{"id":1774,"date":"2021-07-27T00:11:41","date_gmt":"2021-07-27T00:11:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1774"},"modified":"2022-03-21T23:10:55","modified_gmt":"2022-03-21T23:10:55","slug":"combining-and-multiplying-power-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/combining-and-multiplying-power-series\/","title":{"raw":"Combining and Multiplying Power Series","rendered":"Combining and Multiplying 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>Combine power series by addition or subtraction<\/li>\r\n \t<li>Create a new power series by multiplication by a power of the variable or a constant, or by substitution<\/li>\r\n \t<li>Multiply two power series together<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167023772636\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Combining Power Series<\/h2>\r\n<p id=\"fs-id1167023781951\">If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of <em data-effect=\"italics\">x<\/em> or evaluate a power series at [latex]{x}^{m}[\/latex] for a positive integer <em data-effect=\"italics\">m<\/em> to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can find power series representations for related functions, such as<\/p>\r\n\r\n<div id=\"fs-id1167023767309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{3x}{1-{x}^{2}}\\text{ and }y=\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023743578\">In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at [latex]x=0[\/latex]. Similar results hold for power series centered at [latex]x=a[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167023714096\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Combining Power Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167023707168\">Suppose that the two power series [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 a common interval <em data-effect=\"italics\">I<\/em>.<\/p>\r\n\r\n<ol id=\"fs-id1167023719728\" type=\"i\">\r\n \t<li>The power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}\\pm {d}_{n}{x}^{n}\\right)[\/latex] converges to [latex]f\\pm g[\/latex] on <em data-effect=\"italics\">I<\/em>.<\/li>\r\n \t<li>For any integer [latex]m\\ge 0[\/latex] and any real number <em data-effect=\"italics\">b<\/em>, the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }b{x}^{m}{c}_{n}{x}^{n}[\/latex] converges to [latex]b{x}^{m}f\\left(x\\right)[\/latex] on <em data-effect=\"italics\">I<\/em>.<\/li>\r\n \t<li>For any integer [latex]m\\ge 0[\/latex] and any real number <em data-effect=\"italics\">b<\/em>, the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(b{x}^{m}\\right)}^{n}[\/latex] converges to [latex]f\\left(b{x}^{m}\\right)[\/latex] for all <em data-effect=\"italics\">x<\/em> such that [latex]b{x}^{m}[\/latex] is in <em data-effect=\"italics\">I<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167023750235\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\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<p id=\"fs-id1167023721354\">We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series.<\/p>\r\n\r\n<div id=\"fs-id1167023715788\" data-type=\"example\">\r\n<div id=\"fs-id1167023722987\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023722989\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Combining Power Series<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/section>[\/hidden-answer]\r\n\r\nWatch the following video to see the worked solution to Example: Combining Power Series.\r\n\r\n<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\"><\/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>.\r\n<div id=\"fs-id1167023762725\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023762728\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023765375\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023765375\" data-type=\"problem\">\r\n<p id=\"fs-id1167023765377\">Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] has an interval of convergence of [latex]\\left(-1,1\\right)[\/latex]. Find the interval of convergence of [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(\\frac{x}{2}\\right)}^{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-id1167023787253\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023797425\">Find the values of <em data-effect=\"italics\">x<\/em> such that [latex]\\frac{x}{2}[\/latex] is in the interval [latex]\\left(-1,1\\right)[\/latex].<\/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-id1167023758868\" data-type=\"solution\">\r\n<p id=\"fs-id1167023758870\">Interval of convergence is [latex]\\left(-2,2\\right)[\/latex].<\/p>\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<p id=\"fs-id1167023720105\">In the next example, we show how to use Combining Power Series and the power series for a function <em data-effect=\"italics\">f<\/em> to construct power series for functions related to <em data-effect=\"italics\">f<\/em>. Specifically, we consider functions related to the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and we use the fact that<\/p>\r\n\r\n<div id=\"fs-id1167023913652\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023771403\">for [latex]|x|&lt;1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167023806689\" data-type=\"example\">\r\n<div id=\"fs-id1167023806691\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023806693\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Constructing Power Series from Known Power Series<\/h3>\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<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\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]\r\n<div id=\"fs-id1167023922300\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023922304\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023922306\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023922306\" data-type=\"problem\">\r\n<p id=\"fs-id1167023914570\">Use the series for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] on [latex]|x|&lt;1[\/latex] to construct a series for [latex]\\frac{1}{\\left(1-x\\right)\\left(x - 2\\right)}[\/latex]. Determine the interval of convergence.