qedhere是个好东西

01实数和数列极限.tex

@@ -114,7 +114,7 @@
     \end{equation*}
     于是$\forall n > n_0$，都有$n \geq n_0 + 1 > \dfrac{2}{\varepsilon^2} + 1$，那么
     \begin{equation*}
-        \left| \sqrt[n]{n} \right| < \sqrt{\dfrac{2}{n-1}} < \varepsilon \eqper
+        \left| \sqrt[n]{n} \right| < \sqrt{\dfrac{2}{n-1}} < \varepsilon \eqper \qedhere
     \end{equation*}
 \end{proof}
 
@@ -448,7 +448,7 @@
 
     其次，因为$A$是$S$的上确界，因此有$\forall \varepsilon > 0$，$\exists n_0, a_{n_0} > A - \varepsilon$。而$\{a_n\}$单调增，$\forall n > n_0$，$a_n \geq a_{n_0} > A - \varepsilon$。
 
-    因此，$\toinf a_n = A$
+    因此，$\toinf a_n = A$。
 \end{proof}
 
 \section{自然对数底e}
@@ -497,23 +497,23 @@
 
 \begin{proof}
     由上证
-    \[\begin{aligned}
+    \begin{align*}
         a_n & = 2 + \dfrac{1}{2!}\left(1 - \dfrac{1}{n}\right) + \cdots + \dfrac{1}{k!}\left(1 - \dfrac{1}{n}\right) \cdots \left(1 - \dfrac{k-1}{n}\right) + \cdots\\
         & + \dfrac{1}{n!}\left(1-\dfrac{1}{n}\right) \cdots \left(1 - \dfrac{n-1}{n}\right)\\
         & \leq 2 + \dfrac{1}{2!}\left(1 - \dfrac{1}{n+1}\right) + \cdots + \dfrac{1}{k!}\left(1 - \dfrac{1}{n+1}\right) \cdots \left(1 - \dfrac{k-1}{n+1}\right) + \cdots \\
         & + \dfrac{1}{n!}\left(1-\dfrac{1}{n+1}\right) \cdots \left(1 - \dfrac{n-1}{n+1}\right) + \dfrac{1}{(n+1)!}\left(1 - \dfrac{1}{n+1}\right)\cdots \left(1 - \dfrac{n}{n+1}\right)\\
-        & = a_{n+1}
+        & = a_{n+1} \eqper\qedhere
-    \end{aligned}\]
+    \end{align*}
 \end{proof}
 
 在此，我们可以证明定理\ref{Definition of e}的正确性。
 \begin{proof}
     由极限的保序性，引理\ref{e lemma 2} $\Rightarrow a \leq b$。
-    \[\begin{aligned}
+    \begin{align*}
         a_n & = 2 + \dfrac{1}{2!}\left(1 - \dfrac{1}{n}\right) + \cdots + \dfrac{1}{k!}\left(1 - \dfrac{1}{n}\right) \cdots \left(1 - \dfrac{k-1}{n}\right) + \cdots\\
         & + \dfrac{1}{n!}\left(1-\dfrac{1}{n}\right) \cdots \left(1 - \dfrac{n-1}{n}\right)\\
         & \geq 2 + \dfrac{1}{2!}\left(1 - \dfrac{1}{n}\right) + \cdots + \dfrac{1}{k!}\left(1 - \dfrac{1}{n}\right)
-    \end{aligned}\]
+    \end{align*}
     对$k \leq n$成立。固定$k$，令$n \to \infty$，得
     \[a \geq 2 + \dfrac{1}{2!} + \cdots + \dfrac{1}{k!} = b_k\]
     再令$k \to \infty$，有
@@ -534,7 +534,7 @@
 
