Timeline for The direction that gets me closest to a given point in $\mathbb{R}^n$
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2021 at 1:20 | vote | accept | Lostsoul | ||
| Mar 10, 2021 at 23:35 | answer | added | Yaakov Baruch | timeline score: 2 | |
| Mar 5, 2021 at 7:31 | history | edited | Yaakov Baruch |
Added the inequalities tag
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| Mar 5, 2021 at 0:06 | history | reopened |
Yaakov Baruch Will Sawin David C Yemon Choi David E Speyer |
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| Mar 4, 2021 at 23:13 | comment | added | Lostsoul | @YemonChoi I have added a link to this question on MSE post if that might help. | |
| Mar 4, 2021 at 22:01 | comment | added | Yemon Choi | @YaakovBaruch I agree. Perhaps the votes to close were influenced by the fact that the question was asked on MSE, but I think that as long as a note is left on the MSE question pointing to this one, then there is no risk of duplicated effort. | |
| Mar 4, 2021 at 18:26 | review | Reopen votes | |||
| Mar 5, 2021 at 0:10 | |||||
| Mar 4, 2021 at 18:17 | comment | added | Yaakov Baruch | I would add that if the question deserved to be closed, it would be common courtesy on this site to give some reason and pointer in the right direction when closing (unless are dealing with an outrageously inappropriate question - definitely not the case here!). | |
| Mar 4, 2021 at 18:05 | comment | added | Yaakov Baruch | The question is not trivial. It may follow from well known results, but I don't see a trivial proof, despite having worked many year with vcv matrices, linear regressions, t-values and all that. I think the decision to close was at the least a rushed one. | |
| Mar 4, 2021 at 17:48 | history | closed |
Ben McKay abx David Handelman Neil Strickland Thomas Rot |
Not suitable for this site | |
| Mar 4, 2021 at 17:16 | comment | added | Yaakov Baruch | I deleted several comments that are superceded by my answer below. | |
| Mar 4, 2021 at 17:14 | answer | added | Yaakov Baruch | timeline score: 2 | |
| Mar 3, 2021 at 14:17 | comment | added | Lostsoul | That's a nice observation - I will see if I can generalise it. | |
| Mar 1, 2021 at 23:04 | comment | added | Lostsoul | Yes, it is quite obvious for n=2 and not so for higher dimensions. Would be awesome if someone else has any ideas on this. In any case, thanks very much for your input. | |
| Mar 1, 2021 at 21:21 | comment | added | Yaakov Baruch | Erranda: where I wrote $C\ge \dots$ should be $C^2\ge \dots$ of course. I seem to forget squares and square roots a lot. | |
| Mar 1, 2021 at 21:20 | comment | added | Lostsoul | Right. Was wondering if my condition on vectors being $\theta$ apart could help with this because you said it is equivalent to $det X \neq 0$ but It's a stronger statement than just linear independence. | |
| Mar 1, 2021 at 21:02 | comment | added | Yaakov Baruch | As $x\cdot x=|x|^2$ you get $p\cdot p-\frac{(p\cdot x)^2}{x\cdot x}\le C^2 p\cdot p$, so $C\ge 1-\frac{(p\cdot x)^2}{(x\cdot x)(p\cdot p)}$. You want $x$ to be the $x_i$ that maximizes $\frac{(p\cdot x_i)^2}{(x_i\cdot x_i)(p\cdot p)}$, or equivalently $\frac{p\cdot x_i}{|x_i|}$. (What I stated earlier about maximizing $\frac{p\cdot x_i}{x_i\cdot x_i}$ was wrong.) | |
| Feb 28, 2021 at 14:42 | comment | added | Lostsoul | Thanks, I agree. How can I show that for such $i$, $|\frac{p \cdot x_{i}}{x_i \cdot x_i }x_i - p| \leq C |p|$ with $C \in (0,1)$ though? Sorry if this is very obvious, I am not too sure what you meant by standard regression stuff - I'd appreciat some references that might help with this. | |
| Feb 28, 2021 at 13:34 | review | Close votes | |||
| Mar 4, 2021 at 17:55 | |||||
| Feb 28, 2021 at 12:40 | history | edited | Lostsoul | CC BY-SA 4.0 |
added 10 characters in body
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| Feb 28, 2021 at 12:08 | history | asked | Lostsoul | CC BY-SA 4.0 |