diff --git a/images/legal_standard_00076.png b/images/legal_standard_00076.png
index 71a90a1549af96dadf433f34a4e8e33dee71921c..2fcca3f2e0c45dc469f67d568681c62d2d6bc36d 100644
--- a/images/legal_standard_00076.png
+++ b/images/legal_standard_00076.png
@@ -1,3 +1,3 @@
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-oid sha256:962b436db2e3c9fb81626f60431889a04c28b7aef452bd5c86408995ec222367
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diff --git a/markdowns/education_chemistry_00011.md b/markdowns/education_chemistry_00011.md
index 6bc83593624ffd03dae03dbf243dbf42e21cf1d2..7b51e50b070506d460bb6730c99b1fb0f71f66fd 100644
--- a/markdowns/education_chemistry_00011.md
+++ b/markdowns/education_chemistry_00011.md
@@ -136,11 +136,11 @@
 
 对侧链 R 基不解离的中性氨基酸来说,其等电点是它的 $pK_{a1}$ 和 $pK_{a2}$ 的算术平均值;
 
-$pI = \frac{1}{2}(pK_{a1} + pK_{a2}).$
+$$pI = \frac{1}{2}(pK_{a1} + pK_{a2}).$$
 
 这可由氨基酸的解离公式推导出来。将前面所列的 $K_{a1}$ 和 $K_{a2}$ 的等式，
 
-$K_{s1}=\frac{[A^{*}][H^{*}]}{[A^{*}]}$
+$$K_{s1}=\frac{[A^{*}][H^{*}]}{[A^{*}]}$$
 
-$K_{a2} = \frac{[A^-][H^+]}{[A^*]}$
+$$K_{a2} = \frac{[A^-][H^+]}{[A^*]}$$
 
diff --git a/markdowns/education_chemistry_00019.md b/markdowns/education_chemistry_00019.md
index 57c57407ab47b12427ff0282d15f7817088fe88d..4ec163f53675bab1b9292184b014cfa9d0d2fb09 100644
--- a/markdowns/education_chemistry_00019.md
+++ b/markdowns/education_chemistry_00019.md
@@ -14,11 +14,11 @@ A. 当 $E_L \ne E_R$ 时，左、右圆偏振光叠合成椭圆偏振光。虚
 
 圆二色性还能用于估算蛋白质中 $\alpha$ 螺旋、$\beta$ 折叠片和无规卷曲的含量。假设蛋白质分子全由这 3 种构象单元组成，它们所含的残基数占蛋白质分子的总残基数的百分数分别为 $f_{\alpha}, f_{\beta}$ 和 $f_{R}$，则
 
-$f_{*}+f_{\beta}+f_{R}=1\quad(1)$
+$$f_{*}+f_{\beta}+f_{R}=1\tag{1}$$
 
 再假设蛋白质分子中的各种构象单元在波长处的椭圆率也可以加和，则
 
-$\left[\theta\right]_{\lambda}=f_{\nu}\left[\theta\right]_{\nu,\lambda}+f_{\mu}\left[\theta\mid_{\lambda,\lambda}+f_{\mathrm{R}}\left[\theta\right]_{\mathrm{R},\lambda}\right.\quad(2)$
+$$\left[\theta\right]_{\lambda}=f_{\nu}\left[\theta\right]_{\nu,\lambda}+f_{\mu}\left[\theta\mid_{\lambda,\lambda}+f_{\mathrm{R}}\left[\theta\right]_{\mathrm{R},\lambda}\right.\tag{2}$$
 
 式中$\left[\theta\right]$为实验样品CD曲线在波长$\lambda$处的摩尔椭圆率；$\left[\theta\right]_{0.5}$、$\left[\theta\right]_{p.k}$和$\left[\theta\right]_{R.k}$分别为100% $\alpha$螺旋、100% $\beta$折叠片和100%无规卷曲构象在波长$\lambda$处的摩尔椭圆率，这些数据可由人工合成的多聚氨基酸获得（图5-6）。因此利用公式（2），理论上只要选择3个不同波长的$\left[\theta\right]$，即可得到一组三元一次方程，并由此解出未知数$f_{a}$、$f_{b}$和$f_{R}$，现在实际上都是用现成的程序在计算机上完成的。
 
diff --git a/markdowns/education_chemistry_00024.md b/markdowns/education_chemistry_00024.md
index 9afa131148da3ef7323fc819a3d9a241f06dd55e..bf6a6d145ed441f6626cf7a7525753dee09938f4 100644
--- a/markdowns/education_chemistry_00024.md
+++ b/markdowns/education_chemistry_00024.md
@@ -1,24 +1,24 @@
 式中 $MbO_2$ 代表氧合肌红蛋白, $Mb$ 代表去氧肌红蛋白。根据生物化学中的习惯, 把氧合平衡看成解离平衡, 并用 $K$ 代表解离平衡常数:
 
-$K = \frac{[Mb][O_2]}{[MbO_2]}$ $\quad(6-1)$
+$$K = \frac{[Mb][O_2]}{[MbO_2]}\tag{6-1}$$
 
 或
 
-$\frac{[MbO_{2}]}{[Mb]} = \frac{[O_{2}]}{K} \quad (6-2)$
+$$\frac{[MbO_{2}]}{[Mb]} = \frac{[O_{2}]}{K} \tag{6-2}$$
 
 所以，MbO$_{2}$ 与 Mb 浓度之比恰好与 [O$_{2}$] 成正比。
 
 由于 Mb 和 $MbO_{2}$ 的浓度难于测定，因此需要引进一个新的参数 Y 以便消去 Mb 和 $MbO_{2}$ 两个参数。Y 被定义为在给定的氧压下肌红蛋白的氧分数饱和度（fractional saturation），即 $MbO_{2}$ 分子数占肌红蛋白（Mb和 $MbO_{2}$） 分子总数的百分数：
 
-$Y=\frac{[MbO_{2}]}{[MbO_{2}]+[Mb]}\quad(6-3)$
+$$Y=\frac{[MbO_{2}]}{[MbO_{2}]+[Mb]}\tag{6-3}$$
 
 把方程$(6-1)$改写为$[MbO_2] = [Mb][O_2]/K$并代入方程$(6-2)$得：
 
-$Y = \frac{[O_2]}{[O_2] + K}$ $\quad(6-4)$
+$$Y = \frac{[O_2]}{[O_2] + K}\tag{6-4}$$
 
 根据 henry 定律，溶于液体的任一气体的浓度与液体上面的该气体分压成正比。因此 $\left[O_{2}\right]$ 可用分压 $p(\mathrm{O}_{2})$ 表示，则方程(6-3)可改写为：
 
-$Y=\frac{p(O_{2})}{p(O_{2})+K}\quad(6-5)$
+$$Y=\frac{p(O_{2})}{p(O_{2})+K}\tag{6-5}$$
 
 实验中 $p(\mathrm{O}_{2})$ 值可以进行调节和测量，氧分压常用 $torr^{*}$ 作单位；Y 值可用分光光度计法测定，因为肌红蛋白氧合时卟啉环中电子位移引起吸收光谱改变。Y 对 $p(\mathrm{O}_{2})$ 作图所得的曲线称为氧结合曲线或解离曲线。如图 6-6A 所示，肌红蛋白的氧合曲线为一双曲线，它的两条渐近线是 Y=1 和 $p(\mathrm{O}_{2})=-K$。
 
diff --git a/markdowns/education_chemistry_00026.md b/markdowns/education_chemistry_00026.md
index d578d35c854ac85ef54cab79bdc15a60e626dbcf..82071297fbeff44b04d1590c8214edfe27079c27 100644
--- a/markdowns/education_chemistry_00026.md
+++ b/markdowns/education_chemistry_00026.md
@@ -6,12 +6,12 @@ $$0\approx n\left[\log p(O_{2})\right]-\log K\approx n\left(\log P_{50}\right)-\
 
 因而，
 
-$ \log K = n (\log P_{S_{0}}) \text{ 或 } K = (P_{S_{0}})^{n} \quad (6-14) $
+$$ \log K = n (\log P_{S_{0}}) \text{ 或 } K = (P_{S_{0}})^{n} \tag{6-14}$$
 
 将 $K = (P_{50})^n$ 代入方程(6-11)得：
 
-$\frac{Y}{1-Y}=\left[\frac{p\left(O_{2}\right)}{P_{50}}\right]^{n}$
-$\quad(6-15)$
+$$\frac{Y}{1-Y}=\left[\frac{p\left(O_{2}\right)}{P_{50}}\right]^{n}\tag{6-15}$$
+
 
 也即肌红蛋白和血红蛋白的情况不同，对 $HbP_{50}$ 和 $K$ 不相等，因为 $Hb$ 是具有多个的 $O_2$ 结合部位的别构蛋白质。
 
@@ -27,7 +27,12 @@ $\quad(6-15)$
 
 组织中的代谢作用既产生$H^{+}$，也产生$CO_{2}$。代谢越旺盛的组织，需要的氧越多，产生的$H^{+}$和$CO_{2}$也越多。细胞呼吸的终产物$CO_{2}$在体内被水合为碳酸氢盐：
 
-$\begin{aligned} CO_2 + H_2O \xrightleftharpoons{碳酸酐酶} &H_2CO_3 \rightleftharpoons H^+ + HCO_3^-\\&碳酸\end{aligned}$
+$$
+\begin{aligned}
+\ce{CO2 + H2O} \xrightleftharpoons{\text{碳酸酐酶}} &\ce{H2CO3 <=> H+ + HCO3^-} \\
+&\text{碳酸}
+\end{aligned}
+$$
 
 
 此反应受碳酸酐酶(carbonic anhydrase)催化，该酶在红细胞中特别丰富。$CO_{2}$在水中的溶解度不大，如果不转变为碳酸氢盐，将在组织和血中形成气泡。从上面的方程可以看到，$CO_{2}$水合的结果将增加组织中的$H^{+}$浓度(pH下降)。氧与血红蛋白结合深受pH和$CO_{2}$浓度的影响。去氧血红蛋白对$H^{+}$的亲和力比氧
diff --git a/markdowns/education_chemistry_00030.md b/markdowns/education_chemistry_00030.md
index 65e09f229aa3ea903b73e16b8a907af65c0e03e3..0da44e61cb20238509b403fdad74c1e6e728e441 100644
--- a/markdowns/education_chemistry_00030.md
+++ b/markdowns/education_chemistry_00030.md
@@ -62,7 +62,7 @@
 
 9. 抗原与抗体的结合方式与血红蛋白的氧结合相似。假设抗原是一价，抗体是 $n$ 价，即抗体分子有 $n$ 个结合部位，且各结合部位的结合常数 $K_a$ 值是相同的，则可证明当游离抗原浓度为 $[L]$ 时，结合到抗体上的抗原浓度 $[L_p]$ 与抗体的总浓度 $[P_t]$ 之比值：
 
-$\hat{N}=\frac{[L_{P}]}{[P_{T}]}=\frac{nK_{a}[L]}{1+K_{a}[L]}$
+$$\hat{N}=\frac{[L_{P}]}{[P_{T}]}=\frac{nK_{a}[L]}{1+K_{a}[L]}$$
 
 $\hat{N}$ 实际上表示被一个抗体分子结合的抗原分子平均数。
 
diff --git a/markdowns/education_chemistry_00031.md b/markdowns/education_chemistry_00031.md
index cccbc00330cd989453389ac08fc1507a9b6ef73e..9745762ac079b47997013fc03149d47964fbc66c 100644
--- a/markdowns/education_chemistry_00031.md
+++ b/markdowns/education_chemistry_00031.md
@@ -14,29 +14,29 @@
 
 在离心场中蛋白质颗粒发生沉降时,它将受到三种力的作用;
 
-$F_{c}$(离心力) = $m_{\nu}\omega^{2}x$
+$$F_{c}(离心力) = m_{\nu}\omega^{2}x$$
 
-$F_{b}$(浮力)$=V_{p}\rho\omega^{2}x=m_{p}vp\omega^{2}x$
+$$F_{b}(浮力)=V_{p}\rho\omega^{2}x=m_{p}vp\omega^{2}x$$
 
-$F_{f}(\text{摩擦力}) = fv = f \frac{\mathrm{d}x}{\mathrm{d}t}$
+$$F_{f}(\text{摩擦力}) = fv = f \frac{\mathrm{d}x}{\mathrm{d}t}$$
 
 这里，$m_{p}$ 是分子颗粒的质量(g)；$\omega$ 是转头的角速度(rad/s)；x 是旋转中心至界面的径向距离(cm)；$\omega^{2}x$ 是离心加速度（也称离心场强度或离心场，是单位质量的力）；$V_{p}$ 是分子颗粒的体积；$\rho$ 是溶剂的密度($g/cm^{3}$)；v 是蛋白质的偏微比容（partial specific volume），偏微比容的定义是：当加入1克干物质于无限大体积的溶剂中时溶液体积的增量；$V_{p}\rho$ 或 $m_{p}\bar{\nu}\rho$ 是被分子颗粒排开的溶剂质量；f 是摩擦系数；v 是沉降速率，即 $\frac{dx}{dt}$。离心力减去浮力为分子颗粒所受到的净离心力：
 
-$F_{c}-F_{b}=m_{\rho}\omega^{2}x-m_{\rho}\nu\rho\ \omega^{2}x
+$$F_{c}-F_{b}=m_{\rho}\omega^{2}x-m_{\rho}\nu\rho\ \omega^{2}x
 
-=m_{\rho}\omega^{2}x(1-\nu\rho)$
+=m_{\rho}\omega^{2}x(1-\nu\rho)$$
 
 式中，$(1-\bar{v}_{p})$ 为浮力因子(buoyancy factor)。当分子颗粒以恒定速度移动时，净离心力与摩擦力(阻力)处于稳态平衡中：
 
-$F_{c} - F_{b} = F_{f}$
+$$F_{c} - F_{b} = F_{f}$$
 
 即
 
-$m_{\mu}\omega^{2}x(1-\bar{\nu}\rho)=f\frac{\mathrm{d}x}{\mathrm{d}t}$
+$$m_{\mu}\omega^{2}x(1-\bar{\nu}\rho)=f\frac{\mathrm{d}x}{\mathrm{d}t}$$
 
 或
 
-$\frac{d.r/dt}{\omega^{2}x}=\frac{m_{p}(1-\upsilon p)}{f}$
+$$\frac{d.r/dt}{\omega^{2}x}=\frac{m_{p}(1-\upsilon p)}{f}$$
 
 可见单位离心场的沉降速度是个定值。称为沉降系数(sedimentation coefficient)或沉降常数,用$s$(小写)
 
diff --git a/markdowns/education_chemistry_00035.md b/markdowns/education_chemistry_00035.md
index c1b989f7718520b8bf5c5f8ded6637c565abfc27..0913b26b269d1f85d7a4dd16667045eec25f1aeb 100644
--- a/markdowns/education_chemistry_00035.md
+++ b/markdowns/education_chemistry_00035.md
@@ -14,11 +14,13 @@
 
 1913年 Michaelis 和 Menten 在前人工作的基础上，根据酶反应的中间复合物学说：
 
-$E + S \xrightleftharpoons{k_{5}} ES \xrightarrow{k} E + P \quad (9-10)$
+$$
+\ce{E + S <=>[$k_5$] ES ->[$k$] E + P} \tag{9-10}
+$$
 
 假定 $E + S \rightleftharpoons ES$ 迅速建立平衡，底物浓度远远大于酶浓度下，ES 分解成产物的逆反应忽略不计，以“快速平衡法”，推导出一个数学方程式，表示了底物浓度与酶反应速率之间的定量关系，通常称为米氏方程。
 
-$v=\frac{V_{\max}\cdot[S]}{K_{s}+[S]}\quad(9-11)$
+$$v=\frac{V_{\max}\cdot[S]}{K_{s}+[S]}\tag{9-11}$$
 
 式中 $v$ 为反应速率，$V_{\max}$ 为酶完全被底物饱和时的最大反应速率，$[S]$ 为底物浓度，$K_s$ 为 ES 的解离常数（底物常数）。
 
@@ -26,12 +28,12 @@ $v=\frac{V_{\max}\cdot[S]}{K_{s}+[S]}\quad(9-11)$
 
 第一步:酶与底物作用,形成酶-底物复合物:
 
-$E + S \underset{k_{2}}{\overset{k_{1}}{\rightleftharpoons}} ES \quad (9-12)$
+$$E + S \underset{k_{2}}{\overset{k_{1}}{\rightleftharpoons}} ES \tag{9-12}$$
 
 第二步:ES复合物分解形成产物,释放出游离酶;
 
-$\mathrm{FS} \underset{k_{4}}{\overset{k_{3}}{\rightleftharpoons}} \mathrm{P}+\mathrm{E}$
-(9 - 13)
+$$\mathrm{FS} \underset{k_{4}}{\overset{k_{3}}{\rightleftharpoons}} \mathrm{P}+\mathrm{E}\tag{9 - 13}$$
+
 
 这两步反应都是可逆的。它们的正反应与逆反应的速率常数分别为 $k_1, k_2, k_3, k_4$。
 
@@ -39,7 +41,7 @@ $\mathrm{FS} \underset{k_{4}}{\overset{k_{3}}{\rightleftharpoons}} \mathrm{P}+\m
 
 所谓稳态是指反应进行一段时间后，系统的复合物ES浓度，由零逐渐增加到一定数值，在一定时间内，尽管底物浓度和产物浓度不断地变化，复合物ES也在不断地生成和分解，但是当反应系统中ES的生成速率和ES的分解速率相等时，络合物ES浓度保持不变的这种反应状态称为稳态，即：
 
-$\frac{d[ES]}{dt}=0$
+$$\frac{d[ES]}{dt}=0$$
 
 图 9—8 表示实验所得各种浓度对时间的曲线，表示了底物浓度降低，产物形成及 ES 稳态过程。
 
diff --git a/markdowns/education_chemistry_00061.md b/markdowns/education_chemistry_00061.md
index 8dee636973d9f69b42b96bbf852369ad6e8f4e2f..758a8a214b67b9f8684526f19c14318737f9f9e8 100644
--- a/markdowns/education_chemistry_00061.md
+++ b/markdowns/education_chemistry_00061.md
@@ -4,13 +4,13 @@
 
 定温下,气体在液体中的溶解度 $x_B$ 与该气体在液面上的平衡分压 $p_B$ 成正比,其定量关系为
 
-$p_{B}=kx_{B}\quad(x_{B}\rightarrow0)\quad(4.3.1)$
+$$p_{B}=kx_{B}\quad(x_{B}\rightarrow0)\tag{4.3.1}$$
 
 式中 $x_B$ 为被溶解的气体物质在溶液中的摩尔分数，比例常数 $k$ 称为 Henry 常数，与温度有关，是溶质及溶剂性质共同的表现。
 
 在稀溶液时，二元溶液溶质的摩尔分数 $x_{B}$ 、质量摩尔浓度 $m_{B}$ 和物质的量浓度 $c_{B}$ 间存在下列关系，
 
-$x_{\mathrm{B}}=M_{\mathrm{A}} m_{\mathrm{B}}=\left(M_{\mathrm{A}} / \rho_{0}\right) c_{\mathrm{B}}$ $\quad(4.3.2)$
+$$x_{\mathrm{B}}=M_{\mathrm{A}} m_{\mathrm{B}}=\left(M_{\mathrm{A}} / \rho_{0}\right) c_{\mathrm{B}}\tag{4.3.2}$$
 
 因此,Henry定律可表示为
 
@@ -18,7 +18,7 @@ $$p_{\mathrm{B}}=k x_{\mathrm{B}}=k_{m}m_{\mathrm{B}}=k_{c}c_{\mathrm{B}}\tag{4.
 
 式中
 
-$k=(1/M_{A})k_{m}=(\rho_{0}/M_{A})k_{c}\quad(4.3.4)$
+$$k=(1/M_{A})k_{m}=(\rho_{0}/M_{A})k_{c}\tag{4.3.4}$$
 
 例 当潜水员由深水急速上升到水面，氮的溶解度降低，在血液中形成气泡阻塞血液流通，这就是“潜函病”。假设氮在血液中的溶解度与水中相同，在$p=101.3\ \mathrm{kPa}$时，$c(\mathrm{N_2})=1.39\times 10^{-5}\ \mathrm{kg(N_2)/kg(H_2O)}$，一个人身体中有$3\ \mathrm{kg}$血。在$20^\circ \mathrm{C}$时，人从$60\ \mathrm{m}$深水中急速上升，请求在人的血液中形成的氮气泡体积有多大？半径为多少？
 
@@ -28,12 +28,17 @@ $p(\mathrm{N}_2) = 101.3\ \mathrm{kPa} \times 0.8 = 81.0\ \mathrm{kPa}$。
 
 根据 $p(\mathrm{N}_2) = k_c c(\mathrm{N}_2)$，可得 Henry 常数 $k_c$：
 
-$k_{c} = \frac{p(\mathrm{N}_{2})}{c(\mathrm{N}_{2})} = \frac{1.013 \times 10^{5} \, \mathrm{Pa} \times 0.8}{1.39 \times 10^{-5} \, \mathrm{kg}(\mathrm{N}_{2})/\mathrm{kg}(\mathrm{H}_{2}\mathrm{O})}$
-$= 5.81 \times 10^{9} \, \mathrm{Pa} \cdot \mathrm{kg}(\mathrm{H}_{2}\mathrm{O})/\mathrm{kg}(\mathrm{N}_{2})$
+$$
+\begin{aligned}
+k_{c} &= \frac{p(\mathrm{N}_{2})}{c(\mathrm{N}_{2})}
+= \frac{1.013 \times 10^{5} \, \mathrm{Pa} \times 0.8}{1.39 \times 10^{-5} \, \mathrm{kg}(\mathrm{N}_{2})/\mathrm{kg}(\mathrm{H}_{2}\mathrm{O})} \\
+&= 5.81 \times 10^{9} \, \mathrm{Pa} \cdot \mathrm{kg}(\mathrm{H}_{2}\mathrm{O})/\mathrm{kg}(\mathrm{N}_{2})
+\end{aligned}
+$$
 
 60 m 深水中的压力 $p_B$ 及血液中溶解氮的浓度 $c_B$ 为
 
-$p_{\mathrm{B}}=\rho_{1}g h=60\mathrm{~m}\times998\mathrm{~k g}/\mathrm{m}^{3}\times9.81\mathrm{~m}/\mathrm{s}^{2}=5.87\times10^{5}\mathrm{~P a}$
+$$p_{\mathrm{B}}=\rho_{1}g h=60\mathrm{~m}\times998\mathrm{~k g}/\mathrm{m}^{3}\times9.81\mathrm{~m}/\mathrm{s}^{2}=5.87\times10^{5}\mathrm{~P a}$$
 
-$c_{\mathrm{B}}=p_{\mathrm{B}}/k_{\mathrm{c}}=5.87\times10^{5}\times0.8\ \mathrm{Pa}/5.81\times10^{9}\ \mathrm{Pa\cdot kg(H_{2}O)/kg(N_{2})}$
+$$c_{\mathrm{B}}=p_{\mathrm{B}}/k_{\mathrm{c}}=5.87\times10^{5}\times0.8\ \mathrm{Pa}/5.81\times10^{9}\ \mathrm{Pa\cdot kg(H_{2}O)/kg(N_{2})}$$
 
diff --git a/markdowns/education_chemistry_00063.md b/markdowns/education_chemistry_00063.md
index 321a7d8af27160c4dfeb89b5fce22b557d8c0984..fc4a17157bafb95ba6d0792aab903bd0d56fda80 100644
--- a/markdowns/education_chemistry_00063.md
+++ b/markdowns/education_chemistry_00063.md
@@ -28,23 +28,23 @@
 
 求算 $CO_2$ 在环己醇中溶解的 Henry 常数及 $p = 60 \times 10^5\ Pa$ 时 $CO_2$ 的活度系数。
 
-$[k_{\mathrm{H}}=371\times10^{5}\mathrm{~P a},\gamma_{\mathrm{C O_{2}}}=0.98]$
+$$[k_{\mathrm{H}}=371\times10^{5}\mathrm{~P a},\gamma_{\mathrm{C O_{2}}}=0.98]$$
 
 14. 若令 $Q = G_{\mathrm{m}}^{\mathrm{E}} / RT$，则 $Q = \sum_{i} x_{i} \ln \gamma_{i}$，如果 $\ln \gamma_{i}$ 当作偏摩尔量，则上式即是偏摩尔量集合公式，利用式(2.8.14)及式(2.8.15)，
 
-$L_{1}-L_{1}^{*}=\Delta_{\mathrm{m i x}}L_{\mathrm{m i}}-x_{2}\left(\frac{\partial\Delta_{\mathrm{m i x}}L_{\mathrm{m}}}{\partial x_{2}}\right)_{T,p}$
+$$L_{1}-L_{1}^{*}=\Delta_{\mathrm{m i x}}L_{\mathrm{m i}}-x_{2}\left(\frac{\partial\Delta_{\mathrm{m i x}}L_{\mathrm{m}}}{\partial x_{2}}\right)_{T,p}$$
 
 对于二元混合物,请推导
 
-$\ln\gamma_{A}=Q-x_{B}\left(\frac{\partial Q}{\partial x_{B}}\right)_{T,P}$
+$$\ln\gamma_{A}=Q-x_{B}\left(\frac{\partial Q}{\partial x_{B}}\right)_{T,P}$$
 
-$\ln\gamma_{\mathrm{B}}=Q-x_{\mathrm{A}}\left(\frac{\partial Q}{\partial x_{\mathrm{A}}}\right)_{T,P}$
+$$\ln\gamma_{\mathrm{B}}=Q-x_{\mathrm{A}}\left(\frac{\partial Q}{\partial x_{\mathrm{A}}}\right)_{T,P}$$
 
 若 $Q = \frac{G_{\text{m}}^{\text{E}}}{RT} = Cx_A x_B$，且 $C$ 与组成无关，请求 $\ln \gamma_A$、$\ln \gamma_B$。[$\ln \gamma_A = Cx_B^2$，$\ln \gamma_B = Cx_A^2$]
 
 15. 试证明：恒温恒压下,在二元混合物中当每个组分均选在 $T,p$ 时的纯物质为标准态时,存在下述关系:
 
-$\int_{0}^{1} \ln\left(\frac{\gamma_{A}}{\gamma_{B}}\right) \mathrm{d}x_{A} = 0$
+$$\int_{0}^{1} \ln\left(\frac{\gamma_{A}}{\gamma_{B}}\right) \mathrm{d}x_{A} = 0$$
 
 [提示:从 Gibbs-Duhem 公式出发,推出 $\int_{\ln\gamma_{A}}^{\ln(1/\gamma_{B})}x_{A}\mathrm{d}\ln\frac{\gamma_{A}}{\gamma_{B}}=\int_{\ln1}^{\ln\gamma_{B}}-\mathrm{d}\ln\gamma_{B}$。
 再应用全微分性质 $\int u \, dv = uv - \int v \, du$，即可证得]
diff --git a/markdowns/education_chemistry_00064.md b/markdowns/education_chemistry_00064.md
index 1313dd4fafd9e089396aced2ccc59c5b4612a57a..e91a32389eae40a365ad39093a8191847e63093f 100644
--- a/markdowns/education_chemistry_00064.md
+++ b/markdowns/education_chemistry_00064.md
@@ -12,23 +12,25 @@ $$\begin{equation}
 
 式中 B 表示反应物或生成物的化学式，$\nu_{B}$ 是物质 B 的化学计量数，显然对反应物取负值，对生成物取正值，这与惯例是一致的。因此，式(5.1.1)不能写成 $\sum_{B}\nu_{B}B=0$。应特别强调，$\nu_{B}$ 是无量纲的纯数，可以是整数或简单的分数，只是表示反应过程中各物质转化的比例关系，以合成氨反应为例：
 
-$0=-\frac{1}{2}\mathrm{N}_{2}(\mathrm{~g})-\frac{3}{2}\mathrm{H}_{2}(\mathrm{~g})+\mathrm{NH}_{3}(\mathrm{~g})\quad(R_{1})$
+$$0=-\frac{1}{2}\mathrm{N}_{2}(\mathrm{~g})-\frac{3}{2}\mathrm{H}_{2}(\mathrm{~g})+\mathrm{NH}_{3}(\mathrm{~g})\quad(R_{1})$$
 
-$0 = -\mathrm{N}_{2}(\mathrm{g}) - 3\mathrm{H}_{2}(\mathrm{g}) + 2\mathrm{NH}_{3}(\mathrm{g}) \quad (R_{2})$
+$$
+0 = \ce{-N2(g) - 3H2(g) + 2NH3(g)} \quad (R_2)
+$$
 
 它们所表示的反应中各物质转化的比例关系是相同的，反应方程式中各物质的计量数不是反应过程中各相应物质所转化的物质的量。因此，不能绝对地说反应 $R_{2}$ 是 1 mol $\mathrm{N}_{2}(\mathrm{~g})$ 与 3 mol $\mathrm{H}_{2}(\mathrm{~g})$ 生成 2 mol $\mathrm{NH}_{3}(\mathrm{~g})$。
 
 为了描述反应量,20世纪初比利时科学家 Dekonder 引入了一个“反应进度”$\xi$ 的概念,$\xi$ 定义为
 
-$\xi = \frac{\pi_{\mathrm{B}} - \pi_{\mathrm{B}}^{0}}{\nu_{\mathrm{B}}}$ (5.1.2)
+$$\xi = \frac{\pi_{\mathrm{B}} - \pi_{\mathrm{B}}^{0}}{\nu_{\mathrm{B}}}\tag{5.1.2}$$
 
 以反应$(R_2)$为例
 
-$\xi=\frac{n\left(N_{2}\right)-n^{0}\left(N_{2}\right)}{-1}=\frac{n\left(H_{2}\right)-n^{0}\left(H_{2}\right)}{-3}=\frac{n\left(NH_{3}\right)-n^{0}\left(NH_{3}\right)}{2}$
+$$\xi=\frac{n\left(N_{2}\right)-n^{0}\left(N_{2}\right)}{-1}=\frac{n\left(H_{2}\right)-n^{0}\left(H_{2}\right)}{-3}=\frac{n\left(NH_{3}\right)-n^{0}\left(NH_{3}\right)}{2}$$
 
 或者写成
 
-$n(\mathrm{N_2}) = n^0(\mathrm{N_2}) - \xi, \quad 
+$$n(\mathrm{N_2}) = n^0(\mathrm{N_2}) - \xi, \quad 
 n(\mathrm{H_2}) = n^0(\mathrm{H_2}) - 3\xi, \quad 
-n(\mathrm{NH_3}) = n^0(\mathrm{NH_3}) + 2\xi$
+n(\mathrm{NH_3}) = n^0(\mathrm{NH_3}) + 2\xi$$
 
diff --git a/markdowns/education_chemistry_00065.md b/markdowns/education_chemistry_00065.md
index 29dfed0add4f361dfd94ed1d2928af397f3b0462..0dfc41b5b2eddcd75b99cbdda9460123b2778106 100644
--- a/markdowns/education_chemistry_00065.md
+++ b/markdowns/education_chemistry_00065.md
@@ -2,12 +2,17 @@
 
 将上述数据再代入(3)式,可得
 
-$\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(T) / \mathrm{J} \cdot \mathrm{mol}^{-1}=185.6 \times 10^{3}+115 T \ln (T / \mathrm{K})+4.18 \times 10^{-3}(T / \mathrm{K})^{2}$
-$-5.44 \times 10^5 (1/(T/K)) - 244.0 (T/K)\quad(4)$ 
+$$
+\begin{aligned}
+\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(T) / \left(\mathrm{J}\cdot\mathrm{mol}^{-1}\right)
+&= 185.6 \times 10^{3} + 115 T \ln (T / \mathrm{K}) + 4.18 \times 10^{-3}(T / \mathrm{K})^{2} \\
+&\quad - 5.44 \times 10^{5} \bigl(1/(T/\mathrm{K})\bigr) - 244.0 (T/\mathrm{K}) \tag{4}
+\end{aligned}
+$$
 
 分解温度时，$K_p = (p_{CO_2} / p^\ominus) = (p^\ominus / p^\ominus) = 1$（假设 $CO_2(g)$ 为理想气体）故
 
-$\Delta_{r}G_{m}^{\ominus}(T)=-RT\ln K_{p}=0$
+$$\Delta_{r}G_{m}^{\ominus}(T)=-RT\ln K_{p}=0$$
 
 (4)式是一个较复杂的方程,没有现成的公式求解,只能采用试解法,求得$T_{合}=1176\ K$(即$903\ ^{\circ}C$)。
 
@@ -15,11 +20,11 @@ $\Delta_{r}G_{m}^{\ominus}(T)=-RT\ln K_{p}=0$
 
 例4 由反应各组分之 $\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(298 \mathrm{~K})$、$\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\ominus}(298 \mathrm{~K})$、$C_{p, \mathrm{~m}}$ 数据可得反应
 
-$C_2H_4(g) + H_2O(g) = C_2H_5OH(g)$
+$$C_2H_4(g) + H_2O(g) = C_2H_5OH(g)$$
 
 的 $K_f(T,p)$ 与 $T$ 的关系为
 
-$\ln K_{f} = \frac{37\,447}{RT} - \frac{6.018}{R} \ln(T/K) + \frac{0.0367}{R} T - \frac{52.72 \times 10^{-7}}{R} T^{2} + 3.09$
+$$\ln K_{f} = \frac{37\,447}{RT} - \frac{6.018}{R} \ln(T/K) + \frac{0.0367}{R} T - \frac{52.72 \times 10^{-7}}{R} T^{2} + 3.09$$
 
 求在下列条件下乙烯的平衡转化率。温度为 $523 \, K$，$p = 3.4 \times 10^{6} \, Pa$，$ n(\mathrm{H}_{2}\mathrm{O}) $：
 $ n(\mathrm{C}_{2}\mathrm{H}_{4}) = 5:1 $。
diff --git a/markdowns/education_chemistry_00066.md b/markdowns/education_chemistry_00066.md
index a2363064fc66cad5ddc706dca4ffd7281d8cee01..186df5b226ce2f475ee5180d3956d63bf15c55e0 100644
--- a/markdowns/education_chemistry_00066.md
+++ b/markdowns/education_chemistry_00066.md
@@ -10,11 +10,11 @@
 
 (1) 2NaHCO$_3$(s) =  Na$_2$CO$_3$(s) + H$_2$O(g) + CO$_2$(s)
 
-$\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(1)=(129~076-334.2(T / \mathrm{K})) \mathrm{J} \cdot \mathrm{mol}^{-1}$
+$$\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(1)=(129~076-334.2(T / \mathrm{K})) \mathrm{J} \cdot \mathrm{mol}^{-1}$$
 
 (2) $\mathrm{NH_{4}HCO_{3}(s)}$ == $\mathrm{NH_{3}(g)} + \mathrm{H_{2}O(g)} + \mathrm{CO_{2}(s)}$
 
-$\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(2)=(171~502-476.4(T/\mathrm{K}))\mathrm{J}\cdot\mathrm{mol}^{-1}$
+$$\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\ominus}(2)=(171~502-476.4(T/\mathrm{K}))\mathrm{J}\cdot\mathrm{mol}^{-1}$$
 
 请解答：
 
diff --git a/markdowns/education_chemistry_00068.md b/markdowns/education_chemistry_00068.md
index e4b17b0121aa5f73f3eb29317aed974e1808babd..89902ab3478e132126ab881975403512bdf7aea1 100644
--- a/markdowns/education_chemistry_00068.md
+++ b/markdowns/education_chemistry_00068.md
@@ -14,7 +14,9 @@
 
 今欲使 $1 \, mol \, 373 \, K, p^{\ominus}$ 的水完全变为 $373 \, K, p^{\ominus}$ 的水蒸气，用方程式表示为
 
-$H_2O(l, 373\ K, p^\ominus) \rightarrow H_2O(g, 373\ K, p^\ominus)$
+$$
+\ce{H2O}(l, 373\,\mathrm{K}, p^\ominus) -> \ce{H2O}(g, 373\,\mathrm{K}, p^\ominus)
+$$
 
 当然可采用多种方式进行,我们只讨论下面两种:
 
diff --git a/markdowns/education_chemistry_00069.md b/markdowns/education_chemistry_00069.md
index 795deb42f5fbf1169c100d9ffe2131c12a333a9b..d3aebf51e4ecf716971518f8afbda1155fe92db6 100644
--- a/markdowns/education_chemistry_00069.md
+++ b/markdowns/education_chemistry_00069.md
@@ -1,16 +1,16 @@
-$\begin{array}{r l r l r l}&{=\left(\frac{\partial\mu_{2}^{\beta}}{\partial T}\right)_{p,x_{1}^{\beta}}\mathrm{d}T+\left(\frac{\partial\mu_{2}^{\beta}}{\partial p}\right)_{T,x_{1}^{\beta}}\mathrm{d}p+\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}\mathrm{d}x_{1}^{\beta}}&{\quad(4)}\end{array}$
+$$\begin{array}{r l r l r l}&{=\left(\frac{\partial\mu_{2}^{\beta}}{\partial T}\right)_{p,x_{1}^{\beta}}\mathrm{d}T+\left(\frac{\partial\mu_{2}^{\beta}}{\partial p}\right)_{T,x_{1}^{\beta}}\mathrm{d}p+\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}\mathrm{d}x_{1}^{\beta}}&{\tag{4}}\end{array}$$
 
 根据$(\partial \mu / \partial T)_p = -S_m, (\partial \mu / \partial p)_T = V_m$，则(3)式、(4)式可改写为
 
-$\begin{array}{r l}{-\;S_{1}^{\alpha}\mathsf{d}T+\;V_{1}^{\alpha}\mathsf{d}p+\left(\frac{\partial\mu_{1}^{\alpha}}{\partial x_{1}^{\alpha}}\right)_{T,p}\mathsf{d}x_{1}^{\alpha}=-\;S_{1}^{\beta}\mathsf{d}T+\;V_{1}^{\beta}\mathsf{d}p+\left(\frac{\partial\mu_{1}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}\mathsf{d}x_{1}^{\beta}}&{{}\quad(5)}\end{array}$
+$$\begin{array}{r l}{-\;S_{1}^{\alpha}\mathsf{d}T+\;V_{1}^{\alpha}\mathsf{d}p+\left(\frac{\partial\mu_{1}^{\alpha}}{\partial x_{1}^{\alpha}}\right)_{T,p}\mathsf{d}x_{1}^{\alpha}=-\;S_{1}^{\beta}\mathsf{d}T+\;V_{1}^{\beta}\mathsf{d}p+\left(\frac{\partial\mu_{1}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}\mathsf{d}x_{1}^{\beta}}&{{}\tag{5}}\end{array}$$
 
