Mathematical Physics Vol 1

5.10 Examples

293

and thus

xE 2 , 2 ( − x ( − x 2 ) k Γ ( 2 k + 2 ) . Given that Γ ( 2 k + 2 )=( 2 k + 1 ) ! the previous equation can be expressed as 2 )= x ∞ ∑ k = 0

∞ ∑ k = 0

( − 1 ) k x 2 k + 1 ( 2 k + 1 ) !

2 )=

xE 2 , 2 ( − x

The series on the right hand side is the Maclaurin series for the sinus function

∞ ∑ ℓ = 0

( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! ,

sin x =

and thus

2 )= sin x .

xE 2 , 2 ( − x

Problem 209 Let α > 0 i x ∈ R . Prove the validity of the so called duplication formula 1 2 E α ( √ x )+ E α ( − √ x )= E 2 α ( x ) = E 2 α ( x ) .

Solution According to the definition

( ± √ x ) k Γ ( α k + 1 ) ,

∞ ∑ k = 0

√ x )=

E α ( ±

and the required sum is thus

( √ x ) k +(

√ x ) k

( √ x ) k ( 1 +(

∞ ∑ k = 0

∞ ∑ k = 0

− 1 ) k (

− 1 ) k )

E α ( √ x )+ E α (

√ x )=

−

=

.

Γ ( 2 α k + 1 )

Γ ( 2 α k + 1 )

Given that

( − 1 ) k = (

1 , if k = 2 n , 0 , if k = 2 n + 1 ,

it follows that

∞ ∑ n = 0

x n Γ ( 2 α n + 1 )

E α ( √ x )+ E α (

√ x )= E 2 α ( x )= 2

= E 2 α ( x ) .

−