用途:可用它研究和导出P(x)的其他 性质。称为母函数法 二、递推公式 (1+(x2+(x)+(30() (2)(2+)9(x)=P1(x)-P(x)(2) 证明思路: ①.可用 Legendre表达式证,比较繁琐
用途：可用它研究和导出 的其他 性质。称为母函数法。 P x l( ) 二、递推公式 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 -1 . l +1 Pl+ x - 2l +1 xPl l x + = lP x 0 1 (2).(2l + = 1) Pl ( x) Pl l ¢ ¢ +1 ( x) - 2 P x -1 ( ) ( ) 证明思路： ①.可用Legendre表达式证，比较繁琐
②.可用母函数法,比较简便 证明:(1)4(*) x-t ∑P(x) 2xt+1 两边都乘上(-2x+),并再用() (x-)∑2(x)=(2x+2)∑B(x)
②.可用母函数法，比较简便 证明：(1) ( ) ( ) -1 3 2 0 2 - 1- 2 l l l x t P x lt xt t ¥ = = å + ( ) 2 两边都乘上 1- 2xt t + ，并再用(*) ( ) ( ) ( ) ( ) 2 -1 0 0 - 1- 2 l l l l l l x t P x t xt t lP x t ¥ ¥ = = å å = + d dt (*)
即:x∑P(x)1-∑P(x)14= ∑P(x)-2x∑(x)t+∑P(x) =0 1:xP(x)-B(x)=(+1)1-2xB+(1-1)P 即(7+1)P1-(21+1)xP(x)+P1(x)=0 问:如何证(2)?答: d(*) 2xt+t )32
( ) ( ) 1 0 0 - l l l l l l x P x t P x t ¥ ¥ + = = 即: å å = ( ) ( ) ( ) -1 1 0 0 0 - 2 l l l l l l l l l lP x t x lP x t lP x t ¥ ¥ ¥ + = = = å å å+ ( ) ( ) ( ) ( ) -1 1 -1 : - 1 - 2 -1 l l l l l l t xP x P x = l + + P+ xlP l P 即(l +1) Pl+1 - (2l +1 0 ) xPl l ( x) + = lP x -1 ( ) 问：如何证(2)？答： ( ) ( ) 3 2 0 2 1 - 2 l l l t P x t xt t ¥ = = å ¢ + d dt (*)
1()=(-2x+)P(x) :P(x)=21(x2xP(x)+Pm(x)(3) dx ():(+1)m(x)(1+1)(x)-(21+)xP( 17(x)=0 (4) 由(4)xP(x)=P() 2l+112/+1 代入(3)即证(2)
( ) ( ) ( ) 2 0 0 1- 2 l l l l l t P x t xt t P x t l ¥ ¥ = = å å = + ¢ ( ) ( ) ( ) ( ) ( ) 1 1 -1 : - 2 3 l l l l l t P x P x xP x P x + + = + ¢ ¢ ¢ (l +1) Pl ¢ ¢ +1 ( x) - (2l + + 1) Pl l ( x) - (2 1 l ) xP x( ) ( ) ( ) -1 0 4 l + = lP x ¢ 由(4) ( ) ( ) 1 -1 1 - - - 2 1 2 1 l l l l l l xP x P x P P l l + + ¢ = ¢ ¢ + + 代入(3)即证(2) ( ) 1 : d dx