Mathematical Physics Vol 1

5.10 Examples

283

Comparing the real coefficients next to t n from the previous and the initial equation, for x = cos θ we obtain P n ( cos θ )= Re " n ∑ k = 0 − 1 / 2 n − 1 / 2 n − k e ( 2 k − n ) i θ # = = n ∑ k = 0 − 1 / 2 n − 1 / 2 n − k cos ( 2 k − n ) θ . (5.259) This equation can be expressed in the following form

n ∑ k = 0

( 2 n − 2 k − 1 ) !! ( 2 n − 2 k ) !!

( 2 k − 1 ) !! ( 2 k ) !!

P n ( cos θ )=

cos ( n − 2 k ) θ .

Special forms of (5.259) are used in practice for even values of n P n ( cos θ )= − 1 / 2 n 2 + 2 n 2 − 1 ∑ k = 0 − 1 / 2 n − 1 / 2 n − k

cos ( 2 k − n ) θ ,

that is, for odd values of n

n − 1 2 ∑ k = 0

n

− 1 / 2 n − k

− 1 / 2

P n ( cos θ )= − 2

cos ( 2 k − n ) θ .

Laguerre polynomials

Problem 198 Prove that Laguerre polynomials are solutions of Laguerre differential equation xy ′′ +( 1 − x ) y ′ + ny = 0 .

Solution Laguerre polynomials are generated by the function

∞ ∑ n = 0

t n n !

1 1 − t

xt 1 − t =

e −

G ( x , t )=

L n ( x )

(5.260)

.

If both sides of the equation are differentiated by t , and then multiplied by 1 − t 2 , we obtain e − xt 1 − t − x 1 − t e − xt 1 − t =( 1 − t 2 ) ∞ ∑ n = 1 L n ( x ) t n − 1 ( n − 1 ) ! . By applying (5.260) we further obtain

∞ ∑ n = 0

∞ ∑ n = 0

∞ ∑ n = 1

t n n ! −

t n n !

t n − 1 ( n − 1 ) !

=( 1 − t 2 )

( 1 − t )

L n ( x )

x

L n ( x )

L n ( x )

,