Suppose we have two real-valued functions $f(x)$ and $g(x)$, both equal to their Newton series expansion:
$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$
$$g(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k g\left (0\right)$$
Is their composition $F(x)=f(g(x))$ also equal to its Newton series expansion (if it converges)?
$$F(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k F\left (0\right)$$
Here goes, as promised.
Let $f$ be entire of order less than $1$, so $|f(z)|\le Ce^{|z|^p}$, $p<1$. Write the Newton polynomial $$ P(x)=\sum_{k=0}^n\Delta^kf(0) {x \choose k} $$ Note that $g(k)=f(k)-P(k)=0$ for $k=0,1,\dots,n$. On the other hand, we can crudely estimate $|g|$ in a disk of radius $R>2n$ by $Ce^{R^p}+\sum_{k=0}^n(2R)^k\frac 1{k!}|\Delta^k f(0)|$. Now, $\frac 1{k!}|\Delta^k f(0)|\le \max_{[0,R/2]}\frac{|f^{(k)}|}{k!}\le (2/R)^k Ce^{R^p}$ by Cauchy, so we finally get $$ |g|\le C 4^n e^{R^p} $$ in the disk of radius $R$ centered at the origin.
Now, for $|x|<n$, each corresponding Blaschke factor $\frac{R(x-k)}{R^2-kx}$ is at most $\frac{3n}R$ in absolute value, so $$ |g(x)|\le C\left(\frac{12}{R/n}\right)^n e^{R^p} $$ Choosing $R=n^{1/p}$, we get $|g(x)|\le \left(12en^{-\frac{1-p}p}\right)^n\to 0$ as $n\to\infty$.
Another counter-example is extractable from Gerald Edgar's answer to this question, where he shows that $\sin (ax)$ is discrete analytic for $a \in (-\pi/3, \pi/3)$. So take $ f(x) = \sin ((\pi/4) x)$ (for example) and $g(x) = 9 x$ (for example). Then $f$ and $g$ are discrete analytic but the Newton series for $(f \circ g)(x) = \sin ((9 \pi/4) x)$ will converge to $\sin( (\pi/4) x)$, which equals $\sin ((9 \pi/4) x)$ for integer $x$, but not in general.
