Timeline for A natural sum over multisets (expectation over multinomial)
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2015 at 19:00 | comment | added | esg | In "A recurrence related to trees", Proc. of the AMS (1989), Knuth and Pittel have investigated the ``tree polynomials" $t_n(y)=n![z^n]1/(1+T(z))^y$ (where $T(z)=-W(-z)$) and in praticular given the first two terms of the asymptotic expansion for fixed $y$ | |
| Dec 6, 2014 at 19:40 | comment | added | usul | @LiamBaker, it took me a little while to see that $F(y)$ and $f(x)^k$ really were the same generating function, but now I think I understand. Thanks! | |
| Dec 6, 2014 at 10:04 | comment | added | Liam Baker | @usul, if $k$ is constant and we denote the $n$th quantity in question by $f_n$, then the generating function $F(y)=\sum_n f_n y^n/n!$ is equal to $1/(1+W(-y))^k$, so the coefficients of $F(y)$ can be estimated using aforementioned standard techniques. | |
| Dec 5, 2014 at 23:53 | history | edited | usul | CC BY-SA 3.0 |
doesn't appear to be too basic, given sparsity of responses
|
| Dec 5, 2014 at 23:51 | comment | added | usul | @IraGessel, thanks, it is probably my naivete, but can you say how exactly $f(x)$ connects to the question (i.e. how this is useful)? I will see if I can hunt down the book, thanks. | |
| Dec 5, 2014 at 18:52 | comment | added | Ira Gessel | The generating function $f(x)=\sum_n n^n x^n/n!$ is equal to $1/(1+W(-x))$, where $W$ is the Lambert $W$-function. The denominator has a zero at $x=1/e$ which should enable you to get a good approximation to the coefficients of $f(x)^k$ by standard techniques (see, e.g., Flajolet and Sedgewick's Analytic Combinatorics). | |
| Dec 5, 2014 at 17:04 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced tag 'multinomial' with 'probability-distributions'; added tag 'asymptotics'
|
| Dec 5, 2014 at 16:28 | history | asked | usul | CC BY-SA 3.0 |