fix 11.14, 8.3

docs/chapter11/chapter11.md

@@ -147,9 +147,9 @@ $$
 
    a. 假设$x^i<0$，则$\operatorname{sign}(x^i)=-1$，那么有$x^i=z^i+\frac{\lambda}{L}>0$与假设矛盾；
 
-   b. 假设$x^i>0$，则$\operatorname{sign}(x^i)=1$，那么有$x^i=z^i-\frac{\lambda}{L}<0$和假设相符和，下面来检验$x^i=z^i-\frac{\lambda}{L}$是否是使函数$g(x^i)$的取得最小值。当$x^i<0$时，
+   b. 假设$x^i>0$，则$\operatorname{sign}(x^i)=1$，那么有$x^i=z^i-\frac{\lambda}{L}<0$和假设相符和，下面来检验$x^i=z^i-\frac{\lambda}{L}$是否是使函数$g(x^i)$的取得最小值。当$x^i>0$时，
    $$
-   \frac{d g\left(x^{i}\right)}{d x^{i}}=L\left(x^{i}-z^{i}\right)-\lambda
+   \frac{d g\left(x^{i}\right)}{d x^{i}}=L\left(x^{i}-z^{i}\right)+\lambda
    $$
    在定义域内连续可导，则$g(x^i)$的二阶导数
    $$

docs/chapter8/chapter8.md

@@ -20,24 +20,30 @@ $$
 \begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned}
 $$
 
-[推导]：由基分类器相互独立，假设随机变量$X$为$T$个基分类器分类正确的次数，因此$\mathrm{X} \sim \mathrm{B}(\mathrm{T}, 1-\mathrm{\epsilon})$，设$x_i$为每一个分类器分类正确的次数，则$x_i\sim \mathrm{B}(1, 1-\mathrm{\epsilon})（i=1，2，3，...，\mathrm{T}）$，那么有$$\mathrm{X}=\sum_{i=1}^{\mathrm{T}} x_i，\mathbb{E}(X)=\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i)$$
-证明过程如下：
+[推导]：由基分类器相互独立，假设随机变量$X$为$T$个基分类器分类正确的次数，因此$\mathrm{X} \sim \mathrm{B}(\mathrm{T}, 1-\mathrm{\epsilon})$，设$x_i$为每一个分类器分类正确的次数，则$x_i\sim \mathrm{B}(1, 1-\mathrm{\epsilon})（i=1，2，3，...，\mathrm{T}）$，那么有
+$$
+\begin{aligned}
+\mathrm{X}&=\sum_{i=1}^{\mathrm{T}} x_i\\
+\mathbb{E}(X)&=\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i)=(1-\epsilon)T
+\end{aligned}
+$$
+则：
 $$
 \begin{aligned} P(H(x) \neq f(x))=& P(X \leq\lfloor T / 2\rfloor) \\ & \leqslant P(X \leq T / 2)
-\\ & =P\left[X-(1-\varepsilon) T \leqslant \frac{T}{2}-(1-\varepsilon) T\right] 
+\\ & =P\left[X-(1-\epsilon) T \leqslant \frac{T}{2}-(1-\epsilon) T\right] 
 \\ & =P\left[X-
-(1-\varepsilon) T \leqslant -\frac{T}{2}\left(1-2\varepsilon\right)]\right]
+(1-\epsilon) T \leqslant -\frac{T}{2}\left(1-2\epsilon\right)]\right]
 \\ &=P\left[\sum_{i=1}^{\mathrm{T}} x_i-
-\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i) \leqslant -\frac{T}{2}\left(1-2\varepsilon\right)]\right]
+\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i) \leqslant -\frac{T}{2}\left(1-2\epsilon\right)]\right]
 \\ &=P\left[\frac{1}{\mathrm{T}}\sum_{i=1}^{\mathrm{T}} x_i-\frac{1}{\mathrm{T}}
-\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i) \leqslant -\frac{1}{2}\left(1-2\varepsilon\right)]\right]
+\sum_{i=1}^{\mathrm{T}}\mathbb{E}(x_i) \leqslant -\frac{1}{2}\left(1-2\epsilon\right)]\right]
  \end{aligned}
 $$
 根据Hoeffding不等式知
 $$
-P\left(\frac{1}{m} \sum_{i=1}^{m} x_{i}-\frac{1}{m} \sum_{i=1}^{m} \mathbb{E}\left(x_{i}\right) \leqslant -\epsilon\right) \leqslant  \exp \left(-2 m \epsilon^{2}\right)
+P\left(\frac{1}{m} \sum_{i=1}^{m} x_{i}-\frac{1}{m} \sum_{i=1}^{m} \mathbb{E}\left(x_{i}\right) \leqslant -\delta\right) \leqslant  \exp \left(-2 m \delta^{2}\right)
 $$
-令$\varepsilon=\frac {(1-2\epsilon)}{2},m=T$得
+令$\delta=\frac {(1-2\epsilon)}{2},m=T$得
 $$
 \begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned}
 $$