Timeline for Why are these sets divisible by n?
Current License: CC BY-SA 4.0
14 events
when toggle format | what | by | license | comment | |
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Mar 24, 2021 at 21:07 | answer | added | Joe Silverman | timeline score: 2 | |
Mar 24, 2021 at 16:51 | answer | added | Asvin | timeline score: 1 | |
Mar 24, 2021 at 14:07 | comment | added | Vlad Matei | I think this conjecture is due to Odoni. See here matwbn.icm.edu.pl/ksiazki/aa/aa93/aa9317.pdf for a general discussion of irreducibility of iterates of quadratic polynomials | |
Mar 24, 2021 at 5:33 | history | edited | Asvin | CC BY-SA 4.0 |
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Mar 24, 2021 at 5:31 | history | undeleted | Asvin | ||
Mar 24, 2021 at 5:31 | history | deleted | Asvin | via Vote | |
Mar 24, 2021 at 2:29 | comment | added | Asvin | Sorry, $g_n(c) = f^{(n)}(c,0)$, I shouldn't have set that equal to 0. | |
Mar 24, 2021 at 2:28 | history | edited | Asvin | CC BY-SA 4.0 |
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Mar 24, 2021 at 2:25 | comment | added | LSpice | I'm still confused: $g_n(c) = f^{(n)}(0, c) = 0$ seems to me to say that $g_n(c) = 0$. I guess it's some implicit definition: $g_n(c)$ is the value such that $f^{(n)}(0, c) = 0$? But the value of what? Anyway, maybe it's just me, and someone more familiar with these sort of iterated systems will understand the notation immediately. | |
Mar 24, 2021 at 2:16 | comment | added | Asvin | I made a minor edit but what's going is that $f^{(n)}(z,c)$ is a 2 variable polynomial. We set $z = 0$ and $f^{(n)}(0,c) = 0$ to define $g_n(c)$ as a 1 variable polynomial. I hope that helps - the definition is definitely confusing at first sight. | |
Mar 24, 2021 at 2:15 | history | edited | Asvin | CC BY-SA 4.0 |
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Mar 24, 2021 at 2:13 | comment | added | LSpice | I hope it's just me, but I can't understand the definition of $g_n$. $f$ starts off as a 2-variable polynomial, then becomes a 1-variable polynomial, and then we define $g_n(c)$ by a formula that doesn't seem to involve $c$ at all. | |
Mar 24, 2021 at 2:11 | history | edited | LSpice | CC BY-SA 4.0 |
TeXing
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Mar 24, 2021 at 2:08 | history | asked | Asvin | CC BY-SA 4.0 |