{"id":950,"date":"2025-06-20T17:24:37","date_gmt":"2025-06-20T17:24:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=950"},"modified":"2025-09-10T13:20:52","modified_gmt":"2025-09-10T13:20:52","slug":"taylor-and-maclaurin-series-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-fresh-take\/","title":{"raw":"Taylor and Maclaurin Series: Fresh Take","rendered":"Taylor and Maclaurin Series: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Learn how to find Taylor polynomials of a given order for a function<\/li>\r\n \t<li>Estimate the remainder when using a Taylor series to approximate a function<\/li>\r\n \t<li>Determine when a Taylor series converges to the original function<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin 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\">Taylor series provide a systematic method for finding power series representations of functions using derivatives. Instead of clever algebraic tricks, you can now represent almost any \"nice\" function as a power series by following a straightforward formula.<\/p>\r\n<p class=\"whitespace-normal break-words\">If a function [latex]f[\/latex] has a power series representation at [latex]x = a[\/latex], then the coefficients must be determined by the function's derivatives at that point. By matching the function value and all its derivatives at [latex]x = a[\/latex], we get:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]c_n = \\frac{f^{(n)}(a)}{n!}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Taylor series formula:<\/strong> For a function with derivatives of all orders at [latex]x = a[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\sum_{n=0}^{\\infty} \\frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\frac{f'''(a)}{3!}(x-a)^3 + \\cdots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Maclaurin series:<\/strong> When [latex]a = 0[\/latex], this becomes the Maclaurin series:<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\sum_{n=0}^{\\infty} \\frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3 + \\cdots[\/latex]<\/p>\r\nSince power series representations are unique, if a function has any power series representation at [latex]x = a[\/latex], it must be the Taylor series. This means Taylor series are not just one way to find power series - they're the only way for functions that can be represented this way.\r\n\r\n<strong>The key requirement:<\/strong> The function must have derivatives of all orders at the point of expansion. Functions like [latex]e^x[\/latex], [latex]\\sin x[\/latex], [latex]\\cos x[\/latex], and [latex]\\ln(1+x)[\/latex] all satisfy this condition and have Taylor series representations.\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit the MacTutor History of Mathematics archive to <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Taylor\/\" target=\"_blank\" rel=\"noopener\">read a biography of Brook Taylor<\/a> and a <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Maclaurin\/\" target=\"_blank\" rel=\"noopener\">biography of\u00a0Colin Maclaurin<\/a> and how they developed the concepts named after them.<\/section>\r\n<h2 data-type=\"title\">Taylor Polynomials<\/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\">Taylor polynomials are the finite \"building blocks\" of Taylor series. Instead of working with an infinite sum, you can use just the first few terms to create polynomial approximations of your function that become increasingly accurate as you add more terms.<\/p>\r\n<p class=\"whitespace-normal break-words\">The [latex]n[\/latex]th Taylor polynomial [latex]p_n(x)[\/latex] is simply the first [latex]n+1[\/latex] terms of the Taylor series:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]p_n(x) = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots + \\frac{f^{(n)}(a)}{n!}(x-a)^n[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">What each polynomial captures:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]p_0(x) = f(a)[\/latex]: Matches the function value at [latex]x = a[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]p_1(x)[\/latex]: Matches function value and slope<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]p_2(x)[\/latex]: Matches function value, slope, and curvature<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]p_3(x)[\/latex]: Matches function value, slope, curvature, and rate of curvature change<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Key patterns for common functions:<\/strong><\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]e^x[\/latex]<\/strong>: All derivatives equal 1 at [latex]x = 0[\/latex], giving [latex]p_n(x) = \\sum_{k=0}^{n} \\frac{x^k}{k!}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\sin x[\/latex]<\/strong>: Only odd powers appear, with alternating signs<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\cos x[\/latex]<\/strong>: Only even powers appear, with alternating signs<\/li>\r\n<\/ul>\r\nAs [latex]n[\/latex] increases, [latex]p_n(x)[\/latex] provides a better approximation of [latex]f(x)[\/latex] in a neighborhood around [latex]x = a[\/latex]. The graphs show how higher-degree polynomials \"hug\" the original function more closely over larger intervals.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\r\n<p id=\"fs-id1167025019203\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] at [latex]x=1[\/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-id1167025112903\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025118432\">Find the first three derivatives of [latex]f[\/latex] and evaluate them at [latex]x=1[\/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-id1167024971078\" data-type=\"solution\">\r\n<p id=\"fs-id1167024971080\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1 - 2\\left(x - 1\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}-4{\\left(x - 1\\right)}^{3}[\/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-id1167025235550\" data-type=\"problem\">\r\n<p id=\"fs-id1167025235553\">Find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{1+x}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial. Write your anwer using sigma notation.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167025102143\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025102151\">Evaluate the first four derivatives of [latex]f[\/latex] and look for a pattern.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167025168862\" data-type=\"solution\">\r\n<p id=\"fs-id1167025168865\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1-x[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1-x+{x}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{n}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}+\\cdots +{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}{x}^{k}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Taylor\u2019s Theorem with Remainder<\/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 use a Taylor polynomial to approximate a function, you need to know how accurate your approximation is. Taylor's theorem with remainder gives you both the exact form of the error and a way to bound it.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The remainder formula:<\/strong> The error when using the [latex]n[\/latex]th Taylor polynomial is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R_n(x) = f(x) - p_n(x) = \\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]c[\/latex] is some unknown point between [latex]a[\/latex] and [latex]x[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Even though you can't find the exact value of [latex]c[\/latex], you can still bound your error. If you know that [latex]|f^{(n+1)}(x)| \\leq M[\/latex] on your interval, then:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|R_n(x)| \\leq \\frac{M}{(n+1)!}|x-a|^{n+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Practical applications:<\/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>Proving convergence:<\/strong> If [latex]\\lim_{n \\to \\infty} R_n(x) = 0[\/latex], then the Taylor series converges to [latex]f(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Error bounds:<\/strong> You can determine how many terms you need for a desired accuracy<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Quality control:<\/strong> You know the maximum possible error in your approximation<\/li>\r\n<\/ul>\r\n<strong>Strategy for finding bounds:<\/strong> Look for the maximum value of [latex]|f^{(n+1)}(x)|[\/latex] on your interval. For functions like [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex], all derivatives have absolute value at most 1, making the bound especially clean.\r\n\r\nAs you move further from the center [latex]a[\/latex], the term [latex]|x-a|^{n+1}[\/latex] grows larger, requiring more terms for the same accuracy. Taylor polynomials work best near their center point.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\r\n<p id=\"fs-id1167024969443\">Find the first and second Taylor polynomials for [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex]. Use these polynomials to estimate [latex]\\sqrt{6}[\/latex]. Use Taylor\u2019s theorem to bound the error.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558849\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1167025098843\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025098850\">Evaluate [latex]f\\left(4\\right),{f}^{\\prime }\\left(4\\right)[\/latex], and [latex]f^{\\prime\\prime} \\left(4\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1167024969487\" data-type=\"solution\">\r\n<p id=\"fs-id1167024969489\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)-\\frac{1}{64}{\\left(x - 4\\right)}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(6\\right)=2.5[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(6\\right)=2.4375[\/latex];<\/p>\r\n<p id=\"fs-id1167025098788\" style=\"text-align: center;\">[latex]|{R}_{1}\\left(6\\right)|\\le 0.0625;|{R}_{2}\\left(6\\right)|\\le 0.015625[\/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-id1167025131376\" data-type=\"problem\">\r\n<p id=\"fs-id1167025131379\">Use the fourth Maclaurin polynomial for [latex]\\cos{x}[\/latex] to approximate [latex]\\cos\\left(\\frac{\\pi }{12}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558819\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1167025131419\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025131426\">The fourth Maclaurin polynomial is [latex]{p}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1167025131413\" data-type=\"solution\">\r\n<p id=\"fs-id1167025131415\">0.96593<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Convergence of Taylor 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\">Finding a Taylor series is only half the battle - you also need to prove it actually converges to the function you want. Just because you can write down the series doesn't guarantee it represents your function.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The key question:<\/strong> Does [latex]\\lim_{n \\to \\infty} p_n(x) = f(x)[\/latex]? In other words, do the Taylor polynomials actually approach the function as you add more terms?<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The convergence test:<\/strong> A Taylor series converges to [latex]f(x)[\/latex] if and only if the remainder approaches zero: [latex]\\lim_{n \\to \\infty} R_n(x) = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">\u00a0Use Taylor's theorem to bound the remainder:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|R_n(x)| \\leq \\frac{M}{(n+1)!}|x-a|^{n+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]M[\/latex] bounds all [latex](n+1)[\/latex]th derivatives on your interval.