\chapter{微分与Taylor定理}
\section{微分的概念与应用}
考虑函数的逼近问题：设$f:I \to \realnum$，$x_0 \in I$（$I$为区间）。考虑用一个多项式在$x_0$附近逼近$f(x)$，比如用0次多项式$P_0(x)$逼近：
\[f(x_0 + \Delta x) \approx P_0(\delx) = a \quad \text{（$\delx$比较小时）}\]
即我们希望
\[\tolim{\Delta x}{0} [f(x_0 + \Delta x) - P_0(\Delta x)] = 0\]
如果$f$在$x_0$处连续，那么上式导出$a = f(x_0)$，即$P_0(\Delta x) = f(x_0)$。

进一步，考虑用一次多项式$P_1(\Delta x) = a + bx$逼近$f(x)$：
\[f(x + \Delta x) \approx P_1(\Delta x) \quad \text{（$\delx$比较小时）}\]
即我们希望
\[\tolim{\Delta x}{0} [f(x_0 + \Delta x) - P_1(\Delta x)] = 0\]
和刚才一样，我们还能得到$a = f(x_0)$。那么$b$应满足什么要求？

\begin{definition}[微分]
    设$f$在$x_0$点附近有定义，如果存在实数$\lambda$使得
    \begin{equation*}\label{微分定义}\tag{$\ast$}
        f(x_0 + \Delta x) = f(x_0) + \lambda \Delta x + o(\Delta x) \quad (\Delta x \to 0)
    \end{equation*}
    则称$f$在$x_0$点可微，记$f$在$x_0$点的微分为$\dif f(x_0) = \lambda \Delta x$，$\lambda$称为$f$在$x_0$点的微分系数。
\end{definition}

\begin{remark}
    式\eqref{微分定义}说明$f$在$x_0$点附近有一次多项式近似
    \[f(x_0 + \Delta x) \approx f(x_0) + \lambda \Delta x \quad \text{（$\Delta x$很小）}\eqper\]
\end{remark}

回忆记号$o(\Delta x)$的含义，\eqref{微分定义}实际上表示的是
\[\frac{f(x_0 + \Delta x) - f(x_0) - \lambda \Delta x}{\Delta x} = \frac{o(\Delta x)}{\Delta x} \to 0\]
即
\[\tolim{\Delta x}{0} \left[\frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} - \lambda\right] = 0\]
从此我们可以看到$\lambda = \deriv{f}(x_0)$，这也就是刚才的$b$。

\begin{corollary}[可微与可导的关系]
    函数$f$在$x_0$点可微的充分必要条件是$f$在$x_0$点可导。这时微分系数$\lambda = \deriv{f}(x_0)$，即$\dif f(x_0) = \deriv{f}(x_0) \Delta x$。
\end{corollary}

\begin{remark}
    注意到$\dif x = \deriv{(x)}\Delta x = \Delta x$，因此我们可以将$\dif f(x) = \deriv{f}(x) \Delta x$改写成$\dif f(x) = \deriv{f}(x) \dif x$。
\end{remark}

\begin{remark}
    回忆Leibniz记号，对于函数$y = f(x)$
    \[\frac{\dif y}{\dif x} = \deriv{f}(x)\]
    即
    \[\dif y = \frac{\dif y}{\dif x} \dif x\eqper\]
\end{remark}

\begin{theorem}[微分的四则运算]
    关于函数四则运算运算的微分，有下列法则：
    \begin{enumerate}
        \item $\dif (f \pm g) = \dif f \pm \dif g$；
        \item $\dif (fg) = g \dif f + f \dif g$；
        \item $\dif \left(\dfrac{f}{g}\right) = \dfrac{g \dif f - f \dif g}{g^2}$，其中$g \neq 0$。
    \end{enumerate}
\end{theorem}

\begin{theorem}[复合函数的微分]
    设函数$x = \varphi(t)$在$t$点可微，函数$y = f(x)$在$x = \varphi(t)$点可微，则复合函数$y = (f \circ \varphi)(t) = f(\varphi(t))$在$t$点可微，且
    \[\dif y = \deriv{(f \circ \varphi)}(t) \dif t = \deriv{f}(\varphi(t))\deriv{\varphi}(t) \dif t \eqper\]
\end{theorem}

\begin{proof}
    利用微分与导数的关系及复合函数求导法则即得。
\end{proof}

\begin{remark}
    回忆Leibniz记号，对于函数$y = f(x), x = \varphi(t)$
    \[\frac{\dif y}{\dif t} = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t}\]
    因此有
    \[\dif y = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t} \dif t\eqper\]
\end{remark}

