diff --git a/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex
index c39a2da..c728806 100644
--- a/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex
+++ b/advanced-math/knowledge/1-function-and-limit/function-and-limit.tex
@@ -1,9 +1,9 @@
 \documentclass[UTF8, 12pt]{ctexart}
 \usepackage{xcolor}
 % 使用颜色
-\definecolor{orange}{RGB}{255,127,0} 
-\definecolor{violet}{RGB}{192,0,255} 
-\definecolor{aqua}{RGB}{0,255,255} 
+\definecolor{orange}{RGB}{255,127,0}
+\definecolor{violet}{RGB}{192,0,255}
+\definecolor{aqua}{RGB}{0,255,255}
 \usepackage{geometry}
 \setcounter{tocdepth}{5}
 \setcounter{secnumdepth}{5}
@@ -526,7 +526,7 @@ \subsubsection{极限情况总结}
         过程 & $n\to\infty$ & $x\to\infty$ & $x\to+\infty$ & $x\to-\infty$ \\ \hline
         时刻 & \multicolumn{4}{c|}{$N$} \\ \hline
         从此时刻以后 & $n>N$ & $\vert x\vert>N$ & $x>N$ & $x<-N$ \\ \hline
-        $f(x)$ & \multicolumn{4}{c|}{$\vert f(x)-A\vert<\xi$} \\ 
+        $f(x)$ & \multicolumn{4}{c|}{$\vert f(x)-A\vert<\xi$} \\
         \hline
     \end{tabular}
 \end{center}
@@ -537,7 +537,7 @@ \subsubsection{极限情况总结}
         过程 & $x\to x_0$ & $x\to x_0^+$ & $x\to x_0^-$ \\ \hline
         时刻 & \multicolumn{3}{c|}{$\delta$} \\ \hline
         从此时刻以后 & $0<\vert x-x_0\vert<\delta$ & $0<x-x_0<\delta$ & $-\delta<x-x_0<0$\\ \hline
-        $f(x)$ & \multicolumn{3}{c|}{$\vert f(x)-A\vert<\xi$} \\ 
+        $f(x)$ & \multicolumn{3}{c|}{$\vert f(x)-A\vert<\xi$} \\
         \hline
     \end{tabular}
 \end{center}
@@ -692,7 +692,7 @@ \subsection{函数极限}
         \end{array}
     \right.$
     \item 若$f(x)\geqslant g(x)$，则$A\geqslant B$。
-    \item 若$y=f[g(x)]$由$y=f(u)$与$u=g(x)$复合而成，且$\lim\limits_{x\to x_0}g(x)=u_0$且$\lim\limits_{u\to u_0}f(u)=a$，当$x\in\mathring{U}(x_0,\delta_0)$时，$g(x)\neq u_0$，则$\lim\limits_{x\to x_0}f[g(x)]=a$。 
+    \item 若$y=f[g(x)]$由$y=f(u)$与$u=g(x)$复合而成，且$\lim\limits_{x\to x_0}g(x)=u_0$且$\lim\limits_{u\to u_0}f(u)=a$，当$x\in\mathring{U}(x_0,\delta_0)$时，$g(x)\neq u_0$，则$\lim\limits_{x\to x_0}f[g(x)]=a$。
 \end{enumerate}
 
 对于结论7必须\textcolor{orange}{注意}$g(x)\neq u_0$。
@@ -904,7 +904,7 @@ \subsection{\texorpdfstring{$\lim\limits_{x\to\infty}\left(1+\dfrac{1}{x}\right)
 
 证明：$\lim\limits_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=\lim\limits_{x\to\infty}e^{\ln(1+\frac{1}{x})^x}=\lim\limits_{x\to\infty}e^{x\ln(1+\frac{1}{x})}=e^{\lim\limits_{x\to\infty}x\ln(1+\frac{1}{x})}=e^{\lim\limits_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}}$
 
-$=e^{\lim\limits_{x\to\infty}\frac{\left(\frac{1}{1+\frac{1}{x}}\right)\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}}=e^{\lim\limits_{x\to\infty}\frac{1}{1+x}}=e$\medskip
+$=e^{\lim\limits_{x\to\infty}\frac{\left(\frac{1}{1+\frac{1}{x}}\right)\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}}=e^{\lim\limits_{x\to\infty}\frac{1}{1+\frac{1}{x}}}=e$\medskip
 
 从而$\lim\limits_{\Delta\to\infty}\left(1+\dfrac{1}{\Delta}\right)^\Delta=e$与$\lim\limits_{\Delta\to 0}\left(1+\Delta\right)^{\frac{1}{\Delta}}=e(\Delta\neq 0)$。
 
@@ -1202,7 +1202,7 @@ \subsection{零点与介值定理}
 
 \textbf{例题：}证明方程$x=a\sin x+b(a>0,b>0)$中至少有一个正根，并且不超过$a+b$。
 
-证明：令$f(x)=x-a\sin x-b$，其中$f(0)=-b<0$，$f(a+b)=a+b=a\sin(a+b)-b=a[1-\sin(a+b)]\geqslant 0$。
+证明：令$f(x)=x-a\sin x-b$，其中$f(0)=-b<0$，$f(a+b)=a+b-a\sin(a+b)-b=a[1-\sin(a+b)]\geqslant 0$。
 
 若$\sin(a+b)=1$，则根为$a$，结论成立。
 
diff --git a/advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex b/advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex
index a045817..99d6753 100644
--- a/advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex
+++ b/advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex
@@ -2,9 +2,9 @@
 % UTF8编码，ctexart现实中文
 \usepackage{xcolor}
 % 使用颜色
-\definecolor{orange}{RGB}{255,127,0} 
-\definecolor{violet}{RGB}{192,0,255} 
-\definecolor{aqua}{RGB}{0,255,255} 
+\definecolor{orange}{RGB}{255,127,0}
+\definecolor{violet}{RGB}{192,0,255}
+\definecolor{aqua}{RGB}{0,255,255}
 \usepackage{geometry}
 \setcounter{tocdepth}{4}
 \setcounter{secnumdepth}{4}
@@ -567,7 +567,7 @@ \subsection{对幂指函数}
         $C$ & $0$ & $n^x$ & $n^x\ln n$ \\ \hline
         $\log_ax$ & $\dfrac{1}{x\ln a}$ & $\ln x=\ln\vert x\vert$ & $\dfrac{1}{x}$ \\ \hline
         $x^n$ & $nx^{n-1}$ & $\sqrt[n]{x}$ & $\dfrac{x^{-\frac{n-1}{n}}}{n}$ \\ \hline
-        $\dfrac{1}{x^n}$ & $-\dfrac{n}{x^{n+1}}$ & & \\ 
+        $\dfrac{1}{x^n}$ & $-\dfrac{n}{x^{n+1}}$ & & \\
         \hline
     \end{tabular}
 \end{center}
@@ -608,7 +608,7 @@ \subsection{双曲与反双曲函数}
         原函数 & 导函数 & 原函数 & 导函数\\ \hline
         $\textrm{sinh}\,x$ & $\textrm{cosh}\,x$ & $\textrm{cosh}\,x$ & $\textrm{sinh}\,x$ \\ \hline
         $\textrm{tanh}\,x$ & $\dfrac{1}{\textrm{cosh}\,x^2}$ & $\textrm{arcsinh}\,x$ & $\dfrac{1}{\sqrt{x^2+1}}$ \\ \hline
-        $\textrm{arccosh}\,x$ & $\dfrac{1}{\sqrt{x^2-1}}$ & $\textrm{arctan}\,x$ & $\dfrac{1}{1-x^2}$ \\
+        $\textrm{arccosh}\,x$ & $\dfrac{1}{\sqrt{x^2-1}}$ & $\textrm{arctanh}\,x$ & $\dfrac{1}{1-x^2}$ \\
         \hline
     \end{tabular}
 \end{center}
diff --git a/advanced-math/knowledge/3-applications-of-derivatives/applications-of-derivatives.tex b/advanced-math/knowledge/3-applications-of-derivatives/applications-of-derivatives.tex
index 05b510f..7c6daaa 100644
--- a/advanced-math/knowledge/3-applications-of-derivatives/applications-of-derivatives.tex
+++ b/advanced-math/knowledge/3-applications-of-derivatives/applications-of-derivatives.tex
@@ -2,9 +2,9 @@
 % UTF8编码，ctexart现实中文
 \usepackage{xcolor}
 % 使用颜色
-\definecolor{orange}{RGB}{255,127,0} 
-\definecolor{violet}{RGB}{192,0,255} 
-\definecolor{aqua}{RGB}{0,255,255} 
+\definecolor{orange}{RGB}{255,127,0}
+\definecolor{violet}{RGB}{192,0,255}
+\definecolor{aqua}{RGB}{0,255,255}
 \usepackage{geometry}
 \setcounter{tocdepth}{4}
 \setcounter{secnumdepth}{4}
@@ -100,7 +100,7 @@ \subsubsection{定义}
         \item $f(x)$在$(a,b)$内可导。
         \item $f(a)=f(b)$。
     \end{enumerate}
-    
+
     则$\exists\,\xi\in(a,b)$，使得$f'(\xi)=0$。
 \end{minipage}
 \hfill
@@ -396,7 +396,7 @@ \subsection{曲线凹凸性与拐点}
 