<\/p>\r\n<strong>Finding intervals of convergence:<\/strong> Use standard tests (ratio test, root test) just like with power series. The Taylor series will have some radius of convergence, but proving it converges to the actual function requires the remainder analysis.\r\n\r\nA Taylor series might converge to some function, but that doesn't mean it converges to the function you started with. Only by showing [latex]R_n(x) \\to 0[\/latex] can you guarantee the series represents your original function.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\r\n<p id=\"fs-id1167024969277\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{2x}[\/latex] at [latex]x=2[\/latex] and determine its interval of convergence.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558699\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1167025232134\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025232142\" style=\"text-align: center;\">[latex]{f}^{\\left(n\\right)}\\left(2\\right)=\\frac{{\\left(-1\\right)}^{n}n\\text{!}}{{2}^{n+1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1167024969311\" data-type=\"solution\">\r\n<p id=\"fs-id1167024969313\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2-x}{{2}^{n+2}}\\right)}^{n}[\/latex]. The interval of convergence is [latex]\\left(0,4\\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-id1167024999410\" data-type=\"problem\">\r\n<p id=\"fs-id1167024999413\">Find the Maclaurin series for [latex]f\\left(x\\right)=\\cos{x}[\/latex]. Use the ratio test to show that the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Show that the Maclaurin series converges to [latex]\\cos{x}[\/latex] for all real numbers [latex]x[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558399\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558399\"]\r\n<div id=\"fs-id1167025241321\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167025241328\">Use the Maclaurin polynomials for [latex]\\cos{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558499\"]\r\n<div id=\"fs-id1167025241158\" data-type=\"solution\">\r\n<p id=\"fs-id1167025241160\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\r\n<p id=\"fs-id1167025241218\">By the ratio test, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Since [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex], the series converges to [latex]\\cos{x}[\/latex] for all real <em data-effect=\"italics\">x<\/em>.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Learn how to find Taylor polynomials of a given order for a function<\/li>\n<li>Estimate the remainder when using a Taylor series to approximate a function<\/li>\n<li>Determine when a Taylor series converges to the original function<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Taylor series provide a systematic method for finding power series representations of functions using derivatives. Instead of clever algebraic tricks, you can now represent almost any &#8220;nice&#8221; function as a power series by following a straightforward formula.<\/p>\n<p class=\"whitespace-normal break-words\">If a function [latex]f[\/latex] has a power series representation at [latex]x = a[\/latex], then the coefficients must be determined by the function&#8217;s derivatives at that point. By matching the function value and all its derivatives at [latex]x = a[\/latex], we get:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]c_n = \\frac{f^{(n)}(a)}{n!}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Taylor series formula:<\/strong> For a function with derivatives of all orders at [latex]x = a[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\sum_{n=0}^{\\infty} \\frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\frac{f'''(a)}{3!}(x-a)^3 + \\cdots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Maclaurin series:<\/strong> When [latex]a = 0[\/latex], this becomes the Maclaurin series:<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\sum_{n=0}^{\\infty} \\frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \\frac{f''(0)}{2!}x^2 + \\frac{f'''(0)}{3!}x^3 + \\cdots[\/latex]<\/p>\n<p>Since power series representations are unique, if a function has any power series representation at [latex]x = a[\/latex], it must be the Taylor series. This means Taylor series are not just one way to find power series &#8211; they&#8217;re the only way for functions that can be represented this way.<\/p>\n<p><strong>The key requirement:<\/strong> The function must have derivatives of all orders at the point of expansion. Functions like [latex]e^x[\/latex], [latex]\\sin x[\/latex], [latex]\\cos x[\/latex], and [latex]\\ln(1+x)[\/latex] all satisfy this condition and have Taylor series representations.<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit the MacTutor History of Mathematics archive to <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Taylor\/\" target=\"_blank\" rel=\"noopener\">read a biography of Brook Taylor<\/a> and a <a href=\"https:\/\/mathshistory.st-andrews.ac.uk\/Biographies\/Maclaurin\/\" target=\"_blank\" rel=\"noopener\">biography of\u00a0Colin Maclaurin<\/a> and how they developed the concepts named after them.<\/section>\n<h2 data-type=\"title\">Taylor Polynomials<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Taylor polynomials are the finite &#8220;building blocks&#8221; of Taylor series. Instead of working with an infinite sum, you can use just the first few terms to create polynomial approximations of your function that become increasingly accurate as you add more terms.<\/p>\n<p class=\"whitespace-normal break-words\">The [latex]n[\/latex]th Taylor polynomial [latex]p_n(x)[\/latex] is simply the first [latex]n+1[\/latex] terms of the Taylor series:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]p_n(x) = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots + \\frac{f^{(n)}(a)}{n!