# 反向传导算法

 Revision as of 14:59, 9 April 2013 (view source)Kandeng (Talk | contribs)← Older edit Revision as of 02:35, 15 April 2013 (view source)Kandeng (Talk | contribs) Newer edit → Line 102: Line 102: = \frac{\partial}{\partial z^{n_l-1}_i}\frac{1}{2} \left\|y - h_{W,b}(x)\right\|^2 = \frac{\partial}{\partial z^{n_l-1}_i}\frac{1}{2} \left\|y - h_{W,b}(x)\right\|^2 = \frac{\partial}{\partial z^{n_l-1}_i}\frac{1}{2} \sum_{j=1}^{S_{n_l}}(y_j-a_j^{(n_l)})^2 \\ = \frac{\partial}{\partial z^{n_l-1}_i}\frac{1}{2} \sum_{j=1}^{S_{n_l}}(y_j-a_j^{(n_l)})^2 \\ - &= \frac{1}{2} \sum_{j=1}^{S_{n_l}}\frac{\partial}{\partial z^{n_l-1}_i}(y_i-a_j^{(n_l)})^2 + &= \frac{1}{2} \sum_{j=1}^{S_{n_l}}\frac{\partial}{\partial z^{n_l-1}_i}(y_j-a_j^{(n_l)})^2 - = \frac{1}{2} \sum_{j=1}^{S_{n_l}}\frac{\partial}{\partial z^{n_l-1}_i}(y_i-f(z_j^{(n_l)})^2 \\ + = \frac{1}{2} \sum_{j=1}^{S_{n_l}}\frac{\partial}{\partial z^{n_l-1}_i}(y_j-f(z_j^{(n_l)}))^2 \\ &= \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)}) \cdot \frac{\partial}{\partial z_i^{(n_l-1)}}f(z_j^{(n_l)}) &= \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)}) \cdot \frac{\partial}{\partial z_i^{(n_l-1)}}f(z_j^{(n_l)}) = \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)}) \cdot  f'(z_j^{(n_l)}) \cdot \frac{\partial z_j^{(n_l)}}{\partial z_i^{(n_l-1)}} \\ = \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)}) \cdot  f'(z_j^{(n_l)}) \cdot \frac{\partial z_j^{(n_l)}}{\partial z_i^{(n_l-1)}} \\ Line 113: Line 113: [/itex] [/itex] - 将上式中的$\textstyle n_l-1$与$\textstyle n_l$的关系替换为$\textstyle l$与$\textstyle l-1$的关系，就可以得到： + 将上式中的$\textstyle n_l-1$与$\textstyle n_l$的关系替换为$\textstyle l$与$\textstyle l+1$的关系，就可以得到： : [itex] : [itex] \delta^{(l)}_i = \left( \sum_{j=1}^{s_{l+1}} W^{(l)}_{ji} \delta^{(l+1)}_j \right) f'(z^{(l)}_i) \delta^{(l)}_i = \left( \sum_{j=1}^{s_{l+1}} W^{(l)}_{ji} \delta^{(l+1)}_j \right) f'(z^{(l)}_i)