Timeline for If $f$ is infinitely differentiable then $f$ coincides with a polynomial
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2016 at 22:14 | comment | added | matgaio | @AndreaFerretti of course, thank you. | |
| Sep 4, 2016 at 10:09 | comment | added | Andrea Ferretti | E_n is the interior of A_n. For a point in E_n you have a whole interval where the nth derivative vanishes identically, hence all subsequent derivatives vanish | |
| Sep 4, 2016 at 6:03 | comment | added | matgaio | @AndreaFerretti, just for me to understanding well: are you proving that if for all $x$ there is a natural $n_x$ such that $n\geq n_x$ implies $f^{(n)}(x)=0$ then $f$ is polynomial? I just want to realize what implies $E_m\subset E_n$ in your proof. | |
| Nov 3, 2011 at 18:54 | comment | added | Andrea Ferretti | Those functions have all derivatives 0 in a point, not on a whole interval | |
| Nov 1, 2011 at 5:38 | comment | added | Will Sawin | In step $3$, what about functions of the form $e^{-1/x}$. They can have a derivative $0$ on an interval and all future ones zero on the boundary. | |
| Aug 1, 2010 at 16:21 | comment | added | Andrea Ferretti | Yes, of course. The proof is the same. | |
| Aug 1, 2010 at 13:21 | comment | added | C.S. | Hi-- Thanks a lot. Now, does this remain true if we replace $[0,1]$ by $\mathbb{R}$ or $[a,b]$ | |
| Aug 1, 2010 at 1:55 | history | answered | Andrea Ferretti | CC BY-SA 2.5 |