[seqfan] Re: n-th derivative of Zeta function

Olivier Gerard olivier.gerard at gmail.com
Fri Jan 2 15:46:26 CET 2009

Dear Vladimir,

This is simply not true.

Just compute more precisely the first few cases. If you use Mathematica
(as you seem to do) you will see for instance that the value at s=1/2 of the
second derivative of Zeta is not an integer.



or following your formulation

(Derivative[2][Zeta][1/2]/2 + 2^(2 + 1) )



this is not zero and this is not a computation artefact.

What you can try to prove instead, using classical analysis, is that

d_(k+1) Zeta(1/2)   /   d_(k) Zeta(1/2)    converges to  2(k+1)  when k -> oo

relatively quickly

All (mathematics) related inquiries are welcome on the seqfan list, but
I prefer that they be in direct relation to integer sequences.

If there is no clear link to integer sequences or the content of the OEIS,
consider posting this on other math oriented mailing lists or newsgroups.

Also note that the art of conjecture requires a minimum of resistance from
the conjectured property (that is: a minimum of force in the initial
attack from the author
of the conjecture) and at least an apparent void in the available means
of proof.


On Fri, Jan 2, 2009 at 14:47, Vladimir Reshetnikov
<v.reshetnikov at gmail.com> wrote:
> Hi
> Table[Round[-Derivative[k][Zeta][1/2]/k!] - 2^(k + 1), {k, 1, 50}]
> Conjecture: this sequence consists of all zeros.
> Can anybody prove it?
> Thanks
> Vladimir

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