Timeline for Bound on $L^2$ norm of $1/\zeta(1+i t)$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2021 at 8:38 | comment | added | H A Helfgott | I am using Trudgian's bounds on $1/\zeta(1+it)$ within the zero-free region. Kadiri gave bounds on the shape of the zero-free region itself (and later bounds are based on hers). Is it possible to use only the zero-free region itself, rather than bounds on $1/\zeta(1+it)$ in it? Or am I reading too much into a reference? | |
| Dec 8, 2021 at 8:36 | comment | added | H A Helfgott | Just as I (and @Lucia) thought, making @Lucia's second method explicit turns out to be clean and straightforward. | |
| Dec 4, 2021 at 14:09 | comment | added | Lucia | @HAHelfgott: Sure that's fine. | |
| Dec 4, 2021 at 12:11 | comment | added | H A Helfgott | I was confused by the exponent $3$. Even if we just take $x = \exp(C (\log T)^2)$, we save $\exp(-C (\log T)) = T^{-C}$, which is more than enough to get the bounds above. | |
| Sep 11, 2021 at 1:21 | comment | added | Lucia | I'm thinking of just using Perron's formula and shifting contours inside the zero free region. At this point $x=\exp((\log T)^3)$, we'd save $x^{-c/\log T}$ which is extremely big. Obviously you can take a smaller stopping point and save less. My point here is that where you stop almost doesn't matter in the mean-square. So go out very far, so that any calculations needed for the approximation become pretty much trivial. | |
| Sep 11, 2021 at 1:02 | comment | added | H A Helfgott | Thanks again! Excuse me if I am not seeing something, but why do you need to go as far up as $\exp((\log T)^3)$? | |
| Sep 10, 2021 at 21:53 | vote | accept | H A Helfgott | ||
| Sep 10, 2021 at 21:03 | history | edited | Lucia | CC BY-SA 4.0 |
added 870 characters in body
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| Sep 10, 2021 at 20:58 | comment | added | H A Helfgott | Thanks! Is the use of zero-density results essential? | |
| Sep 10, 2021 at 17:47 | history | answered | Lucia | CC BY-SA 4.0 |