Timeline for Is this entire function a square?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2023 at 3:09 | review | Close votes | |||
| Jul 17, 2023 at 6:30 | |||||
| Jul 12, 2023 at 21:19 | comment | added | Noam D. Elkies | The function $f(z)$ has zeros very near to $$ z = \pm 7.497676277776385498272325 \pm 2.768678282987321532495314i $$ which are necessarily simple because (as already noted) $f'(z) = 1 - \cos z$ has only real zeros. Therefore $f$ does not have an analytic square root. | |
| Jul 12, 2023 at 12:11 | history | edited | Qfwfq | CC BY-SA 4.0 |
(changed the title to make it, I hope, more clear)
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| Jul 11, 2023 at 19:38 | history | edited | LSpice | CC BY-SA 4.0 |
`\label`
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| Jul 11, 2023 at 18:11 | history | became hot network question | |||
| Jul 11, 2023 at 17:53 | answer | added | Christian Remling | timeline score: 16 | |
| Jul 11, 2023 at 17:32 | comment | added | Christian Remling | I'm slightly puzzled by the poor reception of this question (currently 2 downvotes, 1 close vote). While it may be true, as explained by Aleksei, that no seems much more likely as the answer than yes, I don't see why this would make the question invalid. It still seems perfectly reasonable to ask for a proof of this fact. | |
| Jul 11, 2023 at 12:49 | answer | added | Mikhail Katz | timeline score: 3 | |
| Jul 11, 2023 at 10:47 | review | Close votes | |||
| Jul 11, 2023 at 13:18 | |||||
| Jul 11, 2023 at 10:26 | comment | added | Aleksei Kulikov | That would be the most incredible coincidence in mathematics then :) but alas, most probably it is not true and here's how you can check it: find (numerically) some zero of $f$ off the real line, and then take a small circle around it and compute the logarithmic integral $\frac{f'(z)}{f(z)}dz$ along this circle. If the roots were doubled then we should get an even mutliple of $2\pi i$, but you most likely get something around $2\pi i$ so the root is simple and so there's no such $g$. | |
| Jul 11, 2023 at 10:22 | comment | added | HenrikRüping | In case the notation $g^2$ refers to $x\mapsto g(x)^2$ and not $x\mapsto g(g(x))$: Then I would write down the Taylor series for $f$, write $g(x)=\sum a_i x^2$, try to solve for the $a_i$. I would bet that these $a_i$'s still go to zero fast enough, so that $g$ is defined on the whole of $\mathbb{C}$. | |
| Jul 11, 2023 at 10:11 | history | asked | Bazin | CC BY-SA 4.0 |