{"id":947,"date":"2025-06-20T17:24:28","date_gmt":"2025-06-20T17:24:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=947"},"modified":"2025-09-12T16:16:04","modified_gmt":"2025-09-12T16:16:04","slug":"taylor-and-maclaurin-series-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-learn-it-3\/","title":{"raw":"Taylor and Maclaurin Series: Learn It 3","rendered":"Taylor and Maclaurin Series: Learn It 3"},"content":{"raw":"

Taylor\u2019s Theorem with Remainder<\/h2>\r\nWhen we use a Taylor polynomial [latex]p_n(x)[\/latex] to approximate a function [latex]f(x)[\/latex], we need to know how good our approximation is. The remainder<\/strong> [latex]R_n(x)[\/latex] tells us exactly that\u2014it's the difference between the actual function value and our polynomial approximation:\r\n
[latex]{R}_{n}(x)=f(x)-{p}_{n}(x)[\/latex].<\/div>\r\nThink of it this way: if you're using a Taylor polynomial to estimate [latex]f(x)[\/latex], the remainder is your error. The smaller the remainder, the better your approximation.\r\n\r\nFor a Taylor series to converge to [latex]f[\/latex], we need two things:\r\n
    \r\n \t
  • the sequence of Taylor polynomials [latex]{p_n}[\/latex] must converge<\/li>\r\n \t
  • the remainder [latex]R_n[\/latex] must approach zero as [latex]n[\/latex] increases.<\/li>\r\n<\/ul>\r\nIf [latex]R_n[\/latex] doesn't go to zero, your Taylor series might converge\u2014but not to the function you want! To determine if [latex]R_{n}[\/latex]\u00a0converges to zero, we introduce Taylor\u2019s theorem with remainder<\/strong>. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the [latex]n[\/latex]th Taylor polynomial approximates the function.\r\n

    Let's start with the simplest case: the [latex]0[\/latex]th-degree Taylor polynomial. Here, [latex]p_0(x) = f(a)[\/latex], so:<\/p>\r\n

    [latex]R_0(x) = f(x) - f(a)[\/latex]<\/p>\r\n

    If [latex]f[\/latex] is differentiable on an interval containing both [latex]a[\/latex] and [latex]x[\/latex], the Mean Value Theorem tells us there's some value [latex]c[\/latex] between [latex]a[\/latex] and [latex]x[\/latex] where:<\/p>\r\n\r\n

    \r\n
    \r\n

    [latex]R_0(x) = f'(c)(x - a)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    Using similar reasoning with higher derivatives, we can show that for an [latex]n[\/latex]-times differentiable function:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    [latex]R_n(x) = \\frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    where [latex]c[\/latex] is some unknown value between [latex]a[\/latex] and [latex]x[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    The [latex]c[\/latex] in the remainder formula is NOT the center [latex]a[\/latex]\u2014it's an unknown value somewhere between [latex]a[\/latex] and [latex]x[\/latex]. We usually can't find its exact value, but we can still use this formula to bound our error.<\/section>Since we don't know the exact value of [latex]c[\/latex], we can't calculate [latex]R_n(x)[\/latex] exactly. However, if we know that [latex]|f^{(n+1)}(x)| \\leq M[\/latex] for all [latex]x[\/latex] in our interval, then:\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n

    [latex]|R_n(x)| \\leq \\frac{M}{(n+1)!}|x - a|^{n+1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    This bound is incredibly useful\u2014it tells us the maximum possible error when using a Taylor polynomial.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    <\/div>\r\n<\/div>\r\n

    We now state Taylor\u2019s theorem, which provides the formal relationship between a function [latex]f[\/latex] and its [latex]n[\/latex]th degree Taylor polynomial [latex]{p}_{n}\\left(x\\right)[\/latex]. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for [latex]f[\/latex] converges to [latex]f[\/latex].<\/p>\r\n\r\n

