Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

282

that is

I mn 

P m ( x ) P n ( x ) d x 

1 Z

 =

 = 0

for m̸ = n ,

− 1

and the polynomials are thus orthogonal.

Problem 197 Prove that the following equation holds for Legendre polynomials

n ∑ k = 0

n

− 1 / 2 n − k

− 1 / 2

P n ( cos θ )=

cos ( 2 k − n ) θ .

Solution: Let us start from the generating function of Legendre polynomials G ( x , t )=( 1 − 2 xt + t 2 ) − 1 2 = ∞ ∑ n = 0 P n ( x ) t n . Given that 2cos θ = cos θ + i sin θ + cos θ − i sin θ = e i θ + e − i θ , a e i θ · e − i θ = 1, it follows that 1 − 2 t cos θ + t 2 = 1 − t ( e i θ + e − i θ )+ t 2 e i θ · e − i θ =( 1 − te i θ )( 1 − te − i θ ) . If we now introduce the substitution x = cos θ into the expression ( 1 − 2 xt + t 2 ) − 1 2 we obtain ( 1 − 2 t cos θ + t 2 ) − 1 2 =( 1 − te i θ ) − 1 / 2 ( 1 − te − i θ ) − 1 / 2 . For | t | < 1 the following expansions hold ( 1 − te i θ ) − 1 / 2 = ∞ ∑ n = 0 ( − 1 ) n − 1 / 2 n e in θ t n = ∞ ∑ n = 0 a n t n , ( 1 − te − i θ ) − 1 / 2 = ∞ ∑ n = 0 ( − 1 ) n − 1 / 2 n e − in θ t n = ∞ ∑ n = 0 b n t n , and from there it follows that ( 1 − 2 xt + t 2 ) − 1 2 = ∞ ∑ n = 0 a n t n ! ∞ ∑ n = 0 b n t n ! . Coefficients next to t n in this expansion can be represented by the sum n ∑ k = 0 a k b n − k = n ∑ k = 0 ( − 1 ) 2 n − 1 / 2 n − 1 / 2 n − k e ( 2 k − n ) i θ . Thus we obtain ( 1 − 2 t cos θ + t 2 ) − 1 2 = ∞ ∑ n = 0 t n n ∑ k = 0 − 1 / 2 n − 1 / 2 n − k e ( 2 k − n ) i θ .