-$-S_{2}^{\alpha}dT+V_{2}^{\alpha}dp+\left(\frac{\partial\mu_{2}^{\alpha}}{\partial x_{1}^{\alpha}}\right)_{T,p}dx_{1}^{\alpha}=-S_{2}^{\beta}dT+V_{2}^{\beta}dp+\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}dx_{1}^{\beta}\quad(6)$
+$$-S_{2}^{\alpha}dT+V_{2}^{\alpha}dp+\left(\frac{\partial\mu_{2}^{\alpha}}{\partial x_{1}^{\alpha}}\right)_{T,p}dx_{1}^{\alpha}=-S_{2}^{\beta}dT+V_{2}^{\beta}dp+\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}dx_{1}^{\beta}\tag{6}$$
 
 (5) 式、(6) 式也可以 Gibbs-Duhum 公式直接写出，此处主要是为了介绍微元法。
 
 根据偏摩尔量(此处即化学势)微商相关性公式
 
-$x_{1}^{e}\left(\frac{\partial \mu_{1}^{e}}{\partial x_{1}^{e}}\right)_{T, \rho} + x_{2}^{e}\left(\frac{\partial \mu_{2}^{e}}{\partial x_{1}^{e}}\right)_{T, \rho} = 0\quad(7)$
+$$x_{1}^{e}\left(\frac{\partial \mu_{1}^{e}}{\partial x_{1}^{e}}\right)_{T, \rho} + x_{2}^{e}\left(\frac{\partial \mu_{2}^{e}}{\partial x_{1}^{e}}\right)_{T, \rho} = 0\tag{7}$$
 
 $$x_{1}^{\beta}\left(\frac{\partial\mu_{1}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}+x_{2}^{\beta}\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}=0 \tag{8}$$
 
@@ -22,9 +22,16 @@ $$\left(\frac{\partial\mu_{2}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}=-\fr
 
 将(9)式、(10)式代入(6)式,可得
 
-$-S_{2}^{e}dT+V_{2}^{e}dp-\frac{x_{1}^{e}}{x_{2}^{e}}\left(\frac{\partial\mu_{1}^{e}}{\partial x_{1}^{e}}\right)_{T,p}dx_{1}^{e}$
-
-$= - S_2^\beta \mathrm{d}T + V_2^\beta \mathrm{d}p - \frac{x_1^\beta}{x_2^\beta} \left( \frac{\partial \mu_1^\beta}{\partial x_1^\beta} \right)_{T,p} \mathrm{d}x_1^\beta \quad (11)$
+$$
+\begin{aligned}
+- S_{2}^{e}\,\mathrm{d}T + V_{2}^{e}\,\mathrm{d}p
+- \frac{x_{1}^{e}}{x_{2}^{e}}\left(\frac{\partial \mu_{1}^{e}}{\partial x_{1}^{e}}\right)_{T,p}\mathrm{d}x_{1}^{e} \\
+= - S_{2}^{\beta}\,\mathrm{d}T + V_{2}^{\beta}\,\mathrm{d}p
+- \frac{x_{1}^{\beta}}{x_{2}^{\beta}}
+\left(\frac{\partial \mu_{1}^{\beta}}{\partial x_{1}^{\beta}}\right)_{T,p}\mathrm{d}x_{1}^{\beta}
+\tag{11}
+\end{aligned}
+$$
 
 (11) 式乘以 $x_2^a/x_1^a$, 得到
 
diff --git a/markdowns/education_chemistry_00070.md b/markdowns/education_chemistry_00070.md
index 4fad069d50c6566924924129521b1dbae86d3c07..57b064d02c643cdb5051ed53d9fe606e6da606b3 100644
--- a/markdowns/education_chemistry_00070.md
+++ b/markdowns/education_chemistry_00070.md
@@ -26,11 +26,11 @@
 
 由表 6.9.1 可知,平衡法用了最简捷的办法导出了依数性的公式,再经过合理的近似处理,可得到四种依数性之间的关系。
 
-$- \ln x_1 = x_2 = \frac{\Delta_{\mathrm{f}}^{\mathrm{g}} H_{\mathrm{m}}^{*}}{R T_{\mathrm{b}}^{2}} \Delta T_{\mathrm{b}} = \frac{\Delta_{\mathrm{v}}^{\mathrm{l}} H_{\mathrm{m}}^{*}}{R T_{\mathrm{f}}^{2}} \Delta T = \frac{V_{\mathrm{A}}^{* - 1}}{R T} = \frac{\Delta p_{\mathrm{A}}}{p_{\mathrm{A}}^{*}} \quad (6.9.1)$
+$$- \ln x_1 = x_2 = \frac{\Delta_{\mathrm{f}}^{\mathrm{g}} H_{\mathrm{m}}^{*}}{R T_{\mathrm{b}}^{2}} \Delta T_{\mathrm{b}} = \frac{\Delta_{\mathrm{v}}^{\mathrm{l}} H_{\mathrm{m}}^{*}}{R T_{\mathrm{f}}^{2}} \Delta T = \frac{V_{\mathrm{A}}^{* - 1}}{R T} = \frac{\Delta p_{\mathrm{A}}}{p_{\mathrm{A}}^{*}} \tag{6.9.1}$$
 
 式(6.9.1)反应的稀溶液四种依数性都是溶液内在性质在不同条件下的表现,由平衡条件和化学势等温式不难推得。
 
-$V_{A}\Pi = RT\ln(p_{A}^{*}/p_{A})$ (6.9.2)
+$$V_{A}\Pi = RT\ln(p_{A}^{*}/p_{A})\tag{6.9.2}$$
 
 式(6.9.2)说明：渗透压是为使溶剂A之蒸气压提高到与纯A相同时所需施加于溶液之外压力。
 
diff --git a/markdowns/education_chemistry_00071.md b/markdowns/education_chemistry_00071.md
index 8e9b291aae465d9671b7569c0697bac1bb4bf950..f2da86e1d50b561791009ee32228198278ccce4e 100644
--- a/markdowns/education_chemistry_00071.md
+++ b/markdowns/education_chemistry_00071.md
@@ -8,7 +8,7 @@ $T_{\mathrm{m}}$,现证明如下。
 
 物态方程：$\varepsilon_{\mathrm{m}} - nmc^{2} = \frac{3}{2} nkT_{\mathrm{m}}$
 
-$p_{\mathrm{m}}=nkT_{\mathrm{m}}$
+$$p_{\mathrm{m}}=nkT_{\mathrm{m}}$$
 
 辐射
 
@@ -25,33 +25,33 @@ $$
 \end{aligned}
 $$
 
-$d(R^3 \epsilon_m) = -nkT_md(R^3)$
+$$d(R^3 \epsilon_m) = -nkT_md(R^3)$$
 
-$d(R^3nmc^2) + d(R^3 \cdot \frac{3}{2}nkT_m) = -nkT_m\,d(R^3)$
+$$d(R^3nmc^2) + d(R^3 \cdot \frac{3}{2}nkT_m) = -nkT_m\,d(R^3)$$
 
 因为粒子数守恒 $dN = d\left(\frac{4}{3}\pi R^3 n\right) = 0$
 
 故 $n \propto R^{-3}$, 且 $m$、$c$ 为常数, 可得
 
-$\frac{3}{2}R^{3}nd(T_{m})=-nT_{m}d(R^{3})$
+$$\frac{3}{2}R^{3}nd(T_{m})=-nT_{m}d(R^{3})$$
 
-$\frac{3}{2}\frac{\mathrm{d}T_{\mathrm{m}}}{T_{\mathrm{m}}}=-\frac{\mathrm{d}R^{3}}{R^{3}}$
+$$\frac{3}{2}\frac{\mathrm{d}T_{\mathrm{m}}}{T_{\mathrm{m}}}=-\frac{\mathrm{d}R^{3}}{R^{3}}$$
 
-$T_{m}\propto\frac{1}{R^{2}}\quad(7.10.2)$
+$$T_{m}\propto\frac{1}{R^{2}}\quad(7.10.2)$$
 
 绝热膨胀下的辐射：
 
-$dU_{r}=d\left(\frac{4}{3}\pi R^{3}\epsilon_{r}\right)=-p_{r}d\left(\frac{4}{3}\pi R^{3}\right)$
+$$dU_{r}=d\left(\frac{4}{3}\pi R^{3}\epsilon_{r}\right)=-p_{r}d\left(\frac{4}{3}\pi R^{3}\right)$$
 
-$R^3 d\epsilon_r + \epsilon_r d(R^3) = -\frac{1}{3} \epsilon_r d(R^3)$
+$$R^3 d\epsilon_r + \epsilon_r d(R^3) = -\frac{1}{3} \epsilon_r d(R^3)$$
 
-$\frac{\mathrm{d}\epsilon_{r}}{\epsilon_{r}}=-\frac{4}{3}\frac{\mathrm{d}(R^{3})}{R^{3}}$
+$$\frac{\mathrm{d}\epsilon_{r}}{\epsilon_{r}}=-\frac{4}{3}\frac{\mathrm{d}(R^{3})}{R^{3}}$$
 
-$\varepsilon_{t} \propto \frac{1}{R^{4}}$
+$$\varepsilon_{t} \propto \frac{1}{R^{4}}$$
 
 因为 $\varepsilon_{r}\propto T_{r}^{4}$ (辐射热力学)
 
-$T_{r}\propto\frac{1}{R}\quad(7.10.3)$
+$$T_{r}\propto\frac{1}{R}\quad(7.10.3)$$
 
 根据式(7.10.2)及式(7.10.3),随着宇宙的膨胀, $R$ 增大,辐射温度将降低,且反比于宇宙的尺度因子 $R$, 粒子的温度 $T_{\mathrm{m}}$ 也将降低,但反比于宇宙的尺度因子 $R$ 的平方。
 
diff --git a/markdowns/education_chemistry_00072.md b/markdowns/education_chemistry_00072.md
index f77e78a31718c5de13faf851df3f497cee50c0a7..87d3c86142f966059fcff28d475b7ab90a60b747 100644
--- a/markdowns/education_chemistry_00072.md
+++ b/markdowns/education_chemistry_00072.md
@@ -44,20 +44,20 @@
 
 设一维谐振子的质量为 $m$，选 $x$ 轴与振动方向重合，振子平衡点（不受作用力的点）作为坐标 $x$ 的原点，这样，作用在振子上的 Hooke 力为
 
-$F = -fx$
+$$F = -fx$$
 
 其中 f 称为(弹)力常数，若选取平衡点的势能为零，则振子的势能函数为
 
-$V(x)=\frac{1}{2}f x^{2}$
+$$V(x)=\frac{1}{2}f x^{2}$$
 
 振子的振动频率 $\nu$ 与力常数 $f$ 的关系为
 
-$\nu=\frac{1}{2\pi}\sqrt{\frac{f}{m}}\qquad(\omega=2\pi\nu)$
+$$\nu=\frac{1}{2\pi}\sqrt{\frac{f}{m}}\qquad(\omega=2\pi\nu)$$
 
 求解一维谐振子的 Schrödinger 方程,即得
 
-$\psi_{n}(x)=\left[\frac{\alpha}{\sqrt{\pi}2^{n}n!}\right]^{\frac{1}{2}}e^{-\frac{1}{2}\alpha^{2}x^{2}}H_{n}(\alpha x)$
+$$\psi_{n}(x)=\left[\frac{\alpha}{\sqrt{\pi}2^{n}n!}\right]^{\frac{1}{2}}e^{-\frac{1}{2}\alpha^{2}x^{2}}H_{n}(\alpha x)$$
 
 其中 $\alpha = \sqrt{m\omega/\hbar}$，$H _ {n}(\alpha x)$ 为 Hermite 多项式，n称为振动量子数，它的取值为
 
-$n=0,1,2,\cdots$
+$$n=0,1,2,\cdots$$
diff --git a/markdowns/education_chemistry_00073.md b/markdowns/education_chemistry_00073.md
index 3b4b5209c3f097ad65c909d70ee4a6a363d17385..a692f44efba739a27d925eb089d625b2b11c2cb1 100644
--- a/markdowns/education_chemistry_00073.md
+++ b/markdowns/education_chemistry_00073.md
@@ -31,19 +31,19 @@
 
 则有
 
-$Q = \Omega e^{-\frac{E}{k_{\mathrm{B}}T}}$ (10.3.10)
+$$Q = \Omega e^{-\frac{E}{k_{\mathrm{B}}T}}(10.3.10)\tag{10.3.10}$$
 
-$Q_{\text{定域子}} = q^N \quad(10.3.11)$
+$$Q_{\text{定域子}} = q^N \tag{10.3.11}$$
 
-$Q_{\text{离域子}} = \left( \frac{qe}{N} \right)^N \quad (10.3.12)$
+$$Q_{\text{离域子}} = \left( \frac{qe}{N} \right)^N \tag {10.3.12}$$
 
 这样，用 Q 表示热力学函数，其表观形式对定域子或离域子就无差别，当然这时 Q 是不同的。
 
 在这一节的最后，我们还有条件对于任何独立子体系证明 Lagrange 乘因子 $\beta$ 的取值。在 §9.8 中，我们曾以求单原子一维平动的平均能为例证明 $\beta = \frac{1}{k_B T}$。对于任何独立子（即 $E = \sum_i n_i \varepsilon_i$）体系的封闭且只做体积功的微变过程，热力学指出
 
-$dS = \frac{1}{T}dE + \frac{p}{T}dV$
+$$dS = \frac{1}{T}dE + \frac{p}{T}dV$$
 
 因此
 
-$\left({\frac{\partial S}{\partial E}}\right)_{V}={\frac{1}{T}}$
+$$\left({\frac{\partial S}{\partial E}}\right)_{V}={\frac{1}{T}}$$
 
diff --git a/markdowns/education_chemistry_00074.md b/markdowns/education_chemistry_00074.md
index 4f37024edc45e7771fa5f3773e350d5909d07cb5..ff37681bb3e417fc3420eef8dc54a5e01859c7bd 100644
--- a/markdowns/education_chemistry_00074.md
+++ b/markdowns/education_chemistry_00074.md
@@ -43,7 +43,7 @@ $$\begin{aligned} (C_V)_\nu &= (C_p)_\nu = 2Nk_B T \frac{d \ln q_\nu}{dT} + Nk_B
 
 前面已经讨论了选平衡位置为能量零点(这就是选势能曲线最低点为能量零点)的配分函数。显然若选振动基态为能量零点,这时振动能级公式为
 
-$\epsilon_{\nu} = nh\nu, \quad n=0,1,2,3,\cdots$
+$$\epsilon_{\nu} = nh\nu, \quad n=0,1,2,3,\cdots$$
 
 相应于该能量零点的振动配分函数为
 
diff --git a/markdowns/education_chemistry_00075.md b/markdowns/education_chemistry_00075.md
index ad5916210796fe9146e17591f97c95e055c9e1aa..7f7cc3f64bb96138bddddf34dd93022e70f0c30e 100644
--- a/markdowns/education_chemistry_00075.md
+++ b/markdowns/education_chemistry_00075.md
@@ -3,11 +3,11 @@
 
 解 根据公式$ (22.11.5) $，在298 K时可得：
 
-$$\mathrm { p H } ( 2 ) = \frac { E _ { 2 } + E _ { 1 } } { 0 . 0 5 9 \ 1 6 \ \mathrm { V } } + \mathrm { p H } ( 1 ) = 8 . 6 4 \qquad \qquad ( 1 )$$
+$$\mathrm { p H } ( 2 ) = \frac { E _ { 2 } + E _ { 1 } } { 0 . 0 5 9 \ 1 6 \ \mathrm { V } } + \mathrm { p H } ( 1 ) = 8 . 6 4 \qquad \tag{1}$$
 
 根据(1)式，
 
-$E_3 = E_1 + 0.059\ 16\ \text{V}[\text{pH}(3) - \text{pH}(1)] = 0.023\ 3\ \text{V}$
+$$E_3 = E_1 + 0.059\ 16\ \text{V}[\text{pH}(3) - \text{pH}(1)] = 0.023\ 3\ \text{V}$$
 
 ## 参考资料及课外阅读资料
 
@@ -37,7 +37,7 @@ $E_3 = E_1 + 0.059\ 16\ \text{V}[\text{pH}(3) - \text{pH}(1)] = 0.023\ 3\ \text{
 
 (3) 计算醋酸银 $AgAc$ 的溶度积 $K_{sp}$。
 
-$[-71.769\ \text{kJ} \cdot \text{mol}^{-1},\ -60.293\ \text{kJ} \cdot \text{mol}^{-1},\ 38.6\ \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1},\ 2.07 \times 10^{-3}]$
+$$[-71.769\ \text{kJ} \cdot \text{mol}^{-1},\ -60.293\ \text{kJ} \cdot \text{mol}^{-1},\ 38.6\ \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1},\ 2.07 \times 10^{-3}]$$
 
 2. 电池 $Pt|H_2(p^\Theta)|NaCl(0.01\ mol\cdot kg^{-1})|AgCl|Ag$，已知 $\kappa(AgCl\ 饱和溶液)=2.68\times10^{-4}\ \Omega^{-1}\cdot m^{-1}$，$\kappa(H_2O)=0.84\times10^{-4}\ \Omega^{-1}\cdot m^{-1}$，$U_m^\Theta(Ag^+)=6.42\times10^{-8}\ m^2\cdot V^{-1}\cdot s^{-1}$，$U_m^\Theta(Cl^-)=7.92\times10^{-8}\ m^2\cdot V^{-1}\cdot s^{-1}$，$\varphi^\Theta(Ag^+/Ag)=0.799\ V$，求 298 K 电池的 $E$。[0.757 V]
 
diff --git a/markdowns/education_chemistry_00081.md b/markdowns/education_chemistry_00081.md
index c4c320a62fe0301101353111f40445827c8dc770..595a9bcd0e2e9d1b7f956b611fc8418289348df4 100644
--- a/markdowns/education_chemistry_00081.md
+++ b/markdowns/education_chemistry_00081.md
@@ -4,7 +4,9 @@
 
 聚乙烯是一种常用塑料，是由乙烯在一定条件下，通过加成聚合反应（简称加聚反应）得到的高分子化合物，结构呈线型。高分子化合物又称为聚合物。
 
-$nCH_2—CH_2 \longrightarrow [CH_2—CH_2]_n$
+$$
+\ce{nCH2-CH2 -> [-CH2-CH2-]_{n}}
+$$
 
 乙烯     聚乙烯
 
diff --git a/markdowns/education_chemistry_00091.md b/markdowns/education_chemistry_00091.md
index 08da2442813808a1a87fc1b5980c921665d7b397..28e0c762bd4f0f5e73f8b97221de7e62c0ef520a 100644
--- a/markdowns/education_chemistry_00091.md
+++ b/markdowns/education_chemistry_00091.md
@@ -1,34 +1,48 @@
 酸。由于烷基的推电子效应，增加了醇分子中氧原子周围的电子云密度，O—H键上的氢原子受到的束缚力加大，使得氢不易被取代。邻近羟基碳原子上的烷基增多，氧原子上的电子云密度增大，O—H键的极性减小，酸性随之减弱，与钠的反应活性也随之降低。故醇的酸性及与活泼金属反应活性的顺序为；
 
-$H_2O>CH_3OH>RCH_2OH>R_2CHOH>R_3COH$
+$$
+\ce{H2O > CH3OH > RCH2OH > R2CHOH > R3COH}
+$$
 
 醇的酸性比水小，其共轭碱 RO⁻ 的碱性比 OH⁻ 大。醇钠遇水立即水解生成醇和氢氧化钠：
 
-$RCH_2ONa\xrightarrow{H_2O}RCH_2OH+NaOH$
+$$
+\ce{RCH2ONa} \xrightarrow{\ce{H2O}} \ce{RCH2OH + NaOH}
+$$
 
 醇钠在有机合成中用作碱性试剂，也常用作分子中引入烃氧基（RO—）的亲核试剂。
 
 2. 与氢卤酸的反应 氢卤酸与醇反应生成卤代烷和水，这是制备卤代烃的重要方法；
 
-$R-OH \xrightarrow{HX} R-X + H_2O (X=Cl、Br、I）$
+$$
+\ce{R-OH} \xrightarrow{\ce{HX}} \ce{R-X + H2O} \quad (X=\mathrm{Cl},\mathrm{Br},\mathrm{I})
+$$
 
 反应中醇的 C—O 键断裂，羟基被卤原子取代，属于亲核取代反应。C—OH 键断裂较难，需要酸的催化，使羟基质子化后以水分子的形式离去。醇的反应活性是烯丙式（或苄基式）醇 > 3°醇 > 2°醇 > 1°醇；氢卤酸的反应活性是 HI > HBr > HCl。例如，一级醇与氢碘酸（47%）一起加热就可以生成碘代烃；与氢溴酸（48%）作用时必须在硫酸作用下加热才能反应；与浓盐酸作用时必须有氯化锌存在并加热才能反应。而 3°醇和烯丙式（或苄基式）醇在室温下和浓盐酸一起震荡就可以反应：
 
-$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow[\Delta]{\text{HBr},\ \text{H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$
+$$
+\ce{CH3CH2CH2CH2OH} \xrightarrow[\Delta]{\ce{HBr},\ \ce{H2SO4}} \ce{CH3CH2CH2CH2Br}
+$$
 
-$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow[\Delta]{\text{HCl, ZnCl}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$
+$$
+\ce{CH3CH2CH2CH2OH} \xrightarrow[\Delta]{\ce{HCl, ZnCl2}} \ce{CH3CH2CH2CH2Cl}
+$$
 
-$(CH_{3})_{3}COH\xrightarrow{浓HCl}(CH_{3})_{3}CCl$
+$$
+\ce{(CH3)3COH} \xrightarrow{\ce{浓HCl}} \ce{(CH3)3CCl}
+$$
 
 实验室常用卢卡斯（Lucas）试剂（浓盐酸和无水氯化锌配成的溶液）来区别不同级别的醇。低级一元醇（$C_{6}$ 以下）能溶于卢卡斯试剂中，其氯代物不溶。从出现浑浊所需的时间可以鉴别出伯、仲、叔醇。
 
 伯醇、仲醇、叔醇
 
-$\xrightarrow{\text{卢卡斯试剂}}$
+$$\xrightarrow{\text{卢卡斯试剂}}$$
 
 加热后出现浑浊；静置几分钟后出现浑浊；很快出现浑浊
 
 醇与氢卤酸的反应是酸催化下的亲核取代反应，一般认为烯丙式（或苄基式）醇、叔醇、仲醇是按$S_N1$反应机理进行：
 
-$(CH_3)_3C-OH + HX \rightleftharpoons (CH_3)_3C-\overset{+}{O}H_2 + X^-$
+$$
+\ce{(CH3)3C-OH + HX <=> (CH3)3C-OH2+ + X-}
+$$
 
diff --git a/markdowns/education_chemistry_00100.md b/markdowns/education_chemistry_00100.md
index c19a03976b583f0ad7839f9a3e918a11c9e9814d..1f7597b3a7c2ef4a31b3b52bd222488c8c08dec4 100644
--- a/markdowns/education_chemistry_00100.md
+++ b/markdowns/education_chemistry_00100.md
@@ -38,10 +38,18 @@
 
 • 碱金属元素原子的最外层都有1个电子，它们的化学性质相似，正如上述实验所示，它们都能与氧气等非金属单质以及水反应。例如：
 
-$4Li + O_2 \xlongequal{\Delta} 2Li_2O$
+$$
+\ce{4Li + O2} \xlongequal{\Delta} \ce{2Li2O}
+$$
 
-$2Na + O_2 \xlongequal{\Delta} Na_2O_2$
+$$
+\ce{2Na + O2} \xlongequal{\Delta} \ce{Na2O2}
+$$
 
-$2Na + 2H_{2}O \xlongequal{} 2NaOH + H_{2}\uparrow$
+$$
+\ce{2Na + 2H2O} \xlongequal{} \ce{2NaOH + H2 ^}
+$$
 
-$2K + 2H_{2}O \xlongequal{} 2KOH + H_{2}\uparrow$
+$$
+\ce{2K + 2H2O} \xlongequal{} \ce{2KOH + H2 ^}
+$$
diff --git a/markdowns/education_english_00076.md b/markdowns/education_english_00076.md
index 98ced5bc51a4cb73774e32bb06e589f388c7128c..148db9ac35a6b602635bfdcd13f7398a28febc9d 100644
--- a/markdowns/education_english_00076.md
+++ b/markdowns/education_english_00076.md
@@ -12,7 +12,9 @@
 
 (7) The family moved to California.
 
-( )—( )—( )—( )—( )—( )—( )
+$$
+(\ )-(\ )-(\ )-(\ )-(\ )-(\ )-(\ )-(\ )
+$$
 
 ## 2. Match the words with their definitions.
 
diff --git a/markdowns/education_math_00001.md b/markdowns/education_math_00001.md
index 7ede85799d09fd76f561ac8f9f050ed94342b236..b026625f7d30d95a1c0a22c2ce00760a70b58278 100644
--- a/markdowns/education_math_00001.md
+++ b/markdowns/education_math_00001.md
@@ -21,11 +21,10 @@ A. 0 个          B. 1 个          C. 2 个          D. 3 个
 
 15. 如果命题“$\neg p$”为真，命题“$p \land q$”为假，那么（）
 
-$$\begin{array}{ll}
-\text{A. } q \text{ 为假} & \text{B. } q \text{ 为真} \\
-\text{C. } p \text{ 或 } q \text{ 为真} & \text{D. } p \text{ 或 } q \text{ 不一定为真}
-\end{array}
-$$
+A. q 为假
+B. q 为真
+C. p 或 q 为真
+D. p 或 q 不一定为真
 
 16. 不等式$(|x|+2)(1-x^{2})\leq0$ 的解集是（）
 
diff --git a/markdowns/education_math_00005.md b/markdowns/education_math_00005.md
index bcfd0997618d235b84af18cce0298831ffe65f9e..10ed80661477d70e3f3691eaaac7403ff688bf82 100644
--- a/markdowns/education_math_00005.md
+++ b/markdowns/education_math_00005.md
@@ -16,12 +16,12 @@ $\therefore$ 所求椭圆方程为 $\frac{x^{2}}{6}+\frac{y^{2}}{3}=1$.
 
 25. $\frac{7}{2}$ __解析__：$|F_1F_2|=2\sqrt{2}$，$|AF_1|+|AF_2|=6$，$|AF_2|=6-|AF_1|$.
 
-$|AF_{2}|^{2}=|AF_{1}|^{2}+|F_{1}F_{2}|^{2}-2|AF_{1}|\cdot|F_{1}F_{2}|\cos45^{\circ}
-\\=|AF_{1}|^{2}-4|AF_{1}|+8=(6-|AF_{1}|)^{2},$
+$$|AF_{2}|^{2}=|AF_{1}|^{2}+|F_{1}F_{2}|^{2}-2|AF_{1}|\cdot|F_{1}F_{2}|\cos45^{\circ}
+\\=|AF_{1}|^{2}-4|AF_{1}|+8=(6-|AF_{1}|)^{2},$$
 
 $\therefore |AF_{1}| = \frac{7}{2}.$
 
-$S=\frac{1}{2}\times\frac{7}{2}\times2\sqrt{2}\times\frac{\sqrt{2}}{2}=\frac{7}{2}.$
+$$S=\frac{1}{2}\times\frac{7}{2}\times2\sqrt{2}\times\frac{\sqrt{2}}{2}=\frac{7}{2}.$$
 
 26. __解__: 设圆 $C$ 的方程为 $x^2 + y^2 + Dx + Ey + F = 0$，
 
@@ -44,7 +44,7 @@ $$
 
 由①②③解得
 
-$D=-4,E=6,F=8.$
+$$D=-4,E=6,F=8.$$
 
 $\therefore$ 圆 $C$ 的方程为 $x^{2}+y^{2}-4x+6y+8=0$.
 
diff --git a/markdowns/education_math_00006.md b/markdowns/education_math_00006.md
index 8022c93e0b58968ad44480ab75686ef0bac31a3c..8cd540b522be3a7cb2228ebea4e30f8f13c4b2c2 100644
--- a/markdowns/education_math_00006.md
+++ b/markdowns/education_math_00006.md
@@ -50,7 +50,7 @@
 
 5. 解:20%与30%的两种酒精溶液按3:1的比例混合后得到溶液的浓度为
 
-$\begin{aligned}&(3\times20\%+1\times30\%)\div(3+1)\\&=(60\%+30\%)\div4\\&=22.5\%\end{aligned}$
+$$\begin{aligned}&(3\times20\%+1\times30\%)\div(3+1)\\&=(60\%+30\%)\div4\\&=22.5\%\end{aligned}$$
 
 浓度为 45% 的酒精溶液用量为
 
diff --git a/markdowns/education_math_00009.md b/markdowns/education_math_00009.md
index 5680646978c50a1a93a869afc6bba8baf006727d..f96ce09a976986b9487ad6ae358f135cf99bcf58 100644
--- a/markdowns/education_math_00009.md
+++ b/markdowns/education_math_00009.md
@@ -20,12 +20,10 @@ A. 1:200          B. 1:400000         C. 1:10000          D. 1:2000
 
 5. 一个正方形广场占地面积约为 2250000 平方米, 若按照比例尺 1:12000 缩小后, 其面积比较接近( )。
 
-$$
-\begin{array}{ll}
-\text{A. 一个篮球场的面积} & \text{B. 一张乒乓球台面的面积} \\
-\text{C. 华商报一个版面的面积} & \text{D. 数学课本封面的面积}
-\end{array}
-$$
+A. 一个篮球场的面积
+B. 一张乒乓球台面的面积
+C. 华商报一个版面的面积
+D. 数学课本封面的面积
 
 ## 二、填空题
 
diff --git a/markdowns/education_math_00013.md b/markdowns/education_math_00013.md
index 0b5195c7ff7d09f2b6aabcedb6251d66758069b6..869731ca2de450c0b8cc5e65b1e81b92e6cc649a 100644
--- a/markdowns/education_math_00013.md
+++ b/markdowns/education_math_00013.md
@@ -1,6 +1,6 @@
 4. 设 $f(x)$ 是周期为 $2\pi$ 的周期函数，它在 $[- \pi, \pi)$ 上的表达式为
 
-$f(x)=\left\{\begin{array}{ll}-\frac{\pi}{2}, & -\pi \leqslant x<-\frac{\pi}{2}, \\ x, & -\frac{\pi}{2} \leqslant x<\frac{\pi}{2}, \\ \frac{\pi}{2}, & \frac{\pi}{2} \leqslant x<\pi,\end{array}\right.$
+$$f(x)=\left\{\begin{array}{ll}-\frac{\pi}{2}, & -\pi \leqslant x<-\frac{\pi}{2}, \\ x, & -\frac{\pi}{2} \leqslant x<\frac{\pi}{2}, \\ \frac{\pi}{2}, & \frac{\pi}{2} \leqslant x<\pi,\end{array}\right.$$
 
 将 $f(x)$ 展开成傅里叶级数.
 
@@ -15,17 +15,17 @@ b_n &= \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin nx \, dx = \frac{2}{\pi} \left[ \i
 
 因 $f(x)$ 满足收敛定理的条件，而在 $x=(2k+1)\pi(k\in\mathbb{Z})$ 处间断，故
 
-$f(x)=\sum\limits_{n=1}^{\infty}\left[\frac{(-1)^{n+1}}{n}+\frac{2}{n^{2}\pi}\sin\frac{n\pi}{2}\right]\sin nx,x\neq(2k+1)\pi(k\in\mathbb{Z}).$
+$$f(x)=\sum\limits_{n=1}^{\infty}\left[\frac{(-1)^{n+1}}{n}+\frac{2}{n^{2}\pi}\sin\frac{n\pi}{2}\right]\sin nx,x\neq(2k+1)\pi(k\in\mathbb{Z}).$$
 
 5. 将函数 $f(x)=\frac{\pi-x}{2}(0\leqslant x\leqslant\pi)$ 展开成正弦级数.
 
 解 作
 
-$\varphi(x)=\begin{cases}f(x),&x\in(0,\pi],\\0,&x=0,\\-f(-x),&x\in(-\pi,0),\end{cases}$
+$$\varphi(x)=\begin{cases}f(x),&x\in(0,\pi],\\0,&x=0,\\-f(-x),&x\in(-\pi,0),\end{cases}$$
 
 $\varphi(x)$ 是 $f(x)$ 的奇延拓．令 $\Phi(x)$ 是 $\varphi(x)$ 的周期延拓，则 $\Phi(x)$ 满足收敛定理的条件，而在 $x = 2k\pi (k \in \mathbb{Z})$ 处间断，又在 $(0, \pi]$ 上，$\Phi(x) \equiv f(x)$，因此 $\Phi(x)$ 的傅里叶级数在 $(0, \pi]$ 上收敛于 $f(x)$.
 
-$a_{n}=0(n=0,1,2,\cdots);$
+$$a_{n}=0(n=0,1,2,\cdots);$$
 
 $$\begin{align*}
 b_n &= \frac{2}{\pi} \int_{0}^{\pi} \frac{\pi - x}{2} \sin nx \, dx = \frac{2}{\pi} \left[ \frac{x - \pi}{2n} \cos nx - \frac{1}{2n^2} \sin nx \right]_{0}^{\pi} \\
diff --git a/markdowns/education_math_00014.md b/markdowns/education_math_00014.md
index c32d007b6c64b56aa1deac833f901bb927b18a4b..d580ae64ffadc528e8352ba8a7497afd5634dd17 100644
--- a/markdowns/education_math_00014.md
+++ b/markdowns/education_math_00014.md
@@ -2,11 +2,11 @@ $$\begin{aligned} & =-\frac{1}{2}\cdot\frac{1-n\pi i}{1+(n\pi)^{2}}\left(\mathrm
 
 故
 
-$f(x)=\sum\limits_{n=-\infty}^{\infty}(-1)^{n}\frac{e-e^{-1}}{2}\frac{1-n\pi i}{1+n^{2}\pi^{2}}\cdot e^{in\pi  x},x\in\mathbb{R}\setminus\{2k+1\mid k\in\mathbb{Z}\}.$
+$$f(x)=\sum\limits_{n=-\infty}^{\infty}(-1)^{n}\frac{e-e^{-1}}{2}\frac{1-n\pi i}{1+n^{2}\pi^{2}}\cdot e^{in\pi  x},x\in\mathbb{R}\setminus\{2k+1\mid k\in\mathbb{Z}\}.$$
 
 *4. 设 $u(t)$ 是周期为 $T$ 的周期函数. 已知它的傅里叶级数的复数形式为(参阅本节例题)
 
-$u(t) = \frac{h\tau}{T} + \frac{h}{\pi} \displaystyle\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} \frac{1}{n} \sin \frac{n\pi\tau}{T} e^{i\frac{2n\pi t}{T}} \quad (-\infty < t < +\infty),$
+$$u(t) = \frac{h\tau}{T} + \frac{h}{\pi} \displaystyle\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} \frac{1}{n} \sin \frac{n\pi\tau}{T} e^{i\frac{2n\pi t}{T}} \quad (-\infty < t < +\infty),$$
 
 试写出 $u(t)$ 的傅里叶级数的实数形式(即三角形式).
 
