Mathematical Physics Vol 1

5.10 Examples

273

where

u n ( x )= x

2 − 1

n

,

and also

u n ( ± 1 )= u ′ ( ± 1 )= 0 . The previous integral, by way of partial integration, becomes + 1 Z − 1 d n u n ( x ) d x n d n u n ( x ) d x n d x = n ( ± 1 )= ··· = u ( n − 1 ) n

+ 1

n ( x ) d x n − 1

n ( x ) d x n + 1

d n − 1 u

+ 1 Z

d n + 1 u

( n ) n ( x ) · u

( n − 1 ) n

u

( x )

+( − 1 )

d x ⇒

=

− 1

− 1

d n u

n ( x ) d x n

n ( x ) d x n

+ 1 Z

+ 1 Z

d n u

d 2 n u

n ( x )

d x =( − 1 ) n

u n ( x )

.

d x 2 n

− 1

− 1

Given that

= x 2 n +

n 1

( − 1 )+

n 2

u n = x

2 − 1

x 2

x 2

n

n − 1

n − 2

( − 1 ) 2 + ···

for the derivative within the integral we obtain

d 2 n d x 2 n

d 2 n d x 2 n

x 2 n = 2 n ( 2 n − 1 ) ··· 2 · 1 =( 2 n ) ! .

( u n )=

The square of the norm can now be represented as

+ 1 Z

1 ( 2 n n ! )

2

2 ( 2 n ) ! ( − 1 ) n

∥ P n ( x ) ∥

u n d x .

=

− 1

However, given that

n =( − 1 ) n 1 − x 2 n ,

u n = x

2 − 1

the previous expression can be rewritten in the form

+ 1 Z − 1

1 ( 2 n n ! )

1 − x 2

n

2

∥ P n ( x ) ∥

2 ( 2 n ) !

d x .

=