{"id":946,"date":"2025-06-20T17:24:26","date_gmt":"2025-06-20T17:24:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=946"},"modified":"2025-09-12T14:10:13","modified_gmt":"2025-09-12T14:10:13","slug":"taylor-and-maclaurin-series-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-learn-it-2\/","title":{"raw":"Taylor and Maclaurin Series: Learn It 2","rendered":"Taylor and Maclaurin Series: Learn It 2"},"content":{"raw":"

Taylor Polynomials<\/h2>\r\n

The [latex]n[\/latex]th partial sum of the Taylor series for a function [latex]f[\/latex] at [latex]a[\/latex] is called the [latex]n[\/latex]th Taylor polynomial<\/strong>.<\/p>\r\n\r\n

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The first few Taylor polynomials are:<\/p>\r\n\r\n

    \r\n \t
  • [latex]p_0(x) = f(a)[\/latex]<\/li>\r\n \t
  • [latex]p_1(x) = f(a) + f'(a)(x-a)[\/latex]<\/li>\r\n \t
  • [latex]p_2(x) = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2[\/latex]<\/li>\r\n \t
  • [latex]p_3(x) = f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\frac{f'''(a)}{3!}(x-a)^3[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>These are the [latex]0[\/latex]th, [latex]1[\/latex]st, [latex]2[\/latex]nd, and [latex]3[\/latex]rd Taylor polynomials of [latex]f[\/latex] at [latex]a[\/latex], respectively. When [latex]a = 0[\/latex], these are called Maclaurin polynomials<\/strong>. We now provide a formal definition of Taylor and Maclaurin polynomials for a function [latex]f[\/latex].\r\n\r\n
    \r\n

    Taylor polynomial<\/h3>\r\n

    If [latex]f[\/latex] has [latex]n[\/latex] derivatives at [latex]x = a[\/latex], then the [latex]n[\/latex]th Taylor polynomial for [latex]f[\/latex] at [latex]a[\/latex] is:<\/p>\r\n\r\n

    [latex]{p}_{n}\\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}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}[\/latex].<\/div>\r\n \r\n

    The [latex]n[\/latex]th Taylor polynomial for [latex]f[\/latex] at [latex]0[\/latex] is known as the [latex]n[\/latex]th\u00a0Maclaurin polynomial for [latex]f[\/latex].<\/p>\r\n\r\n<\/section>Let's see how to use this definition by finding several Taylor polynomials for [latex]f(x) = \\ln x[\/latex] at [latex]x = 1[\/latex].\r\n\r\n

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    \r\n

    Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] for [latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]x=1[\/latex]. Use a graphing utility to compare the graph of [latex]f[\/latex] with the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n

    \r\n

    To find these Taylor polynomials, we need to evaluate [latex]f[\/latex] and its first three derivatives at [latex]x=1[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\text{ln}x\\hfill & & & \\hfill f\\left(1\\right)& =\\hfill & 0\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & \\frac{1}{x}\\hfill & & & \\hfill {f}^{\\prime }\\left(1\\right)& =\\hfill & 1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & -\\frac{1}{{x}^{2}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(1\\right)& =\\hfill & -1\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{2}{{x}^{3}}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)& =\\hfill & 2\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

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

    [latex]\\begin{array}{ccc}\\hfill {p}_{0}\\left(x\\right)& =\\hfill & f\\left(1\\right)=0,\\hfill \\\\ \\hfill {p}_{1}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)=x - 1,\\hfill \\\\ \\hfill {p}_{2}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}=\\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2},\\hfill \\\\ \\hfill {p}_{3}\\left(x\\right)& =\\hfill & f\\left(1\\right)+{f}^{\\prime }\\left(1\\right)\\left(x - 1\\right)+\\frac{f^{\\prime\\prime}\\left(1\\right)}{2}{\\left(x - 1\\right)}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(1\\right)}{3\\text{!}}{\\left(x - 1\\right)}^{3}\\hfill \\\\ & =\\hfill & \\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2}+\\frac{1}{3}{\\left(x - 1\\right)}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    The graphs of [latex]y=f\\left(x\\right)[\/latex] and the first three Taylor polynomials are shown in Figure 1.<\/p>\r\n\r\n

    <\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 1. The function [latex]y=\\text{ln}x[\/latex] and the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[\/latex] and [latex]{p}_{3}[\/latex] at [latex]x=1[\/latex] are plotted on this graph.[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    Watch the following video to see the worked solution to the above example.