}(x-a)^n[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">What each polynomial captures:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]p_0(x) = f(a)[\/latex]: Matches the function value at [latex]x = a[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]p_1(x)[\/latex]: Matches function value and slope<\/li>\n<li class=\"whitespace-normal break-words\">[latex]p_2(x)[\/latex]: Matches function value, slope, and curvature<\/li>\n<li class=\"whitespace-normal break-words\">[latex]p_3(x)[\/latex]: Matches function value, slope, curvature, and rate of curvature change<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Key patterns for common functions:<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>[latex]e^x[\/latex]<\/strong>: All derivatives equal 1 at [latex]x = 0[\/latex], giving [latex]p_n(x) = \\sum_{k=0}^{n} \\frac{x^k}{k!}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\sin x[\/latex]<\/strong>: Only odd powers appear, with alternating signs<\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\cos x[\/latex]<\/strong>: Only even powers appear, with alternating signs<\/li>\n<\/ul>\n<p>As [latex]n[\/latex] increases, [latex]p_n(x)[\/latex] provides a better approximation of [latex]f(x)[\/latex] in a neighborhood around [latex]x = a[\/latex]. The graphs show how higher-degree polynomials &#8220;hug&#8221; the original function more closely over larger intervals.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167025019201\" data-type=\"problem\">\n<p id=\"fs-id1167025019203\">Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] at [latex]x=1[\/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-id1167025112903\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025118432\">Find the first three derivatives of [latex]f[\/latex] and evaluate them at [latex]x=1[\/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-id1167024971078\" data-type=\"solution\">\n<p id=\"fs-id1167024971080\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1 - 2\\left(x - 1\\right)[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1 - 2\\left(x - 1\\right)+3{\\left(x - 1\\right)}^{2}-4{\\left(x - 1\\right)}^{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167025235550\" data-type=\"problem\">\n<p id=\"fs-id1167025235553\">Find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\frac{1}{1+x}[\/latex]. Find a formula for the <em data-effect=\"italics\">n<\/em>th Maclaurin polynomial. Write your anwer using sigma notation.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025102143\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025102151\">Evaluate the first four derivatives of [latex]f[\/latex] and look for a pattern.<\/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=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025168862\" data-type=\"solution\">\n<p id=\"fs-id1167025168865\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{0}\\left(x\\right)=1[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=1-x[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=1-x+{x}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{3}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{n}\\left(x\\right)=1-x+{x}^{2}-{x}^{3}+\\cdots +{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}{x}^{k}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2>Taylor\u2019s Theorem with Remainder<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When you use a Taylor polynomial to approximate a function, you need to know how accurate your approximation is. Taylor&#8217;s theorem with remainder gives you both the exact form of the error and a way to bound it.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The remainder formula:<\/strong> The error when using the [latex]n[\/latex]th Taylor polynomial is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R_n(x) = f(x) - p_n(x) = \\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]c[\/latex] is some unknown point between [latex]a[\/latex] and [latex]x[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Even though you can&#8217;t find the exact value of [latex]c[\/latex], you can still bound your error. If you know that [latex]|f^{(n+1)}(x)| \\leq M[\/latex] on your interval, then:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|R_n(x)| \\leq \\frac{M}{(n+1)!}|x-a|^{n+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Practical applications:<\/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>Proving convergence:<\/strong> If [latex]\\lim_{n \\to \\infty} R_n(x) = 0[\/latex], then the Taylor series converges to [latex]f(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Error bounds:<\/strong> You can determine how many terms you need for a desired accuracy<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Quality control:<\/strong> You know the maximum possible error in your approximation<\/li>\n<\/ul>\n<p><strong>Strategy for finding bounds:<\/strong> Look for the maximum value of [latex]|f^{(n+1)}(x)|[\/latex] on your interval. For functions like [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex], all derivatives have absolute value at most 1, making the bound especially clean.<\/p>\n<p>As you move further from the center [latex]a[\/latex], the term [latex]|x-a|^{n+1}[\/latex] grows larger, requiring more terms for the same accuracy. Taylor polynomials work best near their center point.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167024969441\" data-type=\"problem\">\n<p id=\"fs-id1167024969443\">Find the first and second Taylor polynomials for [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]x=4[\/latex]. Use these polynomials to estimate [latex]\\sqrt{6}[\/latex]. Use Taylor\u2019s theorem to bound the error.