@@ -14,19 +14,19 @@ __解__ 由题设知 $c_n = \frac{h}{n\pi} \sin \frac{n\pi\tau}{T}$ ($n = \pm 1,
 
 因
 
-$c_{n}=\frac{a_{n}-ib_{n}}{2},c_{-n}=\frac{a_{n}+ib_{n}}{2}=\overline{c_{n}}\quad(n=1,2,\cdots),$
+$$c_{n}=\frac{a_{n}-ib_{n}}{2},c_{-n}=\frac{a_{n}+ib_{n}}{2}=\overline{c_{n}}\quad(n=1,2,\cdots),$$
 
 可见
 
-$a_n = \operatorname{Re}(2\overline{c_n}), b_n = \operatorname{Im}(2\overline{c_n}).$
+$$a_n = \operatorname{Re}(2\overline{c_n}), b_n = \operatorname{Im}(2\overline{c_n}).$$
 
 而 $c_{n}$ 为实数，故
 
-$a_{n}=\frac{2h}{n\pi}\sin\frac{n\pi\tau}{T},b_{n}=0\quad(n=1,2,\cdots),$
+$$a_{n}=\frac{2h}{n\pi}\sin\frac{n\pi\tau}{T},b_{n}=0\quad(n=1,2,\cdots),$$
 
 故
 
-$u(t) = \frac{h\tau}{T} + \frac{2h}{\pi} \sum\limits_{n=1}^{\infty} \frac{1}{n} \sin \frac{n\pi\tau}{T} \cdot \cos \frac{2n\pi t}{T} \quad (-\infty < t < +\infty).$
+$$u(t) = \frac{h\tau}{T} + \frac{2h}{\pi} \sum\limits_{n=1}^{\infty} \frac{1}{n} \sin \frac{n\pi\tau}{T} \cdot \cos \frac{2n\pi t}{T} \quad (-\infty < t < +\infty).$$
 
 ## 总习题十二
 
diff --git a/markdowns/education_math_00015.md b/markdowns/education_math_00015.md
index 298bd45af1ab7917c5156d8528794732ba1b8acc..da542e384c75fc7f2a6c81461492be9ec46e49c8 100644
--- a/markdowns/education_math_00015.md
+++ b/markdowns/education_math_00015.md
@@ -10,7 +10,7 @@ __解__ 两已知直线的方向向量分别为 $s_1=(1,2,1)$ 和 $s_2=(0,1,1)$
 
 3. (1991. I, II) 已知两条直线的方程是
 
-$L_1:\frac{x-1}{1}=\frac{y-2}{0}=\frac{z-3}{-1}, L_2:\frac{x+2}{2}=\frac{y-1}{1}=\frac{z}{1},$
+$$L_1:\frac{x-1}{1}=\frac{y-2}{0}=\frac{z-3}{-1}, L_2:\frac{x+2}{2}=\frac{y-1}{1}=\frac{z}{1},$$
 
 则过 $L_1$ 且平行于 $L_2$ 的平面方程是______．
 
diff --git a/markdowns/education_math_00016.md b/markdowns/education_math_00016.md
index fdf4905e3309e7013b17cc9bd7ec160edb8b6425..517e97c8c0ebd387343676cc0b9bb877eecfa3f6 100644
--- a/markdowns/education_math_00016.md
+++ b/markdowns/education_math_00016.md
@@ -18,24 +18,26 @@ $$
 
 设 $l_0$ 绕 $y$ 轴旋转一周所得的旋转曲面为 $S，(x,y,z)$ 为 $S$ 上任意一点。则该点由直线 $l_0$ 上的一点 $(x_0,y_0,z_0)$ 绕 $y$ 轴旋转而得，于是有关系：$y = y_0$，
 
-$x^{2} + z^{2} = x_{0}^{2} + z_{0}^{2} = (2y_{0})^{2} + \left[-\frac{1}{2}(y_{0}-1)\right]^{2} = 4y^{2} + \frac{1}{4}(y-1)^{2},$
+$$x^{2} + z^{2} = x_{0}^{2} + z_{0}^{2} = (2y_{0})^{2} + \left[-\frac{1}{2}(y_{0}-1)\right]^{2} = 4y^{2} + \frac{1}{4}(y-1)^{2},$$
 
 从而得 $S$ 的方程为
 
-$4x^2 - 17y^2 + 4z^2 + 2y - 1 = 0.$
+$$4x^2 - 17y^2 + 4z^2 + 2y - 1 = 0.$$
 
 __解法二__ 将直线 $l$ 的方程改写为一般方程：$\begin{cases} x - y - 1 = 0, \\ y + z - 1 = 0. \end{cases}$ 过 $l$ 的平面束方程为
 
-$(x-y-1)+\lambda(y+z-1)=0,$
+$$(x-y-1)+\lambda(y+z-1)=0,$$
 
 即
 
-$x + (\lambda - 1)y + \lambda z - (1 + \lambda) = 0.$
+$$x + (\lambda - 1)y + \lambda z - (1 + \lambda) = 0.$$
 
 现确定 $\lambda$ 的值，使向量 $(1,\lambda-1,\lambda)$ 与平面 $\pi$ 的法向量 $\boldsymbol{n}=(1,-1,2)$ 垂直，即令 $1-(\lambda-1)+2\lambda=0$ ，解得 $\lambda=-2$ 。从而得过 l 且垂直于 $\pi$ 的平面方程为 $x-3y-2z+1=0$ .(下同解法一)
 
 __解法三__ 经过$l$且垂直于平面$\pi$的平面$\pi_1$的法向量$n_1$可取为$(1,1,-1)\times(1,-1,2)=(1,-3,-2)$. 又$\pi_1$通过$l$上的点$(1,0,1)$，故$\pi_1$的方程为
 
-$(x-1)-3y-2(z-1)=0$ , 即 $x-3y-2z+1=0.$
+$$(x-1)-3y-2(z-1)=0, 即 x-3y-2z+1=0.$$
+
+（下同解法一）
+
 
-（下同解法一）
\ No newline at end of file
diff --git a/markdowns/education_math_00017.md b/markdowns/education_math_00017.md
index ff23d9caa0eb964b418d1a71e84224c7004a8a9f..bb7b107d739a2e4f196c815ff10cf6e4e0c4609f 100644
--- a/markdowns/education_math_00017.md
+++ b/markdowns/education_math_00017.md
@@ -7,9 +7,12 @@
 (C) 不连续、偏导数存在.  
 (D) 不连续、偏导数不存在.  
 
-$$\textbf{解}\quad f_x(0,0) = \lim_{\Delta x \to 0} \frac{f(0+\Delta x, 0) - f(0,0)}{\Delta x} = 0, \text{同理}, f_y(0,0) = 0, \text{故偏导数存在. 又当} (x,y) \text{沿} y=kx \text{趋向于} (0,0) \text{时}$$
+解 $f _ {x}(0,0)=\lim _ {\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=0$ , 同理，$f _ {y}(0,0)=0$ ，故偏导数存在．又当 $(x,y)$ 沿 y=kx 趋向于 $(0,0)$ 时
 
-$\lim\limits_{\substack{(x,y)\to(0,0)\\y=kx}}f(x,y)=\lim\limits_{x\to0}\frac{kx^{2}}{x^{2}+(kx)^{2}}=\frac{k}{1+k^{2}}$.
+
+$$
+\lim_{\begin{subarray}{c}(x,y)\to(0,0)\\ y=kx\end{subarray}}f(x,y)=\lim_{x\to 0}\frac{kx^{2}}{x^{2}+(kx)^{2}}=\frac{k}{1+k^{2}}.
+$$
 
 随着 k 的不同,该极限值也不同,所以极限 $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ 不存在, $f(x,y)$ 在 $(0,0)$ 不连续. 应选(C).
 
@@ -17,7 +20,7 @@ $\lim\limits_{\substack{(x,y)\to(0,0)\\y=kx}}f(x,y)=\lim\limits_{x\to0}\frac{kx^
 
 解 在所给方程两端分别取全微分, 得
 
-$yz\,dx + xz\,dy + xy\,dz + \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}(xdx + ydy + zdz) = 0,$
+$$yz\,dx + xz\,dy + xy\,dz + \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}(xdx + ydy + zdz) = 0,$$
 
 因此，在点$(1,0,-1)$处 $dz=dx-\sqrt{2}dy.$
 
@@ -25,9 +28,22 @@ $yz\,dx + xz\,dy + xy\,dz + \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}(xdx + ydy + z
 
 解
 
-$\frac{\partial z}{\partial x} = f_u \frac{\partial u}{\partial x} + f_x = f_u \cdot e^y + f_x,$
-
-$\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(f_{u} \cdot e^{y}+f_{x}\right)=\left(f_{u u} \cdot \frac{\partial u}{\partial y}+f_{u y}\right) e^{y}+f_{u} \cdot e^{y}+f_{x u} \frac{\partial u}{\partial y}+f_{x y}$
-$=f_{u u} \cdot x e^{2 y}+f_{u y} \cdot e^{y}+f_{u} \cdot e^{y}+f_{x u} \cdot x e^{y}+f_{x y}.$
+$$\frac{\partial z}{\partial x} = f_u \frac{\partial u}{\partial x} + f_x = f_u \cdot e^y + f_x,$$
+
+$$
+\begin{aligned}
+\frac{\partial^{2} z}{\partial x \partial y}
+&= \frac{\partial}{\partial y}\left(f_{u}\cdot e^{y}+f_{x}\right)
+= \left(f_{uu}\cdot \frac{\partial u}{\partial y}+f_{uy}\right)e^{y}
++ f_{u}\cdot e^{y}
++ f_{xu}\frac{\partial u}{\partial y}
++ f_{xy} \\
+&= f_{uu}\cdot x e^{2y}
++ f_{uy}\cdot e^{y}
++ f_{u}\cdot e^{y}
++ f_{xu}\cdot x e^{y}
++ f_{xy}.
+\end{aligned}
+$$
 
 4. (1995. I, II) 设 $u = f(x, y, z)$, $\varphi(x^2, e^y, z) = 0$, $y = \sin x$, 其中 $f, \varphi$ 都具有一阶连续偏导数, 且 $\frac{\partial \varphi}{\partial z} \neq 0$, 求 $\frac{du}{dx}$.
diff --git a/markdowns/education_math_00018.md b/markdowns/education_math_00018.md
index dd3143df8b6bc75070348ce1e3fb13f808ca26b6..14deffc3cfc298741564f031eba51ef62081ca76 100644
--- a/markdowns/education_math_00018.md
+++ b/markdowns/education_math_00018.md
@@ -9,9 +9,9 @@ $$\begin{aligned}\oint _ { C } { \frac { \varphi ( y ) d x + 2 x y d y } { 2 x ^
 
 （Ⅱ）解 设 $P = \frac{\varphi(y)}{2x^2 + y^4}$，$Q = \frac{2xy}{2x^2 + y^4}$，$P$、$Q$ 在单连通区域 $x > 0$ 内具有一阶连续偏导数。由（Ⅰ）知，曲线积分 $\int_L \frac{\varphi(y)dx + 2xydy}{2x^2 + y^4}$ 在该区域内与路径无关，故当 $x > 0$ 时，恒有 $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$.
 
-$\frac{\partial Q}{\partial x}=\frac{2y(2x^{2}+y^{4})-4x\cdot2xy}{(2x^{2}+y^{4})^{2}}=\frac{-4x^{2}y+2y^{5}}{(2x^{2}+y^{4})^{2}},$
+$$\frac{\partial Q}{\partial x}=\frac{2y(2x^{2}+y^{4})-4x\cdot2xy}{(2x^{2}+y^{4})^{2}}=\frac{-4x^{2}y+2y^{5}}{(2x^{2}+y^{4})^{2}},$$
 
-$\frac{\partial P}{\partial y} = \frac{\varphi^{\prime}(y)(2x^{2} + y^{4}) - 4\varphi(y)y^{3}}{(2x^{2} + y^{4})^{2}} = \frac{2x^{2}\varphi^{\prime}(y) + \varphi^{\prime}(y)y^{4} - 4\varphi(y)y^{3}}{(2x^{2} + y^{4})^{2}}$,
+$$\frac{\partial P}{\partial y} = \frac{\varphi^{\prime}(y)(2x^{2} + y^{4}) - 4\varphi(y)y^{3}}{(2x^{2} + y^{4})^{2}} = \frac{2x^{2}\varphi^{\prime}(y) + \varphi^{\prime}(y)y^{4} - 4\varphi(y)y^{3}}{(2x^{2} + y^{4})^{2}},$$
 
 比较上列两式的右端,要使它们恒等,需有
 
@@ -19,7 +19,7 @@ $\varphi'(y) = -2y$ 和 $\varphi'(y)y^4 - 4\varphi(y)y^3 = 2y^5$.
 
 由 $\varphi'(y) = -2y$ 得 $\varphi(y) = -y^{2} + C$，代入第二式得
 
-$\varphi^{\prime}(y)y^{4}-4\varphi(y)y^{3}=-2y^{5}+4y^{5}-4Cy^{3}=2y^{5},$
+$$\varphi^{\prime}(y)y^{4}-4\varphi(y)y^{3}=-2y^{5}+4y^{5}-4Cy^{3}=2y^{5},$$
 
 因此 $C=0$，从而得 $\varphi(y) = -y^2$.
 
@@ -29,8 +29,6 @@ $\varphi^{\prime}(y)y^{4}-4\varphi(y)y^{3}=-2y^{5}+4y^{5}-4Cy^{3}=2y^{5},$
 
 图研 7-6
 
-解
-
-$\overrightarrow{OP}=x\boldsymbol{i}+y\boldsymbol{j},$
+解 $\overrightarrow{OP}=x\boldsymbol{i}+y\boldsymbol{j},$
 
 按题意，$|F|=|\overrightarrow{OP}|=\sqrt{x^2+y^2}$.
diff --git a/markdowns/education_math_00020.md b/markdowns/education_math_00020.md
index 78f4227e3bae2a26785fb3f7fea92c3b5a7c1c52..b6487cdb2f84c57b03a7bc9566a64651e8dadb3d 100644
--- a/markdowns/education_math_00020.md
+++ b/markdowns/education_math_00020.md
@@ -4,7 +4,7 @@ $$\begin{aligned} M & = \iiint\limits_{\Omega} z \, dV = \iint\limits_{D} dxdy \
 
 令 $\boldsymbol{\tau} \cdot \boldsymbol{n} = 0$，得 $t_{1} = 0, t_{2} = 1$，即
 
-$\tau_{1}=(1,0,0),\tau_{2}=(1,-2,3).$
+$$\tau_{1}=(1,0,0),\tau_{2}=(1,-2,3).$$
 
 所求切线方程为
 
@@ -12,4 +12,4 @@ $\frac{x}{1} = \frac{y}{0} = \frac{z}{0} \text{ 与 } \frac{x-1}{1} = \frac{y+1}
 
 (2) 取 $M_{1}(0,0,0)$, $M_{2}=(1,-1,1)$, $\overrightarrow{M_{1}M_{2}}=(1,-1,1)$. 所求距离为
 
-$d=\frac{\left|(\tau_{1}\times\tau_{2})\cdot\overrightarrow{M_{2}M_{1}}\right|}{\left|\tau_{1}\times\tau_{2}\right|}=\frac{1}{\sqrt{13}}$
+$$d=\frac{\left|(\tau_{1}\times\tau_{2})\cdot\overrightarrow{M_{2}M_{1}}\right|}{\left|\tau_{1}\times\tau_{2}\right|}=\frac{1}{\sqrt{13}}$$
diff --git a/markdowns/education_math_00022.md b/markdowns/education_math_00022.md
index 1388c35b8f02edc9cb6b72f552b50ed6bdb76dc7..a2b4d0047e4c0469b0b96e5fab8479ccf1bd3a3c 100644
--- a/markdowns/education_math_00022.md
+++ b/markdowns/education_math_00022.md
@@ -1,12 +1,12 @@
 解 设直角三角形的两直角边之长分别为 $x, y$，则周长
 
-$S = x + y + l \quad (0 < x < l, 0 < y < l).$
+$$S = x + y + l \quad (0 < x < l, 0 < y < l).$$
 
 本题是求周长 $S$ 在 $x^2 + y^2 = l^2$ 条件下的条件极值问题.
 
 作拉格朗日函数
 
-$L(x,y) = x + y + l + \lambda(x^2 + y^2 - l^2).$
+$$L(x,y) = x + y + l + \lambda(x^2 + y^2 - l^2).$$
 
 令
 
@@ -27,7 +27,7 @@ $$
 
 __解__ 设水池的长为 $a$，宽为 $b$，高为 $c$，则水池的表面积为
 
-$A=ab+2ac+2bc(a>0,b>0,c>0).$
+$$A=ab+2ac+2bc(a>0,b>0,c>0).$$
 
 约束条件 $abc = k$.
 
diff --git a/markdowns/education_math_00023.md b/markdowns/education_math_00023.md
index df7bbdf70d6bbb68241fdcdf24952afc2b2875c1..09ba8df34a0874c7ff9ff5889dd35417a9dc9fd0 100644
--- a/markdowns/education_math_00023.md
+++ b/markdowns/education_math_00023.md
@@ -8,29 +8,29 @@ $$\iint \limits_ { D } [ \ln ( x + y ) ] ^ { 2 } d \sigma \geqslant \iint \limit
 
 (2) $I = \iint\limits_D \sin^2 x \sin^2 y \, d\sigma$, 其中 $D = \{(x,y) \mid 0 \leq x \leq \pi, 0 \leq y \leq \pi\}$;
 
-(3) $$ I = \iint\limits_{D} (x + y + 1) \, d\sigma ，其中  D = \{(x, y) \mid 0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 2\} ;$$
+(3) $ I = \iint\limits_{D} (x + y + 1) \, d\sigma ，其中  D = \{(x, y) \mid 0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 2\} ;$
 
-(4) $$ I = \iint\limits_{D} (x^{2} + 4y^{2} + 9) \, d\sigma ，其中  D = \{(x, y) \mid x^{2} + y^{2} \leqslant 4\}. $$
+(4) $ I = \iint\limits_{D} (x^{2} + 4y^{2} + 9) \, d\sigma ，其中  D = \{(x, y) \mid x^{2} + y^{2} \leqslant 4\}. $
 
 解 （1）在积分区域 $D$ 上，$0 \leqslant x \leqslant 1$，$0 \leqslant y \leqslant 1$，从而 $0 \leqslant xy(x+y) \leqslant 2$ 又 $D$ 的面积等于 1，因此
 
-$0 \leqslant \iint_{D} xy (x + y) \, d\sigma \leqslant 2.$
+$$0 \leqslant \iint_{D} xy (x + y) \, d\sigma \leqslant 2.$$
 
 (2) 在积分区域 D 上，$0 \leqslant \sin x \leqslant 1, 0 \leqslant \sin y \leqslant 1$，从而 $0 \leqslant \sin^{2} x \sin^{2} y \leqslant 1$，又 D 的面积等于 $\pi^{2}$，因此
 
-$0 \leqslant \iint_{D} \sin^{2} x \sin^{2} y \, d\sigma \leqslant \pi^{2}.$
+$$0 \leqslant \iint_{D} \sin^{2} x \sin^{2} y \, d\sigma \leqslant \pi^{2}.$$
 
 (3) 在积分区域 D 上有 $1 \leqslant x + y + 1 \leqslant 4$，D 的面积等于 2，因此
 
-$2 \leqslant \iint\limits_{D}(x+y+1) \mathrm{d} \sigma \leqslant 8.$
+$$2 \leqslant \iint\limits_{D}(x+y+1) \mathrm{d} \sigma \leqslant 8.$$
 
 (4) 因为在积分区域 D 上有 $0 \leqslant x^{2} + y^{2} \leqslant 4$，所以有
 
-$9 \leqslant x^{2} + 4y^{2} + 9 \leqslant 4(x^{2} + y^{2}) + 9 \leqslant 25.$
+$$9 \leqslant x^{2} + 4y^{2} + 9 \leqslant 4(x^{2} + y^{2}) + 9 \leqslant 25.$$
 
 又 D 的面积等于 $4\pi$，因此
 
-$36\pi\leqslant\iint_{D}(x^{2}+4y^{2}+9)\,d\sigma\leqslant100\pi.$
+$$36\pi\leqslant\iint_{D}(x^{2}+4y^{2}+9)\,d\sigma\leqslant100\pi.$$
 
 ## 习题 10-2 二重积分的计算法
 
diff --git a/markdowns/education_math_00024.md b/markdowns/education_math_00024.md
index 75496da2503bd9c9e0b6f34e3b1b64bfb1ad8f5e..38bf09e86c0ba1d0699606be8d3b29547d4fe708 100644
--- a/markdowns/education_math_00024.md
+++ b/markdowns/education_math_00024.md
@@ -49,7 +49,7 @@
 
 5. 解: 20% 与 30% 的两种酒精溶液按 3:1 的比例混合后得到溶液的浓度为
 
-$\begin{aligned}&(3\times20\%+1\times30\%)\div(3+1)\\&=(60\%+30\%)\div4\\&=22.5\%\end{aligned}$
+$$\begin{aligned}&(3\times20\%+1\times30\%)\div(3+1)\\&=(60\%+30\%)\div4\\&=22.5\%\end{aligned}$$
 
 浓度为 45% 的酒精溶液用量为
 
diff --git a/markdowns/education_math_00032.md b/markdowns/education_math_00032.md
index 77996a9948e8985d9cfb1c03c8efaf2c6612ff67..0183eba9a91fa0ece637dc4d18c30d473bef54a8 100644
--- a/markdowns/education_math_00032.md
+++ b/markdowns/education_math_00032.md
@@ -1,8 +1,8 @@
 (5)$\frac{2}{3}-\frac{1}{2}=$
 
-（6）$\frac{1}{4}+\frac{1}{3}=$
+(6)$\frac{1}{4}+\frac{1}{3}=$
 
-$(7) 8 \div \frac{1}{4} =$
+(7)$ 8 \div \frac{1}{4} =$
 
 (8)$\frac{7}{9}\times\frac{3}{7}$
 
diff --git a/markdowns/education_math_00034.md b/markdowns/education_math_00034.md
index ee35fac09a92541f12ef6b2270e9dd39da3b3438..8a50653283f96c3b1291a05e126d423c4747accf 100644
--- a/markdowns/education_math_00034.md
+++ b/markdowns/education_math_00034.md
@@ -40,5 +40,5 @@
 
 【详解】在甲商店购买:
 
-$40\times120\times78\%$
+$$40\times120\times78\%$$
 
diff --git a/markdowns/education_math_00035.md b/markdowns/education_math_00035.md
index 1eabe498555c8e26d4d64fb85762cb9e98188fcc..be6e370c68a6774db3ffe85e90421c7c6d1be6fb 100644
--- a/markdowns/education_math_00035.md
+++ b/markdowns/education_math_00035.md
@@ -1,4 +1,4 @@
-$=30$
+$$=30$$
 
 $\begin{aligned}&(3)\left(\frac{7}{12}+\frac{3}{8}-\frac{23}{24}\right)\times24\\&=\frac{7}{12}\times24+\frac{3}{8}\times24-\frac{23}{24}\times24\\&=14+9-23\\&=23-23\\&=0\end{aligned}$
 
diff --git a/markdowns/education_math_00062.md b/markdowns/education_math_00062.md
index b7ab109b4e0560bf612ad011fe31bfde3b047ba0..6a348f19621b59c632649712d10fe97d0c543423 100644
--- a/markdowns/education_math_00062.md
+++ b/markdowns/education_math_00062.md
@@ -17,7 +17,7 @@ $$(cv)' = cv' \tag{3.5}$$
 
 公式(3.4) 可以推广到有限多个函数的乘积的情况，即如果 $y = u_{1}u_{2}\cdots u_{n}$，则
 
-$(u_1u_2\cdots u_n)'=u_1'u_2\cdots u_n+u_1u_2'u_3\cdots u_n+\cdots+u_1\cdots u_{n-1}u_n'$
+$$(u_1u_2\cdots u_n)'=u_1'u_2\cdots u_n+u_1u_2'u_3\cdots u_n+\cdots+u_1\cdots u_{n-1}u_n'$$
 
 例2 求函数 $y = (1 + 2x)(3x^3 - 2x^2)$ 的导数.
 
@@ -31,13 +31,13 @@ $= 24x^3 - 3x^2 - 4x$
 
 如果 $u$，$v$ 都是 $x$ 的可导函数，且 $v \neq 0$，则函数 $y = \frac{u}{v}$ 也是 $x$ 的可导函数，并且
 
-$y' = \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$
+$$y' = \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$$
 
 证：由 $\Delta y = \frac{u + \Delta u}{v + \Delta v} - \frac{u}{v} = \frac{v \Delta u - u \Delta v}{v(v + \Delta v)}$
 
 得
 
-$\frac{\Delta y}{\Delta x} = \frac{v \frac{\Delta u}{\Delta x} - u \frac{\Delta v}{\Delta x}}{v(v + \Delta v)}$
+$$\frac{\Delta y}{\Delta x} = \frac{v \frac{\Delta u}{\Delta x} - u \frac{\Delta v}{\Delta x}}{v(v + \Delta v)}$$
 
 因为当 $\Delta x \to 0$ 时，$u$ 与 $v$ 的值不变，而 $\Delta v \to 0$，所以
 
diff --git a/markdowns/education_math_00067.md b/markdowns/education_math_00067.md
index 69e1aa076fb024f39faddbb6b33bd2707220ea10..002819a2c44acac7cdd5a7a648948f57b27457f8 100644
--- a/markdowns/education_math_00067.md
+++ b/markdowns/education_math_00067.md
@@ -2,7 +2,7 @@
 
 傅里叶级数有非常明确的物理含义. 事实上, 在(8.1)式中, 令
 
-$A_{0}=\frac{a_{0}}{2},\quad A_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}},\quad\cos\theta_{n}=\frac{a_{n}}{A_{n}},\quad\sin\theta_{n}=\frac{-b_{n}}{A_{n}}\quad(n=1,2,\cdots),$
+$$A_{0}=\frac{a_{0}}{2},\quad A_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}},\quad\cos\theta_{n}=\frac{a_{n}}{A_{n}},\quad\sin\theta_{n}=\frac{-b_{n}}{A_{n}}\quad(n=1,2,\cdots),$$
 
 则(8.1)式变为
 
@@ -17,9 +17,9 @@ $$\begin{aligned} f_T(t) & = A_0 + \sum_{n=1}^{+\infty} A_n (\cos \theta_n \cos
 
 图 8.1
 
-$c_{0}=A_{0},\arg c_{n}=-\arg c_{-n}=\theta_{n},$
+$$c_{0}=A_{0},\arg c_{n}=-\arg c_{-n}=\theta_{n},$$
 
-$|c_n| = |c_{-n}| = \frac{1}{2} \sqrt{a_n^2 + b_n^2} = \frac{A_n}{2} \quad (n = 1, 2, \cdots).$
+$$|c_n| = |c_{-n}| = \frac{1}{2} \sqrt{a_n^2 + b_n^2} = \frac{A_n}{2} \quad (n = 1, 2, \cdots).$$
 
 因此 $c_n$ 作为一个复数,其模与辐角正好反映了信号 $f_T(t)$ 中频率为 $n\omega_0$ 的简谐波的振幅与相位,其中振幅 $A_n$ 被平均分配到正负频率上,而
 
@@ -27,5 +27,5 @@ $|c_n| = |c_{-n}| = \frac{1}{2} \sqrt{a_n^2 + b_n^2} = \frac{A_n}{2} \quad (n =
 
 例 8.1 求以 $T$ 为周期的函数
 
-$ f_{T}(t)= \begin{cases} 0, & -T/2 < t < 0,\\2,&0< t < T/2 \end{cases}$
+$$f_{T}(t)= \begin{cases} 0, & -T/2 < t < 0,\\2,&0< t < T/2 \end{cases}$$
 
diff --git a/markdowns/education_math_00068.md b/markdowns/education_math_00068.md
index 08dcc9a99290cbaed05dd2cb404dadbba67323cc..06315404c581636b038d9692f4653bf1857763c8 100644
--- a/markdowns/education_math_00068.md
+++ b/markdowns/education_math_00068.md
@@ -1,10 +1,10 @@
-$N_{c} = \int_{c} A_{n} \, \mathrm{d}s$
+$$N_{c} = \int_{c} A_{n} \, \mathrm{d}s$$
 
 称为向量场 $A$ 通过闭或不闭的曲线 $C$ 的流量，其中 $ds$ 是曲线 $C$ 的弧元素，$A_n$ 是向量 $A(x,y)$ 在曲线 $C$ 上点 $(x,y)$ 处法线的正方向上的投影。
 
 设 $dn$ 为向量, 其方向与曲线 C 上点 $(x, y)$ 处的法线一致, 大小等于切线方向向量 $ds = idx + jd y$ 的模, 则有
 
-$dn = \pm (i \, dy - j \, dx)$,
+$$dn = \pm (i \, dy - j \, dx),$$
 
 将这等式右边括号前的两个符号选定一个之后，我们就规定了法线的正方向，以后我们规定选取“+”号。在这种规定下，对于逆时针方向绕行的闭路 C 来说，向量 dn 指向闭路 C 的外法线方向（图 7.2）.
 
@@ -14,29 +14,29 @@ $dn = \pm (i \, dy - j \, dx)$,
 
 由于 $A_{n}\mathrm{d}s=A\cdot\mathrm{d}n=A_{x}(x,y)\mathrm{d}y-A_{y}(x,y)\mathrm{d}x$，所以
 
-$N_{c} = \int_{c} A_{n} \, ds = \int_{c} \mathbf{A} \cdot d\mathbf{n} = \int_{c} A_{x}(x,y) \, dy - A_{y}(x,y) \, dx.$
+$$N_{c} = \int_{c} A_{n} \, ds = \int_{c} \mathbf{A} \cdot d\mathbf{n} = \int_{c} A_{x}(x,y) \, dy - A_{y}(x,y) \, dx.$$
 
 当流入 $C$ 的内部的流体少于流出的流体时, 即 $N_C > 0$ 时, 有源; 当流入 $C$ 的内部的流体多于流出的流体时, 即 $N_C < 0$ 时, 有汇.
 
 假定闭曲线 $C$ 在区域 $D$ 内，并且在 $D$ 内的流体既无源又无汇，即在 $D$ 内任何部分，都无流体放出，也无流体吸入，这时通过 $C$ 的流量为
 
-$N_{C} = \int_{C} A_{x}(x, y) \, dy - A_{y}(x, y) \, dx = 0.$
+$$N_{C} = \int_{C} A_{x}(x, y) \, dy - A_{y}(x, y) \, dx = 0.$$
 
 对于在 $D$ 内的任意简单闭曲线 $C$ 成立.
 
 假定 $D$ 是单连通区域, 且假定 $A_x$ 及 $A_y$ 在 $D$ 内有连续的偏导数. 由微积分所学内容知
 
-$\text{div } \mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} = 0,$
+$$\text{div } \mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} = 0,$$
 
 即
 
-$\frac{\partial A_{x}}{\partial x} = -\frac{\partial A_{y}}{\partial y}. \quad (7.2)$
+$$\frac{\partial A_{x}}{\partial x} = -\frac{\partial A_{y}}{\partial y}. \quad (7.2)$$
 
 从而可知$-A_{y}(x,y)\mathrm{d}x+A_{x}(x,y)\mathrm{d}y$是某一个二元函数$v(x,y)$的全微分，即
 
-$d[v(x,y)] = -A_y dx + A_x dy.$
+$$d[v(x,y)] = -A_y dx + A_x dy.$$
 
 由此得
 
-$\frac{\partial v}{\partial x} = -A_y, \quad \frac{\partial v}{\partial y} = A_x.$
+$$\frac{\partial v}{\partial x} = -A_y, \quad \frac{\partial v}{\partial y} = A_x.$$
 
diff --git a/markdowns/education_math_00069.md b/markdowns/education_math_00069.md
index e311c68a3c47da26632defaa6c264bc88d9530f5..a0c7579e634162cf606f55670422aa0379649203 100644
--- a/markdowns/education_math_00069.md
+++ b/markdowns/education_math_00069.md
@@ -1,6 +1,6 @@
 因此所求映射为
 
-$w = \left( \frac{1 + i e^{-\frac{\pi i}{2z}}}{1 - i e^{-\frac{\pi i}{2z}}} \right)^2.$
+$$w = \left( \frac{1 + i e^{-\frac{\pi i}{2z}}}{1 - i e^{-\frac{\pi i}{2z}}} \right)^2.$$
 
 ## 本章小结
 
diff --git a/markdowns/education_math_00070.md b/markdowns/education_math_00070.md
index ad0c17345f17d317b2a16d954d28aaf05a5fcc58..0c375bf4dd246d07740743dd4346ddaf86b1f140 100644
--- a/markdowns/education_math_00070.md
+++ b/markdowns/education_math_00070.md
@@ -17,11 +17,11 @@
 
 现假设函数 $ w = f(z) $ 在区域 D 内解析，$z_{0} \in D$，且 $ f'(z_{0}) \neq 0 $。采用前面的记号，并由导数的定义可得
 
-$f'(z_0) = \lim_{\Delta z \to 0} \frac{\Delta w}{\Delta z} = \lim_{\Delta z \to 0} \frac{|\Delta w| \, e^{i\varphi}}{|\Delta z| \, e^{i\theta}} = \lim_{\Delta z \to 0} \frac{|\Delta w|}{|\Delta z|} e^{i(\varphi - \theta)}, \quad (6.1)$
+$$f'(z_0) = \lim_{\Delta z \to 0} \frac{\Delta w}{\Delta z} = \lim_{\Delta z \to 0} \frac{|\Delta w| \, e^{i\varphi}}{|\Delta z| \, e^{i\theta}} = \lim_{\Delta z \to 0} \frac{|\Delta w|}{|\Delta z|} e^{i(\varphi - \theta)}, \quad (6.1)$$
 
 因此有
 
-$|f'(z_0)| = \lim_{\Delta z \to 0} \frac{|\Delta w|}{|\Delta z|}.$
+$$|f'(z_0)| = \lim_{\Delta z \to 0} \frac{|\Delta w|}{|\Delta z|}.$$
 
 根据伸缩率的概念可知,导数的模 $|f'(z_0)|$ 实际上就是曲线 $C$ 经函数 $w=f(z)$ 映射后在 $z_0$ 处的伸缩率. 由于函数 $w=f(z)$ 可导,因此 $|f'(z_0)|$ 只与 $z_0$ 有关,而与曲线 $C$ 本身的形状和方向无关,即对经过 $z_0$ 点的任何曲线 $C$,经 $w=f(z)$ 映射后在 $z_0$ 点均有相同的伸缩率. 因此称这种映射具有伸缩率不变性.
 
diff --git a/markdowns/education_math_00071.md b/markdowns/education_math_00071.md
index 6a2e225c39fa3c0447af19e16db9aae1fec13276..a72b56d20344f7a9f85c4caae1a461ce581a1e25 100644
--- a/markdowns/education_math_00071.md
+++ b/markdowns/education_math_00071.md
@@ -10,15 +10,15 @@ $(3)\ \frac{d}{dt}[f_1(t)*f_2(t)] = \frac{d}{dt}f_1(t)*f_2(t) = f_1(t)*\frac{d}{
 
 8.14 设
 
-$f_{1}(t)=\begin{cases}0,&t<0,\\1,&t\geqslant0;\end{cases}$
+$$f_{1}(t)=\begin{cases}0,&t<0,\\1,&t\geqslant0;\end{cases}$$
 
-$f_{2}(t) = \begin{cases} 0, & t < 0, \\ e^{-1}, & t \geq 0, \end{cases}$
+$$f_{2}(t) = \begin{cases} 0, & t < 0, \\ e^{-1}, & t \geq 0, \end{cases}$$
 
 求 $f_1(t) * f_2(t)$.
 
 8.15 设 $F_1(\omega) = \mathscr{F}[f_1(t)]$, $F_2(\omega) = \mathscr{F}[f_2(t)]$, 证明：
 
-$\mathcal{F}\left[f_{1}(t) \cdot f_{2}(t)\right] = \frac{1}{2\pi} F_{1}(\omega) * F_{2}(\omega).$
+$$\mathcal{F}\left[f_{1}(t) \cdot f_{2}(t)\right] = \frac{1}{2\pi} F_{1}(\omega) * F_{2}(\omega).$$
 
 8.16 求下列函数的傅氏变换：
 
diff --git a/markdowns/education_math_00073.md b/markdowns/education_math_00073.md
index 68163e6bf17702cd30e527656ce185c3b073947f..696a48eae0110b646e231a2bf4047b51a5e93209 100644
--- a/markdowns/education_math_00073.md
+++ b/markdowns/education_math_00073.md
@@ -1,13 +1,12 @@
-$X(s) = \frac{F_{0}}{ms^{2}}.$
+$$X(s) = \frac{F_{0}}{ms^{2}}.$$
 
 求拉氏逆变换得物体的运动方程为 $x(t)=\frac{F_0}{m}t$。
 
 注 本例中关于δ函数的拉氏变换涉及拉氏变换本身的定义问题. 一般说来, 若函数$f(t)$在$t=0$附近有界, 则$f(0)$的取值对拉氏变换没有影响, 但若$f(t)$在$t=0$时刻包含了冲激函数, 则有必要考察一下拉氏变换中积分限的设定. 对积分下限分别取$0^+$和$0^-$, 可得下面两种形式的拉氏变换:
 
-$\mathscr{L}_{+}[f(t)] = \int_{0^{+}}^{+\infty} f(t) e^{-st} dt,$ $\quad$
-(9.18)
+$$\mathscr{L}_{+}[f(t)] = \int_{0^{+}}^{+\infty} f(t) e^{-st} dt,\quad(9.18)$$
 
-$\mathscr{L}_{-}[f(t)] = \int_{0^{-}}^{+\infty} f(t) e^{-st} dt.$ （9.19）
+$$\mathscr{L}_{-}[f(t)] = \int_{0^{-}}^{+\infty} f(t) e^{-st} dt.\quad(9.19)$$
 
 对于在 $t=0$ 不含冲激函数的 $f(t)$，有 $\mathscr{L}_+[f(t)] = \mathscr{L}_-[f(t)]$，因此以前的讨论不受影响. 但对于 $\delta$ 函数而言，则有 $\mathscr{L}_+[\delta(t)] = 0$，$\mathscr{L}_-[\delta(t)] = 1$. 考虑到 $\delta$ 函数的傅氏变换为 1，为统一起见，我们推荐使用后一种方式. 此时有关公式要作相应的修改.
 
