# Recommend a Journal

1. Sep 21, 2005

### eljose

hello i would like to open this post to recommend the english journal "Journal of Physics:A General mathematics", they have published my article in spite of me being a non-english speaker,they accept paper in .doc (Microsoft Word) format, when i give me the web address for my article i will put it in the forum...in fct is about RH i find an operator

$$H=cD^{2}+V(x)$$ so all its eigenvalues are the roots of $$\zeta(a+is)$$ where the potential can be written to first order in the form:

$$V(x)=\int_{-\infty}^{\infty}dnR(n,x)\delta{E(n)}$$

with E_n the roots of $$\zeta(a+is)$$ so the potential will depend also on a,for a different from 1/2 we have complex energies in the form $$E*_{n}+(2a-1)i$$ so the potential is complex and a complex potential can not have real roots then necessarily all the roots of $$\zeta(a+is)$$ have real part a=1/2

Last edited: Sep 21, 2005
2. Sep 21, 2005

### matt grime

wow, you can expect that million dollars then, if it is indeed prestigious.

no chance that it's a bad jounral? i can name two such vanity journals, one now defunct called the SWAPM, as well as that one run by the society of egomaniacs.
that was incredibly quickly reviewd by the way; usually it takes a months if not a year or two. have they accpeted submission or are you saying that they are gonig to publish it?

and if your result is as stated then it is false (you never explain why the trivial zeroes are magically not detected), as we keep pointing out to you.

edit: looks genuine; i know someone at IoP.

edit: nope, that was a differernt journal.

Last edited: Sep 21, 2005
3. Sep 21, 2005

### eljose

i don,t want to enter in discussion with you as surely Hurkyl would close the thread in that case,(please Hurkyl i promise not to insult anybody and to answer all the critics politely)..

As i said the non-trivial zeros satisfy that if $$\zeta(1/2+is)$$ is a zero,then also $$\zeta(1/2+is*)$$ is also a zero (and this only happens for the non-trivial zeros),try with a trivial zero,it will of the form: (2n+1/2)i being n an integer (sorry if the sign is not correct at all) then you have that its conjugate (2n+1/2)(-i) is not a zero..the example is that -2 is a zero but 3 is not,-4 is a zero but 5 not and so on....

and as for the potential i have calculated in the integral form given....

$$V(x)=\int_{R}dnR(n,x)\delta{E(n)}$$

i don,t know if the journal is serious or not..they have referees..and yes they have rejected some of my manuscripts,i think you should visit the journal by yourself to raise an opinion.

i don,t think i will earn the 1000000\$ prize but perhaps due to have an article published my math and physics teachers will take me into consideration and give the opportunity to do my PhD... i think that the prize should be distributed among those like me have made a real proof of RH.

4. Sep 22, 2005

### matt grime

that is the one i found. your paper wasn't listed on those published this month, nor on the list of those to be published. if it is published i look forward to reading it.

so, they have accepted it for publication, not just for review?

your comment on the non trivial zeroes doesn't answer anything by the way. you have merely by fiat declared that the zeroes not on the critical line are the trivial ones and the trivial ones alone. I need a reason why it picks out exactly the non-trivial zeroes in terms of its definition as a fucntion of zeta.

As i have said before this method is well known already (an approach of Berry et al, using Random Matrix Theory for example), so i have no problem with this being theoretically possible; I have issues with the fact that you've never explained it in way that can be understood.

the journal title is, incidentally, Journal of Physics A: Mathematical and General. Not the title you gave, hence the confusion.

5. Sep 22, 2005

### eljose

As for the zeroes as i explained before if s is a zero so $$\zeta(1/2+is)=0$$ then also $$\zeta(1/2+is*)=0$$ but this argument is only valid for the non-trivial zeros because s=(2n+1/2)i is a zero but s*=(2n+1/2)(-i) is not.

if you think this journal is not "respectable" i wil send my manuscript to "Physical review letters" journal, ah and i have scanned the copyright agreement for the "Journal of Physics" i will put it in an attached file when i receive it...

As for the method of mine being used , as i said before if the method of Random Matrices or other quantum or math method used to prove RH is true then i wouldn,t mind sharing the prize in case they consider it that all method exposed are valid.

6. Sep 22, 2005

### matt grime

IoP is a very respectable institution.

Your explanation for the non-trivial zeroes is still flawed. From first reading It presumes that all the non-trivial zeroes lie on the critical line, which is what the Riemann hypthesis asks you to prove. Why must the non-trivial zeroes satsify your criterion for s and s*? YOu've given no reason for this, still.

