Does Newton's forward difference formula / Newton's interpolation formula yield well-defined functions (in one variable) if it is convergent?

Question

Suppose we are given a sequence $a_{0}, a_{1}, a_{2}, \ldots$ of real numbers and we define $F:\mathbb{N}\times[0, \infty)\rightarrow\mathbb{R}$ by

$$ F(k, \alpha) = \sum_{n=0}^{\infty} \frac{\alpha^{\underline{n}}}{n!} \Delta^{n} a_{k} = \sum_{n=0}^{\infty} \binom{\alpha}{n} \Delta^{n} a_{k} $$

where $\Delta a_{k} = a_{k+1} - a_{k}$, $\Delta^{2} a_{k} = a_{k+2} - 2a_{k+1} + a_{k}$, etc.

Suppose further the formula converges for all $k, \alpha$. Do we necessarily have the desirable property that

$$ F(k, \alpha) = F(0, k+\alpha)? $$

Is there a sequence $a_n$ for which $F$ is well-defined but it does not have the property (so we can't define $f(k+\alpha) = F(k, \alpha)$ unambiguously)?

Answer 1

Let's consider the case of $k=1$. First, let's manipulate $F(1,\alpha)$:

$$F(1,\alpha)=\sum_{n=0}^{\infty}\binom{\alpha}{n}\Delta^{n}a_1=\sum_{n=0}^{\infty}\binom{\alpha}{n}(\Delta^{n}a_0+\Delta^{n+1}a_0)=F(0,\alpha)+\sum_{n=0}^{\infty}\binom{\alpha}{n}\Delta^{n+1}a_0=F(0,\alpha)+\sum_{n=1}^{\infty}\binom{\alpha}{n-1}\Delta^{n}a_0=F(0,\alpha)+\sum_{n=1}^{\infty}\binom{\alpha}{n}\frac{n}{\alpha-n+1}\Delta^{n}a_0$$

Then, let's manipulate $F(0,\alpha+1)$

$$F(0,\alpha+1)=\sum_{n=0}^{\infty}\binom{\alpha+1}{n}\Delta^{n}a_0=\sum_{n=0}^{\infty}\binom{\alpha}{n}\frac{\alpha+1}{\alpha-n+1}\Delta^{n}a_0=\sum_{n=0}^{\infty}\binom{\alpha}{n}(1+\frac{n}{\alpha-n+1})\Delta^{n}a_0=\sum_{n=0}^{\infty}\binom{\alpha}{n}(\Delta^{n}a_0+\frac{n}{\alpha-n+1}\Delta^{n}a_0)=F(0,\alpha)+\sum_{n=0}^{\infty}\binom{\alpha}{n}\frac{n}{\alpha-n+1}\Delta^{n}a_0$$

So we basically got the same result. The fact that the remaining sum in the first result starts from $1$ and in the second from $0$ doesn't matter because the term for $n=0$ is equal to $0$ because of the $n$ in the numerator. Then I think you can prove this for arbitrary $k$ with induction.

Answer 2

Yes, $F(k,\alpha)=F(0,k+\alpha)$.

Hint 1. Consider the shift operator $(Sf)(n)=f(n+1)$. Observe $(S^kf)(n)=f(k+n)$. Writing $I$ for the identity operator $(If)(n)=f(n)$, we have $S=\Delta+I$, so by the binomial theorem,

$$ S^k ~=~ \sum_{s=0}^k\binom{k}{s}\Delta^s \tag{$\ast$}$$

Hint 2. Vandermonde convolution applies even if $\alpha$ is not an integer:

$$ \binom{\alpha+k}{n}=\sum_{r+s=n}\binom{\alpha}{r}\binom{k}{s} \tag{$\circ$}$$

Indeed, if we treat $\binom{\alpha}{n}$ as a function of integer $\alpha$ and apply $(\ast)$ we get $(\circ)$; since both sides are polynomials in $\alpha$ equal at all integers, an infinite number of values, they must be identical.

Can you put these two hints together?