Timeline for Is a bounded polynomial constant? [duplicate]
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11 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2022 at 12:04 | vote | accept | learningmaths | ||
| Apr 3, 2022 at 11:26 | comment | added | Ian | That's right. Once $|x|$ exceeds 1 and exceeds that ratio by at least $c$ you get $|p| \geq c|x|^{n-1}$. (Note for $n=1$ you'll need to make this $c$ big too). | |
| Apr 3, 2022 at 3:49 | comment | added | learningmaths | Ok, now I understand that first inequality. So, first assuming that $|x|>1$, we obtain $|p(x)|\geq |x|^{n-1}(|a_n||x|-\sum_{i=0}^{n-1}|a_i|)$. This means that the second threshold shold be $|x|>\left(\sum_{i=0}^{n-1}|a_i|\right)/|a_n|$. From here we can conclude that $|p(x)|$ is unbounded. Correct me if I am wrong :) Thanks for your help! | |
| Apr 2, 2022 at 15:41 | comment | added | Ian | This is the "reverse triangle inequality", which comes from writing $a_n x^n$ as $p(x) - \sum_{i=0}^{n-1} a_i x^i$ and then applying the normal triangle inequality. Anyway, the first threshold is $1$; if you exceed that threshold then you can conclude $|p(x)| \geq |a_n||x|^n - |x|^{n-1} \sum_{i=0}^{n-1} |a_i|=|x|^{n-1}(|a_n||x| - \sum_{i=0}^{n-1} |a_i|)$. | |
| Apr 2, 2022 at 4:21 | comment | added | learningmaths | Do you get that inequality from the fact that $|a_n||x|^n\leq \sum_{i=0}^{n-1}|a_i| |x|^i $? My first guess for a threshold is to consider an expression like $(\text{something here})/|a_n|$, so when I plug it in $|a_n||x|^n$ I get an expression like $(\text{something here})/|a_n|^{n-1}$. | |
| Apr 2, 2022 at 3:47 | history | closed | Martin R inequality Users with the inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed. | Duplicate of $f(x)$ is a bounded polynomial. Prove that $f$ must be constant. | |
| Apr 2, 2022 at 3:37 | answer | added | Vitor | timeline score: 0 | |
| Apr 2, 2022 at 3:33 | answer | added | Joe Shmo | timeline score: 1 | |
| Apr 2, 2022 at 3:28 | comment | added | Ian | Hint: $|p(x)| \geq |a_n| |x|^n - \sum_{i=0}^{n-1} |a_i| |x|^i$; from this bound you can conclude that $|p(x)|$ is unbounded by considering large enough $|x|$. ("Large enough" in this context means it needs to exceed two different thresholds...can you figure out what they are?) | |
| Apr 2, 2022 at 3:28 | answer | added | Sundar | timeline score: -1 | |
| Apr 2, 2022 at 3:20 | history | asked | learningmaths | CC BY-SA 4.0 |