 \begin{proof}
     令$\toinf a_n = a$。则$\forall \varepsilon > 0$，$\exists n_0 \in \naturalnum$，使$\forall n, n^\prime > n_0$，都有$\vert a_n - a \vert < \dfrac{\varepsilon}{2}, \vert a_{n^\prime} - a \vert < \dfrac{\varepsilon}{2}$。由三角不等式
-    \[\vert a_n - a_{n^\prime} = \vert a_n - a - (a_{n^\prime} - a) \vert \leq \vert a_n - a \vert + \vert a_{n^\prime} - a \vert < \varepsilon\]
+    \[\vert a_n - a_{n^\prime} = \vert a_n - a - (a_{n^\prime} - a) \vert \leq \vert a_n - a \vert + \vert a_{n^\prime} - a \vert < \varepsilon \eqper \qedhere\]
 \end{proof}
 
 \begin{theorem}[Cauchy收敛原理]\label{cauchy principle of convergence}
@@ -594,7 +594,7 @@
     又因为$\lim \limits_{k \to \infty} (M_k - m_k) = \lim \limits_{k \to \infty} \dfrac{M_1 - m_1}{2^{k-1}} = 0$，因此
     \[\lim \limits_{k \to \infty} M_k = \lim \limits_{k \to \infty} m_k = A\eqco\]
     从而
-    \[\lim \limits_{k \to \infty} a_{n_k} = A \eqper\]
+    \[\lim \limits_{k \to \infty} a_{n_k} = A \eqper \qedhere\]
 \end{proof}
 
 证法二：利用确界概念选子数列。
@@ -634,7 +634,7 @@
     任取数列$\{a_n\}$，定义其``龙头项''为：固定$k \in \naturalnum$，若$\forall n > k, a_k > a_n$，则称$a_k$为一个``龙头项''。那么一个数列必属于下列两种情况之一：
     \begin{enumerate}
         \item $\{a_n\}$有无穷多个``龙头项''，依次记为$a_{k_1}, a_{k_2}, \cdots , a_{k_n}, \cdots$。注意$k_1 < k_2 < \cdots < k_n < \cdots$，因此$a_{k_1} > a_{k_2} > \cdots > a_{k_n} > \cdots$，这时$\{a_n\}$中有严格单调减子列$\{a_{k_n}\}$；
-        \item ``龙头项''只有有限多，那么$\exists n_0 \in \naturalnum$，$\forall n \geq n_0$，$a_n$不是``龙头项''，取$a_{k_1} = a_{n_0}$，那么$a_{k_1}$不是``龙头项''，$\exists k_2 > k_1, a_{k_2} \geq a_{k_1}$，而$a_{k_2}$也不是``龙头项''，$\exists k_3 > k_2, a_{k_3} \geq a_{k_2}$，……依此类推，得到单调增子列$\{a_{k_n}\}$。
+        \item ``龙头项''只有有限多，那么$\exists n_0 \in \naturalnum$，$\forall n \geq n_0$，$a_n$不是``龙头项''，取$a_{k_1} = a_{n_0}$，那么$a_{k_1}$不是``龙头项''，$\exists k_2 > k_1, a_{k_2} \geq a_{k_1}$，而$a_{k_2}$也不是``龙头项''，$\exists k_3 > k_2, a_{k_3} \geq a_{k_2}$，……依此类推，得到单调增子列$\{a_{k_n}\}$。\qedhere
     \end{enumerate}
 \end{proof}
 
@@ -660,7 +660,7 @@
     \end{equation*}
 
     综上，当$n > n_0 = \max{\{n_1, n_2\}} \in \naturalnum$，因为$k_n \geq n > n_0$，因此\eqref{cauchy principle of convergence eq1}与\eqref{cauchy principle of convergence eq2}都成立，
-    \[\vert a_n - a \vert \leq \vert a_n - a_{k_n} \vert + \vert a_{k_n} - a \vert < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} = \varepsilon \eqper\]
+    \[\vert a_n - a \vert \leq \vert a_n - a_{k_n} \vert + \vert a_{k_n} - a \vert < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} = \varepsilon \eqper \qedhere\]
 \end{proof}
 
 基本列的定义中不涉及极限的具体值，这是与收敛数列定义的根本区别。
@@ -706,16 +706,14 @@
 \end{example}
 
 \begin{proof}
-    $\vert x_n - x_{n+p} \vert \leq \dfrac{p}{n^2} \eqco \forall p, n \in \naturalnum$，则$\vert x_n - x_{n+1} \leq \dfrac{1}{n^2} \eqco \forall n$。于是$\forall p, n > 1$
+    $\vert x_n - x_{n+p} \vert \leq \dfrac{p}{n^2} \eqco \forall p, n \in \naturalnum$，则$\vert x_n - x_{n+1} \leq \dfrac{1}{n^2} \eqco \forall n$。于是$\forall p, n > 1$，
-    \[
+    \begin{align*}
-        \begin{aligned}
+        \vert x_n - x_{n+p} \vert & \leq \vert x_n - x_{n+1} \vert + \vert x_{n+1} - x_{n+2} \vert + \cdots + \vert x_{n+p-1} - x_{n+p} \vert \\
-            \vert x_n - x_{n+p} \vert & \leq \vert x_n - x_{n+1} \vert + \vert x_{n+1} - x_{n+2} \vert + \cdots + \vert x_{n+p-1} - x_{n+p} \vert \\
+        & \leq \frac{1}{n^2} + \cdots + \frac{1}{(n+p-1)^2}\\
-            & \leq \frac{1}{n^2} + \cdots + \frac{1}{(n+p-1)^2}\\
+        & \leq \frac{1}{n(n-1)} + \cdots + \frac{1}{(n+p-1)(n+p-2)}\\
-            & \leq \frac{1}{n(n-1)} + \cdots + \frac{1}{(n+p-1)(n+p-2)}\\
+        & = \frac{1}{n-1} - \frac{1}{n+p-1}\\
-            & = \frac{1}{n-1} - \frac{1}{n+p-1}\\
+        & < \frac{1}{n-1} \eqper \qedhere
-            & < \frac{1}{n-1} \eqper
+    \end{align*}
-        \end{aligned}
-    \]
 \end{proof}
 
 证明的核心思想是通过放缩把$n, p$两个任意的数只留下一个，这样就可以说明他在什么情况下一定小于一个$\varepsilon$。
@@ -767,7 +765,7 @@
     由闭区间套定理，$\exists ! \xi \in \bigcap \limits_{n \geq 1} [a_n, b_n]$，且$\toinf a_n = \toinf b_n = \xi$。
 
     $[a_1, b_1]$中包含\setname{X_n}的无穷多项，因此$\exists x_{n_1} \in [a_1, b_1]$。$[a_2, b_2]$中包含\setname{X_n}的无穷多项，因此$\exists X_{n_2} \in [a_1, b_2]$且$n_2 > n_1$。依次做下去，$\exists x_{n_{k+1}} \in [a_{k+1}, b_{k+1}]$且$n_{k+1} > n_k$。由此可以得到\setname{X_n}的子列\setname{X_{n_k}}，满足$\forall k, a_k \leq x_{n_k} \leq b_k$。令$k \to \infty$，由夹逼原理得
-    \[\lim \limits_{k \to \infty} x_{n_k} = \lim \limits_{k \to \infty} a_k = \lim \limits_{k \to \infty} b_k = \xi \eqper\]
+    \[\lim \limits_{k \to \infty} x_{n_k} = \lim \limits_{k \to \infty} a_k = \lim \limits_{k \to \infty} b_k = \xi \eqper \qedhere\]
 \end{proof}
 
 \begin{theorem}[有限覆盖定理]