@@ -23,13 +22,13 @@ $\mathscr{L}_{-}[f(t)] = \int_{0^{-}}^{+\infty} f(t) e^{-st} dt.$ （9.19）
 
 图 9.3
 
-$\begin{aligned} f(t) &= (1 - t)u(t) + (t - 1)u(t - 1) \\ &= u(t) - tu(t) + (t - 1)u(t - 1). \end{aligned}$
+$$\begin{aligned} f(t) &= (1 - t)u(t) + (t - 1)u(t - 1) \\ &= u(t) - tu(t) + (t - 1)u(t - 1). \end{aligned}$$
 
 由于
 
-$\mathscr{L}\left[ u(t) \right] = \frac{1}{s}, \quad \mathscr{L}\left[ tu(t) \right] = \frac{1}{s^2},$
+$$\mathscr{L}\left[ u(t) \right] = \frac{1}{s}, \quad \mathscr{L}\left[ tu(t) \right] = \frac{1}{s^2},$$
 
 所以
 
-$\mathscr{L}\left[f(t)\right]=\frac{1}{s}-\frac{1}{s^{2}}+\frac{1}{s^{2}}e^{-s}.$
+$$\mathscr{L}\left[f(t)\right]=\frac{1}{s}-\frac{1}{s^{2}}+\frac{1}{s^{2}}e^{-s}.$$
 
diff --git a/markdowns/education_math_00074.md b/markdowns/education_math_00074.md
index cd87602af4b000b072744ee6693bb1f7b2be3343..d21e6a78c1f48a0702618e537d653a7bd02916b7 100644
--- a/markdowns/education_math_00074.md
+++ b/markdowns/education_math_00074.md
@@ -1,30 +1,37 @@
-$\begin{aligned} F(\omega) &= \int_{-\infty}^{+\infty} f(t) e^{-j\omega t} \, dt = \int_{-\infty}^{+\infty} e^{-|t|} \cos t e^{-j\omega t} \, dt \\&= \int_{0}^{+\infty} e^{-t} \cos t e^{-j\omega t} \, dt + \int_{-\infty}^{0} e^{t} \cos t e^{-j\omega t} \, dt. \end{aligned}$
+$$\begin{aligned} F(\omega) &= \int_{-\infty}^{+\infty} f(t) e^{-j\omega t} \, dt = \int_{-\infty}^{+\infty} e^{-|t|} \cos t e^{-j\omega t} \, dt \\&= \int_{0}^{+\infty} e^{-t} \cos t e^{-j\omega t} \, dt + \int_{-\infty}^{0} e^{t} \cos t e^{-j\omega t} \, dt. \end{aligned}$$
 
 对上式第二个积分作变量代换 $t_1 = -t$，有
 
-$$F ( \omega ) = \int _ { 0 } ^ { + \infty } \mathrm { e } ^ { - t } \cos t \, \mathrm { e } ^ { - \mathrm { j } \omega t } \mathrm { d } t + \int _ { 0 } ^ { + \infty } \mathrm { e } ^ { - t _ { 1 } } \cos t _ { 1 } \, \mathrm { e } ^ { - \mathrm { j } ( - \omega ) t _ { 1 } } \mathrm { d } t _ { 1 }$$
-$$= \int _ { - \infty } ^ { + \infty } \mathrm { e } ^ { - t } u ( t ) \cos t \, \mathrm { e } ^ { - \mathrm { j } \omega t } \mathrm { d } t + \int _ { - \infty } ^ { + \infty } \mathrm { e } ^ { - t } u ( t ) \cos t \, \mathrm { e } ^ { \mathrm { j } \omega t } \mathrm { d } t .$$
+$$
+\begin{aligned}
+F(\omega)
+&= \int_{0}^{+\infty} e^{-t}\cos t \, e^{-j\omega t}\,\mathrm{d}t
+ + \int_{0}^{+\infty} e^{-t_1}\cos t_1 \, e^{-j(-\omega)t_1}\,\mathrm{d}t_1 \\
+&= \int_{-\infty}^{+\infty} e^{-t}u(t)\cos t \, e^{-j\omega t}\,\mathrm{d}t
+ + \int_{-\infty}^{+\infty} e^{-t}u(t)\cos t \, e^{j\omega t}\,\mathrm{d}t.
+\end{aligned}
+$$
 
 令 $f_{1}(t)=e^{-t}u(t)\cos t$，$F_{1}(\omega)=\mathscr{F}[f_{1}(t)]$，则
 
-$F(\omega) = F_1(\omega) + \overline{F_1(\omega)} = 2\mathrm{Re}\, F_1(\omega).$
+$$F(\omega) = F_1(\omega) + \overline{F_1(\omega)} = 2\mathrm{Re}\, F_1(\omega).$$
 
 由例 8.16 可知
 
-$F_{1}(\omega) = \frac{1 + j\omega}{(1 + j\omega)^{2} + 1}$ ,
+$$F_{1}(\omega) = \frac{1 + j\omega}{(1 + j\omega)^{2} + 1},$$
 
 因此，$F(\omega)=2\mathrm{Re}$  $F_{1}(\omega)=2\frac{\omega^{2}+2}{\omega^{4}+4}$。对 $F(\omega)$ 求逆变换有
 
-$\begin{aligned}\mathscr{F}^{-1}\left[F(\omega)\right] &=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} 2 \frac{\omega^{2}+2}{\omega^{4}+4} e^{\mathrm{j} \omega t} \mathrm{~d} \omega \\ 
-& =\frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{\omega^{2}+2}{\omega^{4}+4}(\cos \omega t+j \sin \omega t) \mathrm{d} \omega \\ &=\frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{\omega^{2}+2}{\omega^{4}+4} \cos \omega t \mathrm{~d} \omega=f(t). \end{aligned}$
+$$\begin{aligned}\mathscr{F}^{-1}\left[F(\omega)\right] &=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} 2 \frac{\omega^{2}+2}{\omega^{4}+4} e^{\mathrm{j} \omega t} \mathrm{~d} \omega \\ 
+& =\frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{\omega^{2}+2}{\omega^{4}+4}(\cos \omega t+j \sin \omega t) \mathrm{d} \omega \\ &=\frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{\omega^{2}+2}{\omega^{4}+4} \cos \omega t \mathrm{~d} \omega=f(t). \end{aligned}$$
 
 即得
 
-$\frac{2}{\pi}\int_{0}^{+\infty}\frac{\omega^{2}+2}{\omega^{4}+4}\cos\omega t\,d\omega=e^{-|t|}\cos t.$
+$$\frac{2}{\pi}\int_{0}^{+\infty}\frac{\omega^{2}+2}{\omega^{4}+4}\cos\omega t\,d\omega=e^{-|t|}\cos t.$$
 
 例 8.20 已知 $\int_{-\infty}^{+\infty} e^{-t^2} dt = \sqrt{\pi}$，求 $f(t) = e^{-t^2}$ 的傅氏变换.
 
 解 设 $ F(\omega) = \mathcal{F}[f(t)] $，则
 
-$F(2\omega)=\int_{-\infty}^{+\infty}e^{-t^{2}}e^{-2j\omega t}\,dt=e^{-\omega^{2}}\int_{-\infty}^{+\infty}e^{-(t+j\omega)^{2}}dt,$
+$$F(2\omega)=\int_{-\infty}^{+\infty}e^{-t^{2}}e^{-2j\omega t}\,dt=e^{-\omega^{2}}\int_{-\infty}^{+\infty}e^{-(t+j\omega)^{2}}dt,$$
 
diff --git a/markdowns/education_math_00076.md b/markdowns/education_math_00076.md
index 25a75effa8aba8078b7b653dc66d60d37d648aef..b1a56698ec98aac388438470fc1b7dda69e40060 100644
--- a/markdowns/education_math_00076.md
+++ b/markdowns/education_math_00076.md
@@ -4,18 +4,32 @@
 
 图 4.2
 
-$f(z) = \frac{1}{2\pi i} \oint_{c} \frac{f(\zeta)}{\zeta - z} d\zeta.$ $\quad$(4.5)
+$$f(z) = \frac{1}{2\pi i} \oint_{c} \frac{f(\zeta)}{\zeta - z} d\zeta.\quad(4.5)$$
 
 由于$\zeta$在$C$上，而$z$在$C$内，所以$\left|\frac{z-z_0}{\zeta-z_0}\right|< 1$，根据例4.1，就有展开式
 
-$\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}}$
+$$
+\begin{aligned}
+\frac{1}{\zeta-z}
+&= \frac{1}{(\zeta-z_{0})-(z-z_{0})}
+= \frac{1}{\zeta-z_{0}} \cdot \frac{1}{1-\frac{z-z_{0}}{\zeta-z_{0}}} \\
+&= \frac{1}{\zeta-z_{0}}
+\left[
+1+\frac{z-z_{0}}{\zeta-z_{0}}
++\left(\frac{z-z_{0}}{\zeta-z_{0}}\right)^{2}
++\cdots
++\left(\frac{z-z_{0}}{\zeta-z_{0}}\right)^{n}
++\cdots
+\right] \\
+&= \sum_{n=0}^{\infty} \frac{(z-z_{0})^{n}}{(\zeta-z_{0})^{n+1}} \, .
+\end{aligned}
+$$
+
 
-$$= \frac { 1 } { \zeta - z _ { 0 } } \left[ 1 + \frac { z - z _ { 0 } } { \zeta - z _ { 0 } } + \left( \frac { z - z _ { 0 } } { \zeta - z _ { 0 } } \right) ^ { 2 } + \cdots + \left( \frac { z - z _ { 0 } } { \zeta - z _ { 0 } } \right) ^ { n } + \cdots \right]$$
-$$= \sum _ { n = 0 } ^ { \infty } \frac { 1 } { ( \zeta - z _ { 0 } ) ^ { n + 1 } } ( z - z _ { 0 } ) ^ { n } ,$$
 
 将上式代入(4.5)式,有
 
-$f(z)=\frac{1}{2 \pi i} \oint_{C}\left[\sum_{n=0}^{N-1} \frac{f(\zeta) \mathrm{d} \zeta}{\left(\zeta-z_{0}\right)^{n+1}}\right]\left(z-z_{0}\right)^{n}+R_{N}(z),$ $\quad(4.6)$
+$$f(z)=\frac{1}{2 \pi i} \oint_{C}\left[\sum_{n=0}^{N-1} \frac{f(\zeta) \mathrm{d} \zeta}{\left(\zeta-z_{0}\right)^{n+1}}\right]\left(z-z_{0}\right)^{n}+R_{N}(z),\quad(4.6)$$
 
 其中
 
@@ -25,7 +39,7 @@ $$R _ { N } ( z ) = \frac { 1 } { 2 \pi \, \mathrm { i } } \oint _ { C } \left[
 
 由于$f(z)$在$D$内解析,从而在$C$上连续,因此,存在一个正常数$M$,在$C$上$\mid f(\zeta)\mid\leqslant M$,又由于
 
-$\left| \frac{z - z_0}{\zeta - z_0} \right| = \frac{|z - z_0|}{r} = q < 1,$
+$$\left| \frac{z - z_0}{\zeta - z_0} \right| = \frac{|z - z_0|}{r} = q < 1,$$
 
 于是有
 
diff --git a/markdowns/education_math_00077.md b/markdowns/education_math_00077.md
index 38e27d489e80bbaf2e0ce98536dade9ea6538a8c..c0ad9595ba52350cf0f3df4f73b7895e10047b9b 100644
--- a/markdowns/education_math_00077.md
+++ b/markdowns/education_math_00077.md
@@ -1,18 +1,27 @@
-$d(u,v):=a(u,D^{\gamma}\partial v)-a_{P}(D^{\gamma}u,v)$
-$=\sum_{\substack{|\alpha|\leq m\\ |\beta|\leq m-2}}\int_{\mathbb{R}^{n}}a_{\alpha\beta}(D^{\alpha}u)(D^{\beta+\gamma}\partial v)\,dx$
-$\quad-\sum_{\substack{|\alpha|\leq m\\ |\beta|\leq m-1}}\int_{\mathbb{R}^{n}}[D^{\gamma}(a_{\alpha\beta}D^{\alpha}u)](D^{\beta}\partial v)\,dx$
-$+\sum_{\substack{|\alpha|\leq m\\ |\beta|=m}}\int_{\mathbb{R}^{n}}[(D^{\gamma}\partial a_{\alpha\beta})(D^{\alpha}u(x-\tfrac{h}{2}\mathbf{e}_{i})) \\$
-$\quad+( \partial a_{\alpha\beta}) ( D^{\alpha+\gamma}u(x-\tfrac{h}{2}\mathbf{e}_{i}))\\$
-$\quad+(D^{\gamma}a_{\alpha\beta}(x+\tfrac{h}{2}\mathbf{e}_{i})D^{\alpha}\partial u)\\$
-$\quad+(a_{\alpha\beta}(x+\tfrac{h}{2}\mathbf{e}_{i})-a_{\alpha\beta}(x))D^{\alpha+\gamma}\partial u](D^{\beta}v)\,dx.$
+$$
+\begin{aligned}
+d(u,v) &:= a(u,D^{\gamma}\partial v)-a_{P}(D^{\gamma}u,v) \\
+&= \sum_{\substack{|\alpha|\leq m\\ |\beta|\leq m-2}}
+\int_{\mathbb{R}^{n}} a_{\alpha\beta}(D^{\alpha}u)(D^{\beta+\gamma}\partial v)\,dx \\
+&\quad - \sum_{\substack{|\alpha|\leq m\\ |\beta|\leq m-1}}
+\int_{\mathbb{R}^{n}} \bigl[D^{\gamma}(a_{\alpha\beta}D^{\alpha}u)\bigr](D^{\beta}\partial v)\,dx \\
+&\quad + \sum_{\substack{|\alpha|\leq m\\ |\beta|=m}}
+\int_{\mathbb{R}^{n}} \Bigl[
+(D^{\gamma}\partial a_{\alpha\beta})(D^{\alpha}u(x-\tfrac{h}{2}\mathbf{e}_{i})) \\
+&\qquad\qquad + (\partial a_{\alpha\beta})(D^{\alpha+\gamma}u(x-\tfrac{h}{2}\mathbf{e}_{i})) \\
+&\qquad\qquad + (D^{\gamma}a_{\alpha\beta}(x+\tfrac{h}{2}\mathbf{e}_{i})D^{\alpha}\partial u) \\
+&\qquad\qquad + (a_{\alpha\beta}(x+\tfrac{h}{2}\mathbf{e}_{i})-a_{\alpha\beta}(x))D^{\alpha+\gamma}\partial u
+\Bigr](D^{\beta}v)\,dx .
+\end{aligned}
+$$
 
 As in the first part of the proof, from $|d(u,v)|\leq C_{d}|u|_{m+1}|v|_{m}$ and
 
-$|(f,D^{\gamma}\partial v)_{0}|\leq|f|_{2-m}|D^{\gamma}\partial v|_{m-2}\leq|f|_{2-m}|v|_{m}$
+$$|(f,D^{\gamma}\partial v)_{0}|\leq|f|_{2-m}|D^{\gamma}\partial v|_{m-2}\leq|f|_{2-m}|v|_{m}$$
 
 follows
 
-$|D^{\gamma}\partial_{h,i}u|_m \leq [|f|_{-m+2} + (C_d + C_K)|u|_{m+1}]/C_E$
+$$|D^{\gamma}\partial_{h,i}u|_m \leq [|f|_{-m+2} + (C_d + C_K)|u|_{m+1}]/C_E$$
 
 for all $|\gamma|=1$, h>0, $1\leq i\leq n$. Thus, the differences $|\partial_{h,i}u|_{m+1}$ are uniformly bounded so that Lemma 6.2.24 proves $u\in H^{m+2}(\mathbb{R}^{n})$ and $|u|_{m+2}\leq C||f|_{-m+2}+|u|_{m+1}$. If one uses the estimate $|u|_{m+1}\leq C||f|_{-m+1}+|u|_{m}\leq C||f|_{-m+2}+|u|_{m}$ proved in (a), (8) follows for k=2.
 
@@ -22,7 +31,7 @@ Corollary 9.1.4. The coercivity in Theorem 3 can be replaced (a) by the sufficie
 
 Corollary 9.1.5. If, in addition, $a(\cdot,\cdot)$ is $H^m(\mathbb{R}^n)$-elliptic or if $a(\cdot,\cdot)$ satisfies condition (6.5.4a,b), then in Theorem 3 the estimate (8) should be replaced by
 
-$|u|_{m+k} \leq C_k |f|_{-m+k}. \quad (9.1.11)$
+$$|u|_{m+k} \leq C_k |f|_{-m+k}. \quad (9.1.11)$$
 
 PROOF. By Remark 2.
 
diff --git a/markdowns/education_math_00078.md b/markdowns/education_math_00078.md
index 1e4f59090fca6f3ef9b44ccc18c64e9f868f429b..432f87a40eb41e0e3df2ae363c8e67438f52a347 100644
--- a/markdowns/education_math_00078.md
+++ b/markdowns/education_math_00078.md
@@ -1,16 +1,18 @@
 Since (11) holds for all $ f \in H^{-m}(\Omega) $ and $ u = L^{-1}f \left( \operatorname{cf.}(9') \right) $, inequality (11) is equivalent to
 
-$\|L^{-1}\|_{H_{0}^{m}(\Omega)\to H^{-m}(\Omega)}\leq C:=1/C_{E}.\quad(7.2.11')$
+$$\|L^{-1}\|_{H_{0}^{m}(\Omega)\to H^{-m}(\Omega)}\leq C:=1/C_{E}.\quad(7.2.11')$$
 
 The term “variational problem” for (9) goes back to the following statement, inferred from Theorem 6.5.12:
 
 Theorem 7.2.9. Let $a(\cdot,\cdot)$ be an $H_0^m(\Omega)$-elliptic and symmetric bilinear form. Then (9) is equivalent to the variational problem
 
-find $u \in H_0^m(\Omega)$, such that $J(u) \leq J(v)$ for all $v \in H_0^m(\Omega)$, $\quad (7.2.12a)$
-
+$$
+\text{find } u \in H_0^m(\Omega), \text{ such that } J(u) \leq J(v)
+\text{ for all } v \in H_0^m(\Omega). \qquad (7.2.12a)
+$$
 where
 
-$J(v) := \frac{1}{2} a(v, v) - f(v). \quad (7.2.12b)$
+$$J(v) := \frac{1}{2} a(v, v) - f(v). \quad (7.2.12b)$$
 
 Attention. If $a(\cdot,\cdot)$ is either not $H_0^m(\Omega)$-elliptic or not symmetric, Problem (9) remains meaningful although the solution does not minimize the functional $J(u)$.
 
@@ -25,7 +27,7 @@ PROOF. We write $L$ as $L = L_I + L_{II}$, with $L_I$ satisfying the conditions
 
 Lemma 7.2.12. Let $a(\cdot,\cdot)=a'(\cdot,\cdot)+a''(\cdot,\cdot)$ be decomposed such that $a'(\cdot,\cdot)$ is $H_{0}^{m}(\Omega)$-elliptic, or perhaps only $H_{0}^{m}(\Omega)$-coercive, while
 
-$a''(u,u)=\sum\limits_{\substack{|\alpha|,|\beta|\leq m\\|\alpha|+|\beta|<2m}}\int_{\Omega}a_{\alpha\beta}D^{\alpha}uD^{\beta}u\,dx$
+$$a''(u,u)=\sum\limits_{\substack{|\alpha|,|\beta|\leq m\\|\alpha|+|\beta|<2m}}\int_{\Omega}a_{\alpha\beta}D^{\alpha}uD^{\beta}u\,dx$$
 
 with $a_{\alpha\beta} \in L^{\infty}(\Omega)$ contains only derivatives of order $\leq 2m - 1$. Then $a(\cdot, \cdot)$ is also $H_{0}^{m}(\Omega)$-coercive.
 
diff --git a/markdowns/education_math_00079.md b/markdowns/education_math_00079.md
index 7e8a7ef42e8f3ed2be514d019e6a8ffcb3278aa9..d11d3c947429f520d5a9f7a46005524782cfd916 100644
--- a/markdowns/education_math_00079.md
+++ b/markdowns/education_math_00079.md
@@ -8,23 +8,23 @@ The matrix $L_h$ is singular because the system $L_h u_h = q_h$, like the contin
 
 Theorem 4.7.3. The system of equations (5a) is solvable if and only if
 
-$-h^2 \sum_{x \in \Omega_h} f(x) = h \sum_{x \in \Gamma'_h} \varphi(x). \quad (4.7.6)$
+$$-h^2 \sum_{x \in \Omega_h} f(x) = h \sum_{x \in \Gamma'_h} \varphi(x). \quad (4.7.6)$$
 
 Any two solutions of (5a) can only differ by a constant: $u_h^1 - u_h^2 = c\mathbf{1}, c \in \mathbb{R}$.
 
 PROOF. Evidently, $L_{h}1=0$ holds, i.e., $\mathbf{1}\in\operatorname{kernel}(L_{h})$. Furthermore, Theorem 4 will then imply $\dim(\operatorname{kernel}(L_{h}))=1$. This proves
 
-$kernel(L_h) = \{c1: c \in R\} \quad (4.7.7)$
+$$kernel(L_h) = \{c1: c \in R\} \quad (4.7.7)$$
 
 and thus the second part of the assertion. (5a) is solvable if and only if the scalar product $\langle v, q_{h}\rangle$ vanishes for all $v \in$ kernel $(L_{h}^{\top}) = \text{range}(L_{h})^{\perp}$. Because of $L_{h}^{T} = L_{h}$ and (7)
 
-$\langle 1, q_h \rangle = 0 \quad \text{or} \quad \sum_{x \in \Omega_h} q_h(x) = 0 \quad (4.7.6')$
+$$\langle 1, q_h \rangle = 0 \quad \text{or} \quad \sum_{x \in \Omega_h} q_h(x) = 0 \quad (4.7.6')$$
 
 is sufficient and necessary. According to Definition (5c), (6) and (6′) agree.
 
 Let condition (6) be satisfied. System (5a) can be solved as follows. Select an arbitrary $x_0 \in \Omega_h$ and normalise the solution $u_h$ (determined except for one constant) by
 
-$u_{h}(x_{0})=0.\quad(4.7.8)$
+$$u_{h}(x_{0})=0.\quad(4.7.8)$$
 
 Let $\hat{u}_h$ be the vector $u_h$ without the component $u_h(x_0)$. Let $\hat{L}_h$ be the submatrix of $L_h$ in which the row and column with index $x_0$ have been left out. Let $\hat{q}_h$ be constructed likewise. Then
 
diff --git a/markdowns/education_math_00080.md b/markdowns/education_math_00080.md
index 7c8afdf16b68a78d42441db09dd617c792922385..323d06aab3e02ff02c6ebf5e1a33767b5cb0a738 100644
--- a/markdowns/education_math_00080.md
+++ b/markdowns/education_math_00080.md
@@ -2,27 +2,27 @@
 
 This section concerns the question as to whether $\lambda_h \to \lambda$ and $e^h \to e$ for $h \to 0$. The rate of convergence will be discussed in Section 11.2.3. The basic assumptions are the following:
 
-$Let  a(\cdot,\cdot):  V \times V \to R  be  V-coercive, \quad (11.2.2a)$
+$$Let  a(\cdot,\cdot):  V \times V \to R  be  V-coercive, \quad (11.2.2a)$$
 
 Let $V \subset L^2(\Omega)$ be continuously, densely, and compactly embedded,  
   (11.2.2b)
 
 so that the Riesz-Schauder theory is applicable (Theorem 11.1.2). Furthermore, let a sequence of subspaces $V_{h_i}$ ($h_i \to 0$) be given which increasingly approximate $V$ (cf. (8.2.4a)):
 
-$\lim_{h \to 0} \inf\{\|u - u^h\|_V : u^h \in V_h\} = 0 \quad \text{for all } u \in V.$ $\quad (11.2.2c)$
+$$\lim_{h \to 0} \inf\{\|u - u^h\|_V : u^h \in V_h\} = 0 \quad \text{for all } u \in V.\quad (11.2.2c)$$
 
 We define
 
-$a_{\lambda}(\cdot,\cdot):V\times V\to C,\quad a_{\lambda}(u,v):=a(u,v)-\lambda(u,v)_{0},\quad(11.2.3a)$
+$$a_{\lambda}(\cdot,\cdot):V\times V\to C,\quad a_{\lambda}(u,v):=a(u,v)-\lambda(u,v)_{0},\quad(11.2.3a)$$
 
-$\omega(\lambda) := \inf_{\substack{u \in V \\ \|u\|_V = 1}} \sup_{\substack{v \in V \\ \|v\|_V = 1}} |a_\lambda(u, v)|, \quad (11.2.3b)$
+$$\omega(\lambda) := \inf_{\substack{u \in V \\ \|u\|_V = 1}} \sup_{\substack{v \in V \\ \|v\|_V = 1}} |a_\lambda(u, v)|, \quad (11.2.3b)$$
 
 $$\omega_h(\lambda) := \inf_{\substack{u \in V_h \\ \|u\|_V = 1}} \sup_{\substack{v \in V_h \\ \|v\|_V = 1}} |a_\lambda(u, v)|. \tag{11.2.3c}$$
 
 Exercise 11.2.4. Let L and $L_{h}$ be the operators associated with $a(\cdot,\cdot):V\times V\to\mathbb{R}$ and $a(\cdot,\cdot):V_{h}\times V_{h}\to\mathbb{R}$ (cf. (8.1.10a)). Show that  
 (i) If $\lambda$ is not an eigenvalue we have
 
-$\omega(\lambda)=\|(L-\lambda I)^{-1}\|_{V\leftarrow V^{\prime}}^{-1},\quad\omega_{h}(\lambda)=\|(L_{h}-\lambda I)^{-1}\|_{V_{h}\leftarrow V_{h}^{\prime}}^{-1}\qquad(11.2.4)$
+$$\omega(\lambda)=\|(L-\lambda I)^{-1}\|_{V\leftarrow V^{\prime}}^{-1},\quad\omega_{h}(\lambda)=\|(L_{h}-\lambda I)^{-1}\|_{V_{h}\leftarrow V_{h}^{\prime}}^{-1}\qquad(11.2.4)$$
 
 (cf. Lemma 6.5.3 and Exercise 8.1.16).
 
diff --git a/markdowns/education_math_00081.md b/markdowns/education_math_00081.md
index 598055f20f94ab84ebb96011113767a7c1ea6671..b1cd92b71ad13d341e6571b87347a28bac923fd2 100644
--- a/markdowns/education_math_00081.md
+++ b/markdowns/education_math_00081.md
@@ -10,12 +10,18 @@ All the spaces of finite elements, $V_h$, constructed so far are useless for equ
 
 As a model problem we introduce the one-dimensional biharmonic equation
 
-$u''''(x) = g(x)$ for $0 < x < 1$, $u(0) = u'(0) = u(1) = u'(1) = 0$
+$$u''''(x) = g(x)\quad for\quad0 < x < 1, u(0) = u'(0) = u(1) = u'(1) = 0$$
 
 which becomes in its weak formulation
 
-$a(u,v)=f(v)$ for all $v\in H_0^2(0,1)$, where  
-$a(u,v)=\int_0^1 u''v''\,dx,\quad f(v)=\int_0^1 gv\,dx.$ $\quad (8.5.4)$
+$$
+a(u,v)=f(v)\quad \text{for all } v\in H_0^2(0,1).
+$$
+
+$$
+a(u,v)=\int_0^1 u''v''\,dx,\qquad
+f(v)=\int_0^1 gv\,dx. \qquad (8.5.4)
+$$
 
 Divide the interval $\Omega=(0,1)$ into equal subintervals of length h. Piecewise linear functions (cf. Figure 8.3.1) can be viewed as linear spline functions, so that it is natural to define $V_{h}$ as the space of cubic splines (with $u=u'=0$ for x=0 and x=1) (cf. Stoer [1, §2.4] and Stoer-Bulirsch [1, §2.4]). We can take as basis functions B-splines, whose supports in general consist of four subintervals (cf. Figure 1). Since cubic spline functions belong to $C^{2}(0,1)$, they are not just in $H_{0}^{2}(0,1)$, but even in $H^{3}(0,1)\cap H_{0}^{2}(0,1)$.
 
@@ -25,7 +31,11 @@ Figure 8.5.1. B-spline
 
 Simpler yet than working with spline functions is to use cubic Hermite interpolation:
 
-$V_{h}:=\{u\in C^{1}(0,1)\colon u \text{ cubic on each subinterval}$,
+$$
+V_h := \{\, u \in C^1(0,1) \colon u \text{ is cubic on each subinterval},
+$$
 
-$u(0) = u'(0) = u(1) = u'(1) = 0$. $\quad$ (8.5.5)
+$$
+u(0)=u'(0)=u(1)=u'(1)=0 \,\}. \qquad (8.5.5)
+$$
 
diff --git a/markdowns/education_math_00083.md b/markdowns/education_math_00083.md
index 43ffb702341a3ec4ecf27b3818a7ef6689257c88..64827910b0a3afbc89e94b9f8cd98cf570a84e6c 100644
--- a/markdowns/education_math_00083.md
+++ b/markdowns/education_math_00083.md
@@ -1,4 +1,4 @@
-$\|(I-\hat{S}_h)^*\|_{H_*^{2-t}(\Omega)'\leftarrow V'} = \|I-\hat{S}_h\|_{V\leftarrow H_*^{2-t}(\Omega)} \leq C_\beta h^{1-t}.$
+$$\|(I-\hat{S}_h)^*\|_{H_*^{2-t}(\Omega)'\leftarrow V'} = \|I-\hat{S}_h\|_{V\leftarrow H_*^{2-t}(\Omega)} \leq C_\beta h^{1-t}.$$
 
 As in (19d) one shows that $\|I-S_h\|_{H^t(\Omega)\leftarrow V}\leq Ch^{1-t}$. Combining this with $\|I-S_h\|_{V\leftarrow H^s(\Omega)\cap V}\leq Ch^{s-1}$ (cf. (11$^\prime$)) there follows (21), and thus (20). $\blacksquare$
 
@@ -12,19 +12,19 @@ The estimates for the errors for linear functions on triangular elements proved
 
 For quadratic elements (cf. Section 8.3.4) one expects a correspondingly improved order of convergence. In general one can show the following: If the ansatz function is, in each $T_{i} \in \tau$, a polynomial of degree $k \geq 1$ (i.e., $ u(\mathbf{x}) = \sum_{|\nu| \leq k} a_{\nu} \mathbf{x}^{\nu} $ on $T_{i}$), then
 
-$d(u,V_h)=\inf\{|u-v|_1:v\in V_h\}\le Ch^k|u|_{k+1}\text{ for all }u\in H^{k+1}(\Omega)\cap V\quad(8.5.1)$
+$$d(u,V_h)=\inf\{|u-v|_1:v\in V_h\}\le Ch^k|u|_{k+1}\text{ for all }u\in H^{k+1}(\Omega)\cap V\quad(8.5.1)$$
 
 (cf. Ciarlet [1, Theorem 3.2.1]). If one uses parallelograms as a basis and uses ansatz functions that are polynomials of degree at least k, then (1) also holds. For example, biquadratic elements and quadratic ansatz functions of the serendipity class fulfil this requirement for k=2. The result corresponding to Theorem 8.4.6 follows from Theorem 8.2.1:
 
 Theorem 8.5.1. Let $V_h$ fulfil (1), and let the bilinear form satisfy (1.2) and (1.12) with $\epsilon_n := \epsilon_h \ge \tilde{\epsilon} > 0$. Assume problem (1.1) has a solution $u \in V \cap H^{k+1}(\Omega)$. Then the finite-element solution $u^h \in V_h$ satisfies the inequality
 
-$|u - u^h|_1 \leq Ch^k |u|_{k+1}.$ (8.5.2)
+$$|u - u^h|_1 \leq Ch^k |u|_{k+1}. (8.5.2)$$
 
 The Ritz projection satisfies $\|I-S_{h}\|_{V\leftarrow H^{k+1}(\Omega)\cap V}\leq Ch^{k}$
 
 The estimate (4.9) now holds for $s \in [1, k]$ if $u \in H^s(\Omega) \cap V$. With suitable regularity conditions, there are, as in Theorem 8.4.11, the error estimates
 
-$|u - u^h|_0 \leq Ch^{k+1}|u|_{k+1}, \quad \text{(8.5.3a)}$
+$$|u - u^h|_0 \leq Ch^{k+1}|u|_{k+1}, \quad \text{(8.5.3a)}$$
 
-$||u-u^{h}||_{H^{k-1}(\Omega^{\prime})}\leq C h^{2k}|u|_{k+1}. \quad \text{(8.5.3b)}$
+$$||u-u^{h}||_{H^{k-1}(\Omega^{\prime})}\leq C h^{2k}|u|_{k+1}. \quad \text{(8.5.3b)}$$
 
diff --git a/markdowns/education_math_00084.md b/markdowns/education_math_00084.md
index be5ea003cb8079c6be191599b4edcf59fac0cf8f..10ba9be9fc33ea50dcfdaa310a58e1e268b8fdac 100644
--- a/markdowns/education_math_00084.md
+++ b/markdowns/education_math_00084.md
@@ -6,16 +6,22 @@
 
 (bc) According to Remark 14, for each $\epsilon > 0$ there exists a $\tilde{u}_\epsilon \in C^\infty(\mathbb{R}_+)$ with finite support such that $|u - \tilde{u}_\epsilon|_1 \leq \eta/2$. For each $\eta > 0$ there exists $\varphi_\eta \in C^\infty(\mathbb{R}_+)$ with
 
-$\varphi_{\eta}(x)=0\quad\text{for }x\leq\eta/2,\quad\varphi_{\eta}(x)=1\quad\text{for }x\geq\eta,$
+$$\varphi_{\eta}(x)=0\quad\text{for }x\leq\eta/2,\quad\varphi_{\eta}(x)=1\quad\text{for }x\geq\eta,$$
 
-$|\varphi_{\eta}'(x)| \leq 3/\eta, \quad 0 \leq \varphi_{\eta}(x) \leq 1.$
+$$|\varphi_{\eta}'(x)| \leq 3/\eta, \quad 0 \leq \varphi_{\eta}(x) \leq 1.$$
 
 The function $ u_{\epsilon} := \tilde{u}_{\epsilon} - (1 - \varphi_{1})\tilde{u}_{\epsilon}(0) $ satisfies $ u_{\epsilon} \in C^{\infty}(\mathbb{R}_{+}) $, $ u_{\epsilon}(0) = 0 $, $ \text{supp}(u_{\epsilon}) $ finite, and $ |u - u_{\epsilon}|_{1} \leq C \epsilon $. Thus $ X := \{ v \in C^{\infty}(\mathbb{R}_{+}) : v(0) = 0 $, $ \text{supp}(v) $ finite is dense in $ \{ u \in H^{1}(\mathbb{R}_{+}) : u(0) = 0 \} $.
 
 (bd) The statement would be proved if $C_0^\infty(\mathbb{R}_+)$ were also dense in $X$ with respect to $|\cdot|_1$. Let $v\in X$. Evidently, $v_\eta:=\varphi_\eta v\in C_0^\infty(\mathbb{R}_+)$ holds for all $\eta>0$. Set $\Theta(\eta):=\|v'\|_{L^2(0,\eta)}$ and note that $\Theta(\eta)\to 0$ for $\eta\to 0$. Since $v(x)=v_\eta(x)$ for $x\geq \eta$, it remains to estimate the variables $\|v-v_\eta\|_{L^2(0,\eta)}$ and $\|v'-v_\eta'\|_{L^2(0,\eta)}$. Because of $v_\eta'-v'=\varphi_\eta'v+(\varphi_\eta-1)v'$, one obtains
 
-$\|v_{\eta}^{\prime}-v^{\prime}\|_{L^{2}(0,\eta)}\leq\|\varphi_{n}^{\prime}\|_{L^{\infty}(0,\eta)}\|v\|_{L^{2}(0,\eta)}+\|\varphi_{n}-1\|_{L^{\infty}(0,\eta)}\|v^{\prime}\|_{L^{2}(0,\eta)}$
-$\leq(3/\eta)\|v\|_{L^{2}(0,\eta)}+\Theta(\eta).$
+$$
+\begin{aligned}
+\|v_{\eta}^{\prime}-v^{\prime}\|_{L^{2}(0,\eta)}
+&\leq \|\varphi_{n}^{\prime}\|_{L^{\infty}(0,\eta)}\|v\|_{L^{2}(0,\eta)}
++\|\varphi_{n}-1\|_{L^{\infty}(0,\eta)}\|v^{\prime}\|_{L^{2}(0,\eta)} \\
+&\leq \frac{3}{\eta}\|v\|_{L^{2}(0,\eta)}+\Theta(\eta).
+\end{aligned}
+$$
 
 Since $v(x) = \int_0^x v'(\xi)\,d\xi$, (5b) implies the estimate $|v(x)| \leq \sqrt{\eta}\Theta(\eta)$ for all $0 \leq x \leq \eta$, and hence $\|v\|_{L^2(0,\eta)} \leq \eta\Theta(\eta)$. Then the statement follows from $|v - v_n|_1^2 \leq \|v - v_n\|_{L^2(0,\eta)}^2 + \|v' - v_n'\|_{L^2(0,\eta)}^2 \leq C\Theta^2(\eta) \to 0$.$\quad\blacksquare$
 
@@ -23,8 +29,13 @@ When $\Omega \in C^1$, the normal direction $\mathbf{n}$ exists at all boundary
 
 Corollary 6.2.43. For $\Omega \in C^1$ and $k \in \mathbb{N}$ holds
 
-$H_0^k(\Omega) = \{u \in H^k(\Omega): \partial^l u/\partial n^l|_\Gamma = 0 \text{ for } 0 \leq l \leq k-1\}$
-$= \{u \in H^k(\Omega): D^\alpha u|_\Gamma = 0 \text{ for } 0 \leq |\alpha| \leq k-1\}.$
+$$
+\begin{aligned}
+H_0^k(\Omega)
+&= \left\{ u \in H^k(\Omega) : \left.\frac{\partial^l u}{\partial n^l}\right|_{\Gamma} = 0 \text{ for } 0 \le l \le k-1 \right\} \\
+&= \left\{ u \in H^k(\Omega) : \left.D^\alpha u\right|_{\Gamma} = 0 \text{ for } 0 \le |\alpha| \le k-1 \right\}.
+\end{aligned}
+$$
 
 ## 6.3 Dual Spaces
 
@@ -32,5 +43,5 @@ $= \{u \in H^k(\Omega): D^\alpha u|_\Gamma = 0 \text{ for } 0 \leq |\alpha| \leq
 
 Let $X$ be a normed, linear space over $\mathbb{R}$. As a dual space, $X'$ denotes the space of all bounded, linear mappings of $X$ onto $\mathbb{R}$:
 
-$X' := L(X, \mathbb{R}).$
+$$X' := L(X, \mathbb{R}).$$
 
diff --git a/markdowns/education_math_00085.md b/markdowns/education_math_00085.md
index 4f30b8efc75de39f39177c7eaa0a189bdaa7b0f4..3f5bb3924727f60105234e32b1f01c950bf09995 100644
--- a/markdowns/education_math_00085.md
+++ b/markdowns/education_math_00085.md
@@ -8,14 +8,19 @@
 
 形如
 
-$a_1x_1 + a_2x_2 + \cdots + a_nx_n = b$
+$$a_1x_1 + a_2x_2 + \cdots + a_nx_n = b$$
 
 的方程称为含有 $n$ 个未知量的线性方程，其中 $a_1$，$a_2$，$\cdots$，$a_n$ 和 $b$ 为实数，$x_1$，$x_2$，$\cdots$，$x_n$ 称为变量. 含有 $m$ 个方程和 $n$ 个未知量的线性方程组定义为
 
-$$a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} = b_{1} \\
-a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} = b_{2} \\
-\vdots \\
-a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n} = b_{m} \tag{1}$$
+$$
+\begin{aligned}
+a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} &= b_{1}, \\
+a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} &= b_{2}, \\
+&\vdots \\
+a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n} &= b_{m}.
+\tag{1}
+\end{aligned}
+$$
 
 其中 $a_{ij}$ 及 $b_{i}$ 均为实数。(1)称为 $m \times n$ 的线性方程组。下面是几个线性方程组的例子：
 
@@ -37,8 +42,8 @@ $x_1 = 4$
 
 若有序 $n$ 元组 $(x_1, x_2, \cdots, x_n)$ 满足方程组中的所有方程，则称其为 $m \times n$ 的方程组的解．例如，有序数对 $(1, 2)$ 为方程组 (a) 的解，因为
 
-$1\cdot(1)+2\cdot(2)=5$
+$$1\cdot(1)+2\cdot(2)=5$$
 
-$2\cdot(1)+3\cdot(2)=8$
+$$2\cdot(1)+3\cdot(2)=8$$
 
 有序三元组(2，0，0)为方程组(b)的解，因为
diff --git a/markdowns/education_math_00086.md b/markdowns/education_math_00086.md
index 4c0dd5e00ce469e9773fba2e55288604dd40e6ea..da2e55d46482b891be5293648e794679e7432d8e 100644
--- a/markdowns/education_math_00086.md
+++ b/markdowns/education_math_00086.md
@@ -2,7 +2,7 @@
 
 定义 若存在一个非奇异的矩阵 $X$ 和一个对角矩阵 $D$，使得 $n \times n$ 矩阵 $A$ 满足
 
-$X^{-1}AX = D$
+$$X^{-1}AX = D$$
 
 则称 $A$ 为可对角化的(diagonalizable). 称 $X$ 将 $A$ 对角化(diagonalize).
 
