Timeline for For what real $t$ is $\{n^t : n \geq 1\}$ linearly independent over $\mathbb{Q}$?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2015 at 0:12 | vote | accept | jordanbell2357 | ||
| Apr 28, 2015 at 10:56 | comment | added | Emil Jeřábek | Better yet, each relation is a generalized polynomial in $e^t$, hence it rules out a finite set. | |
| Apr 28, 2015 at 0:36 | comment | added | Will Sawin | @DouglasZare In fact each relation rules out a countable set, because any linear combinations of $n^t$ is a nonconstant analytic function, so has finitely many zeros in an interval. | |
| Apr 27, 2015 at 19:38 | comment | added | Douglas Zare | There are countably many possible relations, which each rule out a set of measure $0$, so the set of reals for which the collection is linearly independent has full measure. | |
| Apr 27, 2015 at 19:29 | answer | added | Will Sawin | timeline score: 19 | |
| Apr 27, 2015 at 19:01 | answer | added | Robert Israel | timeline score: 6 | |
| Apr 27, 2015 at 17:24 | history | edited | GH from MO | CC BY-SA 3.0 |
added 2 characters in body; edited tags
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| Apr 27, 2015 at 17:02 | history | edited | jordanbell2357 | CC BY-SA 3.0 |
linearly independent
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| Apr 27, 2015 at 17:00 | comment | added | Emil Jeřábek | Yes, you’re right, it can’t be algebraic. Though I gather $\{n^t:n\ge1\}$ is linearly independent iff $\{p^t:p\text{ prime}\}$ is algebraically independent. | |
| Apr 27, 2015 at 16:56 | comment | added | Robert Israel | Has to be linearly. $4^t = (2^t)^2$. | |
| Apr 27, 2015 at 16:55 | comment | added | Emil Jeřábek | Independent how: linearly? Algebraically? | |
| Apr 27, 2015 at 16:44 | history | edited | Loïc Teyssier | CC BY-SA 3.0 |
LaTeX
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| Apr 27, 2015 at 16:39 | history | asked | jordanbell2357 | CC BY-SA 3.0 |