# Is Zeta function discrete-analytic?

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 • When you write $\zeta(s,q)$, do you mean the Hurwitz zeta function, or some other function with similar notation? – S. Carnahan Apr 16 '12 at 5:50
• @Mrc Plm, what is standard? Why you think that if there no poles the function and expansion coincide? Why do you think the expansion converges? If you know, can you please also answer this question? mathoverflow.net/questions/90744/… – Anixx Apr 16 '12 at 9:38
• Note, that in this graphic you can see three meromorphic functions that have the same Newton series expansion, but they are not equal! – Anixx Apr 16 '12 at 9:41
• @Carnahan, yes. – Anixx Apr 16 '12 at 9:47
• – Anixx Apr 16 '12 at 10:36

$$\zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(n+q)^s} \tag{1}$$ 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 $q>1/2$.
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.
• In this argument, $a>1$. Required because (1) fails for $s \le 1$. – Gerald Edgar Apr 27 '12 at 14:01