Timeline for The $n$-th derivative has $n$ zeros. Can such a function be unbounded?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Sep 29, 2017 at 0:46 | vote | accept | M. Winter | ||
| Sep 29, 2017 at 0:45 | comment | added | M. Winter | @AaronBergman I think we can divide the "bell-shaped" functions from my question into the two classes bounded and unbounded and the latter ones are exactly the ones you want to characterize. When the function has a minimum at $x=0$ then $f''(0)>0$ and because there are zeros of $f''$ left and right of $x=0$ the second derivative is negative far from the origin. This means $f$ is finally concave, hence sub-linear. I am missing something? | |
| Sep 29, 2017 at 0:16 | comment | added | Aaron Bergman | For what it's worth, I think basically the same proof works for $log(1+x^2)$. I wonder if one can characterize a class of functions that look like these inverted bell shapes with slow (i.e., less than linear) growth to infinity. | |
| Sep 28, 2017 at 15:52 | history | edited | js21 | CC BY-SA 3.0 |
added 7 characters in body
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| Sep 28, 2017 at 15:46 | history | answered | js21 | CC BY-SA 3.0 |