Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

276

is satisfied, we can arbitrarily group the terms in relation (5.247). Coefficient next to t n in relation (5.247) is determined by equating it with the coefficient next to t n in the expansion

n ∑ k = 0

( 2 k ) ! 2 2 k k ! k !

( 2 xt − t 2 ) k ,

that is

n ∑ k = 0

( 2 n − 2 k ) ! 2 2 n − 2 k ( n − k ) ! ( n − k ) !

( 2 xt − t 2 ) n − k ,

(5.249)

because

n ∑ k = 0

n ∑ k = 0

f ( k )= f ( n − k ) . From (5.249) the coefficient next to t n is obtained in the form

( 2 n − 2 k ) ! 2 2 n − 2 k ( n − k ) ! ( n − k ) ! ( 2 n − 2 k ) ! 2 n ( n − k ) ! ( n − k ) !

k k

[ n / 2 ] ∑ k = 0

n − k

( − 1 ) k

( 2 x ) n − 2 k ,

that is

[ n / 2 ] ∑ k = 0

n − k

x n − 2 k ,

( − 1 ) k

where k takes the values from k = 0 to k =[ n / 2 ] , because the binomial coefficient n − k k for k > [ n / 2 ] is equal to zero. From this it follows that P n ( x )= [ n / 2 ] ∑ k = 0 ( − 1 ) k ( 2 n − 2 k ) ! 2 n ( n − k ) ! ( n − k ) ! n − k k x n − 2 k . (5.250) This proof has been developed under the assumption that − 1 ≤ x ≤ 1. The function G ( x , t )=( 1 − 2 xt + t 2 ) − 1 2 (5.251)

is called the generating function of Legendre polynomial.

Problem 193 Prove the Bonnet recursive formula

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

x )= 0 ,

(5.252)

if the Legendre polynomial P n ( x ) is expressed in terms of the function G ( x , t )

∞ ∑ n = 0

1

n .

G ( x , t )=

P n ( x ) t

√ 1

=

− 2 xt − t 2