{"id":938,"date":"2025-06-20T17:23:55","date_gmt":"2025-06-20T17:23:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=938"},"modified":"2025-09-10T13:15:06","modified_gmt":"2025-09-10T13:15:06","slug":"operations-with-power-series-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/operations-with-power-series-fresh-take\/","title":{"raw":"Operations with Power Series: Fresh Take","rendered":"Operations with Power Series: Fresh Take"},"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<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Once you have power series representations for basic functions, you can combine and manipulate them to create power series for more complex functions. This is like having a toolkit where each known power series becomes a building block for constructing new ones.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The three key operations:<\/strong> For power series that converge on a common interval, you can:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Addition\/Subtraction:<\/strong> [latex]\\sum c_n x^n \\pm \\sum d_n x^n = \\sum (c_n \\pm d_n) x^n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar and Power Multiplication:<\/strong> [latex]bx^m \\sum c_n x^n = \\sum bx^m c_n x^n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Composition:<\/strong> Replace [latex]x[\/latex] with [latex]bx^m[\/latex] in your series: [latex]\\sum c_n (bx^m)^n[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Since [latex]\\frac{1}{1-x} = \\sum_{n=0}^{\\infty} x^n[\/latex] for [latex]|x| &lt; 1[\/latex], you can use substitution to create new series. For example:<\/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\">Replace [latex]x[\/latex] with [latex]-x^2[\/latex]: [latex]\\frac{1}{1+x^2} = \\sum_{n=0}^{\\infty} (-x^2)^n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Multiply by [latex]3x[\/latex]: [latex]\\frac{3x}{1+x^2} = 3x \\sum_{n=0}^{\\infty} (-x^2)^n[\/latex]<\/li>\r\n<\/ul>\r\nWhen combining series, your new series converges on the <strong>intersection<\/strong> of the original convergence intervals. If one series works on [latex](-1,1)[\/latex] and another on [latex](-2,2)[\/latex] , their combination works on [latex](-1,1)[\/latex] .\r\n\r\n<strong>Strategy for compositions:<\/strong> When you substitute [latex]bx^m[\/latex] for [latex]x[\/latex] in a series that converges for [latex]|x| &lt; R[\/latex], the new series converges when [latex]|bx^m| &lt; R[\/latex]. Solve this inequality to find your new interval.\r\n\r\nYou can also work backwards. If you see [latex]\\sum 2^n x^n = \\sum (2x)^n[\/latex], recognize this as [latex]\\frac{1}{1-2x}[\/latex] with convergence when [latex]|2x| &lt; 1[\/latex], or [latex]|x| &lt; \\frac{1}{2}[\/latex].\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Multiplication of Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When you multiply two power series together, you're essentially multiplying two \"infinite polynomials.\" Just like with regular polynomial multiplication, you distribute each term and collect like powers of [latex]x[\/latex] to form new coefficients.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The multiplication process:<\/strong> To multiply [latex]\\sum c_n x^n[\/latex] and [latex]\\sum d_n x^n[\/latex], you need to find all the ways to get each power [latex]x^n[\/latex] in the product. For the coefficient of [latex]x^n[\/latex], collect all combinations where the powers add to [latex]n[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]e_n = c_0 d_n + c_1 d_{n-1} + c_2 d_{n-2} + \\cdots + c_n d_0[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is called the <strong>Cauchy product<\/strong>. Each new coefficient [latex]e_n[\/latex] comes from all possible ways to pair terms from the two original series that multiply to give [latex]x^n[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">If both original series converge on interval [latex]I[\/latex], then their product series also converges on [latex]I[\/latex] and represents the product of the two functions.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Practical strategy:<\/strong> Instead of using the general formula, you can often work out the first several terms by hand:<\/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\">Multiply each term in the first series by each term in the second series<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Group terms with the same power of [latex]x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Add up the coefficients for each power<\/li>\r\n<\/ul>\r\nWhen you see a rational function that factors into simpler pieces, consider whether you can find power series for the individual factors and then multiply them together.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2>Differentiating and Integrating Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">One of the most powerful features of power series is that you can differentiate and integrate them term-by-term, just like polynomials. This transforms complex calculus problems into straightforward arithmetic operations.<\/p>\r\n<p class=\"whitespace-normal break-words\">The process is simple:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>To differentiate:<\/strong> [latex]\\frac{d}{dx}\\sum c_n x^n = \\sum nc_n x^{n-1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>To integrate:<\/strong> [latex]\\int \\sum c_n x^n dx = C + \\sum c_n \\frac{x^{n+1}}{n+1}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">You apply the power rule to each term individually. For [latex]c_n x^n[\/latex], the derivative becomes [latex]nc_n x^{n-1}[\/latex] and the integral becomes [latex]c_n \\frac{x^{n+1}}{n+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Both the differentiated and integrated series have the same radius of convergence as the original series. However, endpoint behavior might change - always check endpoints separately if needed.<\/p>\r\n<p class=\"whitespace-normal break-words\">This technique lets you find power series for functions that would be difficult to obtain directly:<\/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\">Start with [latex]\\frac{1}{1-x} = \\sum x^n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Differentiate to get [latex]\\frac{1}{(1-x)^2} = \\sum (n+1)x^n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate [latex]\\frac{1}{1+x} = \\sum (-1)^n x^n[\/latex] to get [latex]\\ln(1+x) = \\sum (-1)^{n+1}\\frac{x^n}{n}[\/latex]<\/li>\r\n<\/ul>\r\nWhen integrating, use initial conditions to find [latex]C[\/latex]. For example, if [latex]f(0) = 0[\/latex], then [latex]C = 0[\/latex].\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023778779\" data-type=\"problem\">\r\n<p id=\"fs-id1167023778781\">Differentiate the series [latex]\\frac{1}{{\\left(1-x\\right)}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){x}^{n}[\/latex] term-by-term to find a power series representation for [latex]\\frac{2}{{\\left(1-x\\right)}^{3}}[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1167023911140\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023911147\">Write out the first several terms and apply the power rule.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167023911084\" data-type=\"solution\">\r\n<p id=\"fs-id1167023911086\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+2\\right)\\left(n+1\\right){x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167023788087\" data-type=\"problem\">\r\n<p id=\"fs-id1167023788089\">Integrate the power series [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}[\/latex] term-by-term to evaluate [latex]\\displaystyle\\int \\text{ln}\\left(1+x\\right)dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558809\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558809\"]\r\n<div id=\"fs-id1167023804519\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167023804526\">Use the fact that [latex]\\frac{{x}^{n+1}}{\\left(n+1\\right)n}[\/latex] is an antiderivative of [latex]\\frac{{x}^{n}}{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558819\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1167023804458\" data-type=\"solution\">\r\n<p id=\"fs-id1167023804461\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{n}}{n\\left(n - 1\\right)}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Uniqueness of Power Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When you find a power series representation for a function, you might wonder: \"Could there be another, completely different power series for the same function?\" The answer is no - power series representations are unique.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The uniqueness principle:<\/strong> If two power series [latex]\\sum c_n (x-a)^n[\/latex] and [latex]\\sum d_n (x-a)^n[\/latex] both converge and represent the same function on an open interval containing [latex]a[\/latex], then their coefficients must be identical: [latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Think of power series as \"infinite polynomials.\" Just as two polynomials can only be equal if they have identical coefficients, the same principle applies to power series. If [latex]c_0 + c_1 x + c_2 x^2 + \\cdots = d_0 + d_1 x + d_2 x^2 + \\cdots[\/latex] for all [latex]x[\/latex] in some interval, then [latex]c_0 = d_0[\/latex], [latex]c_1 = d_1[\/latex], [latex]c_2 = d_2[\/latex], and so on.<\/p>\r\n<p class=\"whitespace-normal break-words\">This uniqueness means you can use any valid method to find a power series - algebraic manipulation, differentiation, integration, or others we'll learn later. No matter which approach you use, you'll always get the same series for a given function centered at a specific point.<\/p>\r\n\r\n<\/div>","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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you have power series representations for basic functions, you can combine and manipulate them to create power series for more complex functions. This is like having a toolkit where each known power series becomes a building block for constructing new ones.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The three key operations:<\/strong> For power series that converge on a common interval, you can:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Addition\/Subtraction:<\/strong> [latex]\\sum c_n x^n \\pm \\sum d_n x^n = \\sum (c_n \\pm d_n) x^n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar and Power Multiplication:<\/strong> [latex]bx^m \\sum c_n x^n = \\sum bx^m c_n x^n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Composition:<\/strong> Replace [latex]x[\/latex] with [latex]bx^m[\/latex] in your series: [latex]\\sum c_n (bx^m)^n[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Since [latex]\\frac{1}{1-x} = \\sum_{n=0}^{\\infty} x^n[\/latex] for [latex]|x| < 1[\/latex], you can use substitution to create new series. For example:<\/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\">Replace [latex]x[\/latex] with [latex]-x^2[\/latex]: [latex]\\frac{1}{1+x^2} = \\sum_{n=0}^{\\infty} (-x^2)^n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Multiply by [latex]3x[\/latex]: [latex]\\frac{3x}{1+x^2} = 3x \\sum_{n=0}^{\\infty} (-x^2)^n[\/latex]<\/li>\n<\/ul>\n<p>When combining series, your new series converges on the <strong>intersection<\/strong> of the original convergence intervals. If one series works on [latex](-1,1)[\/latex] and another on [latex](-2,2)[\/latex] , their combination works on [latex](-1,1)[\/latex] .<\/p>\n<p><strong>Strategy for compositions:<\/strong> When you substitute [latex]bx^m[\/latex] for [latex]x[\/latex] in a series that converges for [latex]|x| < R[\/latex], the new series converges when [latex]|bx^m| < R[\/latex]. Solve this inequality to find your new interval.\n\nYou can also work backwards. If you see [latex]\\sum 2^n x^n = \\sum (2x)^n[\/latex], recognize this as [latex]\\frac{1}{1-2x}[\/latex] with convergence when [latex]|2x| < 1[\/latex], or [latex]|x| < \\frac{1}{2}[\/latex].\n\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558893\">Hint<\/button><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Show Solution<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Hint<\/button><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Show Solution<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<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<\/section>\n<h2 data-type=\"title\">Multiplication of Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When you multiply two power series together, you&#8217;re essentially multiplying two &#8220;infinite polynomials.&#8221; Just like with regular polynomial multiplication, you distribute each term and collect like powers of [latex]x[\/latex] to form new coefficients.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The multiplication process:<\/strong> To multiply [latex]\\sum c_n x^n[\/latex] and [latex]\\sum d_n x^n[\/latex], you need to find all the ways to get each power [latex]x^n[\/latex] in the product. For the coefficient of [latex]x^n[\/latex], collect all combinations where the powers add to [latex]n[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]e_n = c_0 d_n + c_1 d_{n-1} + c_2 d_{n-2} + \\cdots + c_n d_0[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is called the <strong>Cauchy product<\/strong>. Each new coefficient [latex]e_n[\/latex] comes from all possible ways to pair terms from the two original series that multiply to give [latex]x^n[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">If both original series converge on interval [latex]I[\/latex], then their product series also converges on [latex]I[\/latex] and represents the product of the two functions.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Practical strategy:<\/strong> Instead of using the general formula, you can often work out the first several terms by hand:<\/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\">Multiply each term in the first series by each term in the second series<\/li>\n<li class=\"whitespace-normal break-words\">Group terms with the same power of [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Add up the coefficients for each power<\/li>\n<\/ul>\n<p>When you see a rational function that factors into simpler pieces, consider whether you can find power series for the individual factors and then multiply them together.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558869\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Show Solution<\/button><\/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<\/section>\n<h2>Differentiating and Integrating Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">One of the most powerful features of power series is that you can differentiate and integrate them term-by-term, just like polynomials. This transforms complex calculus problems into straightforward arithmetic operations.