Timeline for A polynomial identity involving Wick ordering of a complex power
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 17 at 12:49 | history | became hot network question | |||
| May 17 at 10:22 | vote | accept | Student | ||
| May 17 at 8:31 | answer | added | Peter Taylor | timeline score: 7 | |
| May 17 at 7:41 | comment | added | მამუკა ჯიბლაძე | @PeterTaylor Right. | |
| May 17 at 7:28 | comment | added | Peter Taylor | Up to sign. The sign correction looks like $$\sum_{m \ge 0} \frac{P_m(x)}{m!}t^m = \frac{1}{1+2t} \exp\left(\frac{xt}{1+2t}\right)$$ | |
| May 17 at 7:14 | comment | added | მამუკა ჯიბლაძე | The polynomials $P_m$ go like $P_0(x)=1$, $P_1(x)=x-2$, $P_2(x)=x^2-8x+8$, $P_3(x)=x^3-18x^2+72x-48$, etc. Seems to be A286724. | |
| May 17 at 7:01 | history | edited | Student | CC BY-SA 4.0 |
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| May 17 at 6:55 | history | edited | Student | CC BY-SA 4.0 |
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| May 17 at 6:48 | history | edited | Student | CC BY-SA 4.0 |
added 36 characters in body
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| May 17 at 6:47 | comment | added | Student | @CarloBeenakker Sorry, you are right, a binomial prefactor is missing. Now the problem is corrected. | |
| May 17 at 6:27 | comment | added | Carlo Beenakker | I think I found the error: the correct identity is $$\sum_{\ell=0}^m {m\choose \ell} H_{2\ell}(a) H_{2(m-\ell)}(b)=P_m(a^2+b^2)$$ I don't have a proof, but at least this one checks out if I try it for small $m$ | |
| May 17 at 6:23 | comment | added | Carlo Beenakker | I understand that these are the "probabilist" Hermite polynomials, but I am not able to verify the identity for $m=2$; then I find for the right-hand-side: $a^4-\left(a^2+7\right) x^2+x^4+7$ with $x^2=a^2+b^2$, so this is not a function of only $a^2+b^2$. | |
| May 17 at 4:47 | history | asked | Student | CC BY-SA 4.0 |