\section{带Peano余项的Taylor公式}
受引入微分概念的启发，如果我们想更精确低用$n$次多项式逼近$f(x)$，即我们希望用
\[P_n(\Delta x) = a_0 + a_1 \Delta x + \cdots + a_n \Delta x^n\]
逼近$f(x)$，那么它应该满足
\[f(x_0 + \Delta x) = P_n(\Delta x) + o(\Delta x^n)\eqper\]

以二次多项式的逼近为例，令
\[P_2(\Delta x) = a + b \Delta x + c \Delta x^2\]
满足
\[f(x_0 + \Delta x) = P_2(\Delta x) + o(\Delta x^2)\]
即
\[f(x_0 + \Delta x) - a - b \Delta x - c \Delta x^2 = o(\Delta x^2)\]
考虑$\Delta x \to 0$，则有
\begin{equation*}
    \begin{aligned}
        f(x_0 + \Delta x) - a \to 0 & \Rightarrow a = f(x_0)\\
        \frac{f(x_0 + \Delta x) - a}{\Delta x} - b \to 0 & \Rightarrow b = \deriv{f}(x_0)\\
        \frac{f(x_0 + \Delta x) - a - b \Delta x}{\Delta x^2} - c \to 0 & \Rightarrow c = \frac{f^{\prime \prime}(x_0)}{2}
    \end{aligned}
\end{equation*}
这说明只要$f^{\prime \prime}(x_0)$存在就有
\[f(x_0 + \Delta x) = f(x_0) + \deriv{f}(x_0)\Delta x + \frac{f^{\prime \prime}(x_0)}{2}\Delta x^2 + o(\Delta x^2)\eqper\]

将其推广，我们有
\begin{definition}[Taylor多项式与Maclaurin多项式]
    设$f(x)$在$x_0$附近有定义，且$f^{(n)}(x_0)$存在，引入多项式
    \[P_n(\Delta x) = f(x_0) + \frac{\deriv{f}(x_0)}{1!}\Delta x + \frac{f^{\prime \prime}(x_0)}{2!}\Delta x^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}\Delta x^n\]
    或
    \[T_n(f, x_0;x) = f(x_0) + \frac{\deriv{f}(x_0)}{1!}(x-x_0) + \frac{f^{\prime \prime}(x_0)}{2!}(x-x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\]
    称为$f(x)$在$x_0$点的$n$次Taylor多项式。

    特别地，在$x_0 = 0$时，Taylor多项式称为Maclaurin多项式：
    \[P_n = f(0) + \frac{\deriv{f}(0)}{1!}(x) + \frac{f^{\prime \prime}(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(x)}{n!}x^n\eqper\]
\end{definition}

\begin{theorem}[Taylor公式]
    设$f(x)$在$x_0$附近有定义，且$f^{(n)}(x_0)$存在，则
    \[f(x_0 + \Delta x) = P_n(\Delta x) + o(\Delta x^n)\]
    等价地
    \[f(x) = P_n(x - x_0) + o((x-x_0)^n)\]
    称为$f(x)$在$x_0$点（带Peano型余项）的$n$阶/$n$次Taylor多项式展开。其中$o(\Delta x^n)$项称为Peano型余项。
\end{theorem}

\begin{proof}
    注意到对$k = 1, 2, \cdots n$，
    \[P_n^{(k)}(\Delta x) = f^{(k)}(x_0) + \frac{f^{(k+1)}(x_0)}{1!} \Delta x + \cdots + \frac{f^{(n)}(x_0)}{(n-k)!}\Delta x^{n-k}\]
    因此有$P_n^{(n-1)}(\Delta x) = f^{(n-1)}(x_0) + f^{(n)}(x_0)\Delta x$。

