Timeline for Polynomial positive on an interval
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2010 at 17:02 | comment | added | Bill Thurston | Yes, I too rechecked, including the i=0 term and being more careful about rounding, and got a degree 2 polynomial. I understand now: the expression necessarily involves polynomials of arbitrarily high degree, even though the answer is quadratic, and no such expression is possible for $p$ that is nonnegative with an interior 0. It would be interesting to plot the polyhedral subdivision of coefficients of quadratic polynomials given by the degree 2 or 3 polynomials expressible in the above form, as $n$ varies. | |
| Sep 19, 2010 at 16:53 | comment | added | Sergei Ivanov | I rechecked your example (using computer algebra program 'maxima') with corrected summation range and got degree 15 too. What a surprise! Then I replaced 0.1 by 1/10 and got a nice short answer: $\frac{14}{15}x^2-\frac{29}{15}+\frac{11}{10}$. Damn stupid machine! | |
| Sep 19, 2010 at 16:30 | comment | added | Sergei Ivanov | Sorry for the typos. Yes all sums should start with $i=0$. Fixed now. | |
| Sep 19, 2010 at 16:29 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
fixed typos
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| Sep 19, 2010 at 16:23 | comment | added | Bill Thurston | I had trouble following, and had doubts because of my negative answer, so I tried the approximation in your formula for the polynomial q(x) = .1 + (x-.1)^2, correcting assumed typo $x^i(1-x)^i \to x^i(1-x)^{n-i}$. For the degree 15 approximation, I get degree 15: $ 1.51667 x - 10.6167 x^2 + 50.05 x^3 - 150.15 x^4 + 330.33 x^5 - 550.55 x^6 + 707.85 x^7 - 707.85 x^8 + 550.55 x^9 - 330.33 x^10 + 150.15 x^11 - 50.05 x^12 + 11.55 x^13 - 1.65 x^14 + 0.11 x^15 $. It approximates well except near the 0 endpoint --- should the summation have started from 0? The extra term still gives degree 15. | |
| Sep 19, 2010 at 15:48 | comment | added | Keivan Karai | Very nice. As a comment, there is an alternative way to show why the $\deg q_n \le \deg q$. In fact, $q_n={\mathbb E}[q(\frac{X_1+ \cdots X_n}{n})]$ where each $X_i$ are i.i.d. with Bin. distribution with parameter $x$. Now assume $q(x)=x^k$. Now expand $(X_1+\cdots +X_n)^k$ and use the fact that all of the moments ${\mathbb E}[X^k]=x$ for all $k$ when $X$ has Binomial distribution. So using independence of $X_i$, each summand has degree at most $k$. | |
| Sep 19, 2010 at 13:54 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |