Page 54 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

2.4 Recurrence Relations for Legendre Polynomials Notes

I. Prove that

(2n 1)xP n ( ) = (n 1)P n 1 ( ) nP n 1 ( )
x
x
x
We now have from generating function
n
(1 2hx h 2 ) 1/2 = h P n ( )
x
n 0
Differentiating both sides w.r.t. h we have
1 2 3/2 n 1
x
( 2x 2 )(1 2hx h ) = nh P n ( )
h
2
n 0
Multiplying both sides by (1 2hx h 2 ); we get

(x h )(1 2hx h 2 ) 1/2 = (1 2hx h 2 ) nh n 1 P n ( )
x
n 0
or

n
x
x
(x h ) h P ( ) = (1 2hx h 2 ) nh n 1 P ( )
n n
n 0 n 0
Expanding
2
2
x
)
x
x
(x h P 0 ( ) h P 1 ( ) h P 2 ( ) ... (1 2hx h 2 ) P 1 ( ) 2h P 2 ( ) 3h P 3 ( ) ...
x
x
x
n
Comparing the coefficients of h on both sides, we have
x
x
x
x
x
xP n ( ) P n 1 ( ) = (n 1)P n 1 ( ) 2xnP n ( ) (n 1)P n 1 ( )
Rearranging terms, we have
x
x
xP ( ) 2xnP ( ) = (n 1)P ( ) (n 1 1)P ( )
x
x
n n n 1 n 1
or
x
x
x
(2n 1) P n ( ) = (n 1)P n 1 ( ) nP n 1 ( )
II. Prove that
x
nP n ( ) = xP n ( ) P n 1 ( )
x
x
Proof:
Consider the relation
x
x
(1 2hx h 2 ) 1/2 = h P n ( ) ...(A)
n 0
Differentiating w.r.t. h, we have
1 2 3/2 n 1
h
x
( 2x 2 )(1 2xh h ) = nh P n ( )
2
n 0
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