    已知条件中已经隐含了$f$的$n$阶一下导数在$x_0$附近都存在，因此下面反复对``$\dfrac{0}{0}$型''极限运用L'Hospital法则：
    \begin{align*}
        \tolim{\Delta x}{0}\frac{f(x_0+\Delta x) - P_n(\Delta x)}{\Delta x^n} & = \tolim{\Delta x}{0} \frac{\deriv{f}(x_0 + \Delta x) - \deriv{P}_n(\Delta x)}{n\Delta x^{n-1}}\\
        & = \cdots = \tolim{\Delta x}{0} \frac{f^{(n-1)}(x_0 + \Delta x) - P_n^{(n-1)}(\Delta x)}{n(n-1)\cdots 2\cdot \Delta x}\\
        & = \tolim{\Delta x}{0} \frac{f^{(n-1)}(x_0 + \Delta x) - f^{(n-1)}(x_0) - f^{(n)}(x_0)\Delta x}{n!\Delta x}\\
        & = \frac{1}{n!} \tolim{\Delta x}{0} \left[\frac{f^{(n-1)}(x_0 + \Delta x) - f^{(n-1)}(x_0)}{\Delta x} - f^{(n)}(x_0) \right]\\
        & = 0 \eqper \qedhere
    \end{align*}
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = e^x$。
\end{example}

\begin{proof}[解]
    $f^{(k)}(x) = e^x$，$f^{(k)}(0) = 1$，$k = 1, 2, 3 \cdots$。因此
    \[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper \qedhere\]
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = \sin x$。
\end{example}

\begin{proof}[解]
    $f^{(k)}(x) = \sin \left(x + \dfrac{k\pi}{2}\right)$，$f^{(k)}(0) = \sin \dfrac{k\pi}{2}$，$k = 0, 1, 2, \cdots$。因此
    \begin{equation*}
        f^{(k)} = 
        \begin{cases}
            0, \quad k = 2m\\
            (-1)^m, \quad k = 2m + 1
        \end{cases}
        ,\quad m = 0, 1, 2\cdots
    \end{equation*}
    所以
    \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper \qedhere\]
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = \cos x$。
\end{example}

\begin{proof}[解]
    类似上面，
    \begin{equation*}
        f^{(k)} = 
        \begin{cases}
            (-1)^m, \quad k = 2m\\
            0, \quad k = 2m + 1
        \end{cases}
        ,\quad m = 0, 1, 2\cdots
    \end{equation*}
    所以
    \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper \qedhere\]
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = \ln (1 + x)$。
\end{example}

\begin{proof}[解]
    $f^{(k)}(x) = (-1)^{k-1}(k-1)!(1+x)^{k-1}$，因此$f^{(k)}(0) = (-1)^{k-1}(k-1)!$，$k = 1, 2, \cdots$。
    所以
    \[\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)\eqper\qedhere\]
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = \arctan x$。
\end{example}

\begin{proof}
    \begin{equation*}
        f^{(k)}(0)
        \begin{cases}
            0, \quad k = 2m\\
            (-1)^m(2m)!, \quad k = 2m + 1
        \end{cases}
        m = 0, 1, 2, \cdots
    \end{equation*}
    因此
    \[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper \qedhere\]
\end{proof}

\begin{example}
    计算Maclaurin展开：$f(x) = (1 + x)^\alpha$，$\alpha \in \realnum$。
\end{example}

\begin{proof}
    $f^{(k)}(0) = \alpha (\alpha - 1)\cdots(\alpha - k + 1),\quad k = 1, 2, \cdots$，因此
    \[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper \qedhere\]
\end{proof}

\begin{remark}
    上式的几个例子：
    \begin{align*}
        \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots + (-1)^n x^n + o(x^n) & \quad (\alpha = -1)\\
        \sqrt{1 + x} 1 + \frac{1}{2}x - \frac{1}{8}x^2 + o(x^2) & \quad (\alpha = \frac{1}{2})\\
        \frac{1}{\sqrt{1 + x}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 + o(x^2) & \quad (\alpha = -\frac{1}{2})
    \end{align*}
\end{remark}

