Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The $\zeta$ function maybe written as Euler Product: $$ \zeta(s)=\prod_{p} \frac{1}{1-p^{-s}}=\prod_p e_p(s). $$ Now let's substitute $s$ with $\rho_k$, the $k$th root of $\zeta$, and have a look at the individual factors. There are 2 options: 1.$|e_p(\rho_k)|<1$ or 2. $|e_p(\rho_k)|\ge1$ (e.g. $|e_{23}(\rho_1)|\approx 1.2404$, for more values, see here) and obviously the values less $1$ must be infintely many more than the others, otherwise it would not converge to $0$. So we can write it as: $$ \zeta(\rho_k)=\prod_{\color{blue}{||<1}} e_p(\rho_k) \times \prod_{\color{red}{||\geq 1}} e_p(\rho_k), $$

So my questions are:

  1. Do these values $e_p(\rho_k)$ have a special meaning or a straight forward interpretation?
  2. For a given $\rho_k$, how do these values distribute? Plots for $\rho_1$ and $\rho_2$ (not shown) show a spiral around $1+0i$: rho_1 color ($x$ is real axis, $y$ the imaginary, the line indicates $||=1$)
  3. obsolete Are there finitely or infinitely many primes $p$, where $|e_p(\rho_k)|\ge1$? Infinitely.
  4. How does that distribution behave, if $\Re(s)\neq \frac{1}{2}$? Some values of $e_p(\varepsilon + \rho_k)$ should move outwards, such that $\prod_{||<1} e_p(\rho_k)$ doesn't become $0$ anylonger, if Riemann's Hypothesis is true. $s=1/8+14.134725i$ is shown in the plot: enter image description here

Thanks for your help/comments/plots/answers...

share|cite|improve this question
That there must be infinitely many more cases of 1. than of 2. is not only not obvious but false. There may well be infinitely many of either kind; they could even alternate and still the product might converge to zero. That also means that you can't split the product up like you did, since the two parts might both contain infinitely many factors and might not converge individually though the original product converges. (At least generally; there might be a reason why this can't happen for this particular product, but your argument from convergence is general.) – joriki Mar 18 '12 at 22:50
@joriki So the whole ansatz is blur? – draks ... Mar 18 '12 at 22:52
Sorry, I don't know what "blur" means in this context. – joriki Mar 18 '12 at 22:53
@joriki I mean wrong. – draks ... Mar 18 '12 at 22:54
Just like the Dirichlet series, the Euler product for $\zeta$ converges for $\text{Re}(s) > 1$. The zeros are in a region where the Euler product should diverge. – Robert Israel Mar 19 '12 at 6:33
up vote 1 down vote accepted

In this MO question, I showed that, if $s = \sigma+i t$ with $\sigma \in (0,1)$ and $t$ nonzero, then the partial products of the Euler product oscillate extremely wildly.

Let $\Pi_P$ be $\prod_{p \leq P} e_p(s)$. I show that, for any $M> 1$, there are arbitrarily many $P_1 < P_2$ such that $|\Pi_{P_2}| > M | \Pi_{P_1} |$ and infinitely many $P_1 < P_2$ such that $|\Pi_{P_2}| < M^{-1} | \Pi_{P_1} |$. In terms of your metaphor of driving inwards and driving outwards, the product is driven in both directions for very long trips.

This means that numerical data on these partial products is untrustworthy. I suspect that they do not approach zero even at the zeroes of $\zeta$, although I can't prove that. GH (in the same question) proved that, assuming RH, the partial products don't approach zero when $\sigma>1/2$.

It would probably be good to nail down what these products do on the critical line, but it looks to me like they don't have much to do with the behavior of the $\zeta$ function.

share|cite|improve this answer
I just had a look at the numerics and the oscillate as you proposed. Thanks for answer. – draks ... Mar 21 '12 at 14:46
@David : You talk about GH , do you mean GH Hardy or some user of MO or MSE ? – mick Oct 8 '12 at 12:59
2 I assume. – mick Oct 8 '12 at 13:00
@mick You presume correctly. – David Speyer Oct 8 '12 at 14:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.