Timeline for An inequality on some pairs of orthogonal vectors
Current License: CC BY-SA 4.0
11 events
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| Oct 2, 2020 at 19:25 | comment | added | fedja | @LSpice I posted the $\pi$ bound for $k=2$. I'm still reluctant to post my arguments for higher dimensions because they give constants rather short of the requested $3$, so they may be merely dead-ends that can divert somebody else from the right path :-) | |
| Oct 2, 2020 at 19:18 | comment | added | LSpice | I edited to add links. I tried to link to @fedja's observation that you mention, but I couldn't figure out which of their comments here makes that observation. | |
| Oct 2, 2020 at 19:17 | history | edited | LSpice | CC BY-SA 4.0 |
Links and very light proofreading
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| Oct 2, 2020 at 19:01 | comment | added | Ivan Meir | @fedga One idea I did have was to try more complex configurations other than a triangle and try to prove that a configuration say for >3 points had a maximal sum of squared sines. For 4 points this might be a regular tetrahedron. Unfortunately this doesn't give you a better result than using the triangle as the inter-vertex angles are pretty close to 90 degrees. | |
| Oct 2, 2020 at 18:50 | comment | added | Ivan Meir | @fedja No I can only do this for $k=2$. The approach I use for $k=2$ relies on the fact that for a given set of vectors you can find another set with precisely double the angles between them and this easy in the plane but impossible except obviously on a great circle for higher dimensions. I did wonder if there was some generalisation of this idea that could give an approximate "doubling" in higher dimensions and hence allow one to use the same method with a poorer bound but I don't see anything obvious unfortunately. | |
| Oct 2, 2020 at 18:43 | comment | added | fedja | @IvanMeir Mistakes are a normal part of the process :-) However, my second question still remains unanswered: are you sure that you can really prove the $2\sqrt 2$ lower bound in all dimensions? That would be wonderful but I'm still not convinced. | |
| Oct 2, 2020 at 18:37 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Added explicit value for C(k,2)
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| Oct 2, 2020 at 18:34 | comment | added | Ivan Meir | @fedja Yes you are right I actually get 4 rather than $2\sqrt{2}$ in my example, embarrassing mistake. I'll update my answer. Appreciate the correction, thank you. | |
| Oct 2, 2020 at 18:12 | comment | added | Giorgio Metafune | It seems that you get $4$ instead of $2\sqrt 2$ when $4$ divides $n$ (there is another inequality in Cauchy Schwartz, I guess). | |
| Oct 2, 2020 at 18:09 | comment | added | fedja | Something is fishy. First of all, if the unit vectors $a_i$ on the plane have just $2$ directions (say $n_1$ vectors in one direction and $n_2$ in the other direction), then, clearly, the average scalar product $|a_i\cdot b_j|$ is $\frac {2n_1n_2}{n^2}\le \frac 12$, so we are rather far from the factor $2\sqrt 2$ or even $\pi$ in the original inequality: we get $4$ instead). Second, while you can arrange $a_i\cdot b_j\ge 0$, for any fixed $i,j$, you may fail to do it for all $i,j$ simultaneously. Am I missing something? | |
| Oct 2, 2020 at 17:12 | history | answered | Ivan Meir | CC BY-SA 4.0 |