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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...

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    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.)2012-03-18
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    @joriki So the whole _ansatz_ is blur?2012-03-18
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    Sorry, I don't know what "blur" means in this context.2012-03-18
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    @joriki I mean wrong.2012-03-18
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    I see now that your third question is whether there are finitely or infinitely many cases 2. I don't understand how that fits in with your earlier claim that there must obviously be infinitely many more cases 1, since that would imply finitely many cases 2.2012-03-18
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    @joriki So can't it be that we have infinitely many cases of 2, but still (infinitely) more cases of 1? I didn't want to exclude this case, but if can show that this is wrong, I'll edit my question...2012-03-18
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    Then you'd have to explain in which sense you're using "(infinitely) more". Under the standard usage of terms like "infinitely many" and "infinitely more", they refer to cardinals and cardinal arithmetic. Since the cardinality of all terms is the least infinite cardinal, that of the natural numbers, there can't be infinitely many of both cases *and* infinitely more of one case, since that would require at least one of the cardinalities to be an infinite cardinal less than the least infinite cardinal. In other words, infinitely many natural numbers are always as many as all natural numbers.2012-03-18
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    @joriki ok, if this solves question 3. and therefore we have only a finite number of cases 2, can I then use the product splitting?2012-03-18
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    From what are you concluding that we have only a finite number of cases 2? I was saying that so far you haven't given an argument why there should be a finite or infinite number of them. That there has to be a finite number would follow if we knew there were infinitely many more cases 1 than cases 2, but you haven't given an argument for that either. The convergence of the product doesn't imply that.2012-03-18
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2825/discussion-between-draks-and-joriki)2012-03-18
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    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.2012-03-19
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    The issue @Robert Israel raises has been discussed several times on MO. See http://mathoverflow.net/questions/63714 and http://mathoverflow.net/questions/63787 . I showed that the Euler product does not converge to a nonzero value anywhere in the critical step (except on the real line), and GH showed that the Euler product does not converge to $0$ or $\infty$ in the region $s+it$, $s>1/2$, $t \neq 0$. We never nailed down whether converges to zero on $s=1/2$, although we expect not.2012-03-19

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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.

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    I just had a look at the numerics and the oscillate as you proposed. Thanks for answer.2012-03-21
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    @David : You talk about GH , do you mean GH Hardy or some user of MO or MSE ?2012-10-08
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    http://mathoverflow.net/users/11919/gh I assume.2012-10-08
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    @mick You presume correctly.2012-10-08