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The multivariate Lagrange inversion formula, which I found in a couple of papers (such as this and this), is as follows. If $f_i=t_ig_i(f)$, $1\le i\le k$, then $$ [t^n]h(f(t))=\frac{1}{n_1n_2\cdots n_k}[x^{n-1}]\sum_T \frac{\partial (h,g_1^{n_1},...,g_k^{n_k})}{\partial T},$$ where $t^n=t_1^{n_1}\cdots t_k^{n_k}$ and the derivative is taken with respect to some trees (as discussed in those papers).

Not one of the papers in question has addressed the question of how this formula is to be used when some of the powers are zero, $n_j=0$, something that does not happen in the one variabe case (due to the assumption that $g(0)=0$) but can happen in the multivariable one.

  • $\begingroup$ Proof of Theorem 2 at emis.ams.org/journals/EJC/Volume_5/PDF/v5i1r33.pdf , which is referenced in the 2nd of your links. $\endgroup$ Dec 11 '19 at 22:26
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    $\begingroup$ @MarkL.Stone As far as I can tell, that proof assumes that all $n$'s are positive. I am sure it can be adapted to treat this case, but I need help with that. I know this is probably not quite research level, but I asked the question in MSE and got nothing. $\endgroup$
    – thedude
    Dec 12 '19 at 17:36
  • $\begingroup$ The most straightforward solution (which can be justified) is to regard $n_i$ as real, let $n_i\to 0$, and apply l'Hôpital's rule (or just the definition of the derivative). $\endgroup$ Feb 8 '20 at 16:25
  • $\begingroup$ As a continuation of my previous comment, the one-variable case is Exercise 5.56(a) of Enumerative Combinatorics, vol. 1, second ed. I give three proofs and some references. This result is also stated in my solution to mathoverflow.net/questions/350882. $\endgroup$ Feb 8 '20 at 17:06

The formula cannot be applied for multinomial terms where an $n_i$ is $0$.

For $n=0$, Lagrange inversion formula and Lagrange-Bürmann formula have particular formulas.

You have to use another multivariate Lagrange inversion formula.

See Rosenkranz, M: Lagrange Inversion. Diploma thesis, RISC Linz 1997:
page 40: "It turns out that the inversion formulas in their second form can be generalized to the multivariate case in a very natural way. (For the first form of the inversion formulas, the multivariate generalizations are very complicated.)"
See theorem 42 at page 38, corollary 43 at page 39, and theorem 47 at page 41.

Corollary 43 (univariate case) and theorem 47 (multivariate case) present formulas for the general series coefficients without the factor $\frac{1}{\boldsymbol{n}}$.


Part 2 of A combinatorial proof of the multivariable Lagrange inversion formula (also available at Academia.edu) shows how the multivariate Lagrange inversion formula with $n_j\in\mathbb{N}$ can be generalized to $n_j\in\mathbb{Z}$. (Theorem 2, attributed to I.J. Good.)

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    $\begingroup$ The theorem you mention is equation (1) in the paper cited by Mark Stone in his comment. It is not the formula involving trees. It doesn't have the $n$'s in the denominator and doesn't have the $[x^{n-1}]$ part. So this theorem is fine for zero $n$, but it doesn't really help me. $\endgroup$
    – thedude
    Dec 13 '19 at 18:37

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