9

As an undergrad, I sometimes toy with the Riemann Hypothesis (RH) to A) humble myself and B) better myself in complex analysis and proof writing. Given its infamy and my relative mathematical immaturity, I obviously don't expect to make any significant results any time soon. Hence, I apologize for fulfilling the often annoying trope of "young ambitious undergrad attempting millennium prize problems."

That said, after some playing around with the zeta function for the past few years, I believe I have come across a proof that $$\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1\implies \Re(s)=\frac{1}{2}$$ for all $s$ within the critical strip.

Assuming this is true, I am curious if this result is meaningful to RH. Is this result already known, or a triviality of the functional identity? If not, is it meaningful / significant to the RH?

Any insight as to its potential (non)significance, how this may be false, or how this is already known would be greatly appreciated.

Edit: As pointed out, this 'result' (or conjecture, if you will) fails near the origin. I'll have to overlook my proof again, but fortunately I believe it should still hold for $|\Im(s)|\ggg 0$, at the least.

Graviton
  • 4,462
  • 3
    No which I illustrated in my question at https://math.stackexchange.com/questions/3008597/questions-related-to-the-riemann-zeta-function-where-zetas-zeta1-s . – Steven Clark Aug 05 '22 at 19:37
  • @StevenClark: maybe you have already thought about this. It should not be too difficult to prove that for any $\sigma=\text{Re}(s)\in\left(\frac{1}{2},1\right)$ there are exactly two opposite $\tau$s such that $|\zeta(s)|/|\zeta(1-s)|=1$ at $s=\sigma\pm i\tau$. You have already noticed that $\sigma=f(\tau)$ is almost constant for $\sigma\in(0,1)$. – Jack D'Aurizio Aug 05 '22 at 20:22
  • @StevenClark: are non-trivial zeroes of the $\zeta$ function forced to fulfill $\sigma=\frac{1}{2}$ or $\tau=\pm f(\sigma)$? If so, this can actually lead to a proof of RH. Indeed by the integral representation the non-trivial zero closest to the origin has to fulfill $\tau > 10$, while the values of $f$ are in a tight neighbourhood of $2\pi\ll 10$. – Jack D'Aurizio Aug 05 '22 at 20:25
  • @JackD'Aurizio I hadn't thought about proving the conjecture in your first comment, but both of your comments seem related to my question (5) in the link above which was of primary interest to me. The comment by user reuns at the question I linked above seemed to imply there was no connection between my investigation and the RH. – Steven Clark Aug 05 '22 at 20:53
  • @StevenClark: I am not so sure, please refer to my last comment under my answer. Of course in the Weierstrass factorization of $\zeta(s)/\zeta(1-s)$ the exceptional zeroes of the $\zeta$ function cancel out; yet the limit of $\zeta(s)/\zeta(1-s)$ at an exceptional zero equals $-\zeta'(s)/\zeta'(1-s)$. $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the reflection formula. An accurate description of the range of $\zeta'(s)/\zeta'(1-s)$ and the solutions of $\zeta(s)/\zeta(1-s)=k$ could lead to something non-trivial. – Jack D'Aurizio Aug 05 '22 at 23:31
  • @JackD'Aurizio I'm not totally sure either, but I was discouraged by the comment by user reuns. Since $|f(s)|=\left|2^s\pi^{s-1}\sin\left(\frac{\pi,s}{2}\right),\Gamma (1-s)\right|=1$ for all known non-trivial zeta zeros which are all on the critical line it seemed logical to ask if this property would also hold for any non-trivial zeta zeros potentially off the critical line, and at first it seemed to make some sense to me but then I began to suspect my thinking was perhaps flawed. – Steven Clark Aug 06 '22 at 00:11
  • 1
    I would propose to merge this question and https://math.stackexchange.com/questions/266948/riemann-zeta-function-and-modulus – Jack D'Aurizio Aug 06 '22 at 23:13

1 Answers1

6

Intro: I am sorry for the following lines since your result would have been an outstanding achievement towards RH.
However...

By the reflection formula $$ \rho(s)=\frac{\zeta(s)}{\zeta(1-s)} = 2^s \pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s) $$ and $$ R(\nu)=\rho\left(\frac{1}{2}+\nu\right) =(2\pi)^{\nu} \sqrt{\frac{2}{\pi}}\sin\left(\frac{\pi \nu}{2}+\frac{\pi}{4}\right)\Gamma\left(\frac{1}{2}-\nu\right)$$ is a real analytic function on $(-1/2,1/2)$ which is also increasing on such interval and equal to $1$ at the origin. For any $\nu\in(-1/2,1/2)$ $$\|R(\nu+i\tau)\|^2 = R(\nu+i\tau)R(\nu-i\tau) = (2\pi)^{2\nu}\left(1+\frac{\sin(\pi\nu)}{\cosh(\pi \tau)}\right)\left\|\frac{\Gamma\left(\tfrac{1}{2}-\nu-i\tau\right)}{\Gamma\left(\tfrac{1}{2}-i\tau\right)}\right\|^2$$ is an even function of the $\tau$ variable. Assuming $\nu > 0$ it is also a Schwartz function.
On the other hand, by just considering some concrete instances, we may realize that your conjecture does not hold. For $\nu=\frac{1}{10}$ we have $\|R(\nu)\|^2>1.72$ and $\|R(\nu+i\tau)\|=1$ for $|\tau|\approx 6.29 $. In particular there are points on the line $\text{Re}(s)=\frac{3}{5}$ such that $|\zeta(s)|=|\zeta(1-s)|$. But exactly two of them.

