# 4个等式的证明

## 等式一

证明：$$\delta^L\_j = \frac{\partial C}{\partial a^L\_j} \sigma'(z^L\_j)$$

$$
\begin{eqnarray}
\delta^L\_j = & \frac{\partial C}{\partial z^L\_j}.
\tag{36}  \\
\= & \sum\_k \frac{\partial C}{\partial a^L\_k} \frac{\partial a^L\_k}{\partial z^L\_j},
\tag{37} \\
\= & \frac{\partial C}{\partial a^L\_j} \frac{\partial a^L\_j}{\partial z^L\_j}.
\tag{38}  \\
\= & \frac{\partial C}{\partial a^L\_j} \sigma'(z^L\_j),
\tag{39}
\end{eqnarray}
$$

等式（1）得证

## 等式二

证明：$$\delta^l = ((w^{l+1})^T \delta^{l+1}) \odot \sigma'(z^l)$$

$$
\begin{eqnarray}
\delta^l\_j & = & \frac{\partial C}{\partial z^l\_j} \tag{40}\\
& = & \sum\_k \frac{\partial C}{\partial z^{l+1}\_k} \frac{\partial z^{l+1}\_k}{\partial z^l\_j} \tag{41}\\
& = & \sum\_k \frac{\partial z^{l+1}\_k}{\partial z^l\_j} \delta^{l+1}*k,
\tag{42} \\
& = & \sum\_k w^{l+1}*{kj}  \delta^{l+1}\_k \sigma'(z^l\_j).
\tag{45}
\end{eqnarray}
$$

等式二得证

公式（45）说明：

$$
\begin{eqnarray}
z^{l+1}*k = \sum\_j w^{l+1}*{kj} a^l\_j +b^{l+1}*k = \sum\_j w^{l+1}*{kj} \sigma(z^l\_j) +b^{l+1}\_k.
\tag{43} \\
\frac{\partial z^{l+1}*k}{\partial z^l\_j} = w^{l+1}*{kj} \sigma'(z^l\_j).
\tag{44}
\end{eqnarray}
$$

## 等式三

证明：$$\frac{\partial C}{\partial b^l\_j} = \delta^l\_j$$

本章在4个等式的形式和证明过程中，下标有些混乱，给理解公式带来障碍，这里把下标的含义重新申请一下：\
j：当前层的神经元的下标\
k：下一层神经元的下标\
i：上一层神经元的下标

书上已经写给了$$z^{l+1}*k$$与$$w^{l+1}*{kj}$$、$$b^{l+1}\_k$$的关系为：

$$
\begin{eqnarray}
z^{l+1}*k = \sum\_j w^{l+1}*{kj} a^l\_j +b^{l+1}\_k
\tag{43}\end{eqnarray}
$$

同理可写出$$z^{l}*j$$与$$w^{l}*{ji}$$、$$b^{l}\_j$$的关系为：

$$
\begin{eqnarray}
z^{l}*j = \sum\_i w^{l}*{ji} a^{l-1}\_i +b^{l}\_j
\tag{43.1}\end{eqnarray}
$$

$$
\begin{eqnarray}
\frac{\partial C}{\partial b^l\_j} & = & \frac{\partial C}{\partial z^l\_j} \frac{\partial z^l\_j}{\partial b^l\_j} = \delta^l\_j
\end{eqnarray}
$$

等式三得证

## 等式四

证明：$$\frac{\partial C}{\partial w^l\_{jk}} = a^{l-1}\_k \delta^l\_j$$

根据等式三中关于下标的定义，等式四应调整为：

$$
\frac{\partial C}{\partial w^l\_{ji}} = a^{l-1}\_i \delta^l\_j
$$

$$
\begin{eqnarray}
\frac{\partial C}{\partial w^l\_{ji}} = \frac{\partial C}{\partial z^l\_j} \frac{\partial z^l\_j}{\partial w^l\_{ji}} = a^{l-1}\_k \delta^l\_j
\end{eqnarray}
$$

等式三得证