@@ -27,11 +27,11 @@ $$
 
 由于 $X$ 有 $n$ 个线性无关的列向量，可得 $X$ 为非奇异的，因此
 
-$D = X^{-1} X D = X^{-1} A X$
+$$D = X^{-1} X D = X^{-1} A X$$
 
 反之，假设 $A$ 为可对角化的，则存在一个非奇异矩阵 $X$，使得 $AX = XD$。若 $x_1, x_2, \cdots, x_n$ 为 $X$ 的列向量，则对每一 $j$，
 
-$Ax_j = \lambda_j x_j \qquad (\lambda_j = d_{ij})$
+$$Ax_j = \lambda_j x_j \qquad (\lambda_j = d_{ij})$$
 
 因此，对每一 $j$，$\lambda_j$ 为 $A$ 的特征值，且 $x_j$ 为 $A$ 的属于 $\lambda_j$ 的特征向量．由于 $X$ 的列向量是线性无关的，因此 $A$ 有 $n$ 个线性无关的特征向量．
 
@@ -45,10 +45,21 @@ $Ax_j = \lambda_j x_j \qquad (\lambda_j = d_{ij})$
 
 从注 4 中可得
 
-$A^2 = (XDX^{-1})(XDX^{-1}) = XD^2X^{-1}$
+$$A^2 = (XDX^{-1})(XDX^{-1}) = XD^2X^{-1}$$
 
 且一般地，
 
-$A^k = XD^kX^{-1}$
-$\\\quad= X \begin{bmatrix} (\lambda_1)^k & & & \\ & (\lambda_2)^k & & \\ & & \ddots & \\ & & & (\lambda_n)^k \end{bmatrix} X^{-1}$
+$$
+\begin{aligned}
+A^k &= XD^kX^{-1} \\
+&\quad= X
+\begin{bmatrix}
+(\lambda_1)^k &        &        &        \\
+              & (\lambda_2)^k &        &        \\
+              &        & \ddots &        \\
+              &        &        & (\lambda_n)^k
+\end{bmatrix}
+X^{-1}
+\end{aligned}
+$$
 
diff --git a/markdowns/education_math_00087.md b/markdowns/education_math_00087.md
index 65fc83097480e8445248f0a2afb3b0e256fd0786..359bdffe938d0c00a943263be941e74ef2300c71 100644
--- a/markdowns/education_math_00087.md
+++ b/markdowns/education_math_00087.md
@@ -6,62 +6,71 @@ Modeling and Text Retrieval, SIAM, Philadelphia, 1999.
 
 1. 设
 
-$A = \begin{bmatrix}{{{3}}}&{{{1}}}&{{{4}}} \\{{{-2}}}&{{{0}}}&{{{1}}} \\{{{1}}}&{{{2}}}&{{{2}}}\end{bmatrix} \quad 和 \quad B = \begin{bmatrix}{{{1}}}&{{{0}}}&{{{2}}} \\{{{-3}}}&{{{1}}}&{{{1}}} \\{{{2}}}&{{{-4}}}&{{{1}}}\end{bmatrix}$
+$A = \begin{bmatrix}{{{3}}}&{{{1}}}&{{{4}}} \\{{{-2}}}&{{{0}}}&{{{1}}} \\{{{1}}}&{{{2}}}&{{{2}}}\end{bmatrix} \quad  和 \quad  B = \begin{bmatrix}{{{1}}}&{{{0}}}&{{{2}}} \\{{{-3}}}&{{{1}}}&{{{1}}} \\{{{2}}}&{{{-4}}}&{{{1}}}\end{bmatrix}$
 
 求：
 
-(a) $2A$
+(a)$2A$
 
 (b)$A+B$
 
-(c) $2A - 3B$
+(c)$2A-3B$
 
 (d)$(2A)^{\mathrm{T}}-(3B)^{\mathrm{T}}$
 
-(e) $AB$
+(e)$AB$
 
-(f)BA
+(f)$BA$
 
-$(g)A^{\mathrm{T}}B^{\mathrm{T}}$
+(g)$A^{T}B^{T}$
 
 (h)$(BA)^{\mathrm{T}}$
 
-2. 对下列每一对矩阵，确定是否可以用第一个矩阵乘以第二个矩阵，如果可以，求它们的乘积.
+2. 对下列每一对矩阵，确定是否可以用第一个矩阵乘以第二个矩阵。如果可以，求它们的乘积。
 
 (a)$\begin{bmatrix}3&5&1\\ -2&0&2\end{bmatrix}\begin{bmatrix}2&1\\ 1&3\\ 4&1\end{bmatrix}$
 
-(b)$\begin{bmatrix}4&-2\\ 6&-4\\ 8&-6\end{bmatrix}$ $\begin{bmatrix}1&2&3\end{bmatrix}$
+(b)$\begin{bmatrix}4&-2\\ 6&-4\\ 8&-6\end{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix}$
 
-(c)$\begin{bmatrix}1&4&3\\ 0&1&4\\ 0&0&2\end{bmatrix}$ $\begin{bmatrix}3&2\\ 1&1\\ 4&5\end{bmatrix}$
+(c)$\begin{bmatrix}1&4&3\\ 0&1&4\\ 0&0&2\end{bmatrix}\begin{bmatrix}3&2\\ 1&1\\ 4&5\end{bmatrix}$
 
-(d)$\begin{bmatrix}4&6\\ 2&1\end{bmatrix}$ $\begin{bmatrix}3&1&5\\ 4&1&6\end{bmatrix}$
+(d)$\begin{bmatrix}4&6\\ 2&1\end{bmatrix}\begin{bmatrix}3&1&5\\ 4&1&6\end{bmatrix}$
 
-(e)$\begin{bmatrix}4&6&1\\ 2&1&1\end{bmatrix}$ $\begin{bmatrix}3&1&5\\ 4&1&6\end{bmatrix}$
+(e)$\begin{bmatrix}4&6&1\\ 2&1&1\end{bmatrix}\begin{bmatrix}3&1&5\\ 4&1&6\end{bmatrix}$
 
-(f)$\begin{bmatrix}2\\-1\\3\end{bmatrix}$$\begin{bmatrix}3&2&4&5\end{bmatrix}$
+(f)$\begin{bmatrix}2\\-1\\3\end{bmatrix}\begin{bmatrix}3&2&4&5\end{bmatrix}$
 
 3. 对练习 2 中的每一对矩阵，是否可以用第二个矩阵乘以第一个矩阵？它们的乘积的维数是多少？
-4. 将下列方程组写为矩阵方程的形式。
-
-(a) $3x_1 + 2x_2 = 1$
-
-$2x_1 - 3x_2 = 5$
-
-(b) $x_1 + x_2 = 5$
-
-$2x_1 + x_2 - x_3 = 6$
-
-$3x_1 - 2x_2 + 2x_3 = 7$
-
-(c) $2x_1 + x_2 + x_3 = 4$
-
-$x_1 - x_2 + 2x_3 = 2$
-
-$3x_1 - 2x_2 - x_3 = 0$
+4. 将下列方程组写为矩阵方程的形式.
+
+$$
+\begin{aligned}
+(a)\quad &3x_{1}+2x_{2}=1, \\
+&2x_{1}-3x_{2}=5.
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+(b)\quad &x_{1}+x_{2}=5, \\
+&2x_{1}+x_{2}-x_{3}=6, \\
+&3x_{1}-2x_{2}+2x_{3}=7.
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+(c)\quad &2x_{1}+x_{2}+x_{3}=4, \\
+&x_{1}-x_{2}+2x_{3}=2, \\
+&3x_{1}-2x_{2}-x_{3}=0.
+\end{aligned}
+$$
 
 5. 设
 
-$A = \begin{bmatrix} 3 & 4 \\ 1 & 1 \\ 2 & 7 \end{bmatrix}$
+$$
+A=\begin{bmatrix}3&4\\ 1&1\\ 2&7\end{bmatrix}
+$$
 
 验证：
 
@@ -69,11 +78,11 @@ $A = \begin{bmatrix} 3 & 4 \\ 1 & 1 \\ 2 & 7 \end{bmatrix}$
 
 (b)$6A=3(2A)$
 
-(c) $(A^T)^T = A$
+(c)$(A^T)^T = A$
 
 6. 设
 
-$A=\begin{bmatrix}{{{4}}}&{{{1}}}&{{{6}}} \\{{{2}}}&{{{3}}}&{{{5}}}\end{bmatrix}\quad及\quad B=\begin{bmatrix}{{{1}}}&{{{3}}}&{{{0}}} \\{{{-2}}}&{{{2}}}&{{{-4}}}\end{bmatrix}$
+$A=\begin{bmatrix}{{{4}}}&{{{1}}}&{{{6}}} \\{{{2}}}&{{{3}}}&{{{5}}}\end{bmatrix}\quad 及\quad  B=\begin{bmatrix}{{{1}}}&{{{3}}}&{{{0}}} \\{{{-2}}}&{{{2}}}&{{{-4}}}\end{bmatrix}$
 
 验证：
 
@@ -85,11 +94,11 @@ $A=\begin{bmatrix}{{{4}}}&{{{1}}}&{{{6}}} \\{{{2}}}&{{{3}}}&{{{5}}}\end{bmatrix}
 
 7. 设
 
-$A = \begin{bmatrix}{{{2}}}&{{{1}}} \\{{{6}}}&{{{3}}} \\{{{-2}}}&{{{4}}}\end{bmatrix} \quad 及 \quad B = \begin{bmatrix}{{{2}}}&{{{4}}} \\{{{1}}}&{{{6}}}\end{bmatrix}$
+$A = \begin{bmatrix}{{{2}}}&{{{1}}} \\{{{6}}}&{{{3}}} \\{{{-2}}}&{{{4}}}\end{bmatrix} \quad  及 \quad  B = \begin{bmatrix}{{{2}}}&{{{4}}} \\{{{1}}}&{{{6}}}\end{bmatrix}$
 
 验证：
 
 (a) $3(AB) = (3A)B = A(3B)$
 
-(b)$(AB)^{\top}=B^{\top}A^{\top}$
+(b)$(AB)^{\mathrm{T}}=B^{\mathrm{T}}A^{\mathrm{T}}$
 
diff --git a/markdowns/education_math_00088.md b/markdowns/education_math_00088.md
index b1277d325170cb21c4b85a5224b60566d93971ac..d1242a62190831fbaa81f4cb64bfe159bb7b4683 100644
--- a/markdowns/education_math_00088.md
+++ b/markdowns/education_math_00088.md
@@ -2,13 +2,13 @@
 
 定理 7.7.1 若 A 是一个秩为 r < n 的 $m \times n$ 矩阵，其奇异值分解为 $U \Sigma V^{T}$，则向量
 
-$x = A^+ b = V \Sigma^+ U^\top b$
+$$x = A^+ b = V \Sigma^+ U^\top b$$
 
 最小化 $\| b - Ax \|_2^2$。此外，若 $z$ 为任何其他使得 $\| b - Ax \|_2^2$ 最小化的向量，则 $\| z \|_2 > \| x \|_2$。
 
 证 令 x 为 $R^{n}$ 中的一个向量，并定义
 
-$c = U^{T} b = \begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} \quad 和 \quad y = V^{T} x = \begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix}$
+$$c = U^{T} b = \begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} \quad 和 \quad y = V^{T} x = \begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix}$$
 
 其中 $c_{1}$ 和 $y_{1}$ 为 $R^{r}$ 中的向量．由于 $U^{T}$ 是正交的，可得
 
@@ -24,11 +24,11 @@ $$
 
 由于 $c_2$ 和 $x$ 线性无关，可得 $\| b - Ax \|^2$ 取得最小值的充要条件为
 
-$\| c_1 - \Sigma_1 y_1 \| = 0$
+$$\| c_1 - \Sigma_1 y_1 \| = 0$$
 
 因此 $x$ 为最小二乘问题的解的充要条件是 $x = Vy$，其中 $y$ 为一个向量，形如
 
-$\begin{bmatrix} \Sigma_1^{-1} c_1 \\ y_2 \end{bmatrix}$
+$$\begin{bmatrix} \Sigma_1^{-1} c_1 \\ y_2 \end{bmatrix}$$
 
 特别地，
 
@@ -36,11 +36,11 @@ $$\begin{aligned} \boldsymbol { x } & = V \left[ \begin{array} { c } { { \Sigma
 
 是一个解．若 $z$ 为任何其他的解，则 $z$ 必形如
 
-$z = V y = V \begin{bmatrix} \Sigma_{1}^{-1} c_{1} \\ y_{2} \end{bmatrix}$
+$$z = V y = V \begin{bmatrix} \Sigma_{1}^{-1} c_{1} \\ y_{2} \end{bmatrix}$$
 
 其中 $y_2 \neq 0$. 由此可得
 
-$\| \mathbf{z} \|^{2} = \| \mathbf{y} \|^{2} = \| \Sigma_{1}^{-1} \mathbf{c}_{1} \|^{2} + \| \mathbf{y}_{2} \|^{2} > \| \Sigma_{1}^{-1} \mathbf{c}_{1} \|^{2} = \| \mathbf{x} \|^{2}$
+$$\| \mathbf{z} \|^{2} = \| \mathbf{y} \|^{2} = \| \Sigma_{1}^{-1} \mathbf{c}_{1} \|^{2} + \| \mathbf{y}_{2} \|^{2} > \| \Sigma_{1}^{-1} \mathbf{c}_{1} \|^{2} = \| \mathbf{x} \|^{2}$$
 
 若 A 的奇异值分解 $U \Sigma V^{T}$ 是已知的，则求最小二乘问题的解是一个简单的问题。若
 
diff --git a/markdowns/education_math_00089.md b/markdowns/education_math_00089.md
index d49a59a9e3d800b79f5ca1a4dfa525ad76aad871..4954f40679a731bdb9fc630531a06166ddd04c82 100644
--- a/markdowns/education_math_00089.md
+++ b/markdowns/education_math_00089.md
@@ -2,15 +2,15 @@ $$Q = \begin{bmatrix} 0.000 & 0.594 & 0.327 & 0.000 & 0.100 & 0.000 & 0.147 & 0.
 
 若令 $y = Q^{\top} x$，则
 
-$y_i = q_i^\top x = \cos\theta_i$
+$$y_i = q_i^\top x = \cos\theta_i$$
 
 其中 $\theta_{i}$ 为单位向量 x 和 $q_{i}$ 之间的夹角．对我们的例子，
 
-$\boldsymbol{y}=(0.000,0.229,0.567,0.331,0.635,0.577,0.000,0.535)^{\mathrm{T}}$
+$$\boldsymbol{y}=(0.000,0.229,0.567,0.331,0.635,0.577,0.000,0.535)^{\mathrm{T}}$$
 
 由于 $y_5 = 0.635$ 为 $y$ 中最接近 1 的元素，这说明搜索向量 $x$ 最接近 $q_5$ 的方向，因此，模块 5 是最匹配搜索标准的一个。接下来最好的匹配为模块 6（$y_6 = 0.577$）和 3（$y_3 = 0.567$）。若一个文档不包含任何搜索单词，则对应的数据库矩阵的列向量将与搜索向量正交。注意到，模块 1 和 7 不包含三个搜索关键字中的任何一个，因此有
 
-$y_1 = q_1^{\mathrm{T}} x = 0$ 和 $y_7 = q_7^{\mathrm{T}} x = 0$
+$$y_1 = q_1^{\mathrm{T}} x = 0 \quad和\quad y_7 = q_7^{\mathrm{T}} x = 0$$
 
 这个例子说明了一些数据库搜索的基本思想．利用现代矩阵方法，可以将搜索过程显著改进．我们可以加快搜索速度，并同时纠正由于多义词和同义词而出现的错误．这些新的技术称为潜语义索引(latent semantic indexing，LSI)，它依赖于矩阵分解，奇异值分解(singular value decomposition)，这方面的内容将在6.5节中讨论.
 
diff --git a/markdowns/education_math_00090.md b/markdowns/education_math_00090.md
index 041b91b83e01dd16c02461edc53cf4c8d3a03533..e192f8bde440e5db055b4bcc18f9bf63ba86b93e 100644
--- a/markdowns/education_math_00090.md
+++ b/markdowns/education_math_00090.md
@@ -2,7 +2,7 @@
 
 例如，
 
-$\vec{x} = (x_{1}, x_{2}, x_{3}, x_{4}), \boldsymbol{y} = \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{bmatrix}$
+$$\vec{x} = (x_{1}, x_{2}, x_{3}, x_{4}), \boldsymbol{y} = \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{bmatrix}$$
 
 分别是有 4 项的行向量和列向量.
 
@@ -10,11 +10,11 @@ $\vec{x} = (x_{1}, x_{2}, x_{3}, x_{4}), \boldsymbol{y} = \begin{bmatrix} y_{1}
 
 设 $A$ 为一 $m \times n$ 矩阵，则 $A$ 的行向量为
 
-$\vec{a}_{i}=(a_{i1},a_{i2},\cdots,a_{in}),\quad i=1,\cdots,m$
+$$\vec{a}_{i}=(a_{i1},a_{i2},\cdots,a_{in}),\quad i=1,\cdots,m$$
 
 同时，列向量表示为
 
-$a_{j}=\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{mj}\end{bmatrix},\quad j=1,\cdots,n$
+$$a_{j}=\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{mj}\end{bmatrix},\quad j=1,\cdots,n$$
 
 矩阵 A 可以用它的列向量或者行向量表示.
 
@@ -22,17 +22,17 @@ $A = (a_{1}, a_{2}, \cdots, a_{n}) \quad 或 \quad A = \begin{bmatrix} \vec{a}_{
 
 类似地，如果 $B$ 为一 $n \times r$ 矩阵，则
 
-$B=(b_1,\ {b}_2,\ \cdots,\ {b}_r)=\begin{bmatrix} \vec{b}_1 \\ \vec{b}_2 \\ \vdots \\ \vec{b}_n \end{bmatrix}$
+$$B=(b_1,\ {b}_2,\ \cdots,\ {b}_r)=\begin{bmatrix} \vec{b}_1 \\ \vec{b}_2 \\ \vdots \\ \vec{b}_n \end{bmatrix}$$
 
 ▶例 1 如果
 
-$A=\begin{bmatrix}3&2&5\\ -1&8&4\end{bmatrix}$
+$$A=\begin{bmatrix}3&2&5\\ -1&8&4\end{bmatrix}$$
 
 则
 
-$\boldsymbol{a}_{1}=\begin{bmatrix}3\\-1\end{bmatrix},\boldsymbol{a}_{2}=\begin{bmatrix}2\\8\end{bmatrix},\boldsymbol{a}_{3}=\begin{bmatrix}5\\4\end{bmatrix}$
+$$\boldsymbol{a}_{1}=\begin{bmatrix}3\\-1\end{bmatrix},\boldsymbol{a}_{2}=\begin{bmatrix}2\\8\end{bmatrix},\boldsymbol{a}_{3}=\begin{bmatrix}5\\4\end{bmatrix}$$
 
-$\vec{a}_1 = (3,2,5), \vec{a}_2 = (-1,8,4)$
+$$\vec{a}_1 = (3,2,5), \vec{a}_2 = (-1,8,4)$$
 
 ## 相等
 
diff --git a/markdowns/education_math_00091.md b/markdowns/education_math_00091.md
index 6f19f2bc21ef861e301d9ebe0b6f34dbcfbe874c..00d151d95cca770e4af9cfe5f747a994b7a7db72 100644
--- a/markdowns/education_math_00091.md
+++ b/markdowns/education_math_00091.md
@@ -4,47 +4,71 @@ $m \times n$ 的方程组与此类似. $m \times n$ 的方程组可能相容，
 
 考虑两个方程组
 
-(a)$3x_{1}+2x_{2}-x_{3}=-2$
-
-$x_2 = 3$
-
-$2x_3 = 4$
-
-(b) $3x_{1}+2x_{2}-x_{3}=-2$
-
-$-3x_1 - x_2 + x_3 = 5$
-
-$3x_1 + 2x_2 + x_3 = 2$
+$$
+\begin{aligned}
+(a)\quad &3x_1 + 2x_2 - x_3 = -2, \\
+&x_2 = 3, \\
+&2x_3 = 4.
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+(b)\quad &3x_1 + 2x_2 - x_3 = -2, \\
+&-3x_1 - x_2 + x_3 = 5, \\
+&3x_1 + 2x_2 + x_3 = 2.
+\end{aligned}
+$$
 
 显然，方程组(a)容易求解．因为，由后两个方程容易得到$x_{2}=3$ 和$x_{3}=2$．将这些值代入第一个方程，可得
 
-$3x_1 + 2 \cdot 3 - 2 = -2$
+$$
+3x_1 + 2\cdot 3 - 2 = -2
+$$
 
-$x_1 = -2$
+$$
+x_1 = -2
+$$
 
 于是，方程组(a)的解为$(-2,3,2)$。求解方程组(b)似乎不是很容易。其实，方程组(b)与方程组(a)有相同的解。为看清这一点，首先将其前两个方程相加：
 
-$3x_1 + 2x_2 - x_3 = -2$
+$$
+3x_1 + 2x_2 - x_3 = -2
+$$
 
-$\frac{-3x_{1}-x_{2}+x_{3}=5}{x_{2}=3}$
+$$
+\frac{-3x_1 - x_2 + x_3 = 5}{x_2 = 3}
+$$
 
 若$(x_1, x_2, x_3)$为(b)的解，则它必满足方程组中的所有方程．因此，它必然满足任意两个方程相加后得到的新方程．因此，$x_2$必为3．类似地，$(x_1, x_2, x_3)$必满足第三个方程减去第一个方程后得到的新方程：
 
-$3x_1 + 2x_2 + x_3 = 2$
+$$
+3x_1 + 2x_2 + x_3 = 2
+$$
 
-$\frac{3x_{1}+2x_{2}-x_{3}=-2}{2x_{3}=4}$
+$$
+\dfrac{3x_1 + 2x_2 - x_3 = -2}{2x_3 = 4}
+$$
 
 因此，方程组(b)的解必为方程组(a)的解. 通过类似的讨论，可以证明方程组(a)的解也是方程组(b)的解. 由(a)中的第二个方程减去第一个方程得
 
-$x_{2}=3$
+$$
+x_2 = 3
+$$
 
-$\frac{3x_{1}+2x_{2}-x_{3}=-2}{-3x_{1}-x_{2}+x_{3}=5}$
+$$
+\dfrac{3x_1 + 2x_2 - x_3 = -2}{-3x_1 - x_2 + x_3 = 5}
+$$
 
 然后，将其第一个方程与第三个方程相加：
 
-$3x_1 + 2x_2 - x_3 = -2$
+$$
+3x_1 + 2x_2 - x_3 = -2
+$$
 
-$\frac{2x_3 = 4}{3x_1 + 2x_2 + x_3 = 2}$
+$$
+\dfrac{2x_3 = 4}{3x_1 + 2x_2 + x_3 = 2}
+$$
 
 因此，$(x_1, x_2, x_3)$为(b)的解的充要条件是，它是方程组(a)的解. 即方程组(a)和方程组(b)有相同的解集$\{(-2, 3, 2)\}$.
 
diff --git a/markdowns/education_math_00092.md b/markdowns/education_math_00092.md
index 64b1f53c522eca142badb4a0414ecec5558977c5..ea4fa74945651d30a5584bf961bd39664b641876 100644
--- a/markdowns/education_math_00092.md
+++ b/markdowns/education_math_00092.md
@@ -1,4 +1,4 @@
-$\Sigma^{+}=\left[\begin{array}{c|c} \Sigma_{1}^{-1} & O \\ \hline O & O \end{array}\right]=\left[\begin{array}{ccc|c} \frac{1}{\sigma_{1}} & & & \\ & \ddots & & O \\ & & \frac{1}{\sigma_{r}} & \\ \hline & O & & O \end{array}\right]$
+$$\Sigma^{+}=\left[\begin{array}{c|c} \Sigma_{1}^{-1} & O \\ \hline O & O \end{array}\right]=\left[\begin{array}{ccc|c} \frac{1}{\sigma_{1}} & & & \\ & \ddots & & O \\ & & \frac{1}{\sigma_{r}} & \\ \hline & O & & O \end{array}\right]$$
 
 方程(1)给出了一个矩阵逆的自然推广。(1)定义的矩阵$A^{+}$称为A的伪逆(pseudoinverse).
 
@@ -34,19 +34,23 @@ $Y = YAY \quad (2) \\
 
 为看到伪逆如何用于求解最小二乘问题, 首先考虑 $A$ 是一个秩为 $n$ 的 $m\times n$ 矩阵的情形. 因此, $\Sigma$ 形如
 
-$\Sigma=\begin{bmatrix}\Sigma_{1}\\ O\end{bmatrix}$
+$$\Sigma=\begin{bmatrix}\Sigma_{1}\\ O\end{bmatrix}$$
 
 其中 $\Sigma_1$ 为一个非奇异 $n \times n$ 对角矩阵. 矩阵 $A^T A$ 为非奇异的, 且
 
-$(A^T A)^{-1} = V(\Sigma^T \Sigma)^{-1} V^T$
+$$(A^T A)^{-1} = V(\Sigma^T \Sigma)^{-1} V^T$$
 
 正规方程的解为
 
-$x = (A^T A)^{-1} A^T b$
-$\\\quad= V(\Sigma^T \Sigma)^{-1} V^T V \Sigma^T U^T b$
-$\\\quad= V(\Sigma^T \Sigma)^{-1} \Sigma^T U^T b$
-$\\\quad= V \Sigma^+ U^T b$
-$\\\quad= A^+ b$
+$$
+\begin{aligned}
+x &= (A^T A)^{-1} A^T b \\
+&\quad = V(\Sigma^T \Sigma)^{-1} V^T V \Sigma^T U^T b \\
+&\quad = V(\Sigma^T \Sigma)^{-1} \Sigma^T U^T b \\
+&\quad = V \Sigma^{+} U^T b \\
+&\quad = A^{+} b
+\end{aligned}
+$$
 
 因此，若 A 为满秩的，则 $A^{+}b$ 为最小二乘问题的解。若 A 的秩 r<n 时如何呢？此
 
diff --git a/markdowns/education_math_00093.md b/markdowns/education_math_00093.md
index 3038b012471dc7fe6a98495c748884b31a485143..51023572ea5dee4e3c62eafad84b3fad1e8e8936 100644
--- a/markdowns/education_math_00093.md
+++ b/markdowns/education_math_00093.md
@@ -1,42 +1,59 @@
-$\mathbf{Y} = X e^{tD} X^{-1} \mathbf{Y}_0$
+$$\mathbf{Y} = X e^{tD} X^{-1} \mathbf{Y}_0$$
 
 因此
 
-$$\mathbf { Y } = X \mathrm { e } ^ { t D } \mathbf { c } = ( x _ { 1 } , x _ { 2 } , \cdots , x _ { n } ) { \left[ \begin{array} { l } { c _ { 1 } \, \mathrm { e } ^ { \lambda _ { 1 } t } } \\ { c _ { 2 } \, \mathrm { e } ^ { \lambda _ { 2 } t } } \\ { \vdots } \\ { c _ { n } \, \mathrm { e } ^ { \lambda _ { n } t } } \end{array} \right] }$$
-$$= c _ { 1 } \, \mathrm { e } ^ { \lambda _ { 1 } t } \mathbf { x } _ { 1 } + \cdots + c _ { n } \, \mathrm { e } ^ { \lambda _ { n } t } \mathbf { x } _ { n }$$
+$$
+\begin{aligned}
+\mathbf{Y}
+&= X \mathrm{e}^{tD} \mathbf{c}
+= (x_1, x_2, \cdots, x_n)
+\begin{bmatrix}
+c_1 \mathrm{e}^{\lambda_1 t} \\
+c_2 \mathrm{e}^{\lambda_2 t} \\
+\vdots \\
+c_n \mathrm{e}^{\lambda_n t}
+\end{bmatrix} \\
+&= c_1 \mathrm{e}^{\lambda_1 t} \mathbf{x}_1 + \cdots + c_n \mathrm{e}^{\lambda_n t} \mathbf{x}_n
+\end{aligned}
+$$
 
 综上所述，初值问题(7)的解为
 
-$\mathbf{Y} = e^{tA}\mathbf{Y}_0$
+$$\mathbf{Y} = e^{tA}\mathbf{Y}_0$$
 
 若 $A$ 为可对角化的，这个解又可写为
 
-$\mathbf{Y} = X e^{tD} X^{-1} \mathbf{Y}_0$
-$\\\quad= c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + \cdots + c_n e^{\lambda_n t} \mathbf{x}_n \quad (c = X^{-1} \mathbf{Y}_0)$
+$$
+\begin{aligned}
+\mathbf{Y} &= X e^{tD} X^{-1} \mathbf{Y}_0 \\
+&\quad = c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + \cdots + c_n e^{\lambda_n t} \mathbf{x}_n,
+\qquad (c = X^{-1} \mathbf{Y}_0)
+\end{aligned}
+$$
 
 ▶例 7 用矩阵指数求解初值问题
 
-$\mathbf{Y}' = A\mathbf{Y}, \quad \mathbf{Y}(0) = \mathbf{Y}_0$
+$$\mathbf{Y}' = A\mathbf{Y}, \quad \mathbf{Y}(0) = \mathbf{Y}_0$$
 
 其中
 
-$A = \begin{bmatrix} 3 & 4 \\ 3 & 2 \end{bmatrix}, \quad \mathbf{Y}_0 = \begin{bmatrix} 6 \\ 1 \end{bmatrix}$
+$$A = \begin{bmatrix} 3 & 4 \\ 3 & 2 \end{bmatrix}, \quad \mathbf{Y}_0 = \begin{bmatrix} 6 \\ 1 \end{bmatrix}$$
 
 （这个问题在 6.2 节中的例 1 中已经求出。）
 
 解 $A$ 的特征值为 $\lambda_1 = 6$ 和 $\lambda_2 = -1$，其特征向量为 $x_1 = (4, 3)^\mathrm{T}$ 和 $x_2 = (1, -1)^\mathrm{T}$. 因此
 
-$A = XDX^{-1} = \begin{bmatrix} 4 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \frac{1}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix}$
+$$A = XDX^{-1} = \begin{bmatrix} 4 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \frac{1}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix}$$
 
 且其解为
 
-$Y=e^{tA}Y_{0}\\\quad=Xe^{tD}X^{-1}Y_{0}\\\quad=\begin{bmatrix}4&1\\ 3&-1\end{bmatrix}\begin{bmatrix}e^{6t}&0\\ 0&e^{-t}\end{bmatrix}\begin{bmatrix}\frac{1}{7}&\frac{1}{7}\\ \frac{3}{7}&-\frac{4}{7}\end{bmatrix}\begin{bmatrix}6\\ 1\end{bmatrix}\\\quad=\begin{bmatrix}4e^{6t}+2e^{-t}\\ 3e^{6t}-2e^{-t}\end{bmatrix}$
+$$Y=e^{tA}Y_{0}\\\quad=Xe^{tD}X^{-1}Y_{0}\\\quad=\begin{bmatrix}4&1\\ 3&-1\end{bmatrix}\begin{bmatrix}e^{6t}&0\\ 0&e^{-t}\end{bmatrix}\begin{bmatrix}\frac{1}{7}&\frac{1}{7}\\ \frac{3}{7}&-\frac{4}{7}\end{bmatrix}\begin{bmatrix}6\\ 1\end{bmatrix}\\\quad=\begin{bmatrix}4e^{6t}+2e^{-t}\\ 3e^{6t}-2e^{-t}\end{bmatrix}$$
 
 比较它和 6.2 节例 1 中的解。
 
 ▶例 8 用矩阵指数求解初值问题
 
-$\mathbf{Y}' = A\mathbf{Y}, \quad \mathbf{Y}(0) = \mathbf{Y}_0$
+$$\mathbf{Y}' = A\mathbf{Y}, \quad \mathbf{Y}(0) = \mathbf{Y}_0$$
 
 其中
 
diff --git a/markdowns/education_math_00094.md b/markdowns/education_math_00094.md
index 4e937543f97f9a8e378e58dee164bd38050c266b..31e58c5eb04c432549148a5414bc1dd55b14c507 100644
--- a/markdowns/education_math_00094.md
+++ b/markdowns/education_math_00094.md
@@ -1,24 +1,31 @@
-$\int_{E_{n,k}} \left| f_n(x) \right|^p \, dx < \frac{\varepsilon}{2^{p+2}} \quad (n \in \mathbf{N}, k \geq N).$
+$$\int_{E_{n,k}} \left| f_n(x) \right|^p \, dx < \frac{\varepsilon}{2^{p+2}} \quad (n \in \mathbf{N}, k \geq N).$$
 
 于是我们有
 
-$\begin{aligned} \int_{E_{n,k}} &|f_{n}(x)-f_{k}(x)|^{p}\,\mathrm{d}x \\
-&\leqslant2^{p}\int_{E_{n,k}}|f_{n}(x)|^{p}\,\mathrm{d}x+2^{p}\int_{E_{n,k}}|f_{k}(x)|^{p}\,\mathrm{d}x \\
-&\leqslant2^{p}\left(\frac{\varepsilon}{2^{p+2}}+\frac{\varepsilon}{2^{p+2}}\right)=\frac{\varepsilon}{2}\quad(n\in\mathbf{N},k\geqslant N). \end{aligned}$
+$$
+\begin{aligned}
+\int_{E_{n,k}} |f_n(x)-f_k(x)|^p \,\mathrm{d}x
+&\leq 2^p \int_{E_{n,k}} |f_n(x)|^p \,\mathrm{d}x
+   + 2^p \int_{E_{n,k}} |f_k(x)|^p \,\mathrm{d}x \\
+&\leq 2^p \left( \frac{\varepsilon}{2^{p+2}} + \frac{\varepsilon}{2^{p+2}} \right)
+ = \frac{\varepsilon}{2},
+\qquad (n \in \mathbf{N},\ k \geq N).
+\end{aligned}
+$$
 
 再根据$(**)$式, 即知$\{f_n(x)\}$是$L^p(E)$中的Cauchy列.
 
 (B) $ m(E) = +\infty $ 时. 作分解 $ E = \bigcup_{k=1}^{\infty} E_{k} $: $ m(E_{k}) < +\infty $ ($ k \in N $), 并记 $ A_{j} = \bigcup_{k=j}^{\infty} E_{k} $ ($ j = 1, 2, \cdots $), $\{ A_{j} \}$ 是递减可测集合列, 使得 $\bigcap_{j=1}^{\infty} A_{j} = \varnothing$. 根据 $\Gamma$ 具有积分一致绝对连续性, 可知存在 $j_{0}$, 使得
 
-$\int_{A_{j_0}} |f_n(x)|^p \, dx < \frac{\varepsilon}{2^{p+2}} \quad (n = 1, 2, \cdots).$
+$$\int_{A_{j_0}} |f_n(x)|^p \, dx < \frac{\varepsilon}{2^{p+2}} \quad (n = 1, 2, \cdots).$$
 
 从而我们有
 
-$\int_{A_{J_{0}}}|f_{n}(x)-f_{m}(x)|^{p}\,dx\\\quad\leqslant2^{p}\int_{A_{J_{0}}}|f_{n}(x)|^{p}\,dx+2^{p}\int_{A_{J_{0}}}|f_{m}(x)|^{p}\,dx<\varepsilon.\quad(*)$
+$$\int_{A_{J_{0}}}|f_{n}(x)-f_{m}(x)|^{p}\,dx\\\quad\leqslant2^{p}\int_{A_{J_{0}}}|f_{n}(x)|^{p}\,dx+2^{p}\int_{A_{J_{0}}}|f_{m}(x)|^{p}\,dx<\varepsilon.\quad(*)$$
 
 注意到 $m(A_{j_{0}}^{c})\leqslant\sum_{j=1}^{j_{0}-1}m(E_{j})<+\infty$ ，根据(A)就有
 
-$\int_{A_{j_{0}}^{\epsilon}} |f_{n}(x) - f_{m}(x)|^{p} \, dx < \varepsilon \quad (充分大的 n, m). \quad (**)$
+$$\int_{A_{j_{0}}^{\epsilon}} |f_{n}(x) - f_{m}(x)|^{p} \, dx < \varepsilon \quad (充分大的 n, m). \quad (**)$$
 
 综合$(*)$，$(**)$式，说明$\{f_{n}(x)\}$是$L^{p}(E)$中Cauchy列.
 
diff --git a/markdowns/education_math_00095.md b/markdowns/education_math_00095.md
index 7d7e231d72461cf428c13085114747fb8ff628e2..a47e5654504f56f30b12e503580d735e550e71ff 100644
--- a/markdowns/education_math_00095.md
+++ b/markdowns/education_math_00095.md
@@ -1,24 +1,24 @@
 $N_1(x), g(x) = P_2(x) - N_2(x)$, 则
 
-$f(x)g(x)=\left[P_{1}(x)P_{2}(x)+N_{1}(x)N_{2}(x)\right]\\\quad-\left[P_{1}(x)N_{2}(x)+N_{1}(x)P_{2}(x)\right].$
+$$f(x)g(x)=\left[P_{1}(x)P_{2}(x)+N_{1}(x)N_{2}(x)\right]\\\quad-\left[P_{1}(x)N_{2}(x)+N_{1}(x)P_{2}(x)\right].$$
 
 从而得
 
-$\bigvee_{a}^{b}(fg)\leqslant[P_{1}(b)P_{2}(b)+N_{1}(b)N_{2}(b)]\\\quad+[P_{1}(b)N_{2}(b)+N_{1}(b)P_{2}(b)]\\\quad=[P_{1}(b)+N_{1}(b)][P_{2}(b)+N_{2}(b)]\\\quad=\bigvee_{a}^{b}(f)\cdot\bigvee_{a}^{b}(g).$
+$$\bigvee_{a}^{b}(fg)\leqslant[P_{1}(b)P_{2}(b)+N_{1}(b)N_{2}(b)]\\\quad+[P_{1}(b)N_{2}(b)+N_{1}(b)P_{2}(b)]\\\quad=[P_{1}(b)+N_{1}(b)][P_{2}(b)+N_{2}(b)]\\\quad=\bigvee_{a}^{b}(f)\cdot\bigvee_{a}^{b}(g).$$
 
 （6）依题设知存在 $M>0$，使得
 
-$\left|f(x)-f(y)\right|\leqslant M\left|x-y\right|\quad(a\leqslant x,y\leqslant b).$
+$$\left|f(x)-f(y)\right|\leqslant M\left|x-y\right|\quad(a\leqslant x,y\leqslant b).$$
 
 由此知对$a,b$的任一分划$\Delta:a=x_{0}< x_{1}<\cdots< x_{n}=b$，均有
 
-$v_{\Delta} = \sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| \leqslant M \sum_{i=1}^{n} |x_{i} - x_{i-1}| = M (b-a).$
+$$v_{\Delta} = \sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| \leqslant M \sum_{i=1}^{n} |x_{i} - x_{i-1}| = M (b-a).$$
 
 这说明 $\bigvee_a^b(f)\leqslant M(b-a)$.
 
