{"id":1775,"date":"2021-07-27T00:11:30","date_gmt":"2021-07-27T00:11:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1775"},"modified":"2022-03-21T23:11:16","modified_gmt":"2022-03-21T23:11:16","slug":"differentiating-and-integrating-power-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/differentiating-and-integrating-power-series\/","title":{"raw":"Differentiating and Integrating Power Series","rendered":"Differentiating and Integrating 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>Differentiate and integrate power series term-by-term<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167023795149\">Consider a power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots [\/latex] that converges on some interval <em data-effect=\"italics\">I<\/em>, and let [latex]f[\/latex] be the function defined by this series. Here we address two questions about [latex]f[\/latex].<\/p>\r\n\r\n<ul id=\"fs-id1167023795233\" data-bullet-style=\"bullet\">\r\n \t<li>Is [latex]f[\/latex] differentiable, and if so, how do we determine the derivative [latex]{f}^{\\prime }?[\/latex]<\/li>\r\n \t<li>How do we evaluate the indefinite integral [latex]\\displaystyle\\int f\\left(x\\right)dx?[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1167023761836\">We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Recall the general expression of the power rules for derivatives and integrals of polynomials.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Power rule for Derivatives and Integrals<\/h3>\r\n<ol>\r\n \t<li>[latex] \\frac{d}{dx} \\left (x^n \\right) = nx^{n-1} [\/latex]<\/li>\r\n \t<li>[latex] \\int{x^n\\: dx} = \\frac{1}{n+1} x^{n+1} + C \\:(\\text{for}\\: n \\ne -1 [\/latex])<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn this section we show that we can take advantage of the simplicity of integrating and differentiating polynomials to do the same thing for convergent power series. That is, if\r\n<div id=\"fs-id1167023761842\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)={c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023761906\">converges on some interval <em data-effect=\"italics\">I<\/em>, then<\/p>\r\n\r\n<div id=\"fs-id1167023761914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{f}^{\\prime }\\left(x\\right)={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023761570\">and<\/p>\r\n\r\n<div id=\"fs-id1167023761573\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right)dx=C+{c}_{0}x+{c}_{1}\\frac{{x}^{2}}{2}+{c}_{2}\\frac{{x}^{3}}{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n\r\nEvaluating the derivative and indefinite integral in this way is called <span data-type=\"term\">term-by-term differentiation of a power series<\/span> and <span data-type=\"term\">term-by-term integration of a power series<\/span>, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can differentiate term-by-term to find the power series for [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex]. Similarly, using the power series for [latex]g\\left(x\\right)=\\frac{1}{1+x}[\/latex], we can integrate term-by-term to find the power series for [latex]G\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], an antiderivative of <em data-effect=\"italics\">g<\/em>. We show how to do this in the next two examples.\u00a0First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.\r\n<div id=\"fs-id1167023789599\" 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: Term-by-Term Differentiation and Integration for Power Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167023789606\">Suppose that the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] converges on the interval [latex]\\left(a-R,a+R\\right)[\/latex] for some [latex]R&gt;0[\/latex]. Let <em data-effect=\"italics\">f<\/em> be the function defined by the series<\/p>\r\n\r\n<div id=\"fs-id1167023789693\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{c}_{n}{\\left(x-a\\right)}^{n}\\hfill \\\\ &amp; ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167023788277\">for [latex]|x-a|&lt;R[\/latex]. Then <em data-effect=\"italics\">f<\/em> is differentiable on the interval [latex]\\left(a-R,a+R\\right)[\/latex] and we can find [latex]{f}^{\\prime }[\/latex] by differentiating the series term-by-term:<\/p>\r\n\r\n<div id=\"fs-id1167023790437\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)&amp; ={\\displaystyle\\sum _{n=1}^{\\infty}}n{c}_{n}{\\left(x-a\\right)}^{n - 1}\\hfill \\\\ &amp; ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023916491\">for [latex]|x-a|&lt;R[\/latex]. Also, to find [latex]\\displaystyle\\int f\\left(x\\right)dx[\/latex], we can integrate the series term-by-term. The resulting series converges on [latex]\\left(a-R,a+R\\right)[\/latex], and we have<\/p>\r\n\r\n<div id=\"fs-id1167023916569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int f\\left(x\\right)dx}&amp; =C+{\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}\\frac{{\\left(x-a\\right)}^{n+1}}{n+1}}\\hfill \\\\ &amp; =C+{c}_{0}\\left(x-a\\right)+{c}_{1}\\frac{{\\left(x-a\\right)}^{2}}{2}+{c}_{2}\\frac{{\\left(x-a\\right)}^{3}}{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023808074\">for [latex]|x-a|&lt;R[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167023766601\">The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series\u00a0guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.