On a positive note, let us try to salvage things.

Even if your conjecture does not hold, it leaves some room for an interesting attack.
Let us set $s=\sigma+i\tau$ for $\sigma\in (0,1)$.

Jack's conjecture 1: The locus of points of the critical strip such that $|\zeta(s)|=|\zeta(1-s)|$ is made by the line $\sigma=\frac{1}{2}$ and by the union of two graphs, symmetric with respect to the real line, given by $\tau=\pm f(\sigma)$.

Jack's conjecture 2: In order to prove RH it is sufficient to show that there are no zeroes of the $\zeta$ function on the curve $\tau=f(\sigma)$.

Numerically it seems that $f$ is almost constant, taking values between $6.28$ and $6.3$.
And the non-trivial zero closest to the origin is well-known to fulfill $\tau > 10$ (this just follows from integral representations for $\zeta(s)$).

If this actually works we'll share the prize, deal? :)

Update. As shown in the comments, there is no hope to prove RH through this naive approach, but I am still convinced that something (like improving the shape of the zero-free region) can be done. Let us assume that all the exceptional zeroes in the critical strip are simple. Then for any one of them the limit of $\frac{\zeta(s)}{\zeta(1-s)}$ equals $-\frac{\zeta'(s)}{\zeta'(1-s)}$, so they have to lie on the region defined by the equality between $f(s)=\left|\frac{\zeta(s)}{\zeta(1-s)}\right|$ and $g(s)=\left|\frac{\zeta'(s)}{\zeta'(1-s)}\right|$. $f(s)$ essentially is an elementary function and its behavior for a fixed $\sigma\in\left(\frac{1}{2},1\right)$ and a varying $\tau\in\mathbb{R}$ is Gaussian-like. $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the derivative of the reflection formula, but the ratio $\frac{\zeta'(s)}{\zeta'(1-s)}$ is less elementary, depending on bounds for $|\zeta(s)|$. Numerical experiments reveal that the behaviour of $g(s)$ for a fixed $\sigma$ and a varying $\tau$ still is essentially Gaussian, but for $\tau$ close to zero $g(s)$ is much larger than $f(s)$. If we prove that the curve given by $f(s)=g(s),\tau > 0$ lies in the currently known zero-free region, we have proven RH under the assumption that all the exceptional zeroes are simple.

Jack D'Aurizio
  • 353,855
  • 1
    According to my version of Mathematica $f(0)$ is a tad larger than $2\pi$, but pretty close indeed :) – Jack D'Aurizio Aug 05 '22 at 20:12
  • 1
    Really grateful for the counterexample. I'll review my work with your points in mind and see where I went wrong. Fortunately, as you also suggested, It may be salvageable, as I had a hint that my conjecture may fail close to the origin. I believe the "proof" should still work for some significant subset of the critical strip, at the least. (also, deal ;) – Graviton Aug 05 '22 at 20:42
  • 1
    note that if $\rho$ is a non critical line zero (counterexample to RH), there is no particular reason to have $|f(\rho)| =1$ as we can easily give counterxamples of (entire, conjugate invariant) functions $g$ satisfying $g(s)=f(s)g(1-s)$ with $f$ having no zero, but for which $|f(\rho)|$ is arbitrary (non zero) at a zero of $g$ so where $g(\rho)=g(1-\rho)=0$; similarly there is no particular reason to have any relation between a zero of $g$ and some specific behavior of $f$ there – Conrad Aug 05 '22 at 20:54
  • @Conrad: of course you are right, if at some $s_0=\sigma+i\tau$ with $\sigma\in(0,1)\setminus{1/2}$ we have $\zeta(s_0)=0$, then $\lim_{s\to s_0}\left|\frac{\zeta(s)}{\zeta(1-s)}\right|$ simply equals the modulus of $\zeta'(s_0)/\zeta'(1-s_0)$. On the other hand $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the (derivative of the) reflection formula, so it does not sound completely insane that we can get something by studying the ratio $\zeta(s)/\zeta(1-s)$. – Jack D'Aurizio Aug 05 '22 at 23:14
  • the ratio $\zeta(s)/\zeta(1-s)$ is really well understood so not sure if there is anything new to say about it (if you look at the same ratio for the Davenport Heilbronn function, it is quite similar, but of course, the behavior of the Davenport Heilbronn function is very different as zeroes go) – Conrad Aug 05 '22 at 23:38
  • My point is that for small values of $|\tau|$ the modulus of $\zeta'(s)/\zeta'(1-s)$ is usually much larger than the modulus of $\zeta(s)/\zeta(1-s)$, so exceptional zeroes cannot fall too close to the real line. I doubt this is enough to improve the bounds for the zero-free region, but the idea deserves to be investigated imho. – Jack D'Aurizio Aug 05 '22 at 23:46
  • we already know that the first umpteen zeroes are on the critical line, so any exceptional zero would have an imaginary part beyond our computational range (not sure of the current record, but shouldn't be hard to google); on theoretical grounds, the first exceptional zero (it if exists) should be in the range of $10^{10^{500}}$ or higher, where the argument of zeta goes in the hundreds - although we know that $\arg \zeta(1/2+it)$ is unbounded, all the current computations (that found all those umpteen zeroes) cannot get a value higher than $4$ or $5$ afaik, so long way to go computationally – Conrad Aug 05 '22 at 23:55
  • But we can also go the other way around. Assume that all the exceptional zeroes of the $\zeta$ function are simple. Then they have to lie on the curve(s) described by $\left|\zeta(s)/\zeta(1-s)\right|=\left|\zeta'(s)/\zeta'(1-s)\right|$. If we prove that the curve in the upper half-plane lies below the countour of the zero-free region we have proven RH. – Jack D'Aurizio Aug 06 '22 at 00:00
  • I do not see why $|\zeta(s)/\zeta(1-s)|=|\zeta'(s)/\zeta'(1-s)|$ is that different than $|D(s)/D(1-s)|=|D'(s)/D'(1-s)|$ for $D$ the Davenport Heilbronn function and that has arbitrary high (imaginary part) roots in the critical strip, since $D$ has lots and lots of critical strip zeroes that are not on the critical line; this being said, one of course never knows, but imho trying to make any advance on RH using only the functional equation is highly unlikely as $D$ shows; the Euler product and the analyticity of $\log \zeta, 1/\zeta4$in $\Re s >1$ are highly likely to come in – Conrad Aug 06 '22 at 00:07
  • Still $D$ and $\zeta$ are different functions, and one might be happy in just improving the shape of the zero-free region. I need time to perform numerical experiments and fully grasp the behaviour of $\zeta'(s)/\zeta'(1-s)$ up to the current knowledge. I am mildly optimistic that something can be done. – Jack D'Aurizio Aug 06 '22 at 00:16
  • 1
    @JackD'Aurizio truly, your optimism and ardent analysis of my question is inspiring. If you'd like, (although I lack as much of a toolset as you do) I'd be delighted to share more of my work / ideas on RH as per the email in my profile. I've taken a very geometric approach of the partial sums of the zeta function which would be nice to flesh out with more experienced eyes. – Graviton Aug 06 '22 at 00:43
  • Contour plots of $\left| \frac{\zeta(\nu +i \tau)}{\zeta(1-(\nu +i \tau))}\right| =\left| \frac{\zeta'(\nu +i \tau)}{\zeta'(1-(\nu +i \tau))}\right|$ reveal there's sort of an ellipse in the region $-1.5<\nu<1.5$ and $-1<\tau<1$ centered at $(\nu,\tau)=\left(\frac{1}{2},0\right)$ where this property holds (as well as along the critical line (see https://i.stack.imgur.com/MFXgr.jpg ). – Steven Clark Aug 07 '22 at 18:04
  • But this contour plot (blue) does not seem to correlate with $\frac{\zeta(\nu +i \tau)}{\zeta(1-(\nu +i \tau))}=-\frac{\zeta'(\nu +i \tau)}{\zeta'(1-(\nu +i \tau))}$ which I determined via contour plots of $\Re\left(\frac{\zeta(\nu +i \tau)}{\zeta(1-(\nu +i \tau))}\right)=-\Re\left(\frac{\zeta'(\nu +i \tau)}{\zeta'(1-(\nu +i \tau))}\right)$ (orange) and $\Im\left(\frac{\zeta(\nu +i \tau)}{\zeta(1-(\nu +i \tau))}\right)=-\Im\left(\frac{\zeta'(\nu +i \tau)}{\zeta'(1-(\nu +i \tau))}\right)$ (green) except for the end points on the real axis. (see https://i.stack.imgur.com/z1pPa.jpg ) – Steven Clark Aug 07 '22 at 21:25
  • I'm not sure if it provides any additional insight, but the non-trivial zeros of $\zeta(s)$ are a subset of the non-trivial zeros of $g(s)=\zeta '(s)+2^s\ \pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta'(1-s)$ which has at least one extra pair of non-trivial zeros at $s=\frac{1}{2}\pm\lambda$ where $\pm\lambda$ corresponds to the locations of the minima and maxima of the Riemann-Siegel theta function $\vartheta(t)$. – Steven Clark Aug 28 '22 at 19:29