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)?


$$F(x) = \sum\_{k=0}^\infty \binom{x}k \Delta^k F\left (0\right)$$



If two functions are equal to their Newton series, is their composition also equal to its Newton series?
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 $n\to\infty$.


