Timeline for Problem with definition of tangent vectors as derivations
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| Oct 21 at 20:47 | comment | added | Iosif Pinelis | @SebastianGoette : This is a good point. Let me think about this. | |
| Oct 21 at 19:49 | comment | added | Sebastian Goette | The only proof of your first formula that I know uses functions similar to $g(x)=\begin{cases}(f(x)-f(0))/x&x\ne0\\f'(0)&x=0\end{cases}$ in dimension $1$. But $g$ has one derivative less than $f$, so one cannot apply $D$ to it. But maybe you do this differently? | |
| Oct 21 at 17:41 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |