Let us consider the operator \(D\) that maps a function to its derivative \[ D:f\to f' \] Notice that this is a linear operator on the space of smooth functions. Now consider the exponential of this operator, defined \[ e^{tD}=\mathrm{id}+tD+\frac{t^2}{2!}D^2+\frac{t^3}{3!}D^3+\cdots \] Notice that this is again a linear operator. Let us see how this operator acts over a smooth function \(f\): \[ e^{tD}f=f+tf'+\frac{t^2}{2!}f''+\frac{t^3}{3!}f'''+\cdots \] This is a function that can be evaluated at \(0\), giving: \[ \left(e^{tD}f\right)(0)=f(t) \] Thus, the Taylor series says that the exponential of \(D\) is a shift.