# 推导3

$$
\begin{aligned}
L(a\_2) = f(1,1)+2f(1,2)+f(2,2) \\
+2\sum\_{j=3}^Nf(1,j) +2\sum\_{j=3}^Nf(2,j) \\
-a\_1 - a\_2 + \text{常数项}&& {1}
\end{aligned}
$$

其中：

$$
\begin{aligned}
f(i, j) = a\_ia\_jy\_iy\_jK\_{ij}   \\
a\_1 = \frac {-\xi - a\_2y\_2}{y\_1}
\end{aligned}
$$

$$K\_{ij}$$是预先计算好的值，也看作是常数

公式（3）对$$a\_2$$求导，并令导数为0，得到$$a\_2$$的值

已知：

$$
\begin{aligned}
\frac{\partial a\_1}{\partial a\_2} = \frac{\partial \frac {-\xi - a\_2y\_2}{y\_1}}{\partial a\_2} = -\frac{y\_2}{y\_1} && {2}  \\
\frac{\partial f(1,1)}{\partial a\_2} = \frac{\partial a\_1^2K\_{11}}{\partial a\_1}\frac{\partial a\_1}{\partial a\_2} = -2K\_{11}a\_1\frac{y\_2}{y\_1} \\
\= 2K\_{11}y\_2\xi + 2K\_{11}a\_2    && {3} \\
f(1,2) = a\_1a\_2y\_1y\_2K\_{12} = \frac {-\xi - a\_2y\_2}{y\_1}a\_2y\_1y\_2K\_{12} = -\xi y\_2K\_{12}a\_2 - K\_{12}a\_2^2   \\
\frac{\partial f(1,2)}{\partial a\_2} = \frac{\partial (-\xi y\_2K\_{12}a\_2 - K\_{12}a\_2^2))}{\partial a\_2} = -\xi y\_2K\_{12} - 2K\_{12}a\_2 && {4} \\
\frac{\partial f(2,2)}{\partial a\_2} = \frac{\partial a\_2^2K22}{\partial a\_2} = 2K\_{22}a\_2 && {5} \\
\frac{\partial \sum\_{j=3}^Nf(1,j)}{\partial a\_2} = \frac{\partial \sum\_{j=3}^Na\_1a\_jy\_1y\_jK\_{1j}}{\partial a\_1}\frac{\partial a\_1}{\partial a\_2} \\
\= \sum\_{j=3}^Na\_jy\_1y\_jK\_{1j} (-\frac{y\_2}{y\_1}) \\
\= -y\_2\sum\_{j=3}^Na\_jy\_jK\_{1j} && {6} \\
\frac{\partial \sum\_{j=3}^Nf(2,j)}{\partial a\_2} = \frac{\partial \sum\_{j=3}^Na\_2a\_jy\_2y\_jK\_{2j}}{\partial a\_2}
\= \sum\_{j=3}^Na\_jy\_2y\_jK\_{2j} && {7}
\end{aligned}
$$

得：

$$
\begin{aligned}
\frac{\partial L(a\_2)}{\partial a\_2} = \frac{\partial f(1,1)}{\partial a\_2} + 2\frac{\partial f(1,2)}{\partial a\_2} + \frac{\partial f(2,2)}{\partial a\_2} + 2\frac{\partial \sum\_{j=3}^Nf(1,j)}{\partial a\_2} + 2\frac{\partial \sum\_{j=3}^Nf(2,j)}{\partial a\_2} - \frac{\partial a\_1}{\partial a\_2} - 1
\= {...} && {8}
\end{aligned}
$$

令公式（8）等于0，解得的$$a\_2$$为更新后的$$a\_2^{new}$$

$$
a\_2^{new} = a\_2^{old} + \frac{y\_2(E\_1-E\_2)}{K\_{11}+K\_{12}-2K\_{12}}
$$