Mathematical Physics Vol 1

5.10 Examples

285

Hermite polynomials

Problem 199 Prove the recurrence formula

2 xH n ( x ) − 2 nH n x )= H n + 1 ( x ) , bearing in mind that the function generating Hermite polynomials H n ( x ) has the following form G ( x , t )= e 2 xt − t 2 = ∞ ∑ n = 0 H n ( x ) t n n ! . − 1 (

Solution Let us start from the equation

∞ ∑ n = 0

t n n !

G ( x , t )= e 2 xt − t 2

H n ( x )

=

.

Differentiating by t yields

∞ ∑ n = 1

t n − 1 n !

( 2 x − 2 t ) e 2 xt − t 2

H n ( x ) n

=

,

and from there it follows that

∞ ∑ n = 0

∞ ∑ n = 1

t n n !

t n − 1 ( n − 1 ) !

( 2 x − 2 t )

H n ( x )

H n ( x )

=

∞ ∑ n = 0 ∞ ∑ n = 0 ∞ ∑ n = 0

∞ ∑ n = 0 ∞ ∑ n = 0 ∞ ∑ n = 1

∞ ∑ n = 0

t n n ! − t n n ! − t n n ! −

t n + 1 n !

t n n !

2 x

H n ( x )

2

H n ( x )

H n + 1 ( x )

=

∞ ∑ n = 0

t n + 1 ( n + 1 ) !

t n n !

2 x

H n ( x )

2

H n ( x )( n + 1 )

H n + 1 ( x )

=

∞ ∑ n = 0

t n n !

t n n !

2 x

H n ( x )

2

H n

x ) n

H n + 1 ( x )

− 1 (

=

.

Equating coefficients next to t n yields

2 xH n ( x ) − 2 nH n

x )= H n + 1 ( x ) .

− 1 (

Bessel polynomials

Problem 200

Prove

1 2

z ) − J n + 1 ( z ))= J ′ n ( z ) ,

( J n

− 1 (