推导3

L(a2)=f(1,1)+2f(1,2)+f(2,2)+2j=3Nf(1,j)+2j=3Nf(2,j)a1a2+常数项1\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:

f(i,j)=aiajyiyjKija1=ξa2y2y1\begin{aligned} f(i, j) = a_ia_jy_iy_jK_{ij} \\ a_1 = \frac {-\xi - a_2y_2}{y_1} \end{aligned}

KijK_{ij}是预先计算好的值,也看作是常数

公式(3)对a2a_2求导,并令导数为0,得到a2a_2的值

已知:

a1a2=ξa2y2y1a2=y2y12f(1,1)a2=a12K11a1a1a2=2K11a1y2y1=2K11y2ξ+2K11a23f(1,2)=a1a2y1y2K12=ξa2y2y1a2y1y2K12=ξy2K12a2K12a22f(1,2)a2=(ξy2K12a2K12a22))a2=ξy2K122K12a24f(2,2)a2=a22K22a2=2K22a25j=3Nf(1,j)a2=j=3Na1ajy1yjK1ja1a1a2=j=3Najy1yjK1j(y2y1)=y2j=3NajyjK1j6j=3Nf(2,j)a2=j=3Na2ajy2yjK2ja2=j=3Najy2yjK2j7\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:

L(a2)a2=f(1,1)a2+2f(1,2)a2+f(2,2)a2+2j=3Nf(1,j)a2+2j=3Nf(2,j)a2a1a21=...8\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a2a_2为更新后的a2newa_2^{new}

a2new=a2old+y2(E1E2)K11+K122K12a_2^{new} = a_2^{old} + \frac{y_2(E_1-E_2)}{K_{11}+K_{12}-2K_{12}}

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