Timeline for antiderivative always exists? [closed]
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 17, 2020 at 7:20 | history | closed |
David Handelman Noah Schweber Tobias Fritz Max Horn Stefan Waldmann |
Not suitable for this site | |
| Mar 11, 2020 at 18:19 | history | became hot network question | |||
| Mar 11, 2020 at 17:56 | comment | added | Noah Schweber | This is a good question, but more appropriate at MSE. | |
| Mar 11, 2020 at 17:13 | answer | added | Bazin | timeline score: 2 | |
| Mar 11, 2020 at 13:19 | comment | added | Mateusz Kwaśnicki | No, it is perfectly sufficient to have $f'$ strictly monotone (by assumption) and continuous (by monotonicity and intermediate value property). This becomes fairly straightforward if written as the limit of Riemann–Stieltjes sums. | |
| Mar 11, 2020 at 13:13 | comment | added | Hair80 | Writing $df'(s)$ in the first line does not imply that $f'$ is differentiable? | |
| Mar 11, 2020 at 11:23 | comment | added | Mateusz Kwaśnicki | (1) $f(x) = 0$ for $x \ne 0$, $f(0) = 1$ has no antiderivative (and it is Riemann-integrable, by the way). (2) Every continuous function clearly has an anti-derivative. (3) Using the Riemann–Stieltjes integral and integration by parts, we have $$\begin{aligned} \int (f')^{-1}(s) ds & = \int x df'(x) \\& = xf'(x) - \int f'(x) dx = xf'(x) - f(x) + C \\& = s(f')^{-1}(s)-f((f')^{-1}(s)) + C.\end{aligned}$$ | |
| Mar 11, 2020 at 11:00 | history | edited | Hair80 | CC BY-SA 4.0 |
added 321 characters in body
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| Mar 11, 2020 at 10:47 | vote | accept | Hair80 | ||
| Mar 11, 2020 at 10:50 | |||||
| Mar 11, 2020 at 10:40 | review | Close votes | |||
| Mar 17, 2020 at 7:20 | |||||
| Mar 11, 2020 at 9:57 | answer | added | A. Bailleul | timeline score: 14 | |
| Mar 11, 2020 at 9:50 | history | asked | Hair80 | CC BY-SA 4.0 |