Timeline for Does iterating the derivative infinitely many times give a smooth function whenever it converges?
Current License: CC BY-SA 4.0
10 events
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| Jan 4, 2023 at 1:38 | history | edited | LSpice | CC BY-SA 4.0 |
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| Oct 14, 2022 at 17:39 | history | edited | Christian Remling | CC BY-SA 4.0 |
I removed a mildly confusing use of the symbol $g$ for two different functions (the limit and the dominating function).
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| Jan 6, 2022 at 6:48 | history | edited | username | CC BY-SA 4.0 |
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| Jan 6, 2022 at 6:42 | history | edited | username | CC BY-SA 4.0 |
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| Jan 6, 2022 at 6:04 | comment | added | username | @IosifPinelis Yes, I agree, it was silly. I started trying to do something like exp(-x)f and changed mid-flight... | |
| Jan 5, 2022 at 22:00 | comment | added | Iosif Pinelis | If you assume that the sequence $f^{(n)}$ is uniformly bounded (even locally), then you do not even need to integrate by parts to get the desired result -- just use dominated convergence for the integrals $\int_0^x f^{(n)}(t)\,dt$. | |
| Jan 5, 2022 at 21:43 | comment | added | username | @PaulCusson what it does is that it gives a strong indication that you find as expected the exponential and nothing more. | |
| Jan 5, 2022 at 21:33 | comment | added | Paul Cusson | Very nice, though of course in the general case that Fedor does $f$ need not be bounded at all. I'm still struggling to fully understand his argument and the consequence, in particular is the value of $c$ the same at every point where it works? I can see that it works on a dense set, does this imply it works everywhere as well? | |
| Jan 5, 2022 at 21:28 | history | edited | username | CC BY-SA 4.0 |
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| Jan 5, 2022 at 21:22 | history | answered | username | CC BY-SA 4.0 |