On 13 November 2012 03:25, Dima Pasechnik <dimp...@gmail.com> wrote:
> On 2012-11-13, Nils Bruin <nbr...@sfu.ca> wrote:
>> On Nov 12, 6:06 pm, Dima Pasechnik <dimp...@gmail.com> wrote:
>>> sage: e.cremona_label()
>>> '457532830151317a1'
>>> sage: e.analytic_rank(leading_coefficient=True)
>>> (4, -2.50337480324368498e-9)
>>>
>>> here is what I got after some hours of running.
>>> e-9 does not look as suspiciosuly small to me...
>>
>> That depends on what the bounds on the error in that number are. If
>> it's
>> -2.5e-9 +- 1e-3
>> then it can very well be an approximation to 0 (which we know it is).
>> It's the accuracy of the approximation that matters--not its size.
>
> looking at the slide 22 of
> http://homepages.warwick.ac.uk/~masgaj/papers/bsd50.pdf
> seems to tell that we are not talking about a guaranteed computation,
> but just about a heuristic, no more than that. Unless I miss something.

No, you do not miss anything.  As someone has already reminded us (and
I say clearly in the talk for which those are the slides), there is no
known way of proving that the 3rd (or higher) derivative of an
elliptic curve L-function is 0.  Any computations of higher derivative
are always based on an assumption that some lower derivatives which
have been computed to some approximation and which look as if they are
zero, really are zero.  In this example, one's view of this is
possibly biassed by the fact that we know that the curve has algebraic
rank 8, so that if the k'th derivaitve was non-zero for any k<8 then
we would have a counterexample to BSD.

The second point, just as important here, is that the complexity of
the computation of these L-values goes up with the conductor, and
curves of high rank (as this one) must have high conductor.  So in
this example, even if you did know that the lower derivatives were
exactly 0, to compute the 8th derivative would require many more terms
of the L-series than are being used here.

John Cremona

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