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

Marks : 3

Year : DEC 2015

By definition $Ef(x)=f(x+h)------(1) and \\ \triangle f(x)=f(x+h)-f(x)---------(2)$

Substituting (1) and (2) we get

$$\triangle f(x)=Ef(x)-f(x) \\ \triangle f(x)+f(x)=Ef(x)$$

$\therefore (1+\triangle)f(x)=Ef(x)\\ \therefore (1+\triangle)f(x)=Ef(x)\\ \therefore E=1+ \triangle $