Page 46 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Also it can be written as Notes
n ( 1) (2n 2 )!
r
r
P ( ) = x n 2r ...(E)
x
n n
r
r 0 2 (n r )!(n 2 )!
Example 1: From the relation
1
n
(1 2xh h 2 ) 2 = h P n ( )
x
n 0
x
x
Obtain P 0 ( ), P 1 ( ), P 2 ( ), P 3 ( ) and P 4 ( ).
x
x
x
i.e.
Prove
x
P 0 ( ) = 1, P 1 ( ) 1
x
1 2 1 3
P 2 ( ) = (3x 1), P 3 ( ) (5x 3 )
x
x
x
2 2
1 4 2
P ( ) = (35x 30x 3).
x
4
8
Example 2: Express ( )P x x 4 2x 3 2x 2 x 3 in terms of Legendre’s polynomials.
Solution: From Example 1, we have
(3x 2 1)
x
P 0 ( ) = 1, P 1 ( ) x P 2 ( ) ,
x
x
,
2
(5x 3 3 ) 35x 4 30x 2 3
x
x
P ( ) = ,P ( )
x
3 4
2 8
1 4 2
from P 4 ( ) = (35x 30x 3),
x
8
8 6 2 3
x
x 4 = P 4 ( ) x ,
35 7 35
1 3 3 2 3
x
x
from P 3 ( ) = (5x 3 ), x P 3 ( ) , x
x
2 5 5
1 2 2 2 1
x
x
from P 2 ( ) = (3x 1), x P 2 ( )
2 5 3
x
and x = P 1 ( ); 1 P 0 ( )
x
Substituting these values, we have
8 6 2 3 3 2
x
P ( ) = P 4 ( ) x 2x 2x x 3
x
35 7 35
8 3 20 2 108
x
= P 4 ( ) 2x x x
35 7 35
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