{"id":200,"date":"2019-08-27T21:08:43","date_gmt":"2019-08-27T21:08:43","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=200"},"modified":"2020-07-28T16:56:58","modified_gmt":"2020-07-28T16:56:58","slug":"taylors-theorem","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/taylors-theorem\/","title":{"rendered":"Taylor&#8217;s Theorem"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Taylor&#8217;s Theorem &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nSuppose we&#8217;re working with a function $f(x)$ that is continuous and\nhas $n+1$ continuous derivatives on an interval about $x=0$.  We can\napproximate $f$ near $0$ by a polynomial $P_n(x)$ of degree $n$:\n\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> For $n=0$, the best constant approximation near $0$ is \n\t\t\\[P_0(x)=f(0)\\]\n\t\twhich matches $f$ at $0$.\n\t<br><br>\n\t<\/li><li> For $n=1$, the best linear approximation near $0$ is \n\t\t\\[P_1(x)=f(0)+f'(0)x.\\]\n\t\tNote that $P_1$ matches $f$ at $0$ and $P_1&#8217;$ matches $f&#8217;$ at $0$.\n\t<br><br>\n\t<\/li><li> For $n=2$, the best quadratic approximation near $0$ is\n\t\t\\[P_2(x)=f(0)+f'(0)x+\\frac{f&#8221;(0)}{2!}x^2.\\]\n\t\tNote that $P_2$, $P_2&#8217;$, and $P_2&#8221;$ match $f$, $f&#8217;$, and $f&#8221;$,\n\t\trespectively, at $0$.\n<\/li><\/ul>\n\n\n\n<p> Continuing this process,  \\[P_n(x)=f(0)+f'(0)x+\\frac{f&#8221;(0)}{2!}x^2+\\ldots +\\frac{f^{(n)}(0)}{n!}x^n.\\] This is the <b>Taylor polynomial of degree $n$ about $0$<\/b> (also called the <b>Maclaurin series of degree $n$<\/b>).  More generally, if $f$ has $n+1$ continuous derivatives at $x=a$, the <b>Taylor series of degree $n$ about $a$<\/b> is \\[\\sum_{k=0}^n \\frac{f^{(k)}(a)}{k!}(x-a)^k=f(a)+f'(a)(x-a)+\\frac{f&#8221;(a)}{2!}(x-a)^2+\\ldots + \\frac{f^{(n)}(a)}{n!}(x-a)^n.\\] This formula approximates $f(x)$ near $a$.  Taylor&#8217;s Theorem gives bounds for the error in this approximation: <\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Taylor&#8217;s Theorem<\/h4>\n\n\n\n<p>\n\nSuppose $f$ has $n+1$ continuous derivatives on an open interval\ncontaining $a$.  Then for each $x$ in the interval,\n\\[f(x)=\\left[\\sum_{k=0}^n\n\\frac{f^{(k)}(a)}{k!}(x-a)^k\\right]+R_{n+1}(x)\\]\nwhere the error term $R_{n+1}(x)$ satisfies $\\displaystyle\nR_{n+1}(x)=\\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$\nbetween $a$ and $x$.\n\n<\/p>\n\n\n\n<p>\nThis form for the error $R_{n+1}(x)$, derived in 1797 by Joseph\nLagrange, is called the Lagrange formula for the remainder.  The\n<i>infinite<\/i> Taylor series converges to $f$,\n\\[f(x)=\\sum_{k=0}^{\\infty}\\frac{f^{(k)}(a)}{k!}(x-a)^k,\\]\nif and only if $\\displaystyle \\lim_{n\\to\\infty} R_n(x)=0$.\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Examples of Taylor Series about $0$<\/h4>\n\n\n\n<ol class=\"wp-block-list\"><li> For $f(x)=e^x$,\n\t\t\\[f^{(k)}(x)=e^x \\quad\\Longrightarrow\\quad f^{(k)}(0)=1.\\]\n\t\tSo\n\t\t\\begin{eqnarray*}\n\t\t\te^x&amp;=&amp;1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\ldots\\\\\n\t\t\t&amp;=&amp;\\sum_{k=0}^{\\infty}\\frac{x^k}{k!}\n\t\t\\end{eqnarray*}\n\t\twhich converges for all $x$ since $\\displaystyle\\lim_{n\\to\\infty}\n\t\tR_n(x)=\\lim_{n\\to\\infty} \\frac{e^cx^{(n+1)}}{(n+1)!}=0$ for all $c$\n\t\tbetween $0$ and $x$.\n\t<br><br>\n\t<\/li><li> For $f(x)=\\ln (1+x)$,\n\t\t\\begin{eqnarray*}\n\t\t\t\\left.