# 推导1

$$
\begin{aligned}
L(a\_1, a\_2) = \sum\_{i=1}^N\sum\_{j=1}^Na\_ia\_jy\_iy\_jK(x\_i, x\_j) - \sum\_{i=1}^Na\_i && {1}
\end{aligned}
$$

为了简化公式，令：

$$
\begin{aligned}
f(i, j) = a\_ia\_jy\_iy\_jK(x\_i,x\_j)  && {2}
\end{aligned}
$$

公式（1）简化为：

$$
\begin{aligned}
L(a\_1, a\_2) = \sum\_{i=1}^N\sum\_{j=1}^Nf(i,j) - \sum\_{i=1}^Na\_i && {3}
\end{aligned}
$$

在公式（3）中，将$$a\_1$$、$$a\_2$$是变量，其它参数a是常量，把公式（3）分成包含变量的部分和常量部分：

$$
\begin{aligned}
L(a\_1, a\_2) = f(1,1)+f(1,2)+\sum\_{j=3}^Nf(1,j) \\
+f(2,1)+f(2,2)+\sum\_{j=3}^Nf(2,j) \\
+\sum\_{i=3}^Nf(i,1)+\sum\_{i=3}^Nf(i,2) + +\sum\_{i=3}^N\sum\_{j=3}^Nf(i,j) \\
-a\_1 - a\_2 - \sum\_{i=3}^Na\_i && {4}
\end{aligned}
$$

要L(a\_1, a\_2)对a\_1和a\_2求导，公式（4）中的常数部分对求导结果不影响，直接合并为一个不需要关心具体内容的常数项，得：

$$
\begin{aligned}
L(a\_1, a\_2) = f(1,1)+f(1,2)+\sum\_{j=3}^Nf(1,j) \\
+f(2,1)+f(2,2)+\sum\_{j=3}^Nf(2,j) \\
+\sum\_{i=3}^Nf(i,1)+\sum\_{i=3}^Nf(i,2)  \\
-a\_1 - a\_2 + \text{常数项}&& {5}
\end{aligned}
$$

根据f(i,j)的定义可知，f具有对称性，得：

$$
\begin{aligned}
L(a\_1, a\_2) = f(1,1)+2f(1,2)+f(2,2) \\
+2\sum\_{j=3}^Nf(1,j) +2\sum\_{j=3}^Nf(2,j) \\
-a\_1 - a\_2 + \text{常数项}&& {6}
\end{aligned}
$$


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