{ "cells": [ { "cell_type": "markdown", "id": "81b1d2f0", "metadata": {}, "source": [ "---\n", "title: \"Taylor Remainder Theorem\"\n", "subtitle: \"\"\n", "abstract: \"\"\n", "date: 2026-02-06\n", "format:\n", " html:\n", " other-links:\n", " - text: This notebook\n", " href: App1.ipynb\n", "---" ] }, { "cell_type": "code", "execution_count": null, "id": "db6fe346", "metadata": {}, "outputs": [], "source": [ "#| echo: false\n", "\n", "using Plots, LaTeXStrings" ] }, { "cell_type": "markdown", "id": "aeac01f4", "metadata": {}, "source": [ "## Single variable case\n", "\n", "
⚠ Intermediate Value Theorem \n", "\n", "Suppose that $f : [a,b] \\to \\mathbb R$ is continuous. Then, $f$ is bounded on $[a,b]$ and for all $y \\in [\\min_{[a,b]} f, \\max_{[a,b]} f ]$ there exists $x\\in[a,b]$ such that $f(x) = y$. \n", "\n", "
\n", "\n", "
⚠ Mean Value Theorem \n", "\n", "Suppose that $f \\colon [a,b] \\to \\mathbb R$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then, there exists $c\\in[a,b]$ for which \n", "\n", "\\begin{align*}\n", " f(b) - f(a) = f'(c) (b-a)\n", "\\end{align*}\n", "\n", "
\n", "\n", "*Proof:* Consider $g(x) := f(x) - \\frac{f(b)-f(a)}{b-a} x$, a function for which $g(a) = g(b)$. You can show that there exists $c\\in [a,b]$ for which $g'(c) = 0$ (this is known as Rolle's theorem). This follows from the intermediate value theorem - either the max or min is achieved at $a$ and $b$ (and the function is constant), or there is an extreme point in $(a,b)$. In the latter case, you can show this point is stationary: i.e. $g'$ vanishes here. \n", "\n", "Cauchy's Mean Value Theorem (need this for the proof of Taylor remainder theorem)\n", "\n", "Suppose that $f, g \\colon [a,b] \\to \\mathbb R$ are continuous on $[a,b]$ and differentiable on $(a,b)$. Then, there exists $c\\in[a,b]$ for which \n", "\n", "\\begin{align*}\n", " \\big( f(b) - f(a) \\big) g'(c) = \\big(g(b) - g(a) \\big) f'(c)\n", "\\end{align*}\n", "\n", "*Proof:* $h(x) = \\big( g(b) - g(a) \\big) f(x) - \\big( f(b) - f(a) \\big) g(x)$ is such that $h(a) = h(b)$ and so there exists $c \\in [a,b]$ such that $h'(c) = 0$.\n", "\n", "
⚠ Taylor Remainder Theorem \n", "\n", "Suppose that $f\\colon [a,b] \\to \\mathbb R$ is a function with $n$ times continuously differentiable. Then, there exists $c\\in[a,b]$ for which \n", "\n", "\\begin{align*}\n", " f(x) = f(a) + f'(a) (x - a) + \\dots + \\frac{f^{(n-1)}(a)}{(n-1)!} (x - a)^{n-1} + \\frac{f^{(n)}(c)}{n!} (x-a)^{n}\n", "\\end{align*}\n", "\n", "
\n", "\n", "\n", "*Proof:* Define $F(x) = f(x) - \\sum_{k=0}^{n-1} \\frac{f^{(k)}(a)}{k!} (x - a)^k$ and $G(x) = (x-a)^n$ and note that $F(a) = F'(a) = \\cdots = F^{(n-1)}(a) = 0$ and $G(a) = \\cdots = G^{(n-1)}(a) = 0$ and $G^{(n)}(x) = n!