作业四 证明勒让德函数的正交性

作业四 证明勒让德函数的正交性

证明：（1）由勒让德方程0)1('2")1(2=++--y n n xy y x 即[(1?x 2)y ′]′+n n +1 y =0可得：

1?x 2 P m ′(x) ′+m m +1 P m x =0 [1]

1?x 2 P n ′(x) ′+n n +1 P n x =0 [2]

方程 1 ×P n ? 2 ×P m 在求其在-1到1上的积分可得： 1 ×P n ? 2 ×P m 1

?1

dx

= m m +1 ?n n +1 P n x P m (x)dx 1?1+ 1?x 2 P m ′ x ′P n 1?1

x dx

? 1?x 2 P n ′ x ′P m 1?1

x dx =0

1?x 2 P m ′ x ′P n 1?1 x dx

= 1?

x 2 P m ′ x P n x 1?1? 1?x 2 1?1P m ′ x P n ′(x )dx =? 1?x 2 1?1

P m ′ x P n ′(x )dx

同理可得：

1?

x 2 P n ′ x ′P m

1?1 x dx =? 1?x 2 1?1P m ′ x P n ′(x)dx 故有： m m +1 ?n n +1 P n x P m (x)dx 1?1=0

当n m ≠时 P n x P m (x)dx 1

?1=0

（2）的证明不妨先证明勒让德函数的递推公式之一： n +1 P n+1(x)? 2n +1 xP n (x)+nP n ?1(x)=0 由母函数：)(211

),(02x P t t tx t x w n n n ∑∞==+-=

ln w(x,t)=?12

ln(1?2xt +t 2) 对t 求导得：

w ′(x,t)=?1?2x +2t 2=x ?t 2

即w x,t x ?t =w ′ x,t (1?2xt +t 2)

又母函数直接对t 求导得：

w ′(x,t)= P n (x)t n ?1∞

n=1

n

带入上式可得：

x ?t P n (x)t n ∞n=0

= P n x t n ?1∞n=1n 1?2xt +t 2 = P n+1 x t n ∞n=0

(n +1) 1?2xt +t 2

xP n (x)t n ∞n=0

? P n (x)t n+1∞n=0

= P n+1 x t n ∞n=0 n +1 ?2x P n+1 x t n+1∞n=0 n +1 + P n+1 x t n+2∞n=0(n +1) 移项合并可得：

t n n +1 P n+1(x)? 2n +1 xP n (x)+nP n ?1(x)=0∞

n=1

n +1 P n+1(x)? 2n +1 xP n (x)+nP n ?1(x)=0得证 nP n x ? 2n ?1 xP n ?1 x +(n ?1)P n ?2(x)=0

P n 21

?1 x dx =1 P n 1?1

x [ 2n ?1 xP n ?1 x ?(n ?1)P n ?2(x)]dx =2n ?1 xP n x P n ?1(x)dx 1?1?n ?1 P n x P n ?2(x)dx 1

?1

因为当n m ≠时 P n x P m (x)dx 1

?1=0

P n x P n ?2(x)dx 1?1=0

P n 21?1

x dx =

2n ?1n xP n x P n ?1(x)dx 1?1 将xP n x =

12n+1[ n +1 P n+1(x)+nP n ?1(x)]带入上式得： P n 21?1

x dx

=2n ?1

n 12n +1 P n ?1 x [ n +1 P n+1(x)+nP n ?1(x)]dx 1?1

=2n ?12n +1 P n ?1

21?1 x dx =2n ?12n +12n ?32n ?1……13 P 021?1

x dx =1 1dx 1?1

=22n +1

即 P n x P m (x)dx 1?1=0当n m ≠时

P n 21?1 x dx =22n+1命题得证！