revert optimize

bad idea....

week1.md

@@ -313,23 +313,21 @@ $\begin{align*} & \text{repeat until convergence:} \; \lbrace \newline \; &{{\th
 
 直接将线性回归模型公式带入梯度下降公式可得出公式
 
-![](image/20180106_203726.png)
+![](image/20180106_203726.png)当 $j = 0, j = 1$ 时，**线性回归中代价函数求导的推导过程：**
 
->  当 $j = 0, j = 1$ 时，**线性回归中代价函数求导的推导过程：**
->
-> $\frac{\partial}{\partial\theta_j} J(\theta_1, \theta_2)=\frac{\partial}{\partial\theta_j} \left(\frac{1}{2m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}} \right)=$
->
-> $\left(\frac{1}{2m}*2\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} \right)*\frac{\partial}{\partial\theta_j}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} =$
->
-> $\left(\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} \right)*\frac{\partial}{\partial\theta_j}{{\left(\theta_0{x_0^{(i)}} + \theta_1{x_1^{(i)}}-{{y}^{(i)}} \right)}}$
->
-> 所以当 $j = 0$ 时：
->
-> $\frac{\partial}{\partial\theta_0} J(\theta)=\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} *x_0^{(i)}$
->
-> 所以当 $j = 1$ 时：
->
-> $\frac{\partial}{\partial\theta_1} J(\theta)=\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} *x_1^{(i)}$
+$\frac{\partial}{\partial\theta_j} J(\theta_1, \theta_2)=\frac{\partial}{\partial\theta_j} \left(\frac{1}{2m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}} \right)=$
+
+$\left(\frac{1}{2m}*2\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} \right)*\frac{\partial}{\partial\theta_j}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} =$
+
+$\left(\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} \right)*\frac{\partial}{\partial\theta_j}{{\left(\theta_0{x_0^{(i)}} + \theta_1{x_1^{(i)}}-{{y}^{(i)}} \right)}}$
+
+所以当 $j = 0$ 时：
+
+$\frac{\partial}{\partial\theta_0} J(\theta)=\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} *x_0^{(i)}$
+
+所以当 $j = 1$ 时：
+
+$\frac{\partial}{\partial\theta_1} J(\theta)=\frac{1}{m}\sum\limits_{i=1}^{m}{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}} *x_1^{(i)}$
 
 
 

week2.md

@@ -147,33 +147,33 @@ $$
 [^1]: 一般来说，当 $n$ 超过 10000 时，对于正规方程而言，特征量较大。
 [^2]: 梯度下降算法的普适性好，而对于特定的线性回归模型，正规方程是很好的替代品。
 