<\/p>\r\n\r\n<div id=\"fs-id1167023766613\" data-type=\"example\">\r\n<div id=\"fs-id1167023766615\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023766617\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Differentiating Power Series<\/h3>\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">Use the power series representation<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167023766633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)&amp; =\\frac{1}{1-x}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ &amp; =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex] to find a power series representation for<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023770529\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\non the interval [latex]\\left(-1,1\\right)[\/latex]. Determine whether the resulting series converges at the endpoints.\r\n\r\nUse the result of part a. to evaluate the sum of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{n+1}{{4}^{n}}[\/latex].\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167023770631\" data-type=\"solution\">\r\n<ol id=\"fs-id1167023770634\" type=\"a\">\r\n \t<li>Since [latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex] is the derivative of [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can find a power series representation for <em data-effect=\"italics\">g<\/em> by differentiating the power series for <em data-effect=\"italics\">f<\/em> term-by-term. The result is<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167023778480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill g\\left(x\\right)&amp; =\\frac{1}{{\\left(1-x\\right)}^{2}}\\hfill \\\\ &amp; =\\frac{d}{dx}\\left(\\frac{1}{1-x}\\right)\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{d}{dx}\\left({x}^{n}\\right)\\hfill \\\\ &amp; =\\frac{d}{dx}\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\hfill \\\\ &amp; =0+1+2x+3{x}^{2}+4{x}^{3}+\\cdots \\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}\\left(n+1\\right){x}^{n}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints [latex]x=\\pm 1[\/latex]. Note that this is the same result found in the previous example.\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li>From part a. we know that<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div id=\"fs-id1167023784746\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){x}^{n}=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023784817\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{n+1}{{4}^{n}}&amp; ={\\displaystyle\\sum _{n=0}^{\\infty}} \\left(n+1\\right){\\left(\\frac{1}{4}\\right)}^{n}\\hfill \\\\ &amp; =\\frac{1}{{\\left(1-\\frac{1}{4}\\right)}^{2}}\\hfill \\\\ &amp; =\\frac{1}{{\\left(\\frac{3}{4}\\right)}^{2}}\\hfill \\\\ &amp; =\\frac{16}{9}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Differentiating Power Series.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/eW-UMc7THf0?controls=0&amp;start=0&amp;end=256&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\/6.2.4_0to256_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.2.4\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167023778772\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023778776\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023778779\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167023911153\" data-type=\"example\">\r\n<div id=\"fs-id1167023911155\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023911158\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Integrating Power Series<\/h3>\r\n<div id=\"fs-id1167023911158\" data-type=\"problem\">\r\n<p id=\"fs-id1167023779742\">For each of the following functions <em data-effect=\"italics\">f<\/em>, find a power series representation for <em data-effect=\"italics\">f<\/em> by integrating the power series for [latex]{f}^{\\prime }[\/latex] and find its interval of convergence.<\/p>\r\n\r\n<ol id=\"fs-id1167023779763\" type=\"a\">\r\n \t<li>[latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<ol id=\"fs-id1167023779830\" type=\"a\">\r\n \t<li>For [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], the derivative is [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{1+x}[\/latex]. We know that<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167023807109\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+x}&amp; =\\frac{1}{1-\\left(\\text{-}x\\right)}\\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(\\text{-}x\\right)}^{n}\\hfill \\\\ &amp; =1-x+{x}^{2}-{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex]. To find a power series for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], we integrate the series term-by-term.