The Zeta-function can be written as the following infinite Hadamard product of its non-trivial zeroes:

$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$

this also implies that:

$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$

Take the 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:

$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$

that can be rewritten as:

$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$

This equation has zeros for $\rho = s$ as long as $2\rho-1 \ne 0$.

The ρ's could obviously lie anywhere in the already proven strip between $0<\Re(\rho)<1$ and I just take them as 'givens' wherever they might be located.

I started to experiment with solving s for different numbers of terms as follows:

$x=\frac13$ (e.g.)

$\prod_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$

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).

Does anybody see why that is (or must be) the case?

  • 3
    $\begingroup$ Your fourth display is an identity expressing the following fact discovered by Riemann: if $\rho$ is a zero of the Riemann zeta function then so is $1-\rho$. In particular, it makes no sense to "solve the equation for $s$", as the identity is valid for any $s$. $\endgroup$
    – GH from MO
    Jul 15, 2011 at 19:46
  • $\begingroup$ In my previous comment I only meant nontrivial zeros. $\endgroup$
    – GH from MO
    Jul 15, 2011 at 19:56
  • $\begingroup$ GH: 'it makes no sense to "solve the equation for s", as the identity is valid for any s'. Agree. But we could also reverse that argument and (boldly) say that all real numbers (except 1) are actually solutions of this equation and that the ρ's are the key building blocks for any complex/real number (as the primes are for composites). But unlike the primes, that are only capable of producing integers and therefore never can even come close to generating 'themselves', the ρ's do generate numbers that will come infinitely close to themselves. That's when the zeros appear (as limits). $\endgroup$
    – Agno
    Jul 15, 2011 at 22:34
  • 1
    $\begingroup$ @Agno: The $\rho$'s are no building blocks for any complex/real number, they are just an encoded version of the prime numbers. Of course such a statement is philosophy, hence off-topic here. Back to mathematics: I only wanted to point out that you try to deduce something interesting about the $\rho$'s from a relation that merely says that the $\rho$'s come in pairs (the pair of $\rho$ is $1-\rho$). $\endgroup$
    – GH from MO
    Jul 16, 2011 at 6:28
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    $\begingroup$ @GH: The question has indeed nothing to do with Riemann Zeta (I see that now, even though that was how I landed on the equation). What I meant with 'random' is that the $\rho$'s really could be any complex number. I have experimented a lot today with various values for $\rho$ and different (finite) numbers of the product terms. The solutions for s obviously grow with the number of terms, but only with all terms of the form $\frac12 + n i$, the outcomes are always real (i.e. whatever n I pick; even tried the first 100 non-trivial zero values for n...). I just don't see why this is the case. $\endgroup$
    – Agno
    Jul 17, 2011 at 17:19

4 Answers 4


Using the notation $s=u+1/2$ your conjecture can be reformulated and generalized as follows.

Proposition. 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.

Proof. 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

  • $\begingroup$ I believe the proof can be quite easily extended towards the lower critical line (i.e. real part 1/2 and imaginary part $v_n$ negative). Take $s=\frac12−u$. This results in $\prod_{n=1}^N \frac{-v_ni+u}{-v_ni-u} = 1$. Multiply both nominator and denominator by $-1$ and the already proven equation reappears again. $\endgroup$
    – Agno
    Jul 20, 2011 at 21:54
  • $\begingroup$ Yes, but you cannot extend it to the case when $v_n$'s of both signs are present. $\endgroup$
    – GH from MO
    Jul 20, 2011 at 22:26
  • $\begingroup$ Not convinced about that, GH. The following alternating signs cases: $\prod_{n=1}^N \frac{(-1)^n v_ni-u}{v_ni+u} = 1$ and $\prod_{n=1}^N \frac{v_ni-u}{(-1)^n v_ni+u} = 1$ also seem to have only real solutions. $\endgroup$
    – Agno
    Jul 22, 2011 at 17:59
  • $\begingroup$ I am not saying you cannot come up with more general conditions under which the conclusion holds. I am only saying that for arbitrary real $v_n$'s the conclusion often fails. At least this is what I would expect. $\endgroup$
    – GH from MO
    Jul 22, 2011 at 18:38
  • $\begingroup$ We are aligned. Following my 'trial and error' attempts in Maple with this equation (I know, that is not proper math!), these appear to be the only conditions (i.e. $s = \frac12 \pm u$ with $v_n$ all positive, all negative or the nominator/denominator having a regular alternating sign), that always produce real-only solutions. So far, all other constructs I've tried, seem to only produce complex outcomes. $\endgroup$
    – Agno
    Jul 22, 2011 at 18:56

I would like to go back to my original question one more time.

$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$

that can be rewritten as:

$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$

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.

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 $\rho$.

This gives the following equation:

$\prod_{n=1}^N \frac{x_n-s}{x_n + s -1} = 1$

Experimenting with various input values for $s$, I now dare to conjecture the following:

1) $s \in \mathbb{R}$ always produces $x_n = \frac12 \pm yi$; $y \in \mathbb{R}$. EDIT : Additional note:

  • for $s \in \mathbb{R}, s \not \in \mathbb{Z}$, all solutions are perfectly symmetrical ie.: $x_n = \frac12 \pm yi$.
  • for $s \in \mathbb{Z}$, (most) solutions are of the form $x_n = \frac12 + yi$, otherwise symmetrical.

