$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Let $U\subseteq\mathbb{R}^n$ be open, and $f\in C^p(I, \mathbb{R})$ for $p\geq 0$, and $\vec{a}, \vec{x}\in U$, and $P_{p, \vec{a}}^f(\vec{x})$ be the pth order Taylor polynomial for $f$ at $\vec{a}$. The Taylor remainder of order $P$ is $$\begin{align*} R_{p, \vec{a}}^f(\vec{x}):=f(\vec{x})-P_{p, \vec{a}}^f(\vec{x}) \end{align*}$$

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Let $U\subseteq\mathbb{R}^n$ be open, and $f\in C^p(I, \mathbb{R})$ for $p\geq 0$, and $\vec{a}, \vec{x}\in U$, and $P_{p, \vec{a}}^f(\vec{x})$ be the pth order Taylor polynomial for $f$ at $\vec{a}$. The Taylor remainder of order $P$ is $$\begin{align*} R_{p, \vec{a}}^f(\vec{x}):=f(\vec{x})-P_{p, \vec{a}}^f(\vec{x}) \end{align*}$$

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definition
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FullPage
definition
concepts
used in
hypothesis
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Let $U\subseteq\mathbb{R}^n$ be open, and $f\in C^p(I, \mathbb{R})$ for $p\geq 0$, and $\vec{a}, \vec{x}\in U$, and $P_{p, \vec{a}}^f(\vec{x})$ be the pth order Taylor polynomial for $f$ at $\vec{a}$. The Taylor remainder of order $P$ is $$\begin{align*} R_{p, \vec{a}}^f(\vec{x}):=f(\vec{x})-P_{p, \vec{a}}^f(\vec{x}) \end{align*}$$

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Let $U\subseteq\mathbb{R}^n$ be open, and $f\in C^p(I, \mathbb{R})$ for $p\geq 0$, and $\vec{a}, \vec{x}\in U$, and $P_{p, \vec{a}}^f(\vec{x})$ be the pth order Taylor polynomial for $f$ at $\vec{a}$. The Taylor remainder of order $P$ is $$\begin{align*} R_{p, \vec{a}}^f(\vec{x}):=f(\vec{x})-P_{p, \vec{a}}^f(\vec{x}) \end{align*}$$

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