Discrete-analytic functions
#1
I do not know if such concept already exists but lets consider functions which are equal to its Newton series.

We know that functions which are equal to their Taylor series are called analytic, so lets call functions that are equal to their Newton series "discrete analytic".

The formula is alalogious to Taylor series but uses finite differences instead of dirivatives, so for any discrete-analytic function:

\( f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right) \)

It is known that for a functional equation \(\Delta f=F\, \)there are infinitely many solutions which differ by any 1-periodic function. But it appears that there is only one (up to a constant) discrete-analytic solution, i.e. all discrete-analytic solutions differ only by a constant term.

Thus I have the following questions:

- Do discrete-analytic functions express special properties on the complex plane?

- Is there a method to extend the notion of discrete analiticity to a range of functions for which Newton series does not converge (so to make it possible to choose the distinguished solution to the abovementioned equation)?

For the second part of the question, as I know there is at least one one similar attempt, the Mueller's formula:

If \( \lim_{x\to{+\infty}}\Delta f(x)=0 \)
then
\( f(x)=\sum_{n=0}^\infty\left(\Delta f(n)-\Delta f(n+x)\right) \)

although it seems not to be universal and I do not now whether it is always useful.
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#2
but there is no such function for sexp ...

and it cannot hold everywhere for slog ...

regards

tommy1729
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#3
There is such function for sexp where the regular iteration method works. Actually, by adding the requirement for "discrete-analiticity" we fix the solution.
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#4
but ?? sexp grows way faster than exponential and hence does not satisfy uniqueness according to carlson's theorem ? Huh
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#5
ofcourse for bases =< eta carlson's theorem applies and we have uniqueness ...

but that was already pointed out by mike 3.
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