Timeline for What are the generating partitions of the odometer?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2016 at 19:49 | history | edited | Anthony Quas | CC BY-SA 3.0 |
include the author's comments
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| Feb 1, 2016 at 19:24 | vote | accept | Stéphane Laurent | ||
| S Feb 1, 2016 at 19:18 | history | suggested | Stéphane Laurent | CC BY-SA 3.0 |
include the author's comments
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| Feb 1, 2016 at 18:08 | comment | added | Stéphane Laurent | Dear Anthony, I took the liberty to edit your answer in order to include the reason for which the partition is generating. I hope this is fine for you. | |
| Feb 1, 2016 at 18:07 | review | Suggested edits | |||
| S Feb 1, 2016 at 19:18 | |||||
| Jan 30, 2016 at 2:35 | comment | added | Stéphane Laurent | That looks nice. I gonna check that. | |
| Jan 30, 2016 at 2:24 | comment | added | Anthony Quas | Code a sequence by a's and b's according to whether it's in A or not. By looking at 4 consecutive terms, you can determine $x_1$; by looking at 8 consecutive terms, you can determine $x_2$ etc. The associated process is Toeplitz: it has a's every 2 terms; then b's every 4 terms filling some of the gaps; then a's every 8 terms filling some of the gaps between these b's etc. | |
| Jan 30, 2016 at 2:15 | comment | added | Stéphane Laurent | I generally index my stochastic processes with negative integers, so I'm not the one who will blame you :-) Ok, I took a look at this partition. Why do you say it is nice ? Is there something special about the associated stationary process on $\{0,1\}$ ? And how do you know this partition is generating ? | |
| Jan 30, 2016 at 2:12 | comment | added | Anthony Quas | I write my odometers backwards compared to the rest of the world, so that carries go to the left (as usual in addition) rather than to the right. | |
| Jan 30, 2016 at 1:44 | comment | added | Stéphane Laurent | Yes, this is the so-called dyadic adic machine. Ok I see now, you mean the number of 0's before the first one. I'll come back after thinking about this partition. | |
| Jan 30, 2016 at 1:21 | comment | added | Anthony Quas | I think of the odometer as the set of left-infinite strings of 0s and 1s. The odometer transformation is adding 1, with carry to the left (just like addition of binary numbers, except these have infinitely many digits). A point in the odometer is therefore $(\ldots,x_2,x_1)$ with the $x_i$'s in $\{0,1\}$. My set is $\{x\colon \exists n\colon x_1=\ldots=x_{2n}=0; x_{2n+1}=1\}$. | |
| Jan 30, 2016 at 0:51 | comment | added | Stéphane Laurent | Excuse-me, what do you mean by the number of terminal 0's ? | |
| Jan 30, 2016 at 0:45 | history | answered | Anthony Quas | CC BY-SA 3.0 |