# is it possible to get the Riemann zeros

since we know that the number of Riemann zeros on the interval $(0,E)$ is given by $N(E) = \frac{1}{\pi}\operatorname{Arg}\xi(1/2+iE)$

is then possible to get the inverse function $N(E)^{-1}$ so with this inverse we can evaluate the Riemann zeros $\rho$ ??

i mean the Riemann zeros are the inverse function of $\arg\xi(1/2+ix)$

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I'm not entirely sure what you are asking but it seems to be related to what I'm about to say. It is entirely possible to calculate actual zeros of the Riemann-zeta function (up to some given height in the complex plane) and verify that they all(up to that height) lie on the critical line. This has been studied extensively, notably, by Odlyzko: dtc.umn.edu/~odlyzko All the information you would need to implement some algorithms and do some calculations yourself are in this well known and cheap book: amazon.com/Riemanns-Zeta-Function-Harold-Edwards/dp/0486417409 – Sam Jones Apr 28 '12 at 16:25
How are we to understand a formula that has $E$ on one side but no $x$, and $x$ on the other side but no $E$? Is $E$ related to $x$ in some way you aren't disclosing? – Gerry Myerson Apr 30 '12 at 4:28
The inverse function would be denoted $N^{-1}(E)$. If by "get" you mean "obtain explicit formula for," then the answer is no, we have no such formula for obtaining the $n$th zero on the upper critical line. – anon Apr 30 '12 at 9:24
and if we plot $N(E)$ and invert it numerically could we then obtain the zeros ?? , – Jose Garcia Apr 30 '12 at 9:26
Sounds like it, yes. Not sure about how that fits into the scheme of things concerning the efficiency of numerical computation of the zeros though. – anon Apr 30 '12 at 9:57

No, your formula is wrong. $N(E)= \frac{1}{\pi} Arg \xi (1/2+iE)$ + a nonzero term coming from the integration along the lines $\Im s =E$ (you are applying an argument prinicple).
Besides, any function $N: \mathbb{R} \rightarrow\mathbb{Z}$ can't be injective for cardinality considerations.