Re: peculiar zeta() behavior

[Karim BELABAS <Karim.Belabas@math.u-psud.fr>]

On Tue, 18 Feb 2003, Igor Schein wrote:
> ? for(k=70,100,print(k" "zeta(1-1/2^k-2^18)))
[...]
> 79 9.206091509744517478696623193 E1097349
>   ***   user interrupt after 23,496 ms.
>
> looks like bad things start happening near a large negative odd
> integer.

Well, when the input becomes undistinguishable from an integer, zeta starts
using the standard formula zeta(k) = -B_(1-k)/(1-k). Unfortunately, this
triggers the computation of the whole table of Bernoulli numbers, instead of
using the asymptotic approximation for B_n ( \pm 2 n!/(2Pi)^n, n even ),
which is precise enough in this range...  [ using approximate values,
the computation is done using the functional equation, with the approximation
zeta(n) = 1 if n is large, so no problem occurs... ]

Fixed.

> This is very related:
>
> http://www.math.mcgill.ca/goren/ZetaValues/zeta.html#explanations

If you can reproduce some of the bugs he found out, I'm interested.

    Karim.
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Karim Belabas                    Tel: (+33) (0)1 69 15 57 48
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