Shane Hsu
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 Nov 4 comment Find $\det(C_n)$ where $c_{ij} = 1$ unless $i-j=\pm 1$ But I fail to see how this can lead to recurrence or a general form. Nov 4 asked Find $\det(C_n)$ where $c_{ij} = 1$ unless $i-j=\pm 1$ Nov 4 asked LU decomposition on 5 by 3 matrix. Sep 24 awarded Autobiographer Oct 19 asked Minimize $ab+bc+ca$ under three second degree constraints Sep 29 comment Am I missing anything when doing this taylor expansion? @njguliyev I'll go kill myself, I'm sorry for posting stupid question here. You wanna post it as an answer or delete it? Sep 29 asked Am I missing anything when doing this taylor expansion? Sep 22 comment How to define “closer to proportion” I see. Thanks a lot! Sep 19 awarded Scholar Sep 19 comment How to define “closer to proportion” So the idea is to turn my proportion into vector for calculation. And use Uniform Norm? Sep 19 accepted How to define “closer to proportion” Sep 19 comment How to define “closer to proportion” @copper.hat sum of absolute difference, I will think about that. But how? For example 1 : 3 : 5 and 3 : 4 : 5, one can say their absolute difference is 2 + 1 + 0 = 3 but if we take 1 : 3 : 5 and multiply it by 4/3, we get 4/3 : 4 : 20/3, and their difference will become 5/3 + 0 + 5/3 = 10 / 3, a bit larger than 3. So there will be cases where there are different difference for two proportions. Sep 19 asked How to define “closer to proportion” Jan 2 awarded Commentator Jan 2 comment Why is $ab+bc+ac = 0$ in some situation? @RossMillikan Well, thanks anyway. Jan 2 comment Why is $ab+bc+ac = 0$ in some situation? @RossMillikan That's what I mean by "Hell, it's still wrong" I mean my calculation is correct, but it only happens in VERY VERY special situation. So, yes, the idea is wrong. Jan 2 awarded Citizen Patrol Jan 2 comment Why is $ab+bc+ac = 0$ in some situation? @RossMillikan But hell, it's still wrong, and I think the question itself is not constructive. LOL Jan 2 comment Why is $ab+bc+ac = 0$ in some situation? @RossMillikan Here's where I think I am right about organizing it: $(a+b-c)^2 = a^2+b^2+c^2+2ab-2bc-2ac$ And $2m^2n^2-2c^2=2ab-2bc-2ac$ I can simply substitute it. $(a+b-c)^2 = a^2+b^2+c^2+2m^2n^2-2c^2$ And here you go, $2m^2n^2-2c^2$ should be zero, which means $2ab-2bc-2ac$ should also be zero. Jan 2 comment Why is $ab+bc+ac = 0$ in some situation? I think I should start to put this under trash.