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readme.md

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@@ -667,7 +667,11 @@ def predict(Theta1,Theta2,X):
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根据向量夹角公式推导一下即可。![\cos \theta = \frac{{\overrightarrow {\text{u}} \overrightarrow v }}{{|\overrightarrow {\text{u}} ||\overrightarrow v |}}](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=%5Ccos%20%5Ctheta%20%20%3D%20%5Cfrac%7B%7B%5Coverrightarrow%20%7B%5Ctext%7Bu%7D%7D%20%5Coverrightarrow%20v%20%7D%7D%7B%7B%7C%5Coverrightarrow%20%7B%5Ctext%7Bu%7D%7D%20%7C%7C%5Coverrightarrow%20v%20%7C%7D%7D)
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- 前面说过,当`C`越大时,`margin`也就越大,我们的目的是最小化代价函数`J(θ)`,当`margin`最大时,`C`的乘积项![\sum\limits_{i = 1}^m {[{y^{(i)}}\cos {t_1}({\theta ^T}{x^{(i)}}) + (1 - {y^{(i)}})\cos {t_0}({\theta ^T}{x^{(i)}})} ]](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=%5Csum%5Climits_%7Bi%20%3D%201%7D%5Em%20%7B%5B%7By%5E%7B%28i%29%7D%7D%5Ccos%20%7Bt_1%7D%28%7B%5Ctheta%20%5ET%7D%7Bx%5E%7B%28i%29%7D%7D%29%20%2B%20%281%20-%20%7By%5E%7B%28i%29%7D%7D%29%5Ccos%20%7Bt_0%7D%28%7B%5Ctheta%20%5ET%7D%7Bx%5E%7B%28i%29%7D%7D%29%7D%20%5D)要很小,所以近似为:
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![J(\theta ) = C0 + \frac{1}{2}\sum\limits_{j = 1}^{\text{n}} {\theta _j^2} = \frac{1}{2}\sum\limits_{j = 1}^{\text{n}} {\theta _j^2} = \frac{1}{2}(\theta _1^2 + \theta _2^2) = \frac{1}{2}{\sqrt {\theta _1^2 + \theta _2^2} ^2}](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=J%28%5Ctheta%20%29%20%3D%20C0%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits_%7Bj%20%3D%201%7D%5E%7B%5Ctext%7Bn%7D%7D%20%7B%5Ctheta%20_j%5E2%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits_%7Bj%20%3D%201%7D%5E%7B%5Ctext%7Bn%7D%7D%20%7B%5Ctheta%20_j%5E2%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Ctheta%20_1%5E2%20%2B%20%5Ctheta%20_2%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7B%5Csqrt%20%7B%5Ctheta%20_1%5E2%20%2B%20%5Ctheta%20_2%5E2%7D%20%5E2%7D)
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![J(\theta ) = C0 + \frac{1}{2}\sum\limits_{j = 1}^{\text{n}} {\theta _j^2} = \frac{1}{2}\sum\limits_{j = 1}^{\text{n}} {\theta _j^2} = \frac{1}{2}(\theta _1^2 + \theta _2^2) = \frac{1}{2}{\sqrt {\theta _1^2 + \theta _2^2} ^2}](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=J%28%5Ctheta%20%29%20%3D%20C0%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits_%7Bj%20%3D%201%7D%5E%7B%5Ctext%7Bn%7D%7D%20%7B%5Ctheta%20_j%5E2%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits_%7Bj%20%3D%201%7D%5E%7B%5Ctext%7Bn%7D%7D%20%7B%5Ctheta%20_j%5E2%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Ctheta%20_1%5E2%20%2B%20%5Ctheta%20_2%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7B%5Csqrt%20%7B%5Ctheta%20_1%5E2%20%2B%20%5Ctheta%20_2%5E2%7D%20%5E2%7D)
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我们最后的目的就是求使代价最小的`θ`
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![\left\{ {\begin{array}{c} {{\theta ^T}{x^{(i)}} \geqslant 1} \\ {{\theta ^T}{x^{(i)}} \leqslant - 1} \end{array} } \right.\begin{array}{c} {({y^{(i)}} = 1)} \\ {({y^{(i)}} = 0)} \end{array} ](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=%5Cleft%5C%7B%20%7B%5Cbegin%7Barray%7D%7Bc%7D%20%20%20%20%7B%7B%5Ctheta%20%5ET%7D%7Bx%5E%7B%28i%29%7D%7D%20%5Cgeqslant%201%7D%20%5C%5C%20%20%20%20%7B%7B%5Ctheta%20%5ET%7D%7Bx%5E%7B%28i%29%7D%7D%20%5Cleqslant%20%20-%201%7D%20%20%5Cend%7Barray%7D%20%7D%20%5Cright.%5Cbegin%7Barray%7D%7Bc%7D%20%20%20%20%7B%28%7By%5E%7B%28i%29%7D%7D%20%3D%201%29%7D%20%5C%5C%20%20%20%20%7B%28%7By%5E%7B%28i%29%7D%7D%20%3D%200%29%7D%20%20%5Cend%7Barray%7D%20)可以得到:
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![\left\{ {\begin{array}{c} {{p^{(i)}}||\theta || \geqslant 1} \\ {{p^{(i)}}||\theta || \leqslant - 1} \end{array} } \right.\begin{array}{c} {({y^{(i)}} = 1)} \\ {({y^{(i)}} = 0)} \end{array} ](http://chart.apis.google.com/chart?cht=tx&chs=1x0&chf=bg,s,FFFFFF00&chco=000000&chl=%5Cleft%5C%7B%20%7B%5Cbegin%7Barray%7D%7Bc%7D%20%20%20%20%7B%7Bp%5E%7B%28i%29%7D%7D%7C%7C%5Ctheta%20%7C%7C%20%5Cgeqslant%201%7D%20%5C%5C%20%20%20%20%7B%7Bp%5E%7B%28i%29%7D%7D%7C%7C%5Ctheta%20%7C%7C%20%5Cleqslant%20%20-%201%7D%20%20%5Cend%7Barray%7D%20%7D%20%5Cright.%5Cbegin%7Barray%7D%7Bc%7D%20%20%20%20%7B%28%7By%5E%7B%28i%29%7D%7D%20%3D%201%29%7D%20%5C%5C%20%20%20%20%7B%28%7By%5E%7B%28i%29%7D%7D%20%3D%200%29%7D%20%20%5Cend%7Barray%7D%20)`p`即为`x``θ`上的投影
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