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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

alt text

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  • $\begingroup$ When you write $\zeta(s,q)$, do you mean the Hurwitz zeta function, or some other function with similar notation? $\endgroup$
    – S. Carnahan
    Apr 16, 2012 at 5:50
  • $\begingroup$ @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/… $\endgroup$
    – Anixx
    Apr 16, 2012 at 9:38
  • $\begingroup$ Note, that in this graphic you can see three meromorphic functions that have the same Newton series expansion, but they are not equal! $\endgroup$
    – Anixx
    Apr 16, 2012 at 9:41
  • $\begingroup$ @Carnahan, yes. $\endgroup$
    – Anixx
    Apr 16, 2012 at 9:47
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    $\begingroup$ Mrc Plm, en.wikipedia.org/wiki/Finite_difference#Newton.27s_series $\endgroup$
    – Anixx
    Apr 16, 2012 at 10:36

1 Answer 1

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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.

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  • $\begingroup$ By the way, what is for s < 0? $\endgroup$
    – Anixx
    Apr 27, 2012 at 6:33
  • $\begingroup$ In this argument, $a>1$. Required because (1) fails for $s \le 1$. $\endgroup$ Apr 27, 2012 at 14:01

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