02函数及其连续性.tex

@@ -268,7 +268,7 @@
     所以
     \[\lim \limits_{x \to 0} \frac{\sin x}{x} = 1\]
     同时也得到
-    \[\lim \limits_{x \to 0} \cos x = 1 \eqper\]
+    \[\lim \limits_{x \to 0} \cos x = 1 \eqper \qedhere\]
 \end{proof}
 
 \begin{proposition}[单调收敛原理]
@@ -364,7 +364,7 @@
 
 \begin{proof}
     \begin{equation*}
-        \toxzero u(x)^{v(x)} = \toxzero e^{v(x) \ln u(x)} = e^{\toxzero \left(v(x) \ln u(x)\right)} = e^{\toxzero v(x) \cdot \toxzero \ln u(x)} = e^{b \ln a} = a^b \eqper
+        \toxzero u(x)^{v(x)} = \toxzero e^{v(x) \ln u(x)} = e^{\toxzero \left(v(x) \ln u(x)\right)} = e^{\toxzero v(x) \cdot \toxzero \ln u(x)} = e^{b \ln a} = a^b \eqper \qedhere
     \end{equation*}
 \end{proof}
 
@@ -824,7 +824,7 @@ $x = 0$是$f$的第二类间断点。
     令$s = x$得
     \[\vert B_n(f)(x) - f(x) \vert \leq \frac{\varepsilon}{2}+\frac{2M}{n\delta^2}\left(x-x^2\right) \leq \frac{\varepsilon}{2} + \frac{M}{2n\delta^2}\eqco \forall x \in [0,1]\]
     任意取定$n > \dfrac{M}{\delta^2 \varepsilon}$，有
-    \[\vert B_n(f)(x) - f(x) \vert = \left|\sum_{k=0}^n f\left(\frac{k}{n}\right)C_n^k x^k (1-x)^{n-k} - f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \eqco x \in [0,1]\]
+    \[\vert B_n(f)(x) - f(x) \vert = \left|\sum_{k=0}^n f\left(\frac{k}{n}\right)C_n^k x^k (1-x)^{n-k} - f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \eqco x \in [0,1]\eqper \qedhere\]
 \end{proof}
 
 \begin{remark}

03函数的导数.tex

@@ -54,7 +54,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
 
 \begin{proof}[解]
     对于任意的$x$，有
-    \[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper\]
+    \[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -116,7 +116,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     为此，考虑
     \[\tolim{\delx}{0}\frac{1}{\delx}\left[\frac{1}{g(x+\delx)}-\frac{1}{g(x)}\right] = \tolim{\delx}{0} -\frac{g(x+\delx)g(x)}{\delx} \cdot \frac{1}{g(x+\delx)g(x)} = - \frac{\deriv{g}(x)}{g^2(x)}\]
     对于一般的$f$可由乘法求导公式得到
-    \[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper\]
+    \[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -173,7 +173,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     令$\Delta t \to 0$，则$\alpha (\Delta x) \to 0 \quad (\Delta x \to 0)$，式\eqref{链式法则式2}可以化为
     \begin{align*}
         \frac{\Delta y}{\Delta x} & = \deriv{f}(x) - \deriv{\varphi}(b)\\
-        & = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper
+        & = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper \qedhere
     \end{align*}
 \end{proof}
 
@@ -243,7 +243,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     \[y = \psi[\invertfunc{\varphi}(x)]\]
     
     再利用复合函数和反函数微分法，得
-    \[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)}\]
+    \[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)} \eqper \qedhere\]
 \end{proof}
 
 \subsection{基本导数公式}
@@ -367,7 +367,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
 
 \begin{proof}
     构造函数$F(x) = f(x) - \frac{f(b) - f(a)}{b-a}(x-a)$，$F(x) \in C[a,b]$。注意到$F(b)  f(b) - [f(b) - f(a)] = f(a) = F(a)$，应用Rolle定理，$\exists c \in (a,b)$使得$\deriv{F}(c) = 0$，即
-    \[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper\]
+    \[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper \qedhere\]
 \end{proof}
 
 \begin{remark}
@@ -464,7 +464,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
                 x \in (x_0, x_0 + \delta),\ \deriv{f}(x) \leq 0,\ \therefore f(x) \leq f(x_0)
             \end{aligned}
         \right\}
-        \text{极大值}
+        \text{极大值}\qedhere
     \end{equation*}
 \end{proof}
 