 （7）由题设知存在 $M>0$，使得
 
-$|f(x) - f(y)| \leqslant M |x - y| \quad (-\infty < x, y < +\infty).$
+$$|f(x) - f(y)| \leqslant M |x - y| \quad (-\infty < x, y < +\infty).$$
 
 从而对$[a,b]$的任一分划 $\Delta$: $a=x_0< x_1<\cdots< x_n=b$, 我们有
 
@@ -28,10 +28,10 @@ $$\begin{aligned} v_{ \Delta } & = \sum _ { i = 1 } ^ { n } | \varphi ( f ( x _
 
 (1) 若 $f \in \mathrm{BV}([a, b])$，则 $f(x)$ 几乎处处可微，且
 
-$\frac{\mathrm{d}}{\mathrm{d}x}\left( \bigvee_a^x (f) \right) = |f'(x)|, \quad \text{a.\,e.} \, x \in [a,b].$
+$$\frac{\mathrm{d}}{\mathrm{d}x}\left( \bigvee_a^x (f) \right) = |f'(x)|, \quad \text{a.\,e.} \, x \in [a,b].$$
 
 (2) 若 $f \in BV([a,b])$，则(弧长) $l_f \geq \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$.
 
 证明 （1）（i）根据全变差的定义可知，对任给 $\varepsilon > 0$，存在分划
 
-$\Delta: a = x_0 < x_1 < \cdots < x_k = b,$
+$$\Delta: a = x_0 < x_1 < \cdots < x_k = b,$$
diff --git a/markdowns/education_math_00096.md b/markdowns/education_math_00096.md
index c249ae2387ec0f9673916c0de9f56733bb97d130..fb440d7422441da4e703dc795e0676071b8e55ba 100644
--- a/markdowns/education_math_00096.md
+++ b/markdowns/education_math_00096.md
@@ -32,6 +32,6 @@ $$\begin{aligned} m ^ { * } ( T ) \leqslant & m ^ { * } ( T \cap E ) + m ^ { * }
 
 (4) 由题设知
 
-$\begin{aligned} m(A_{1})=&m^{*}(A_{2})=m^{*}(A_{1}\cup(A_{2}\setminus A_{1}))\\=&m^{*}(A_{1})+m^{*}(A_{2}\setminus A_{1}),\end{aligned}$
+$$\begin{aligned} m(A_{1})=&m^{*}(A_{2})=m^{*}(A_{1}\cup(A_{2}\setminus A_{1}))\\=&m^{*}(A_{1})+m^{*}(A_{2}\setminus A_{1}),\end{aligned}$$
 
 故可得 $m^*(A_2 \setminus A_1) = 0$. 从而 $A_2 \setminus A_1$ 是可测集, 即 $A_2$ 是可测集.
diff --git a/markdowns/finance_research_report_00015.md b/markdowns/finance_research_report_00015.md
index 17f01355909dfde564c0edf1e48227aa57b49d64..f790c5e87bc56bbc3d4054c0d8af90f5040d2577 100644
--- a/markdowns/finance_research_report_00015.md
+++ b/markdowns/finance_research_report_00015.md
@@ -1,14 +1,10 @@
 In addition to considering the uncertainty adjustment factor in equation (4), one can decompose the contributions of the three effects thus far considered: (i) the effect on the expected stochastic discount factor $E[SD F_t] = E[e^{-(\rho + \eta g_t)t}]$ (or equivalently, on the certainty-equivalent discount rate); (ii) the contribution of the covariance of marginal damages and consumption; and (iii) the effect on expected marginal damages.$^7$ Specifically, the uncertainty adjustment can be factored as follows:
 
 $$
-\underbrace{e^{\frac{1}{2}(\beta - \eta)^2 \sigma_t^2 t^2}}_{\text{Effect of Uncertainty}} = e^{\frac{1}{2}(\beta^2 - 2\beta\eta + \eta^2) \sigma_t^2 t^2} =
-$$
-$$
-\underbrace{e^{\frac{1}{2}\eta^2 \sigma_t^2 t^2}}_{\text{Effect on } E[SDF_t]} \quad
-\underbrace{e^{-\beta\eta \sigma_t^2 t^2}}_{\text{Risk Adjustment}} \quad
-\underbrace{e^{\frac{1}{2}\beta^2 \sigma_t^2 t^2}}_{\text{Effect on } E[MD_t]} \quad (5)
+\begin{array}{l}\underbrace{e^{\frac{1}{2}(\beta-\eta)^{2}\sigma_{t}^{2}t^{2}}}_{ \textit{Effect of Uncertainty}}\ =\ e^{\frac{1}{2}(\beta^{2}-2\beta\eta+\eta^{2})\sigma_{t}^{2}t^{2}}\ =\\ \underbrace{e^{\frac{1}{2}\eta^{2}\sigma_{t}^{2}t^{2}}}_{\textit{Effect on E[SDFt]}} \ \underbrace{e^{-\beta\eta\sigma_{t}^{2}t^{2}}}_{\textit{Risk Adjustment}}\ \underbrace{e^{\frac{1}{2}\beta^{2}\sigma_{t}^{2}t^{2}}}_{\textit{Effect on E[MDt]}} \end{array}\quad(5)
 $$
 
+
 The three terms in equation (5) illustrate how the effect of uncertainty in equation (4) explicitly embodies the effects identified in equation (1) on the risk-free discount rate, equation (2) on the risk adjustment to the discount rate, and equation (3) on expected marginal damages. The first two bracketed terms on the right-hand side of equation (5) correspond to the effect of uncertainty on the discount rate, which, as previously demonstrated, serves to increase the risk-adjusted discount rate and therefore decrease the SCC if $\beta > \frac{1}{2}\eta$, a condition that typically holds in standard IAMs. This is the adjustment made by Barrage and Nordhaus (2023), for example. Overall, however, as the growth uncertainty increases the SCC on net, the increase in expected marginal damages represented in the third term of the right-hand side of equation (5) must dominate the first two terms. Hence including the effect of such uncertainty on expected marginal damages is essential for accurately representing the sign of growth uncertainty’s effect on the SCC, let alone its magnitude.
 
 ## 5. Implications of Small or Negative $\beta$
diff --git a/markdowns/legal_standard_00076.md b/markdowns/legal_standard_00076.md
new file mode 100644
index 0000000000000000000000000000000000000000..bdeefd6401e91ff4e5b168367d57208b4f17a006
--- /dev/null
+++ b/markdowns/legal_standard_00076.md
@@ -0,0 +1,38 @@
+```
+"victimIP": "2.2.2.2"
+
+"victimPort": "1443"
+
+"encryption": "1"
+
+"devID:" NGFW4000-UF "
+
+"devVer": "NGFW4000v2.0"
+
+"devVenodor": "XXX公司"
+
+"alarmArea": "互联网"
+
+"sourceIP": "51.1.1.1"
+
+"sourcePort": "443"
+
+"protocol": "TCP"
+
+"state": "3"
+
+"programName":
+
+"filePath": "/etc/..."
+
+"fileMd5": "6512bd43d9caa6e02c990b0a82652dca"
+
+"family":
+
+"fileSha1": "17ba0791499db908433b80f37c5fbc89b870084b"
+"fileSha256": "4fc82b26aecb47d2868c4efbe3581732a3e7cbc6c2efb32062c08170a05eeb8"
+
+"fileSM3": "d5744897e47fb6d78b726e9ff0c9fa70e0013a0d4f0a757af8ec0b812664b828"
+
+"organization": "匿名者"
+```
\ No newline at end of file
diff --git a/markdowns/manufacturing_form_receipt_00054.md b/markdowns/manufacturing_form_receipt_00054.md
index 0f8003d4af72c1c628a9a514a0bea9938b9b2839..4af986e8c511fc5baaddafcfc63f48e1882483ea 100644
--- a/markdowns/manufacturing_form_receipt_00054.md
+++ b/markdowns/manufacturing_form_receipt_00054.md
@@ -14,7 +14,7 @@
 
 大气中污染物浓度的表示方法有两种：一种是以单位体积内所含的污染物的质量数表示，我国规定的最高容许浓度单位是mg/m³；另一种是对于气体或蒸气用ppm或ppb作浓度单位，ppm单位表示100万体积空气中含有有害气体或蒸气的体积数，ppb是ppm的千分之一。两个单位可用下式换算。
 
-$X(ppm)\fbox{?} \frac{22.4}{M} \fbox{?} A$
+$$X(ppm)\fbox{?} \frac{22.4}{M} \fbox{?} A$$
 
 式中 A——以$mg/m^3$表示的气体浓度；
 
diff --git a/markdowns/manufacturing_operation_sop_00021.md b/markdowns/manufacturing_operation_sop_00021.md
index 4221c698ba20ce4f50b113ff04a3f472ebc05d7d..a039c2543758add4fb9794a746daafe90af66abf 100644
--- a/markdowns/manufacturing_operation_sop_00021.md
+++ b/markdowns/manufacturing_operation_sop_00021.md
@@ -68,11 +68,11 @@ by calculation
 
 Frequency deviation at one particular temperature (Tx) with respect to the frequency at +25°C is
 
-$\Delta\mathrm{f}/\mathrm{f}\left(\mathrm{p p m}\right)=-0.04\ast\left(\mathrm{T x}-25\right)^{2}$
+$$\Delta\mathrm{f}/\mathrm{f}\left(\mathrm{p p m}\right)=-0.04\ast\left(\mathrm{T x}-25\right)^{2}$$
 
 Example: at +60°C, the frequency drifts -49 ppm.
 
-$\Delta\mathrm{f}/\mathrm{f}(\mathrm{p p m})=-0.04*(60-25)^{2}=- 49\mathrm{p p m}$
+$$\Delta\mathrm{f}/\mathrm{f}(\mathrm{p p m})=-0.04*(60-25)^{2}=- 49\mathrm{p p m}$$
 
 ## HOW TO ORDER
 
diff --git a/markdowns/research_academic_paper_00001.md b/markdowns/research_academic_paper_00001.md
index eebb20b7c23647b5173b4e3cb4bcac887d6fe06b..b5482910bdc25ccb45904e995211eb2655d50f2a 100644
--- a/markdowns/research_academic_paper_00001.md
+++ b/markdowns/research_academic_paper_00001.md
@@ -2,7 +2,7 @@
 
 Multi-agent cooperation relies on the transform between agents. In the proposed system, the Ego calculates the relative transform $\mathcal{F}^{(1)} = \mathbf{R}^{(1)} \cdot (*) + \mathbf{t}^{(1)}$ using the position and pose data of both Ego and cooperative CAV. Its rotation and translation component is
 
-$\mathbf{R}^{(1)}=\mathbf{R}_{x}(\Delta\theta)\cdot\mathbf{R}_{y}(\Delta\psi)\cdot\mathbf{R}_{z}(\Delta\phi)\qquad(2)$
+$$\mathbf{R}^{(1)}=\mathbf{R}_{x}(\Delta\theta)\cdot\mathbf{R}_{y}(\Delta\psi)\cdot\mathbf{R}_{z}(\Delta\phi)\qquad(2)$$
 
 and
 
@@ -35,13 +35,30 @@ Once the cost matrix is defined, the task is to find the optimal assignment matr
 
 assignment probability on $\overline{C}_{i,j}$, then we have the following modified form of the optimal transport problem:
 
-$$\begin{array}{r l r}{\underset{\overline{{\bf P}}}{\operatorname*{m i n}}}&{\displaystyle\sum_{i,j}-\overline{{\bf P}}_{i,j}\overline{{\bf C}}_{i,j}}\\ {s.t.}&{\overline{{\bf P}}\mathbf{1}_{n+1}=\left[\mathbf{1}_{m}^{T},n\right]^{T},}\\ &{\overline{{\bf P}}^{T}\mathbf{1}_{m+1}=\left[\mathbf{1}_{n}^{T},m\right]^{T}}\end{array}\tag{7}$$
+$$
+\underset{\overline{\mathbf{P}}}{\operatorname*{min}}
+\;\sum_{i,j} -\overline{\mathbf{P}}_{i,j}\,\overline{\mathbf{C}}_{i,j}
+$$
+
+$$
+\begin{aligned}
+\mathrm{s.t.}\quad
+&\overline{\mathbf{P}}\mathbf{1}_{n+1}
+=
+\left[\mathbf{1}_{m}^{T},\,n\right]^{T}, \\
+&
+\overline{\mathbf{P}}^{T}\mathbf{1}_{m+1}
+=
+\left[\mathbf{1}_{n}^{T},\,m\right]^{T}.
+\end{aligned}
+\tag{7}
+$$
 
 The equality constraint is a relaxation of the original element-wise inequality constraint ($\leq$), which allows each point to be matched with at most one point or dustbin, whereas each dustbin could be matched with all points at most. This relaxation makes the problem computationally efficient to solve using the Sinkhorn algorithm on GPU [11].
 
 Assignment matrix $\mathbf{P}$ is constructed by dropping the last row and column of $\overline{\mathbf{P}}$, points $\mathcal{X}_i$ and $\mathcal{Y}_j$ are associated as an object pair being added into $\hat{\mathcal{M}}$, if
 
-$\mathbf{P}_{i,j} = \arg\max \mathbf{P}(i,:) = \arg\max \mathbf{P}(:,j)$
+$$\mathbf{P}_{i,j} = \arg\max \mathbf{P}(i,:) = \arg\max \mathbf{P}(:,j)$$
 
 where $i \in [1, m]$ and $j \in [1, n]$.
 
@@ -55,7 +72,7 @@ $$\mathcal{F}_{s}^{(2)}(*) = \mathbf{R}^{(2)} \cdot (*) + \mu_{x} - \mathbf{R}^{
 
 where
 
-$\mathbf{R}^{(2)} = \mathbf{U} \mathrm{diag}\Big(1, 1, \det\left[\mathbf{U}\mathbf{V}^T\right]\Big) \mathbf{V}^T,$
+$$\mathbf{R}^{(2)} = \mathbf{U} \mathrm{diag}\Big(1, 1, \det\left[\mathbf{U}\mathbf{V}^T\right]\Big) \mathbf{V}^T,$$
 
 matrices $\mathbf{U}$ and $\mathbf{V}$ are obtainable by taking the Singular Value Decomposition on $\mathbf{Y}^{(1)}\mathbf{X}^T = \mathbf{U}\Lambda\mathbf{V}^T$.
 
diff --git a/markdowns/research_academic_paper_00005.md b/markdowns/research_academic_paper_00005.md
index e94b721dbac016441c82dd56dea1fe11264f9503..b52240bb21d51477e89b9f15108ff3267c5fb3fb 100644
--- a/markdowns/research_academic_paper_00005.md
+++ b/markdowns/research_academic_paper_00005.md
@@ -10,47 +10,47 @@ The rest of this paper is organized as follows. In Section 2 we introduce the ad
 
 In standard supervisory control (Ramadge & Wonham, 1987; Wonham and Cai, 2018), the plant to be controlled is modeled by a finite state automaton
 
-$\mathbf{G}:=(Q,\Sigma,\delta,q_{0},Q_{m})$
+$$\mathbf{G}:=(Q,\Sigma,\delta,q_{0},Q_{m})$$
 
 where $Q$ is the finite state set, $q_0 \in Q$ the initial state, $Q_m \subseteq Q$ the set of marker states, $\Sigma$ the finite event set, and $\delta : Q \times \Sigma \to Q$ the (partial) state transition function. Letting $\Sigma^*$ denote the set of all finite-length strings of events in $\Sigma$, we extend $\delta$ such that $\delta : Q \times \Sigma^* \to Q$ and write $\delta(q,s)!$ to mean that $\delta(q,s)$ is defined.
 
 The closed behavior of G is the set of all strings that can be generated by G:
 
-$L(\mathbf{G}):=\{s\in\Sigma^{*}|\delta(q_{0},s)!\}.$
+$$L(\mathbf{G}):=\{s\in\Sigma^{*}|\delta(q_{0},s)!\}.$$
 
 On the other hand, the marked behavior of G is the subset of strings that can reach a marker state:
 
-$L_{m}(\mathbf{G}):=\{s\in L(\mathbf{G})|\delta(q_{0},s)\in Q_{m}\}\subseteq L(\mathbf{G}).$
+$$L_{m}(\mathbf{G}):=\{s\in L(\mathbf{G})|\delta(q_{0},s)\in Q_{m}\}\subseteq L(\mathbf{G}).$$
 
 __G__ is nonblocking if $L(\mathbf{G}) = \overline{L_{m}(\mathbf{G})} (\overline{\cdot}$  means prefix closure), namely every string in the closed behavior may be completed to a string in the marked behavior.
 
 The event set $\Sigma$ of $\mathbf{G}$ is partitioned into a subset $\Sigma_c$ of controllable events and a subset $\Sigma_u$ of uncontrollable events. A language $E \subseteq \Sigma^*$ is said to be controllable (with respect to $\mathbf{G}$) if $\overline{E}\Sigma_u \cap L(\mathbf{G}) \subseteq \overline{E}$, i.e.
 
-$(\forall s\in\varSigma^{*},\forall\sigma\in\varSigma)s\in\overline{{E}},\sigma\in\varSigma_{u},s\sigma\in L(\mathbf{G})\Rightarrow s\sigma\in\overline{{E}}.$
+$$(\forall s\in\varSigma^{*},\forall\sigma\in\varSigma)s\in\overline{{E}},\sigma\in\varSigma_{u},s\sigma\in L(\mathbf{G})\Rightarrow s\sigma\in\overline{{E}}.$$
 
 Let $ K \subseteq L_{m}(\mathbf{G}) $ be a specification language imposed on the plant G. Denote by $ C(K) $ the family of controllable sublanguages of K, i.e.
 
-$C(K):=\{K^{\prime}\subseteq K\mid\overline{{K^{\prime}}}\Sigma_{u}\cap L(\mathbf{G})\subseteq\overline{{K^{\prime}}}\}.$
+$$C(K):=\{K^{\prime}\subseteq K\mid\overline{{K^{\prime}}}\Sigma_{u}\cap L(\mathbf{G})\subseteq\overline{{K^{\prime}}}\}.$$
 
 Then the supremal controllable sublanguage of K exists and is given by $\sup C(K)=\cup\{K^{\prime}\mid K^{\prime}\in C(K)\}$. Let $\mathbf{SUP}$ be a (nonblocking) automaton such that $L_{m}(\mathbf{SUP})=\sup C(K)$. We call $\mathbf{SUP}$ the supervisor for plant __G__ that enforces $\sup C(K)\subseteq L_{m}(\mathbf{G})$. The control action of $\mathbf{SUP}$ after an arbitrary string $s\in L(\mathbf{G})$ is to enable the events in
 
-$\gamma:=\{\sigma\in\Sigma_{u}\mid s\sigma\in L(\mathbf{G})\}\cup\{\sigma\in\Sigma_{c}\mid s\sigma\in L(\mathbf{S U P})\}.$
+$$\gamma:=\{\sigma\in\Sigma_{u}\mid s\sigma\in L(\mathbf{G})\}\cup\{\sigma\in\Sigma_{c}\mid s\sigma\in L(\mathbf{S U P})\}.$$
 
 To introduce online supervisory control based on limited lookahead policy, we need an operation that ‘truncate’ an automaton from a specified state to limited step ahead. Given an automaton $\mathbf{G} = (Q, \Sigma, \delta, q_0, Q_m)$, a state $q \in Q$, and the step number $N \ge 1$, the ‘truncation’ operation $f^N(\mathbf{G}, q)$ on $\mathbf{G}$ at state $q$ defines a new automaton
 
-$f^N(\mathbf{G}, q) = \mathbf{G}^N(q) := (Q^N, \Sigma^N, \delta^N, q_0^N, Q_m^N)$
+$$f^N(\mathbf{G}, q) = \mathbf{G}^N(q) := (Q^N, \Sigma^N, \delta^N, q_0^N, Q_m^N)$$
 
 where
 
-$Q^N = \{q' \in Q \mid (\exists s \in \Sigma^*) q' = \delta(q, s) \& |s| \leq N\}$
+$$Q^N = \{q' \in Q \mid (\exists s \in \Sigma^*) q' = \delta(q, s) \& |s| \leq N\}$$
 
-$\Sigma^N = \{\sigma \in \Sigma \mid (\exists q' \in Q^N)\delta(q',\sigma)! \text{ \& } \delta(q',\sigma) \in Q^N\}$
+$$\Sigma^N = \{\sigma \in \Sigma \mid (\exists q' \in Q^N)\delta(q',\sigma)! \text{ \& } \delta(q',\sigma) \in Q^N\}$$
 
-$\delta^N = \{(q_1, \sigma, q_2) \mid q_1, q_2 \in Q^N \& \sigma \in \Sigma^N \& \delta(q_1, \sigma) = q_2\}$
+$$\delta^N = \{(q_1, \sigma, q_2) \mid q_1, q_2 \in Q^N \& \sigma \in \Sigma^N \& \delta(q_1, \sigma) = q_2\}$$
 
-$q_{0}^{N}=q$
+$$q_{0}^{N}=q$$
 
-$Q_m^N = Q_m \cap Q^N.$
+$$Q_m^N = Q_m \cap Q^N.$$
 
 In the above definition of $Q^N$, $|s|$ denotes the length of string $s$. Note also that the marker state set $Q_m^N$ may be empty.
 
diff --git a/markdowns/research_academic_paper_00006.md b/markdowns/research_academic_paper_00006.md
index 0edb908f260b39868ec3ec87918d31fb2c13def2..dbcd1523784a8d943c71086f51fe529206e8f925 100644
--- a/markdowns/research_academic_paper_00006.md
+++ b/markdowns/research_academic_paper_00006.md
@@ -6,49 +6,52 @@ Let $ s \in L(\mathbf{G}) $ and plant G is at state $q := \delta(q_0, s)$. Since
 
 (1) For each component agent $ \mathbf{G}_{k} \left( k \in \{1, \ldots, n\} \right) $, generate the N-step-ahead behavior from the current state $q_{k}$ by applying the truncation operation
 
-$\mathbf{G}_{k}^{N}(q_{k})=f^{N}(\mathbf{G}_{k},q_{k}).$
+$$\mathbf{G}_{k}^{N}(q_{k})=f^{N}(\mathbf{G}_{k},q_{k}).$$
 
 (2) Compute the plant model $\mathbf{G}^{N}(q)$ as the synchronous product of all truncated component models $\mathbf{G}_{k}^{N}(q_{k})$:
 
-$\mathbf{G}^{N}(q)=\mathbf{G}_{1}^{N}(q_{1})\|\cdots\|\mathbf{G}_{n}^{N}(q_{n}).$
+$$\mathbf{G}^{N}(q)=\mathbf{G}_{1}^{N}(q_{1})\|\cdots\|\mathbf{G}_{n}^{N}(q_{n}).$$
 
 (3) The states that are exactly $N$-step-ahead from $q$ in $\mathbf{G}^N(q)$ are called pending states. A pending state may transition to an undesirable state by an uncontrollable event, or it may not be able to reach a marker state. There are two opposite attitudes as to how to deal with pending states, for which we follow Chung et al. (1992).
 
 (a) Conservative attitude: all pending states are deemed undesirable state and are removed, i.e.
 
-$\mathbf{G}(q) = f^{N-1}(\mathbf{G}^N(q), q) = (Q^N, \Sigma^N, \delta^N, q, Q_m^N).$
+$$\mathbf{G}(q) = f^{N-1}(\mathbf{G}^N(q), q) = (Q^N, \Sigma^N, \delta^N, q, Q_m^N).$$
 
 (b) Optimistic attitude: all pending states are deemed desirable states and are treated as marker states, i.e.
 
-$\mathbf{G}^{\prime}(q)=f^{N}(\mathbf{G}^{N}(q),q)=(Q^{N},\Sigma^{N},\delta^{N},q,Q_{m}^{^{\prime}N});$
+$$\mathbf{G}^{\prime}(q)=f^{N}(\mathbf{G}^{N}(q),q)=(Q^{N},\Sigma^{N},\delta^{N},q,Q_{m}^{^{\prime}N});$$
 
-$\mathbf{G}(q) = (Q^N, \Sigma^N, \delta^N, q, Q_m^N)$
+$$\mathbf{G}(q) = (Q^N, \Sigma^N, \delta^N, q, Q_m^N)$$
 
 where
 
-$Q_m^N := Q_m^{'N} \cup \{q' \in Q^N \mid (\exists s \in \Sigma^*) q' = \delta^N(q, s) \& |s| = N\}.$
+$$Q_m^N := Q_m^{'N} \cup \{q' \in Q^N \mid (\exists s \in \Sigma^*) q' = \delta^N(q, s) \& |s| = N\}.$$
 
 (4) Let $ K \subseteq L_{m}(\mathbf{G}) $ be a specification, and $ \mathbf{K} = (P, \Sigma, \pi, p_{0}, P_{m}) $ be an automaton such that $ L_{m}(\mathbf{K}) = K $. Since $ s \in L_{m}(\mathbf{G}) $ has occurred in $\mathbf{G}$, the current state of $\mathbf{K}$ is $ p := \pi(p_{0}, s) $. The truncated specification for $ \mathbf{G}(q) $ is determined again according to two opposite attitudes.
 
 (a) Conservative attitude:
 
-$\mathbf{K}(p)=f^{N-1}(\mathbf{K}^{N}(p),p)=(P^{N},\Sigma^{N},\pi^{N},p,P_{m}^{N}).$
+$$\mathbf{K}(p)=f^{N-1}(\mathbf{K}^{N}(p),p)=(P^{N},\Sigma^{N},\pi^{N},p,P_{m}^{N}).$$
 
 (b) Optimistic attitude:
 
-$\mathbf{K}'(p) = f^{N}(\mathbf{K}^{N}(p), p) = (P^{N}, \Sigma^{N}, \pi^{N}, p, P_{m}^{'N});$
+$$\mathbf{K}'(p) = f^{N}(\mathbf{K}^{N}(p), p) = (P^{N}, \Sigma^{N}, \pi^{N}, p, P_{m}^{'N});$$
 
-$\mathbf{K}(p)=(P^{N},\Sigma^{N},\pi^{N},p,P_{m}^{N})$
+$$\mathbf{K}(p)=(P^{N},\Sigma^{N},\pi^{N},p,P_{m}^{N})$$
 
 where
 
-$P_m^N := P_m^{'N} \cup \{p' \in P^N \mid (\exists s \in \Sigma^*) p' = \pi^N(p, s) \& |s| = N\}.$
+$$P_m^N := P_m^{'N} \cup \{p' \in P^N \mid (\exists s \in \Sigma^*) p' = \pi^N(p, s) \& |s| = N\}.$$
 
 (5) Compute the supremal controllable sublanguage $\sup C(L_{m}(\mathbf{K}(p)))$ with respect to $\mathbf{G}(q)$, and let $\mathbf{SUP}(q) = (X^{N}, \Sigma^{N}, \xi^{N}, q, X_{m}^{N})$ be the supervisor for $\mathbf{G}(q)$ that enforces $\sup C(L_{m}(\mathbf{K}(p)))$, i.e. $L_{m}(\mathbf{SUP}(q)) = \sup C(L_{m}(\mathbf{K}(p)))$. The control action of $\mathbf{SUP}(q)$ at state q is to enable the events in
 
-$\gamma(q):=\{\sigma\in\Sigma_{c}\mid\xi^{N}(q,\sigma)\in L(\mathbf{S U P}(q))\}$
-
-$\cup\{\sigma\in\Sigma_{u}\mid\delta^{N}(q,\sigma)\in L(\mathbf{G}(q))\}.$
+$$
+\begin{aligned}
+\gamma(q) :=\; &\{\sigma \in \Sigma_{c} \mid \xi^{N}(q,\sigma) \in L(\mathbf{SUP}(q))\} \\
+&\cup \{\sigma \in \Sigma_{u} \mid \delta^{N}(q,\sigma) \in L(\mathbf{G}(q))\}.
+\end{aligned}
+$$
 
 One event $\sigma\in\gamma(q)$ will be executed such that the plant G transitions to the next state $\delta(q,\sigma)$. Correspondingly, component agents $G_{k}$ will transition to the next state $\delta_{k}(q_{k},\sigma)$ (provided $\delta_{k}(q_{k},\sigma)!$). The above procedure repeats with string $s\sigma$.
 
diff --git a/markdowns/research_academic_paper_00008.md b/markdowns/research_academic_paper_00008.md
index 04bbf56046e9614093d24c3de42441e48f93b0fc..7293986487211b2fbeae36f31d60790a03864066 100644
--- a/markdowns/research_academic_paper_00008.md
+++ b/markdowns/research_academic_paper_00008.md
@@ -6,7 +6,7 @@ Eq. (9) shows the memory update strategy for the ID $i$. And the label set $y_B$
 
 *3) The Supervision under GSAM:* In this part, we show how to utilize GSAM during training. GSAM can be used as global prior information to supervise the training process through a typical Cross-Entropy loss. Besides, to further ensure the compactness in the feature space, GSAM can also be used to constrain the intra-class variance. The loss function that corresponds to GSAM is defined as:
 
-$\mathcal{L}_{G S A M}=\mathcal{L}_{C E}(F(X)\mathcal{M}^{T}/\tau,Y)+(F(X)-\overline{{F(X)}})^{2},\quad(10)$
+$$\mathcal{L}_{G S A M}=\mathcal{L}_{C E}(F(X)\mathcal{M}^{T}/\tau,Y)+(F(X)-\overline{{F(X)}})^{2},\quad(10)$$
 
 where the operation between $ F(x) $ and M is matrix multiplication. $\tau$ denotes the temperature to balance the feature distribution, and $ F(x) $ denotes the averaged feature along batch dimension.
 
diff --git a/markdowns/research_academic_paper_00016.md b/markdowns/research_academic_paper_00016.md
index d15ef0e022ace6a11262bb9d890b2a8421845956..d4a982bcf05a13a4217c5f737d25ea98c7d92ec8 100644
--- a/markdowns/research_academic_paper_00016.md
+++ b/markdowns/research_academic_paper_00016.md
@@ -30,7 +30,7 @@ EIS was carried out in the range of 0.1 Hz to 100 kHz.
 
 For Ph-NO$_2$ ERR, the electrolyte was collected and extracted with ethyl acetate after $i$-$t$ tests. The extracted products were determined by comparing the GC retention times and mass spectra. The Ph-NO$_2$ conversion and Ph-NH$_2$ selectivity were acquired via the GC results analysis and calculated based on the following Eqs. (7) and (8):
 
-$\text{Conversion} = \frac{\text{mol of the consumed Ph-NO}_{2}}{\text{mole of the added Ph-NO}_{2}} \times 100\%$   (7)
+$$\text{Conversion} = \frac{\text{mol of the consumed Ph-NO}_{2}}{\text{mole of the added Ph-NO}_{2}} \times 100\%  \tag{7}$$
 
 $$Selectivity = \frac{\text{mol of the as-formed Ph-NH}_2}{\text{mole of the added Ph-NO}_2} \times 100\% \tag{8}$$
 
diff --git a/markdowns/research_academic_paper_00022.md b/markdowns/research_academic_paper_00022.md
index 43f89f6f433dcea34e941411c5b1b35eb314ef68..359f065e0b3065e632c17b0b7dd911206b5adec3 100644
--- a/markdowns/research_academic_paper_00022.md
+++ b/markdowns/research_academic_paper_00022.md
@@ -75,7 +75,7 @@ $$D_k \cdot D_k \cdot M \cdot D_F \cdot D_F + M \cdot N \cdot D_F \cdot D_F \tag
 
 By comparing depthwise separable convolution and standard convolution, we can get a ratio of the computational cost.
 
-$\frac{D_{k}\cdot D_{k}\cdot M\cdot D_{F}\cdot D_{F}+M\cdot N\cdot D_{F}\cdot D_{F}}{D_{k}\cdot D_{k}\cdot M\cdot D_{F}\cdot D_{F}\cdot N}=\frac{1}{N}+\frac{1}{D_{k}^{2}}$ (8)
+$$\frac{D_{k}\cdot D_{k}\cdot M\cdot D_{F}\cdot D_{F}+M\cdot N\cdot D_{F}\cdot D_{F}}{D_{k}\cdot D_{k}\cdot M\cdot D_{F}\cdot D_{F}\cdot N}=\frac{1}{N}+\frac{1}{D_{k}^{2}}\tag{8}$$
 
 Because the number of output channels is generally large, it can be ignored. We assume that using $3 \times 3$ convolutions separately, depthwise separable convolutions can reduce the computational cost by 8 to 9 times compared to standard convolutions.
 
diff --git a/markdowns/research_academic_paper_00023.md b/markdowns/research_academic_paper_00023.md
index c140042aa2eba639bcf2746470f4bf8c9a9d23a4..985488b4dfabdb8e18ce3ccf245a73739e4e7bb5 100644
--- a/markdowns/research_academic_paper_00023.md
+++ b/markdowns/research_academic_paper_00023.md
@@ -9,7 +9,7 @@ Propensity score matching (PSM).
 
 Panel A of Table 2 reports the t-test of mean differences in covariates between treated and control firms in the pre-S-135 period and Panel B of Table 2 shows a probit model for PSM as per the following specification:
 
-$Treated_{i} = \alpha + X_{it}.\beta' + \vartheta_{j} + \varepsilon_{it}$
+$$Treated_{i} = \alpha + X_{it}.\beta' + \vartheta_{j} + \varepsilon_{it}$$
 
 where $Treated_i$ is a categorical variable that takes the value of one if the firm is affected by S-135 and zero otherwise. $X_{it}$ is the vector of covariates used for matching that comprises Size, Leverage, Cash, CapEx, and IO, all as defined in Table 1. $\theta_j$ is the industry fixed effects using the Fama-French 17 industries classification. Model [1] presents a probit model predicting the likelihood of being a treated firm from the entire sample of firms with no missing covariates in the pre-S-135 period. Model [2] presents the probit likelihood model for matched treated and comparison firms using PSM with replacement. Heteroskedasticity robust t-stats are presented in parentheses. *, **, and *** denote statistical significance at the 10%, 5% and 1% significance levels, respectively. Data source: CIQ and CMIE database.
 
diff --git a/markdowns/research_academic_paper_00024.md b/markdowns/research_academic_paper_00024.md
index 1db29d4f9a511359bf0651bdde2b791f8ed046fe..5c644033e038d80cf50e82347c6f3fe7bcd591d6 100644
--- a/markdowns/research_academic_paper_00024.md
+++ b/markdowns/research_academic_paper_00024.md
@@ -216,7 +216,7 @@ Cross-sectional heterogeneity.
 
 Table 9 reports the multivariate PSM-DiD regression results using the subsample of firms (as indicated) as per the following specification:
 
-$S M L_{i t}=\alpha+\beta.(Treated_{i}\times After_{t})+\pmb{X}_{i t-1}.\pmb{\delta}^{\prime}+\gamma_{i}+\tau_{t}+\varepsilon_{i t}$
+$$S M L_{i t}=\alpha+\beta.(Treated_{i}\times After_{t})+\pmb{X}_{i t-1}.\pmb{\delta}^{\prime}+\gamma_{i}+\tau_{t}+\varepsilon_{i t}$$
 
 where $SML_{it}$ is the stock market liquidity, proxied by Amihud, as defined in Table 1, of firm $i$'s stock in year $t$. Treated, and After, are categorical variables, as defined in Table 4. The DiD is the interaction between Treated, and After, dummies. $X_{it-1}$ is a vector of one year lagged covariates that include Size, Leverage, Cash, CapEx, and IO and stock market variables that consist of Analyst, Volatility, and Turnover, all as defined in Table 1. $\gamma_i$ and $\tau_t$ are the firm and year fixed effects, respectively. Standard errors are clustered at the firm level and t-stats are presented in parenthesis. *, **, and *** denote statistical significance at 10%, 5% and 1% significance levels, respectively. The overall sample period ranges from 2012 to 2017. Data sources: CIQ and CMIE database.
 
diff --git a/markdowns/research_academic_paper_00025.md b/markdowns/research_academic_paper_00025.md
index ca6e4e2a27606bcf12af2b5b72b5ef66bfe84de7..67af8ce89446f6e9ee2ddb8df0ec7fb33e67c627 100644
--- a/markdowns/research_academic_paper_00025.md
+++ b/markdowns/research_academic_paper_00025.md
@@ -5,7 +5,7 @@ Table A2 (continued)
 
 Table A2 reports the multivariate DiD regression results using the PSM matched treated and control group firms as per the following specification:
 
-$SML_{it} = \alpha + \beta. (Treated_i \times After_t) + X_{it-1} \delta' + \gamma_i + \tau_t + \varepsilon_{it}$
+$$SML_{it} = \alpha + \beta. (Treated_i \times After_t) + X_{it-1} \delta' + \gamma_i + \tau_t + \varepsilon_{it}$$
 
 where $SML_{it}$ is the stock market liquidity, proxied by Amihud, Zeros, HL, or FHT, all as defined in Table 1, of firm i’s stock in year t. Treatedi and Aftert are categorical variables, as defined in Table 4. The DID is the interaction between $Treated_{i}$ and Aftert dummies. $X_{it-1}$ is a vector of one year lagged covariates that include Size, Leverage, Cash, CapEx, IO, Goodwill (taken as firm’s goodwill scaled by total assets), and Age (taken as the natural logarithm of firm age in years), and stock market variables that consist of Analyst, Volatility, and Turnover, all as defined in Table 1. $\gamma_i$ and $\tau_t$ are the firm and year fixed effects, respectively. Standard errors are clustered at the firm level and t-stats are presented in parenthesis. *, **, and *** denote statistical significance at 10%, 5% and 1% significance levels, respectively. The overall sample period ranges from 2012 to 2017. Data sources: CIQ and CMIE database.
 