$. Applying Cauchy's Mean Value Theorem $n$ times, we can conclude there exists $c_1,\\dots,c_n$ between $a$ and $x$ such that\n", "\n", "\\begin{align*}\n", " \\frac{F(x)}{G(x)} &= \\frac{F(x) - F(a)}{G(x) - G(a)}\n", " = \\frac{F'(c_1)}{G'(c_1)} \\nonumber\\\\ \n", " &= \\frac{F'(c_1) - F'(a)}{G'(c_1) - G'(a)}\n", " = \\frac{F''(c_2)}{G''(c_2)} \\nonumber\\\\ \n", " &= \\cdots = \\frac{F^{(n)}(c_n)}{G^{(n)}(c_n)}\n", " %\n", " = \\frac\n", " {f^{(n)}(c_n)}\n", " { n! }\n", "\\end{align*}" ] }, { "cell_type": "markdown", "id": "4bc36b9c", "metadata": {}, "source": [ "## Multivariable case\n", "\n", "In the course, we used the fact that, for smooth $f:[0,T] \\times \\mathbb R \\to \\mathbb R$, we have\n", "\n", "\\begin{align}\n", " f(t,u) = f(t_0,u_0) &+ \\frac{\\partial f}{\\partial t}(t_0,u_0) (t-t_0) \n", " %\n", " + \\frac{\\partial f}{\\partial u}(t_0,u_0) (u-u_0) \\nonumber\\\\\n", " %\n", " &+ O( |t-t_0|^2 + |t-t_0|^2 ).\n", "\\end{align}\n", "\n", "This result follows from Taylor expansion in multiple variables. In the following, we will use the notation $\\bm x = (x_1,\\dots,x_d) \\in \\mathbb R^d$ with $d \\in \\mathbb N$:\n", "\n", "::: {#thm-1}\n", "# Multivariable Taylor Remainder Theorem\n", "\n", "Suppose that $f: \\mathbb R^d \\to \\mathbb R$ is $n+1$ times continuously differentiable in each of the variables and fix $\\bm x_0 \\in \\mathbb R^d$. Then, for all $\\bm x \\in \\mathbb R^d$ there exists $\\bm \\xi = t \\bm x + (1-t) \\bm x_0 \\in \\mathbb R^d$ for some $t\\in [0,1]$ such that \n", "\n", "\\begin{align}\n", " f(\\bm x) = f(\\bm x_0) &+ \\sum_{i=1}^d \\frac{\\partial f(\\bm x_0)}{\\partial x_i} (x-x_0)_i \\nonumber\\\\\n", " %\n", " &+ \\frac{1}{2} \\sum_{i,j=1}^d \\frac{\\partial^2 f(\\bm x_0)}{\\partial x_i \\partial x_j} (x-x_0)_i (x-x_0)_j \n", " %\n", " + \\cdots \\nonumber\\\\\n", " %\n", " &+ \\frac{1}{n!} \\sum_{i_1,\\dots,i_n = 1}^d\\frac{\\partial^n f(\\bm x_0)}{\\partial x_{i_1} \\cdots \\partial x_{i_n}} (x-x_0)_{i_1} \\cdots (x-x_0)_{i_n} \\nonumber\\\\\n", " %\n", " &+ \\frac{1}{(n+1)!} \\sum_{i_1,\\dots,i_{n+1}=1}^d\\frac{\\partial^{n+1} f(\\bm \\xi)}{\\partial x_{i_1} \\cdots \\partial x_{i_{n+1}}} (x-x_0)_{i_1} \\cdots (x-x_0)_{i_{n+1}}.\n", "\\end{align}\n", "\n", ":::\n", "\n", "
\n", "Proof. \n", "\n", "We can define $F(t) := f\\big( t \\bm x + (1-t) \\bm x_0 \\big)$, a real valued function of one variable, and apply the Taylor theorem that you know. Since $F$ is $n+1$ times continuously differentiable, there exists $\\xi \\in [0,1]$ such that \n", "\n", "\\begin{align}\n", " F(1) = F(0) + F'(0) + \\frac{1}{2} F''(0) + \\dots + \\frac{1}{n!} F^{(n)}(0) + \\frac{1}{(n+1)!} F^{(n+1)}(\\xi).\n", "\\end{align}\n", "\n", "We conclude by noting that, by the chain rule\n", "\n", "\\begin{align}\n", " F'(t) &= \\sum_{i=1}^d \\frac{\\partial f}{\\partial x_i}\\big( t \\bm x + (1-t) \\bm x_0 \\big) (x-x_0)_i \\nonumber\\\\\n", " F''(t) &= \\sum_{i,j=1}^d \\frac{\\partial^2 f}{\\partial x_i \\partial x_j}\\big( t \\bm x + (1-t) \\bm x_0 \\big) (x-x_0)_i (x-x_0)_j,\n", "\\end{align}\n", "\n", "etc.\n", "\n", "
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