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 Yearling
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  • 0 posts edited
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  • 16 votes cast
Apr
16
asked Is there anything known about the value where the Euler and Hadamard products for $\zeta(s)$ are equal?
Apr
9
answered How could I get access to more than the first 2 mln non-trivial zeros of $\zeta(s)$?
Apr
9
revised How could I get access to more than the first 2 mln non-trivial zeros of $\zeta(s)$?
Turned update into an answer.
Apr
9
revised How could I get access to more than the first 2 mln non-trivial zeros of $\zeta(s)$?
Added solution to my problem.
Apr
8
asked How could I get access to more than the first 2 mln non-trivial zeros of $\zeta(s)$?
Jan
2
comment How are Zeta function values calculated from within the Critical Strip?
Or through this extension: $$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n}{(n+1)^{s}}} + \frac{2\,s-1}{n^s} - {\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad 0<\Re(s)<1$$ we get an even simpler (I believe the simplest) series-expression for: $$\displaystyle \zeta\left(\frac12\right) = \sum _{n=1}^{\infty } \left(\sqrt{n-1} -{\frac {n}{\sqrt{n+1}}}\right)$$
Dec
29
revised Analytic continuation of the Dirichlet $\eta(s)$ series to $\Re(s) \gt -1$. Why does this work?
Simplified the question (restricted to eta only) and included an addition.
Dec
29
comment Analytic continuation of the Dirichlet $\eta(s)$ series to $\Re(s) \gt -1$. Why does this work?
@Timbuc. That is indeed my question. Why does taking the simple average between two series induce the analytic continuation? The only two ways I found to continue the domain of $\eta(s)$ are on the Wiki page en.wikipedia.org/wiki/Riemann_zeta_function under the sections "Rising Factorial" and "Globally convergent series", but these are quite different continuations.
Dec
29
asked Analytic continuation of the Dirichlet $\eta(s)$ series to $\Re(s) \gt -1$. Why does this work?
Dec
28
revised Are these known telescoping series for $\zeta\left(\frac12\right)$?
Added explanation how I derived the second series from the first.
Dec
27
revised An infinite series that gives $f(s)=s$. How could it be explained more easily?
Added a possible explanation.
Dec
26
asked An infinite series that gives $f(s)=s$. How could it be explained more easily?
Dec
23
revised Are these known telescoping series for $\zeta\left(\frac12\right)$?
Added a simplified (more elegant) formula in the last section.
Dec
23
awarded  Yearling
Dec
23
asked Are these known telescoping series for $\zeta\left(\frac12\right)$?
Sep
2
revised Question about the zeros of $\zeta_{H}(s,a) \pm \zeta_{H}(s,1-a)$.
[Edit removed during grace period]
Sep
2
revised Question about the zeros of $\zeta_{H}(s,a) \pm \zeta_{H}(s,1-a)$.
Added meromorphic to be more precise about the type functions I am after.
Sep
2
asked Question about the zeros of $\zeta_{H}(s,a) \pm \zeta_{H}(s,1-a)$.
Aug
20
comment Do all complex zeros of $Li_s(z)\,- \, Li_{1-s}(z)$ get the shape $s=\dfrac12 + \dfrac{k \, \pi }{\,\ln(2)}\,i$ when $z \rightarrow 0^{-}$?
Thanks Antonio! That's it. Subtracting the power series at $z \rightarrow 0$ just reduces to $z^2(2^{āˆ’s}āˆ’2^{sāˆ’1})$. The result then immediately follows (and it also follows that these are the only roots and also none exist outside the strip). Much simpler than I thought...
Aug
19
asked Do all complex zeros of $Li_s(z)\,- \, Li_{1-s}(z)$ get the shape $s=\dfrac12 + \dfrac{k \, \pi }{\,\ln(2)}\,i$ when $z \rightarrow 0^{-}$?