0
$\begingroup$

During a computation, the following came up, and I was wondering if there is a generalized multinomial formula which can handle expressions of the following form:

Let $n\in \mathbb{N}_+$ and $w_1,\dots,w_n>0$ and $W_1,\dots,W_n\in \mathbb{N}_+$; set $W:=(W_i)_{i=1}^n$ and $w:=(w_i)_{i=1}^n$.

I'm looking for a simplified expression on the following function: $$ \label{1} \tag{1} [0,\infty)^n\ni (x_1,\dots,x_n)\mapsto \sum_{\sum_{i=1}^n\, W_i \cdot k_i = n} \frac{ n! }{ \prod_{i=1}^n\, k_i!\,w_i^{k_i} } \, \prod_{i=1}^n\, x_i^{k_i} . $$

When $W_1=\dots=W_n=1$ and $w_1=\dots,w_n=1$ yields the usual binomial formula yields the nice expression for the right-hand side of~\eqref{1}; namely $(x_1+\dots+x_n)^n$.

Is there a similar expression one can obtain in this general case?

If not, are there reasonable upper and lower bounds on the function in terms of the classical multinomial function?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

In general this function equals $n!$ times the coefficient of $z^n$ in $$\exp\big( \sum_{i\geq 1} \frac{x_i}{w_i} z^{W_i}\big).$$

When $W_i = w_i = 1$, the function is simply $\exp( z \sum_{i\geq 1} x_i )$ with the coefficient of $z^n$ equal $\frac{(\sum_{i\geq 1} x_i)^n}{n!}$ as expected.

Improve this answer
$\endgroup$
7
  • $\begingroup$ May I ask for a bit more detail, as I'm not vary familiar with generating function manipulation in combinatorics. For instance, is the sum taken over the positive integers or until $n$? If the former case, then what are $w_i,W_i$ for $i>n$? Also, do you have a reference for this so I can read more/cite? $\endgroup$
    – ABIM
    Commented Apr 30 at 22:17
  • $\begingroup$ If you have just $n$ values of $x_i$, then you can restrict summation to just $i=1..n$. But it is more general to assume that there are infinitely many of them with just $x_{n+1} = x_{n+2} = \dots = 0$ (and it does not matter what are $w_i$ and $W_i$ for $i>n$ as $x_i$ nullify the corresponding summands anyway). The formula is general in the sense that it holds for both finitely or infinitely many nonzero $x$'s. $\endgroup$
    – Max Alekseyev
    Commented Apr 30 at 22:40
  • 1
    $\begingroup$ @ABIM: I do not think there is a simpler general expression, but it can be simplified for some given values of $w_i, W_i$ (like in the example of all ones). I'm not sure what you refer to as "this". $\endgroup$
    – Max Alekseyev
    Commented Apr 30 at 23:04
  • 1
    $\begingroup$ @ABIM: for a very beginner to generating functions, the keyword you want to look for here is "exponential generating function"/"the exponential formula." $\endgroup$
    – Sam Hopkins
    Commented May 1 at 0:42
  • 1
    $\begingroup$ @ABIM: As Sam pointed out, it follows directly from the exponential formula. E.g. see Wikipedia - en.wikipedia.org/wiki/Exponential_formula $\endgroup$
    – Max Alekseyev
    Commented May 1 at 3:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .