vague conjecture
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From: master1729 on 2 Mar 2010 08:24 there is no 'simple' infinite product expansion(*) that doesnt use the zeta zero's in the critical strip , of zeta(s) valid exactly in R(z) >= 0. (*) means that the product expansion is a function of one variable only : thus f(z). i know , 'simple' is vaguely defined , hence the title : vague conjecture. however even without ' simple ' id like to see your results. tommy1729
From: David Bernier on 2 Mar 2010 20:23 master1729 wrote: > there is no 'simple' infinite product expansion(*) that doesnt use the zeta zero's in the critical strip , of zeta(s) valid exactly in R(z) >= 0. > > (*) means that the product expansion is a function of one variable only : thus f(z). > > i know , 'simple' is vaguely defined , hence the title : > vague conjecture. > > however even without ' simple ' id like to see your results. > > tommy1729 What I know about this is that if Re(z) > 1, then there's the famous Euler product formula zeta(z) = [ an infinite product involving all primes ]. The infinite product in question unfortunately diverges when Re(z) <= 1 , from what I remember. A good place to look is the book "The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso Alike" co-authored by Peter Borwein, Stephen Choi, Brendan Rooney and Andrea Weirathmueller. You require R(z) >= 0, which I assume to mean Re(z) >= 0 (correcting a typo.). So AFAIK the Euler product won't work and I don't have any real "results" for your requirements. David Bernier
From: master1729 on 2 Mar 2010 23:39 David Bernier wrote : > master1729 wrote: > > there is no 'simple' infinite product expansion(*) > that doesnt use the zeta zero's in the critical strip > , of zeta(s) valid exactly in R(z) >= 0. > > > > (*) means that the product expansion is a function > of one variable only : thus f(z). > > > > i know , 'simple' is vaguely defined , hence the > title : > > vague conjecture. > > > > however even without ' simple ' id like to see your > results. > > > > tommy1729 > > What I know about this is that if Re(z) > 1, then > there's the famous > Euler product formula > > zeta(z) = [ an infinite product involving all primes > ]. > > The infinite product in question unfortunately > diverges > when Re(z) <= 1 , from what I remember. that is correct , you remembered well. > > A good place to look is the book > "The Riemann Hypothesis: A Resource for the > Afficionado and Virtuoso Alike" > co-authored by Peter Borwein, Stephen Choi, Brendan > Rooney and Andrea > Weirathmueller. since you didnt solve the problem , and i assume you got that book , i am tempted to conclude the solution isnt in that book ... i like Borwein though. > > You require R(z) >= 0, which I assume to mean Re(z) > >= 0 (correcting > a typo.). So AFAIK the Euler product won't work and > I don't have > any real "results" for your requirements. > > David Bernier yes indeed , i mean 'Re(z)' ... we can argue if its a typo , because R(z) is sometimes used instead , though usually with a ' fancy R ' but i cant type weird symbols in this forum. and yes , you dont have a real " result " but thanks for posting politely anyways. regards tommy1729
From: David Bernier on 3 Mar 2010 15:54 master1729 wrote: > David Bernier wrote : > >> master1729 wrote: >>> there is no 'simple' infinite product expansion(*) >> that doesnt use the zeta zero's in the critical strip >> , of zeta(s) valid exactly in R(z) >= 0. >>> (*) means that the product expansion is a function >> of one variable only : thus f(z). >>> i know , 'simple' is vaguely defined , hence the >> title : >>> vague conjecture. >>> >>> however even without ' simple ' id like to see your >> results. >>> tommy1729 >> What I know about this is that if Re(z) > 1, then >> there's the famous >> Euler product formula >> >> zeta(z) = [ an infinite product involving all primes >> ]. >> >> The infinite product in question unfortunately >> diverges >> when Re(z) <= 1 , from what I remember. > > that is correct , you remembered well. > >> A good place to look is the book >> "The Riemann Hypothesis: A Resource for the >> Afficionado and Virtuoso Alike" >> co-authored by Peter Borwein, Stephen Choi, Brendan >> Rooney and Andrea >> Weirathmueller. > > since you didnt solve the problem , and i assume you got that book , i am tempted to conclude the solution isnt in that book ... > > i like Borwein though. > >> You require R(z) >= 0, which I assume to mean Re(z) >>> = 0 (correcting >> a typo.). So AFAIK the Euler product won't work and >> I don't have >> any real "results" for your requirements. >> >> David Bernier > > yes indeed , i mean 'Re(z)' ... > > we can argue if its a typo , because R(z) is sometimes used instead , though usually with a ' fancy R ' but i cant type weird symbols in this forum. Yes, I know about the fancy "R". I was thinking about your question last night. You ask for an infinite product valid for Re(z) > = 0. There are infinitely many zeros that aren't easy to find (the non-trivial ones) in the critical strip. The critical strip is sometimes described as where 0 < Re(z) < 1, and other times (I think) as 0 <= Re(z) <= 1. Either way is fine by me, since it was proved (Hadamard, de la Vallee-Poussin and independent of each other) that if Re(z) = 1, then zeta(z) =/= 0. By the functional equation for zeta and a few other facts, it follows that if Re(z) = 0, the zeta(z) =/= 0. If you don't believe in infinity, then you can say that 10^13 non-trivial zeros with Re(z) = 1/2 have been located in a rectangle 1 by H, where H is basically the imaginary part of the highest zero found in that lot of 10^13 zeros. The rectangle has vertices at 0, 1, H*i and 1+ H*i. All the zeta zeros in that rectangle have a real part equal to 1/2 . For a finite product A(z) B(z) C(z) D(z) of analytic functions in some region, if the product A(z) B(z) C(z) D(z) = 0, then at least one of A(z), B(z), C(z) and D(z) is zero. As for what one can say about your question of an infinite product for zeta(z) valid at least in the critical strip and that doesn't use the zeta zeros in the critical strip, I'd have to think more. Infinite products are more complicated. David Bernier
From: master1729 on 6 Mar 2010 07:58
David Bernier wrote : > If you don't believe in infinity, then you can say > that 10^13 non-trivial... i do believe in infinity. > I'd have to think more. Infinite products are more > complicated. > > David Bernier given up ? regards tommy1729 |