{"id":933,"date":"2025-06-20T17:23:28","date_gmt":"2025-06-20T17:23:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=933"},"modified":"2025-09-11T16:50:09","modified_gmt":"2025-09-11T16:50:09","slug":"operations-with-power-series-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/operations-with-power-series-learn-it-1\/","title":{"raw":"Operations with Power Series: Learn It 1","rendered":"Operations with Power Series: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Add and subtract power series<\/li>\r\n \t<li><span data-sheets-root=\"1\">Multiply two power series together<\/span><\/li>\r\n \t<li>Find derivatives of power series term by term<\/li>\r\n \t<li>Integrate power series term by term<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Combining Power Series<\/h2>\r\n<p class=\"whitespace-normal break-words\">In the previous section, we learned how to represent certain functions using power series. Now we'll explore how to manipulate these power series\u2014combining, modifying, and transforming them to create new power series representations.<\/p>\r\n<p class=\"whitespace-normal break-words\">This ability to work with power series is incredibly valuable for two key reasons:<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>First<\/strong>, it lets us find power series for functions that might otherwise be difficult to work with directly. For example, if we know the power series for [latex]f(x) = \\frac{1}{1-x}[\/latex], we can use calculus operations to find the power series for [latex]f'(x) = \\frac{1}{(1-x)^2}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Second<\/strong>, we can define entirely new functions that can't be expressed using elementary functions. This becomes especially important when solving differential equations that have no elementary solutions.<\/p>\r\n\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Basic Operations with Power Series<\/h3>\r\n<p class=\"whitespace-normal break-words\">When you have two power series that converge on the same interval, you can perform several operations to create new power series.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>theorem: combining power series<\/h3>\r\n<p id=\"fs-id1167023707168\">Suppose the power series [latex]\\sum_{n=0}^{\\infty} c_n x^n[\/latex] and [latex]\\sum_{n=0}^{\\infty} d_n x^n[\/latex] both converge to functions [latex]f[\/latex] and [latex]g[\/latex] respectively on a common interval [latex]I[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167023719728\" type=\"i\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Addition\/Subtraction<\/strong>: [latex]\\sum_{n=0}^{\\infty} (c_n \\pm d_n) x^n[\/latex] converges to [latex]f \\pm g[\/latex] on [latex]I[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar and Power Multiplication<\/strong>: [latex]\\sum_{n=0}^{\\infty} bx^m c_n x^n[\/latex] converges to [latex]bx^m f(x)[\/latex] on [latex]I[\/latex] for any integer [latex]m \\geq 0[\/latex] and real number [latex]b[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Composition<\/strong>: [latex]\\sum_{n=0}^{\\infty} c_n (bx^m)^n[\/latex] converges to [latex]f(bx^m)[\/latex] for all [latex]x[\/latex] where [latex]bx^m[\/latex] is in [latex]I[\/latex]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167023762555\">We prove i. in the case of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex]. Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex] converge to the functions <em data-effect=\"italics\">f<\/em> and <em data-effect=\"italics\">g<\/em>, respectively, on the interval <em data-effect=\"italics\">I<\/em>. Let <em data-effect=\"italics\">x<\/em> be a point in <em data-effect=\"italics\">I<\/em> and let [latex]{S}_{N}\\left(x\\right)[\/latex] and [latex]{T}_{N}\\left(x\\right)[\/latex] denote the <em data-effect=\"italics\">N<\/em>th partial sums of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex], respectively. Then the sequence [latex]\\left\\{{S}_{N}\\left(x\\right)\\right\\}[\/latex] converges to [latex]f\\left(x\\right)[\/latex] and the sequence [latex]\\left\\{{T}_{N}\\left(x\\right)\\right\\}[\/latex] converges to [latex]g\\left(x\\right)[\/latex]. Furthermore, the <em data-effect=\"italics\">N<\/em>th partial sum of [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1167023750821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{N}} \\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)&amp; ={\\displaystyle\\sum _{n=0}^{N}} {c}_{n}{x}^{n}+{\\displaystyle\\sum _{n=0}^{N}}{d}_{n}{x}^{n}\\hfill \\\\ &amp; ={S}_{N}\\left(x\\right)+{T}_{N}\\left(x\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023709048\">Because<\/p>\r\n\r\n<div id=\"fs-id1167023910996\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc} \\hfill {\\underset{N\\to\\infty}\\lim} \\left({S}_{N}\\left(x\\right)+{T}_{N}\\left(x\\right)\\right) &amp; ={\\underset{N\\to\\infty}\\lim} {S}_{N}\\left(x\\right)+{\\underset{N\\to\\infty}\\lim} {T}_{N}\\left(x\\right)\\hfill \\\\ &amp; =f\\left(x\\right)+g\\left(x\\right),\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023732635\">we conclude that the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex] converges to [latex]f\\left(x\\right)+g\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1167023721350\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">These operations open up powerful possibilities. Since we know the power series representation for [latex]f(x) = \\frac{1}{1-x}[\/latex], we can now find power series representations for related functions like:<\/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]y = \\frac{3x}{1-x^2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]y = \\frac{1}{(x-1)(x-3)}[\/latex]<\/li>\r\n<\/ul>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Why the Same Interval?<\/strong> The interval of convergence restriction is crucial. Power series behave nicely only within their convergence intervals, so combining series requires working within the intersection of their individual convergence intervals.<\/section>\r\n<p class=\"whitespace-normal break-words\">We'll examine products of power series in a later theorem. For now, let's focus on applying these combining techniques and determining intervals of convergence for the new power series we create.<\/p>\r\n<p class=\"whitespace-normal break-words\">When you combine power series using the operations above, the new series will generally have the same interval of convergence as the original series. However, there are some important considerations to keep in mind when working with compositions and transformations.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023722989\" data-type=\"problem\">\r\n<p id=\"fs-id1167023724363\">Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a power series whose interval of convergence is [latex]\\left(-1,1\\right)[\/latex], and suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{b}_{n}{x}^{n}[\/latex] is a power series whose interval of convergence is [latex]\\left(-2,2\\right)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1167023806652\" type=\"a\">\r\n \t<li>Find the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\\right)[\/latex].<\/li>\r\n \t<li>Find the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{3}^{n}{x}^{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167023725316\" data-type=\"solution\">\r\n<ol id=\"fs-id1167023920675\" type=\"a\">\r\n \t<li style=\"text-align: left;\">Since the interval [latex]\\left(-1,1\\right)[\/latex] is a common interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{b}_{n}{x}^{n}[\/latex], the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\\right)[\/latex] is [latex]\\left(-1,1\\right)[\/latex].<\/li>\r\n \t<li>Since [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a power series centered at zero with radius of convergence 1, it converges for all <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-1,1\\right)[\/latex]. By Combining Power Series, the series<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167023810302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{3}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(3x\\right)}^{n}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nconverges if 3<em data-effect=\"italics\">x<\/em> is in the interval [latex]\\left(-1,1\\right)[\/latex]. Therefore, the series converges for all <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-\\frac{1}{3},\\frac{1}{3}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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 src=\"\/\/plugin.3playmedia.com\/show?mf=6724894&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=g4k9i7MSy3w&amp;video_target=tpm-plugin-oe3hpwbg-g4k9i7MSy3w\" 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\/6.2.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"6.2.1\" here (opens in new window)<\/a>.<\/section>\r\n<p id=\"fs-id1167023720105\">In the next example, we show how to use combining power series and the power series for a function [latex]f[\/latex] to construct power series for functions related to [latex]f[\/latex]. Specifically, we consider functions related to the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and we use the fact that [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\\cdots [\/latex] for [latex]|x|&lt;1[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023806693\" data-type=\"problem\">\r\n<p id=\"fs-id1167023780822\">Use the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.<\/p>\r\n\r\n<ol id=\"fs-id1167023767384\" type=\"a\">\r\n \t<li>[latex]f\\left(x\\right)=\\frac{3x}{1+{x}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)=\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1167023803086\" data-type=\"solution\">\r\n<ol id=\"fs-id1167023803088\" type=\"a\">\r\n \t<li>First write [latex]f\\left(x\\right)[\/latex] as\r\n<div id=\"fs-id1167023762803\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=3x\\left(\\frac{1}{1-\\left(\\text{-}{x}^{2}\\right)}\\right)[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUsing the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and parts ii. and iii. of Combining Power Series, we find that a power series representation for <em data-effect=\"italics\">f<\/em> is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023795119\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }3x{\\left(\\text{-}{x}^{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }3{\\left(-1\\right)}^{n}{x}^{2n+1}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the interval of convergence of the series for [latex]\\frac{1}{1-x}[\/latex] is [latex]\\left(-1,1\\right)[\/latex], the interval of convergence for this new series is the set of real numbers <em data-effect=\"italics\">x<\/em> such that [latex]|{x}^{2}|&lt;1[\/latex]. Therefore, the interval of convergence is [latex]\\left(-1,1\\right)[\/latex].