# Thoughts needed for this "proof" of riemann s hypothesis

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• Aug 14th 2010, 01:56 PM
chiph588@
Quote:

Originally Posted by jlb
In the case of equation (4), I believe the Laplacian ends up performing a similar "trick" by adding two infinite series for which the terms in $1/p^a$ cancel out due to the fact that the Poisson kernels have vanishing Laplacian.

Could you explain a little more in depth? I'm a little too busy to sit and think. Also have you addressed the point I made about the zeros forming a cusp?
• Aug 14th 2010, 03:00 PM
Vlasev
Quote:

Originally Posted by jlb
I believe the Laplacian ends up performing a similar "trick" by adding two infinite series for which the terms in $1/p^a$ cancel out due to the fact that the Poisson kernels have vanishing Laplacian.

You cannot just believe something like that. You need to rigorously justify it. For now it's wrong. As for the trick itself, this is what I mean.

You are taking the laplacian of a sum that does not converge absolutely (EDIT: uniformly) on the range 1/2 <Re(z) < 1. I don't think the trick could work because you are doing 2nd order derivatives on the function and THEN you are adding up the results. First you need to show absolute (EDIT: uniformly) convergence for the series for the given range so that you can change the order of differentiation and then once you find the 1st order derivatives, you need to assure absolute (EDIT: uniform) convergence for the results on the interval so that you can change the order again. The trick you are talking about comes after you have made a total of 4 illegal changes of order. It is very unlikely to work.

The Dirichlet series of Zeta is derived by adding two series, yes. However, you are hoping that differentiating would do the trick? I highly doubt it.
• Aug 16th 2010, 03:23 AM
jlb
Quote:

Originally Posted by Vlasev
You cannot just believe something like that. You need to rigorously justify it. For now it's wrong. As for the trick itself, this is what I mean.

You are taking the laplacian of a sum that does not converge absolutely (EDIT: uniformly) on the range 1/2 <Re(z) < 1. I don't think the trick could work because you are doing 2nd order derivatives on the function and THEN you are adding up the results. First you need to show absolute (EDIT: uniformly) convergence for the series for the given range so that you can change the order of differentiation and then once you find the 1st order derivatives, you need to assure absolute (EDIT: uniform) convergence for the results on the interval so that you can change the order again. The trick you are talking about comes after you have made a total of 4 illegal changes of order. It is very unlikely to work.

The Dirichlet series of Zeta is derived by adding two series, yes. However, you are hoping that differentiating would do the trick? I highly doubt it.

Poor choice of words on my part using "quick reply" late at night... (I should have skipped the "I believe"). Here is a different way to look at the problem which avoids the "illegals" you mention, followed by the justification for my statement about the terms in $1/p^a$ cancelling out:

1) Assume that I defined a real function of two real variables a, b.

$F \left( a, b \right)
=\zeta \left( 2a \right) \prod^{\infty}_{p=primes}
\left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right)$

-zeta is the real function zeta of a.

It is known that:
-F(a,b) converges absolutely for a>1, and diverges for a<1/2.
-The status of convergence of F(a,b) (divergent? conditionally convergent? convergent?) is as yet unknown for 1/2 < a < 1 (do not underestimate how difficult the status is to find in that range, for the infinite product)
-F(a, b) happens to equal the value of $\left( | \zeta \left( a, ib \right) | ^2 \right)$ everywhere for a>1.

2) Applying the Laplacian to $log( \frac{F(a,b)}{\zeta(2a)})$ in the range a>1 is a perfectly "legal" operation (since both F(a, b) and zeta(2a) are real and always positive in that range) but happens to yield an infinite series over all primes that absolutely converges for a>1/2. This is what puzzles me. If, as you stated, the function F(a, b) clearly diverged in the range 1/2<a<1, then performing a simple step like taking the Laplacian of its logarithm should not give an infinite series that converges in that range, yet it does.

More importantly though, while the series converges in the range 1/2<a<1, does it still equal $\nabla^2 \ln \left( \frac{| \zeta \left( a, ib \right) | ^2 } { \zeta \left( 2a \right) } \right)$ in that range? (as it does for a>1). This I already should have tried numerically but I lack proper knowledge/software to do it inside the critical strip. If the results don t match, then obviously the problem is over.

With respect to my statement as to "why" the Laplacian of the log converges for a>1/2 (instead of simply converging for a>1), it is precisely because the operation cancels out the terms in $1/p^a$. This is shown by using:

$\nabla^2 \ln \left( f(a, b) \right) = - \frac{(\frac{\partial f(a,b)}{ \partial da})^2+(\frac{\partial f(a,b)}{ \partial b})^2}{(f(a,b))^2} + \frac {\nabla^2 \left( f(a, b) \right) }{f(a,b)}
$
(5)

where f(a, b) is real and positive.

Let $f(a, b) = \left(\frac {p^{2a}-1}{p^{2a}-2 p^{a}\cos(b\ln{p})+1} \right) = \left(1 + 2 \sum^{\infty}_{j=1} \frac {\cos \left( jb \ln {p} \right)}{p^{ja}} \right)$

(See eq 2.16 in attachment for the identity on the RHS, the proof is straightforward but tedious in latex.)

If you compute $\nabla^2 \ln \left( f(a, b) \right)$, you will find that the only terms left of order $1/p^a$ in equation (5) come from the second order derivatives in $\frac {\nabla^2 \left( f(a, b) \right) }{f(a,b)}$. Fortunately these cancel out because $\nabla^2 \left( f(a, b) \right) = 0$.
• Aug 18th 2010, 12:31 PM
jlb
"problem" solved...

The laplacian of Log (Zeta Zeta*) vanishes to 0 when one uses the Dirichlet Eta function (being careful not to interchange terms in the series since it is conditionally convergent), leaving only the laplacian of the real function Zeta(2a), expressed as a series which converges for a> 1/2, and is independant of the imaginary part, of course...a trivial equation, as it should be.

Thanks for you help and energy.
• Aug 18th 2010, 01:51 PM
chiph588@
Quote:

Originally Posted by jlb
"problem" solved...

The laplacian of Log (Zeta Zeta*) vanishes to 0 when one uses the Dirichlet Eta function (being careful not to interchange terms in the series since it is conditionally convergent), leaving only the laplacian of the real function Zeta(2a), expressed as a series which converges for a> 1/2, and is independant of the imaginary part, of course...a trivial equation, as it should be.

Thanks for you help and energy.

What about the zeros of $|\zeta(s)|$ forming a cusp in three space? This is an issue that needs to be addressed.
• Aug 18th 2010, 01:53 PM
chiph588@
Quote:

Originally Posted by jlb
"problem" solved...

The laplacian of Log (Zeta Zeta*) vanishes to 0 when one uses the Dirichlet Eta function (being careful not to interchange terms in the series since it is conditionally convergent), leaving only the laplacian of the real function Zeta(2a), expressed as a series which converges for a> 1/2, and is independant of the imaginary part, of course...a trivial equation, as it should be.

Thanks for you help and energy.

What about the zeros of $|\zeta(s)|$ forming a cusp in three space? This is an issue that needs to be addressed.
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