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023809849\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {f}^{\\prime }\\left(x\\right)dx}&amp; ={\\displaystyle\\int \\left(1-x+{x}^{2}-{x}^{3}+\\cdots \\right)dx}\\hfill \\\\ &amp; =C+x-\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{4}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] is an antiderivative of [latex]\\frac{1}{1+x}[\/latex], it remains to solve for the constant <em data-effect=\"italics\">C<\/em>. Since [latex]\\text{ln}\\left(1+0\\right)=0[\/latex], we have [latex]C=0[\/latex]. Therefore, a power series representation for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023775238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\text{ln}\\left(1+x\\right)&amp; =x-\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{4}+\\cdots \\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=1}^{\\infty}}{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at [latex]x=1[\/latex] the series is the alternating harmonic series, which converges. Also, at [latex]x=-1[\/latex], the series is the harmonic series, which diverges. It is important to note that, even though this series converges at [latex]x=1[\/latex], Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to [latex]\\text{ln}\\left(2\\right)[\/latex]. In fact, the series does converge to [latex]\\text{ln}\\left(2\\right)[\/latex], but showing this fact requires more advanced techniques. (Abel\u2019s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is [latex]\\left(-1,1\\right][\/latex].\r\n<ul>\r\n \t<li>The derivative of [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] is [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{1+{x}^{2}}[\/latex]. We know that<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ul>\r\n&nbsp;\r\n<div id=\"fs-id1167023787852\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+{x}^{2}}&amp; =\\frac{1}{1-\\left(\\text{-}{x}^{2}\\right)}\\hfill \\\\ &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<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex]. To find a power series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex], we integrate this series term-by-term.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023811464\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {f}^{\\prime }\\left(x\\right)dx}&amp; ={\\displaystyle\\int \\left(1-{x}^{2}+{x}^{4}-{x}^{6}+\\cdots \\right)dx}\\hfill \\\\ &amp; =C+x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5}-\\frac{{x}^{7}}{7}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]{\\tan}^{-1}\\left(0\\right)=0[\/latex], we have [latex]C=0[\/latex]. Therefore, a power series representation for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167023802567\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\tan}^{-1}x&amp; =x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5}-\\frac{{x}^{7}}{7}+\\cdots \\hfill \\\\ &amp; ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{2n+1}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor [latex]|x|&lt;1[\/latex]. Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at [latex]x=1[\/latex] and [latex]x=-1[\/latex]. As discussed in part a., using Abel\u2019s theorem, it can be shown that the series actually converges to [latex]{\\tan}^{-1}\\left(1\\right)[\/latex] and [latex]{\\tan}^{-1}\\left(-1\\right)[\/latex] at [latex]x=1[\/latex] and [latex]x=-1[\/latex], respectively. Thus, the interval of convergence is [latex]\\left[-1,1\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Integrating Power Series.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/eW-UMc7THf0?controls=0&amp;start=257&amp;end=598&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\/6.2.4_257to598_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip \"6.2.4\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167023788082\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167023788085\" data-type=\"exercise\">\r\n<div id=\"fs-id1167023788087\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167023804579\">Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function <em data-effect=\"italics\">f<\/em> and a power series for <em data-effect=\"italics\">f<\/em> at <em data-effect=\"italics\">a<\/em>, is it possible that there is a different power series for <em data-effect=\"italics\">f<\/em> at <em data-effect=\"italics\">a<\/em> that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if<\/p>\r\n\r\n<div id=\"fs-id1167023777259\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots ={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023777332\">for all values <em data-effect=\"italics\">x<\/em> in some open interval <em data-effect=\"italics\">I<\/em> about zero, then the coefficients <em data-effect=\"italics\">c<sub>n<\/sub><\/em> should equal <em data-effect=\"italics\">d<sub>n<\/sub><\/em> for [latex]n\\ge 0[\/latex]. We now state this result formally in Uniqueness of Power Series.