Non trivial zeros of the Zeta function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:24:30Z http://mathoverflow.net/feeds/question/70460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70460/non-trivial-zeros-of-the-zeta-function Non trivial zeros of the Zeta function Agno 2011-07-15T19:34:44Z 2011-08-11T07:51:36Z <p>The Zeta-function can be written as the following infinite Hadamard product of its non-trivial zeroes: </p> <p>$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$</p> <p>this also implies that:</p> <p>$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$</p> <p>Take the reflection formula: </p> <p>$\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$</p> <p>and substitute the Hadamard products for $\zeta(s)$ and $\zeta(1-s)$. The result is that:</p> <p>$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$</p> <p>that can be rewritten as:</p> <p>$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$</p> <p>This equation has zeros for $\rho = s$ as long as $2\rho-1 \ne 0$.</p> <p>The ρ's could obviously lie anywhere in the already proven strip between $0&lt;\Re(\rho)&lt;1$ and I just take them as 'givens' wherever they might be located. </p> <p>I started to experiment with solving s for different numbers of terms as follows:</p> <p>$x=\frac13$ (e.g.)</p> <p>$\prod_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$</p> <p>Whatever value I pick for x between 0 and 1, the solution is always a complex number (ignoring $s=\frac12$, that is always a solution). However the exception occurs when $x = \frac12$ that always seems to only produce real numbers as solution(s). </p> <p>Does anybody see why that is (or must be) the case?</p> http://mathoverflow.net/questions/70460/non-trivial-zeros-of-the-zeta-function/70774#70774 Answer by GH for Non trivial zeros of the Zeta function GH 2011-07-19T19:40:33Z 2011-07-19T19:46:36Z <p>Using the notation $s=u+1/2$ your conjecture can be reformulated and generalized as follows. </p> <p><strong>Proposition.</strong> Let $v_1,v_2,\dots,v_N$ be arbitrary positive numbers, then all solutions of the equation $$\prod_{n=1}^N \frac{v_ni-u}{v_ni+u} = 1$$ are real.</p> <p><strong>Proof.</strong> The degree of the polynomial $\prod_{n=1}^N(v_ni-u)-\prod_{n=1}^N(v_ni+u)$ is $N$ or $N-1$ depending on whether $N$ is odd or even (the polynomial is always odd). Therefore it suffices to show that there are the same number of real solutions to the displayed equation. As $u$ grows from $-\infty$ to $\infty$, each fraction under the product traverses the unit circle continuously in the positive direction, starting from and arriving back to $-1$. Using ideas similar to how one proves that the fundamental group of the unit circle is $\mathbb{Z}$, we see that the product traverses the unit circle $N$ times in the positive direction, starting from and arriving back to $(-1)^N$. In particular, the product passes $1$ exactly $N$ or $N-1$ times depending on whether $N$ is odd or even. QED</p> http://mathoverflow.net/questions/70460/non-trivial-zeros-of-the-zeta-function/71078#71078 Answer by Agno for Non trivial zeros of the Zeta function Agno 2011-07-23T19:03:01Z 2011-07-25T20:22:02Z <p>I would like to go back to my original question one more time.</p> <p>$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$</p> <p>that can be rewritten as:</p> <p>$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$</p> <p>The insight I gained from the various comments (thanks GH), is that this equation can not be used to 'solve' $s$ by taking the $\rho$'s as given inputs. The equation is just valid for all $s \in \mathbb{C}$ and it is incorrect to reverse the argument by for instance assuming that all numbers can be produced as solutions from the $\rho$'s.</p> <p>So, another approach is to assume that the $s$ is a given input (arriving via the $\zeta(s)$) and that the $\rho$'s are the solutions for any $s$ chosen from $\mathbb{C}$. To avoid a too strong link with the non-trivial zeros (that encode specific info about the prime numbers only and therefore are expected to be a very specific subset of all solutions for this equation), I use the variable $x$ instead of the $\rho$.</p> <p>This gives the following equation:</p> <p>$\prod_{n=1}^N \frac{x_n-s}{x_n + s -1} = 1$</p> <p>Experimenting with various input values for $s$, I now dare to conjecture the following:</p> <p>1) $s \in \mathbb{R}$ always produces $x_n = \frac12 \pm yi$; $y \in \mathbb{R}$. <strong>EDIT : Additional note:</strong></p> <ul> <li>for $s \in \mathbb{R}, s \not \in \mathbb{Z}$, all solutions are perfectly symmetrical ie.: $x_n = \frac12 \pm yi$.</li> <li>for $s \in \mathbb{Z}$, (most) solutions are of the form $x_n = \frac12 + yi$, otherwise symmetrical.</li> </ul> <p>2) $s \in \mathbb{C}$ and $s= a \pm yi$ and $a \ne \frac12$ always produces $x_n = \frac12 \pm (w + yi)$; $w \in \mathbb{R}$</p> <p>3) $s \in \mathbb{C}$ and $s= \frac12 \pm yi$ always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$</p> <p>The last outcome is easy to prove by taking GH's proof from the post above and assuming $s=\frac12 + v_n i$ (or $-v_n i$) and $x = \frac12 + u$ (or $-u$).</p> <p>Proving the first and middle outcomes is much more difficult, however I believe a potential approach for the first result could be to reverse GH's proof and maybe assume a bijective relationship like this: $x_n = \frac12 + y i \rightarrow s \in \mathbb{R}$, and therefore: $s \in \mathbb{R} \rightarrow x_n = \frac12 + y i$.</p> <p>These maybe just some lose observations about the types of solutions for $x_n$ (of which the $\rho$'s are assumed to be a subset), however it does yield a direct conflict with the Riemann hypothesis. Assuming it is proven that $s= \frac12 \pm yi$ always produces a real result for $x_n$ (and therefore $\rho_n$), the $\rho$ could then never become equal to $s$ anymore (and turn the left or right term in the original equation and thereby $\zeta(s)$ into zero). But maybe the way around this dilemma is to assume that the zeros only occur in a limit situation of e.g. $\lim_{a \to \frac12} x_n = a \pm yi$) </p> <p>But I guess I'm doing something wrong here and would appreciate any guidance on where the logic derails. </p> http://mathoverflow.net/questions/70460/non-trivial-zeros-of-the-zeta-function/71518#71518 Answer by Agno for Non trivial zeros of the Zeta function Agno 2011-07-28T21:02:06Z 2011-07-28T21:02:06Z <p>Just wanted to share my latest train of thoughts and leave you with the open question if the logic makes sense or not.</p> <p>The following equation can be derived from combining the Hadamard products of the non-trivial zeros ($\rho$) from the Zeta function and its reflection formula:</p> <p>$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$</p> <p>that can be rewritten as:</p> <p>$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$</p> <p>This equation must be valid for all $s$ (except 1) and to learn more about the $\rho$'s, I tried to solve the following more generic (and less infinite) equation: </p> <p>$\prod_{n=1}^N \left(\frac{x_n-s}{x_n + s -1}\right) = 1$ </p> <p>It can be proven (see above) that when $s= \frac12 \pm yi$ the equation always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$. The outcome is therefore always real and this is in direct conflict with the empirical evidence that at least the first billions of non-trivial zeros lie on the (complex) critical line. The situation that $s$ = $\rho_n$ (i.e. a non-trivial zero, subset of $x_n$) can therefore never be achieved and this makes me suspicious something is wrong here.</p> <p>Therefore the derived equation must be incorrect and the only decent way out of it I see is to assume the following subtle change:</p> <p>$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$</p> <p>Assumption (A):</p> <p>$\rho = a + yi$ and $(1-\rho) = (1-a) - yi$ or $\rho = a - yi$ and $(1-\rho) = (1-a) + yi$.</p> <p>$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)$ </p> <p>that can be dramatically simplified into (note the $s$ dropping out):</p> <p>$\prod_\rho \left(\frac{\rho -1}{\rho} \right) = 1$</p> <p>and more generically to enable experimenting:</p> <p>$\prod_{n=1}^N \left(\frac{x_n-1}{x_n}\right) = 1$</p> <p>Solving this equation for all $x_n$ being equal, the outcome is always $\frac12 \pm y i$, so that's a much more promising outcome for inducing zeros of the Riemann hypothesis at $s=\rho_n$.