Yesterday ago I was reading how the Laplace Transform can be interpreted as the continuous analog of the discrete functional dependance of the power series $f(x) = \sum a(n) x^n$ This is to say, $L\{a(n)\} = f(x)$
Since I find it more natural I will use a the "nucleus" so to call it, $\frac{x^n}{n!}$
Thus we can think about the following. Being $L$ the "discrete Laplace operator".
$L\{1\}(x) = e^x$
$L\{n\}(x) = xe^x$
$L\{n!\}(x) = \frac{1}{1-x}$
$L\left\{\frac{1-(-1)^n}{2}\right\}(x) = \sinh x$
$L\left\{\frac{1+(-1)^n}{2}\right\}(x) = \cosh x$
Similary we can define a "derivative" theorem as follows:
$L\left\{\frac{n}{x}a(n)\right\}(x) = \frac{d}{dx}L\{a(n)\}(x) $
where I leave the $x$ inside to recall we're operating inside the sum, although it'd be the same if we left the $x$ oustide.
This is to say,
$f(x) = \sum a(n) \frac{x^n}{n!}$
$f'(x) =\frac{1}{x} \sum na(n) \frac{x^n}{n!} = \sum a(n) \frac{x^{n-1}}{(n-1)!}$
It is clear that $L\{(\alpha+\beta)(n)\}(x) = L\{\alpha(n)\}(x)+L\{\beta(n)\}(x)$ and that $L\{k\alpha(n)\}(x) = kL\{\alpha(n)\}(x)$ so this $L$ transform is linear.
Is there any theory on this discrete transform? Is there a motivation to use it, or is just not important in discrete mathematics?