Timeline for If $f$ is infinitely differentiable then $f$ coincides with a polynomial
Current License: CC BY-SA 4.0
14 events
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| Apr 10 at 15:23 | comment | added | Giorgio Metafune | Well, I understood that I misunderstood. If $f \in C^n$ and at each point there is a vanishing derivative of order less than $n$, there is an easier proof that $f$ is a polynomial. I was confused by seeing derivatives of order higher than $n$. | |
| Apr 9 at 20:17 | comment | added | Giorgio Metafune | Do you mean that there exists $f \in C^{1000}$ such that at any point one derivative of order $\leq 1000$ vanishes? I do not see it, can you explain? | |
| Mar 5, 2021 at 21:18 | comment | added | Sungjin Kim | If we assume $f\in C^{1000}$, then the sets $E_n$ for $n>1000$ are not guaranteed to be closed. For the example, consider $f(x)=0$ if $x<0$, $f(x) = x^{1001}$ if $x\geq 0$. | |
| Jun 3, 2019 at 0:50 | comment | added | Piotr Hajlasz | @Vincent $\mathbb{R}=(-\infty,\infty)$. I never said that the intervals are finite. | |
| Jun 2, 2019 at 17:42 | comment | added | Vincent | Huh no, now it is even more confusing. I understand that open sets are a union of intervals, but not that they are a union of disjoint intervals. Take the case $\Omega = \mathbb{R}$, I don't see any way of writing this as a union of non-overlapping intervals (let alone a finite number of such intervals as the new formula (1) suggests). But on closer inspection I think the condition $(a_i, b_i) \cap (a_j, b_j) = \emptyset$ for $i \neq j$ which is causing my concern is not being used further down the proof, is it? | |
| Jun 1, 2019 at 12:40 | comment | added | Piotr Hajlasz | @Vincent I edited my proof, see formula (1). Is it okay now? | |
| Jun 1, 2019 at 12:39 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
added 16 characters in body
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| May 31, 2019 at 8:37 | comment | added | Vincent | I like this answer a lot, but I am confused about the sentence "The set $\Omega$ is open so $\Omega = \bigcup_{i=1}^\infty (a_i, b_i)$ where $a_i < b_i$ and $(a_i, b_i) \cap (a_j, b_j) = \emptyset$ for $i \neq j$}. Is the last part (empty intersections of intervals) correct? It seems a bit at odds with the claim further down that $\Omega = \mathbb{R}$. | |
| S May 29, 2019 at 8:03 | history | suggested | Lorenzo | CC BY-SA 4.0 |
I made just a tiny correction
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| May 29, 2019 at 7:46 | review | Suggested edits | |||
| S May 29, 2019 at 8:03 | |||||
| Jun 25, 2018 at 21:57 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
edited body
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| Mar 30, 2018 at 18:41 | history | edited | Piotr Hajlasz | CC BY-SA 3.0 |
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| Mar 30, 2018 at 15:08 | history | edited | Piotr Hajlasz | CC BY-SA 3.0 |
added 4 characters in body
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| Mar 30, 2018 at 14:34 | history | answered | Piotr Hajlasz | CC BY-SA 3.0 |