@@ -183,7 +183,7 @@ Treated firms without extraordinary events.
 
 Table A3 reports the multivariate PSM-DiD regression results without the treated firms subjected to extraordinary events in the post-S-135 period:
 
-$SML_{it} = \alpha + \beta. (Treated_i \times After_t) + X_{it-1} \delta' + \gamma_i + \tau_t + \varepsilon_{it}$
+$$SML_{it} = \alpha + \beta. (Treated_i \times After_t) + X_{it-1} \delta' + \gamma_i + \tau_t + \varepsilon_{it}$$
 
 where $SML_{it}$ is the stock market liquidity, proxied by Amihud, Zeros, HL, or FHT, all as defined in Table 1, of firm i’s stock in year t. Treatedi and Aftert are categorical variables, as defined in Table 4. The DiD is the interaction between $Treated_{i}$ and Aftert, dummies. $X_{it-1}$ is a vector of one year lagged covariates that include Size, Leverage, Cash, CapEx, and IO and stock market variables that consist of Analyst, Volatility, and Turnover, all as defined in Table 1. $\gamma_i$ and $\tau_t$ are the firm and year fixed effects, respectively. Standard errors are clustered at the firm level and t-stats are presented in parenthesis. *, **, and *** denote statistical significance at 10%, 5% and 1% significance levels, respectively. The overall sample period ranges from 2012 to 2017. Data sources: CIQ and CMIE database.
 
diff --git a/markdowns/research_academic_paper_00026.md b/markdowns/research_academic_paper_00026.md
index e270174ab6e77ae3a5f9bc2560451570618d0c12..948b53300e4b350fdbceadd46e3049542eba51e4 100644
--- a/markdowns/research_academic_paper_00026.md
+++ b/markdowns/research_academic_paper_00026.md
@@ -28,7 +28,7 @@ $$T_{\lambda} \equiv \lambda_{\mathrm{m}} \max(\hat{t}_{N} - \hat{t}_{0}), \quad
 
 where $\lambda_m$ is the maximum Lyapunov exponent, and thus, $T_\lambda$ is in Lyapunov time $1/\lambda_m$. The normalized error is defined by
 
-$E(\hat{t}_{N})=\frac{\|\mathbf{v}_{\mathrm{d}}(\hat{t}_{N})-\mathbf{v}(\hat{t}_{N})\|}{\sqrt{\left\langle\mathbf{v}_{\mathrm{d}}^{2}\right\rangle}}$ (17)
+$$E(\hat{t}_{N})=\frac{\|\mathbf{v}_{\mathrm{d}}(\hat{t}_{N})-\mathbf{v}(\hat{t}_{N})\|}{\sqrt{\left\langle\mathbf{v}_{\mathrm{d}}^{2}\right\rangle}} \tag{17}$$
 
 where the mean square of the target is given by
 
diff --git a/markdowns/research_academic_paper_00027.md b/markdowns/research_academic_paper_00027.md
index ea742c0990a1b71cbfc6baa99396172c868f8929..19732898289c642be44bab7c9b3ec773c831f1b3 100644
--- a/markdowns/research_academic_paper_00027.md
+++ b/markdowns/research_academic_paper_00027.md
@@ -8,7 +8,7 @@ and $t = 0, 1, -1, 2, -2, \ldots$ indexes the pixels.
 
 The gradient signal on the input pixels is simply $o = u * v * \cdots * v$, convolving $u$ with $n$ such $v$'s. To compute this convolution, we can use the Discrete Time Fourier Transform to convert the signals into the Fourier domain, and obtain
 
-$U(\omega)=\sum_{t=-\infty}^{\infty}u(t)e^{-j\omega t}=1,\quad V(\omega)=\sum_{t=-\infty}^{\infty}v(t)e^{-j\omega t}=\sum_{m=0}^{k-1}e^{-j\omega m}\quad(2)$
+$$U(\omega)=\sum_{t=-\infty}^{\infty}u(t)e^{-j\omega t}=1,\quad V(\omega)=\sum_{t=-\infty}^{\infty}v(t)e^{-j\omega t}=\sum_{m=0}^{k-1}e^{-j\omega m}\quad(2)$$
 
 Applying the convolution theorem, we have the Fourier transform of $o$ is
 
diff --git a/markdowns/research_academic_paper_00032.md b/markdowns/research_academic_paper_00032.md
index 30ab5d480901586d8e97aea412c699abf550c015..d8d1f64df76dc633b0d8437af038e5a52349a0e1 100644
--- a/markdowns/research_academic_paper_00032.md
+++ b/markdowns/research_academic_paper_00032.md
@@ -2,21 +2,38 @@
 
 According to Eqs. (1) and (15), the nonlinear model of roll motion can be summarized as following equation:
 
-$$\left(I_{xx}+K_{p}\right)\ddot{\phi}+k_{p}\dot{\phi}+K_{p|p|}\dot{\phi}\left|\dot{\phi}\right|$$
-
-$$+\Delta\left(c_{1}\phi+c_{3}\phi^{3}+c_{5}\phi^{5}\right)=\tau_{d}-K_{\alpha}\alpha_{e}$$
-
-$$\alpha_{e}=-\frac{R_{f}p}{V}-\alpha_{m}\tag{16}$$
-
-$$T_{e}\alpha_{m}+\alpha_{m}=K_{d c}\alpha_{c}.$$
+$$
+\begin{aligned}
+\left(I_{xx}+K_{p}\right)\ddot{\phi}
+&+ k_{p}\dot{\phi}
++ K_{p|p|}\dot{\phi}\left|\dot{\phi}\right| \\
+&+ \Delta\left(c_{1}\phi+c_{3}\phi^{3}+c_{5}\phi^{5}\right)
+= \tau_{d}-K_{\alpha}\alpha_{e} \\
+\alpha_{e}
+&= -\frac{R_{f}p}{V}-\alpha_{m} \tag{16}
+\end{aligned}
+$$
+
+$$
+T_{e}\alpha_{m}+\alpha_{m}=K_{dc}\alpha_{c}.
+$$
 
 Also, based on the assumptions listed in section 2.3, it is considered linear model roll motion. A linear state space model to describe the dynamic of the roll motion can be presented by
 
-$$\dot{\text{x}}_1(\text{t}) = \text{x}_2(\text{t})$$
-
-$$\dot{x}_2(t) = \frac{-\Delta GM}{I_{xx} + K_p} x_1(t) + \frac{(k_\alpha R_f / V) - k_p}{I_{xx} + K_p} x_2(t) \\+ \frac{k_\alpha}{I_{xx} + K_p} u(t) + \frac{1}{I_{xx} + K_p} \tau_d(t) \tag{17}$$
-
-$$\dot{\text{x}}_3(t) = \text{k}_2\text{x}_3(t) + \text{k}_1\text{u}(t)$$
+$$
+\begin{aligned}
+\dot{x}_1(t) &= x_2(t), \\
+\dot{x}_2(t)
+&= \frac{-\Delta GM}{I_{xx}+K_p}\,x_1(t)
+ + \frac{(k_\alpha R_f/V)-k_p}{I_{xx}+K_p}\,x_2(t) \\
+&\quad + \frac{k_\alpha}{I_{xx}+K_p}\,u(t)
+ + \frac{1}{I_{xx}+K_p}\,\tau_d(t). \tag{17}
+\end{aligned}
+$$
+
+$$
+\dot{x}_3(t)=k_2 x_3(t)+k_1 u(t).
+$$
 
 where $x(t) = [x_1(t), x_2(t), x_3(t)]^T$ is $[\phi(t), p(t), \alpha_m(t)]^T$ and $u(t)$ is $\alpha_c(t)$. As well as $k_1 = k_{dk}/T_e$ and $k_2 = 1/T_e$. Also, the operational constraints should be considered as:
 
diff --git a/markdowns/research_academic_paper_00033.md b/markdowns/research_academic_paper_00033.md
index 94a66b989c091ccfe66e2de8a782b55d12188cff..5fa2f3cd0b3e1b03c2845f6985452569f27c2448 100644
--- a/markdowns/research_academic_paper_00033.md
+++ b/markdowns/research_academic_paper_00033.md
@@ -61,10 +61,10 @@ Fig. 8(c) shows the profiles of $\Delta A$ versus $\omega$ at the time of maximu
 
 As demonstrated in our previous work [58,59], vector beating solitons are a special case of vector KMSs when the vector plane wave has the equal wavenumber. Here, we consider the eigenvalues of KMSs given by Eq. (48) under the condition $k_1 = -k_2 = k = 0$. Consequently, exact solutions of beating solitons can be written as
 
-$\psi^{(j)} = \phi_D \pm \phi_B,$
+$$\psi^{(j)} = \phi_D \pm \phi_B,$$
 
 $$\phi_{D}=\frac{\epsilon_{1}e^{4x\beta}+\epsilon_{3}e^{2x\beta+2\rho_{K}t}+\epsilon_{5}e^{4\rho_{K}t}}{\gamma(\eta_{1}^{\prime}e^{2x\beta}+\eta_{2}^{\prime}e^{2\rho_{K}})^{\textstyle2}}e^{i\mu(a_{1}^{2}+a_{2}^{2})t},\tag{65}$$
 
-$\phi_{B}=\frac{\epsilon_{2}e^{3x\beta+\rho_{K}t+i\sigma_{K}t}+\epsilon_{4}e^{x\beta+3\rho_{K}t+i\sigma_{K}t}}{\gamma(\eta_{1}^{\prime}e^{2x\beta}+\eta_{2}^{\prime}e^{2\rho_{K}})^{2}}e^{i\mu(a_{1}^{2}+a_{2}^{2})t},$
+$$\phi_{B}=\frac{\epsilon_{2}e^{3x\beta+\rho_{K}t+i\sigma_{K}t}+\epsilon_{4}e^{x\beta+3\rho_{K}t+i\sigma_{K}t}}{\gamma(\eta_{1}^{\prime}e^{2x\beta}+\eta_{2}^{\prime}e^{2\rho_{K}})^{2}}e^{i\mu(a_{1}^{2}+a_{2}^{2})t},$$
 
 where $\phi_D$ denotes the dark soliton component and $\phi_B$ denotes the bright soliton component. It is evident to see from the solution (65) that the beating solitons consist of a linear superposition between a dark and a bright solitons. Parameters $\rho_K$ and $\sigma_K$ are given by Eqs. (51) and (53), respectively. Other parameters are given in Appendix D. The group velocity and the oscillating period of the beating solitons are given by Eqs. (50) and (52), respectively.
\ No newline at end of file
diff --git a/markdowns/research_academic_paper_00036.md b/markdowns/research_academic_paper_00036.md
index ac3f8cd9e0ff5ed2de421eab9d90c91e5cac5680..0709b694f20e60d095f5fb20cb3d763dff6b4e03 100644
--- a/markdowns/research_academic_paper_00036.md
+++ b/markdowns/research_academic_paper_00036.md
@@ -21,7 +21,7 @@ $$E_{i}^{\text{off}}[n] = \delta_{0} P_{i}^{\text{off}}[n] = \frac{\delta_{0} \s
 
 Similarly, the amount of the task bits of the near UE $i$ offloaded to the UAV at the $n$th time slot, denoted by $L_i^{\text{lo}}[n]$, is expressed as
 
-$L_{i}^{\mathrm{l o}}[n]=\delta_{0}B\log_{2}\Bigg(1+\frac{P_{i}^{\mathrm{l o}}[n]h_{i}[n]}{\sigma_{0}^{2}}\Bigg),\forall n\in\{1,2,\ldots,N-1\},i\in\mathcal{R},\quad(11)$
+$$L_{i}^{\mathrm{l o}}[n]=\delta_{0}B\log_{2}\Bigg(1+\frac{P_{i}^{\mathrm{l o}}[n]h_{i}[n]}{\sigma_{0}^{2}}\Bigg),\forall n\in\{1,2,\ldots,N-1\},i\in\mathcal{R},\tag{11}$$
 
 from which, the expression for $P_i^{\text{lo}}[n]$ is obtained as
 
@@ -29,7 +29,7 @@ $$P_{i}^{\mathrm{l o}}[n]=\frac{\sigma_{0}^{2}}{h_{i}[n]}\left(2^{\frac{L_{i}^{\
 
 The energy consumption of the near UE $i$ at the $n$th time slot for offloading its own task is
 
-$\begin{array}{r l r}&{}&{E_{i}^{\mathrm{l o}}[n]=\delta_{0}P_{i}^{\mathrm{l o}}[n]=\cfrac{\delta_{0}\sigma_{0}^{2}}{h_{i}[n]}\left(2^{\frac{L_{i}^{\mathrm{l o}}[n]}{B\delta_{0}}}-1\right),\forall n\in\{1,2,\ldots,N-1\},i\in\mathcal{R}.}\\ &{}&{\quad(13)}\end{array}$
+$$\begin{array}{r l r}&{}&{E_{i}^{\mathrm{l o}}[n]=\delta_{0}P_{i}^{\mathrm{l o}}[n]=\cfrac{\delta_{0}\sigma_{0}^{2}}{h_{i}[n]}\left(2^{\frac{L_{i}^{\mathrm{l o}}[n]}{B\delta_{0}}}-1\right),\forall n\in\{1,2,\ldots,N-1\},i\in\mathcal{R}.}\\ &{}&{\quad(13)}\end{array}$$
 
 ## 2.2. Computing model
 
diff --git a/markdowns/research_academic_paper_00038.md b/markdowns/research_academic_paper_00038.md
index 28ee7d9d3c5a7e43bcdce3113a705266cc2289a8..90e9bcda37fc0f04202602acade96e213a00e1cd 100644
--- a/markdowns/research_academic_paper_00038.md
+++ b/markdowns/research_academic_paper_00038.md
@@ -12,7 +12,7 @@ $$\sum_{n=1}^{N-1} L_i^{\text{lo}}[n] = \sum_{n=1}^{N-1} \left[ B\delta_0 \log_2
 
 From (48),(49),(50), the range of $\sum_{n=1}^{N-1} L_i^{\text{lo}}[n]$ can be obtained as $\sum_{n=1}^{N-1} L_i^{\text{lo}}[n] \in \left[ \bar{L}_{i,\min}^{\text{lo}}, \bar{L}_{i,\max}^{\text{lo}} \right]$, where
 
-$\bar{L}_{i,\min}^{\mathrm{lo}}=I_{i}-\frac{T}{C_{i}}f_{\max},$
+$$\bar{L}_{i,\min}^{\mathrm{lo}}=I_{i}-\frac{T}{C_{i}}f_{\max},$$
 
 $$\bar{L}_{i,\max}^{\mathrm{l o}}=\min\left\{\sum_{n=1}^{N-1}L_{i,\max}^{\mathrm{l o}}[n],(N-1)\frac{\delta}{C_{i}}f_{u,\max}\right\}. \tag{51}$$
 
diff --git a/markdowns/research_academic_paper_00074.md b/markdowns/research_academic_paper_00074.md
index 0928e9a5a795c7a58ee7a7883fefca17876fa271..4b1d1fc34f7b6c06c805e74ff9b1dc0b30151d45 100644
--- a/markdowns/research_academic_paper_00074.md
+++ b/markdowns/research_academic_paper_00074.md
@@ -113,11 +113,11 @@
 
 在建模前需要将燃油流量、压力、温度和转速修正到标准大气压海平面状态下。在此引入温度修正因子 $\theta$ 和压力修正因子 $\delta^{[15]}$
 
-$\theta=\frac{T_{AT}+273.15}{288.15}$ (1)
+$$\theta=\frac{T_{AT}+273.15}{288.15}\tag{1}$$
 
-$\delta=(3.5+0.7Ma^{2})\times(1-2.26\times10^{-5}H)^{5.23585},H\leqslant11000(2)$
+$$\delta=(3.5+0.7Ma^{2})\times(1-2.26\times10^{-5}H)^{5.23585},H\leqslant11000\tag{2}$$
 
-$\delta=(3.5+0.7Ma^{2})\times0.224e^{-\frac{H-11000}{634.62}}$ (3)
+$$\delta=(3.5+0.7Ma^{2})\times0.224e^{-\frac{H-11000}{634.62}}\tag{3}$$
 
 式中：H 为飞行高度，根据 $\theta$ 和 $\delta$ 对参数进行修正。
 温度修正公式为
@@ -127,17 +127,17 @@ $$T_{\mathrm{cm}}=\frac{T_{\mathrm{mw}}}{\theta}\tag{4}$$
 式中：$T_{\text{cor}}$为修正后温度；$T_{\text{raw}}$为修正前温度。
 压力修正公式为
 
-$P_{\mathrm{cor}}=\frac{P_{\mathrm{mw}}}{\delta}$ (5)
+$$P_{\mathrm{cor}}=\frac{P_{\mathrm{mw}}}{\delta}\tag{5}$$
 
 式中：$P_{m}$ 为修正后压力；$P_{m}$ 为修正前压力。
 转速修正公式为
 
-$N_{\mathrm{cr}}=\frac{N_{\mathrm{max}}}{\sqrt{\theta}} \quad (6)$
+$$N_{\mathrm{cr}}=\frac{N_{\mathrm{max}}}{\sqrt{\theta}} \quad (6)$$
 
 式中：$N_{\text{cor}}$ 为修正后转速；$N_{\text{raw}}$ 为修正前转速。
 燃油流量修正公式为
 
-$F_{\mathrm{corr}}=\frac{F_{\mathrm{raw}}}{\delta \sqrt{\theta}} \quad (7)$
+$$F_{\mathrm{corr}}=\frac{F_{\mathrm{raw}}}{\delta \sqrt{\theta}} \quad (7)$$
 
 式中：$F_{\text{cor}}$为修正后燃油流量；$F_{\text{raw}}$为修正前燃油流量。
 
diff --git a/markdowns/research_academic_paper_00079.md b/markdowns/research_academic_paper_00079.md
index bd62157d823401b256a7abbae05bb25deebb7885..4e776ca44572ccab7ce229b5147afd7395c5b141 100644
--- a/markdowns/research_academic_paper_00079.md
+++ b/markdowns/research_academic_paper_00079.md
@@ -6,11 +6,11 @@
 
 叶冠拓扑优化的数学模型为$^{[17]}$
 
-$\begin{cases}
+$$\begin{cases}
 Find:  e=\{e_1,e_2,\cdots,e_n\}^T \\
 Min:  P=\frac{\ln(SE)_{di} \cdot V_{di}}{\ln(SE)_{dD} \cdot V_{dD}} & (1) \\
 s.t.:  (\sigma_{max})_i \leq \frac{(\sigma^*)_i}{s_i}
-\end{cases}$
+\end{cases}$$
 
 式中：$e_{i}$ 为设计变量，表征结构中的单元，取值为0或1，取值为0时代表删除单元，取值为1时代表保留单元，最终得到的e表征当前结构较原结构删除/保留单元的情况；P为无量纲目标函数，优化目标为使结构质量最轻、刚度最大；$V_{d}$为结构体积，$V_{di}$和$V_{d0}$分别为当前和初始结构总体积；$1/\ln(SE)_{d}$为结构刚度，结构刚度越大，在受力时抵抗弹性变形的能力越强，在激振力作用下的振幅越小，其中$(SE)_{di}$和$(SE)_{d0}$分别为当前和初始结构总应变能；$(\sigma_{\max})_r$，为第r个应力约束所表示的最大值；$(\sigma^{*})_r$，为第r个应力约束所表示的许用值；$s_{r}$为相应应力下的安全系数。
 
diff --git a/markdowns/research_academic_paper_00083.md b/markdowns/research_academic_paper_00083.md
index 840773c0eb26fe4bbbe11dacd7038bfafedb2d42..eba023b3255c4a312e2201d032b21237770accd4 100644
--- a/markdowns/research_academic_paper_00083.md
+++ b/markdowns/research_academic_paper_00083.md
@@ -3,14 +3,13 @@
 
 $$\min_{s,h,Q} p(s,N)Q + \mu(s+h)$$
 
-$St:f(b,Q)=C$
+$$St:f(b,Q)=C$$
 
 其 $f(h,Q)$ 中 $h$ 是家庭生产所需投入时间, $Q$ 是投入市场品的数量, $\mu$ 是机会成本。假设家庭能够自由支配家庭生产时间, 建立拉格朗日函数得
 
-$-\frac{\partial p}{\partial s}Q = \mu$ 
-(3)
+$$-\frac{\partial p}{\partial s}Q = \mu\tag{3}$$
 
-$\frac{\partial f}{\partial h} \lambda = \mu$
+$$\frac{\partial f}{\partial h} \lambda = \mu\tag{4}$$
 
 $$\frac{\partial p}{\partial Q}Q+p=\lambda\frac{\partial f}{\partial Q} \tag{5}$$
 
diff --git a/markdowns/research_academic_paper_00092.md b/markdowns/research_academic_paper_00092.md
index b118359c46a0c128132e358a88349b3bbe399349..5ce7b2e1d5606c514c265b390f25093b40d2ed2f 100644
--- a/markdowns/research_academic_paper_00092.md
+++ b/markdowns/research_academic_paper_00092.md
@@ -1,48 +1,48 @@
 的近似公式为
 
-$\overline{D}_{NB} = \nu_0 C^{-1} (\sqrt{2} \sigma_x)^k \Gamma(1 + \frac{k}{2})$ (8)
+$$\overline{D}_{NB} = \nu_0 C^{-1} (\sqrt{2} \sigma_x)^k \Gamma(1 + \frac{k}{2})\tag{8}$$
 
-$\overline{D}_{\mathrm{RC}} \cong \overline{D}_{\mathrm{NB}} \alpha_{2}^{k-1}$ (9)
+$$\overline{D}_{\mathrm{RC}} \cong \overline{D}_{\mathrm{NB}} \alpha_{2}^{k-1}\tag{9}$$
 
 式中：$v_{0}$ 为谱密度函数 $S_{x}(\omega)$ 的平均上跨率
 
-$v_0=\frac{1}{2\pi}\sqrt{\frac{\lambda_2}{\lambda_0}}$ (10)
+$$v_0=\frac{1}{2\pi}\sqrt{\frac{\lambda_2}{\lambda_0}}\tag{10}$$
 
 式中：$\lambda_{m}$ 为单边谱密度函数 $S_{x}(\omega)$ 的谱距
 
-$\lambda_{m}=\int_{0}^{+\infty}\omega^{m}S_{x}(\omega)d\omega,m=0,1,2,\cdots \quad (11)$
+$$\lambda_{m}=\int_{0}^{+\infty}\omega^{m}S_{x}(\omega)d\omega,m=0,1,2,\cdots \quad (11)$$
 
 方差 $\sigma_x^2$ 由 $\lambda_0$ 决定
 
-$\lambda_0 = \sigma_x^2$ (12)
+$$\lambda_0 = \sigma_x^2\tag{12}$$
 
 C和k为S-N曲线中的参数
 
-$S^kN=C$ (13)
+$$S^kN=C\tag{13}$$
 
 权重 b 的选择取决于响应的谱密度函数。D.Benasciutti 和 R.Tovo 给出 b 的表达式为
 
-$b=\frac{(\alpha_{1}-\alpha_{2})}{(\alpha_{2}-1)^{2}}m\quad(14)$
+$$b=\frac{(\alpha_{1}-\alpha_{2})}{(\alpha_{2}-1)^{2}}m\quad(14)$$
 
 其中
 
-$m=1.112(1+\alpha_1\alpha_2·(\alpha_1+\alpha_2))e^{211\alpha_2}+(\alpha_1-\alpha_2)$ (15)
+$$m=1.112(1+\alpha_1\alpha_2·(\alpha_1+\alpha_2))e^{211\alpha_2}+(\alpha_1-\alpha_2)\tag{15}$$
 
 对于谱密度函数 $ S_{x}(\omega) $，$\alpha_{1}/\alpha_{2}$ 为其带宽参数，取值为 $[0,1]^{14}$。
 
-$\alpha_1=\frac{\lambda_1}{\sqrt{\lambda_0\lambda_2}}, \alpha_2=\frac{\lambda_2}{\sqrt{\lambda_0\lambda_4}}$ (16)
+$$\alpha_1=\frac{\lambda_1}{\sqrt{\lambda_0\lambda_2}}, \alpha_2=\frac{\lambda_2}{\sqrt{\lambda_0\lambda_4}}\tag{16}$$
 
 ## 2.2 线性疲劳累计损伤理论
 
 根据线性疲劳累积损伤理论,计算附件机匣的总损伤量从而给出其疲劳寿命$^{[15]}$。
 
-$D=\sum_{i=1}^{N}D_{i}$ (17)
+$$D=\sum_{i=1}^{N}D_{i}\tag{17}$$
 
 式中:D 为总损伤量;D₁ 为单个循环造成的损伤。
 
 线性疲劳累计损伤理论假定损伤量 D=1 时试件将发生疲劳破坏。因此，疲劳寿命为
 
-$T=\frac{1}{D}=\frac{1}{\Sigma D}$ (18)
+$$T=\frac{1}{D}=\frac{1}{\Sigma D}\tag{18}$$
 
 ## 2.3 疲劳寿命计算
 
diff --git a/markdowns/research_patent_00049.md b/markdowns/research_patent_00049.md
index a2f32bff1e79a29f89b8d25a86ba2235c5846e16..5c407aba8caa660c73bbcde8cad4863b15471ba9 100644
--- a/markdowns/research_patent_00049.md
+++ b/markdowns/research_patent_00049.md
@@ -8,9 +8,7 @@ Dirichlet分布中随机采样的一组向量，即$W\sim\mathrm{Dir}(W|d)$。
 
 [0073] 如图5示出客户端基于多模态MRI图像训练集对初始分割模型进行训练的一种场景,如图所示,该初始分割模型可由U-NET网络以及transformer,网络组成,在模型训练过程中,多模态MRI脑肿瘤图像作为模型输入,输入的图像经过U-NET网络编码器的多层卷积神经网络提取5个层次的特征,其中最高层次的特征被用于解码器进行上采样;利用transformer分别对其他4个层次的特征进行融合,并依次与四个阶段的上采样的特征矩阵连接,最后经过两个卷积操作和一个argmax函数得到分割掩码,并将预测分割掩码与多模态MRI图像训练集的groundtruth标签构建损失函数,例如该损失函数可定义为:
 
-[0074]
-
-$L = -\frac{2\sum_{x\in\Omega}F(x)\cdot Y(x)}{\sum_{x\in\Omega}F(x)^2 + \sum_{x\in\Omega}Y(x)^2} - \sum_{x\in\Omega}Y(x)\log\left(F(x) + (1-Y(x))\log(1-F(x))\right] / \sum_{x\in\Omega}1;$
+[0074]$$L = -\frac{2\sum_{x\in\Omega}F(x)\cdot Y(x)}{\sum_{x\in\Omega}F(x)^2 + \sum_{x\in\Omega}Y(x)^2} - \sum_{x\in\Omega}Y(x)\log\left(F(x) + (1-Y(x))\log(1-F(x))\right] / \sum_{x\in\Omega}1;$$
 
 [0075] 其中，$F(x)$ 为模型输出，$Y(x)$ 是 groundtruth 标签，$\Omega$ 表示图像像素区域，即指代 MRI 图像的每一个像素。
 
@@ -20,7 +18,7 @@ $L = -\frac{2\sum_{x\in\Omega}F(x)\cdot Y(x)}{\sum_{x\in\Omega}F(x)^2 + \sum_{x\
 
 [0078] 例如,根据预设的学习率以及损失函数对初始分割模型的初始模型参数进行梯度下降,得到梯度下降参数,其方式可表示如下:
 
-[0079] $\theta_{K}^{'} = \theta_{K} - \eta \nabla L(\theta_{K});$
+[0079] $$\theta_{K}^{'} = \theta_{K} - \eta \nabla L(\theta_{K});$$
 
 [0080] 其中，$\theta_{k}^{\prime}$ 为经过梯度下降后的参数，$\theta_{k}$ 为初始分割模型的初始模型参数，$\eta \eta$ 为预先设置的学习率，$\nabla$ 为梯度（导数），L 为损失函数。
 
diff --git a/markdowns/research_patent_00053.md b/markdowns/research_patent_00053.md
index 19d63cc5435dd0de5cea03a1f1eccbe0d735dce7..eeb47b943fbb592f58a49a13bb154c5c986bfdb1 100644
--- a/markdowns/research_patent_00053.md
+++ b/markdowns/research_patent_00053.md
@@ -14,6 +14,6 @@
 
 [0055] 在收集到多个客户端的本地训练结果，即模型参数差值 $\Delta_{k}$ 之后，可对多个客户端发送的模型参数差值进行聚合，即聚合得到初始聚合梯度 $\Delta^{t}$ 的计算可表示如下：
 
-[0056] $\Delta^{t} = \frac{\sum_{k \in c^{t}} w_{k} \Delta_{K}}{qW};$
+[0056] $$\Delta^{t} = \frac{\sum_{k \in c^{t}} w_{k} \Delta_{K}}{qW};$$
 
 [0057] 其中，$c^{t}$ 指代全体客户端；q 指代客户端的聚合概率，即客户端的本地训练结果（即模型参数差值 $\Delta_{k}$）会有百分之 q 的概率被选中传输到中心服务器进行聚合，以进一步保护客户端的数据隐私。
diff --git a/markdowns/research_patent_00059.md b/markdowns/research_patent_00059.md
index aea1b01ce1030c5f81a5ddfd6eb2310a112ecc31..bf3363e3b29dda7c0d0951faf364e1b784d3253f 100644
--- a/markdowns/research_patent_00059.md
+++ b/markdowns/research_patent_00059.md
@@ -2,11 +2,11 @@
 
 [0046] 当$P(t)\geqslant P_{thresh}$时：电价公式：
 
-[0047] $C(t) = \left[ C_0 + \beta \cdot (P(t) - P_{\text{thresh}})^n \right] \left( 1 - k_2 \cdot \frac{S}{S_{\text{max}}} \right)$
+[0047] $$C(t) = \left[ C_0 + \beta \cdot (P(t) - P_{\text{thresh}})^n \right] \left( 1 - k_2 \cdot \frac{S}{S_{\text{max}}} \right)$$
 
 [004B] 当$P(t)< P_{thresh}$时：电价公式：
 
-[0049] $C(t) = C_0 \cdot \left(1 - \alpha \cdot e^{-\gamma \cdot (P_{\text{thesh}} - P(t))}\right) \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$
+[0049] $$C(t) = C_0 \cdot \left(1 - \alpha \cdot e^{-\gamma \cdot (P_{\text{thesh}} - P(t))}\right) \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$$
 
 [0050] 式中：$P_{thresh}$ 为电网安全风险值；$C_0$ 为基础电价；$\beta$ 为价格增长系数；$n$ 为电价对负荷敏感度的值数；$\alpha$ 为电价减少系数，控制低负荷时电价的下降幅度；$\gamma$ 为低负荷的衰减率。$S$ 为用户当前的信用分数，$S_{max}$ 为信用分数的最大值，$k_s$ 为信用折扣系数。
 
diff --git a/markdowns/research_patent_00060.md b/markdowns/research_patent_00060.md
index 9fac0f464be60dcca559af5bea3334d1461899fb..d307e32a43abbe3622a737fc91483ed87bc7b9b6 100644
--- a/markdowns/research_patent_00060.md
+++ b/markdowns/research_patent_00060.md
@@ -1,32 +1,29 @@
 态约束和电动汽车充放电功率约束,建立约束条件:
 
 [0059] 电动汽车充放电状态约束：
+$$X_{t}^{c}=\begin{cases}1,&EV充电状态\\0,&EV不在充电状态\end{cases}$$
 
 [0060]
+$$X_{t}^{d}=\begin{cases}1,&EV处于放电状态\\0,&EV处于其他状态\end{cases}$$
 
-$X_{t}^{c}=\begin{cases}1,&EV充电状态\\0,&EV不在充电状态\end{cases}$
-
-$X_{t}^{d}=\begin{cases}1,&EV处于放电状态\\0,&EV处于其他状态\end{cases}$
-
-$X_{t}^{c}+X_{t}^{d}\leq1$
+$$X_{t}^{c}+X_{t}^{d}\leq1$$
 
 [0061] 电动汽车荷电状态约束：
 
-[0062] $SOC_{min} \leq SOC_d \leq SOC_{max}$
+[0062] $$SOC_{min} \leq SOC_d \leq SOC_{max}$$
 
-[0063] $SOC_{\min} \leqslant SOC_i \leqslant SOC_{\max}$
+[0063] $$SOC_{\min} \leqslant SOC_i \leqslant SOC_{\max}$$
 
-[0064] $SOC_{l} = SOC_{l} + \frac{\Delta T \sum_{t=T_{c}}^{T_{p}} \left( P_{t}^{c} X_{t}^{c} \eta_{ev}^{c} - P_{t}^{c} X_{t}^{c} \eta_{ev}^{d} \right)}{Cap_{ev}^{bat}}$
+[0064] $$SOC_{l} = SOC_{l} + \frac{\Delta T \sum_{t=T_{c}}^{T_{p}} \left( P_{t}^{c} X_{t}^{c} \eta_{ev}^{c} - P_{t}^{c} X_{t}^{c} \eta_{ev}^{d} \right)}{Cap_{ev}^{bat}}$$
 
 [0065] 式中：$\Delta T$表示充电结束时刻与充电开始时刻的时间间隔，$T_{\beta}$为放电结束时刻；$T_{\text{c}}$为充放电开始时刻；$SOC_{\text{i}}$为电动汽车在入网电网时的荷电容量，$SOC_{\text{d}}$为电动汽车在离网电网时的荷电容量；$SOC_{\text{min}}$与$SOC_{\text{max}}$分别表示电动汽车电池最小与最大的荷电状态，$\eta_{\text{ev}}^{\text{c}}$与$\eta_{\text{n}}^{\text{d}}$为电动汽车充电的效率与放电的效率，$Cap_{\text{w}}^{\text{bart}}$为电动汽车电池额定容量；
 
 [0066] 电动汽车充放电功率约束：
+$$0 \leq P_{t}^{c} \leq P_{\max}^{c}$$
 
 [0067]
 
-$0 \leq P_{t}^{c} \leq P_{\max}^{c}$
-
-$0\leq P_{t}^{d}\leq P_{m a x}^{d}$
+$$0\leq P_{t}^{d}\leq P_{m a x}^{d}$$
 
 [0068] 式中：$P_{\max}^c$为电动汽车最大充电功率，$P_{\max}^d$为电动汽车最大放电功率。
 
@@ -38,7 +35,7 @@ $0\leq P_{t}^{d}\leq P_{m a x}^{d}$
 
 [0072] 步骤6.1:种群初始化:在n维空间中随机产生满足约束条件的M个个体;
 
-[0073] $x_{ij}(0) = rand_{ij}(0,1)(x_{ij}^U - x_{ij}^L) + x_{ij}^L$
+[0073] $$x_{ij}(0) = rand_{ij}(0,1)(x_{ij}^U - x_{ij}^L) + x_{ij}^L$$
 
 [0074] 式中：$x_{ij}(0)$ 为第 i 个个体的第 j 个参数在 t=0 时刻的初始值；$x_{ij}^{U}$ 和 $x_{ij}^{L}$ 分别为第 j 个参数的上界和下界，$\mathrm{rand}_{ij}(0,1)$ 是 [0,1] 之间的随机数。
 
@@ -46,4 +43,4 @@ $0\leq P_{t}^{d}\leq P_{m a x}^{d}$
 
 [0076] 步骤6.2.1. 计算距离矩阵：计算种群中个体之间的欧几里得距离，构建距离矩阵D；
 
-[0077] $D_{ij} = ||x_i - x_j||$
+[0077] $$D_{ij} = ||x_i - x_j||$$
diff --git a/markdowns/research_patent_00063.md b/markdowns/research_patent_00063.md
index af74a37c7265244bcdeff2e888066282aa6b5940..f7b1bfbb4610f9396bb225c3027f4cd6b19252af 100644
--- a/markdowns/research_patent_00063.md
+++ b/markdowns/research_patent_00063.md
@@ -20,16 +20,16 @@ $$f(t_a) = \begin{cases} \dfrac{1}{\sqrt{2\pi}\sigma_a} \exp\left(-\dfrac{(t_a +
 
 所述的充电结束时间模型为：
 
-$f(t_i) = \begin{cases} \frac{1}{\sqrt{2\pi}\sigma_i} \exp\left(-\frac{(t_i - \mu_i)^2}{2\sigma_i^2}\right) & , 0 < t_i \leq u_i + 12 \\ \frac{1}{\sqrt{2\pi}\sigma_i} \exp\left(-\frac{(t_i - 24 - \mu_i)^2}{2\sigma_i^2}\right) & , u_i + 12 < t_i \leq 24 \end{cases}$
+$$f(t_i) = \begin{cases} \frac{1}{\sqrt{2\pi}\sigma_i} \exp\left(-\frac{(t_i - \mu_i)^2}{2\sigma_i^2}\right) & , 0 < t_i \leq u_i + 12 \\ \frac{1}{\sqrt{2\pi}\sigma_i} \exp\left(-\frac{(t_i - 24 - \mu_i)^2}{2\sigma_i^2}\right) & , u_i + 12 < t_i \leq 24 \end{cases}$$
 
 式中：$t_{i}$ 表示充电结束时间；$u_{i}$ 表示充电结束时间 $t_{i}$ 的均值；$\sigma_{a}$ 表示充电结束时间 $t_{i}$ 的标准差；$f(t_{i})$ 表示充电结束时间 $t_{i}$ 的概率函数；
 
