0
20kviews
Prove that $ E = 1 + \Delta = e^{hD}$
1 Answer
0
2.1kviews

By definition

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)

$E=1+\Delta =e^{hD}$

Please log in to add an answer.