Originally Posted by
BobP Call your $\displaystyle x$ values $\displaystyle x_{0},x_{1},x_{2},x_{3}, \dots \equiv x_{0},x_{0}+h,x_{0}+2h,x_{0}+3h,\dots$
and the corresponding function values
$\displaystyle f(x_{0}),f(x_{1}),f(x_{2}),f(x_{3}),\dots \equiv f_{0},f_{1},f_{2},f_{3},\dots,$
then the entries in the first difference column will be
$\displaystyle f_{1}-f_{0},f_{2}-f_{1},f_{3}-f_{2},\dots.$ .
If you were to divide each of these by $\displaystyle h$ you would get a discrete version of the derivative of $\displaystyle f(x).$
If this is to be/represent $\displaystyle e^{x},$ what does that suggest that $\displaystyle f(x)$ should be ? Think calculus.
Make that (obvious) choice for $\displaystyle f(x)$ and construct the first difference column, (a little algebraic simplification is required).
You should find that that doesn't quite work, but after tweaking $\displaystyle f(x)$ slightly it does.