
#55
Sep1405, 03:48 AM

Sci Advisor
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P: 9,398

And I further see no reason why your method picks out exactly the nontrivial zeroes and nothing else. but then the dependence on zeta has never been explained.
so, if you want to make this metho work: take g(z) some function frmo C to C. explain how we may take a restricted set of its zeroes, call them T, and from T create a differential operator [tex]L=\partial^2_x + V_g[/tex] such that the spectrum is exactly T now specify the conditions on g and T that allow us to conclude L is hermitian, and hence T must be a set of real numbers. You have not in any decipherable way done any of those things. I am perfectly willing to believe that it can be done. but I have not seen you do it in a way that convinces me at all. 



#56
Sep1405, 07:45 AM

P: 501

let be perturbation theory:(at first order of the energies)...
[tex]E_{n}E^{0}_{n}=\delta{E(n)}=<\phiV\phi>[/tex] (1) where E_{n} are the "energies" roots of the Riemann Z(1/2+is) and E^0_{n} are the Eneriges of the Hamiltonian H0=P^{2}/2m then (1) provides an integral equation for V we could find a resolvent kernel R for this so: [tex]V(x)=\int_{\infty}^{\infty}dnR(n,x)\delta{E(n)} [/tex] with the resolvent Kernel: [tex]R=\sum_{m=0}^{\infty}b_{m}(KcI)^{m} [/tex] is the Taylor expansion of the operator K^{1} and K^{m} is the mth iterated kernel,I is the identity operator and c is a real constant so the series converge for K<c [tex]\delta{E(n)}=K[V(x)] [/tex] c is a constant so the series converges,the radius of convergence is given by the norm of K operator K<c we have proved that V exist at first order in perturbation theory... with K the kernel given by [tex]K(x,n)=\phi(x,n)^{2} [/tex] phi are the eigenfunctions of the Hamiltonian H0 epending on n i have applied Neumann series to the operator to obtain the resolvent Kernel... 



#57
Sep1405, 11:25 AM

Sci Advisor
P: 41

I'll let Matt continue to point out the circularity of the argument being presented here adn tackle it from a technical viewpoint but will add one other little piece of information to the discussion, just to demonstrate the futility of this in a completely different way.
Supposing this were a valid proof, then credit for it should go to other people who have discovered it long ago. The connection between certain quantum systems and the RH is even wellknown enough to be discussed at length in some of the recent popular books on the Reimann Hypothesis. So supposing this proof were correct, then it has already been demonstrated by numerous other people. But seeing as it's not a valid proof (which all these other people have recognized), there's not much to worry about. None of those people are going to claim you "stole their proof". 



#58
Sep1405, 11:43 AM

P: 501

if they discovered the proof long ago...why did they not publish it?...in a post above i have proof that the potential exist and even calculated it to first order in perturbation theory depending on [tex]\delta{E(n)}[/tex] and the E_{n} and E^0_{n} are known even more if we call Z the inverse of the function [tex]\zeta(s)[/tex] then we could write: [tex]E_{n}=i(1/2Z(0))[/tex] (i have oly inverted the function, and i have used simple integral equation theory to prove the existence of the potential....
The problem with Matt (as happen with most of math teachers) is that they have assumed certain conceptions in math and if you are out of these,you are nothing,i don,t know what argument will now matt grime have to say my maths are wrong,but is only an "approach to the potential" (is would be only correct to first order in perturbation theory, the whole serie of values of energy is perhaps even divergent) and the WKB is also an approach,to say that we can choose some functions that are on L^{2}(R) function space....[tex]\psi=Asen(S(x)/\hbar) [/tex] for example. another question as we are dealing with rigour an other things...can anyone of you brilliant,smart intelligent mathematician to prove the existence of infinitesimals,i,ll put even more easier,write (with numerical value) an infinitesimal... 



#59
Sep1405, 02:51 PM

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P: 1,123

Hyper real numbers have infinitesimals in them:
http://mathforum.org/dr.math/faq/ana...yperreals.html http://en.wikipedia.org/wiki/Hyperreal_number But what has that got to do with this? Anyway, eljose, I am unsure if the mathematics in your post makes any sense or not, it could well do, but I can clearly see it does not show that all s in the critical line of [itex]\zeta (s) = 0[/itex] satisfy [itex]\Re (s) = 1/2[/itex]. And with all due respect, my 12 year old brother could see sentences in your proof that logically made no sense. 



