Timeline for On the nearest integer to $\zeta^{(k)}(1-1/B),B \ge 2$
Current License: CC BY-SA 4.0
12 events
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| Sep 26, 2021 at 16:43 | comment | added | joro | Many thanks for the counterexample! I can reproduce it with mpmath. | |
| Sep 26, 2021 at 8:21 | comment | added | Henri Cohen | My bad. I had divided by $k!$ in my computations. | |
| Sep 26, 2021 at 8:00 | comment | added | Sylvain JULIEN | @Henri Cohen: joro does consider the nearest integer. | |
| Sep 26, 2021 at 1:02 | comment | added | Alexander Kalmynin | @joro, it was not a bug, you just needed to pick a large $k$ (see the update in the answer) | |
| Sep 26, 2021 at 0:50 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Sep 25, 2021 at 22:27 | comment | added | Alexander Kalmynin | @Henri Cohen, besides, even if it was true only for odd $k$, this would imply a similar exponential bound for $f(s)−f(2−2/B−s)$, which is once again false due to the same counterexample. I believe that the experimental data is a kind of the "law of small numbers" and the fact that $\zeta(s)$ grows "slightly" more than exponentially at $s=−n$ for large odd $n$ contributes to this | |
| Sep 25, 2021 at 22:08 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Sep 25, 2021 at 22:04 | comment | added | Alexander Kalmynin | @Henri Cohen, why is it only implied for $k$ odd? The $k$-th derivative of $(s-1)^{-1}$ at $s=1-1/B$ is precisely $k!(-1)^k(s-1)^{-k-1}=-k!(1/(1-s))^{k+1}=-k!B^{k+1}$, the parity of $k$ shouldn't matter, as far as I see. | |
| Sep 25, 2021 at 21:27 | comment | added | Henri Cohen | I believe the above proof is false: the inequality $|f^{(k)}(1-1/B)|\le1$ is implied by the conjecture only for $k$ odd. I believe that using Cauchy it should be easy to prove that the conjectures are both true, replacing integer part by nearest integer. Once this is done, one can ask how the difference behaves, both in sign (is it governed by the complex zeros?) and in magnitude (it tends very fast to $0$, at least experimentally). | |
| Sep 25, 2021 at 20:52 | comment | added | juan | The derivatives of $\zeta(s)$ at a point such as $1-1/B$ behaves in a complicated way. The first few may be dominated by the pole at $0$. So they are near integers if $B$ is an integer. The Taylor series of $f(s)$ is not a good recipe to compute $f(-n)$. Mathematica gives $\zeta^{(4)}(1-1/3)=-5831.99742689039619$ near an integer. mpmath gives -5831.9974268904026. I do not think there is a bug here. | |
| Sep 25, 2021 at 17:59 | comment | added | joro | Thanks. I get experimental support for the conjectures, possibly due to a bug in the libraries I am using. What would be a small counterexample or set of counterexamples to check numerically? | |
| Sep 25, 2021 at 17:39 | history | answered | Alexander Kalmynin | CC BY-SA 4.0 |