2) $s \in \mathbb{C}$ and $s= a \pm yi$ and $a \ne \frac12$ always produces $x_n = \frac12 \pm (w + yi)$; $w \in \mathbb{R}$

3) $s \in \mathbb{C}$ and $s= \frac12 \pm yi$ always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$

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 $-u$).

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: $s \in \mathbb{R} \rightarrow x_n = \frac12 + y i$.

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. $\lim_{a \to \frac12} x_n = a \pm yi$)

But I guess I'm doing something wrong here and would appreciate any guidance on where the logic derails.


Just wanted to share my latest train of thoughts and leave you with the open question if the logic makes sense or not.

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:

$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$

that can be rewritten as:

$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$

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:

$\prod_{n=1}^N \left(\frac{x_n-s}{x_n + s -1}\right) = 1$

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.

Therefore the derived equation must be incorrect and the only decent way out of it I see is to assume the following subtle change:

$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$

Assumption (A):

$\rho = a + yi$ and $(1-\rho) = (1-a) - yi$ or $\rho = a - yi$ and $(1-\rho) = (1-a) + yi$.

$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)$

that can be dramatically simplified into (note the $s$ dropping out):

$\prod_\rho \left(\frac{\rho -1}{\rho} \right) = 1$

and more generically to enable experimenting:

$\prod_{n=1}^N \left(\frac{x_n-1}{x_n}\right) = 1$

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 $s=\rho_n$.

But we know that each $\rho_n$ is different and could lie anywhere in the critical strip $0<\Re(\rho)<1$. 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:

$|\prod_{n=1}^N \left(\frac{\frac12 + ni -1}{\frac12 + ni}\right)| = 1$

N.B: Just briefly like to share a nice byproduct I observed when playing with this equation:

$k_n, a, b, c \in \mathbb{Z}, > |\prod_{n=1}^N \left(\frac{\frac12 + > k_n i -1}{\frac12 + k_n i}\right)| = > |\frac{a}{c} + \frac{b}{c}i| = 1 > \rightarrow a^2 + b^2 = c^2$

Only Pythagorean triples will be produced for random values of $N$ and $k_n$.

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:

$\left(\frac{\frac13 + 6i -1}{\frac13 + 6i}\right)\left(\frac{y -1}{y}\right)=1$ gives $y=\frac23 -6i$.

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.

And now I only have to proof assumption (A) is true...


My last thought on this one (I promise).

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):

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

This means that each term in the product with the same $\rho_n$ can be equated as follows:

$|\left(\dfrac{{\rho_n} - s}{\rho_n} \right)| = |\left(\dfrac{s-\rho_n}{1-\rho_n} \right)|$

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.

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 $0/0$:

$|\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| = |\left(\dfrac{\rho_n}{1-\rho_n} \right)|$ or $|\left(\dfrac{{s-\rho_n}}{\rho_n -s} \right)|= |\left(\dfrac{1-\rho_n}{\rho_n} \right)|$

but by applying L'Hôpital's rule we find,

$\displaystyle \lim_{s \to \rho_n} |\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| =1$

which implies that when $s$ approaches $\rho_n$ infinitely close, the following equation must be true:

$|\left(\dfrac{\rho_n}{1-\rho_n} \right)| = |\left(\dfrac{1-\rho_n}{\rho_n} \right)| =1$

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:

$\displaystyle \prod_{\rho_n} |\left(\frac{1- \rho_n}{\rho_n} \right)| = 1$


$\displaystyle \prod_{\rho_n} |\left(\frac{\rho_n}{1-\rho_n} \right)| = 1$

is equal to $1$ and therefore all $\rho$'s contribute independently from each other to the overall product (and are therefore simple?).

  • 2
    $\begingroup$ You don't need l'Hopital to find $\lim_{s\to\rho_n}|(\rho_n-s)/(s-\rho_n)|=1$. $\endgroup$ Aug 9, 2011 at 23:38
  • $\begingroup$ @Gerry. Agree. It is easy to see that in $\lim_{s\to\rho_n}|(\rho_n-s)/(s-\rho_n)|=1$ both nominator and denominator "crawl" to zero at the same pace. In fact for all $s \neq \rho_n$ and independent of the size of the gap, the solution will be $\Re(\rho_n) =\frac12$. The reason why I introduced the limit is to show that when a non-trivial zero is a limit as well (which I believe it is), the solution still holds (this wouldn't be the case when $s =\rho_n$). However, this all looks a bit too simple and I guess there is a flaw hidden in my logic. Wonder if I did correctly apply the absolutes? $\endgroup$
    – Agno
    Aug 10, 2011 at 21:28
  • $\begingroup$ Found a pretty bad error in my reasoning and added an EDIT in the post to explain it. $\endgroup$
    – Agno
    Aug 11, 2011 at 7:53
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    $\begingroup$ Dear Agno, your enthusiasm for this question makes me happy. However, every time you add an answer or make an edit, this question reappears on the front page. I think it would be best if you used MathOverflow as a place for asking and answering well-formed mathematical questions, instead of a place for publicly working out details. $\endgroup$
    – S. Carnahan
    Aug 11, 2011 at 9:20
  • $\begingroup$ @S.Carnahan. Appreciate your gentle steer. I will stop editing now and allow this thread to calmly sink down into MathOverflow's history :-) $\endgroup$
    – Agno
    Aug 11, 2011 at 9:58

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