11最小逼近函数的解法 (x)=a090+a1+…+un E=a=∑[f(x)-g(x) i=0 E是关于a0a1…,am的连续函数,且E20,所以一定 存在一组数a,a1…an使得E取极小值,只要满足 OE =0(j=0,1,…,m) ∑[agn(x)+a91(x)+…+an9n(x,)-∫(x) J I=0 2∑[aq(x)+aq(x)+…+an9n(x1)-f(x)q(x) i=0 0(j=0,1,…,m)E是关于a0 ,a1 ,…,am的连续函数，且E≥0，所以一定 存在一组数a0 ,a1 ,…,am使得E取极小值，只要满足 即 2 2 2 0 || || ( ) ( ) n i i i E f x g x  = = = −     0 0 1 1 ( ) m m g x a a a = + + +    0 ( 0,1, , ) j E j m a  = =  2 0 0 1 1 0 0 0 1 1 0 ( ) ( ) ( ) ( ) 2 ( ) ( ) ( ) ( ) ( ) 0 ( 0,1, , ) n i i m m i i i j n i i m m i i j i i a x a x a x f x a a x a x a x f x x j m        = =    + + + −    = + + + −      = =   1.1 最小逼近函数的解法