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1167023914328\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023914334\">Use partial fractions to rewrite [latex]\\frac{1}{\\left(1-x\\right)\\left(x - 2\\right)}[\/latex] as the difference of two fractions.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167023914151\" data-type=\"solution\">\r\n<p id=\"fs-id1167023914154\" style=\"text-align: left;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1+\\frac{1}{{2}^{n+1}}\\right){x}^{n}[\/latex]. The interval of convergence is [latex]\\left(-1,1\\right)[\/latex].<\/p>\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<p id=\"fs-id1167023781827\">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>\r\n\r\n<div id=\"fs-id1167023772606\" data-type=\"example\">\r\n<div id=\"fs-id1167023772608\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023772610\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Function Represented by a Given Power Series<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167023772113\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023917754\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023917756\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023917756\" data-type=\"problem\">\r\n<p id=\"fs-id1167023917758\">Find the function represented by the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{{3}^{n}}{x}^{n}[\/latex]. Determine its 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-id1167023801355\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023801362\">Write [latex]\\frac{1}{{3}^{n}}{x}^{n}={\\left(\\frac{x}{3}\\right)}^{n}[\/latex].<\/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<div id=\"fs-id1167023922733\" data-type=\"solution\">\r\n<p id=\"fs-id1167023922735\">[latex]f\\left(x\\right)=\\frac{3}{3-x}[\/latex]. The interval of convergence is [latex]\\left(-3,3\\right)[\/latex].<\/p>\r\n\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[ohm_question]169426[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167023771436\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Multiplication of Power Series<\/h2>\r\n<p id=\"fs-id1167023771441\">We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.<\/p>\r\n<p id=\"fs-id1167023771446\">The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply<\/p>\r\n\r\n<div id=\"fs-id1167023771450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023803417\">and<\/p>\r\n\r\n<div id=\"fs-id1167023803420\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023805856\">It appears that the product should satisfy<\/p>\r\n\r\n<div id=\"fs-id1167023805859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=-0}^{\\infty }{d}_{n}{x}^{n}\\right)&amp; =\\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots \\right)\\cdot \\left({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots \\right)\\hfill \\\\ &amp; ={c}_{0}{d}_{0}+\\left({c}_{1}{d}_{0}+{c}_{0}{d}_{1}\\right)x+\\left({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2}\\right){x}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023913843\">In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex] converge on a common interval <em data-effect=\"italics\">I<\/em>, then we can multiply the series in this way, and the resulting series also converges on the interval <em data-effect=\"italics\">I<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1167023808343\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Multiplying Power Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167023808351\">Suppose that the power series [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 <em data-effect=\"italics\">f<\/em> and <em data-effect=\"italics\">g<\/em>, respectively, on a common interval <em data-effect=\"italics\">I<\/em>. Let<\/p>\r\n\r\n<div id=\"fs-id1167023809633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}_{n}&amp; ={c}_{0}{d}_{n}+{c}_{1}{d}_{n - 1}+{c}_{2}{d}_{n - 2}+\\cdots +{c}_{n - 1}{d}_{1}+{c}_{n}{d}_{0}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{k=0}^{n}{c}_{k}{d}_{n-k}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023777233\">Then<\/p>\r\n\r\n<div id=\"fs-id1167023777236\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left({\\displaystyle\\sum _{n=0}^{\\infty}}{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023914195\">and<\/p>\r\n\r\n<div id=\"fs-id1171361418902\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}\\text{converges to}f\\left(x\\right)\\cdot g\\left(x\\right)\\text{ on }I[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023763107\">The series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}[\/latex] is known as the Cauchy product 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].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167023922191\">We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for<\/p>\r\n\r\n<div id=\"fs-id1167023922196\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(1-{x}^{2}\\right)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023770166\">using the power series representations for<\/p>\r\n\r\n<div id=\"fs-id1167023770169\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{1}{1-x}\\text{and}y=\\frac{1}{1-{x}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167023799559\" data-type=\"example\">\r\n<div id=\"fs-id1167023799561\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023799563\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Multiplying Power Series<\/h3>\r\n<div id=\"fs-id1167023799563\" data-type=\"problem\">\r\n<p id=\"fs-id1167023799568\">Multiply the power series representation<\/p>\r\n\r\n<div id=\"fs-id1167023799572\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-x}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ &amp; =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023783661\">for [latex]|x|&lt;1[\/latex] with the power series representation<\/p>\r\n\r\n<div id=\"fs-id1167023783678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-{x}^{2}}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left({x}^{2}\\right)}^{n}\\hfill \\\\ &amp; =1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023798800\">for [latex]|x|&lt;1[\/latex] to construct a power series for [latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(1-{x}^{2}\\right)}[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1167023764071\" data-type=\"solution\">\r\n<p id=\"fs-id1167023764073\">We need to multiply<\/p>\r\n\r\n<div id=\"fs-id1167023764076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023809752\">Writing out the first several terms, we see that the product is given by<\/p>\r\n\r\n<div id=\"fs-id1167023809755\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)+\\left(x+{x}^{3}+{x}^{5}+{x}^{7}+\\cdots \\right)+\\left({x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\\cdots \\right)+\\left({x}^{3}+{x}^{5}+{x}^{7}+{x}^{9}+\\cdots \\right)\\hfill \\\\ =1+x+\\left(1+1\\right){x}^{2}+\\left(1+1\\right){x}^{3}+\\left(1+1+1\\right){x}^{4}+\\left(1+1+1\\right){x}^{5}+\\cdots \\hfill \\\\ =1+x+2{x}^{2}+2{x}^{3}+3{x}^{4}+3{x}^{5}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023916155\">Since the series for [latex]y=\\frac{1}{1-x}[\/latex] and [latex]y=\\frac{1}{1-{x}^{2}}[\/latex] both converge on the interval [latex]\\left(-1,1\\right)[\/latex], the series for the product also converges on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\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\nWatch the following video to see the worked solution to Example: Multiplying Power Series.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lFb1PrmZKcA?controls=0&amp;start=0&amp;end=128&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\/6.2.3_0to128_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.2.3\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167023797330\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023797333\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023797335\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167023797335\" data-type=\"problem\">\r\n<p id=\"fs-id1167023797337\">Multiply the series [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] by itself to construct a series for [latex]\\frac{1}{\\left(1-x\\right)\\left(1-x\\right)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1167023803277\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023803284\">Multiply the first few terms of [latex]\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1167023803238\" data-type=\"solution\">\r\n<p id=\"fs-id1167023803240\" style=\"text-align: center;\">[latex]1+2x+3{x}^{2}+4{x}^{3}+\\cdots [\/latex]<\/p>\r\n\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[ohm_question]169454[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1167023795144\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Combine power series by addition or subtraction<\/li>\n<li>Create a new power series by multiplication by a power of the variable or a constant, or by substitution<\/li>\n<li>Multiply two power series together<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167023772636\" data-depth=\"1\">\n<h2 data-type=\"title\">Combining Power Series<\/h2>\n<p id=\"fs-id1167023781951\">If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of <em data-effect=\"italics\">x<\/em> or evaluate a power series at [latex]{x}^{m}[\/latex] for a positive integer <em data-effect=\"italics\">m<\/em> to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can find power series representations for related functions, such as<\/p>\n<div id=\"fs-id1167023767309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{3x}{1-{x}^{2}}\\text{ and }y=\\frac{1}{\\left(x - 1\\right)\\left(x - 3\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023743578\">In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at [latex]x=0[\/latex]. Similar results hold for power series centered at [latex]x=a[\/latex].<\/p>\n<div id=\"fs-id1167023714096\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Combining Power Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167023707168\">Suppose that the two power series [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 a common interval <em data-effect=\"italics\">I<\/em>.<\/p>\n<ol id=\"fs-id1167023719728\" type=\"i\">\n<li>The power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left({c}_{n}{x}^{n}\\pm {d}_{n}{x}^{n}\\right)[\/latex] converges to [latex]f\\pm g[\/latex] on <em data-effect=\"italics\">I<\/em>.<\/li>\n<li>For any integer [latex]m\\ge 0[\/latex] and any real number <em data-effect=\"italics\">b<\/em>, the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }b{x}^{m}{c}_{n}{x}^{n}[\/latex] converges to [latex]b{x}^{m}f\\left(x\\right)[\/latex] on <em data-effect=\"italics\">I<\/em>.<\/li>\n<li>For any integer [latex]m\\ge 0[\/latex] and any real number <em data-effect=\"italics\">b<\/em>, the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(b{x}^{m}\\right)}^{n}[\/latex] converges to [latex]f\\left(b{x}^{m}\\right)[\/latex] for all <em data-effect=\"italics\">x<\/em> such that [latex]b{x}^{m}[\/latex] is in <em data-effect=\"italics\">I<\/em>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167023750235\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\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<p id=\"fs-id1167023721354\">We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series.<\/p>\n<div id=\"fs-id1167023715788\" data-type=\"example\">\n<div id=\"fs-id1167023722987\" data-type=\"exercise\">\n<div id=\"fs-id1167023722989\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Combining Power Series<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Combining Power Series.<\/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\"><\/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>.