<\/p>\r\n\r\n<div id=\"fs-id1167023777376\" 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: Uniqueness of Power Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167023777384\">Let [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{\\left(x-a\\right)}^{n}[\/latex] be two convergent power series such that<\/p>\r\n\r\n<div id=\"fs-id1167023913935\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{\\left(x-a\\right)}^{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023914023\">for all <em data-effect=\"italics\">x<\/em> in an open interval containing <em data-effect=\"italics\">a<\/em>. Then [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1167023914065\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1167023763349\">Let<\/p>\r\n\r\n<div id=\"fs-id1167023763352\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)&amp; ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\\\ &amp; ={d}_{0}+{d}_{1}\\left(x-a\\right)+{d}_{2}{\\left(x-a\\right)}^{2}+{d}_{3}{\\left(x-a\\right)}^{3}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023919261\">Then [latex]f\\left(a\\right)={c}_{0}={d}_{0}[\/latex]. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1167023919298\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)&amp; ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; ={d}_{1}+2{d}_{2}\\left(x-a\\right)+3{d}_{3}{\\left(x-a\\right)}^{2}+\\cdots ,\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023801398\">and thus, [latex]{f}^{\\prime }\\left(a\\right)={c}_{1}={d}_{1}[\/latex]. Similarly,<\/p>\r\n\r\n<div id=\"fs-id1167023801435\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f^{\\prime\\prime}\\left(x\\right)&amp; =2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+\\cdots \\hfill \\\\ &amp; =2{d}_{2}+3\\cdot 2{d}_{3}\\left(x-a\\right)+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167023801548\">implies that [latex]f^{\\prime\\prime} \\left(a\\right)=2{c}_{2}=2{d}_{2}[\/latex], and therefore, [latex]{c}_{2}={d}_{2}[\/latex]. More generally, for any integer [latex]n\\ge 0,{f}^{\\left(n\\right)}\\left(a\\right)=n\\text{!}{c}_{n}=n\\text{!}{d}_{n}[\/latex], and consequently, [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\r\n<p id=\"fs-id1167023733734\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169439[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167023733737\">In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.<\/p>\r\n\r\n<\/section><section id=\"fs-id1167023733746\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Differentiate and integrate power series term-by-term<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167023795149\">Consider a power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots[\/latex] that converges on some interval <em data-effect=\"italics\">I<\/em>, and let [latex]f[\/latex] be the function defined by this series. Here we address two questions about [latex]f[\/latex].<\/p>\n<ul id=\"fs-id1167023795233\" data-bullet-style=\"bullet\">\n<li>Is [latex]f[\/latex] differentiable, and if so, how do we determine the derivative [latex]{f}^{\\prime }?[\/latex]<\/li>\n<li>How do we evaluate the indefinite integral [latex]\\displaystyle\\int f\\left(x\\right)dx?[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1167023761836\">We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Recall the general expression of the power rules for derivatives and integrals of polynomials.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Power rule for Derivatives and Integrals<\/h3>\n<ol>\n<li>[latex]\\frac{d}{dx} \\left (x^n \\right) = nx^{n-1}[\/latex]<\/li>\n<li>[latex]\\int{x^n\\: dx} = \\frac{1}{n+1} x^{n+1} + C \\:(\\text{for}\\: n \\ne -1[\/latex])<\/li>\n<\/ol>\n<\/div>\n<p>In this section we show that we can take advantage of the simplicity of integrating and differentiating polynomials to do the same thing for convergent power series. That is, if<\/p>\n<div id=\"fs-id1167023761842\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)={c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023761906\">converges on some interval <em data-effect=\"italics\">I<\/em>, then<\/p>\n<div id=\"fs-id1167023761914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{f}^{\\prime }\\left(x\\right)={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023761570\">and<\/p>\n<div id=\"fs-id1167023761573\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right)dx=C+{c}_{0}x+{c}_{1}\\frac{{x}^{2}}{2}+{c}_{2}\\frac{{x}^{3}}{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Evaluating the derivative and indefinite integral in this way is called <span data-type=\"term\">term-by-term differentiation of a power series<\/span> and <span data-type=\"term\">term-by-term integration of a power series<\/span>, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can differentiate term-by-term to find the power series for [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex]. Similarly, using the power series for [latex]g\\left(x\\right)=\\frac{1}{1+x}[\/latex], we can integrate term-by-term to find the power series for [latex]G\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], an antiderivative of <em data-effect=\"italics\">g<\/em>. We show how to do this in the next two examples.\u00a0First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.