L(a2)=f(1,1)+2f(1,2)+f(2,2)+2j=3∑Nf(1,j)+2j=3∑Nf(2,j)−a1−a2+常数项1 其中:
f(i,j)=aiajyiyjKija1=y1−ξ−a2y2 Kij是预先计算好的值,也看作是常数
公式(3)对a2求导,并令导数为0,得到a2的值
已知:
∂a2∂a1=∂a2∂y1−ξ−a2y2=−y1y2∂a2∂f(1,1)=∂a1∂a12K11∂a2∂a1=−2K11a1y1y2=2K11y2ξ+2K11a2f(1,2)=a1a2y1y2K12=y1−ξ−a2y2a2y1y2K12=−ξy2K12a2−K12a22∂a2∂f(1,2)=∂a2∂(−ξy2K12a2−K12a22))=−ξy2K12−2K12a2∂a2∂f(2,2)=∂a2∂a22K22=2K22a2∂a2∂∑j=3Nf(1,j)=∂a1∂∑j=3Na1ajy1yjK1j∂a2∂a1=j=3∑Najy1yjK1j(−y1y2)=−y2j=3∑NajyjK1j∂a2∂∑j=3Nf(2,j)=∂a2∂∑j=3Na2ajy2yjK2j=j=3∑Najy2yjK2j234567 得:
∂a2∂L(a2)=∂a2∂f(1,1)+2∂a2∂f(1,2)+∂a2∂f(2,2)+2∂a2∂∑j=3Nf(1,j)+2∂a2∂∑j=3Nf(2,j)−∂a2∂a1−1=...8 令公式(8)等于0,解得的a2为更新后的a2new
a2new=a2old+K11+K12−2K12y2(E1−E2)