Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

278

it follows that

2 t − t − x − ( 3 x 2 − 1 ) t = 0 ,

xP 0 + 3 xP 1 t − P 0 t − P 1 − 2 P 2 t = x + 3 x

and from there ∞ ∑ n = 2

∞ ∑ n = 2

∞ ∑ n = 2

n − x

n +

x ) t n = 0 .

( n + 1 ) P n + 1 ( x ) t

( 2 n + 1 ) P n ( x ) t

nP n

− 1 (

Grouping coefficients next to t n we obtain ∞ ∑ n = 2

x )] t n = 0 .

[( n + 1 ) P n + 1 ( x ) − ( 2 n + 1 ) xP n ( x )+ nP n − 1 (

from where it follows

( n + 1 ) P n + 1 ( x ) − ( 2 n + 1 ) xP n ( x )+ nP n − 1 (

x )= 0 .

Problem 194 Prove the Christoffel recurrence formula P ′ n + 1 ( x ) − P ′ n − 1

( x )=( 2 n + 1 ) P n ( x ) ,

(5.254)

if the Legendre polynomial P n ( x ) is defined by the expansion

∞ ∑ n = 0

1

n .

G ( x , t )=

P n ( x ) t

√ 1

=

− 2 xt + t 2

Solution If we start from

∞ ∑ n = 0

1

n ,

G ( x , t )=

P n ( x ) t

√ 1

=

− 2 xt + t 2

and then differentiate both the left and the right hand side by x , we obtain

∞ ∑ n = 0

t ( 1 − 2 xt + t 2 ) − 3 / 2 =

n .

P ′ n ( x ) t

If we substitute the expression ( 1 − 2 xt + t 2 ) − 1 / 2 in this relation by the sum from (5.253), and then multiply both sides by ( 1 − 2 xt + t 2 ) , we obtain t ∞ ∑ n = 0 P n ( x ) t n =( 1 − 2 xt + t 2 ) ∞ ∑ n = 0 P ′ n ( x ) t n , ∞ ∑ n = 0 P n ( x ) t n + 1 = ∞ ∑ n = 0 P ′ n ( x ) t n − 2 x ∞ ∑ n = 0 P ′ n ( x ) t n + 1 + ∞ ∑ n = 0 P ′ n ( x ) t n + 2