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$a_2^{new}$

$$a_2^{new} = a_2^{old} + \frac{y_2(E_1-E_2)}{K_{11}+K_{12}-2K_{12}}$$