Timeline for Prove that a function of two variables is differentiable at $(0,0)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2023 at 8:55 | comment | added | Ryszard Szwarc | I would use $h_1^2h_2^2\le {1\over 2}(h_1^2+h_2^2)^2.$ | |
| Feb 5, 2023 at 1:23 | comment | added | peek-a-boo | @coboy indeed, it’s a very simple, but necessary fix. | |
| Feb 5, 2023 at 1:21 | vote | accept | G2MWF | ||
| Feb 5, 2023 at 1:21 | comment | added | G2MWF | I think I see your point yes. With your decomposition you avoid the division by zero since if one of the $h_i$'s is $0$ the denominator is still defined because it implies the other $h_i$'s to be nonzero.. it's slightly subtle but totally right and it's an important remark, I hadn't thought about the fact of having one of the $h_i$ equal to $0$. Thanks a lot! | |
| Feb 5, 2023 at 1:13 | history | edited | peek-a-boo | CC BY-SA 4.0 | added 242 characters in body |
| Feb 5, 2023 at 1:08 | history | answered | peek-a-boo | CC BY-SA 4.0 |