Legendre多项式推导1
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2022-08-08
21:02:55
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Legendre 方程
( )
( ) ( ) ( )
2
1 1 0
n n
d d
x P x n n P x
dx dx
æ ö
- + - =
ç ÷
è ø
令
( )
0
k
n k
n
P x c x
¥
=
=
å
代入方程得
( )
( )
( )
( ) ( )
( )( ) ( ) ( )
( )
( )
2
0 0
1 2 2
1 2 0
2
1 0 2 0
2 0
1 3 1
1 1 0
2 1 1 1 0
2 2 1 1 1 0
2 1 0
2 6 1 0
2
k k
k k
n n
k k k
k k k
k k k
k k k k
k k k k
k k k n
k
d d
x c x n n c x
dx dx
x kc x x k k c x n n c x
kc x k k c x k k c x n n c x
c n n c
c c n n c
kc k
¥ ¥
= =
¥ ¥ ¥
- -
= = =
¥ ¥ ¥ ¥
+
= = = =
æ ö
- + + =
ç ÷
è ø
æ ö æ ö
- + - - + + =
ç ÷ ç ÷
è ø è ø
- + + + - - + + =
+ + =
- + + + =
- +
å å
å å å
å å å å
( )( ) ( ) ( )
( ) ( )
( )( )
( )
( )
2
2
2
2 0
3 1
2 1 1 1 0, 2
0,
1 1
, 2
2 1
1
2
2 1
6
k k k
k
k k
k c k k c n n c k
c k n
k k n n
c c k
k k
n n
c c
n n
c c
+
+
+
+ + - - + + = ³
= >
+ - +
= ³
+ +
- +
=
- +
=
所以
( ) ( )
( )( )
( )( )
( )( )
2
1 1
2 1
1
=
2 1
k k
k
k k n n
c c
k k
k n k n
c
k k
+
+ - +
=
+ +
- + +
+ +
对
k NÎ
成立。
若令
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