On a limit involving a transform of the chromatic polynomial

Question

I was playing around with the chromatic polynomial (denoted here by $\chi_G(x)$) and I have made the following conjecture.

Let $(G_n)_{n \ge 1}$ be a sequence of graphs with $v(G_n) \to \infty$ ($v(G_n)$ denotes the number of vertices of $G_n$) and $e(G_n) \to \infty$ ($e(G_n)$ denotes the number of edges of $G_n$).

For each $x \neq 0$, let us define the following transform of the chromatic polynomial of $G_n$ $$ \psi_{G_n}(x) = \frac{x^{v(G_n)}}{e(G_n)^{v(G_n)}} \chi_{G_n}\left( \frac{e(G_n)}{x} \right). $$

The conjecture is that for each fixed real number $x \neq 0$, we have $\psi_{G_n}(x) \to \exp(-x)$ as $n$ goes to infinity.

I have checked the conjecture for a few sequences of graphs: for example, $G_n$ being the complete graph $K_n$, for $G_n$ being a tree on $n$ vertices and for $G_n$ being a collection of $n$ independent edges (a matching on $2n$ vertices).

Does anyone know if this is well-known?

PS: I am not sure if the conditions on $v(G_n)$ and $e(G_n)$ are the right one. Any comments on this are welcome as well.

Answer 1

Here is a heuristic argument which perhaps someone can make rigorous. I write $v_n=v(G_n)$ and $e_n=e(G_n)$. Let $$ \chi_{G_n}(x) = x^{v_n}-c_{n,v_n-1} x^{v_n-1}+c_{n,v_n-2}x^{v_n-2}-\cdots. $$ I claim that for fixed $k\geq 0$, $$ \lim_{n\to\infty} \frac{c_{n,v_n-k}}{e_n^k} = \frac{1}{k!}. $$ One can prove this by noting that (by the Broken Circuit Theorem, for instance, which shows that $c_{n,v_n-k}$ increases as we add more edges to $G_n$) $c_{n,v_n-k}$ is bounded below by its value when $G_n$ is a tree, and is bounded above by its value when $G_n$ is a complete graph. The claimed result is easily verified for trees and complete graphs (in the latter case, using known asymptotics for the Stirling numbers of the first kind). Perhaps there is a more direct proof, but in any case, if we don't worry about justifying interchanging limits and sums, we get $$ \lim_{n\to\infty} \frac{x^{v_n}}{e_n^{v_n}}\chi_{G_n}\left( \frac{e_n}{x}\right) = \sum_{k\geq 0} \lim_{n\to\infty} \frac{(-1)^k c_{n,v_n-k}x^k}{e_n^k} $$ $$ \qquad = \sum_{k\geq 0} \frac{(-1)^k x^k}{k!} = \exp(-x). $$