Timeline for Application of polynomials with non-negative coefficients
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2012 at 0:41 | comment | added | Sridhar Ramesh | Surely, the thing to do is to use q = p(1) + 1 rather than q = p(1), so as not to have to treat the case p(q) = q^n on an ad hoc basis. | |
| Nov 9, 2012 at 8:57 | comment | added | Federico Poloni | And, interestingly enough, note that choosing a pair $(a,b)$ and asking for the pair $(p(a),p(b))$ doesn't work instead --- you have to use your knowledge of $p(a)$ to choose $b$. | |
| Nov 9, 2012 at 8:49 | comment | added | Federico Poloni | @Marvis: you have to know $m$ in advance for that trick to work. If you know nothing about the polynomial, I think that the best you can do is asking for $q=p(1)$, and then for either $p(p(1))$ if $q\neq 1$ or $p(2)$ instead. This way you can always reconstruct the polynomial by asking for two evaluations. | |
| Mar 29, 2012 at 6:56 | comment | added | user11000 | @Miroslav One could also take the maximum of the coefficient, say $m$ and evaluate $p(m+1)$ in base $m+1$. | |
| Mar 22, 2012 at 14:49 | comment | added | Miroslav Korbelar | The previous proof works if $p(1)\neq 1$ (i.e. $p(x)=x^n$). But one can just take $p(2)$ and $p(p(2))$ for instance. | |
| Mar 22, 2012 at 2:33 | comment | added | ARupinski | @Igor: $q=p(1)$ gives the sum of the coefficients. Now think of $p(p(1))=p(q)$ written in base $q$; one sees that the "digits" are exactly the coefficients of $p$. The only possible ambiguity comes if $p(q)=q^n$ for some $n$, but since the coefficients sum to $q$, one sees that $p=qx^{n−1}$ in this case. | |
| Mar 22, 2012 at 1:57 | comment | added | Igor Rivin | Why is this statement true? | |
| Mar 22, 2012 at 1:30 | history | answered | Aeryk | CC BY-SA 3.0 |