Re: [sage-support] zeros of the Riemann zeta function
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). 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 -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] zeros of the Riemann zeta function
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 -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org -- William Stein Professor of Mathematics University of Washington http://wstein.org -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] zeros of the Riemann zeta function
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 -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