<\/p>\n<div id=\"fs-id1167023789599\" 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: Term-by-Term Differentiation and Integration for Power Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167023789606\">Suppose that the power series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] converges on the interval [latex]\\left(a-R,a+R\\right)[\/latex] for some [latex]R>0[\/latex]. Let <em data-effect=\"italics\">f<\/em> be the function defined by the series<\/p>\n<div id=\"fs-id1167023789693\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={\\displaystyle\\sum _{n=0}^{\\infty}}{c}_{n}{\\left(x-a\\right)}^{n}\\hfill \\\\ & ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167023788277\">for [latex]|x-a|<R[\/latex]. Then <em data-effect=\"italics\">f<\/em> is differentiable on the interval [latex]\\left(a-R,a+R\\right)[\/latex] and we can find [latex]{f}^{\\prime }[\/latex] by differentiating the series term-by-term:<\/p>\n<div id=\"fs-id1167023790437\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)& ={\\displaystyle\\sum _{n=1}^{\\infty}}n{c}_{n}{\\left(x-a\\right)}^{n - 1}\\hfill \\\\ & ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023916491\">for [latex]|x-a|<R[\/latex]. Also, to find [latex]\\displaystyle\\int f\\left(x\\right)dx[\/latex], we can integrate the series term-by-term. The resulting series converges on [latex]\\left(a-R,a+R\\right)[\/latex], and we have<\/p>\n<div id=\"fs-id1167023916569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int f\\left(x\\right)dx}& =C+{\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}\\frac{{\\left(x-a\\right)}^{n+1}}{n+1}}\\hfill \\\\ & =C+{c}_{0}\\left(x-a\\right)+{c}_{1}\\frac{{\\left(x-a\\right)}^{2}}{2}+{c}_{2}\\frac{{\\left(x-a\\right)}^{3}}{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023808074\">for [latex]|x-a|<R[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023766601\">The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series\u00a0guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.<\/p>\n<div id=\"fs-id1167023766613\" data-type=\"example\">\n<div id=\"fs-id1167023766615\" data-type=\"exercise\">\n<div id=\"fs-id1167023766617\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Differentiating Power Series<\/h3>\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">    <\/ol>\n<\/li>\n<\/ol>\n<div id=\"fs-id1167023766633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& =\\frac{1}{1-x}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ & =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex] to find a power series representation for<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023770529\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\non the interval [latex]\\left(-1,1\\right)[\/latex]. Determine whether the resulting series converges at the endpoints.<\/p>\n<p>Use the result of part a. to evaluate the sum of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{n+1}{{4}^{n}}[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558859\">Show Solution<\/span><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167023770631\" data-type=\"solution\">\n<ol id=\"fs-id1167023770634\" type=\"a\">\n<li>Since [latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex] is the derivative of [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can find a power series representation for <em data-effect=\"italics\">g<\/em> by differentiating the power series for <em data-effect=\"italics\">f<\/em> term-by-term. The result is<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167023778480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill g\\left(x\\right)& =\\frac{1}{{\\left(1-x\\right)}^{2}}\\hfill \\\\ & =\\frac{d}{dx}\\left(\\frac{1}{1-x}\\right)\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{d}{dx}\\left({x}^{n}\\right)\\hfill \\\\ & =\\frac{d}{dx}\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\hfill \\\\ & =0+1+2x+3{x}^{2}+4{x}^{3}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}\\left(n+1\\right){x}^{n}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints [latex]x=\\pm 1[\/latex]. Note that this is the same result found in the previous example.\n\n\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ul>\n<li>From part a. we know that<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167023784746\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){x}^{n}=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023784817\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{n+1}{{4}^{n}}& ={\\displaystyle\\sum _{n=0}^{\\infty}} \\left(n+1\\right){\\left(\\frac{1}{4}\\right)}^{n}\\hfill \\\\ & =\\frac{1}{{\\left(1-\\frac{1}{4}\\right)}^{2}}\\hfill \\\\ & =\\frac{1}{{\\left(\\frac{3}{4}\\right)}^{2}}\\hfill \\\\ & =\\frac{16}{9}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Differentiating Power Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/eW-UMc7THf0?controls=0&amp;start=0&amp;end=256&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\/6.2.4_0to256_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.2.4&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167023778772\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023778776\" data-type=\"exercise\">\n<div