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  • Functions Of Several Real Variables_Theorem_3.1.3:Linear approximation

    $f:\Omega\subseteq\mathbf{R}^n\to \mathbf{R}^m$ is differentiable at $a$ if and only if there is a function $\epsilon(x)$ so that for $x\in\Omega$ we have $$ f(x)=f(a)+T(x-a)+\epsilon(x)||x-a|| $$with $\epsilon(x)\to 0$ as $x\to a$.


    Proof:

    $\Leftarrow$:
    \begin{equation}
    ||f(x)-f(a)-T(x-a)||=||x-a||||\epsilon(x)||
    \end{equation}
    When $x\to a$,$\epsilon(x)\to 0$,so $||\epsilon(x)||\to 0$.Thus
    \begin{equation}
    \lim_{x\to a;x\in\Omega\backslash\{a\}}\frac{||f(x)-f(a)-T(x-a)||}{||x-a||}=0
    \end{equation}
    So $f$ is differentiable at $a$.

    $\Rightarrow$:
    \begin{equation}
    \lim_{x\to a;x\in\Omega\backslash\{a\}}\frac{||f(x)-f(a)-T(x-a)||}{||x-a||}=0
    \end{equation}
    means $\forall \varepsilon>0$,there exists $\delta>0$,such that for all $x'\in\{x\in\Omega\backslash\{a\}:||x-a||\leq \delta\}$,we have
    \begin{equation}\label{eq:4}
    \frac{||f(x')-f(a)-T(x'-a)||}{||x'-a||}\leq \varepsilon
    \end{equation}
    equation \ref{eq:4} is equivalent to
    \begin{equation}\label{eq:5}
    ||f(x')-f(a)-T(x'-a)||\leq \varepsilon||x'-a||
    \end{equation}

    Let $p(x)=f(x)-f(a)-T(x-a)$.Then equation \ref{eq:5} becomes
    \begin{equation}
    ||p(x')||\leq\varepsilon||x'-a||
    \end{equation}
    Now let $\epsilon(x)=\frac{p(x)}{||x-a||}$,then $||\epsilon(x')||\leq\varepsilon$.And it is easy to verify that $f(x)=f(a)+T(x-a)+\epsilon(x)||x-a||$.Done.$\Box$

    注:T(x-a) 叫做全微分.它的意义是自变量进行微小变化后因变量随之而发生的改变.

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  • 原文地址:https://www.cnblogs.com/yeluqing/p/3828293.html
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