第八周。
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04微分与Taylor定理.tex
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04微分与Taylor定理.tex
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\chapter{微分与Taylor定理}
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\section{微分的概念与应用}
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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：
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\[f(x_0 + \Delta x) \approx P_0(\delx) = a \quad \text{（$\delx$比较小时）}\]
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即我们希望
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\[\tolim{\Delta x}{0} [f(x_0 + \Delta x) - P_0(\Delta x)] = 0\]
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如果$f$在$x_0$处连续，那么上式导出$a = f(x_0)$，即$P_0(\Delta x) = f(x_0)$。
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进一步，考虑用一次多项式$P_1(\Delta x) = a + bx$逼近$f(x)$：
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\[f(x + \Delta x) \approx P_1(\Delta x) \quad \text{（$\delx$比较小时）}\]
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即我们希望
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\[\tolim{\Delta x}{0} [f(x_0 + \Delta x) - P_1(\Delta x)] = 0\]
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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？
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\begin{definition}[微分]
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设$f$在$x_0$点附近有定义，如果存在实数$\lambda$使得
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\begin{equation*}\label{微分定义}\tag{$\ast$}
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f(x_0 + \Delta x) = f(x_0) + \lambda \Delta x + o(\Delta x) \quad (\Delta x \to 0)
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\end{equation*}
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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。
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\end{definition}
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\begin{remark}
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式\eqref{微分定义}说明$f$在$x_0$点附近有一次多项式近似
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\[f(x_0 + \Delta x) \approx f(x_0) + \lambda \Delta x \quad \text{（$\Delta x$很小）}\eqper\]
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\end{remark}
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回忆记号$o(\Delta x)$的含义，\eqref{微分定义}实际上表示的是
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\[\frac{f(x_0 + \Delta x) - f(x_0) - \lambda \Delta x}{\Delta x} = \frac{o(\Delta x)}{\Delta x} \to 0\]
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即
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\[\tolim{\Delta x}{0} \left[\frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} - \lambda\right] = 0\]
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从此我们可以看到$\lambda = \deriv{f}(x_0)$，这也就是刚才的$b$。
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\begin{corollary}[可微与可导的关系]
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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$\dif f(x_0) = \deriv{f}(x_0) \Delta x$。
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\end{corollary}
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\begin{remark}
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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$\dif f(x) = \deriv{f}(x) \dif x$。
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\end{remark}
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\begin{remark}
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回忆Leibniz记号，对于函数$y = f(x)$
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\[\frac{\dif y}{\dif x} = \deriv{f}(x)\]
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即
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\[\dif y = \frac{\dif y}{\dif x} \dif x\eqper\]
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\end{remark}
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\begin{theorem}[微分的四则运算]
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关于函数四则运算运算的微分，有下列法则：
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\begin{enumerate}
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\item $\dif (f \pm g) = \dif f \pm \dif g$；
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\item $\dif (fg) = g \dif f + f \dif g$；
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\item $\dif \left(\dfrac{f}{g}\right) = \dfrac{g \dif f - f \dif g}{g^2}$，其中$g \neq 0$。
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[复合函数的微分]
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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
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\[\dif y = \deriv{(f \circ \varphi)}(t) \dif t = \deriv{f}(\varphi(t))\deriv{\varphi}(t) \dif t \eqper\]
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\end{theorem}
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\begin{proof}
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利用微分与导数的关系及复合函数求导法则即得。
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\end{proof}
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\begin{remark}
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回忆Leibniz记号，对于函数$y = f(x), x = \varphi(t)$
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\[\frac{\dif y}{\dif t} = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t}\]
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因此有
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\[\dif y = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t} \dif t\eqper\]
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\end{remark}
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\section{带Peano余项的Taylor公式}
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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
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\[P_n(\Delta x) = a_0 + a_1 \Delta x + \cdots + a_n \Delta x^n\]
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逼近$f(x)$，那么它应该满足
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\[f(x_0 + \Delta x) = P_n(\Delta x) + o(\Delta x^n)\eqper\]
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以二次多项式的逼近为例，令
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\[P_2(\Delta x) = a + b \Delta x + c \Delta x^2\]
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满足
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\[f(x_0 + \Delta x) = P_2(\Delta x) + o(\Delta x^2)\]
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即
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\[f(x_0 + \Delta x) - a - b \Delta x - c \Delta x^2 = o(\Delta x^2)\]
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考虑$\Delta x \to 0$，则有
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\begin{equation*}
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\begin{aligned}
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f(x_0 + \Delta x) - a \to 0 & \Rightarrow a = f(x_0)\\
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\frac{f(x_0 + \Delta x) - a}{\Delta x} - b \to 0 & \Rightarrow b = \deriv{f}(x_0)\\
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\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}
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\end{aligned}
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\end{equation*}
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这说明只要$f^{\prime \prime}(x_0)$存在就有
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\[f(x_0 + \Delta x) = f(x_0) + \deriv{f}(x_0)\Delta x + \frac{f^{\prime \prime}(x_0)}{2}\Delta x^2 + o(\Delta x^2)\eqper\]
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将其推广，我们有
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\begin{definition}[Taylor多项式与Maclaurin多项式]
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设$f(x)$在$x_0$附近有定义，且$f^{(n)}(x_0)$存在，引入多项式
