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Question: Do there exist any Riemann zeta $\zeta(s)$ like functions $f_N(s)$ that may have all nontrivial zeros (verified via numerical calculation) on the critical line but only involve integers up to $N$?

The Riemann zeta function $\zeta(s)$ involves all the integers (for $s>1$) and thus all the primes. I would think that if only integers up to $N$ are involved in these functions $f_N(s)$, then only finite number of primes up to $N$ can be involved.

We would also require that $$\lim_{N\to\infty} f_N(s)=\zeta(s)$$

Thus we may study the interaction of prime $2$ with prime $3$ if we set $N=3$ in $f_N(s)$. We may understand better how and at what $N$ the chaotic behavior (if it exists) of the normalized spacing between adjacent zeros of $f_N(s)$ kicks in.

Of course the Euler product may not exist anymore for these functions.

thanks- mike

mike
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    What do you mean by "real zeros on the critical line"? I don't understand your definition of $f_N(s)$? Do you mean the analytic continuation of $\displaystyle\prod_{p\le N}\dfrac{1}{1-p^{-s}}$? – Daniel R Aug 01 '14 at 16:32
  • I changed it to "nontrivial zeros on the critical line". I mean the continuation of $\sum_{n=1}^{N}\frac{1}{n^s}$ for $s>1$. But $\prod_{p\le N}\dfrac{1}{1-p^{-s}}$ could be another option. – mike Aug 01 '14 at 16:58
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    Ah, ok. And, my bad, you wouldn't need to analytically continue $f_N(s)$, since it obviously doesn't diverge being a finite sum. – Daniel R Aug 01 '14 at 17:24
  • Thanks. Eventually I want to take the limit of $N\to\infty$. So I need to deal with the continuation problem somewhere. Does my reasoning make sense? – mike Aug 02 '14 at 00:44
  • So is the answer not known? Why hasn't this been answered? – tox123 Mar 22 '15 at 00:58

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