Mathematical Physics Vol 1

5.10 Examples

271

When k = m the previous integral becomes

ℓ Z

ℓ Z

1 − cos 2 k π x ℓ 2

k π x ℓ

k π x

sin

d x =

d x =

sin

ℓ ·

− ℓ

− ℓ =

ℓ

ℓ

2 k π x

x 2 −

ℓ 4 k π

sin

= ℓ.

− ℓ

We can see that this set is not normalized, as ℓ̸ = 1.

Problem 187 The function set

1 , cos x , sin x , cos2 x , sin2 x , ..., cos nx , sin nx ,...,

(5.242)

is orthogonal, for x ∈ [ − ℓ, ℓ ] .

Solution As in the previous case, it can be shown that the condition for orthogonality (5.220), p. 263, stands.

Problem 188 Legendre polynomials

d n d x n h

n i

1 2 n n !

x 2 − 1

P n ( x )=

,

are orthogonal on the interval x ∈ [ − 1 , + 1 ] .

Solution To prove this, we will start from the differential equation satisfied by these polyno mials 1 − x 2 y ′′ − 2 xy ′ + n ( n + 1 ) y = 0 . This equation can be rewritten in a form more suitable for future work h 1 − x 2 y ′ i ′ + n ( n + 1 ) y = 0 . If the polynomials P n ( x ) and P m ( x ) are solutions of this differential equation, they satisfy the following equations h 1 − x 2 P ′ n i ′ + n ( n + 1 ) P n = 0 , h 1 − x 2 P ′ m i ′ + m ( m + 1 ) P m = 0 .