Timeline for Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2020 at 15:35 | comment | added | Nik Weaver | @AlexanderBetts Fedor is a genius. Your answer is good too. | |
| Sep 18, 2020 at 15:25 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 1 character in body
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| Sep 18, 2020 at 13:32 | vote | accept | DUO Labs | ||
| Oct 5, 2020 at 0:41 | |||||
| Sep 18, 2020 at 10:45 | comment | added | Alexander Betts | I think this answer really gets to the point of what's going on here. I think it might be helpful to remark that both Fedor's solution and the one I gave revolve around similar ideas: using the first condition to bound the polynomial $p$, and deriving a contradiction by playing off this bound against the value of $p(0)$ using the second condition. However, in the proof I gave, these two steps are bound up together in the choosing to normalise everything such that $|p(0)|=1$, so I think Fedor's proof makes this underlying structure clearer. | |
| Sep 18, 2020 at 9:53 | history | answered | Fedor Petrov | CC BY-SA 4.0 |