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).
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