</p> <p>But we know that each $\rho_n$ is different and could lie anywhere in the critical strip $0&lt;\Re(\rho)&lt;1$. Obviously by simply assuming that all $\rho$'s (as a subset of $x_n$) are lying on the critical line, it is simple to proof that the equation nicely holds, since each (absolute) term in the product will be equal to 1:</p> <p>$|\prod_{n=1}^N \left(\frac{\frac12 + ni -1}{\frac12 + ni}\right)| = 1$</p> <blockquote> <p>N.B: Just briefly like to share a nice byproduct I observed when playing with this equation: </p> <p>$k_n, a, b, c \in \mathbb{Z}, |\prod_{n=1}^N \left(\frac{\frac12 + k_n i -1}{\frac12 + k_n i}\right)| = |\frac{a}{c} + \frac{b}{c}i| = 1 \rightarrow a^2 + b^2 = c^2$ </p> <p>Only Pythagorean triples will be produced for random values of $N$ and $k_n$.</p> </blockquote> <p>So what will happen when some $\rho$'s are lying off the critical line? This would make at least two terms not being 1 anymore (one that causes it and one complementary to make the total product equal to 1 again). Example:</p> <p>$\left(\frac{\frac13 + 6i -1}{\frac13 + 6i}\right)\left(\frac{y -1}{y}\right)=1$ gives $y=\frac23 -6i$.</p> <p>But here's the trick: assumption (A) above now prohibits the $\rho$'s to switch signs in the product (only $(1-\rho)$ does that). This therefore implies that only $\rho_n = \frac12 + yi$ or $\rho_n = \frac12 - yi$ can be valid solutions for the equation. </p> <p>And now I only have to proof assumption (A) is true... </p> http://mathoverflow.net/questions/70460/non-trivial-zeros-of-the-zeta-function/72513#72513 Answer by Agno for Non trivial zeros of the Zeta function Agno 2011-08-09T20:20:45Z 2011-08-11T07:51:36Z <p>My last thought on this one (I promise).</p> <p>To be more precise, I indexed the non-trivial zeros in the formula in the opening post but now changed the $\rho$ in the second product into $1-\rho$ (that is in line with Riemann's observation that when a $\rho$ is a non-trivial zero, also $1-\rho$ must be one):</p> <p>$\displaystyle \prod_{\rho_1}^{\rho_\infty} |\left(1- \dfrac{s}{\rho_n} \right)|=\prod_{\rho_1}^{\rho_\infty} | \left(1- \dfrac{1-s}{1-\rho_n} \right)|$</p> <p>This means that each term in the product with the same $\rho_n$ can be equated as follows:</p> <p>$|\left(\dfrac{{\rho_n} - s}{\rho_n} \right)| = |\left(\dfrac{s-\rho_n}{1-\rho_n} \right)|$</p> <p><strong>EDIT: This step is cleary not allowed and implicitly already assumes $\Re(\rho_n) =\frac12$. The terms in both products can be different. Dividing out all terms with equal $\rho_n$ is allowed (making the infinite product $1$) as I did in my previous post, but this doesn't yield any additional info about the individual $\rho_n$. So, back to the drawing board.</strong></p> <p>This equation is valid for all $s$ when $\Re(\rho_n) =\frac12$, but it is also valid for all $s=\rho_n$. And that could be for any complex number $\rho_n$ in the critical strip. However, to better see what happens when $s$ approaches $\rho_n$, the following equation gives an indeterminate form of type $0/0$: </p> <p>$|\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| = |\left(\dfrac{\rho_n}{1-\rho_n} \right)|$ or $|\left(\dfrac{{s-\rho_n}}{\rho_n -s} \right)|= |\left(\dfrac{1-\rho_n}{\rho_n} \right)|$</p> <p>but by applying L'Hôpital's rule we find, </p> <p>$\displaystyle \lim_{s \to \rho_n} |\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| =1$</p> <p>which implies that when $s$ approaches $\rho_n$ infinitely close, the following equation must be true:</p> <p>$|\left(\dfrac{\rho_n}{1-\rho_n} \right)| = |\left(\dfrac{1-\rho_n}{\rho_n} \right)| =1$</p> <p>And this equation only has solutions when $\Re(\rho_n) =\frac12$. It also implies (if the logic is correct) that each term in the following infinite products: </p> <p>$\displaystyle \prod_{\rho_n} |\left(\frac{1- \rho_n}{\rho_n} \right)| = 1$</p> <p>or</p> <p>$\displaystyle \prod_{\rho_n} |\left(\frac{\rho_n}{1-\rho_n} \right)| = 1$</p> <p>is equal to $1$ and therefore all $\rho$'s contribute independently from each other to the overall product (and are therefore simple?).</p>