\section{带Lagrange余项的Taylor公式}
\begin{theorem}
    设$f$在$(a, b)$内$n+1$阶可导，$\forall x_0$，$x_0 + \theta x \in (a,b)$，$\exists \theta \in (0,1)$满足
    \[f(x_0 + \Delta x) = P_n(\Delta x) + \frac{f^{(n+1)}(x_0 + \theta \Delta x)}{(n+1)!} \Delta x^{n+1}\]
    其中
    \[R_n(\Delta x) := \frac{f^{(n+1)}(x_0 + \theta \Delta x)}{(n+1)!} \Delta x^{n+1}\]
    称为Lagrange余项。
\end{theorem}

\begin{proof}
    要证明的式子等价于证明$\exists \theta \in (0,1)$使得对于$\Delta \neq 0$有
    \[\frac{f(x_0 + \Delta x) - P_n(\Delta x)}{\Delta x^{n+1}} = \frac{f^{(n+1)}(x_0 + \theta \Delta x)}{(n+1)!}\eqper\]

    回忆Cauchy中值定理：
    设$F(t), G(t) \in C[0,1]$且在$(0,1)$内可导，且$\deriv{G}(t) \neq 0$，则$\exists \theta \in (0,1)$满足
    \[\dfrac{F(1) - F(0)}{G(1) - G(0)} = \dfrac{\deriv{F}(\theta)}{\deriv{G}(\theta)}\eqper\]

    那么现在取$F(t) = f(x_0 + t\Delta x) - P_n(t\Delta x), G(t) = (t\Delta x)^{n+1} \in C^{n+1}[0,1]$那么$F(0) = G(0) = 0$且
    \begin{align*}
        \deriv{F}(t) & = \Delta x(\deriv{f}(x_0 + t \Delta x) - \deriv{P_n}(t\Delta x))\\
        \deriv{G}(t) & = \Delta x(n+1)(t\Delta x)^n
    \end{align*}
    那么应用Cauchy中值定理得到$\exists \theta_1 \in (0,1)$满足
    \[\frac{f(x_0 + \Delta x) - P_n(\Delta x)}{\Delta x^{n+1}} = \frac{\deriv{f}(x_0 + \theta_1 \Delta x) - \deriv{P_n}(\theta_1 \Delta x)}{(n+1)(\theta_1 \Delta x)^n}\]
    再对上式左侧应用Cauchy中值定理可得$\exists \theta_2 \in (0,1)$满足
    \[\frac{\deriv{f}(x_0 + \theta_1 \Delta x) - \deriv{P_n}(\theta_1 \Delta x)}{(n+1)(\theta_1 \Delta x)^n} = \frac{f^{\prime \prime}(x_0 + \theta_1 \theta_2 \Delta x) - P_n^{\prime \prime}(\theta_1 \theta_2 \Delta x)}{(n+1)n(\theta_1 \theta_2 \Delta x)^n}\]
    重复此过程继续得到$\theta_3, \theta_4, \cdots, \theta_n, \theta_{n+1} \in (0,1)$使得
    \begin{align*}
        \frac{f(x_0 + \Delta x) - P_n(\Delta x)}{\Delta x^{n+1}} & = \cdots = \frac{f^{(n)}(x_0 + \theta_1\cdots\theta_n\Delta x) - P_n^{(n)}(\theta_1\cdots\theta_n\Delta x)}{(n+1)!(\theta_1\cdots\theta_n\Delta x)}\\
        & = \frac{f^{(n+1)}(x_0 + \theta_1\cdots\theta_n\theta_{n+1}\Delta x)}{(n+1)!}\eqper \qedhere
    \end{align*}
\end{proof}

\begin{remark}
    带Lagrange余项的Taylor公式也常常写作：设$f$在$(a,b)$内$n+1$阶可导，$\forall x_0, x \in (a,b)$，$\exists \xi$在$x_0$与$x$之间满足
    \[f(x) = P_n(x - x_0) + \frac{f^{(n+1)}(\xi)}{(n+1)!}\Delta x^{n+1}\eqper\]
\end{remark}