Timeline for $\zeta^{(k)}(s) < 0$ for $s\in (0,1)$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2017 at 13:37 | vote | accept | H A Helfgott | ||
| Oct 30, 2017 at 3:41 | history | edited | GH from MO |
edited tags
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| Oct 29, 2017 at 22:57 | answer | added | Ralph Furman | timeline score: 14 | |
| Oct 29, 2017 at 22:26 | answer | added | Salvo Tringali | timeline score: 8 | |
| Oct 29, 2017 at 21:03 | comment | added | user41593 | A rigorous but not quick way: write the Abel-Plana formula as $$\zeta(s)=\frac{1}{2}+\frac{1}{s-1}+\int_0^\infty \frac{2\sin(s \arctan x)}{(1+x^2)^{s/2}(e^{2\pi x}-1)} dx$$ and bound the modulus in an appropriate way... | |
| Oct 29, 2017 at 20:39 | comment | added | H A Helfgott | That's nice, but I wonder how to check the bound $|\zeta(s)|\leq 9$ rigorously and quickly. Incidentally, there is a paper by Apostol (ams.org/journals/mcom/1985-44-169/S0025-5718-1985-0771044-5/…) that gives numerical values for $\zeta^{(n)}(0)/n!$ for $n<=18$ (but, oddly does not seem to give a simple bound on the rate of convergence to $-1$, though the data very strongly suggest such a convergence). | |
| Oct 29, 2017 at 20:20 | comment | added | Terry Tao | For instance, it seems that $|\zeta(s)| \leq 9$ for $|s| = 10$, which seems to handle all $n \geq 1$, leaving only the classical $\zeta(0) = -1/2$. | |
| Oct 29, 2017 at 20:04 | comment | added | Terry Tao | One can make Emanuele's argument quantitative by observing from the residue theorem that $\frac{\zeta^{(n)}(0)}{n!} = -1 + \frac{1}{2\pi i} \int_\gamma \frac{\zeta(s)}{s^{n+1}}\ ds$ for any contour $\gamma$ going anticlockwise around both $0$ and $1$, e.g. the circle of radius $2$. One can use numerical bounds on $\zeta$ on such a contour to get an exponentially decaying bound for the integral which should suffice to obtain the claim for all but a small number of $n$ (perhaps just the ones listed by Gerald, in fact, given how fast the coefficients seem to converge to $-1$). | |
| Oct 29, 2017 at 19:59 | comment | added | user41593 | They are indeed eventually negative: write $$\zeta(1-s)+s^{-1}=\sum_n (\zeta^{(n)}(0)/n!+1)(1-s)^n$$ which converges around $s=0$ and is $O(1)$ at $s=0$, thus $\zeta^{(n)}(0)/n!+1 \rightarrow 0$ as $n \rightarrow \infty$. | |
| Oct 29, 2017 at 19:51 | comment | added | H A Helfgott | Sure, but how do you do that? (Does it follow easily from the functional equation?) | |
| Oct 29, 2017 at 19:42 | comment | added | Gerald Edgar | The Taylor series for $\zeta(z)$ at $z=0$ has radius of convergence $1$, so it suffices to show all coefficients of it are negative... $\zeta(z) =- 0.5000000000- 0.9189385335\,z- 1.003178229\,{z}^{2}- 1.000785195\,{ z}^{3}- 0.9998793011\,{z}^{4}- 1.000001942\,{z}^{5}+\dots$ | |
| Oct 29, 2017 at 19:04 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
typo in the title
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| Oct 29, 2017 at 18:33 | history | asked | H A Helfgott | CC BY-SA 3.0 |