Mathematical Physics Vol 1

5.10 Examples

279

Reducing all to sums of t n + 1 we obtain ∞ ∑ n = 0 P n ( x ) t n + 1 = ∞ ∑ n = − 1 P ′ n + 1 ( x ) t

∞ ∑ n = 0

∞ ∑ n = 1

n + 1 − 2 x

n + 1 +

( x ) t n + 1

P ′ n ( x ) t

P ′ n

− 1

Given that these sums start from different values of n , we shall separate the terms for n = − 1 and n = 0, which yields P 0 t + ∞ ∑ n = 1 P n ( x ) t n + 1 = P ′ 0 + P ′ 1 t + ∞ ∑ n = 1 P ′ n + 1 ( x ) t n + 1 − 2 xP ′ 0 t − 2 x ∞ ∑ n = 1 P ′ n ( x ) t n + 1

∞ ∑ n = 1

( x ) t n + 1 .

P ′ n

+

− 1

Substituting the Legendre polynomial P 0 = 1 and the derivatives of Legendre polyno mials P ′ 0 = 0 i P ′ 1 = 1, we obtain

∞ ∑ n = 1

∞ ∑ n = 1

∞ ∑ n = 1

n + 1 = 0 + t +

n + 1 − 0 − 2 x

n + 1

P ′ n + 1 ( x ) t

P ′ n ( x ) t

t +

P n ( x ) t

∞ ∑ n = 1

( x ) t n + 1 .

P ′ n

+

− 1

From there, after grouping coefficients next to t n + 1 , it follows P n ( x )= P ′ n + 1 ( x ) − 2 xP ′ n ( x )+ P ′ n − 1 ( x ) .

(5.255)

Further, by differentiating the Bonnet formula (equations 5.252) by x , we obtain

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

( x )= 0 .

(5.256)

Eliminating P ′ n from equations (5.255) and (5.256) we obtain the required recurrent formula. P ′ n + 1 ( x ) − P ′ n − 1 ( x )=( 2 n + 1 ) P n ( x ) .

Problem 195 Prove that Legendre polynomials satisfy the Legendre differential equation ( x 2 − 1 ) y ′′ + 2 xy ′ − n ( n + 1 ) y = 0 .

Solution If we start from

∞ ∑ n = 0

1

n ,

G ( x , t )=

P n ( x ) t

√ 1

=

− 2 xt + t 2