Timeline for zeros of a complex function defined by integers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2017 at 19:46 | comment | added | hyportnex | Thank you for taking my question seriously. Originally, I was hoping to learn about an infinite series with all zeros on a line on the right half plane but I have no idea who downvoted your attempt to solve it. If you and @NoamD.Elkies and others in this forum do not have a ready answer then may surmise that this problem has not been explored? This is actually quite surprising. | |
| Oct 20, 2017 at 19:05 | comment | added | Igor Rivin | Whoever downvoted this, why? | |
| Oct 20, 2017 at 1:41 | comment | added | Noam D. Elkies | Of course the first two zeros have the same real part because they're complex conjugates. But in general the real parts can fall anywhere between $-1$ and the positive root $.7878849\ldots$ of $2^{-\sigma} + 3^{-\sigma} = 1$. Applying Newton's method to the first $10$ odd multiples of $\pi i \, / \log 3$ finds complex roots with real parts approximately $$ 0.4544,\; -.9406,\; .7332,\; -.1326,\; -.4565,\; .7798,\; -.7860,\; .2357,\; .6106,\; -.9992 \;. $$ | |
| Oct 19, 2017 at 23:55 | history | edited | Igor Rivin | CC BY-SA 3.0 |
more pics
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| Oct 19, 2017 at 23:52 | comment | added | Igor Rivin | @NoamD.Elkies No, I am not sure, and in fact, the picture I am about to add seems to indicate that maybe these are only approximate truths. | |
| Oct 19, 2017 at 23:50 | comment | added | Noam D. Elkies | Are you sure? That seems impossible for the same reason: given $\sigma$, there are at most two complex numbers at distance $2^{-\sigma}$ from $0$ and $3^{-\sigma}$ from $1$, and each of them can arise only once, again because $\log 2$ and $\log 3$ are incommensurable. | |
| Oct 19, 2017 at 23:29 | comment | added | Igor Rivin | @NoamD.Elkies I was unclear - they are on a vertical line but not purely imaginary. | |
| Oct 19, 2017 at 23:27 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed more confusion.
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| Oct 19, 2017 at 22:51 | comment | added | Noam D. Elkies | Wait, I don't think $1 + 2^{-s} + 3^{-s}$ can have any purely imaginary zeros at all. If it did then since $2^{-s}$ and $3^{-s}$ are both on the unit circle they'd have to be the cube roots of unity $-\frac12 \pm \frac{\sqrt{3}}{2} i$, and that can't happen because $\log_2 3$ is irrational. | |
| Oct 19, 2017 at 22:43 | comment | added | reuns | Did you mean $1 + \frac{1}{2^s} + \frac{1}{3^s}+\frac{1}{6^s}$ ? I think in "the structure of the Selberg class" they proved such Dirichlet polynomials need to be of the form $\alpha\sum_{d |n} d^{c-s} (a_d+d^b a_{n/d})$ (if the zeros of a Dirichlet polynomial are all on $\Re(s) = \sigma$ then $\log P(s)$ is almost invariant under $\sigma+s \to \sigma-s$. How can it be the case for $1+2^{-s}+3^{-s}$ ?) | |
| Oct 19, 2017 at 22:41 | comment | added | Noam D. Elkies | ( . . . as reuns noted a few seconds before me in the case $(m,n)=(2,3)$; but Igor Rivin probably does mean just $\{1,2,3\}$, not $\{1,2,3,6\}$.) | |
| Oct 19, 2017 at 22:40 | comment | added | Noam D. Elkies | and a more obvious fact is that it holds for $\{1,m,n,mn\}$ (with $1<m<n$). | |
| Oct 19, 2017 at 22:32 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added experimental wisdom.
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| Oct 19, 2017 at 22:25 | history | answered | Igor Rivin | CC BY-SA 3.0 |