#60
Sep1405, 03:51 PM

P: 501

hyperreal numbers including infinitesimal..but could you write "an" infinitesimal?..
i have shown that the potential (and calculated it too) for [tex]\zeta(1/2+is)[/tex] is real,(the proof is that for any existing E_{n} also E*{n}=E_{k} is also an energy from this we deduce using the expected value of the Hamiltonian that V=V* so V is real,as you can see this only happens with a=1/2 the other cases there are complex energies in the form E*{n}+(2a1)i, but a complex energies will come from a complex potential so [tex]\zeta(a+is)=0[/tex] can not have any real root except a=1/2 that have all the roots real as the potential is real) then i have also calculated (given an integral expression for) the potential upto first order in perturbation theory. I am checking my grammar and spelling to make it the most clearer of possible zurtex,i am not ofended . 



#61
Sep1605, 09:21 AM

P: 60

Oh Jesus Christ and Holy Maria!!!
What a thread what a thread????!!!!! I cant spare my time for this post so I just refer to the last post on V.S.'s "proof" of Fermat. 



#62
Sep1605, 11:41 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

Can I jump in just to point out that you can't prove a theorem in MATHEMATICS by using PHYSICS? The truth of a mathematics statement does not depend upon whether the physics statements are true or not.




#63
Sep1605, 01:33 PM

Sci Advisor
P: 41

People can already calculate all the zeros of the zeta function. But it will take an infinite amount of time, just as your approximate processes might be able to do the same calculation if you do it properly, but it still has nothing to do with a proof because an infinite process will never be considered a proof seeing as you can't complete it. People have been pretty patient with you and tried to help you understand why what you are doing is not proof but you seem to refuse to even learn what "proof" means in mathematics and, instead, complain that mathematicians won't listen to you. Why should anybody listen to you tell them how to do their work when you have demonstrated that you don't understand what their work is? Given that people have been trying to help you, don't you think it is pretty rude of you to refuse to listen or learn? 



#64
Sep1905, 04:20 AM

P: 501

To Hallsoft Ivy:this problem can be seen in a mathematical way as the HilbertPolya conjecture (a selfadjoint operator with real potential that has its eigenvalues as the roots of the function [tex]\zeta(1/2+is)[/tex],that is to prove that exist a real potential and a self adjoint operator with a given V so all its eigenvalue are the roots of the function zeta evaluated at 1/2+is,then i prove this is selfadjoint (as the potential is real and a Hamiltonian with a real potentia has all its eigenvalues real) for the cases a+is with a different from 1/2 there are compplex roots so the potential can not be real...all the steps i made are justified mathematically i have not introduced any physical or empirical proof. (you can view H as a selfadjoint operator,and WKB solutions as the solutions to an equation of the form:
ey``+f(x)y=0 with e an small parameter e<<1 as you can see all is math there To symplectic manifold...if this proof were made by an universtiy smart and snob teacher i,m sure that you would accept it... To David:i have proved that the potential exist and have given an expression for it in the form: [tex]V(x)=\int_{\infty}^{\infty}dnR(n,x)\delta{E(n)}[/tex] (1) (although of course this is only an approximation,you can calculate it in finite time by integrating upto a N finite so you only should need to calculate a finite (but big ) number of roots of Z(1/2+is)). as you can see the potential exists and can be calculated,also for a=1/2 the potential is real i have calculated all the necessary steps to prove RH but if they don,t want to give you the fame and the prize because you are not famous they will invent any excuse....i am sure that if my steps (as i have said before) were made by a famous mathematician from a famous university the RH would have been proved long ago...as i have told before at least Gauss,Euler and others were given an opportunity to publish their ideas....my teachers don,t want to know anything from me or give me the chance to prove RH... From the mathematical point of view my proof of RH is similar to this: a differential operator [tex]H=aD^{2}+V(x) [/tex] where we must choos V so the eigenvalues of this operator are precisely the roots of [tex]\zeta(1/2+is)=0[/tex]. for the potential V knowing some of the energies E_{n} we could obtain V in an integral form as expressed by (1). for a small a<<1 (in our case is m>>h) the WKB solutions are the solutions of a differential equation [tex]ey´´+f(x)y=0[/tex].. i have not made any physical assumptions at all. 



#65
Sep1905, 06:17 AM

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P: 1,123

Let K be the set of all solutions that satisfy [itex]\zeta (s) = 0[/itex]. Let there exist some p such that:
[tex]\frac{\partial^2 p}{\partial i^2} a^3  \Gamma (i^2) = 0[/tex] Now I can prove that these exists a solution [itex]i = f_k (p)[/itex] therefore i exists. If i exists there must be a limit to g(p) and p approaches infinity and therefore p exists. Thus the roots of: [tex]\int_0^\infty \frac{g(s)}{p \Gamma(s)} ds = \sin i[/tex] Are synonyms to the Zeta functions roots and all have roots of [itex]\Re (p) = 1/2[/itex] Therefore RH is proven and no one can come up with a counterexample. 


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