\\begin{array}{l}\n\t\t\t\tf(x)=\\ln(1+x)\\\\\n\t\t\t\tf'(x)=\\frac{1}{1+x}\\\\\n\t\t\t\tf&#8221;(x)=\\frac{-1}{(1+x)^2}\\\\\n\t\t\t\tf&#8221;'(x)=\\frac{2}{(1+x)^3}\\\\\n\t\t\t\tf^{(4)}(x)=\\frac{-3\\cdot 2}{(1+x)^4}\\\\\n\t\t\t\t\\vdots\n\t\t\t\\end{array}\\right\\}\\qquad\\Longrightarrow\\qquad\n\t\t\t\\left\\{ \\begin{array}{l}\n\t\t\t\tf(0)=0\\\\\n\t\t\t\tf'(0)=1\\\\\n\t\t\t\tf&#8221;(0)=-1\\\\\n\t\t\t\tf&#8221;'(0)=2\\\\\n\t\t\t\tf^{(4)}(x)=-6\\\\\n\t\t\t\t\\vdots\n\t\t\t\\end{array} \\right.\n\t\t\\end{eqnarray*}\n\t\tSo\n\t\t\\begin{eqnarray*}\n\t\t\t\\ln (1+x)&amp;=&amp;x-\\frac{x^2}{2}+\\frac{x^3}{3}-\\frac{x^4}{4}+\\ldots\\\\\n\t\t\t&amp;=&amp;\\sum_{k=0}^\\infty (-1)^k\\frac{x^{k+1}}{k+1}\n\t\t\\end{eqnarray*}\n\t\twhich converges only for $-1<x\\leq 1$.=\"\" <=\"\" ol=\"\">\n\n<p>\nThe Taylor Series in $(x-a)$ is the <i>unique<\/i> power series in\n$(x-a)$ converging to $f(x)$ on an interval containing $a$.  For this\nreason,\n\n<\/p><p>\n<\/p><ul>\n\t<li> By Example 1,\n\t\t\\[e^{-2x}=1-2x+2x^2-\\frac{4}{3}x^3+\\ldots\\]\n\t\twhere we have substituted $-2x$ for $x$.\n\t<br><br>\n\t<\/li><li> By Example 2, since $\\displaystyle \\frac{d}{dx}[\\ln\n\t\t(1+x)]=\\frac{1}{1+x}$, we can differentiate the Taylor series for $\\ln\n\t\t(1+x)$ to obtain\n\t\t\\[\\frac{1}{1+x}=1-x+x^2-x^3+\\ldots.\\]\n\t\tSubstituting $-x$ for $x$,\n\t\t\\[\\frac{1}{1-x}=1+x+x^2+x^3+\\ldots.\\]\n<\/li><\/ul>\n\n<p>\nIn the Exploration, compare the graphs of various functions with their\nfirst through fourth degree Taylor polynomials.\n\n<\/p><center>\n<font size=\"+1\"><a href=\"javascript:openWindow('exploration.html','taylors_thm_1')\" ;=\"\">Exploration<\/a><\/font>\n\n<hr color=\"blue\">\n\n<p>\n<\/p><h3>Key Concepts<\/h3>\n<p>\n<\/p><\/center>\n\n<b>Taylor&#8217;s Theorem<\/b>\n\n<p>\nSuppose $f$ has $n+1$ continuous derivatives on an open interval containing $a$. \nThen for each $x$ in the interval,\n\n\\[f(x) = \\left[\\sum_{k=0}^n \\frac{f^{(k)}(a)}{k!} (x-a)^k\\right]+ R_{n+1}(x)\\]\n\n<\/p><p>\nwhere the error term $R_{n+1}(x)$ satisfies $R_{n+1}(x) = \n\\left[\\frac{f^{(n+1)}(c)}{(n+1)!}\\right] (x-a)^{n+1}$ for some $c$ between $a$ and $x$.\n\n<!------------------------>\n\n\n<br>\n\n<\/p><hr>\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ0410\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p><p>\n\n\n\n\n\n<\/p>\n\n\n\n<p><\/p>\n<\/x\\leq><\/li><\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Taylor&#8217;s Theorem &#8211; HMC Calculus Tutorial Suppose we&#8217;re working with a function $f(x)$ that is continuous and has $n+1$ continuous derivatives on an interval about $x=0$. We can approximate $f$ near $0$ by a polynomial $P_n(x)$ of degree $n$: For $n=0$, the best constant approximation near $0$ is \\[P_0(x)=f(0)\\] which matches $f$ at $0$. For&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-200","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/200","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=200"}],"version-history":[{"count":5,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/200\/revisions"}],"predecessor-version":[{"id":1248,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/200\/revisions\/1248"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=200"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}