<\/p>\n<p class=\"whitespace-normal break-words\">The process is simple:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>To differentiate:<\/strong> [latex]\\frac{d}{dx}\\sum c_n x^n = \\sum nc_n x^{n-1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>To integrate:<\/strong> [latex]\\int \\sum c_n x^n dx = C + \\sum c_n \\frac{x^{n+1}}{n+1}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">You apply the power rule to each term individually. For [latex]c_n x^n[\/latex], the derivative becomes [latex]nc_n x^{n-1}[\/latex] and the integral becomes [latex]c_n \\frac{x^{n+1}}{n+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Both the differentiated and integrated series have the same radius of convergence as the original series. However, endpoint behavior might change &#8211; always check endpoints separately if needed.<\/p>\n<p class=\"whitespace-normal break-words\">This technique lets you find power series for functions that would be difficult to obtain directly:<\/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\">Start with [latex]\\frac{1}{1-x} = \\sum x^n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Differentiate to get [latex]\\frac{1}{(1-x)^2} = \\sum (n+1)x^n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Integrate [latex]\\frac{1}{1+x} = \\sum (-1)^n x^n[\/latex] to get [latex]\\ln(1+x) = \\sum (-1)^{n+1}\\frac{x^n}{n}[\/latex]<\/li>\n<\/ul>\n<p>When integrating, use initial conditions to find [latex]C[\/latex]. For example, if [latex]f(0) = 0[\/latex], then [latex]C = 0[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023778779\" data-type=\"problem\">\n<p id=\"fs-id1167023778781\">Differentiate the series [latex]\\frac{1}{{\\left(1-x\\right)}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){x}^{n}[\/latex] term-by-term to find a power series representation for [latex]\\frac{2}{{\\left(1-x\\right)}^{3}}[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558839\">Hint<\/button><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023911140\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023911147\">Write out the first several terms and apply the power rule.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Show Solution<\/button><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023911084\" data-type=\"solution\">\n<p id=\"fs-id1167023911086\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+2\\right)\\left(n+1\\right){x}^{n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167023788087\" data-type=\"problem\">\n<p id=\"fs-id1167023788089\">Integrate the power series [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}[\/latex] term-by-term to evaluate [latex]\\displaystyle\\int \\text{ln}\\left(1+x\\right)dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558809\">Hint<\/button><\/p>\n<div id=\"q44558809\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023804519\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167023804526\">Use the fact that [latex]\\frac{{x}^{n+1}}{\\left(n+1\\right)n}[\/latex] is an antiderivative of [latex]\\frac{{x}^{n}}{n}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558819\">Show Solution<\/button><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023804458\" data-type=\"solution\">\n<p id=\"fs-id1167023804461\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{n}}{n\\left(n - 1\\right)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Uniqueness of Power Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When you find a power series representation for a function, you might wonder: &#8220;Could there be another, completely different power series for the same function?&#8221; The answer is no &#8211; power series representations are unique.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The uniqueness principle:<\/strong> If two power series [latex]\\sum c_n (x-a)^n[\/latex] and [latex]\\sum d_n (x-a)^n[\/latex] both converge and represent the same function on an open interval containing [latex]a[\/latex], then their coefficients must be identical: [latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Think of power series as &#8220;infinite polynomials.&#8221; Just as two polynomials can only be equal if they have identical coefficients, the same principle applies to power series. If [latex]c_0 + c_1 x + c_2 x^2 + \\cdots = d_0 + d_1 x + d_2 x^2 + \\cdots[\/latex] for all [latex]x[\/latex] in some interval, then [latex]c_0 = d_0[\/latex], [latex]c_1 = d_1[\/latex], [latex]c_2 = d_2[\/latex], and so on.<\/p>\n<p class=\"whitespace-normal break-words\">This uniqueness means you can use any valid method to find a power series &#8211; algebraic manipulation, differentiation, integration, or others we&#8217;ll learn later. No matter which approach you use, you&#8217;ll always get the same series for a given function centered at a specific point.<\/p>\n<\/div>\n","protected":false},"author":15,"menu_order":16,"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\/938"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/938\/revisions"}],"predecessor-version":[{"id":2276,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/938\/revisions\/2276"}],"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\/938\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=938"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=938"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=938"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=938"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}