I was reading a text on q,t-Catalan numbers and Diagonal Harmonics by Haglund, where they mention the following $q$-analogue of Lagrange Inversion, taken from Page 53:
Let $e_n, h_n$ denote the elementary symmetric polynomials and complete homogenous symmetric polynomials respectively. Define $H(t) := \sum_{k \ge 0} h_k t^k$. Then for $\vec{c} := (c_1, \ldots, c_k), k > 0$, define $h_n^*(\vec{c}, q)$ for $n \ge 0$ via the following equation:
$$ \sum_{k=0}^{\infty} e_k c_k z^k = \sum_{n=0}^{\infty} q^{-\binom{n}{2}} h_n^*(\vec{c}, q) \prod_{m=1}^{n} H(-q^{-m}z) $$
Then, the result states that $$h_n^*(\vec{c},q) = \sum_{k=0}^{n} c_k \sum_{\pi \in L_{n,n}^{+}} q^{area(\pi)} e_{\beta(\pi)}$$
where $L_{n,n}^{+}$ is the set of paths from $(0,0)$ to $(n,n)$ with steps $(i,j) \to (i,j+1)$ or $(i,j) \to (i+1,j)$, which do not go below the line $y=x$ (intersecting at a point and not crossing is okay). The area of such a path $\pi$ is the number of $1 \times 1$ cells strictly above the $y=x$ line, but weakly beneath $\pi$. Also, $\beta(\pi)$ corresponds to the partition given by the sizes of consecutive vertical steps of $\pi$.
My question is: sure, the book covers details on why this formula is a generalisation of the conventional Lagrange inversion formula, and this can be used to justify the strength of the underlying theory the text really cares for. But are there problems which are hard/impossible to tackle with vanilla Lagrange inversion, and the $q$-analogue helps better with analysing the problem? That is, are there problems (preferably not contrived for the express purpose) where this formula shines above the normal Lagrange inversion formula? Is it 'more useful' anyhow?
Posted 6 days ago on MSE here, but stands without response.