<\/p>\n<div id=\"fs-id1167023762725\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023762728\" data-type=\"exercise\">\n<div id=\"fs-id1167023765375\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023765375\" data-type=\"problem\">\n<p id=\"fs-id1167023765377\">Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] has an interval of convergence of [latex]\\left(-1,1\\right)[\/latex]. Find the interval of convergence of [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(\\frac{x}{2}\\right)}^{n}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023787253\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023797425\">Find the values of <em data-effect=\"italics\">x<\/em> such that [latex]\\frac{x}{2}[\/latex] is in the interval [latex]\\left(-1,1\\right)[\/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=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023758868\" data-type=\"solution\">\n<p id=\"fs-id1167023758870\">Interval of convergence is [latex]\\left(-2,2\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023720105\">In the next example, we show how to use Combining Power Series and the power series for a function <em data-effect=\"italics\">f<\/em> to construct power series for functions related to <em data-effect=\"italics\">f<\/em>. Specifically, we consider functions related to the function [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] and we use the fact that<\/p>\n<div id=\"fs-id1167023913652\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023771403\">for [latex]|x|<1[\/latex].<\/p>\n<div id=\"fs-id1167023806689\" data-type=\"example\">\n<div id=\"fs-id1167023806691\" data-type=\"exercise\">\n<div id=\"fs-id1167023806693\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Constructing Power Series from Known Power Series<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/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<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\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<div id=\"fs-id1167023922300\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023922304\" data-type=\"exercise\">\n<div id=\"fs-id1167023922306\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023922306\" data-type=\"problem\">\n<p id=\"fs-id1167023914570\">Use the series for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex] on [latex]|x|<1[\/latex] to construct a series for [latex]\\frac{1}{\\left(1-x\\right)\\left(x - 2\\right)}[\/latex]. Determine 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=\"q44558893\">Hint<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023914328\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023914334\">Use partial fractions to rewrite [latex]\\frac{1}{\\left(1-x\\right)\\left(x - 2\\right)}[\/latex] as the difference of two fractions.<\/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-id1167023914151\" data-type=\"solution\">\n<p id=\"fs-id1167023914154\" style=\"text-align: left;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1+\\frac{1}{{2}^{n+1}}\\right){x}^{n}[\/latex]. The interval of convergence is [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023781827\">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<div id=\"fs-id1167023772606\" data-type=\"example\">\n<div id=\"fs-id1167023772608\" data-type=\"exercise\">\n<div id=\"fs-id1167023772610\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Function Represented by a Given Power Series<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023772113\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023917754\" data-type=\"exercise\">\n<div id=\"fs-id1167023917756\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023917756\" data-type=\"problem\">\n<p id=\"fs-id1167023917758\">Find the function represented by the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{{3}^{n}}{x}^{n}[\/latex]. Determine its 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-id1167023801355\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023801362\">Write [latex]\\frac{1}{{3}^{n}}{x}^{n}={\\left(\\frac{x}{3}\\right)}^{n}[\/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=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023922733\" data-type=\"solution\">\n<p id=\"fs-id1167023922735\">[latex]f\\left(x\\right)=\\frac{3}{3-x}[\/latex]. The interval of convergence is [latex]\\left(-3,3\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169426\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169426&theme=oea&iframe_resize_id=ohm169426&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167023771436\" data-depth=\"1\">\n<h2 data-type=\"title\">Multiplication of Power Series<\/h2>\n<p id=\"fs-id1167023771441\">We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.<\/p>\n<p id=\"fs-id1167023771446\">The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply<\/p>\n<div id=\"fs-id1167023771450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023803417\">and<\/p>\n<div id=\"fs-id1167023803420\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023805856\">It appears that the product should satisfy<\/p>\n<div id=\"fs-id1167023805859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=-0}^{\\infty }{d}_{n}{x}^{n}\\right)& =\\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots \\right)\\cdot \\left({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots \\right)\\hfill \\\\ & ={c}_{0}{d}_{0}+\\left({c}_{1}{d}_{0}+{c}_{0}{d}_{1}\\right)x+\\left({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2}\\right){x}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023913843\">In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}[\/latex] converge on a common interval <em data-effect=\"italics\">I<\/em>, then we can multiply the series in this way, and the resulting series also converges on the interval <em data-effect=\"italics\">I<\/em>.