@@ -489,7 +489,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
                 x \in (x_0, x_0 + \delta),\ \deriv{f}(x) < 0,\ \therefore f(x) < f(x_0)
             \end{aligned}
         \right\}
-        \text{严格极大值}
+        \text{严格极大值}\qedhere
     \end{equation*}
 \end{proof}
 
@@ -518,7 +518,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     \begin{align*}
         f(\lambda_1x_1 + \lambda_2x_2 + \lambda_3x_3) & \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)f\left(\frac{\lambda_2 x_2 + \lambda_3 x_3}{\lambda_2 + \lambda_3}\right)\\
         & \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)\left(\frac{\lambda_2}{\lambda_2 + \lambda_3}f(x_2) + \frac{\lambda_3}{\lambda_2 + \lambda_3}f(x_3)\right)\\
-        & = \lambda_1 f(x_1) + \lambda_2 f(x_2) + \lambda_3 f(x_3)
+        & = \lambda_1 f(x_1) + \lambda_2 f(x_2) + \lambda_3 f(x_3) \eqper \qedhere
     \end{align*}
 \end{proof}
 
@@ -585,7 +585,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     再利用初等分式不等式
     \[\frac{a_1}{b_1} \leq \frac{a_2}{b_2}\text{且}b_1, b_2 > 0,\ \text{则}\frac{a_1}{b_1} \leq \frac{a_1 + a_2}{b_1 + b_2} \leq \frac{a_2}{b_2}\]
     由此即可得到
-    \[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper\]
+    \[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper \qedhere\]
 \end{proof}
 
 \begin{corollary}[二阶导数判别凸性]
@@ -604,7 +604,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     补充定义$f(a) = g(a) = 0$，从而使得$f,g$在$[a,b)$上连续。利用Cauchy中值定理，对$x \in (a,b)$，存在$a < \xi < x$使得
     \[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)}\]
     那么当$x \to a^+$时，有$\xi \to a^+$，因此
-    \[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
+    \[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere\]
 \end{proof}
 
 \begin{remark}
@@ -624,7 +624,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
     那么
     \begin{align*}
         \tolim{x}{+\infty} & = \tolim{t}{0^+}\frac{f \left(\dfrac{1}{t}\right)}{g \left(\dfrac{1}{t}\right)} = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}\\
-        & = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper
+        & = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere
     \end{align*}
 \end{proof}
 

04微分与Taylor定理.tex

@@ -123,7 +123,7 @@
         & = \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
+        & = 0 \eqper \qedhere
     \end{align*}
 \end{proof}
 
@@ -133,7 +133,7 @@
 
 \begin{proof}[解]
     $f^{(k)}(x) = e^x$，$f^{(k)}(0) = 1$，$k = 1, 2, 3 \cdots$。因此
-    \[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper\]
+    \[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -151,7 +151,7 @@
         ,\quad m = 0, 1, 2\cdots
     \end{equation*}
     所以
-    \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper\]
+    \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -169,7 +169,7 @@
         ,\quad m = 0, 1, 2\cdots
     \end{equation*}
     所以
-    \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper\]
+    \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -196,7 +196,7 @@
         m = 0, 1, 2, \cdots
     \end{equation*}
     因此
-    \[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper\]
+    \[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper \qedhere\]
 \end{proof}
 
 \begin{example}
@@ -205,7 +205,7 @@
 
 \begin{proof}
     $f^{(k)}(0) = \alpha (\alpha - 1)\cdots(\alpha - k + 1),\quad k = 1, 2, \cdots$，因此
-    \[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper\]
+    \[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper \qedhere\]
 \end{proof}
 
 \begin{remark}