Using Excel to play with the Riemann zeta function

[john...@btinternet.com]

I have tried to put the formula for the Riemann zeta function into Excel
using the figure for the second zero, 0.5 + 21.022i in the hope of that when
it all comes out I will get zero for both the real and imaginary parts. But
I have summed about 500 terms and am not getting zero. Does the function
converge for this zero much slower than that?

My formula looks like this
:

1+

Sigma (x=2 to x=500),

(1/(x^0.5))*(cos (-21.022lnx) + isin (-21.022lnx))

Apologies for any infelicities writing out the formula above,
and if I have got the formula wrong please let me know.

Thanks.

[tommy1729]

1) first make sure excel understands "i" or seperate the sums for real and imaginary in two series.

2) try a smaller zero of zeta.

closer to 0.5 + 13i usually works better...

it depends on the version of excel you have of course.

3) if 1) and 2) did not help much ; try an analytic continuation , that often converges better.

regards
tommy1729

[Gottfried Helms]

If you plot the coordinates of the partial sums, you'll notice
that the points form a spiral, which very slowly converges to
a center point.
What I did, was to use the alternating zeta-function (which has
the same zeros) and then Euler-summing the terms of the series.
That accelerated convergence remarkably.

Gottfried Helms

--
---

Gottfried Helms, Kassel

[David Bernier]

From a book on analytic number theory by Hugh Montgomery and Robert
Vaughan,
I read a bit about the Euler-MacLaurin summation formula and the
Riemann-Siegel function Z(t), t real.

Z(t) is real-valued, and |Z(t)| = |zeta(1/2+it)|. The book gives a good
approximation
formula for Z(t), which is asymptotically O(|t|^(-3/4)). The one thing
that I'm not
too familiar with is approximating Gamma(1/4 + it/2), once per value of t.
So I let PARI-gp use its own implementation of the gamma function.
Other than that, only sqrt(), cos(), sin() and log() and the floor()
functions on reals
are used. The formula involves summing floor(sqrt(|t|/(2Pi) ) terms,
and evaluating
a remainder term with no summation.

So, including the remainder term, this is what I get for the zero
nearest 21:

(16:48) gp > zz(21.02203963877155499) // Riemann-Siegel
%96 = 0.0026589607946314576631

(16:48) gp > zeta(1/2 + 21.02203963877155499*I) // zeta() of PARI-gp

%97 = -6.5378117352195749743 E-19 - 2.9157607522157476701 E-18*I

I wonder how to go about getting a higher order remainder term, that
might be O( 1/(t^5/4) )
or so.

David Bernier

[Pink Pig]

<john3...@btinternet.com> wrote:
>...
> Sigma (x=2 to x=500),

The real part of each term in your sum is x^-0.5, thus for the first
omitted term it is 1/sqrt(501), which is greater than 1/23, so you
shouldn't expect much in the way of convergence. To get anything close
to 0, you'd probably need an upper limit of 1000000 or so, which is
why this method isn't practical for calculating values of zeta near
the critical line. The Euler-MacLaurin expansion converges much faster.

[Raymond Manzoni]

john...@btinternet.com a écrit :

Hi,

Here is an older thread that contains the alternate sum and the
Euler-MacLaurin formula (see 28 for example) :
http://groups.google.fr/group/sci.math/browse_frm/thread/155eb74994c4b9a9/23c67398913ae15f

Hoping it helped too!
Raymond

[Gerry Myerson]

In article <C31855E7.8131%john...@btinternet.com>,
<john...@btinternet.com> wrote:

> I have tried to put the formula for the Riemann zeta function

*Which* formula for the Riemann zeta function?

The sum n^(-s) formula? which doesn't converge for Re s < 1?

> into Excel
> using the figure for the second zero, 0.5 + 21.022i

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

[Gottfried Helms]

David -

I used a very simple version of the Euler-summation-method. However,
it allows to use different "orders", even complex orders. Order 2 is
the classical method, which uses the simple binomial-transform of
values. K.Knopp described the method of using such arbitrary orders,
I'll see, if I get the reference today.

For a first impression, you might have a look at some graphs, with
which I played around some monthes ago. I've not much time at the
moment, but I can comment on these graphs (and the Euler-sum-method)
later. I've just uploaded these graphs at

http://go.helms-net.de/math/divers/zetagraphenPartialsummen.htm

The first group of graphs show the proceedings of partial-sums,
if the zeta-terms are Euler-summed with certain different orders
(including complex orders).
The same with the eta-series.

I discussed the first zero r0 = 0.5 + 14.xxx *I here.
To have a better impression of concurring values I have always
three lines, with a real-valued deviation
line1 = r0 - d
line2 = r0
line3 = r0 + d

The interesting aspect here was, that line1 and line3 are
perfectly separated by line2 and seemed never to meet (so
line1 and line3 could not approach the same zero without crossing
line2.

As I said, I'll have more time in the afternoon.

Regards -

Gottfried