Timeline for Projecting the unit cube onto subspaces
Current License: CC BY-SA 2.5
11 events
when toggle format | what | by | license | comment | |
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Apr 6, 2011 at 19:59 | comment | added | Denis Serre | @Emil. That was my first solution. But under some external pressure, I switched to my present construction where the $v_k$'s are all positive. Should be more self-confident... | |
Apr 6, 2011 at 15:56 | comment | added | Emil Jeřábek | I hesitate to post this as an answer as all the work has been done by someone else, but the optimal vector for $d=1$ is $v_{2k+1}=\sqrt{k+1}-\sqrt k$, $v_{2k+2}=-v_{2k+1}$. This gives $\|P_ve_I\|\le a\|e_I\|$ for $a=1/\|v\|\sim\sqrt{2/\log n}$. If $w$ is another vector, then one can show the existence of $I$ such that $\|P_we_I\|\ge a\|e_I\|$ by considering the positive and negative entries of $w$ separately, and applying Seva’s argument to both of them. | |
Apr 6, 2011 at 15:48 | comment | added | Emil Jeřábek | $v'=(-1,\sqrt2-1,1)$ achieves $a=1/\sqrt{5-2\sqrt2}\sim0.679$, which is better than $v=(1,\sqrt2-1,\sqrt3-\sqrt2)$ with its $a\sim0.886$. | |
Apr 6, 2011 at 15:20 | comment | added | Denis Serre | What do you mean Aaron with this new vector $(-1,\sqrt2-1,1)$ ? | |
Apr 6, 2011 at 15:14 | comment | added | Denis Serre | @Aaron. One N in Denis in French. Even if there are two A's in Aaron. | |
Apr 6, 2011 at 14:46 | comment | added | Aaron Meyerowitz | @Dennis Aha, so using $\sqrt{2} \approx 7/5.$ in$(-1,\sqrt{2}-1,1).$ | |
Apr 6, 2011 at 13:22 | comment | added | Denis Serre | @Aaron. See my edits. Now my $(a_1,a_2,a_3)$ is better than $(-5,2,5)$. | |
Apr 6, 2011 at 13:22 | history | edited | Denis Serre | CC BY-SA 2.5 |
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Apr 6, 2011 at 11:48 | comment | added | Aaron Meyerowitz | At least for $n=3,$ Your example $(-a,0,a)$ is not quite as good as $(-5,2,5)$. | |
Apr 6, 2011 at 7:59 | history | edited | Denis Serre | CC BY-SA 2.5 |
added 110 characters in body
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Apr 6, 2011 at 7:38 | history | answered | Denis Serre | CC BY-SA 2.5 |