Timeline for Roots of polynomial-like equations with irrational powers
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2010 at 17:42 | vote | accept | Ganesh | ||
| Nov 6, 2010 at 15:35 | comment | added | Thierry Zell | Terminology continued: here is another example of a recent question using the more generalized meaning of exponential polynomial: mathoverflow.net/questions/45031/… | |
| Nov 6, 2010 at 15:22 | history | edited | Thierry Zell | CC BY-SA 2.5 |
added terminlogy caveat
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| Nov 6, 2010 at 15:21 | comment | added | Thierry Zell | @Mark: Oh I see: it's a terminology issue. Many people use "exponential polynomial" for any polynomial that shows up in Khovanskii theory, i.e. of the form $\exp \langle a \mid y \rangle$, because in that context, there's no reason to distinguish between exponential of a variable and of a linear combination of variables. Sorry for the confusion... | |
| Nov 6, 2010 at 9:17 | comment | added | user6976 | @Thierry: By definition an exponential polynomial is $P(x_1,...,x_k, e^{y_1},...,e^{y_m}$, where $P$ is a polynomial. So $e^{t\sqrt{2}}$ is not an exponential polynomial? Right? | |
| Nov 6, 2010 at 1:56 | history | edited | Thierry Zell | CC BY-SA 2.5 |
Clarified (?) the change of variable issue
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| Nov 6, 2010 at 1:47 | comment | added | Thierry Zell | @Mark: You're right, my original answer is a bit sloppy. I'll fix this. | |
| Nov 6, 2010 at 1:35 | comment | added | user6976 | @Thierry: Perhaps I do not understand something in your answer, but why $x^{\sqrt{2}}$ an exponential polynomial? | |
| Nov 6, 2010 at 1:02 | history | answered | Thierry Zell | CC BY-SA 2.5 |