2
$\begingroup$

For my application, the expression I'm interested in is:

\begin{equation} \prod_{j=1}^{L}(1+xq^{j})=\sum_{k=0}^{L}x^{k}q^{\frac{k(k+1)}{2}}\sum_{n=0}^{k(L-k)}q_{L\geq}(n+\frac{k(k+1)}{2},k)q^{n}, \end{equation}

where $q_{L\geq}(N,m)$ is the partition of $N$ into $m$ distinct parts, each of which is bounded by $L$; combinatorially, $\prod_{j=1}^{L}(1+xq^{j})$ is interpreted as the generating function for $q_{L\geq}(N,m)$. However, I don't want the whole expression; I only want this expansion for a $fixed$ $value$ of $m$, i.e.

\begin{equation} q^{\frac{m(m+1)}{2}}\sum_{n=0}^{m(L-m)}q_{L\geq}(n+\frac{m(m+1)}{2},m)q^{n}. \end{equation}

I did some numerical experiments, and viewing the partitions as a distribution over $N$, I discovered that they follow a Gaussian for large enough values of $L$ and $m$, which I guess is simply a consequence of the Central Limit Theorem. While it would be best to have a closed form for $q^{\frac{m(m+1)}{2}}\sum_{n=0}^{m(L-m)}q_{L\geq}(n+\frac{m(m+1)}{2},m)q^{n}$, I would settle for getting the form for this distribution for large enough $L$ and $m$; however, I don't know how to find the standard deviation for the distribution in that limit. How can this be done?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.