Timeline for The $n$-th derivative has $n$ zeros. Can such a function be unbounded?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2017 at 15:37 | comment | added | Neil Strickland | If we take $f(x)=(1+x^2)^{1/4}$ then numerical calculation suggests that $f^{(n)}(x)$ has precisely $n$ roots, all simple, for $n\leq 20$. | |
| Sep 28, 2017 at 15:24 | comment | added | Mateusz Kwaśnicki | Oh, indeed, then I remembered incorrectly. How about $(1+x^2)^s$ for $s \in (0, \tfrac{1}{2}$)? | |
| Sep 28, 2017 at 15:05 | comment | added | Neil Strickland | The $n$'th derivative is actually a polynomial of degree $n-2$ (not $n$) times $(1+x^2)^{1/2-n}$, provided that $n\geq 2$. This means that $y''$ is potentially bell-shaped, but $y''=(1+x^2)^{-3/2}$, which is bounded. | |
| Sep 28, 2017 at 14:57 | comment | added | M. Winter | Thank you. But a quick check showed no zeros of the second derivative. For your question, do you mean progress since my question on Math.SE? | |
| Sep 28, 2017 at 14:50 | history | answered | Mateusz Kwaśnicki | CC BY-SA 3.0 |