推导2

L(a1,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_1, 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)=aiajyiyjK(xi,xj)f(i, j) = a_ia_jy_iy_jK(x_i,x_j)

对偶问题中的限制条件也可以分成变量和常量两部分:

i=1Naiyi=a1y1+a2y2+i=3Naiyi=02\begin{aligned} \sum_{i=1}^Na_iy_i=a_1y_1+a_2y_2 + \sum_{i=3}^Na_iy_i = 0 && {2} \end{aligned}

令常数项为:

ξ=i=3Naiyi3\begin{aligned} \xi = \sum_{i=3}^Na_iy_i && {3} \end{aligned}

根据公式(3),得到a1a_1a2a_2的关系为:

a1=ξa2y2y1a_1 = \frac {-\xi - a_2y_2}{y_1}

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