Mathquery

The Laplace Transformation :    In recent years , in the solution of differential equations much use has been made of what are known as "Operational Methods". Such methods have wide applications in science and engineering and represent a large field for advanced study .  This method consists of a procedure of solving differential equations where the boundary  or initial conditions are automatically satisfied in the course of the solution . The Laplace transformation is but one of many possible operational methods of solving linear differential equations . Definition :             Given a function f(t) of a real variable t>0 , if we multiply it by e⁻ᵖᵗ and with respect to t between the limits 0 and ∞, the result is a function of p, say f̅(p). This function f̅(p) is called the Laplace Transformation of f(t) , also written as L{f(t)}. Thus                   ...

Integral Representation Of Confluent Hypergeometric Function :   Theorem :             If γ>α>0 , then the function F(α;γ;x) can be expressed as                                1    Γ(γ)/Γ(γ)Γ(γ-α)  ∫ eˣᵗ t^(α-1)(1-t)^(γ-α-1) dt.                                0 Proof :           We know that    B(α+n,γ-α)/B(α,γ-α) =         Γ(α+n)Γ(γ-α)/Γ(γ+n) / Γ(α)Γ(γ-α)/Γ(γ)      = Γ(α+n)/Γ(α) / Γ(γ+n)/Γ(γ) But Γ(α+n)/Γ(α) = (α)ₙ and Γ(γ+n)/Γ(γ) = (γ)ₙ Therefore, (α)ₙ/(γ)ₙ  = B(α+n,γ-α)/Β(α,γ-α)                                   1   = Γ(γ)/Γ(α)Γ(γ-α) ∫ t^(α...

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  ...

Integral Formula For The  Hypergeometric Series : Theorem :              The hypergeometric function F(α,β;γ;x) can be represented by                                  [Γ(γ)/Γ(β)Γ(γ-β)]            1           ∫(1-t)^(γ-β-1) t^(β-1) (1-xt)^-α dt           0 Proof :                We know that   (β)ₙ = β(β+1)...(β+n-1) = Γ(β+n)/Γ(β) Similarly        (γ)ₙ = Γ(γ+n)/Γ(γ) Again           B(m,n) = Γ(m)Γ(n)/Γ(m+n) Consider,    B(β+n,γ-β) / B(β,γ-β)      = [Γ(β+n)Γ(γ-β)/Γ(γ+n)]/[Γ(β)Γ(γ-β)/Γ(γ)    = [Γ(β+n)/Γ(β)] / [Γ(γ+n)/Γ(γ)] Thus    (β)ₙ/(γ)ₙ = B(β+n, γ-β) / B(β, γ-β) ...

Elementary Properties Of The Hyper geometric Function :          In this section we consider some properties of the hyper geometric function which are immediate consequences of its definition by the series           ∞   y₁= Σ (α)ₙ(β)ₙxⁿ/n!(γ)ₙ = F(α,β,;γ,x)         n=0                                    γ≠0,-1,-2......    (9) (i) We observe that the terms of the series do not change if the parameters α and β are permuted (interchanged). Hence we obtain the symmetry property             F(α,β;γ;x) = F(β,α;γ;x)  .........(1) (ii) d/dx[ F(α,β;γ;x) ]= αβ F(α+1,β+1;γ+1;x)/γ Proof :                From equation(9), we have                      ...