Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

284

and from there, equating coefficients next to t n , the recurrent relation L n + 1 ( x )+( x − 2 n − 1 ) L n ( x )+ n 2 L n − 1 ( x )= 0 . If the equation (5.260) is now differentiated by x we obtain

(5.261)

∞ ∑ n = 0

t n n !

t ( 1 − t ) 2

xt 1 − t =

e −

L ′ n ( x )

−

.

Using (5.260) we further obtain

∞ ∑ n = 0

∞ ∑ n = 0

t n n !

t n n !

L ′ n ( x )

t

L n ( x )

+( 1 − t )

= 0 .

Equating coefficients next to t n yields the recurrent formula

x )+ L ′ n ( x ) − nL ′ n

( x )= 0 .

(5.262)

nL n

− 1 (

− 1

If the equation (5.261) is differentiated twice by x , we obtain

L ′′ n + 1 ( x )+( x − 2 n − 1 ) L ′′ n ( x )+ 2 L ′ n ( x )+ n 2 L ′′ n − 1

( x )= 0 .

Further, substituting n by n + 1, yields

L ′′ n + 2 ( x )+( x − 2 n − 3 ) L ′′ n + 1 ( x )+ 2 L ′ n + 1 +( n + 1 ) 2 L ′′

n ( x )= 0 .

(5.263)

From (5.262) it follows that

L ′ n ( x )= nL ′ n

( x ) − nL n

x ) ,

− 1 (

− 1

and then differentiating by x

L ′′ n ( x )= nL ′′ n

( x ) − nL ′ n

( x ) .

− 1

− 1

If we now substitute n by n + 1 and n + 2 respectively, we obtain

L ′ n + 1 ( x )=( n + 1 ) L ′ n ( x ) − ( n + 1 ) L n ( x ) L ′′ n + 1 ( x )=( n + 1 ) L ′′ n ( x ) − ( n + 1 ) L ′ n ( x ) , L ′′ n + 2 ( x )=( n + 2 ) L ′′ n + 1 ( x ) − ( n + 2 ) L ′ n + 1 ( x ) .

(5.264) (5.265) (5.266)

Substituting L ′ n + 1 ( x ) from (5.264) and L ′′ n + 1 ( x ) from (5.265) into (5.266), yields L ′′ n + 2 ( x )=( n + 2 ) { ( n + 1 )[ L ′′ n ( x ) − L ′ n ( x )] − ( n + 1 )[ L ′ n ( x ) − L n ( x )] } = =( n + 1 )( n + 2 ) L ′′ n ( x ) − 2 ( n + 1 )( n + 2 ) L ′ n ( x )+( n + 1 )( n + 2 ) L n ( x ) . Using the obtained results, the equation (5.263) can be reduced to the following form xL ′′ n ( x )+( 1 − x ) L ′ n ( x )+ nL n ( x )= 0 , and we thus conclude that the Laguerre polynomial L n ( x ) is a particular solution of the Laguerre differential equation.