figure style=\"width: 487px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 3.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n","protected":false},"author":15,"menu_order":10,"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\/926"}],"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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/926\/revisions"}],"predecessor-version":[{"id":2399,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/926\/revisions\/2399"}],"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\/926\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=926"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=926"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=926"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=926"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}