I'm going to guess that this has something to do with the known analytic properties of the zeta function (though not known to me; this is one of the common problems in your presentation: you never explain what's going on clearly). That is in the critical region there is some symmetry such that any zeroes are placed with refelective symmetry in the real axis and the line re(z)=1/2. Is that what you're appealing to? If so, state that, that is all you had to do.

Now, given that is true, then how do you conclude that the potential so produced gives an hermitian operator so that the zeroes are necessarily real?

Last edited: Sep 22, 2005
7. Sep 22, 2005

### matt grime

So, to summarize, we have a function,c all it f, here f(s)=zeta(1/2+is), and a (sub)set of its zeroes that is colsed under conjugation, ie s is a zero iff s* is a zero. Now, you magic a potential V(x) so that there is a differential operator

$$L=\partial^2_x + V(x)$$

so that L's sepctrum is precisely these zeroes.

Fine, but now you conclude that L is hermitian so that the s must be real.

However, going frmo memory, there is nothing to stop me doing the same thing to the fucntion $g(s)=\sin(2\pi is)$, certainly (a relevany subset of) its zeroes are closed under conjugation, apply this construction of yours and conclude that all the zeroes are real. However we know that is false. So what properties of the zeta function did you use to conclude that L was hermitian?

8. Sep 22, 2005

### eljose

first of all is proved that the non-trivial zeroes have real part a with 0<a<1 (are on this critical strip) this is a proved result in number theory.

second you say $$g(s)=sin(2i\pi{s})$$ that has its roots as s=ni (they are pure complex numbers) but with the non-trivial zeros does not happen the same as they can not be pure imaginary,the zeros of $$\zeta(1/2+is)$$ can,t be purely imaginary as if s=b(n)i then would exist a b(n) so $$\zeta(1/2-b(n))=0$$ and this only happens with the trivial zeros of Riemann function... but not with the non trivial ones.

and other thing the case that s and s* are roots of $$\zeta(a+is)$$ only happens with a=1/2 for the other cases are complex energies in the form
s*+(2a-1)i so the only possibility of the zeroes being real is a=1/2 for your function try with a function $$f(a+ix)$$ with f and a real,you can,t find any counterexample with x and x* being roots of it and x being imaginary....

ah and in case there are non-trivial roots of $$\zeta(1/2+is)$$ then their complex part must satisfy $$|b(n)|<1/2$$ as the non-trivial roots of the Riemann function $$\zeta(s)$$ have real part 0<Re(s)<1
and for the cases $$\zeta(a+is)$$ we have the expression for the potential: $$V(x)=\int_{\Re}dnR(n,x)\delta{E(n)}$$ we have that for a different from 1/2 are energies of the form s*+(2a-1)i so the potential will be complex..but a complex potential can not have real roots as using Ehrenfrest.s theorem $$i\hbar{d<b>}=<[H,b]>dt$$ x and p do not commute so the expected value of b (the complex part of the potential is never 0 so there are no real energies for the case $$\zeta(a+is)$$.

Last edited: Sep 22, 2005
9. Sep 29, 2005

### matt grime

you really need to improve the standard of your presentation. I am not sufficiently interested to try and decode what you're talking about there (you repeatedly use s to mean different things). I still can't see anything which rules out applying your method to the function sin(is): the zeroes are symmetrically distributed about some line (if it bothers you about them being pure imaginary try a more interesting change of variable).

10. Oct 11, 2005

### eljose

these f**** journal has cheated on me..they said they were going to publish my manuscript and noww i have received a message saying it had been an error and rejecting my manuscript,these snobs don,t want to publish my papers,perhaps if i were a famous teacher the RH would have been solved months ago..

If you are interested for the case a=1/2 in $$\zeta(1/2+is)$$ the REAL potential is:

$$\int_{-\infty}^{\infty}dnR(x,n)[g(n)+g*(n)-E_{n}^0]$$

where g(n) is $$g(n)=\sum_{p}\alpha(n)C(n)\delta(n-p)$$

with $$C(n)=i[1/2-\zeta^{-1}_{n}(0)]$$

the sum "p" is made over integers so g(n) generates the roots a(n)+ib(n) with ab>0 and the case ab=0

and alpha is a function with value 1 if C is complex and 1/2 if C is real.

Last edited: Oct 11, 2005
11. Oct 11, 2005