Mathematical Physics Vol 1

5.9 Orthogonal and normalized functions

263

we obtain

q n + 1 2 1 − q 2 n + 1 q n + 1 2 1 + q 2 n + 1

∞ ∑ n = 0 ∞ ∑ n = 0

2 π kK 2 π kK

sn ( u , k )=

sin [( 2 n + 1 ) v ] ,

(5.217)

cn ( u , k )=

cos [( 2 n + 1 ) v ] ,

∞ ∑ n = 0

q n 1 + q 2 n

2 π K

π K

dn ( u , k )=

cos ( 2 nv ) .

+

Determining the value of the elliptic integral K The function 1 √ 1 − k 2 sin 2 ϕ

can be expanded into a series (following the binomial formula),

which yields

1 q 1 − k 2 sin 2 ϕ

1 2

1 · 3 2 · 4

1 · 3 · 5 2 · 4 · 6

k 2 sin 2 ϕ +

k 4 sin 4 ϕ +

k 6 sin 6 ϕ + ···

= 1 +

This series is uniformly convergent for k 2 < 1. According to the Wallis 15 formula π / 2 Z 0 sin 2 n ϕ d ϕ = 1 · 3 · 5 ··· ( 2 n − 1 ) 2 · 4 · 6 ··· 2 n · π 2 and we obtain

π 2 "

k 6 + ··· # .

1 +

1 2

k 2 +

1 · 3 2 · 4

k 4 +

1 · 3 · 5 2 · 4 · 6

2

2

2

K =

(5.218)

5.9 Orthogonal and normalized functions

Observe the set of integrable functions for x ∈ [ a , b ] , ( a < b ) f 1 ( x ) , f 2 ( x ) ,..., f n ( x ) ,...

(5.219)

Definition The function set (5.219) is said to be orthogonal in the interval [ a , b ] if

b Z a f m ( x ) f n ( x ) d x = 0 , for ∀ m̸ = n , n , m = 1 , 2 ,...

( f m , f n )

(5.220)

where functions f n ( x ) , n = 1 , 2 ,... , are not identically equal to zero in the observed interval.

As f n ( x ) are integrable functions, and a and b are constants, then the following integral obviously also exists b Z a f 2 n ( x ) d x = I n > 0 , where I n is constant. 15 Wallis