My Math Forum Riemann non-trivial zeroes

 Number Theory Number Theory Math Forum

 July 9th, 2011, 01:18 PM #1 Senior Member   Joined: Jan 2011 Posts: 120 Thanks: 2 Riemann non-trivial zeroes I guess there is a flaw in the following logic somewhere, but would appreciate some sharp eyes to help me point out where it goes wrong. The Zeta-function can be written as the following infinite Hadamard product of the non-trivial zeroes: $\zeta(s)= \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$ but this also means that: $\zeta(1-s)= \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$ The first infinite product clearly shows the pole at s = 1, the trivial zeroes at ?2, ?4, ... (Gamma function) and the non-trivial zeroes when s = ?. Please note that this doesn't provide any hint on where these non-trivial zeroes (?'s) are located. Now let's take the famous reflection formula: $\zeta(s)= 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$ and substitute the Hadamard products for $\zeta(s)$ and $\zeta(1-s)$. The result is that all terms nicely balance out into: $\prod_\rho \left(1- \frac{s}{\rho} \right)= \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$ That's the easy part and the obvious next question is what this equation could tell us about the location of the zeroes. The ?'s could lie anywhere in the already proven band between $0<\Re(\rho)<1$ and they are basically to be taken as 'givens' for any further analysis. So, let's therefore look at the equation itself and try to solve it for s. Any information about the solutions for s will tell us more about $\frac{s}{\rho}$ i.e. one of the ?'s needs to match a solution of s to make $\zeta(s)=0$. In this approach the solution for s 'dictates' the location (or the shape) of the ?'s (and not the other way around). I started to experiment with subsequent solutions for s with a different number of terms and $a,b,c, \dots \in \mathbb{C}$: $\left( 1- \frac {s}{a} \right) \left( 1- \frac {s}{b} \right) \dots= \left( 1- \frac {(1-s)}{a} \right) \left( 1- \frac {(1-s)}{b} \right) \dots$ and obviously found that  always is a solution. But since $\zeta(\frac12)$ is not a zero,I just ignore that outcome. I also found that adding subsequent terms to the product, does add more solutions for s, however all these solutions seem to be of the form: $\frac12 \pm x \sqrt(\...)$ for whatever a,b,c,... (in casu the ?'s) I pick. And that is of course an interesting result (if correct). Can anybody find a counter example of these products where such solution for s does not start with $\frac12 \pm \...$ ? P.S. One of the stranger outcomes of my experiments is that when I take a,b,c,... all in the form $\frac12 + y i$, then the outcome is always real ! (i.e. the $i$ disappears from the solution contrary to when I take f.i. $\frac13 + y i$ as input). I find this strange, since one would expect the outcome of the infinite multiplication of all ?'s (if they indeed all are lying on the critical line...) to yield all solutions for s that then are exactly equal to each individual ? (hence I would expect a complex rather than a real number as the result for s).
 July 10th, 2011, 04:06 AM #2 Senior Member   Joined: Jan 2011 Posts: 120 Thanks: 2 Re: Riemann non-trivial zeroes P.S. I know I am probably going way out of line now, but the reasoning above could be extended as follows: If indeed all solutions for s in: $\prod_\rho \left(1- \frac{s}{\rho} \right)= \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$ are of the shape  (thereby making the Riemann hypothesis true) and $\zeta(s)$ being valid for any $s \in \mathbb{C}$ (except s=1), then this implies that all numbers (real or complex) are just solutions of the equation with the "conjugated" shape  (e.g. 3 = 0.5 + 2.5 and its conjugate solution is -2 = 0.5 - 2.5). So, this means that just as the primes are the unique multiplication seeds for any composite integer, the ?'s are the basic building blocks for any complex/real number (s) that emerges as a solution from the infinite product equation above. But unlike the primes, that are only capable to produce integers and therefore never can even come close to generating 'one self', the ?'s do produce solutions for s that will come infinitely close to each them. That is the moment when a non-trivial zero (that is a limit anyway) emerges. It is the point where the ?'s generate a solution for s that is so close to a ? that it 'bites in its own tail'. So, primes or any other number are just higher level "conjugated" manifestations that originate from the ?'s as solutions for s in the formula above. (similar to the Faculty just being a higher level integer instance or 'correspondence limit' of the Gamma-function). And that would make $\frac12$ the start of everything...
 July 10th, 2011, 07:09 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Riemann non-trivial zeroes Does $\prod_\rho \left(1- \frac{s}{\rho} \right)= \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$ mean $\frac{p\,-\,s}{p}\,=\,\frac{p\,-\,1\,+\,s}{p}$ ?
July 10th, 2011, 07:57 AM   #4
Senior Member

