For a function $f(x)$ defined for all $x\geqslant 0$, a constant $c \in \mathbb{R}$ and $n \in \mathbb{N}^+$, is there a way to simplify the following summation? \begin{align} \sum_{i = 0}^{n} f^{(i)}(x) \frac{c^{-i}}{(n - i)!~i!}, \end{align} where $f^{(i)}$ is the $i$th derivative of $f$.
It looks kind of like the Taylor series of $f$ but I don't know where to begin.
Can anyone please give a hint?
We observe that $$ \begin{align} S &= \frac{1}{n!}\sum_{i=0}^n \left({n\choose i}f^{(i)}(x)c^{-i} \right) \\ &= \frac{1}{n!}\exp\left(-\frac{x}{c} \right)\cdot \left(\exp\left(\frac{x}{c} \right)\sum_{i=0}^n \left({n\choose i}f^{(i)}c^{-i} \right)\right) \tag{1}\\ \end{align} $$ And we have
$$\frac{d^n}{dx^n}\left(f(x)\exp\left(\frac{x}{c} \right) \right) = \exp\left(\frac{x}{c} \right)\sum_{i=0}^n \left({n\choose i}f^{(i)}c^{-i} \right)\tag{2}$$
From $(1)$ and $(2)$, we deduce that $$S = \frac{e^{-\frac{x}{c}}}{n!} \frac{d^n}{dx^n}\left(f(x)e^{\frac{x}{c}} \right)$$
You can verify this final formula for $n=1,2,3,...$