Timeline for $n$th root of $(a,b) \mapsto (\operatorname{gm}, \operatorname{am})$
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2019 at 14:55 | comment | added | Yaakov Baruch | Nice argument! I think I see one possible reason for the the lack of second differentiability. The construction I gave in my answer relies on the inverse funtion $T^{-1}(a,b)=\big(b-\sqrt{b^2-a^2},b+\sqrt{b^2-a^2}\big)$. Interestingly $T^{-1}$, unlike $T$, is not symmetric in the arguments. That is, $T$ does not really require $b\ge a$, but $T^{-1}$ does (or else the $\text{max}$ and $\text{sign}$ functions need to be used). That could cause my $T^{1/2}$ to be the collation of 2 functions with compatible first derivatives along the diagonal, but opposite second derivatives. | |
| Oct 20, 2010 at 14:21 | history | edited | Denis Serre | CC BY-SA 2.5 |
changing format of a matrix
|
| Sep 25, 2010 at 17:42 | comment | added | Michael Hardy | Hello. I haven't yet looked this over closely; I've been busy with a few things. I'll get to it. | |
| Sep 24, 2010 at 20:40 | comment | added | Denis Serre | The lack of answer from you suggests to me that my answer is not acceptable. Perhaps because your domain is $0< a< b$ and therefore you don't care about the behaviour of the square root at the boundary $a=b$. My answer uses regularity there, and you might expect a square root that is not regular, at least not twice differentiable at the boundary. | |
| Sep 22, 2010 at 15:18 | comment | added | Denis Serre | @Michael. If you think that this answer solves the problem, please feel free to accept it. | |
| Sep 22, 2010 at 9:28 | history | answered | Denis Serre | CC BY-SA 2.5 |