Fourier expansion for a function:
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\int_{-\infty}^{+\infty}e^{i\omega t}f(t)dt \, d\omega$$
Newton series expansion of a function:
$$f(x)=\sum_{m=0}^{\infty} \binom {x}m(-1)^{-m} \sum_{k=0}^\infty\binom mk(-1)^{k}f(k)$$
Is there an analogy? Is it possible to make with Newton's series the same things as with Fourier transform? Are the Fourier transform and inverse Fourier transform analoguous to difference delta and summation operators?
Best Answer
to give some "exact math", expanding on my comment, here is the Poisson-Mellin-Newton cycle:
from $g(k)$ to $f_n$ via a Newton series: $f_n=\sum_{k=0}^n {n\choose k}(-1)^k g(k)$
from $f_n$ to $f(t)$ via a Poisson generating function: $f(t)=\sum_{n=0}^\infty f_n e^{-t} \frac{t^n}{n!}$
from $f(t)$ to $\hat{f}(s)$ via a Mellin transform: $\hat{f}(s)=\int_0^\infty f(t) t^{s-1}\,dt$
closing the cycle: $\hat{f}(s)=\Gamma(s)g(-s)$.
so if you allow me to substitute your Fourier transform for a Mellin transform, the Poisson generating function allows one to relate that transform to a Newton series, which seems to be the essence of your question.
a related way to make the connection is via the Euler transformation:
We have two involutions, firstly the Newton series (a.k.a. binomial transform)
$$f_n\mapsto g_n=\sum_{k=0}^n {n\choose k}(-1)^k f(k)\mapsto f_n=\sum_{k=0}^n {n\choose k}(-1)^k g(k)$$
and secondly the Euler transformation
$$F(z)\mapsto G(z)=\frac{1}{1-z}F\left(\frac{z}{z-1}\right)\mapsto F(z)=\frac{1}{1-z}G\left(\frac{z}{z-1}\right)$$
The two are related by a Fourier series (or generating function):
$$F(e^{is})=\sum_{n=0}^\infty f_n e^{ins},\;\;G(e^{is})=\sum_{n=0}^\infty g_n e^{ins}$$
Just like for Fourier transforms, there is a correspondence between multiplication and (binomial) convolution, as discussed in this MSE thread.