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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?

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    I asked a similar question here: http://math.stackexchange.com/questions/793550/example-for-finite-dimensional-analog-of-integral-transforms?noredirect=1#comment1642786_793550 - what I find strange though is that while you can also read that the Laplace transform is the continuous analog of the dot product for infinite dimensional vectors (=functions) you still calculate infinitely many terms in the discrete version (and not just over the available dimensions of your vectors as with the dot product) - perhpas you could contribute :-)2014-05-14

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The traditional "discrete laplace transform", that corresponds to $f(x) = \sum a(n) x^n$, is well known in discrete signal processing (with complex variable, and summation over all integers - not only positive) as the "Z-transform". In probability and combinatorics it's also very important, as the standard generating function.

Your variation $f(x) = \sum a(n) x^n /n!$ is also known and used -though less-, as the Exponential generating function.

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    I asked a similar question here: http://math.stackexchange.com/questions/793550/example-for-finite-dimensional-analog-of-integral-transforms?noredirect=1#comment1642786_793550 - what I find strange though is that while you can also read that the Laplace transform is the continuous analog of the dot product for infinite dimensional vectors (=functions) you still calculate infinitely many terms in the discrete version (and not just over the available dimensions of your vectors as with the dot product) - perhpas you could contribute :-)2014-05-14