    \r\n

    theorem: Taylor\u2019s theorem with remainder<\/h3>\r\n

    Let [latex]f[\/latex] be a function that can be differentiated [latex]n+1[\/latex] times on an interval [latex]I[\/latex] containing the real number [latex]a[\/latex]. Let [latex]p_{n}[\/latex]\u00a0be the [latex]n[\/latex]th Taylor polynomial of [latex]f[\/latex] at [latex]a[\/latex]\u00a0and let<\/p>\r\n\r\n

    [latex]{R}_{n}\\left(x\\right)=f\\left(x\\right)-{p}_{n}\\left(x\\right)[\/latex]<\/div>\r\n \r\n

    be the [latex]n[\/latex]th remainder. Then for each [latex]x[\/latex]\u00a0in the interval [latex]I[\/latex], there exists a real number [latex]c[\/latex]\u00a0between [latex]a[\/latex]\u00a0and [latex]x[\/latex]\u00a0such that<\/p>\r\n\r\n

    [latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex].<\/div>\r\n \r\n

    If there exists a real number [latex]M[\/latex]\u00a0such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all [latex]x\\in I[\/latex], then<\/p>\r\n\r\n

    [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex]<\/div>\r\n \r\n

    for all [latex]x[\/latex] in [latex]I[\/latex].<\/p>\r\n\r\n<\/section>

    \r\n

    Proof<\/strong><\/p>\r\n\r\n\r\n

    \r\n

    Fix a point [latex]x\\in I[\/latex] and introduce the function g<\/em> such that<\/p>\r\n\r\n

    [latex]g\\left(t\\right)=f\\left(x\\right)-f\\left(t\\right)-{f}^{\\prime }\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}-\\cdots -\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}-{R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n+1}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\r\n \r\n

    We claim that g<\/em> satisfies the criteria of Rolle\u2019s theorem. Since g<\/em> is a polynomial function (in t<\/em>), it is a differentiable function. Also, g<\/em> is zero at [latex]t=a[\/latex] and [latex]t=x[\/latex] because<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill g\\left(a\\right)& =\\hfill & f\\left(x\\right)-f\\left(a\\right)-{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)-\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}-{R}_{n}\\left(x\\right)\\hfill \\\\ & =\\hfill & f\\left(x\\right)-{p}_{n}\\left(x\\right)-{R}_{n}\\left(x\\right)\\hfill \\\\ & =\\hfill & 0,\\hfill \\\\ g\\left(x\\right)\\hfill & =\\hfill & f\\left(x\\right)-f\\left(x\\right)-0-\\cdots -0\\hfill \\\\ & =\\hfill & 0.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    Therefore, g<\/em> satisfies Rolle\u2019s theorem, and consequently, there exists c<\/em> between a<\/em> and x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. We now calculate [latex]{g}^{\\prime }[\/latex]. Using the product rule, we note that<\/p>\r\n\r\n

    [latex]\\frac{d}{dt}\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]=\\frac{-{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}+\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}[\/latex].<\/div>\r\n \r\n

    Consequently,<\/p>\r\n\r\n

    [latex]\\begin{array}{cc}{g}^{\\prime }\\left(t\\right)\\hfill & = -{f}^{\\prime }\\left(t\\right)+\\left[{f}^{\\prime }\\left(t\\right)-f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)\\right]+\\left[f^{\\prime\\prime}\\left(t\\right)\\left(x-t\\right)-\\frac{f^{\\prime\\prime\\prime}\\left(t\\right)}{2\\text{!}}{\\left(x-t\\right)}^{2}\\right]+\\cdots \\hfill \\\\ & +\\left[\\frac{{f}^{\\left(n\\right)}\\left(t\\right)}{\\left(n - 1\\right)\\text{!}}{\\left(x-t\\right)}^{n - 1}-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}\\right]+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    Notice that there is a telescoping effect. Therefore,<\/p>\r\n\r\n

    [latex]{g}^{\\prime }\\left(t\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(t\\right)}{n\\text{!}}{\\left(x-t\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-t\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex].<\/div>\r\n \r\n

