Mathematical Physics Vol 1

5.10 Examples

291

Solution The derivative of the ML function is

d d x "

x α k Γ ( α k + 1 ) #

d d x

x α k Γ ( α k + 1 )

∞ ∑ k = 0

∞ ∑ k = 1

∞ ∑ k = 1

x α k Γ ( α k + 1 )

d d x

d d x

α

E α ( x

1 +

)=

=

=

=

∞ ∑ k = 0

( α k ) x α k − 1 Γ ( α k + 1 )

=

.

As Γ ( z + 1 )= z Γ ( z ) , i.e. in our case Γ ( α k + 1 )=( α k ) Γ ( α k ) , it follows that the derivative of the ML function is ∞ ∑ k = 0 ( α k ) x α k − 1 Γ ( α k + 1 ) = ∞ ∑ k = 0 ( α k ) x α k − 1 ( α k ) Γ ( α k ) ,

that is

∞ ∑ k = 1

∞ ∑ k = 1

x α k − 1 Γ ( α k )

x α k Γ ( α k )

d d x

1 x

α

E α ( x

)=

=

.

By substituting k = n + 1, we obtain

∞ ∑ n = 0

∞ ∑ n = 0

x α ( n + 1 ) Γ ( α ( n + 1 ))

x α · x α n Γ ( α n + α )

1 x

1 x

=

=

∞ ∑ n = 0

x α n Γ ( α n + α )

α − 1

α − 1 E

α

= x

= x

α , α ( x

) .

Thus, this derivative is given by the expression d d x E α ( x α )= x α − 1 E In the special case, when α = 1, we obtain d E 1 ( x ) d x

α

α , α ( x

) .

x ,

= E 1 , 1 ( x )= e

x , and the derivative of this function is the

which was to be expected, because E 1 ( x )= e

function itself.

R Note that we have used here a property of convergent series (see Properties of uniformly convergent series, p. 225, eq. (5.7)).

Problem 207 Let k ∈ N and E k ( · ) be the ML function. Prove that d d x k E k ( x k )= E k ( x k ) .