Newton series and Fourier transform - is there an analogy? 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?
 A: 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.
A: (This is a revamping of MSE-Q, which has examples of Newton series interpolation for particular functions and the relation to the Mellin transform.)
With $ \; \; \displaystyle \bigtriangledown^{s}_n \; c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n} \; c_n \; \; $  and $\; \;  \displaystyle m_j= \int_{0}^{\infty} h(x) \; x^j \; dx \; \; $,  formally
$$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}=\int_0^\infty h(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx=\int_0^\infty h(x) \frac{x^{s-1}}{(s-1)!} \; dx,$$
so since $ \; \; \displaystyle m_j=j!\; g(j+1)=\int_0^\infty h(x) x^j \; dx \; \;$, 
$$ g(s+1)=\bigtriangledown^{s}_{n} \bigtriangledown^{n}_{j} g(j+1).$$
Identifying $g(x+1)$ with $f(x)$ in your formula, you can see that your Newton series is the Mellin transform in disguise.
Performing the modified inverse Mellin transform
$h(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$ .

For the Fourier transform, the analogous formula is a sinc function interpolation of a bandlimited function
$$\hat{g}(x) = \int_{-W/2}^{W/2} h(\omega) e^{i 2 \pi x \omega} d\omega.$$
Then
$$\hat{g}(x) = \sum_{n=-\infty}^{\infty} \frac{sin[\pi W\; (n/W-x)]}{
\pi W\;(n/W-x)} \hat{g}(n/W)$$
with $$e^{i2 \pi x \omega}=\sum_{n=-\infty}^{\infty} \frac{sin[\pi W \; (n/W-x)]}{\pi W\; (n/W-x)} e^{i2 \pi n \omega /W} $$
for $-W/2<\omega<W/2$ being the analogue of
$$ \frac{x^{s-1}}{(s-1)!}=\bigtriangledown_{n}^{s-1}\bigtriangledown_{j}^{n} \frac{x^j}{j!}$$
valid for $Real(s)>0$. (This is the basis for the Whittaker-Shannon theorem.) 

An aside on a related Newton series and Ramanujan's Master Formula:
Using umbral notation (MSE-Q) for the Taylor series, $f(t)=e^{a.t}$, the normalized Mellin transform formally gives
$$g(s)= \displaystyle \int_{0}^{\infty} e^{-a.t} \frac{t^{s-1}}{(s-1)!}dt$$
$$ = \int_{0}^{\infty} e^{-t}\; e^{(1-a.)t}  \frac{t^{s-1}}{(s-1)!}dt =\sum_{m=0}^{\infty} (-1)^m \binom{-s}{m}\sum_{k=0}^m(-1)^k \binom mk a_k \; .$$
Then $f(t)=e^{a.t}$ with  $a_n=g(-n)$ since, for $s=-n$, the binomial transform here is an involution and the regularization of the normalized Mellin transform (MT) gives the Dirac delta function or its derivatives in the integrand. Both sides of the equation would converge simultaneously only over restricted values of $s$ as noted in the MO-Q link, but could be analytically continued. 
(Use $\binom{s+m-1}{m}=(-1)^m \binom{-s}{m}$ for the MT term-by-term of the Taylor series for $e^{(1-a.)t)}$.)
This is the essence of Ramanujan's Master Formula/Theorem (cf. MO-Q).
So, the Newton interpolation is more akin to the Fourier transform itself; in fact, you can regard the MT here as the interpolation of the coefficients of the Taylor series with the Newton series as one of its avatars. 
The inverse MT gives, for appropriate $ \sigma$ when convergent, 
$$\displaystyle \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(s) \frac{(-t)^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(-n) \frac{t^{n}}{n!} = f(t) \; .$$