Joined: Jan 2011

Posts: 120
Thanks: 2

Re: Riemann non-trivial zeroes

Quote:
 Originally Posted by greg1313 Does $\prod_\rho \left(1- \frac{s}{\rho} \right)= \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$ mean $\frac{p\,-\,s}{p}\,=\,\frac{p\,-\,1\,+\,s}{p}$ ?
Greg,

To be more precise the equation should be:

$\prod_{\rho_1}^{\rho_\infty} \left(1- \frac{s}{\rho_n} \right)=\prod_{\rho_1}^{\rho_\infty} \left(1- \frac{(1-s)}{\rho_n} \right)$

with ${\rho_n}$ being the n-th non-trivial zero of the Zeta-function (it has already been proven that there are an infinite number of these zeroes).

You can of course rewrite this into:

$\prod_{\rho_1}^{\rho_\infty} \left(\frac{{\rho_n} - s}{\rho_n} \right)=\prod_{\rho_1}^{\rho_\infty} \left(1- \frac{\rho_n - 1+ s)}{\rho_n} \right)$

Even though we assume (or hope) that all non-trivial zeroes have a real part of $\frac12$, the $\rho_n$ could be any number with a real part between 0 and 1. So, as an example you can try to solve the equation for e.g. just 3 terms like this:

$\left( 1- \frac{s}{\frac13 + 2i} \right)\left( 1- \frac{s}{\frac13 + 4i} \right)\left( 1- \frac{s}{\frac13 + 6i} \right)=\left( 1- \frac{1-s}{\frac13 + 2i} \right)\left( 1- \frac{1-s}{\frac13 + 4i} \right)\left( 1- \frac{1-s}{\frac13 + 6i} \right)$

This gives 3 solutions for s:



My conjecture is that all solutions for s will be of the shape $\frac12 \pm x \dots$ and thereby dictate that all $\rho_n$ must have the same form (otherwise they would not induce a zero).

 July 19th, 2011, 11:05 AM #5 Senior Member   Joined: Jan 2011 Posts: 120 Thanks: 2 Re: Riemann non-trivial zeroes Whether or not this has anything to do with the Riemann hypothesis, the question remains why this particular equation: $\prod_{n=1}^y \left(1- \frac{s}{(x + ni)} \right) = \prod_{n=1}^y \left(1- \frac{(1-s)}{(x + ni)} \right)$ that can be simplified as: $\prod_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$ only seems to have real solutions for s when $x=\frac12$. Even a single x-term in this sequence deviating from this value or taking $x= \frac{1}{2.0000001}$, immediately turns all solutions for s into imaginary numbers (except for ). Of course the sequence of the n's can be of any order, I even used n=0 for all terms and the result is the same. It always gives only real values as solutions for s when $x= \frac12$. Studying various values (0 < x < 1) and numbers of terms (up to y=200) in more detail, I did note that all terms under the square root of the solutions $\frac12 \pm \sqrt{(...)}$, are always containing sums of terms that have even powers of $i$ (i.e. $i^2, i^4, i^6, \dots$), when $x= \frac12$. This of course does annihilate the complex values, however it doesn't proof that the overall real number (the sum of all values) couldn't become negative (thereby inducing another complex solution when the square root is taken). All values other than $x=\frac12$ always leave at least one 'widowed, oddly powered' $i$ behind, hence the square root could never turn into a real value. EDIT: I also asked the same question at MathOverflow and it seems that an expert (GH) has found an elegant proof for this conjecture http://mathoverflow.net/questions/70...-zeta-function

 Tags nontrivial, riemann, zeroes

### riemann hypothesis how to produce the nontrivial zeroes

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post hopefullearner Complex Analysis 2 January 23rd, 2012 12:31 AM Gustav Number Theory 3 February 27th, 2011 01:58 PM AlexBot Number Theory 2 November 10th, 2010 02:43 PM vampira673 Complex Analysis 0 February 19th, 2010 04:27 PM cheloniophile Real Analysis 1 November 23rd, 2008 05:30 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top