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Riemann Hypothesis
GRH
Statement: for \(L(\chi, s)\) an arbitrary L function, its meromorphic continuation to \({\mathbf{C}}\) has zeros only along the line \(\Re(s) = {1\over 2}\).
Obviously wildly open! Some interesting conditional results:
- Compare to Dirichlet’s theorem on primes in arithmetic progressions. Fix an arithmetic progressions \(A = A(a, d) = \left\{{a, a+d, a+2d,\cdots}\right\}\) for \(a,d\) coprime and define \(\pi_A(x) \coloneqq\left\{{p\in A {~\mathrel{\Big\vert}~}p\leq x, p\text{ prime }}\right\}\); then GRH implies an asymptotic estimate in \(x\): \begin{align*} \pi_A(x) \sim \phi(d)^{-1}\int_2^x {1\over \log t}\,dt+ { \mathsf{O}} \qty{x^{{1\over 2} + {\varepsilon}}} \end{align*}
- GRH implies a weak form of Goldbach.
EGRH
More generally, the Dedekind zeta function \(\zeta_K(s)\) of a number field \(K\), the conjecture is that if \(s\) is a zero with \(\Re(s) \in (0, 1)\), then \(\Re(s) = {1\over 2}\).
Conditional results:
- An effective version of Chebotarev density.
Computing zeta(2)
Compute \(\zeta(2)\): consider \(f(z) \coloneqq\pi \cot(\pi z)\), whence \(f(z) = f(z+1)\), has poles at every \(z\in {\mathbf{Z}}\) with residue 1, and is bounded at \(i\infty\).
Then:
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Notes
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Primes
See Connes operator: 