At what hight is the nth zero Riemann zeta function?
It is a mathematical formula but I can not get under it on the Internet (once I found).
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Duplicated: http://math.stackexchange.com/questions/921987/asymptotics-for-zeta-zeros. – Martín-Blas Pérez Pinilla Feb 10 '15 at 09:07
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It is known that $N(T)$, the number of zeros of the Riemann zeta-function above the real axis with imaginary part (height) at most $T$ has the asymptotic formula $$ N(T) \sim \frac T{2\pi} \log \frac T{2\pi e}. $$ (The error term is very good, namely $O(\log T)$, and can be made explicit if needed.) It follows that the height of the $n$th zero is asymptotically $2\pi n/\log n$; more precisely, it is $$ \sim \frac{2\pi n}{\log n} \bigg( 1 + \frac{\log\log n+1}{\log n} \bigg) $$ (although here the error term isn't as good as in the first formula).
Greg Martin
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Something, however, with the first Fomule probably wrong. For N(1) = -0.45118513581289504; for N(10) = -0.8470652798318857 – user3911255 Feb 10 '15 at 09:02
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