\r\n<ul>\r\n \t<li>To find the power series representation, use partial fractions to write [latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(x - 3\\right)}[\/latex] as the sum of two fractions. We have<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ul>\r\n&nbsp;\r\n<div id=\"fs-id1167023911435\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}&amp; =\\frac{\\text{-}\\frac{1}{2}}{x - 1}+\\frac{\\frac{1}{2}}{x - 3}\\hfill \\\\ &amp; =\\frac{\\frac{1}{2}}{1-x}-\\frac{\\frac{1}{2}}{3-x}\\hfill \\\\ &amp; =\\frac{\\frac{1}{2}}{1-x}-\\frac{\\frac{1}{6}}{1-\\frac{x}{3}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFirst, using part ii. of Combining Power Series, we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023772794\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{2}}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{2}{x}^{n}\\text{for}|x|&lt;1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThen, using parts ii. and iii. of Combining Power Series, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023766046\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{6}}{1-\\frac{x}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{6}{\\left(\\frac{x}{3}\\right)}^{n}\\text{for}|x|&lt;3[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023920590\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{1}{2}-\\frac{1}{6\\cdot {3}^{n}}\\right){x}^{n}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhere the interval of convergence is [latex]\\left(-1,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section>In the previous example, we showed how to find power series for certain functions. In the next example we show how to do the opposite: given a power series, determine which function it represents.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023772610\" data-type=\"problem\">\r\n<p id=\"fs-id1167023772615\">Consider the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex]. Find the function <em data-effect=\"italics\">f<\/em> represented by this series. Determine the interval of convergence of the series.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1167023799497\" data-type=\"solution\">\r\n<p id=\"fs-id1167023766949\">Writing the given series as<\/p>\r\n\r\n<div id=\"fs-id1167023766952\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(2x\\right)}^{n}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023761296\">we can recognize this series as the power series for<\/p>\r\n\r\n<div id=\"fs-id1167023761299\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{1 - 2x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023775609\" style=\"text-align: left;\">Since this is a geometric series, the series converges if and only if [latex]|2x|&lt;1[\/latex]. Therefore, the interval of convergence is [latex]\\left(-\\frac{1}{2},\\frac{1}{2}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Add and subtract power series<\/li>\n<li><span data-sheets-root=\"1\">Multiply two power series together<\/span><\/li>\n<li>Find derivatives of power series term by term<\/li>\n<li>Integrate power series term by term<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Combining Power Series<\/h2>\n<p class=\"whitespace-normal break-words\">In the previous section, we learned how to represent certain functions using power series. Now we&#8217;ll explore how to manipulate these power series\u2014combining, modifying, and transforming them to create new power series representations.<\/p>\n<p class=\"whitespace-normal break-words\">This ability to work with power series is incredibly valuable for two key reasons:<\/p>\n<p class=\"whitespace-normal break-words\"><strong>First<\/strong>, it lets us find power series for functions that might otherwise be difficult to work with directly. For example, if we know the power series for [latex]f(x) = \\frac{1}{1-x}[\/latex], we can use calculus operations to find the power series for [latex]f'(x) = \\frac{1}{(1-x)^2}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Second<\/strong>, we can define entirely new functions that can&#8217;t be expressed using elementary functions. This becomes especially important when solving differential equations that have no elementary solutions.<\/p>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Basic Operations with Power Series<\/h3>\n<p class=\"whitespace-normal break-words\">When you have two power series that converge on the same interval, you can perform several operations to create new power series.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>theorem: combining power series<\/h3>\n<p id=\"fs-id1167023707168\">Suppose the power series [latex]\\sum_{n=0}^{\\infty} c_n x^n[\/latex] and [latex]\\sum_{n=0}^{\\infty} d_n x^n[\/latex] both converge to functions [latex]f[\/latex] and [latex]g[\/latex] respectively on a common interval [latex]I[\/latex].<\/p>\n<ol id=\"fs-id1167023719728\" type=\"i\">\n<li class=\"whitespace-normal break-words\"><strong>Addition\/Subtraction<\/strong>: [latex]\\sum_{n=0}^{\\infty} (c_n \\pm d_n) x^n[\/latex] converges to [latex]f \\pm g[\/latex] on [latex]I[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar and Power Multiplication<\/strong>: [latex]\\sum_{n=0}^{\\infty} bx^m c_n x^n[\/latex] converges to [latex]bx^m f(x)[\/latex] on [latex]I[\/latex] for any integer [latex]m \\geq 0[\/latex] and real number [latex]b[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Composition<\/strong>: [latex]\\sum_{n=0}^{\\infty} c_n (bx^m)^n[\/latex] converges to [latex]f(bx^m)[\/latex] for all [latex]x[\/latex] where [latex]bx^m[\/latex] is in [latex]I[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1167023762555\">We prove i. in the case of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex]. Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex] converge to the functions <em data-effect=\"italics\">f<\/em> and <em data-effect=\"italics\">g<\/em>, respectively, on the interval <em data-effect=\"italics\">I<\/em>. Let <em data-effect=\"italics\">x<\/em> be a point in <em data-effect=\"italics\">I<\/em> and let [latex]{S}_{N}\\left(x\\right)[\/latex] and [latex]{T}_{N}\\left(x\\right)[\/latex] denote the <em data-effect=\"italics\">N<\/em>th partial sums of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex], respectively. Then the sequence [latex]\\left\\{{S}_{N}\\left(x\\right)\\right\\}[\/latex] converges to [latex]f\\left(x\\right)[\/latex] and the sequence [latex]\\left\\{{T}_{N}\\left(x\\right)\\right\\}[\/latex] converges to [latex]g\\left(x\\right)[\/latex]. Furthermore, the <em data-effect=\"italics\">N<\/em>th partial sum of [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex] is<\/p>\n<div id=\"fs-id1167023750821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{N}} \\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)& ={\\displaystyle\\sum _{n=0}^{N}} {c}_{n}{x}^{n}+{\\displaystyle\\sum _{n=0}^{N}}{d}_{n}{x}^{n}\\hfill \\\\ & ={S}_{N}\\left(x\\right)+{T}_{N}\\left(x\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023709048\">Because<\/p>\n<div id=\"fs-id1167023910996\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc} \\hfill {\\underset{N\\to\\infty}\\lim} \\left({S}_{N}\\left(x\\right)+{T}_{N}\\left(x\\right)\\right) & ={\\underset{N\\to\\infty}\\lim} {S}_{N}\\left(x\\right)+{\\underset{N\\to\\infty}\\lim} {T}_{N}\\left(x\\right)\\hfill \\\\ & =f\\left(x\\right)+g\\left(x\\right),\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023732635\">we conclude that the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\\right)[\/latex] converges to [latex]f\\left(x\\right)+g\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1167023721350\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">These operations open up powerful possibilities. Since we know the power series representation for [latex]f(x) = \\frac{1}{1-x}[\/latex], we can now find power series representations for related functions like:<\/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]y = \\frac{3x}{1-x^2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y = \\frac{1}{(x-1)(x-3)}[\/latex]<\/li>\n<\/ul>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Why the Same Interval?<\/strong> The interval of convergence restriction is crucial. Power series behave nicely only within their convergence intervals, so combining series requires working within the intersection of their individual convergence intervals.<\/section>\n<p class=\"whitespace-normal break-words\">We&#8217;ll examine products of power series in a later theorem. For now, let&#8217;s focus on applying these combining techniques and determining intervals of convergence for the new power series we create.<\/p>\n<p class=\"whitespace-normal break-words\">When you combine power series using the operations above, the new series will generally have the same interval of convergence as the original series. However, there are some important considerations to keep in mind when working with compositions and transformations.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023722989\" data-type=\"problem\">\n<p id=\"fs-id1167023724363\">Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a power series whose interval of convergence is [latex]\\left(-1,1\\right)[\/latex], and suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{b}_{n}{x}^{n}[\/latex] is a power series whose interval of convergence is [latex]\\left(-2,2\\right)[\/latex].<\/p>\n<ol id=\"fs-id1167023806652\" type=\"a\">\n<li>Find the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\\right)[\/latex].<\/li>\n<li>Find the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{3}^{n}{x}^{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\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<div id=\"fs-id1167023725316\" data-type=\"solution\">\n<ol id=\"fs-id1167023920675\" type=\"a\">\n<li style=\"text-align: left;\">Since the interval [latex]\\left(-1,1\\right)[\/latex] is a common interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{b}_{n}{x}^{n}[\/latex], the interval of convergence of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\\right)[\/latex] is [latex]\\left(-1,1\\right)[\/latex].<\/li>\n<li>Since [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a power series centered at zero with radius of convergence 1, it converges for all <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-1,1\\right)[\/latex]. By Combining Power Series, the series<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<div id=\"fs-id1167023810302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{3}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(3x\\right)}^{n}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nconverges if 3<em data-effect=\"italics\">x<\/em> is in the interval [latex]\\left(-1,1\\right)[\/latex]. Therefore, the series converges for all <em data-effect=\"italics\">x<\/em> in the interval [latex]\\left(-\\frac{1}{3},\\frac{1}{3}\\right)[\/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\" src=\"\/\/plugin.3playmedia.com\/show?mf=6724894&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=g4k9i7MSy3w&amp;video_target=tpm-plugin-oe3hpwbg-g4k9i7MSy3w\" 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\/6.2.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;6.2.1&#8221; here (opens in new window)<\/a>.<\/section>\n<p id=\"fs-id1167023720105\">In the next example, we show how to use combining power series and the power series for a function [latex]f[\/latex] to construct power series for functions related to [latex]f[\/latex]. Specifically, we consider functions related to the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and we use the fact that [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\\cdots[\/latex] for [latex]|x|<1[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023806693\" data-type=\"problem\">\n<p id=\"fs-id1167023780822\">Use the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.<\/p>\n<ol id=\"fs-id1167023767384\" type=\"a\">\n<li>[latex]f\\left(x\\right)=\\frac{3x}{1+{x}^{2}}[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)=\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023803086\" data-type=\"solution\">\n<ol id=\"fs-id1167023803088\" type=\"a\">\n<li>First write [latex]f\\left(x\\right)[\/latex] as\n<div id=\"fs-id1167023762803\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=3x\\left(\\frac{1}{1-\\left(\\text{-}{x}^{2}\\right)}\\right)[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUsing the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and parts ii. and iii. of Combining Power Series, we find that a power series representation for <em data-effect=\"italics\">f<\/em> is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023795119\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }3x{\\left(\\text{-}{x}^{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }3{\\left(-1\\right)}^{n}{x}^{2n+1}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the interval of convergence of the series for [latex]\\frac{1}{1-x}[\/latex] is [latex]\\left(-1,1\\right)[\/latex], the interval of convergence for this new series is the set of real numbers <em data-effect=\"italics\">x<\/em> such that [latex]|{x}^{2}|<1[\/latex]. Therefore, the interval of convergence is [latex]\\left(-1,1\\right)[\/latex].\n\n\n<ul>\n<li>To find the power series representation, use partial fractions to write [latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(x - 3\\right)}[\/latex] as the sum of two fractions. We have<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167023911435\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}& =\\frac{\\text{-}\\frac{1}{2}}{x - 1}+\\frac{\\frac{1}{2}}{x - 3}\\hfill \\\\ & =\\frac{\\frac{1}{2}}{1-x}-\\frac{\\frac{1}{2}}{3-x}\\hfill \\\\ & =\\frac{\\frac{1}{2}}{1-x}-\\frac{\\frac{1}{6}}{1-\\frac{x}{3}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFirst, using part ii. of Combining Power Series, we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023772794\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{2}}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{2}{x}^{n}\\text{for}|x|<1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThen, using parts ii. and iii. of Combining Power Series, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023766046\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{6}}{1-\\frac{x}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{6}{\\left(\\frac{x}{3}\\right)}^{n}\\text{for}|x|<3[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023920590\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{1}{2}-\\frac{1}{6\\cdot {3}^{n}}\\right){x}^{n}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhere the interval of convergence is [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/section>\n<p>In the previous example, we showed how to find power series for certain functions. In the next example we show how to do the opposite: given a power series, determine which function it represents.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023772610\" data-type=\"problem\">\n<p id=\"fs-id1167023772615\">Consider the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex]. Find the function <em data-effect=\"italics\">f<\/em> represented by this series. Determine the interval of convergence of the series.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023799497\" data-type=\"solution\">\n<p id=\"fs-id1167023766949\">Writing the given series as<\/p>\n<div id=\"fs-id1167023766952\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(2x\\right)}^{n}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023761296\">we can recognize this series as the power series for<\/p>\n<div id=\"fs-id1167023761299\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{1 - 2x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023775609\" style=\"text-align: left;\">Since this is a geometric series, the series converges if and only if [latex]|2x|<1[\/latex]. Therefore, the interval of convergence is [latex]\\left(-\\frac{1}{2},\\frac{1}{2}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":11,"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\/933"}],"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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/933\/revisions"}],"predecessor-version":[{"id":2326,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/933\/revisions\/2326"}],"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\/933\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=933"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=933"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=933"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}