Timeline for Positivity of sequences via generating series
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23, 2010 at 2:26 | comment | added | Wadim Zudilin | Thanks for your final comments! My intuition is really bad in the morning. | |
| Jul 23, 2010 at 1:28 | comment | added | Ryan Reich | Also (and perhaps this should have gone first), you are very welcome and thank you for the interesting problem. | |
| Jul 23, 2010 at 1:28 | comment | added | Ryan Reich | No, it has to be triangular. For example, the 2x2 matrix written in pseudo-LaTeX as [ 0 & 1 \\ 1 & 0 ] is an involution. More generally, I guess, this holds of permutation matrices. I think, actually, that the "positive monomial matrices" which are permutations of positive-entry diagonal matrices are exactly the set of real square matrices which together with their inverses have nonnegative entries. | |
| Jul 23, 2010 at 1:12 | vote | accept | Wadim Zudilin | ||
| Jul 23, 2010 at 1:12 | history | bounty ended | Wadim Zudilin | ||
| Jul 23, 2010 at 1:12 | comment | added | Wadim Zudilin | Thank you very much, Ryan, for your exhaustive solution! So, if $A$ is a square invertible (not necessarily diagonal) matrix with nonnegative entries, then $A^{-1}$ must have a negative entry. | |
| Jul 22, 2010 at 19:05 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Improve case coverage
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| Jul 22, 2010 at 17:34 | comment | added | Gjergji Zaimi | I haven't gone through your solution yet, but I'm wondering where do you lose answers like $p(x)=x^3$ and $r(x)=x^2$? | |
| Jul 22, 2010 at 13:51 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Rewrite and change conclusion
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| Jul 22, 2010 at 13:08 | comment | added | Ryan Reich | It is definitely "no". My mistake (which betrays the fact that I am NOT a combinatorialist) is that I forgot all the binomial coefficients :) | |
| Jul 22, 2010 at 6:43 | comment | added | Wadim Zudilin | No hurry, Ryan, and thanks again. I was thinking of the problem hard before posting it. The expectation is "no" which is probably harder than giving at least one (nontrivial) example. | |
| Jul 22, 2010 at 6:38 | comment | added | Ryan Reich | Hmmm. Looking back over the argument I realize that it is not possible for $F$ to be diagonal, since that would necessitate $p(x) r(x)^n \in \mathbb{Q}$ for all $n$, an impossibility. I will think on this and replace the above answer with something correct tomorrow, when I am more awake. | |
| Jul 22, 2010 at 6:04 | comment | added | Wadim Zudilin | Ryan, thank you very much for your computation (it will take time for me to follow it in details). I took $p_0=r_1=q(x)=1$, so that $p(x)=1+x$, $r(x)=x-x^2$, and computed $p(x)A(r(x))$ for $A=\sum_{j=0}^4a_jx^j$. The coefficient of $x^8$ in the resulting polynomial is $-3a_4<0$. | |
| Jul 22, 2010 at 5:48 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Fix an index
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| Jul 22, 2010 at 5:09 | history | answered | Ryan Reich | CC BY-SA 2.5 |