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Hint<\/button><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025098843\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025098850\">Evaluate [latex]f\\left(4\\right),{f}^{\\prime }\\left(4\\right)[\/latex], and [latex]f^{\\prime\\prime} \\left(4\\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=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024969487\" data-type=\"solution\">\n<p id=\"fs-id1167024969489\" style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(x\\right)=2+\\frac{1}{4}\\left(x - 4\\right)-\\frac{1}{64}{\\left(x - 4\\right)}^{2}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{1}\\left(6\\right)=2.5[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]{p}_{2}\\left(6\\right)=2.4375[\/latex];<\/p>\n<p id=\"fs-id1167025098788\" style=\"text-align: center;\">[latex]|{R}_{1}\\left(6\\right)|\\le 0.0625;|{R}_{2}\\left(6\\right)|\\le 0.015625[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167025131376\" data-type=\"problem\">\n<p id=\"fs-id1167025131379\">Use the fourth Maclaurin polynomial for [latex]\\cos{x}[\/latex] to approximate [latex]\\cos\\left(\\frac{\\pi }{12}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558819\">Hint<\/button><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025131419\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025131426\">The fourth Maclaurin polynomial is [latex]{p}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2\\text{!}}+\\frac{{x}^{4}}{4\\text{!}}[\/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=\"q44558829\">Show Solution<\/button><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025131413\" data-type=\"solution\">\n<p id=\"fs-id1167025131415\">0.96593<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Convergence of Taylor Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding a Taylor series is only half the battle &#8211; you also need to prove it actually converges to the function you want. Just because you can write down the series doesn&#8217;t guarantee it represents your function.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The key question:<\/strong> Does [latex]\\lim_{n \\to \\infty} p_n(x) = f(x)[\/latex]? In other words, do the Taylor polynomials actually approach the function as you add more terms?<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The convergence test:<\/strong> A Taylor series converges to [latex]f(x)[\/latex] if and only if the remainder approaches zero: [latex]\\lim_{n \\to \\infty} R_n(x) = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">\u00a0Use Taylor&#8217;s theorem to bound the remainder:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|R_n(x)| \\leq \\frac{M}{(n+1)!}|x-a|^{n+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]M[\/latex] bounds all [latex](n+1)[\/latex]th derivatives on your interval.<\/p>\n<p><strong>Finding intervals of convergence:<\/strong> Use standard tests (ratio test, root test) just like with power series. The Taylor series will have some radius of convergence, but proving it converges to the actual function requires the remainder analysis.<\/p>\n<p>A Taylor series might converge to some function, but that doesn&#8217;t mean it converges to the function you started with. Only by showing [latex]R_n(x) \\to 0[\/latex] can you guarantee the series represents your original function.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167024969275\" data-type=\"problem\">\n<p id=\"fs-id1167024969277\">Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{2x}[\/latex] at [latex]x=2[\/latex] and 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=\"q44558699\">Hint<\/button><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025232134\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025232142\" style=\"text-align: center;\">[latex]{f}^{\\left(n\\right)}\\left(2\\right)=\\frac{{\\left(-1\\right)}^{n}n\\text{!}}{{2}^{n+1}}[\/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=\"q44558799\">Show Solution<\/button><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167024969311\" data-type=\"solution\">\n<p id=\"fs-id1167024969313\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2-x}{{2}^{n+2}}\\right)}^{n}[\/latex]. The interval of convergence is [latex]\\left(0,4\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1167024999410\" data-type=\"problem\">\n<p id=\"fs-id1167024999413\">Find the Maclaurin series for [latex]f\\left(x\\right)=\\cos{x}[\/latex]. Use the ratio test to show that the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Show that the Maclaurin series converges to [latex]\\cos{x}[\/latex] for all real numbers [latex]x[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558399\">Hint<\/button><\/p>\n<div id=\"q44558399\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025241321\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167025241328\">Use the Maclaurin polynomials for [latex]\\cos{x}[\/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=\"q44558499\">Show Solution<\/button><\/p>\n<div id=\"q44558499\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167025241158\" data-type=\"solution\">\n<p id=\"fs-id1167025241160\" style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\n<p id=\"fs-id1167025241218\">By the ratio test, the interval of convergence is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]. Since [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{{|x|}^{n+1}}{\\left(n+1\\right)\\text{!}}[\/latex], the series converges to [latex]\\cos{x}[\/latex] for all real <em data-effect=\"italics\">x<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":22,"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\/950"}],"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":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/950\/revisions"}],"predecessor-version":[{"id":2277,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/950\/revisions\/2277"}],"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\/950\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=950"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=950"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=950"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=950"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}