Taylor expansion of $\frac{1}{1-D}$, where $D$ is the differential operator

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I understand that we can represent $e^D$ simply as a power series of D. But what about functions of D which are not entire on the complex plane? What if the function has no taylor expansion, or if it is only valid in a certain region.

An interesting example to me is $\frac{1}{1-D}$. The taylor expansion for this function is valid for $|D|<1$ which makes no sense.

If I were to try applying the taylor expansion of this function and then apply $(1-D)$ on the result, I would indeed get the original function, indicating that the power series was valid, which leads me to believe that $|D|$ is indeed less than 1.

What is the meaning of this all, what am I misinterpreting?

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One has to specify a space of functions (or some linear space, if you are not interpreting the elements as functions) on which $D$ acts, and some topology with respect to which $\sum D^n f$ might converge, or not converge, to an element of the space. For example, the series $(1-D)^{-1}$, interpreted as a power series, does always converge and give a meaningful result when applied to the vector space of polynomials in one variable.

Your calculation shows that, if the sum does converge in some specified sense, applying $(1-D)$ to it does recover $f$.