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\[P_n(\Delta x) = f(x_0) + \frac{\deriv{f}(x_0)}{1!}\Delta x + \frac{f^{\prime \prime}(x_0)}{2!}\Delta x^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}\Delta x^n\]
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称为$f(x)$在$x_0$点的$n$次Taylor多项式。
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特别地，在$x_0 = 0$时，Taylor多项式称为Maclaurin多项式：
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\[P_n = f(0) + \frac{\deriv{f}(0)}{1!}(x) + \frac{f^{\prime \prime}(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(x)}{n!}x^n\eqper\]
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\end{definition}
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\begin{theorem}[Taylor公式]
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设$f(x)$在$x_0$附近有定义，且$f^{(n)}(x_0)$存在，则
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\[f(x_0 + \Delta x) = P_n(\Delta x) + o(\Delta x^n)\]
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等价地
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\[f(x) = P_n(x - x_0) + o((x-x_0)^n)\]
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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。
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\end{theorem}
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\begin{proof}
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注意到对$k = 1, 2, \cdots n$，
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\[P_n^{(k)}(\Delta x) = f^{(k)}(x_0) + \frac{f^{(k+1)}(x_0)}{1!} \Delta x + \cdots + \frac{f^{(n)}(x_0)}{(n-k)!}\Delta x^{n-k}\]
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因此有$P_n^{(n-1)}(\Delta x) = f^{(n-1)}(x_0) + f^{(n)}(x_0)\Delta x$。
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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：
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\begin{align*}
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\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}}\\
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& = \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}\\
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& = \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}\\
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& = \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]\\
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& = 0
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\end{align*}
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = e^x$。
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\end{example}
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\begin{proof}[解]
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$f^{(k)}(x) = e^x$，$f^{(k)}(0) = 1$，$k = 1, 2, 3 \cdots$。因此
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\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper\]
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = \sin x$。
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\end{example}
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\begin{proof}[解]
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$f^{(k)}(x) = \sin \left(x + \dfrac{k\pi}{2}\right)$，$f^{(k)}(0) = \sin \dfrac{k\pi}{2}$，$k = 0, 1, 2, \cdots$。因此
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\begin{equation*}
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f^{(k)} =
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\begin{cases}
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0, \quad k = 2m\\
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(-1)^m, \quad k = 2m + 1
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\end{cases}
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,\quad m = 0, 1, 2\cdots
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\end{equation*}
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所以
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\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper\]
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = \cos x$。
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\end{example}
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\begin{proof}[解]
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类似上面，
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\begin{equation*}
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f^{(k)} =
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\begin{cases}
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(-1)^m, \quad k = 2m\\
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0, \quad k = 2m + 1
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\end{cases}
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,\quad m = 0, 1, 2\cdots
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\end{equation*}
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所以
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\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper\]
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = \ln (1 + x)$。
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\end{example}
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\begin{proof}[解]
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$f^{(k)}(x) = (-1)^{k-1}(k-1)!(1+x)^{k-1}$，因此$f^{(k)}(0) = (-1)^{k-1}(k-1)!$，$k = 1, 2, \cdots$。
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所以
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\[\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)\eqper\]
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = \arctan x$。
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\end{example}
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\begin{proof}
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\begin{equation*}
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f^{(k)}(0)
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\begin{cases}
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0, \quad k = 2m\\
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(-1)^m(2m)!, \quad k = 2m + 1
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\end{cases}
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m = 0, 1, 2, \cdots
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\end{equation*}
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因此
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\[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper\]
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\end{proof}
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\begin{example}
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计算Maclaurin展开：$f(x) = (1 + x)^\alpha$，$\alpha \in \realnum$。
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\end{example}
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\begin{proof}
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$f^{(k)}(0) = \alpha (\alpha - 1)\cdots(\alpha - k + 1),\quad k = 1, 2, \cdots$，因此
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\[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper\]
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\end{proof}
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\begin{remark}
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上式的几个例子：
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\begin{align*}
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\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots + (-1)^n x^n + o(x^n) & \quad (\alpha = -1)\\
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\sqrt{1 + x} 1 + \frac{1}{2}x - \frac{1}{8}x^2 + o(x^2) & \quad (\alpha = \frac{1}{2})\\
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\frac{1}{\sqrt{1 + x}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 + o(x^2) & \quad (\alpha = -\frac{1}{2})
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\end{align*}
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\end{remark}
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