Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

280

and then differentiate both sides by x , we obtain

∞ ∑ n = 0

P ′

t ( 1 − 2 xt + t 2 ) − 3 / 2 =

n .

n ( x ) t

Differentiating the initial equation by t yields

∞ ∑ n = 0

( x − t )( 1 − 2 xt + t 2 ) − 3 / 2 =

n − 1 .

P n ( x ) nt

Eliminating ( 1 − 2 xt + t 2 ) − 3 / 2 from these two equations, we obtain ( x − t ) ∞ ∑ n = 0 P ′ n ( x ) t n = t ∞ ∑ n = 0 P n ( x ) nt n − 1 , that is x ∞ ∑ n = 0 P ′ n ( x ) t n − ∞ ∑ n = 0 P ′ n ( x ) t n + 1 = ∞ ∑ n = 0 P n ( x ) nt n , or x ∞ ∑ n = 0 P ′ n ( x ) t n − ∞ ∑ n = 1 P ′ n − 1 ( x ) t n = ∞ ∑ n = 0 P n ( x ) nt n . From here, equating coefficients next to t n yields xP ′ n ( x ) − P ′ n − 1 ( x )= nP n ( x ) . By differentiating the Bonnet formula (equation 5.252) we obtain ( n + 1 ) P ′ n + 1 ( x ) − ( 2 n + 1 ) xP ′ n ( x )+ nP ′ n − 1 ( x )=( 2 n + 1 ) P n ( x ) . After eliminating P ′ n − 1 ( x ) from the two previous equations, we obtain P ′ n + 1 ( x ) − xP ′ n ( x )=( n + 1 ) P n ( x ) , that is, substituting n + 1with n , P ′ n ( x ) − xP ′ n − 1 ( x )= nP n − 1 ( x ) . Eliminating again P ′ n − 1 ( x ) from the last equation and equation (5.257) yields ( x 2 − 1 ) P ′ n ( x ) − nxP n ( x )+ nP n − 1 ( x )= 0 . By differentiating this equation we obtain ( x 2 − 1 ) P ′′ n ( x )+( 2 − n ) xP ′ n ( x ) − nP n ( x )+ nP ′ n − 1 ( x )= 0 . Eliminating once again P ′ n − 1

(5.257)

( x ) form previous equation and equation (5.257), we finally

obtain

( x 2 − 1 ) P ′′

n ( x )+ 2 xP ′

n ( x ) − n ( n + 1 ) P n ( x )= 0 ,

which was to be proved.