Timeline for Lebesgue measure of some set of irrational numbers
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2014 at 3:10 | vote | accept | sokho | ||
| Mar 28, 2014 at 0:54 | vote | accept | sokho | ||
| Mar 28, 2014 at 4:09 | |||||
| Mar 27, 2014 at 14:58 | answer | added | Thierry de la Rue | timeline score: 10 | |
| Mar 27, 2014 at 13:43 | comment | added | user25199 | The $M$ constraint is still at infinitely many values, so the measure will be zero. | |
| Mar 27, 2014 at 13:29 | comment | added | sokho | @Carl I have edited my question. Sorry it was my mistake. what about now can I expect a positive measure? | |
| Mar 27, 2014 at 13:22 | history | edited | sokho | CC BY-SA 3.0 |
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| Mar 27, 2014 at 13:08 | comment | added | user25199 | @user39115 Both sets are uncountable, since the bound of $M$ (unless $M=1$) permits arbitrary infinite sequences of 1 and 2 in the relevant coefficients. The measure is zero; the usual question for such sets is the Hausdorff dimension. | |
| Mar 27, 2014 at 12:08 | comment | added | user39115 | For me will be interesting to define $a_s\leq \log s$ and then ask for the measure of the sets when $M$ was replaced by this function. | |
| Mar 27, 2014 at 11:58 | comment | added | user39115 | It is not clear to me what does it mean $\alpha=[a_1,a_2,\ldots,a_s,\ldots)$? If the meaning is $\alpha=\sum_{i=1}^{\infty} \frac{a_i}{10^i},$ then the condition $a_s<M$ does not make sense for $M>9,$ so probably there will be not lost of generality in assuming $M=2$ for example, in which case i will bet that is not difficult to prove that the answer is a number in (0,1). If the meaning is the continuous fraction expansion then I will bet again that is not difficult to prove that the answer is $0.$ | |
| Mar 27, 2014 at 11:54 | comment | added | user25199 | @user39115 Look up "Continued fraction expansion." | |
| Mar 27, 2014 at 7:22 | comment | added | user25199 | I'm fairly sure that fixing a partial quotient or insisting that it be less than a fixed bound will in each case reduce the measure by a factor bounded away from unity. You have an infinite number of such constraints so it is clear the answer should be zero. Perhaps an expert can confirm and provide more details. | |
| Mar 27, 2014 at 2:22 | history | edited | sokho | CC BY-SA 3.0 |
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| Mar 27, 2014 at 2:17 | comment | added | Kurisuto Asutora | What does the statement of Question 1 mean? Maybe you want to say $\mathcal{I}(n^2,n,M)$? In any case, you may want to reconsider the answers to your question from yesterday, they might give you some advice on today's problem as well. | |
| Mar 27, 2014 at 2:11 | comment | added | Ben Willson | I don't understand how you are defining $\alpha$. | |
| Mar 27, 2014 at 1:53 | history | asked | sokho | CC BY-SA 3.0 |