On Jun 1, 2010, at 11:05 AM, William Stein wrote:
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.
IIRC, the broad idea is to compute sign changes and then perform a
contour integral to prove that you have located all the zeros. If no,
refine the grid and try again. Of course this is a huge
oversimplification, but if there are zeros not on the critical line
than this would simply fail to terminate, and otherwise it would prove
the hypothesis.
- Robert
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