id=\"fs-id1167023778779\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558839\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558849\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167023911153\" data-type=\"example\">\n<div id=\"fs-id1167023911155\" data-type=\"exercise\">\n<div id=\"fs-id1167023911158\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Integrating Power Series<\/h3>\n<div id=\"fs-id1167023911158\" data-type=\"problem\">\n<p id=\"fs-id1167023779742\">For each of the following functions <em data-effect=\"italics\">f<\/em>, find a power series representation for <em data-effect=\"italics\">f<\/em> by integrating the power series for [latex]{f}^{\\prime }[\/latex] and find its interval of convergence.<\/p>\n<ol id=\"fs-id1167023779763\" type=\"a\">\n<li>[latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558829\">Show Solution<\/span><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167023779830\" type=\"a\">\n<li>For [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], the derivative is [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{1+x}[\/latex]. We know that<\/li>\n<\/ol>\n<div id=\"fs-id1167023807109\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+x}& =\\frac{1}{1-\\left(\\text{-}x\\right)}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(\\text{-}x\\right)}^{n}\\hfill \\\\ & =1-x+{x}^{2}-{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex]. To find a power series for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex], we integrate the series term-by-term.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023809849\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {f}^{\\prime }\\left(x\\right)dx}& ={\\displaystyle\\int \\left(1-x+{x}^{2}-{x}^{3}+\\cdots \\right)dx}\\hfill \\\\ & =C+x-\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{4}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] is an antiderivative of [latex]\\frac{1}{1+x}[\/latex], it remains to solve for the constant <em data-effect=\"italics\">C<\/em>. Since [latex]\\text{ln}\\left(1+0\\right)=0[\/latex], we have [latex]C=0[\/latex]. Therefore, a power series representation for [latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex] is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023775238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\text{ln}\\left(1+x\\right)& =x-\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{4}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=1}^{\\infty}}{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at [latex]x=1[\/latex] the series is the alternating harmonic series, which converges. Also, at [latex]x=-1[\/latex], the series is the harmonic series, which diverges. It is important to note that, even though this series converges at [latex]x=1[\/latex], Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to [latex]\\text{ln}\\left(2\\right)[\/latex]. In fact, the series does converge to [latex]\\text{ln}\\left(2\\right)[\/latex], but showing this fact requires more advanced techniques. (Abel\u2019s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is [latex]\\left(-1,1\\right][\/latex].\n\n\n<ul>\n<li>The derivative of [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] is [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{1+{x}^{2}}[\/latex]. We know that<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167023787852\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{1}{1+{x}^{2}}& =\\frac{1}{1-\\left(\\text{-}{x}^{2}\\right)}\\hfill \\\\ & ={\\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><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex]. To find a power series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex], we integrate this series term-by-term.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023811464\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int {f}^{\\prime }\\left(x\\right)dx}& ={\\displaystyle\\int \\left(1-{x}^{2}+{x}^{4}-{x}^{6}+\\cdots \\right)dx}\\hfill \\\\ & =C+x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5}-\\frac{{x}^{7}}{7}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]{\\tan}^{-1}\\left(0\\right)=0[\/latex], we have [latex]C=0[\/latex]. Therefore, a power series representation for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167023802567\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\tan}^{-1}x& =x-\\frac{{x}^{3}}{3}+\\frac{{x}^{5}}{5}-\\frac{{x}^{7}}{7}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{2n+1}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor [latex]|x|<1[\/latex]. Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at [latex]x=1[\/latex] and [latex]x=-1[\/latex]. As discussed in part a., using Abel\u2019s theorem, it can be shown that the series actually converges to [latex]{\\tan}^{-1}\\left(1\\right)[\/latex] and [latex]{\\tan}^{-1}\\left(-1\\right)[\/latex] at [latex]x=1[\/latex] and [latex]x=-1[\/latex], respectively. Thus, the interval of convergence is [latex]\\left[-1,1\\right][\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Integrating Power Series.