    By Rolle\u2019s theorem, we conclude that there exists a number c<\/em> between a<\/em> and x<\/em> such that [latex]{g}^{\\prime }\\left(c\\right)=0[\/latex]. Since<\/p>\r\n\r\n

    [latex]{g}^{\\prime }\\left(c\\right)=-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}[\/latex]<\/div>\r\n \r\n

    we conclude that<\/p>\r\n\r\n

    [latex]-\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{n\\text{!}}{\\left(x-c\\right)}^{n}+\\left(n+1\\right){R}_{n}\\left(x\\right)\\frac{{\\left(x-c\\right)}^{n}}{{\\left(x-a\\right)}^{n+1}}=0[\/latex].<\/div>\r\n \r\n

    Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by [latex]n+1[\/latex], we conclude that<\/p>\r\n\r\n

    [latex]{R}_{n}\\left(x\\right)=\\frac{{f}^{\\left(n+1\\right)}\\left(c\\right)}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex]<\/div>\r\n \r\n

    as desired. From this fact, it follows that if there exists M<\/em> such that [latex]|{f}^{\\left(n+1\\right)}\\left(x\\right)|\\le M[\/latex] for all x<\/em> in I<\/em>, then<\/p>\r\n\r\n

    [latex]|{R}_{n}\\left(x\\right)|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex].<\/div>\r\n \r\n

    [latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/section>\r\n

    This theorem gives us two powerful tools:<\/p>\r\n\r\n

      \r\n \t
    • A way to prove that a Taylor series converges to [latex]f[\/latex]<\/li>\r\n \t
    • A method to bound the error when approximating [latex]f[\/latex] with a Taylor polynomial<\/li>\r\n<\/ul>\r\n

      Notice how the factorial in the denominator grows very quickly as [latex]n[\/latex] increases. This rapid growth often causes the remainder to approach zero, which is exactly what we need for convergence. Not only does Taylor\u2019s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] at [latex]x=8[\/latex] and determine how accurate these approximations are at estimating [latex]\\sqrt[3]{11}[\/latex].<\/p>\r\n\r\n

      \r\n
      \r\n

      Consider the function [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex].<\/p>\r\n\r\n

