From 785f77973e6b79ddf9d39713af8ba5bae08ebfbc Mon Sep 17 00:00:00 2001 From: unlockable Date: Sun, 6 Nov 2022 19:06:06 +0800 Subject: [PATCH] =?UTF-8?q?=E7=AC=AC=E5=85=AB=E5=91=A8=E3=80=82?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 03函数的导数.tex | 4 +- 04微分与Taylor定理.tex | 218 +++++++++++++++++++++++++++++++++++++++++ 高等微积分.tex | 3 +- 3 files changed, 222 insertions(+), 3 deletions(-) create mode 100644 04微分与Taylor定理.tex diff --git a/03函数的导数.tex b/03函数的导数.tex index 1f64fe7..8ed583a 100644 --- a/03函数的导数.tex +++ b/03函数的导数.tex @@ -554,7 +554,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么 \begin{proof} 注意$x_1 < x < x_2$，可以将$x$表示为 - \[x = \lambda x_1 + 1 - \lambda x_2 = \frac{x_2 - x}{x_2 - x_1}x_1 + \frac{x - x_1}{x_2 - x_2}x_2\] + \[x = \lambda x_1 + (1 - \lambda) x_2 = \frac{x_2 - x}{x_2 - x_1}x_1 + \frac{x - x_1}{x_2 - x_2}x_2\] 利用下凸的性质 \[f(x) \leq \frac{x_2 - x}{x_2 - x_1}f(x_1) + \frac{x - x_1}{x_2 - x_2}f(x_2)\] @@ -643,7 +643,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么 曲线作图步骤：对于方程$y = f(x)$， \begin{enumerate} \item 确定函数$f$的定义域 - \item 检查$f$对对称性：奇偶性与周期性 + \item 检查$f$的对称性：奇偶性与周期性 \item 检查曲线的渐近线：垂直、水平斜渐近线 \item 研究导函数$\deriv{f}$：确定函数的增减区间、极值点、临界点 \item 研究二阶到函数$f^{\prime \prime}$：确定曲线的上下凸区间与拐点 diff --git a/04微分与Taylor定理.tex b/04微分与Taylor定理.tex new file mode 100644 index 0000000..d14db91 --- /dev/null +++ b/04微分与Taylor定理.tex @@ -0,0 +1,218 @@ +\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 + \[P_n(\Delta x) = f(x_0) + \frac{\deriv{f}(x_0)}{1!}\Delta x + \frac{f^{\prime \prime}(x_0)}{2!}\Delta x^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}\Delta x^n\] + 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 + \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 + \[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper\] +\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 + \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper\] +\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 + \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper\] +\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 + \[\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)\eqper\] +\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 + \[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper\] +\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 + \[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper\] +\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} \ No newline at end of file diff --git a/高等微积分.tex b/高等微积分.tex index ca34282..0f2629b 100644 --- a/高等微积分.tex +++ b/高等微积分.tex @@ -56,7 +56,7 @@ \date{} % linespread{1.5} -% \includeonly{03函数的导数} +% \includeonly{04微分与Taylor定理.tex} \begin{document} \maketitle @@ -70,4 +70,5 @@ \include{01实数和数列极限.tex} \include{02函数及其连续性.tex} \include{03函数的导数.tex} + \include{04微分与Taylor定理.tex} \end{document} \ No newline at end of file