反向传导算法

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</math>  
</math>  
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{译者注:
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:<math>
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\begin{align}
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\delta^{(n_l-1)}_i &=\frac{\partial}{\partial z^{n_l-1}_i}J(W,b;x,y)
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= \frac{\partial}{\partial z^{n_l-1}_i}\frac{1}{2} \left\|y - h_{W,b}(x)\right\|^2
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= \frac{\partial}{\partial z^{n_1-1}_i}\frac{1}{2} \sum_{j=1}^{S_{n_l}}(y_j-a_j^{(n_l)})^2 \\
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&= \frac{1}{2} \sum_{j=1}^{S_{n_l}}\frac{\partial}{\partial z^{n_l-1}_i}(y_i-a_j^{(n_l)})^2
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= \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 \\
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&= \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)})\frac{\partial}{\partial z_i^{(n_l-1)}}f(z_j^{(n_l)})
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= \sum_{j=1}^{S_{n_l}}(y_j-f(z_j^{(n_l)}) f'(z_j^{(n_l)})\frac{\partial z_j^{(n_l)}}{\partial z_i^{(n_l-1)}} \\
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&= \sum_{j=1}^{S_{n_l}} \delta_j^{(n_l)}\frac{\partial z_j^{(n_l)}}{z_i^{n_l-1}}
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= \sum_{j=1}^{S_{n_l}} \left(\delta_j^{(n_l)}\frac{\partial}{\partial z_i^{n_l-1}}\sum_{k=1}^{S_{n_l-1}}f(z_k^{n_l-1})W_{jk}^{n_l-1}\right) \\
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&= \sum_{j=1}^{S_{n_l}} \delta_j^{(n_l)} W_{ji}^{n_l-1}f'(z_i^{n_l-1})
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= \left(\sum_{j=1}^{S_{n_l-1}}W_{ji}^{n_l-1}\delta_j^{(n_l)}\right)f'(z_i^{n_l-1})
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\end{align}
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</math>
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]
<li>计算我们需要的偏导数,计算方法如下:
<li>计算我们需要的偏导数,计算方法如下:

Revision as of 14:46, 9 April 2013

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