-> **正规方程法的推导过程**：
->
->  $\begin{aligned} & J\left( \theta  \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {h_{\theta}}\left( {x^{(i)}} \right)-{y^{(i)}} \right)}^{2}}}\newline \; & =\frac{1}{2m}||X\theta-y||^2 \newline \; & =\frac{1}{2m}(X\theta-y)^T(X\theta-y) &\newline  \end{aligned}$
->
-> 展开上式可得
->
-> $J(\theta )= \frac{1}{2m}\left( {{\theta }^{T}}{{X}^{T}}X\theta -{{\theta}^{T}}{{X}^{T}}y-{{y}^{T}}X\theta + {{y}^{T}}y \right)$
->
-> 注意到 ${{\theta}^{T}}{{X}^{T}}y$ 与 ${{y}^{T}}X\theta$ 都为标量，实际上是等价的，则
->
-> $J(\theta) = \frac{1}{2m}[X^TX\theta-2\theta^TX^Ty+y^Ty]$
->
-> 接下来对$J(\theta )$ 求偏导，根据矩阵的求导法则:
->
-> $\frac{dX^TAX}{dX}=(A+A^\mathrm{T})X$
->
-> $\frac{dX^TA}{dX}={A}$
->
-> 所以有:
->
-> $\frac{\partial J\left( \theta  \right)}{\partial \theta }=\frac{1}{2m}\left(2{{X}^{T}}X\theta -2{{X}^{T}}y \right)={{X}^{T}}X\theta -{{X}^{T}}y$
->
-> 令$\frac{\partial J\left( \theta  \right)}{\partial \theta }=0$, 则有
-> $$
-> \theta ={{\left( {X^{T}}X \right)}^{-1}}{X^{T}}y
-> $$
->
+**正规方程法的推导过程**：
+
+$\begin{aligned} & J\left( \theta  \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {h_{\theta}}\left( {x^{(i)}} \right)-{y^{(i)}} \right)}^{2}}}\newline \; & =\frac{1}{2m}||X\theta-y||^2 \newline \; & =\frac{1}{2m}(X\theta-y)^T(X\theta-y) &\newline  \end{aligned}$
+
+展开上式可得
+
+$J(\theta )= \frac{1}{2m}\left( {{\theta }^{T}}{{X}^{T}}X\theta -{{\theta}^{T}}{{X}^{T}}y-{{y}^{T}}X\theta + {{y}^{T}}y \right)$
+
+注意到 ${{\theta}^{T}}{{X}^{T}}y$ 与 ${{y}^{T}}X\theta$ 都为标量，实际上是等价的，则
+
+$J(\theta) = \frac{1}{2m}[X^TX\theta-2\theta^TX^Ty+y^Ty]$
+
+接下来对$J(\theta )$ 求偏导，根据矩阵的求导法则:
+
+$\frac{dX^TAX}{dX}=(A+A^\mathrm{T})X$
+
+$\frac{dX^TA}{dX}={A}$
+
+所以有:
+
+$\frac{\partial J\left( \theta  \right)}{\partial \theta }=\frac{1}{2m}\left(2{{X}^{T}}X\theta -2{{X}^{T}}y \right)={{X}^{T}}X\theta -{{X}^{T}}y$
+
+令$\frac{\partial J\left( \theta  \right)}{\partial \theta }=0$, 则有
+$$
+\theta ={{\left( {X^{T}}X \right)}^{-1}}{X^{T}}y
+$$
+
 
 
 ## 4.7 不可逆性正规方程(Normal Equation Noninvertibility)

week3.md

@@ -167,39 +167,39 @@ $\begin{align*}& \text{repeat until convergence:} \; \lbrace \newline \; & \thet
 
 
 
