I came across this maths statement \begin{align*} f(t+h) &= f(t) + h\frac{d}{dt}f(t) + \frac{h^2}{2!}\frac{d^2}{dt^2}f(t) + \frac{h^3}{3!}\frac{d^3}{dt^3}f(t) + \cdots \\ &= f(t) + hDf(t) + \frac{h^2}{2!}D^2f(t) + \frac{h^3}{3!}D^3f(t) + \cdots \\ &= \left(1 + hD + \frac{h^2}{2!}D^2 + \frac{h^3}{3!}D^3 + \cdots \right)f(t) \\ &= e^{hD}f(t). \end{align*} I doubt this is a correct thing since I cannot understand the meaning of $e^{hD}f(t)$ where $D=\dfrac{d}{dt}$ is a differential operator. As an extra information, I found this info when reading something related to Grunwald fractional derivative.
Edit: My question is, can we actually factorise the operator $D$ and take a Taylor series for it as done above? This is because, I imagine that, if $t+h$ is a value on the x-axis, then $f(t+h)$ is a value on the y-axis. On the other hand, $e^{hD}f(t) = e^{h\frac{d}{dt}}f(t)$ is not a value, thus how can $f(t+h)$ be equivalent to $e^{h\frac{d}{dt}}f(t)$?
