Timeline for Numbers with known irrationality measures?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2014 at 3:23 | vote | accept | Stanley Yao Xiao | ||
| Nov 1, 2013 at 22:25 | comment | added | Salvo Tringali | For the record (and for those interested), the result mentioned by Noam in the comments above (as for the irrationality measure of $e^{2/k}$ when $k$ is a non-zero integer) is due to C. S. Davis, see Rational approximations to $e$, J. Austral. Math. Soc. Ser. A 25 (1978), 497-502. | |
| Mar 15, 2012 at 3:54 | answer | added | Jeffrey Shallit | timeline score: 11 | |
| Feb 27, 2012 at 8:45 | comment | added | Jan Jitse Venselaar | The result that for almost all numbers $\mu(x)$ is $2$ is Khinchin's Diophantine approximation theorem. See en.wikipedia.org/wiki/Diophantine_approximation for some details. Schmidt's book on Diophantine approximation has proofs and references. | |
| Feb 27, 2012 at 5:29 | answer | added | Robert Israel | timeline score: 23 | |
| Feb 27, 2012 at 5:22 | answer | added | Gerry Myerson | timeline score: 5 | |
| Feb 27, 2012 at 0:41 | comment | added | Zack Wolske | @Anthony: Consider $x=0$, so that we want $\frac{p}{q} < \frac{1}{q^r}$. This has no solution for any $r \geq 1$ when $p \neq 0$. The statement of the question should say that $\frac{p}{q} \neq x$, then it follows that all rationals have $\mu(x) = 1$. | |
| Feb 26, 2012 at 22:49 | comment | added | Anthony Quas | You probably mean $\mu(q)=\infty$ for a rational $q$? Everything has irrationality measure at least 2 by the pigeonhole principle | |
| Feb 26, 2012 at 21:39 | comment | added | Stanley Yao Xiao | Yes you are right, I must flipped the letters around in my head while typing, thanks for pointing that out. | |
| Feb 26, 2012 at 21:38 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
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| Feb 26, 2012 at 20:01 | comment | added | Gerald Edgar | Your definition is garbled. Perhaps $s=r$. | |
| Feb 26, 2012 at 19:38 | comment | added | Noam D. Elkies | $\mu(e)=2$ follows quickly from the continued-fraction expansion (and generalizes to $e^{2/k}$ if I remember right). – | |
| Feb 26, 2012 at 19:25 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |