Shift operator as an exponential of differentiation

Remarks

The right side is the Taylor series of $f$ at $x$, evaluated at the increment $h$. A power series that converges on all of $\R$ is an analytic function whose Taylor series at every point converges to it on all of $\R$, which gives the identity. For a polynomial of degree $d$ the sum stops at $n = d$, and the identity is exact algebra.

Example: $f(x) = x^2$. Then $f + h f' + \frac{h^2}{2} f'' = x^2 + 2hx + h^2 = (x + h)^2$.

The hypothesis of analyticity is needed. For the smooth non-analytic function $f(x) = e^{-1/x^2}$ for $x \ne 0$, $f(0) = 0$, all derivatives vanish at $0$, so the right side at $x = 0$ is $0$, while $f(h) > 0$ for $h \ne 0$.

Going deeper

For a student

Since $e^{hD} e^{kD} = e^{(h+k)D}$, the identity explains why shifts compose by adding the steps. In the calculus of finite differences, the forward difference $\Delta = T_1 - \id$ becomes $e^D - 1$, and on polynomials this gives formal identities such as $D = \log(1 + \Delta) = \Delta - \Delta^2/2 + \Delta^3/3 - \cdots$, a finite sum on each polynomial. In quantum mechanics the same identity says that the momentum operator generates translations.