Doubts related to Differential Operator of Infinite order Let $$f(s)=\sum_0^\infty c_vs^v$$ be some entire function. We say that the differential operator $f(d/dx)=\sum_0^\infty c_vd^v/dx^v$ is defined in some fundamental space $\varPhi$, if for any $\varphi \in \varPhi$, the series $$f(\frac{d}{dx})\varphi(x)=\sum_0^\infty c_v \varphi^{(v)}(x)$$ is again a fundamental function.
My questions :
1) What guarantees me to represent f as an infinite series as defined here.
2) I know that $f(d/dx)$ is an operator acting on some test function from a fundamental space $\varPhi$, what does the notation $f(d/dx)$ convey in particular ?
3) How am I able to replace $s$ by $d/dx$ ?
Thank you for your help.
Reference: Generalized Functions.. Volume 2 by I.M. Gelfand and G.E. Shilov
 A: 1) $f$ is an entire function, so it has a power series representation centered at $0$ that converges on the whole complex plane.
2) The notation $f(d/dx)$ denotes exactly what you have written: it is an operator defined by
$$
f\left(\frac{d}{dx}\right)\varphi(x) = \sum_{v=0}^\infty c_v \varphi^{(v)}(x)
$$
where as part of the definition the right-hand side is assumed to converge. (The convergence might not hold if we take a different function $g$, but this is beside the point when we're just taking the definition.)
3) Replacing $s$ by $d/dx$ to write
$$
f\left(\frac{d}{dx}\right) = \sum_{v=0}^\infty c_v\frac{d^v}{dx^v}
$$
is purely notation. There is a calculus associated with this sort of notation (the holomorphic functional calculus) that makes it particularly convenient. But you shouldn't read too far into it. The above sum shouldn't be interpreted as literally the partial sums
$$
\sum_{v=0}^N c_v\frac{d^v}{dx^v}
$$
converging; any convergence should be interpreted against a suitable fundamental space of test functions.
