Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

292

Solution The k -th derivative of the ML function can be calculated as follows d d x ( k ) E k ( x k )= d k d x k " ∞ ∑ l = 0 x kl Γ ( kl + 1 ) # = d k d x k " 1 + ∞ ∑ l = 1 x kl Γ ( kl + 1 ) # =

x kl Γ ( kl + 1 ) !

d k d x k

x kl Γ ( kl + 1 )

∞ ∑ l = 1

∞ ∑ l = 1

d k d x k

=

=

=

d k d x k

x kl ,

∞ ∑ l = 1

1 Γ ( kl + 1 )

=

given that d k

d x k

x kl =( kl )( kl − 1 ) ··· ( kl − k + 1 ) · x kl − k =

( kl − k ) !

z { ( kl − k )( kl − k − 1 ) · 2 · 1 · x kl − k }|

( kl )( kl − 1 ) ··· ( kl − k + 1 ) ·

=

=

( kl − k ) !

Γ ( kl + 1 ) Γ ( kl − k + 1 )

( kl ) ! ( kl − k ) !

x kl − k =

x kl − k .

=

The following relations were also used

( kl ) ! = Γ ( kl + 1 ) , ( kl − k ) ! = Γ ( kl − k + 1 ) .

Thus, the k -th derivative can be expressed as d d x ( k ) E k ( x k )= ∞ ∑ l = 1 1 Γ ( kl + 1 )

Γ ( kl + 1 ) Γ ( kl − k + 1 )

∞ ∑ l = 1

x kl − k Γ ( kl − k + 1 ) .

x kl − k =

Introducing the substitution l = n + 1, we finally obtain d d x ( k ) E k ( x k )= ∞ ∑ n = 0 x kn Γ ( kn + 1 )

k ) .

= E k ( x

Problem 208 Let x ∈ R . Determine the value xE 2 , 2 ( − x 2 ) .

Solution

Given that

∞ ∑ k = 0

z k Γ ( α k + β ) ,

E α , β ( z )=

it follows that

∞ ∑ k = 0

( − x 2 ) k Γ ( 2 k + 2 ) ,

2 )=

E 2 , 2 ( − x