On Tue, Jun 1, 2010 at 10:58 AM, Robert Bradshaw
<rober...@math.washington.edu> wrote:
> On Jun 1, 2010, at 8:13 AM, Anne Driver wrote:
>
>> Hello,
>>
>> I am new to this list, and relatively new to Sage. I'm puzzled by the
>> logic of one part of Sage though.
>>
>> Although I don't have access to Mathematica at the minute on this
>> computer, I know if I compute the first zero, I get something like
>>
>> In[1] = ZetaZero[1] //N (to get a numerical value)
>> Out[1] = 1/2 + I*14.134...
>>
>> Trying this in Sage, I get:
>>
>> sage: lcalc.zeros(1)
>> [14.1347251]
>>
>>
>> Why does Sage not do the sensible thing like Mathematica and return the
>> complex number 0.5 + I 14.1347251 ? It would seem much more logical.
>>
>> Of course, it is not proven that the real part is 1/2, so how would the
>> case be handled if a root was not found to have a real part of 1/2 ?
>
> I believe both algorithms assume the Riemann hypothesis in computing them
> (otherwise, for example, it would be ambiguous to talk about the n-th zero
> anyways).

Often such computations actually prove the Riemann hypothesis up to a
given height
(see, e.g., 
http://numbers.computation.free.fr/Constants/Miscellaneous/zetazeros1e13-1e24.pdf)

I've cc'd Mike Rubinstein, so he can respond if he wants, since I'm
not sure lcalc is actually doing
this or not.

 -- William

> I would guess the reason that lcalc returns the imaginary part
> only is that otherwise the first thing one would do to actually do anything
> interesting with this data would be to take the imaginary part, so this just
> saves the effort and overhead.
>
> - Robert
>
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-- 
William Stein
Professor of Mathematics
University of Washington
http://wstein.org

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