zeta zero
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From: master1729 on 22 Sep 2009 14:42 just for fun : what is the simplest way of proving that the first zero of zeta(s) on the critical line has indeed real part 1/2 ?
From: Michael on 22 Sep 2009 19:01 On Sep 22, 3:42 pm, master1729 <tommy1...(a)gmail.com> wrote: > just for fun : > > what is the simplest way of proving that the first zero of zeta(s) on the critical line has indeed real part 1/2 ? @ECHO OFF ECHO Hello Universe! Copywrite(C) 2009 Martin Musatov. All rights reserved. PAUSE @\windows\ie7\iexplore.exe meami.org @"c:\program files\internet exploder\iexplore.exe 'http:// www.meami.org'"@open http:\\www.meami.org" #include <windows.h> void main (void) { ShellExecute (NULL, "open", "http://www.meami.org", NULL, NULL, SW_SHOWNORMAL); }
From: Robert Israel on 23 Sep 2009 01:23 > just for fun : > > what is the simplest way of proving that the first zero of zeta(s) on the > critical line has indeed real part 1/2 ? I think you mean "in the critical strip": the critical line is Re(z) = 1/2 by definition. By the functional equation and zeta(conjugate(s)) = conjugate(zeta(s)), zeros not on the critical line must occur in pairs symmetric about Re(z) = 1/2. Integrate numerically zeta'(z)/zeta(z) dz around a suitable contour with sufficient precision and you'll have shown that there is exactly one zero, not two, inside that contour. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: master1729 on 23 Sep 2009 06:39 Robert Israel wrote : > > > just for fun : > > > > what is the simplest way of proving that the first > zero of zeta(s) on the > > critical line has indeed real part 1/2 ? > > I think you mean "in the critical strip": the > critical line is Re(z) = 1/2 > by definition. yes of course , sorry. > > By the functional equation and zeta(conjugate(s)) = > conjugate(zeta(s)), > zeros not on the critical line must occur in pairs > symmetric about > Re(z) = 1/2. True. although i expect you actually mean the functional equation of the related Xi(z) function. Thus we have symmetry for Re = 1/2 and Im = 0 ( btw i wonder how one proves that a zero of a certain function that has no symmetry (for its zero's) and is not related to primes has a rational real part = 1/2 in a fast way in general ) > Integrate numerically zeta'(z)/zeta(z) > dz around a suitable > contour with sufficient precision and you'll have > shown that there is > exactly one zero, not two, inside that contour. ' integrate " numerically " ' does not sound like a proof. especially the " numerically " part ; more like a heuristic argument. ... with any desired precision around 1/2. but no proof of Re(z) = 1/2 for the first non-trivial zero. or did you leave out some important detail ? in general - under usual conditions - , a contour integral gives the sums of residues ( * 2pi i ) of all the 1/zero's (poles) inside its contour. so what value are we suppose to find when we compute : > Integrate numerically zeta'(z)/zeta(z) > dz around a suitable > contour with sufficient precision without assuming there is only one zero with that residue value ( * 2pi i = contour integral ) btw what suitable contour ? i assume (you meant) a limit of contour integrals where the limit of the contours approaches the zeta zero ? with this method i expect that i need the residue value(s) of the zero's ' still ' in the contours. what if the residues of the poles/zero's are all 0 or cancel out apart from the ones with Re(z) = 1/2. and the 'proof' would be more of an argument in the style of contours numbered lower real part upper real part 1 0.4 0.6 2 0.45 0.55 3 0.47 0.53 ... 0.499999...? 0.500...1 ? oo 1/2 ? 1/2 ? > -- > Robert Israel > israel(a)math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada sorry if im hard on you , but unless you clarify , it seems like an argument rather than a proof. its a good strategy at first sight. not wrong , but somewhat vague. suppose for instance we have 2 zero's off the critical line. and the related Res are + 1 and - 1. or even worse both are 0. contour A a + 1 - 1 contour B a but a = a + 1 - 1 ! similarly a = a + 0 + 0 i hope you understand what im trying to say. regards tommy1729
From: master1729 on 23 Sep 2009 07:24
i wrote : > Robert Israel wrote : > > > > Integrate numerically zeta'(z)/zeta(z) > > dz around a suitable > > contour with sufficient precision and you'll have > > shown that there is > > exactly one zero, not two, inside that contour. > > suppose for instance we have 2 zero's off the > critical line. > > and the related Res are + 1 and - 1. > > or even worse both are 0. > oh wait , Res have to be non-zero because you considered zeta'(s)/zeta(s) i assume. i assume thats why you took zeta'(s)/zeta(s) instead of 1/zeta(s) right ? now if the limit of the contours are taken assymetric around Re = 1/2 and the values of all the integrals are constant ... then we have a proof i assume ? however this reduces to finding the simplest way of proving those integrals to be equal. i suppose the trick is to adapt the contours so that the constant value is most easily proved. now i think i get israels approach. but still no full proof , i assume we need to continue like this ; 1) define contours and 2) prove constant value btw is this then really the easiest way ? did i get the full intention of roberts approach now ? regards tommy1729 |