第六章。

04微分与Taylor定理.tex

@@ -98,6 +98,8 @@
 \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：
@@ -179,7 +181,7 @@
 \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
-    \[\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)\eqper\]
+    \[\ln (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)\eqper\qedhere\]
 \end{proof}
 
 \begin{example}
@@ -215,4 +217,42 @@
         \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
+    \[f(x) = P_n(x - x_0) + \frac{f^{n+1}(x_0 + \theta \Delta x)}{(n+1)!}\Delta x^{n+1}\eqper\]
 \end{remark}