Mathematical Physics Vol 1

5.10 Examples

277

Solution If we start from

∞ ∑ n = 0

1

n ,

G ( x , t )=

P n ( x ) t

(5.253)

√ 1

=

− 2 xt + t 2

and the differentiate both the left an right hand side by t , we obtain the equality

∞ ∑ n = 1

1 2

− 2 x + 2 t p ( 1 − 2 xt + t 2 ) 3

n − 1 ,

P n ( x ) nt

−

=

and from there

∞ ∑ n = 1

x − t 1 − 2 xt + t 2

1

n − 1 .

P n ( x ) nt

√ 1

=

− 2 xt + t 2

Based on (5.253), and after multiplying by 1 − 2 xt + t 2 , we obtain ( x − t ) ∞ ∑ n = 0 P n ( x ) t n =( 1 − 2 xt + t 2 ) ∞ ∑ n = 1 P n ( x ) nt n − 1 , that is

∞ ∑ n = 0

∞ ∑ n = 0 ∞ ∑ n = 1

n −

n + 1 =

x

P n ( x ) t

P n ( x ) t

∞ ∑ n = 1

∞ ∑ n = 1

n − 1 − 2 x

n +

n + 1 .

P n ( x ) nt

P n ( x ) nt

P n ( x ) nt

=

Reducing all powers in the sums to t n we obtain

∞ ∑ n = 0

∞ ∑ n = 1

n −

x ) t n =

x

P n ( x ) t

P n

− 1 (

∞ ∑ n = 0

∞ ∑ n = 1

∞ ∑ n = 2

n − 2 x

n +

x )( n − 1 ) t n .

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

P n ( x ) nt

P n

=

− 1 (

Separating the terms of the sum for n = 0 and n = 1we obtain

∞ ∑ n = 2

∞ ∑ n = 2

n − P

x ) t n =

xP 0 + xP 1 t + x

P n ( x ) t

0 t −

P n

− 1 (

∞ ∑ n = 2

∞ ∑ n = 2

n − 2 xP

n +

= P 1 + 2 P 2 t +

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

1 t − 2 x

nP n ( x ) t

∞ ∑ n = 2

x ) t n ,

( n − 1 ) P n

+

− 1 (

that is

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

∞ ∑ n = 2

∞ ∑ n = 2

∞ ∑ n = 2

n − x

n +

x ) t n .

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

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

nP n

=

− 1 (

As the three first Legendre polynomials are [ 35 ] P 0 ( x )= 1 , P 1 ( x )= x , P 2 ( x )=

1 2

( 3 x 2 − 1 ) ,