Timeline for How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2018 at 19:14 | comment | added | Tom Copeland | What is true is $e^{-xt}f'(x)dx = exp(-f^{-1}(z)t)dz$ for any inverse pair $(z,x)=(f(x),f^{-1}(z))$, so any "solution" has to be a divergent series to circumvent this analytic fact. | |
| Jan 2, 2017 at 19:11 | vote | accept | zeraoulia rafik | ||
| Jan 2, 2017 at 6:43 | comment | added | Denis Serre | The missing argument (understatement ?): because $f'>0$, $f$ must be increasing. Being onto, it thus satisfies $f(\pm\infty)=\pm\infty$. | |
| Jan 2, 2017 at 3:36 | comment | added | user78249 | Oh that integral trick, I never would've thought of that. Nice one. | |
| Jan 2, 2017 at 1:45 | history | answered | Christian Remling | CC BY-SA 3.0 |