In wolfram alpha the values of $$\zeta(0.5+ie^x)$$ closed to zero then How
do I know the real values of $\zeta(0.5+ie^x)$ for large real number $x$ ?
Thank you for any help
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1The paper by Gourdon and Sebah "Numerical evaluation of the Riemann Zeta-function" and this thread may help. – Raymond Manzoni Jun 28 '15 at 09:28
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Choose your favorite analytic continuation and calculate it. That's one way.
For instance, you might use that $$ \zeta(s) = (1 - 2^{1 - s})^{-1} \eta(s)$$ where $$ \eta(s) = \sum_{n \geq 1} \frac{(-1)^{n - 1}}{n^s},$$ which simply converges at values $s = \frac 12 + it$.
If you're asking how others go about it, many use a so-called Approximate Functional Equation (or series accelerations of it or the $\eta$ function). See this MO question for a bit more about the approximate functional equation.
davidlowryduda
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@thank you for your answer, is n(s) able to calculate :n(0.5+ie^x) for large x ? – zeraoulia rafik Jun 26 '15 at 23:48
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1Yes, of course. What I call $t$, you call $e^x$. I'm not promising that it converges quickly, but it converges all the same. – davidlowryduda Jun 27 '15 at 00:22
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but should be know that w.alpha also used n(s) but didn't up to calculate it – zeraoulia rafik Jun 27 '15 at 00:23
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