 不妨设$x_1<x_2$，且$\dfrac{x_1+x_2}{2}=x_0$。
 
-$f(x_1)+f(x_2)-2(x_0)=[f(x_2)-f(x_0)]-[f(x_1)-f(x_0)]$
+$f(x_1)+f(x_2)-2f(x_0)=[f(x_2)-f(x_0)]-[f(x_1)-f(x_0)]$
 
 $\xRightarrow{\text{拉格朗日中值定理}}=f'(\xi_2)(x_2-x_0)-f'(\xi_1)(x_0-x_1)=\dfrac{x_2-x_1}{2}[f'(\xi_2)-f'(\xi_1)]>0$
 
@@ -501,7 +501,7 @@ \subsection{弧微分}
         \filldraw[black](0.95,1.1) node{$y_0$};
         \filldraw[black](0.75,0.35) node{$\Delta x$};
         \filldraw[black](1.1,0.6) node{$\Delta y$};
-    \end{tikzpicture} 
+    \end{tikzpicture}
 \end{minipage}
 \hfill
 \begin{minipage}{0.4\linewidth}
@@ -528,7 +528,7 @@ \subsection{弧微分}
         \filldraw[black](0.95,1.1) node{$y_0$};
         \filldraw[black](0.75,0.35) node{$\textrm{d}x$};
         \filldraw[black](1.1,0.6) node{$\textrm{d}y$};
-    \end{tikzpicture} 
+    \end{tikzpicture}
 \end{minipage}
 \hfill
 \begin{minipage}{0.4\linewidth}
@@ -585,7 +585,7 @@ \subsection{曲率}
         \filldraw[black] (0,0) node[below]{$O$};
         \draw[black, thick,domain=-0.1:1.1] plot (\x, \x*\x);
         \draw[densely dashed](0.5,0.25) -- (0.5, 0) node[below]{$x$};
-        \draw[densely dashed](1,1) -- (1, 0) node[below]{$x+\Delta x$}; 
+        \draw[densely dashed](1,1) -- (1, 0) node[below]{$x+\Delta x$};
         \filldraw[black](0.5,0.35) node{$y$};
         \filldraw[black](0.95,1.1) node{$y_0$};
         \filldraw[black] (1/2,1/4) circle (0.5pt);
@@ -595,7 +595,7 @@ \subsection{曲率}
         \draw[line width=0.1] (0.85,0.7) arc (50:0:0.1);
         \filldraw[black](1,0.8) node{$\Delta\alpha$};
         \filldraw[black](0.5,0.8) node{$\vert\wideparen{yy_0}\vert=\Delta s$};
-    \end{tikzpicture} 
+    \end{tikzpicture}
 \end{minipage}
 \hfill
 \begin{minipage}{0.6\linewidth}
@@ -638,7 +638,7 @@ \subsection{曲率半径}
         \filldraw[black] (-0.75,0) node{$T$};
         \draw[black, thick,domain=-1:1] plot (\x, \x*\x);
         \draw[black] (0,0.5) circle (0.5);
-    \end{tikzpicture} 
+    \end{tikzpicture}
 \end{minipage}
 
