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[This question is copied from math.stackexchange, it didn't get answers so far]

For some exercises with (divergent) summation of the Stieltjes constants,see also MSE I'm trying a formula, which involves derivatives of the $\zeta()$ -function at negative integers; perhaps better formulated as $$ f(x) = \zeta(1-x) + 1/x \qquad x \in \mathbb N $$ and then the scaled derivatives $$ g(x,d) = x^d \cdot {f^{(d)}(x)\over d!} $$.

In Pari/GP there is a very nice procedure for the computation of $\zeta$ implemented, but I don't see anything for the derivatives except to implement numerical differentiation along the line ${f(x+\epsilon/2) - f(x-\epsilon/2)\over \epsilon}$ and so on, but which becomes inaccurate for higher derivatives.
For small x I can use the representation of the function $f(x)$ as a power series involving the Stieltjes-constants themselves, but a suitable computation is then possible only for a small range of x-values if I use only, say 100 or 200 terms of the power series.

Q: What would be a more efficient representation of $f(x)$ and/or $g(x)$ for the actual computation, where -say- x and d both can go up to 100 or 200 ? Note, that for high c (the index for the partial sums, see below) there is the significant effect , that either (x is small and d is high) or converse, such that d+x in the terms of one partial sum is constant.


More context: what I'm finally after is to be able to explore the partial sums and its single terms in
$$ S \underset{\mathfrak E}{=} \lim_{c\to \infty} S(c) = \sum_{k=0}^{c} (-1)^k g( c+1-k ,k) $$ which should be a representation for the (divergent) sum of the Stieltjes constants, taken by some "Eulerian" summation procedure $\mathfrak E$ (see an earlier post here on MO).

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this earlier post might be relevant: mathoverflow.net/questions/120232/… –  Carlo Beenakker Mar 22 '13 at 18:54
    
@Carlo - thank you very much; I'll take a deeper look at it later... –  Gottfried Helms Mar 23 '13 at 18:22
1  
Maybe related: math.stackexchange.com/questions/340718/… –  i707107 Apr 5 '13 at 21:22

1 Answer 1

The Euler-Maclaurin formula works quite well.

I have implemented simultaneous computation of $(\zeta(s,a), \zeta'(s,a) \ldots, \zeta^{(n)}(s,a))$ for arbitrary $s, a \in \mathbb{C}$, with the option to subtract $1/(s-1)$ to remove the singularity at $s = 1$, in my library for ball arithmetic Arb. All the derivatives are computed with rigorous error bounds (as currently described on this page).

As a benchmark, computing the first 1000 Stieltjes constants to 1000 accurate digits (using 1500 digits of precision to give $\gamma_0$ accurate to 1500 digits and $\gamma_{1000}$ accurate to 1000 digits) takes 14 seconds, and computing the first 5000 Stieltjes constants to 5000 digits takes 40 minutes.

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Hi Fredrik, did you mean to apply the Euler-Maclaurin formula directly at the Stieltjes-numbers? –  Gottfried Helms Mar 8 '13 at 18:10
    
Yes (if I understand your question directly), the Euler-Maclaurin formula works directly for zeta at any integer, and gives you the Stieltjes constants at $s = 1$ after just removing the singular $1/(s-1)$ term. –  Fredrik Johansson Mar 8 '13 at 20:59
    
Ah, so I see, there was possibly a misunderstanding. I have the first 500 Stieltjes constants to 1000 digits by a nice, small procedure. What I was trying to do, was to give a sense to the sum-of-all-Stieltjes numbers, by some summation-procedure. That procedure requires to use the derivatives of the zetas in a certain manner in finite sums as written above. That finite sums, having more and more terms according to the increasing parameter c, approximate something like 0.4990749...xyz , but convergence is slow and needs many of the zeta's derivatives... –  Gottfried Helms Mar 9 '13 at 6:20

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