Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

290

In our case, α = 1, β = 2, and thus

∞ ∑ k = 0

∞ ∑ k = 1

x k Γ ( k + 2 )

x 0 Γ ( 2 )

x k Γ ( k + 2 )

E 1 , 2 ( x )=

=

+

.

Given that x 0 = 1 and Γ ( 2 )= 1! = 1, it follows that

∞ ∑ k = 1

x k Γ ( k + 2 )

E 1 , 2 ( x )= 1 +

.

If we introduce the substitution k = n + 1we obtain

∞ ∑ n + 1 = 1

∞ ∑ n + 1 = 1

x n + 1 Γ ( n + 3 )

x · x n Γ ( n + 3 )

E 1 , 2 ( x )= 1 +

= 1 +

=

∞ ∑ n + 1 = 1 x n Γ ( n + 3 ) | {z } E 1 , 3

= 1 + x

which was to be proved.

Problem 205 Find E 0 ( x ) , if | x | < 1.

Solution The ML function, for α = 0, is

∞ ∑ k = 0

∞ ∑ k = 0

x k Γ ( 0 + 1 )

E 0 ( x )= x k = 1 + x + x 2 + x 3 + ··· . Given that Γ ( 0 + 1 )= 1, this series, for | x | < 1, becomes the so called geometric series, the sum of which is ∞ ∑ k = 0 x k = 1 1 − x . Thus E 0 ( x )= ∞ ∑ k = 0 x k = 1 1 − x . =

Problem 206 Determine the first derivative of the function E α ( x

α ) , for α > 0 and x > 0, and then

find its value for α = 1.