 \end{document}
diff --git a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
index d24f052..eb77fff 100644
--- a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
+++ b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
@@ -2,9 +2,9 @@
 % UTF8编码，ctexart现实中文
 \usepackage{color}
 % 使用颜色
-\definecolor{orange}{RGB}{255,127,0} 
-\definecolor{violet}{RGB}{192,0,255} 
-\definecolor{aqua}{RGB}{0,255,255} 
+\definecolor{orange}{RGB}{255,127,0}
+\definecolor{violet}{RGB}{192,0,255}
+\definecolor{aqua}{RGB}{0,255,255}
 \usepackage{geometry}
 \setcounter{tocdepth}{5}
 \setcounter{secnumdepth}{5}
@@ -102,7 +102,7 @@ \subsubsection{直角坐标系}
 
 \begin{minipage}{0.6\linewidth}
     $\sigma=\{(x,y)|a\leqslant x\leqslant b,\psi(x)\leqslant y\leqslant\phi(x)\}$。
-    
+
     也称为上下型区域。
 
     $\iint\limits_Df(x,y)\,\textrm{d}\sigma=\int_a^b\textrm{d}x\int_{\psi(x)}^{\phi(x)}f(x,y)\,\textrm{d}y$。
@@ -242,7 +242,7 @@ \subsubsection{二重积分处理一元积分}
 
 $\textrm{d}\sigma$是第一象限，可以看作一个广义的圆，半径无限大，转换为极坐标系。
 
-$=\int_0^\frac{\pi}{2}\textrm{d}\theta\int_0^{+\infty}e^{-r^2}r\,\textrm{d}r=\displaystyle{\int_0^\frac{\pi}{2}\dfrac{1}{2}\,\textrm{d}\theta}=\dfrac{\pi}{2}$。$\therefore I=\dfrac{\sqrt{\pi}}{2}$。
+$=\int_0^\frac{\pi}{2}\textrm{d}\theta\int_0^{+\infty}e^{-r^2}r\,\textrm{d}r=\displaystyle{\int_0^\frac{\pi}{2}\dfrac{1}{2}\,\textrm{d}\theta}=\dfrac{\pi}{4}$。$\therefore I=\dfrac{\sqrt{\pi}}{2}$。
 
 \subsection{二重积分应用}
 
@@ -471,7 +471,7 @@ \subsubsection{基础方法}
     \item $L:\left\{\begin{array}{c}
         x=\phi(t) \\
         y=\psi(t)
-    \end{array}\right.$（$\alpha\leqslant t\leqslant\beta$），$\int_Lf(x,y)\,\textrm{d}s=\int_\alpha^\beta f[\phi(t),\psi(t)]\\\sqrt{\phi'^2(t)+\psi'^2(t)}\textrm{d}t$。 
+    \end{array}\right.$（$\alpha\leqslant t\leqslant\beta$），$\int_Lf(x,y)\,\textrm{d}s=\int_\alpha^\beta f[\phi(t),\psi(t)]\\\sqrt{\phi'^2(t)+\psi'^2(t)}\textrm{d}t$。
 \end{itemize}
 
 \paragraph{平面} \leavevmode \medskip
@@ -680,7 +680,7 @@ \subsubsection{定积分法}
     \item $L:\left\{\begin{array}{c}
         x=\phi(t) \\
         y=\psi(t)
-    \end{array}\right.$（$t$起于$\alpha$终于$\beta$），$\int_LP(x,y)\,\textrm{d}x+Q(x,y)\,\textrm{d}y=\\\int_\alpha^\beta P[\phi(t),\psi(t)]\phi'(t)\,\textrm{d}t+Q[\phi(t),\psi(t)]\psi'(t)\textrm{d}t$。 
+    \end{array}\right.$（$t$起于$\alpha$终于$\beta$），$\int_LP(x,y)\,\textrm{d}x+Q(x,y)\,\textrm{d}y=\\\int_\alpha^\beta P[\phi(t),\psi(t)]\phi'(t)\,\textrm{d}t+Q[\phi(t),\psi(t)]\psi'(t)\textrm{d}t$。
 \end{itemize}
 
 \subsubsection{二重积分法}