Timeline for Checking the uniform denseness of a set in $C([0, 1], \mathbb{R}^2)$
Current License: CC BY-SA 4.0
11 events
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| Dec 6, 2021 at 5:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 8, 2021 at 4:09 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 10, 2021 at 4:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 11, 2020 at 3:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 11, 2020 at 2:50 | history | edited | potionowner | CC BY-SA 4.0 |
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| Nov 9, 2020 at 23:25 | history | edited | potionowner | CC BY-SA 4.0 |
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| Nov 9, 2020 at 6:16 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Nov 8, 2020 at 22:52 | comment | added | potionowner | @DieterKadelka Sorry I should have been clearer. I’ve edited the OP. I meant that $P$, as a function, can be written as a polynomial of $\lambda$. For example, if we have $\lambda(x)=\cos (x)$, then $P(x)= \cos^2(x)=\lambda^2$ is a polynomial of $\lambda$. | |
| Nov 8, 2020 at 22:50 | history | edited | potionowner | CC BY-SA 4.0 |
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| Nov 8, 2020 at 22:34 | comment | added | Dieter Kadelka | I don't understand. what is meant by "$P(\lambda)$ is a polynomial"? Is $P$ (without $\lambda$) a polynomial? I'm irritated since $\lambda$ is a function, not a parameter. | |
| Nov 8, 2020 at 22:13 | history | asked | potionowner | CC BY-SA 4.0 |