Timeline for Polynomial positive on an interval
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2019 at 7:32 | answer | added | Fedor Petrov | timeline score: 2 | |
| Oct 11, 2012 at 0:39 | answer | added | Markus Schweighofer | timeline score: 2 | |
| Sep 20, 2010 at 8:39 | vote | accept | Manjunath Krishnapur | ||
| Sep 20, 2010 at 1:34 | answer | added | Vicki Powers | timeline score: 20 | |
| Sep 19, 2010 at 14:25 | history | edited | Thierry Zell |
edited tags
|
|
| Sep 19, 2010 at 14:24 | answer | added | Thierry Zell | timeline score: 4 | |
| Sep 19, 2010 at 14:14 | answer | added | Bill Thurston | timeline score: 4 | |
| Sep 19, 2010 at 13:54 | answer | added | Sergei Ivanov | timeline score: 13 | |
| Sep 19, 2010 at 13:28 | comment | added | Charles Matthews | Google on "polynomials positive on an interval", why not? This finds me a paper by Powers and Reznick which apparently covers much more, though. | |
| Sep 19, 2010 at 11:50 | comment | added | Pietro Majer | Keivan: of course the representation is not unique: that's why I say the claim needs to be stated into more precise terms. | |
| Sep 19, 2010 at 11:49 | comment | added | Keivan Karai | So, what I wrote above is not a solution, but maybe someone can find a clever way to modify these polynomials to get the answer. | |
| Sep 19, 2010 at 11:48 | comment | added | Keivan Karai | You can certainly use Bernstein polynomials to approximate your polynomials by positive linear combinations of $x^i (1-x)^{n-i}$. More precisely, if you set $$ P_n(x)=\sum_{i=0}^{n} p(\frac{i}{n}) {n \choose i} x^i (1-x)^{n-1}$$ then it turns out that $p_n$ converges uniformly to $p(x)$ on $[0,1]$. (This is, by the way, true for any continuous function, hence it gives a proof of Weierstrass approximation theorem). I did some calculations which show that the sequence does not stabilize if you start with a polynomial, but it always produces a polynomial of the same degree. | |
| Sep 19, 2010 at 11:36 | comment | added | Helge | Don't forget about the $c_{i,j} \geq 0$ restrictions.... | |
| Sep 19, 2010 at 11:29 | comment | added | Keivan Karai | Pietro: Yes, but you can also write it as $x(1-x)$. The dimension of the space of the functions $x^i(1-x)^{n-i}$ for $0 \le i \le n$ is less than $n+1$, so to write an arbitrary polynomial of degree $n$ you need to use polynomials of degree $n+1$, so the representation will be non-unique. | |
| Sep 19, 2010 at 11:17 | comment | added | Pietro Majer | For further needs: note the usual $ \$ $ for the TeX. As to the question, note that $p(x):=x(1-x)=x^1(1-x)^0-x^2(1-x)^0$, so the claim as you stated has to be corrected/made precise. | |
| Sep 19, 2010 at 11:06 | history | edited | Pietro Majer | CC BY-SA 2.5 |
added 6 characters in body
|
| Sep 19, 2010 at 10:36 | history | asked | Manjunath Krishnapur | CC BY-SA 2.5 |