-> **逻辑回归中代价函数求导的推导过程：**
->
-> $J(\theta) = - \frac{1}{m} \displaystyle \sum_{i=1}^m [y^{(i)}\log (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))]$
->
-> 令 $f(\theta) = {{y}^{(i)}}\log \left( {h_\theta}\left( {{x}^{(i)}} \right) \right)+\left( 1-{{y}^{(i)}} \right)\log \left( 1-{h_\theta}\left( {{x}^{(i)}} \right) \right)$
->
-> 将 $h_\theta(x^{(i)}) = g\left(\theta^{T}x^{(i)} \right)=\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} $ 带入得
->
-> $f(\theta)={{y}^{(i)}}\log \left( \frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} \right)+\left( 1-{{y}^{(i)}} \right)\log \left( 1-\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} \right)$
-> $=-{{y}^{(i)}}\log \left( 1+{{e}^{-{\theta^T}{{x}^{(i)}}}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{{\theta^T}{{x}^{(i)}}}} \right)$
->
-> 根据求偏导的性质，没有 $\theta_j$ 的项求偏导即为 $0$，都消去，则得：
->
-> $\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)=x^{(i)}_j$
->
-> 所以有：
->
-> $\frac{\partial }{\partial {\theta_{j}}}f\left( \theta  \right)=\frac{\partial }{\partial {\theta_{j}}}[-{{y}^{(i)}}\log \left( 1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{{\theta^{T}}{{x}^{(i)}}}} \right)]$
->
-> $=-{{y}^{(i)}}\frac{\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)\cdot{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}{1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)\cdot{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$
->
-> $=-{{y}^{(i)}}\frac{-x_{j}^{(i)}{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}{1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$
-> $={{y}^{(i)}}\frac{x_j^{(i)}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$ 
-> $={\frac{{{y}^{(i)}}x_j^{(i)}-x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}+{{y}^{(i)}}x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}}$
-> $={\frac{{{y}^{(i)}}\left( 1\text{+}{{e}^{{\theta^T}{{x}^{(i)}}}} \right)-{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}x_j^{(i)}}$
-> $={({{y}^{(i)}}-\frac{{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}})x_j^{(i)}}$
-> $={({{y}^{(i)}}-\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}})x_j^{(i)}}$
-> $={\left({{y}^{(i)}}-{h_\theta}\left( {{x}^{(i)}} \right)\right)x_j^{(i)}}$
-> $={\left({h_\theta}\left( {{x}^{(i)}} \right)-{{y}^{(i)}}\right)x_j^{(i)}}$
->
-> 则可得代价函数的导数：
->
-> $\frac{\partial }{\partial {\theta_{j}}}J(\theta) = -\frac{1}{m}\sum\limits_{i=1}^{m}{\frac{\partial }{\partial {\theta_{j}}}f(\theta)}=\frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)} $
+**逻辑回归中代价函数求导的推导过程：**
+
+$J(\theta) = - \frac{1}{m} \displaystyle \sum_{i=1}^m [y^{(i)}\log (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))]$
+
+令 $f(\theta) = {{y}^{(i)}}\log \left( {h_\theta}\left( {{x}^{(i)}} \right) \right)+\left( 1-{{y}^{(i)}} \right)\log \left( 1-{h_\theta}\left( {{x}^{(i)}} \right) \right)$
+
+将 $h_\theta(x^{(i)}) = g\left(\theta^{T}x^{(i)} \right)=\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} $ 带入得
+
+$f(\theta)={{y}^{(i)}}\log \left( \frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} \right)+\left( 1-{{y}^{(i)}} \right)\log \left( 1-\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}} \right)$
+$=-{{y}^{(i)}}\log \left( 1+{{e}^{-{\theta^T}{{x}^{(i)}}}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{{\theta^T}{{x}^{(i)}}}} \right)$
+
+根据求偏导的性质，没有 $\theta_j$ 的项求偏导即为 $0$，都消去，则得：
+
+$\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)=x^{(i)}_j$
+
+所以有：
+
+$\frac{\partial }{\partial {\theta_{j}}}f\left( \theta  \right)=\frac{\partial }{\partial {\theta_{j}}}[-{{y}^{(i)}}\log \left( 1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{{\theta^{T}}{{x}^{(i)}}}} \right)]$
+
+$=-{{y}^{(i)}}\frac{\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)\cdot{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}{1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)}  \right)\cdot{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$
+
+$=-{{y}^{(i)}}\frac{-x_{j}^{(i)}{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}{1+{{e}^{-{\theta^{T}}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$
+$={{y}^{(i)}}\frac{x_j^{(i)}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}-\left( 1-{{y}^{(i)}} \right)\frac{x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}$ 
+$={\frac{{{y}^{(i)}}x_j^{(i)}-x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}+{{y}^{(i)}}x_j^{(i)}{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}}$
+$={\frac{{{y}^{(i)}}\left( 1\text{+}{{e}^{{\theta^T}{{x}^{(i)}}}} \right)-{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}}x_j^{(i)}}$
+$={({{y}^{(i)}}-\frac{{{e}^{{\theta^T}{{x}^{(i)}}}}}{1+{{e}^{{\theta^T}{{x}^{(i)}}}}})x_j^{(i)}}$
+$={({{y}^{(i)}}-\frac{1}{1+{{e}^{-{\theta^T}{{x}^{(i)}}}}})x_j^{(i)}}$
+$={\left({{y}^{(i)}}-{h_\theta}\left( {{x}^{(i)}} \right)\right)x_j^{(i)}}$
+$={\left({h_\theta}\left( {{x}^{(i)}} \right)-{{y}^{(i)}}\right)x_j^{(i)}}$
+
+则可得代价函数的导数：
+
+$\frac{\partial }{\partial {\theta_{j}}}J(\theta) = -\frac{1}{m}\sum\limits_{i=1}^{m}{\frac{\partial }{\partial {\theta_{j}}}f(\theta)}=\frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)} $
 
 ## 6.6 进阶优化(Advanced Optimization)