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No because that's a polynomial. The proofs that involve computing moments already have an analog for the Riemann zeta function, but for zeta(s) they just repre
by rfurmani 10y ago
No because that's a polynomial. The proofs that involve computing moments already have an analog for the Riemann zeta function, but for zeta(s) they just represent zero-density results, while in the Deligne case they give you the full on Riemann Hypothesis.
- tprice7 10y agoI am curious which particular proofs you are referring to, could you please give a link or reference or something?
- rfurmani 10y agoI'm having some difficulty finding it, but Terry Tao had a nice exposition in one of his posts, I believe the crux was computing/bounding traces of powers and then being able to conclude the result. My work is almost purely on the analytic side (hence perhaps the RH skepticism), so I don't know the right reference off-hand.