 日行驶里程符合对数正态分布$f_{d}(x)$为：
 
-$f_{\mathrm{d}}(x)={\frac{1}{x\sigma{\sqrt{2\pi}}}}\exp\quad[{\frac{-(\ln x-\mu_{l})^{2}}{2\sigma_{l}^{2}}}]$
+$$f_{\mathrm{d}}(x)={\frac{1}{x\sigma{\sqrt{2\pi}}}}\exp\quad[{\frac{-(\ln x-\mu_{l})^{2}}{2\sigma_{l}^{2}}}]$$
 
 式中：$x$表示日行驶里程；$\mu_1$表示行驶里程的均值；$\sigma_1$表示行驶里程的标准差；$lnx$表示行驶里程的自然对数；
 
 结合EV用户的出行特性,计算调度EVi充电所需时长$T_{i}$:
 
-$T_{i} = \frac{(SOC_{d} - SOC_{i}) \times S_{e}}{P_{i} \times \eta}$
+$$T_{i} = \frac{(SOC_{d} - SOC_{i}) \times S_{e}}{P_{i} \times \eta}$$
diff --git a/markdowns/research_patent_00064.md b/markdowns/research_patent_00064.md
index 789f5ecc453a78b88bae72a7950166356f82ad46..75b5f3944ceafec4febcbb3474e3021fcf648450 100644
--- a/markdowns/research_patent_00064.md
+++ b/markdowns/research_patent_00064.md
@@ -2,7 +2,7 @@
 
 3. 根据权利要求1所述一种基于改进差分进化算法的电动汽车有序充放电控制方法，其特征在于，所述步骤2中，通过获取电网信息与充电电动汽车的信息分别包括：预计离开时间、期望电量及可调度意愿，剩余荷电状态SOC、EV电池容量，汽车还有基础负荷信息；其所述的需求性分析：
 
-$\left\{ \begin{array}{l} I=1,\text{进行可行性判断} \\ I=0,\text{直接进行无序充电} \end{array} \right.$
+$$\left\{ \begin{array}{l} I=1,\text{进行可行性判断} \\ I=0,\text{直接进行无序充电} \end{array} \right.$$
 
 式中：I=1 表示电动汽车的有调度意愿，I=0 表示电动汽车的无调度意愿；
 电动汽车调度可行性判别：
@@ -22,7 +22,7 @@ $$
 
 考虑电网安全风险值为：
 
-$P_{\text{thresh}} = \frac{1}{T} \int_0^T P(t), dt$
+$$P_{\text{thresh}} = \frac{1}{T} \int_0^T P(t), dt$$
 
 其中$P(t)$为在时刻$t$的该地区的总功率需求；$dt$为微小的时间增量，用于累积$P(t)$在整个时间区间$[0,T]$内的总功率需求；$T$为一天的总时间；
 
@@ -32,11 +32,11 @@ $P_{\text{thresh}} = \frac{1}{T} \int_0^T P(t), dt$
 
 当$P(t) \geq P_{\text{thresh}}$时:电价公式:
 
-$C(t) = [C_0 + \beta \cdot (P(t) - P_{\text{thresh}})^n] \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$
+$$C(t) = [C_0 + \beta \cdot (P(t) - P_{\text{thresh}})^n] \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$$
 
 当$P(t) < P_{thresh}$时：电价公式：
 
-$C(t) = C_0 \cdot \left(1 - \alpha \cdot e^{-\gamma \cdot (P_{\text{thesh}} - P(t))}\right) \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$
+$$C(t) = C_0 \cdot \left(1 - \alpha \cdot e^{-\gamma \cdot (P_{\text{thesh}} - P(t))}\right) \left(1 - k_2 \cdot \frac{S}{S_{\text{max}}}\right)$$
 
 式中：$P_{thresh}$ 为电网安全风险值；$C_0$ 为基础电价；$\beta$ 为价格增长系数；$n$ 为电价对负荷敏感度的值数；$\alpha$ 为电价减少系数，控制低负荷时电价的下降幅度；$\gamma$ 为低负荷的衰减率；$S$ 为用户当前的信用分数，$S_{max}$ 为信用分数的最大值，$k_s$ 为信用折扣系数。
 
diff --git a/markdowns/research_patent_00065.md b/markdowns/research_patent_00065.md
index d2308507b09efb04aeb6b5f0f3aaad1f05d86bfe..ea6ce670552c9d99a3c0989c88fe5e26f24b0f3f 100644
--- a/markdowns/research_patent_00065.md
+++ b/markdowns/research_patent_00065.md
@@ -23,18 +23,18 @@ $$
 
 电动汽车充放电状态约束：
 
-$X_{t}^{c}=\begin{cases}1,&EV充电状态\\0,&EV不在充电状态\end{cases}$
+$$X_{t}^{c}=\begin{cases}1,&EV充电状态\\0,&EV不在充电状态\end{cases}$$
 
-$X_{t}^{d}=\begin{cases}1,&EV处于放电状态\\0,&EV处于其他状态\end{cases}$
+$$X_{t}^{d}=\begin{cases}1,&EV处于放电状态\\0,&EV处于其他状态\end{cases}$$
 
-$X_{t}^{c}+X_{t}^{d}\leq1$
+$$X_{t}^{c}+X_{t}^{d}\leq1$$
 
 电动汽车荷电状态约束：
 
-$SOC_{\min} \leq SOC_d \leq SOC_{\max}$
+$$SOC_{\min} \leq SOC_d \leq SOC_{\max}$$
 
-$SOC_{\min}\leqslant SOC_{i}\leqslant SOC_{\max}$
+$$SOC_{\min}\leqslant SOC_{i}\leqslant SOC_{\max}$$
 
-$SOC_{l}=SOC_{i}+\frac{\Delta T\sum_{t=T_{c}}^{T_{\beta}}\left(P_{t}^{c}X_{t}^{c}\eta_{ev}^{c}-P_{t}^{c}X_{t}^{c}\eta_{ev}^{d}\right)}{Cap_{ev}^{bat}}$
+$$SOC_{l}=SOC_{i}+\frac{\Delta T\sum_{t=T_{c}}^{T_{\beta}}\left(P_{t}^{c}X_{t}^{c}\eta_{ev}^{c}-P_{t}^{c}X_{t}^{c}\eta_{ev}^{d}\right)}{Cap_{ev}^{bat}}$$
 
 式中：$\Delta T$ 表示充放电结束时刻与充放电开始时刻的时间间隔，$T_{p}$ 为充放电结束时刻；$T_{c}$
diff --git a/markdowns/research_patent_00066.md b/markdowns/research_patent_00066.md
index ef0dab263a18824641d957d661742dd767be62ae..2df12f074cfe5960a59a1e7f4cbfc47c8f9a9e90 100644
--- a/markdowns/research_patent_00066.md
+++ b/markdowns/research_patent_00066.md
@@ -2,9 +2,9 @@
 
 电动汽车充放电功率约束：
 
-$0 \leq P_{t}^{c} \leq P_{max}^{c}$
+$$0 \leq P_{t}^{c} \leq P_{max}^{c}$$
 
-$0 \leq P_{t}^{d} \leq P_{\max}^{d}$
+$$0 \leq P_{t}^{d} \leq P_{\max}^{d}$$
 
 式中：$P_{max}^{c}$ 为电动汽车最大充电功率，$P_{max}^{d}$ 为电动汽车最大放电功率。
 
@@ -14,7 +14,7 @@ $0 \leq P_{t}^{d} \leq P_{\max}^{d}$
 
 步骤6.1:种群初始化:在多维空间中随机产生满足约束条件的M个个体;
 
-$x_{ij}(0) = rand_{ij}(0,1)(x_{ij}^U - x_{ij}^L) + x_{ij}^L$
+$$x_{ij}(0) = rand_{ij}(0,1)(x_{ij}^U - x_{ij}^L) + x_{ij}^L$$
 
 式中：$x_{ij}(0)$ 为第 i 个个体的第 j 个参数在 t=0 时刻的初始值；$x_{ij}^{U}$ 和 $x_{ij}^{L}$ 分别为第 j 个参数的上界和下界，$\mathrm{rand}_{ij}(0,1)$ 是 [0,1] 之间的随机数；
 
@@ -22,17 +22,17 @@ $x_{ij}(0) = rand_{ij}(0,1)(x_{ij}^U - x_{ij}^L) + x_{ij}^L$
 
 步骤6.2.1:计算距离矩阵:计算种群中个体之间的欧几里得距离,构建距离矩阵$D_{ij}$;
 
-$D_{ij} = ||x_i - x_j||$
+$$D_{ij} = ||x_i - x_j||$$
 
 步骤6.2.2. 计算局部密度：对每个个体$x_{i}$，计算其局部密度$\rho_{i}$，通常使用k-最近邻来定义局部密度：
 
-$\rho_{i}=\sum_{j\in N_{i}}\exp\left(-\frac{D_{i j}^{2}}{2\sigma^{2}}\right)$
+$$\rho_{i}=\sum_{j\in N_{i}}\exp\left(-\frac{D_{i j}^{2}}{2\sigma^{2}}\right)$$
 
 式中：$N_j$是个体$x_j$的$k$-最近邻集合，$\sigma$是高斯核的尺度参数；
 
 步骤6.2.3. 计算距离峰值: 对每个个体$x_i$, 计算其距离峰值$\delta_i$:
 
-$\delta_{i}=\min\{\mathrm{dist}_{ij}|\rho_{j}>\rho_{j}\}$
+$$\delta_{i}=\min\{\mathrm{dist}_{ij}|\rho_{j}>\rho_{j}\}$$
 
 式中$dist_{ij}$表示个体$x_{i}$到$x_{j}$的距离；$\rho_{i}$为个体$x_{i}$的局部密度；$\rho_{j}$为个体$x_{j}$的局部密度；
 
@@ -42,7 +42,7 @@ $\delta_{i}=\min\{\mathrm{dist}_{ij}|\rho_{j}>\rho_{j}\}$
 
 步骤3:变异操作:从子种群中随机选择3个个体$x_{p1}$、$x_{p2}$和$x_{p3}$且$p_1 \ne p_2 \ne p_3$;
 
-$h_{ij}(t+1) = x_{p1j}(t) + F(x_{p2j}(t) - x_{p3j}(t))$
+$$h_{ij}(t+1) = x_{p1j}(t) + F(x_{p2j}(t) - x_{p3j}(t))$$
 
 式中：$h_{ij}(t+1)$ 为第 $i$ 个个体的第 $j$ 个参数在变异操作后的值；$x_{p1j}(t)$ 为随机选择的第一个个体的第 $j$ 个参数在当前代的值；$x_{p2j}(t) - x_{p3j}(t)$ 为变异项；$F$ 为变异因子，$p_1、p_2、p_3$ 表示随机整数，为个体在种群中的序号；
 
diff --git a/markdowns/research_patent_00067.md b/markdowns/research_patent_00067.md
index f8b47370237689bb48841b0edfa7ca8461017e19..dd72150676fbab1e077d4a81b0b00de97083db87 100644
--- a/markdowns/research_patent_00067.md
+++ b/markdowns/research_patent_00067.md
@@ -1,10 +1,10 @@
-$v_{ij}(t+1)=\begin{cases}h_{ij}(t+1),\text{rand}l_{ij}\leq CR\\x_{ij}(t),\text{rand}l_{ij}>CR\end{cases}$
+$$v_{ij}(t+1)=\begin{cases}h_{ij}(t+1),\text{rand}l_{ij}\leq CR\\x_{ij}(t),\text{rand}l_{ij}>CR\end{cases}$$
 
 式中：$v_{ij}(t+1)$为第$i$个个体的第$j$个参数在交叉操作后的候选值；$x_{ij}(t)$为第$i$个个体的第$j$个参数在当前代的值；$rand1_{ij}$为$[0,1]$之间的随机数，$CR$为交叉概率，$CR\in[0,1]$；
 
 步骤6.5:选择操作:要确定$x_{ij}(t)$是否能成为下一代成员,$v_{ij}(t+1)$和$x_{ij}(t)$对评价函数进行比较;
 
-$x_{ij}(t+1)=\begin{cases}v_{ij}(t+1),\text{if }f(v_{i1}(t+1),\ldots,v_{in}(t+1))>f(x_1(t),\ldots,x_{in}(t))\\x_{ij}(t),\text{if }f(v_{i1}(t+1),\ldots,v_{in}(t+1))\leq f(x_1(t),\ldots,x_{in}(t))\end{cases}$
+$$x_{ij}(t+1)=\begin{cases}v_{ij}(t+1),\text{if }f(v_{i1}(t+1),\ldots,v_{in}(t+1))>f(x_1(t),\ldots,x_{in}(t))\\x_{ij}(t),\text{if }f(v_{i1}(t+1),\ldots,v_{in}(t+1))\leq f(x_1(t),\ldots,x_{in}(t))\end{cases}$$
 
 其中，$x_{ij}(t+1)$ 为第 i 个个体的第 j 个参数在下一代的值；$f(v_{i1}(t+1), \ldots, v_{in}(t+1))$ 为第 i 个个体在新一代的适应度函数值，其中 $v_{i1}(t+1), \ldots, v_{in}(t+1)$ 是该个体所有参数的候选值；
 
@@ -14,13 +14,13 @@ $x_{ij}(t+1)=\begin{cases}v_{ij}(t+1),\text{if }f(v_{i1}(t+1),\ldots,v_{in}(t+1)
 
 步骤6.6.2生成新个体:利用全局最优个体与其他子群的最优个体生成新的个体;
 
-$x_{ij}^{\text{new}}(t+1) = x^{\text{best}}k(t) + F \cdot (x^{\text{global}}best(t) - x_i^{\text{best}}(t))$
+$$x_{ij}^{\text{new}}(t+1) = x^{\text{best}}k(t) + F \cdot (x^{\text{global}}best(t) - x_i^{\text{best}}(t))$$
 
 其中，$x_{ij}^{new}(t+1)$为第$i$个个体的第$j$个参数在信息交流后的新值；$x^{best}k(t)$为第$k$个子群在当前代的最优个体；$x_i^{best}(t)$为其他子群的最优个体；$F$为变异因子，控制着全局最优个体对新个体生成的影响程度；
 
 步骤6.6.3替换操作：将新生成的个体与当前个体进行比较，选择适应度较优者作为下一代成员；
 
-$x_{ij}(t+1)=\begin{cases}x_{ij}^{new}(t+1),\text{ if }f\left(x_{ij}^{new}(t+1)\right)>f\left(x_{ij}(t+1)\right)\\x_{ij}(t+1),\text{ otherwise}\end{cases}$
+$$x_{ij}(t+1)=\begin{cases}x_{ij}^{new}(t+1),\text{ if }f\left(x_{ij}^{new}(t+1)\right)>f\left(x_{ij}(t+1)\right)\\x_{ij}(t+1),\text{ otherwise}\end{cases}$$
 
 步骤6.7: 终止条件: 重复执行上述操作, 直到满足最大迭代次数为止;
 
diff --git a/markdowns/research_patent_00070.md b/markdowns/research_patent_00070.md
index 4b63525d837815e56c0a790573c3594870048195..6047ae73a38f9ade68cb3cb2cad946fe49b2bd1d 100644
--- a/markdowns/research_patent_00070.md
+++ b/markdowns/research_patent_00070.md
@@ -2,13 +2,13 @@
 
 [0027] 日行驶里程符合对数正态分布$f_{d}(x)$为：
 
-[0028] $f_{\mathrm{d}}(x)={\frac{1}{x\sigma{\sqrt{2\pi}}}}\exp\quad[{\frac{-(\ln\quad x-\mu_{l})^{2}}{2\sigma_{l}^{2}}}]$
+[0028] $$f_{\mathrm{d}}(x)={\frac{1}{x\sigma{\sqrt{2\pi}}}}\exp\quad[{\frac{-(\ln\quad x-\mu_{l})^{2}}{2\sigma_{l}^{2}}}]$$
 
 [0029] 式中：x 表示日行驶里程；$\mu_{1}$ 表示行驶里程的均值；$\sigma_{1}$ 表示行驶里程的标准差；lnx 表示行驶里程的自然对数。
 
 [0030] 结合EV用户的出行特性,可以计算调度EVi充电所需时长$T_{i}$;
 
-[0031] $T_{i}=\frac{(SOC_{d}-SOC_{i})\times S_{e}}{P_{i}\times\eta}$
+[0031] $$T_{i}=\frac{(SOC_{d}-SOC_{i})\times S_{e}}{P_{i}\times\eta}$$
 
 [0032] 式中：Ti为第i辆电动汽车的充电时长；$S_{e}$为电动汽车的电池容量；$SOC_{d}$为电动汽车充电结束时电池的荷电状态；$SOC_{i}$为电动汽车充电开始时电池的荷电状态；$P_{i}$为电动汽车的充电功率；$\eta$为电动汽车的充电效率。
 
@@ -16,11 +16,11 @@
 
 [0034] 通过获取电网信息与充电电动汽车的信息分别包括：预计离开时间、期望电量及可调度意愿，剩余荷电状态SOC(State Of Charge)、EV电池容量，汽车还有基础负荷信息。其所述的需求性分析：
 
-[0035] $\left\{\begin{aligned}I&=1, 进行可行性判断 \\ I&=0, 直接进行无序充电\end{aligned}\right.$
+[0035] $$\left\{\begin{aligned}I&=1, 进行可行性判断 \\ I&=0, 直接进行无序充电\end{aligned}\right.$$
 
 [0036] 式中：$I=1$表示电动汽车的有调度意愿，$I=0$表示电动汽车的无调度意愿；电动汽车调度可行性判别：
 
-[0037] $\left\{ \begin{array}{l} if\ T_i \leq T_L - T_a,\ 则可进行调度 \\ if\ T_i \geq T_L - T_a,\ 则不进行调度，\ 在离网前一直充电 \end{array} \right.$
+[0037] $$\left\{ \begin{array}{l} if\ T_i \leq T_L - T_a,\ 则可进行调度 \\ if\ T_i \geq T_L - T_a,\ 则不进行调度，\ 在离网前一直充电 \end{array} \right.$$
 
 [0038] 式中：$T_i$表示电动汽车充电所需时间，$T_L$表示电动汽车离网时间，$T_a$表示电动汽车入网时间。
 
@@ -30,7 +30,7 @@
 
 [0041] 考虑电网安全风险值为：
 
-[0042] $P_{\mathrm{t h r e s h}}={\frac{1}{T}}\int_{0}^{T}P(t),d t$
+[0042] $$P_{\mathrm{t h r e s h}}={\frac{1}{T}}\int_{0}^{T}P(t),d t$$
 
 [0043] 其中$P(t)$为在时刻t的该地区的总功率需求；dt为微小的时间增量，用于累积$P(t)$在整个时间区间$[0,T]$内的总功率需求；T为一天的总时间；
 
diff --git a/markdowns/research_patent_00085.md b/markdowns/research_patent_00085.md
index 29e385f59bdc6e3f0feadfd2a857eef5fc639995..e8a726db59fe7ee9a7b6ad6d30e6339f3789c947 100644
--- a/markdowns/research_patent_00085.md
+++ b/markdowns/research_patent_00085.md
@@ -2,7 +2,7 @@
 
 [0052]
 
-$D_{mn}=\int_{0}^{\Delta}2\pi M_{0}f_{n}\cdot d\Delta=\int_{0}^{\delta}2\pi M_{0}hf_{n}d\delta=\begin{cases}2\pi M_{0}h\left(\frac{4}{9}\delta^{3}+\delta\right),\delta\leq\frac{1}{2}\\2\pi M_{0}h\left(\delta^{2}+\frac{1}{6}\ln\delta+\frac{11}{36}+\frac{\ln2}{6}\right),\delta>\frac{1}{2}\end{cases}(b=1)$
+$$D_{mn}=\int_{0}^{\Delta}2\pi M_{0}f_{n}\cdot d\Delta=\int_{0}^{\delta}2\pi M_{0}hf_{n}d\delta=\begin{cases}2\pi M_{0}h\left(\frac{4}{9}\delta^{3}+\delta\right),\delta\leq\frac{1}{2}\\2\pi M_{0}h\left(\delta^{2}+\frac{1}{6}\ln\delta+\frac{11}{36}+\frac{\ln2}{6}\right),\delta>\frac{1}{2}\end{cases}(b=1)$$
 
 (16a)
 
diff --git a/markdowns/research_patent_00093.md b/markdowns/research_patent_00093.md
index 16569542b927574c88a101751021eae54b01c6e1..cdc493c2a1cdc6515b9ed98ceb46739f738e6b33 100644
--- a/markdowns/research_patent_00093.md
+++ b/markdowns/research_patent_00093.md
@@ -19,18 +19,18 @@
 
 [0023] 可选地,所述定位回归网络的损失函数为:
 
-[0024] $Loss_{total} = GD(\hat{r}, r) + \lambda MSE(\hat{t}, t);$
+[0024] $$Loss_{total} = GD(\hat{r}, r) + \lambda MSE(\hat{t}, t);$$
 
 [0025] 式中，$\lambda$ 表示平移损失的权重，r 表示真实刚性变换中的旋转，t 表示真实刚性变换中的平移，$\hat{r}$ 表示预测的刚性变换中的旋转，$\hat{t}$ 表示预测的刚性变换中的平移，$GD(\hat{r}, r)$ 表示预测的刚性变换中的旋转$\hat{r}$和真实刚性变换中的旋转之间的测地距离误差，$MSE(\hat{t}, t)$ 表示预测的刚性变换中的平移和真实刚性变换中的平移 $\hat{t}$ 之间的均方误差，$Loss_{total}$ 表示用于损失值。
 
 [0026] 可选地,所述查询图像在所述目标3D心脏CT空间中的候选位置表示为:
 
-[0027] $\widehat{p}_{q}^{v_{c}}=RT\times\widehat{p}_{q}^{R};$
+[0027] $$\widehat{p}_{q}^{v_{c}}=RT\times\widehat{p}_{q}^{R};$$
 
 [0028] 式中，$\widehat{p}_{q}^{V_{c}}$ 表示候选位置，RT 表示标准3D心脏图谱和目标3D心脏CT空间的刚性变换，$\widehat{p}_{q}^{R}$ 表示查询图像在图谱空间中的切面位置。
 
 [0029] 可选地,与所述查询图像相似度最高的候选图像切面表示为:
 
-[0030] $\hat{s}_{q}^{v} = \arg\max_{s_{i}^{c}} (SSIM\{s_{i}^{c}, q\}), i = \{1, \cdots, m\};$
+[0030] $$\hat{s}_{q}^{v} = \arg\max_{s_{i}^{c}} (SSIM\{s_{i}^{c}, q\}), i = \{1, \cdots, m\};$$
 
 [0031]    式中，$\hat{s}_{q}^{v}$ 表示与所述查询图像相似度最高的候选图像切面，$s_{i}^{c}$ 表示候选数据库中的第 i 个候选图像切面，m 表示候选图像切面的个数，$SSIM\{s_{i}^{c}, q\}$ 表示第 i 个候选图像切面
\ No newline at end of file
diff --git a/markdowns/research_patent_00094.md b/markdowns/research_patent_00094.md
index b536d487bab0347eae1f83e22ea13ae0d293b0e4..924b73f5cbd1ad54580f925bb3fcc29c383a3a15 100644
--- a/markdowns/research_patent_00094.md
+++ b/markdowns/research_patent_00094.md
@@ -1,4 +1,4 @@
-[0072] $\mathrm{L o s s}_{\mathrm{t o t a l}}=\mathrm{G D}(\hat{\mathrm{r}},\mathrm{r})+\lambda\mathrm{M S E}(\hat{\mathrm{t}},\mathrm{t})$。
+[0072] $$\mathrm{L o s s}_{\mathrm{t o t a l}}=\mathrm{G D}(\hat{\mathrm{r}},\mathrm{r})+\lambda\mathrm{M S E}(\hat{\mathrm{t}},\mathrm{t})。$$
 
 [0073] 式中，$\lambda$表示平移损失的权重，$r$表示真实刚性变换中的旋转，$t$表示真实刚性变换中的平移，$\hat{r}$表示预测的刚性变换中的旋转，$\hat{t}$表示预测的刚性变换中的平移，$GD(\hat{r},r)$表示预测的刚性变换中的旋转$\hat{r}$和真实刚性变换中的旋转$\hat{r}$之间的测地距离误差，$MSE(\hat{t},t)$表示预测的刚性变换中的平移$\hat{t}$和真实刚性变换中的平移$\hat{t}$之间的均方误差，$Loss_{total}$表示用作损失值。
 
@@ -6,7 +6,7 @@
 
 [0075] 步骤501：查询图像q被输入到训练好的定位回归网络PosNet中，得到查询图像q在标准3D心脏图谱$\mathcal{R}$中的位置，记为$\hat{\mathbf{p}}_{q}^{\mathcal{R}}$：
 
-[0076] $\hat{p}_{q}^{\mathcal{R}} = \gamma(q; \mu).$
+[0076] $$\hat{p}_{q}^{\mathcal{R}} = \gamma(q; \mu).$$
 
 [0077] 步骤502:标准3D心脏图谱$\mathcal{R}$以及查询图像q在标准3D心脏图谱$\mathcal{R}$中的位置$\hat{p}_{q}^{\mathcal{R}}$组成查询图像q的图谱提示$\left\{\mathcal{R}, \hat{p}_{q}^{\mathcal{R}}\right\}$。
 
@@ -22,7 +22,7 @@
 
 [0083] 其中,与查询图像相似度最高的候选图像切面表示为:
 
-[0084] $\widehat{s}_{q}^{V}=\operatorname*{argmax}_{s_{i}^{c}}\left(S S I M\{s_{i}^{c},q\}\right),i=\{1,\cdots,m\}$。
+[0084] $$\widehat{s}_{q}^{V}=\operatorname*{argmax}_{s_{i}^{c}}\left(S S I M\{s_{i}^{c},q\}\right),i=\{1,\cdots,m\}。$$
 
 [0085] 式中，$\hat{S}_{q}^{V}$ 表示与查询图像相似度最高的候选图像切面，$S_{i}^{c}$ 表示候选数据库中的第 $i$ 个候选图像切面，$m$ 表示候选图像切面的个数，$SSIM[S_{i}^{c}, q]$ 表示第 $i$ 个候选图像切面 $S_{i}^{c}$ 和查询图像 $q$ 的图像相似度。
 
diff --git a/markdowns/research_programming_book_00059.md b/markdowns/research_programming_book_00059.md
index f66a0a7db0ce4937ecb464266758ebd6e1060d5f..779559c9478a57d7ff0c87b25dccfc3463cd30db 100644
--- a/markdowns/research_programming_book_00059.md
+++ b/markdowns/research_programming_book_00059.md
@@ -23,21 +23,21 @@ returm(2*z);
 
 解题思路：要求第5个学生的年龄，就必须先知道第4个学生的年龄，而第4个学生的年龄也不知道，要求第4个学生的年龄必须先知道第3个学生的年龄，而第3个学生的年龄又取决于第2个学生的年龄，第2个学生的年龄取决于第1个学生的年龄。而且每一个学生的年龄都比其前1个学生的年龄大2。即：
 
-age(5) = age(4) + 2
+$$age(5) = age(4) + 2$$
 
-$age(4) = age(3) + 2$
+$$age(4) = age(3) + 2$$
 
-$age(3) = age(2) + 2$
+$$age(3) = age(2) + 2$$
 
-$age(2) = age(1) + 2$
+$$age(2) = age(1) + 2$$
 
-$age(1) = 10$
+$$age(1) = 10$$
 
 可以用数学公式表述如下：
 
-$\text{age}(n) = 10 \quad (n = 1)$
+$$\text{age}(n) = 10 \quad (n = 1)$$
 
-$\mathrm{age}(n)=\mathrm{age}(n-1)+2\quad(n>1)$
+$$\mathrm{age}(n)=\mathrm{age}(n-1)+2\quad(n>1)$$
 
 可以看到，当 $n \geq 1$ 时，求每位学生的年龄的公式是相同的。因此可以用一个函数表示上述关系。图 7.8 表示求第 5 个学生年龄的过程。
 
diff --git a/markdowns/research_programming_book_00063.md b/markdowns/research_programming_book_00063.md
index 4f0be0352f9b2cdf2aa9ace41e65228a5c9eca04..a523fcfb6d2471e37e13a652faf1d0d9d36d655f 100644
--- a/markdowns/research_programming_book_00063.md
+++ b/markdowns/research_programming_book_00063.md
@@ -33,15 +33,15 @@ c=5; //将 5 赋给 c
 
 解题思路：首先要知道求方程式的根的方法。由数学知识已知：如果 $b^2 - 4ac \geq 0$，则一元二次方程有两个实根：
 
-$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}, \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$
+$$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}, \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$
 
 可以将上面的分式分为两项：
 
-$p=\frac{-b}{2a},\quad q=\frac{\sqrt{b^2-4ac}}{2a}$
+$$p=\frac{-b}{2a},\quad q=\frac{\sqrt{b^2-4ac}}{2a}$$
 
 则
 
-$x_1 = p + q, \quad x_2 = p - q$
+$$x_1 = p + q, \quad x_2 = p - q$$
 
 有了这些式子,只要知道 $a,b,c$ 的值,就能顺利地求出方程的两个根。
 
diff --git a/markdowns/research_programming_book_00077.md b/markdowns/research_programming_book_00077.md
index a744da7ce692eb32d2956ee39f2dd1713cd6ab28..9121a7205662207ae1e521e8222e5b8fa361dab7 100644
--- a/markdowns/research_programming_book_00077.md
+++ b/markdowns/research_programming_book_00077.md
@@ -8,19 +8,19 @@ Adadelta是AdaGrad的另一种变体（11.7节），主要区别在于前者减
 
 以下是Adadelta的技术细节。鉴于参数du jour是$\rho$，我们获得了与11.8节类似的以下泄漏更新：
 
-$\mathbf{s}_{t}=\rho\mathbf{s}_{t-1}+(1-\rho)\mathbf{g}_{t}^{2}.\quad(11.9.1)$
+$$\mathbf{s}_{t}=\rho\mathbf{s}_{t-1}+(1-\rho)\mathbf{g}_{t}^{2}.\quad(11.9.1)$$
 
 与 11.8节的区别在于，我们使用重新缩放的梯度$\mathbf{g}_t'$执行更新，即
 
-$\mathbf{x}_{t}=\mathbf{x}_{t-1}-\mathbf{g}_{t}^{\prime}.\quad(11.9.2)$
+$$\mathbf{x}_{t}=\mathbf{x}_{t-1}-\mathbf{g}_{t}^{\prime}.\quad(11.9.2)$$
 
 那么，调整后的梯度$\mathbf{g}'_t$是什么？我们可以按如下方式计算它：
 
-$\mathbf{g}_t' = \frac{\sqrt{\Delta \mathbf{x}_{t-1} + \epsilon}}{\sqrt{\mathbf{s}_t + \epsilon}} \odot \mathbf{g}_t,$ $\quad(11.9.3)$
+$$\mathbf{g}_t' = \frac{\sqrt{\Delta \mathbf{x}_{t-1} + \epsilon}}{\sqrt{\mathbf{s}_t + \epsilon}} \odot \mathbf{g}_t,\quad(11.9.3)$$
 
 其中$\Delta \mathbf{x}_{t-1}$是重新缩放梯度的平方$\mathbf{g}_t'$的泄漏平均值。我们将$\Delta \mathbf{x}_0$初始化为0，然后在每个步骤中使用$\mathbf{g}_t'$更新它，即
 
-$\Delta\mathbf{x}_{t}=\rho\Delta\mathbf{x}_{t-1}+(1-\rho)\mathbf{g}_{t}^{\prime2},\quad(11.9.4)$
+$$\Delta\mathbf{x}_{t}=\rho\Delta\mathbf{x}_{t-1}+(1-\rho)\mathbf{g}_{t}^{\prime2},\quad(11.9.4)$$
 
 和$\epsilon$（例如$10^{-5}$这样的小值）是为了保持数字稳定性而加入的。
 
diff --git a/markdowns/research_programming_book_00078.md b/markdowns/research_programming_book_00078.md
index 8424c9b415eb93c1fda622a5d610f1ee6cc7700c..0339ed60f46c012b7ba85ace58253b13f62438e7 100644
--- a/markdowns/research_programming_book_00078.md
+++ b/markdowns/research_programming_book_00078.md
@@ -4,7 +4,7 @@
 
 通常，如果我们添加$p_h$行填充（大约一半在顶部，一半在底部）和$p_w$列填充（左侧大约一半，右侧一半），则输出形状将为
 
-$(n_{h}-k_{h}+p_{h}+1)\times(n_{w}-k_{w}+p_{w}+1)\boxtimes\quad(6.3.1)$
+$$(n_{h}-k_{h}+p_{h}+1)\times(n_{w}-k_{w}+p_{w}+1)\boxtimes\quad(6.3.1)$$
 
 这意味着输出的高度和宽度将分别增加$p_{h}$和$p_{w}$。
 
diff --git a/markdowns/research_programming_book_00084.md b/markdowns/research_programming_book_00084.md
index 7ad8ee9d122254d204ec1458686160f89160a18f..44b847d4c300b6890b1646d15a9a7656892b7517 100644
--- a/markdowns/research_programming_book_00084.md
+++ b/markdowns/research_programming_book_00084.md
@@ -4,10 +4,10 @@ $$\begin{aligned}&\begin{aligned}\\ &P(x_{1},\ldots,x_{T})\\&=\sum_{h_{1},\ldots
 
 通常，我们将前向递归（forward recursion）写为：
 
-$\pi_{t+1}(h_{t+1})=\sum_{h_{t}}\pi_{t}(h_{t})P(x_{t}\mid h_{t})P(h_{t+1}\mid h_{t}).\quad(9.4.3)$
+$$\pi_{t+1}(h_{t+1})=\sum_{h_{t}}\pi_{t}(h_{t})P(x_{t}\mid h_{t})P(h_{t+1}\mid h_{t}).\quad(9.4.3)$$
 
 递归被初始化为$\pi_{1}(h_{1})=P(h_{1})$。符号简化，也可以写成$\pi_{t+1}=f(\pi_{t},x_{t})$，其中f是一些可学习的函数。这看起来就像我们在循环神经网络中讨论的隐变量模型中的更新方程。
 
 与前向递归一样，我们也可以使用后向递归对同一组隐变量求和。这将得到：
 
-$\begin{array}{r l}&{\quad P(x_{1},\ldots,x_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T}}P(x_{1},\ldots,x_{T},h_{1},\ldots,h_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T}}\prod_{t=1}^{T-1}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot P(h_{T}\mid h_{T-1})P(x_{T}\mid h_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T-1}}\prod_{t=1}^{T-1}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot\underbrace{\left[\sum_{h_{T}}P(h_{T}\mid h_{T-1})P(x_{T}\mid h_{T})\right]}_{\rho_{T-1}(h_{T-1}){\stackrel{\mathrm{def}}{=}}}\quad(9.4.4)}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T-2}}\prod_{t=1}^{T-2}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot\underbrace{\left[\sum_{h_{T-1}}P(h_{T-1}\mid h_{T-2})P(x_{T-1}\mid h_{T-1})\rho_{T-1}(h_{T-1})\right]}_{\rho_{T-2}(h_{T-2}){\stackrel{\mathrm{def}}{=}}}}\\ &{=\ldots}\\ &{=\displaystyle\sum_{h_{1}}P(h_{1})P(x_{1}\mid h_{1})\rho_{1}(h_{1}).}\end{array}$
+$$\begin{array}{r l}&{\quad P(x_{1},\ldots,x_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T}}P(x_{1},\ldots,x_{T},h_{1},\ldots,h_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T}}\prod_{t=1}^{T-1}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot P(h_{T}\mid h_{T-1})P(x_{T}\mid h_{T})}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T-1}}\prod_{t=1}^{T-1}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot\underbrace{\left[\sum_{h_{T}}P(h_{T}\mid h_{T-1})P(x_{T}\mid h_{T})\right]}_{\rho_{T-1}(h_{T-1}){\stackrel{\mathrm{def}}{=}}}\quad(9.4.4)}\\ &{=\displaystyle\sum_{h_{1},\ldots,h_{T-2}}\prod_{t=1}^{T-2}P(h_{t}\mid h_{t-1})P(x_{t}\mid h_{t})\cdot\underbrace{\left[\sum_{h_{T-1}}P(h_{T-1}\mid h_{T-2})P(x_{T-1}\mid h_{T-1})\rho_{T-1}(h_{T-1})\right]}_{\rho_{T-2}(h_{T-2}){\stackrel{\mathrm{def}}{=}}}}\\ &{=\ldots}\\ &{=\displaystyle\sum_{h_{1}}P(h_{1})P(x_{1}\mid h_{1})\rho_{1}(h_{1}).}\end{array}$$
diff --git a/markdowns/research_programming_book_00089.md b/markdowns/research_programming_book_00089.md
index db9df979866f0969e9c72534a24898756259b9f1..bf6f4975ffff1b193f5210868a8374bdf52381a9 100644
--- a/markdowns/research_programming_book_00089.md
+++ b/markdowns/research_programming_book_00089.md
@@ -10,17 +10,17 @@ if(a < b) {  //如果a比b小,就把b赋值给max(最大值)
 ```
 这样一来，max就是那个较大的位数，假如max为2，那么就需要把2变成100，也就是10×10。同理，如果max为1，那就是10了。具体逻辑如下所示。
 
-$\mathrm{m a x}=1--->10$
+$$\mathrm{m a x}=1--->10$$
 
-$max = 2---> 10 \times 10$
+$$max = 2---> 10 \times 10$$
 
-$\mathrm{m a x}=3{---}>10\times10\times10$
+$$\mathrm{m a x}=3{---}>10\times10\times10$$
 
-$\mathrm{m a x}=4--->10\times10\times10\times10$
+$$\mathrm{m a x}=4--->10\times10\times10\times10$$
 
-$\text{max} = 5 \longrightarrow 10 \times 10 \times 10 \times 10 \times 10$
+$$\text{max} = 5 \longrightarrow 10 \times 10 \times 10 \times 10 \times 10$$
 
-$\text{max} = 6 \longrightarrow 10 \times 10 \times 10 \times 10 \times 10 \times 10$
+$$\text{max} = 6 \longrightarrow 10 \times 10 \times 10 \times 10 \times 10 \times 10$$
 
 如何实现这个逻辑呢？叶小凡思索了一下，决定用for循环完成，于是他右手一挥，一段代码流便凭空出现。
 ```