        \r\n \t
      1. Find the first and second Taylor polynomials for [latex]f[\/latex] at [latex]x=8[\/latex]. Use a graphing utility to compare these polynomials with [latex]f[\/latex] near [latex]x=8[\/latex].<\/li>\r\n \t
      2. Use these two polynomials to estimate [latex]\\sqrt[3]{11}[\/latex].<\/li>\r\n \t
      3. Use Taylor\u2019s theorem to bound the error.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n
        \r\n
          \r\n \t
        1. For [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex], the values of the function and its first two derivatives at [latex]x=8[\/latex] are as follows:\r\n<\/span>\r\n
          [latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\sqrt[3]{x}\\hfill & & & \\hfill f\\left(8\\right)& =\\hfill & 2\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & \\frac{1}{3{x}^{\\frac{2}{3}}}\\hfill & & & \\hfill {f}^{\\prime }\\left(8\\right)& =\\hfill & \\frac{1}{12}\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{-2}{9{x}^{\\frac{5}{3}}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(8\\right)& =\\hfill & -\\frac{1}{144}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThus, the first and second Taylor polynomials at [latex]x=8[\/latex] are given by\r\n<\/span>\r\n
          [latex]\\begin{array}{ccc}\\hfill {p}_{1}\\left(x\\right)& =\\hfill & f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)\\hfill \\\\ & =\\hfill & 2+\\frac{1}{12}\\left(x - 8\\right)\\hfill \\\\ {p}_{2}\\left(x\\right)\\hfill & =\\hfill & f\\left(8\\right)+{f}^{\\prime }\\left(8\\right)\\left(x - 8\\right)+\\frac{f^{\\prime\\prime}\\left(8\\right)}{2\\text{!}}{\\left(x - 8\\right)}^{2}\\hfill \\\\ & =\\hfill & 2+\\frac{1}{12}\\left(x - 8\\right)-\\frac{1}{288}{\\left(x - 8\\right)}^{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThe function and the Taylor polynomials are shown in Figure 5.\r\n<\/span>\r\n
          [caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 5. The graphs of [latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex] and the linear and quadratic approximations [latex]{p}_{1}\\left(x\\right)[\/latex] and [latex]{p}_{2}\\left(x\\right)[\/latex].[\/caption]<\/figure>\r\n<\/li>\r\n \t
        2. Using the first Taylor polynomial at [latex]x=8[\/latex], we can estimate\r\n<\/span>\r\n
          [latex]\\sqrt[3]{11}\\approx {p}_{1}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)=2.25[\/latex].<\/div>\r\n\r\n<\/span>\r\nUsing the second Taylor polynomial at [latex]x=8[\/latex], we obtain\r\n<\/span>\r\n
          [latex]\\sqrt[3]{11}\\approx {p}_{2}\\left(11\\right)=2+\\frac{1}{12}\\left(11 - 8\\right)-\\frac{1}{288}{\\left(11 - 8\\right)}^{2}=2.21875[\/latex].<\/div>\r\n <\/li>\r\n \t
        3. By the Uniqueness of Taylor Series, there exists a c<\/em> in the interval [latex]\\left(8,11\\right)[\/latex] such that the remainder when approximating [latex]\\sqrt[3]{11}[\/latex] by the first Taylor polynomial satisfies\r\n<\/span>\r\n
          [latex]{R}_{1}\\left(11\\right)=\\frac{f^{\\prime\\prime}\\left(c\\right)}{2\\text{!}}{\\left(11 - 8\\right)}^{2}[\/latex].<\/div>\r\n\r\n<\/span>\r\nWe do not know the exact value of c<\/em>, so we find an upper bound on [latex]{R}_{1}\\left(11\\right)[\/latex] by determining the maximum value of [latex]f^{\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex]. Since [latex]f^{\\prime\\prime}\\left(x\\right)=-\\frac{2}{9{x}^{\\frac{5}{3}}}[\/latex], the largest value for [latex]|f^{\\prime\\prime}\\left(x\\right)|[\/latex] on that interval occurs at [latex]x=8[\/latex]. Using the fact that [latex]f^{\\prime\\prime}\\left(8\\right)=-\\frac{1}{144}[\/latex], we obtain\r\n<\/span>\r\n
          [latex]|{R}_{1}\\left(11\\right)|\\le \\frac{1}{144\\cdot 2\\text{!}}{\\left(11 - 8\\right)}^{2}=0.03125[\/latex].<\/div>\r\n\r\n<\/span>\r\nSimilarly, to estimate [latex]{R}_{2}\\left(11\\right)[\/latex], we use the fact that\r\n<\/span>\r\n
          [latex]{R}_{2}\\left(11\\right)=\\frac{f^{\\prime\\prime\\prime}\\left(c\\right)}{3\\text{!}}{\\left(11 - 8\\right)}^{3}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]f^{\\prime\\prime\\prime}\\left(x\\right)=\\frac{10}{27{x}^{\\frac{8}{3}}}[\/latex], the maximum value of [latex]f^{\\prime\\prime\\prime}[\/latex] on the interval [latex]\\left(8,11\\right)[\/latex] is [latex]f^{\\prime\\prime\\prime}\\left(8\\right)\\approx 0.0014468[\/latex]. Therefore, we have\r\n<\/span>\r\n
          [latex]|{R}_{2}\\left(11\\right)|\\le \\frac{0.0011468}{3\\text{!}}{\\left(11 - 8\\right)}^{3}\\approx 0.0065104[\/latex].<\/div>\r\n <\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
          Watch the following video to see the worked solution to the example above.