Timeline for $f : [0,\infty) → \mathbb{R}$ with $f(0) = f'(0) = 0$ and $f(x) < x^2$ and $f',f'',f''' > 0$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2019 at 21:55 | audit | First posts | |||
| Feb 7, 2019 at 21:55 | |||||
| Jan 22, 2019 at 9:05 | audit | First posts | |||
| Jan 22, 2019 at 9:05 | |||||
| Jan 19, 2019 at 16:55 | comment | added | Henry | @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x \gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$. | |
| Jan 19, 2019 at 14:02 | comment | added | D777 | How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$? | |
| Jan 17, 2019 at 20:44 | comment | added | Math-fun | Many thanks for this counter example which together with other very nice hints makes the issue clear to me! | |
| Jan 17, 2019 at 20:44 | vote | accept | Math-fun | ||
| Jan 17, 2019 at 19:15 | comment | added | Henry | @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$ | |
| Jan 17, 2019 at 19:06 | comment | added | Daniel Schepler | Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < \frac{1}{2} x^2$ for $x > 0$. | |
| Jan 17, 2019 at 18:56 | history | edited | Henry | CC BY-SA 4.0 |
graph
|
| Jan 17, 2019 at 18:35 | history | edited | Henry | CC BY-SA 4.0 |
correction
|
| Jan 17, 2019 at 18:32 | history | answered | Henry | CC BY-SA 4.0 |