In "The Potts model and the Tutte Polynomial", D.J.A. Welsh and C. Merino claim on pg. 1135, equation 18, that,
\begin{align\*}
\sum\_{i=0}^{n-1}f\_i t^i = t^{n-1}T\_G(1+\frac{1}{t},1)
\end{align\*}
where $T\_G(x,y)$ is the Tutte polynomial of some graph $G$ and $f\_i$ denotes the number of forests of $G$ with exactly $i$ edges. Their paper is available online [here](http://www.yaroslavvb.com/papers/welsh-potts.pdf).


Welsh and Merino do not give any citations or proof for this claim. Does anyone know of such a proof? Also, is there a more explicit formula for $f\_i$ in terms of $T\_G(x,y)$?


EDIT: I noticed recently that something must be wrong with the claimed formula because if $G$ is the disjoint union of $n$ isolated vertices, then $T\_G(x,y) = 1$, so the righthand side becomes $t^{n-1}$; however the lefthand side is $1$. Implicitly Welsh and Merino are assuming that $G$ is connected or I do not know what is meant by the "size" of a forest.



Number of forests of size $i$ and the Tutte polynomial
Here are two quick ways of proving this: (1) Notice that one of the many equivalent definitions of the Tutte polynomial says
$$T\_G(x,y)=\sum\_{A\subset E}(x-1)^{k(A)-k(E)}(y-1)^{k(A)+|A|-|V|}$$
Where $A$ runs through subsets of the edges of $G$, and the function $k(A)$ measures the number of components of the graph $(V,A)$.


Here you only need consider the terms where $k(A)+|A|=|V|$, which happens precisely when $(A,V)$ is a forest!


(2) Remember that the Tutte polynomial satisfies the deletion-contraction relation
$$T\_{G}(x,y)=T\_{G/e}(x,y)+T\_{G-e}(x,y),$$
when $e$ is neither a loop nor a bridge. So you only need to show that the generating function of forests of a graph also satisfies this recurrence, as well as the "initial conditions" that it agrees with $t^{|E|-1}T\_G(1+\frac{1}{t},1)$ for $G$ a tree.


