Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : MAY 2013

By definition $E.f (x) = f (x + h) \to (1) \\ \triangle f(x)=f(x+h)-f(x)\\ \triangle f(x)=Ef(x)-f(x) \\ \triangle f(x)=(E-1)f(x)\\ \therefore \triangle =E-1\\ \therefore E=1+\triangle ---- (2)\\ Also \space E.f(x)=f(x+h)\\ =f(x)+hf'(x)+\dfrac {h^2}{2!}f'(x)+\dfrac {h^3}{3!}f''(x)+.... \text{ Taylor series } \\ =f(x)+hDf(x) + \dfrac {h^2}{2!}D^2f(x)+ \dfrac {h^3}{3!}D^3f(x)... where \space D^nf(x)=f^n(x)\\ =\Bigg[1+\dfrac {(hD)^1}{1!}+\dfrac {(hD)^2}{2!} + \dfrac {(hD)^3}{3!} +....\Bigg]f(x)\\ \therefore Ef(x)=e^{hd}f(x)\\ \therefore E=e^{hD} ---- (3)\\ From 2 \space and \space 3\\ E=1+\triangle =e^{hD}$