zeta zero
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From: Robert Israel on 23 Sep 2009 15:40 master1729 <tommy1729(a)gmail.com> writes: > 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 ? The Argument Principle: If f(z) is meromorphic in a domain D and C is a simple positively-oriented contour in D, not passing through any pole or zero of f, then int_C f'(z)/f(z) dz = (2 pi i) (n - m) where n is the number of zeros and m the number of poles of f inside C, counted by multiplicity. Since you know the result has to be an integer times 2 pi i, all you have to do is make sure your numerical error in your integral has absolute value less than pi. In this case, we know that there is a pole at z = 1, and it's the only pole. Take the rectangular contour with corners at, say, -i, 2-i, 2+15 i, 15 i. In Maple: q:= unapply(diff(Zeta(z),z)/Zeta(z),z): evalf(Int(q(-I+t),t=0..2) + I*Int(q(2+I*t),t=-1..15) - Int(q(15*I+t),t=0..2) - I*Int(q(I*t),t=-1..15)); .4e-9-.6e-9*I Clearly this should be 0. Since we know there is one pole within the contour, there is one zero as well. If you replaced 15 by 13, Maple would get .4e-9-6.283185307*I for the integral, which should thus be - 2 pi i. Thus there is no zero in the strip with imaginary part from -1 to 13, but there is one with imaginary part from 13 to 15. -- 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 27 Sep 2009 12:13
very nice. but a few more questions if you dont mind. what if the contour by accident goes through a zero ? can we express the imaginary part of a zero by an integral ? what if a zero has a multiplicity of 2 ? i dont believe any zero does have multiplicity 2 but i dont know if that has been proven. and if multiplicity 2 turns out , might that not give the illusion that 2 distinct zero's are found by the contour integral ? are there other good ways to find zero's in a rectangle apart from contours ? can we easily define contours such that they dont find 2 zero's at once ? ( zeta zero's can be close to eachother ) regards tommy1729 |