Page 50 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Notes
2 ( 1) n (n 1) ( 1) = 0
P
P
n
n
1
P
or P ( 1) = ( n n 1) ( 1)
n 2 n
= ( 1) n 1 1 (n 1) [ P ( 1) ( 1) ].
n
. n
2 n
Example 6: Prove that P n (0) 0, for n odd and
( 1) n /2 ! n
,
P n (0) = 2 { /2!} 2 for n even.
n
n
Solution:
(i) We know that

1.3.5...(2n 1) n ( n n 1) n 2 ( n n 1)(n 2)(n 3) n 4
x
P n ( ) = x x x ...
! n 2(2n 1) 2.4(2n 1)(2n 3)
when n is odd, say n (2m 1), then

1.3.5...{2(2m 1) 1} 2m 1 (2m 1)(2m 1 1) 2m 1 2
x
P 2m 1 ( ) = x x ...
(2m 1)! 2.{2(2m 1) 1}
Putting x 0, we get P 2m 1 (0) 0,

i.e., P n (0) = 0 when n is odd.
Also, we have

n
h P n ( ) = (1 2xh h 2 ) 1/2
x
n 0
n
or h P n (0) = (1 h 2 ) 1/2 1 ( h ) 2 1/2
n 0
1 2 1.3 2 2 1.3.5 2 3 1.3.5...(2r 1) 2 r
= 1 .( h ) ( h ) ( h ) ... ( h ) ...
2 2.4 2.4.6 2.4...2r
Hence all powers of h on the R.H.S. are even.
2m
Equating the coefficient of h on both sides, we have
1.3.5...(2m 1) m
P 2m (0) = ( 1)
2.4.6...2m
m
(2 )!
m
= ( 1) 2m 2
2 ( !)
m
i.e. when n = 2 m, then
n
( 1) n !
2
P n (0) = 2
n
2 ( /2)!
n
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