Timeline for concavity of a vector function
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2016 at 12:19 | comment | added | Alex Ravsky | The function $f(x,y)=3xy-x^2-y^2$ is strictly concave at the point $(0,0)$ in every standard direction (either $x=0$ or $y=0$) but is not concave in direction $x=y$. | |
| Sep 10, 2015 at 8:57 | comment | added | Marek Adamczyk | Basically, I wanted to know if this is a well established fact. I think I can come up with a tailored argument using the fact that the function is quite simple, but would prefer to rely on a general property. | |
| Sep 10, 2015 at 8:52 | comment | added | Marek Adamczyk | Sorry, a mistake. It's from $\mathbb{R}$. The original I was considering was from $[0,1]$ because it had a normalization factor. | |
| Sep 10, 2015 at 8:51 | history | edited | Marek Adamczyk | CC BY-SA 3.0 |
added 5 characters in body
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| Sep 10, 2015 at 8:48 | comment | added | Fedor Petrov | Yes, $g$, why do its values belong to $[0,1]$? | |
| Sep 10, 2015 at 8:06 | comment | added | Marek Adamczyk | You mean $g$? It takes $n$ numbers and outputs a product of $n$ numbers. Function $h$ is from $\mathbb{R}^{n(n-1)/2}$. | |
| Sep 10, 2015 at 7:18 | comment | added | Fedor Petrov | First of all, how is it from $\mathbb{R}^n$ to $[0,1]$? | |
| Sep 10, 2015 at 3:23 | history | edited | Marek Adamczyk | CC BY-SA 3.0 |
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| Sep 10, 2015 at 2:24 | history | asked | Marek Adamczyk | CC BY-SA 3.0 |