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/eW-UMc7THf0?controls=0&amp;start=257&amp;end=598&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\/6.2.4_257to598_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip &#8220;6.2.4&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167023788082\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167023788085\" data-type=\"exercise\">\n<div id=\"fs-id1167023788087\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558809\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558819\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167023804579\">Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function <em data-effect=\"italics\">f<\/em> and a power series for <em data-effect=\"italics\">f<\/em> at <em data-effect=\"italics\">a<\/em>, is it possible that there is a different power series for <em data-effect=\"italics\">f<\/em> at <em data-effect=\"italics\">a<\/em> that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if<\/p>\n<div id=\"fs-id1167023777259\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots ={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023777332\">for all values <em data-effect=\"italics\">x<\/em> in some open interval <em data-effect=\"italics\">I<\/em> about zero, then the coefficients <em data-effect=\"italics\">c<sub>n<\/sub><\/em> should equal <em data-effect=\"italics\">d<sub>n<\/sub><\/em> for [latex]n\\ge 0[\/latex]. We now state this result formally in Uniqueness of Power Series.<\/p>\n<div id=\"fs-id1167023777376\" 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: Uniqueness of Power Series<\/h3>\n<hr \/>\n<p id=\"fs-id1167023777384\">Let [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] and [latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{\\left(x-a\\right)}^{n}[\/latex] be two convergent power series such that<\/p>\n<div id=\"fs-id1167023913935\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{\\left(x-a\\right)}^{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023914023\">for all <em data-effect=\"italics\">x<\/em> in an open interval containing <em data-effect=\"italics\">a<\/em>. Then [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1167023914065\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1167023763349\">Let<\/p>\n<div id=\"fs-id1167023763352\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\\\ & ={d}_{0}+{d}_{1}\\left(x-a\\right)+{d}_{2}{\\left(x-a\\right)}^{2}+{d}_{3}{\\left(x-a\\right)}^{3}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023919261\">Then [latex]f\\left(a\\right)={c}_{0}={d}_{0}[\/latex]. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,<\/p>\n<div id=\"fs-id1167023919298\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)& ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={d}_{1}+2{d}_{2}\\left(x-a\\right)+3{d}_{3}{\\left(x-a\\right)}^{2}+\\cdots ,\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023801398\">and thus, [latex]{f}^{\\prime }\\left(a\\right)={c}_{1}={d}_{1}[\/latex]. Similarly,<\/p>\n<div id=\"fs-id1167023801435\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill f^{\\prime\\prime}\\left(x\\right)& =2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+\\cdots \\hfill \\\\ & =2{d}_{2}+3\\cdot 2{d}_{3}\\left(x-a\\right)+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167023801548\">implies that [latex]f^{\\prime\\prime} \\left(a\\right)=2{c}_{2}=2{d}_{2}[\/latex], and therefore, [latex]{c}_{2}={d}_{2}[\/latex]. More generally, for any integer [latex]n\\ge 0,{f}^{\\left(n\\right)}\\left(a\\right)=n\\text{!}{c}_{n}=n\\text{!}{d}_{n}[\/latex], and consequently, [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\n<p id=\"fs-id1167023733734\">[latex]_\\blacksquare[\/latex]<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169439\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169439&theme=oea&iframe_resize_id=ohm169439&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167023733737\">In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.<\/p>\n<\/section>\n<section id=\"fs-id1167023733746\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1775\">\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>6.2.4. <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":8,"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\":\"6.2.4\",\"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-1775","chapter","type-chapter","status-publish","hentry"],"part":161,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1775","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":9,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1775\/revisions"}],"predecessor-version":[{"id":2248,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1775\/revisions\/2248"}],"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\/1775\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1775"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1775"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1775"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}