[net.math] infinite differentiation

don@allegra.UUCP (12/11/83)

Let D be the operator d/dx, and consider the operator exp(D).  Prove
that for any continuous differentiable function, f, the following is
true:

		exp(D) f(x) = f(x + 1)

ags@pucc-k (Seaman) (12/13/83)

More generally, if D is the operator d/dx and h is a real number, then
hD is the operator h d/dx, and

		exp(hD) f(x) = f(x+h)


				Dave Seaman
				..!pur-ee!pucc-k!ags

quiroz@rochester.UUCP (Cesar Quiroz) (12/13/83)

      In a word :

		exp(D) = 1 + D + ... + (D^n)/n! + ... (as operators)

		so:

		exp(D) f(x) = f(x) + Df(x) + D^2 f(x) /2! + ...D^n f(x) / n! ...

			    = sum [ D^i f(x) / i! * ((x+1) -x)^i ] i= 0,1,...

			    = Taylor expansion for f around x, eval. at 1

			    = f(x+1)


							Cesar Quiroz
							UofR Comp Sc Dpt.