<\/p>\n<div id=\"fs-id1167023808343\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Multiplying Power Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167023808351\">Suppose that the power series [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 <em data-effect=\"italics\">f<\/em> and <em data-effect=\"italics\">g<\/em>, respectively, on a common interval <em data-effect=\"italics\">I<\/em>. Let<\/p>\n<div id=\"fs-id1167023809633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {e}_{n}& ={c}_{0}{d}_{n}+{c}_{1}{d}_{n - 1}+{c}_{2}{d}_{n - 2}+\\cdots +{c}_{n - 1}{d}_{1}+{c}_{n}{d}_{0}\\hfill \\\\ & ={\\displaystyle\\sum _{k=0}^{n}{c}_{k}{d}_{n-k}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023777233\">Then<\/p>\n<div id=\"fs-id1167023777236\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left({\\displaystyle\\sum _{n=0}^{\\infty}}{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023914195\">and<\/p>\n<div id=\"fs-id1171361418902\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}\\text{converges to}f\\left(x\\right)\\cdot g\\left(x\\right)\\text{ on }I[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023763107\">The series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}[\/latex] is known as the Cauchy product 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].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023922191\">We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for<\/p>\n<div id=\"fs-id1167023922196\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(1-{x}^{2}\\right)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023770166\">using the power series representations for<\/p>\n<div id=\"fs-id1167023770169\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{1}{1-x}\\text{and}y=\\frac{1}{1-{x}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167023799559\" data-type=\"example\">\n<div id=\"fs-id1167023799561\" data-type=\"exercise\">\n<div id=\"fs-id1167023799563\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Multiplying Power Series<\/h3>\n<div id=\"fs-id1167023799563\" data-type=\"problem\">\n<p id=\"fs-id1167023799568\">Multiply the power series representation<\/p>\n<div id=\"fs-id1167023799572\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-x}& ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ & =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023783661\">for [latex]|x|<1[\/latex] with the power series representation<\/p>\n<div id=\"fs-id1167023783678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-{x}^{2}}& ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left({x}^{2}\\right)}^{n}\\hfill \\\\ & =1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023798800\">for [latex]|x|<1[\/latex] to construct a power series for [latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(1-{x}^{2}\\right)}[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023764071\" data-type=\"solution\">\n<p id=\"fs-id1167023764073\">We need to multiply<\/p>\n<div id=\"fs-id1167023764076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023809752\">Writing out the first several terms, we see that the product is given by<\/p>\n<div id=\"fs-id1167023809755\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)+\\left(x+{x}^{3}+{x}^{5}+{x}^{7}+\\cdots \\right)+\\left({x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\\cdots \\right)+\\left({x}^{3}+{x}^{5}+{x}^{7}+{x}^{9}+\\cdots \\right)\\hfill \\\\ =1+x+\\left(1+1\\right){x}^{2}+\\left(1+1\\right){x}^{3}+\\left(1+1+1\\right){x}^{4}+\\left(1+1+1\\right){x}^{5}+\\cdots \\hfill \\\\ =1+x+2{x}^{2}+2{x}^{3}+3{x}^{4}+3{x}^{5}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023916155\">Since the series for [latex]y=\\frac{1}{1-x}[\/latex] and [latex]y=\\frac{1}{1-{x}^{2}}[\/latex] both converge on the interval [latex]\\left(-1,1\\right)[\/latex], the series for the product also converges on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Multiplying Power Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lFb1PrmZKcA?controls=0&amp;start=0&amp;end=128&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\/6.2.3_0to128_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.2.3&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167023797330\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023797333\" data-type=\"exercise\">\n<div id=\"fs-id1167023797335\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167023797335\" data-type=\"problem\">\n<p id=\"fs-id1167023797337\">Multiply the series [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] by itself to construct a series for [latex]\\frac{1}{\\left(1-x\\right)\\left(1-x\\right)}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Hint<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023803277\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023803284\">Multiply the first few terms of [latex]\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)[\/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=\"q44558879\">Show Solution<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023803238\" data-type=\"solution\">\n<p id=\"fs-id1167023803240\" style=\"text-align: center;\">[latex]1+2x+3{x}^{2}+4{x}^{3}+\\cdots[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169454\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169454&theme=oea&iframe_resize_id=ohm169454&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1167023795144\" data-depth=\"1\"><\/section>\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-1774\">\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>6.2.1. <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><li>6.2.3. <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":7,"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\":\"6.2.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"6.2.3\",\"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-1774","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1774","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":8,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1774\/revisions"}],"predecessor-version":[{"id":2246,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1774\/revisions\/2246"}],"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\/1774\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1774"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1774"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1774"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1774"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}