Confluent Hypergeometric Function

Confluent Hypergeometric Function:

      Let us put x = z/β or z= βx in the hypergeometric equation . Then the equation assumes the form 

 z(1- z/β)d²y/dz²  +

                [γ-(1+α+β)z/β]dy/dz -αy = 0

There are three regular singularities , one at z= 0 , another given by z= β , and also at z=∞.

    When β-->∞ , we get the equation 

z = d²y/dz² + (γ-z)dy/dz -αy = 0   .......(1)

   The equation is known as the Confluent Hypergeometric Equation following to confluence of two singularities β,∞ when      β-->∞  .

  Again putting x= z/β in the hypergeometric series we obtain 

1+ α.βz/1.γβ + α(α+1).β(β+1)z²/1.2.γ(γ+1)β²+..

  = 1+ αz/1.γ + α(α+1)(1+ 1/β)z²/1.2.γ(γ+1)+...

When β-->∞ this series reduces to 

 1+αz/1.γ  + α(α+1)z²/2!γ(γ+1) + ....

which can be put as 

      ∞

      Σ (α)ₖ zᵏ/k! (γ)ₖ   or simply F(α;γ;z) .   ...(2)

   k=0

This function is called Confluent Hypergeometric Function .

Note :

    If we put γ= α and z= x in equation(2) , we have 

                   ∞                           ∞

 F(α;α;x) = Σ (α)ₖxᵏ/k!(α)ₖ = Σ xᵏ/k! = eˣ

                  k=0                       k=0

 From the definition of confluent hypergeometric function it can be shown that

 d/dx  [F(α;γ;x)] = α/γ F(α+1;γ+1;x) ......(3)
                                  ∞
Since F(α;γ;x) = 1+ Σ (α)ₖ xᵏ /k!(γ)ₖ .
                                 k=1

differentiating with respect to x , we get
                           ∞
d/dx F(α;γ;x) = Σ  (α)ₖ xᵏ⁻¹/(k-1)!(γ)ₖ
                         k=1

Writing k-1 = n , we obtain
                            ∞
d/dx  F(α;γ;x) = Σ  (α)ₙ₊₁ xⁿ / n!(γ)ₙ₊₁
                           k=1
                    ∞
                 = Σ α(α+1)...(α+n)xⁿ/n!γ(γ+1)..(γ+n)
                  n=0
                         ∞
              = α/γ  Σ (α+1)ₙxⁿ/n!(γ+1)ₙ
                       n=0

            = α/γ F(α+1;γ+1;x).

Example:

           Prove that 

   αF(α+1;γ+1;x) - γF(α;γ;x) 

                                       = (α-γ)F(α;γ+1;x).

Solution :

             

      

        L.H.S =

            ∞

            Σ [α(α+1)ₖxᵏ/k!(γ+1)ₖ- γ(α)ₖxᵏ/k!(γ)ₖ]

          k=0

    ∞

=  Σ (α)ₖxᵏ[(α+k)-γ(γ+1)ₖ/(γ)ₖ] / k!(γ+1)ₖ

  k=0

Now    γ(γ+1)ₖ/(γ)ₖ = 

                 γ(γ+1)....(γ+k)/ γ(γ+1)..(γ+k-1) =γ+k.

                                 ∞

Hence the L.H.S = Σ (α)ₖxᵏ(α+k-γ-k)/k!(γ+1)ₖ

                               k=0

                     ∞

         = (α-γ)Σ   (α)ₖxᵏ/k!(γ+1)ₖ 

                  k=0

         = (α-γ)F(α;γ+1;x) .(Proved)