Timeline for Modeling decay of a linear system with a mixing term
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 3, 2023 at 22:51 | comment | added | Yaroslav Bulatov | Apparently the $d\to \infty$ case reduces to a kind of "continuous mean-field model", the rank-1 term is the mean-field interaction | |
| Apr 3, 2023 at 21:26 | comment | added | fedja | OK, give me a few days :-) | |
| Apr 3, 2023 at 21:03 | comment | added | Yaroslav Bulatov | $p$ is $\in (1, 2]$. Concrete problems I'm looking at have $p=1.009$, $p=1.064$, $p=1.033$, $p=1.101$, $p=1.4$ (sources) | |
| Apr 3, 2023 at 20:05 | comment | added | fedja | "Entries of h add up to 1, and decay according to a power law with constant p" And $p\in[1,2]$ as before or you want a wider range now? | |
| Apr 2, 2023 at 0:46 | comment | added | Yaroslav Bulatov | @fedya I have the exact formula for $h$ so I was looking for a decent analytic approximation. Entries of h add up to 1, and decay according to a power law with constant $p$ | |
| Apr 1, 2023 at 19:28 | comment | added | fedja | PS If you have your entries in a decreasing order and have an oracle telling $h_i$ by $i$, founding the percentiles can be done by binary search pretty quickly too, so there is no need to guess anything but if you don't, then, as I said, you'd better tell us how we are supposed to just read the data first :-) | |
| Apr 1, 2023 at 19:23 | comment | added | fedja | Just group the entries: if you know them with $1\%$ precision (I'm afraid you cannot claim even that much), then you'll declare those between $0.99^k$ and $0.99^{k+1}$ exactly equal and play with $100\log d$ size, which is manageable up to $\log d=200$ with the previous technique, which gives $d$ not formally infinite, but certainly larger than the number of atoms in the visible universe. If you have an exact formula for $h$, we can think a bit and find a decent analytic approximation that is just a function of 2-4 variables. | |
| Apr 1, 2023 at 2:41 | comment | added | Yaroslav Bulatov | @fedya The difference is that $d$ is assumed near infinite but h has specific form which gives hope for O(1) approach. Your earlier answer was was O(d) and was great in the context of original discussion which was motivated by a different (smaller) problem | |
| Apr 1, 2023 at 2:23 | comment | added | fedja | @YaroslavBulatov Just one question: what is substantially different here from what we discussed before? | |
| Apr 1, 2023 at 2:00 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 23:18 | comment | added | Rodrigo de Azevedo | Somewhat related | |
| Mar 31, 2023 at 22:55 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 22:45 | comment | added | Yaroslav Bulatov | Yes, for $d\times d$ dynamics matrix $A$ we can write $A=\operatorname{diag}[(\mathbf{1}-\mathbf{h})^2]+\mathbf{h}\mathbf{h}^T$ so the question is how $\exp(At)$ behaves when $d$ approaches infinity | |
| Mar 31, 2023 at 22:42 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 22:40 | comment | added | Rodrigo de Azevedo | OK, so you are using $\mbox{diag}$ to build a diagonal matrix with the $(1-h_i)^2$ | |
| Mar 31, 2023 at 22:35 | comment | added | Yaroslav Bulatov | Fully explicit equation is $x_i^{t+1}=(1-h_i)^2 x_i^t + h_i \sum_i x_i^t h_i$ , this form feels verbose | |
| Mar 31, 2023 at 19:13 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 18:56 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 18:46 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 18:32 | comment | added | Yaroslav Bulatov | Ah right, I got too used to frameworks just letting me add $1$ to a vector and various operations applying pointwise by default. Updated for clarity | |
| Mar 31, 2023 at 18:31 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Mar 31, 2023 at 18:11 | comment | added | Robert Israel | I think the letter $h$ may be doing too many different jobs here. | |
| Mar 31, 2023 at 17:29 | history | asked | Yaroslav Bulatov | CC BY-SA 4.0 |