Is Zeta function discrete-analytic?

Question

Let's define discrete-analytic functions as functions that are equal to their Newton series expansion:

$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f(a)$$

My question is whether $\zeta(s,q)$ ($q$=const) is discrete-analytic against $s$?

That is whether its Newton series converges and is equal to the function itself.

For comparison, in the following graphic there are four functions:

• red is the function $\zeta(x,3)$
• blue is $\frac{\cos (\pi x)\psi_b^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi_b$ is the balanced polygamma
• yellow is $\frac{\cos (\pi x)\psi^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi$ is the polygamma as implemented in Mathematica
• green is the partial Newton expansion of the above functions taken at first 20 terms.

The three first functions and the Newton expansion, if it converges, have the same values at non-negative integer arguments.

notation $\zeta(x,q)$ is the Hurwitz zeta function, LINK

Balanced polygamma LINK

Answer 1

I hope this is OK.
$$ \zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(n+q)^s} \tag{1}$$ Let's first consider just one term, $f(s) = 1/(n+q)^{s}$. Then $$ \Delta f(s) = f(s+1) - f(s) = \frac{1}{(n+q)^{s+1}} - \frac{1}{(n+q)^s} =\left(\frac{1}{n+q}-1\right)\frac{1}{(n+q)^s} $$ $$ \Delta^k f(s) = \left(\frac{1}{n+q}-1\right)^k\frac{1}{(n+q)^s} $$ $$\begin{align} \sum_{k=0}^\infty\binom{x-a}{k} \Delta^k f(a) &= \frac{1}{(n+q)^a}\sum_{k=0}^\infty\binom{x-a}{k}\left(\frac{1}{n+q}-1\right)^k \cr &=\frac{1}{(n+q)^a}\left(1+\frac{1}{n+q}-1\right)^{x-a} \cr &= \frac{1}{(n+q)^x}=f(x) \end{align}$$ We applied the binomial theorem, which requires $$ \left|\frac{1}{n+q}-1\right| < 1 $$ so this works for $q>1/2$.

Thus The question is whether the convergence in (1) is good enough that we can interchange two sums and get our conclusion...

But in fact for $q \gt 1/2, a \gt 1, x \gt 1$ we can interchange the sums in $$ \sum_{k=0}^\infty\sum_{n=0}^\infty\frac{1}{(n+q)^a}\binom{x-a}{k}\left(\frac{1}{n+q}-1\right)^k $$ because, for $k \gt x-a, n \gt 1$, all terms have the same sign.