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Prove that $ E = 1 + \Delta = e^{hD}$
1 Answer
written 8.4 years ago by | • modified 8.4 years ago |
E.g
$f (x) = f (x+h) --------- \to(1)$
$\Delta f(x) =f(x+h)-f(x) $
$\Delta f(x) =Ef(x)-f(x) $
$\Delta f(x) =(E-1)f(x) $
$\therefore \Delta =E-1$
$\therefore E= 1+ \Delta ---------------- \to (2)$
Also
E.g
$f (x) = f (x + h)$
$f(x) +hf^1(x)\dfrac {h^2}{2!}f^1(x)+\dfrac{h^3}{3!}f^1(x) + ............$ Taylor series
$f(x) +hDf(x)\dfrac {h^2}{2!}D^2f(x)+\dfrac{h^3}{3!}D^3f(x) + ............$
Where
$D^nf(x)=f^n(x)$
$=\Big[1+\dfrac{(hD)^1}{1!} + \dfrac{(hD)^2}{2!} + \dfrac{(hD)^3}{3!} +....\Big]f(x)$
$\therefore Ef(x)=e^{hd}f(x)$
$\therefore E=e^{hD} ----\to (3)$
From (2) and (3)
$E=1+\Delta =e^{hD}$