Relation between the operators.

 Relation between the operators:

We can express each of Δ,∇,𝛿,μ and D in terms of E. 

1) Δ = E - 1

Proof: By definition of  Δ

 Δf(x) = f(x+h) - f(x)

          =Ef(x) - f(x)         [∵Ef(x) = f(x+h)]

Hence Δ = E -1

2) ∇ = 1 - E⁻¹

Proof: By definition of ∇

∇f(x) = f(x) - f(x-h)

         = f(x) - E⁻¹f(x)   [∵E⁻¹f(x) = f(x-h)]

=> ∇f(x) =  (1 - E⁻¹)f(x)

=> ∇ = 1 - E⁻¹

3) 𝛿 =E ¹/² - E⁻ ¹/²

Proof: By definition of 𝛿

𝛿f(x) = f[x+(h/2)] - f[x-(h/2)]

          = E ¹/²f(x) -  E⁻ ¹/²f(x)

Hence  𝛿 =E ¹/² - E⁻ ¹/²

4) 𝝁 = 1/2 (E ¹/² + E⁻ ¹/²)

Proof: By definition of  𝝁

   𝝁f(x) = (1/2){f[x+(h/2)] + f[x-(h/2)]}

             = (1/2){E ¹/²f(x) +  E⁻ ¹/²f(x)}

             =(1/2){E ¹/² + E⁻ ¹/²}f(x)

Hence     𝝁 = 1/2 (E ¹/² + E⁻ ¹/²)

5) E = e^(hD)

Proof: By definition of E

Ef(x) = f(x+h)

Using Taylor's series, we get

Ef(x) = f(x) + hf'(x) + (h²/2!)f"(x) + .......

         = f(x) + hDf(x) + (h²/2!)D²f(x) + ......  [∵D =d/dx]

       = [1+hD + (h²D²)/2! + .......] f(x)

       = e^(hD) f(x)  [∵ e^x = 1+x + x²/2!+.....]

Hence E = e^(hD).