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   {\smcp the electronic journal of combinatorics 6 (1999), \#R8
    \hfill\folio} \fi} 
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\def\ArqTorus   {2}
\def\Beiss      {3}
\def\BCR        {4}
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\def\BRlifps    {6}
\def\BRpoly     {7}
\def\BRhayman   {8}
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\null\vfill

\centerline{Multivariate Asymptotics}
\centerline{for Products of Large Powers}
\centerline{with Applications to Lagrange Inversion}

\vskip12pt

\centerline{Edward A. Bender}
\centerline{Department of Mathematics}

\centerline{University of California, San Diego}
\centerline{La Jolla, CA 92093-0112, USA}
\centerline{\tt ebender@ucsd.edu}

\medskip

\centerline{L. Bruce Richmond}
\centerline{Department of Combinatorics and Optimization}
\centerline{University of Waterloo}
\centerline{Waterloo, Ontario N2L 3G1, Canada}
\centerline{\tt lbrichmond@watdragon.uwaterloo.ca}


\medskip

\centerline{Submitted: March 27, 1998}
\centerline{Revised: January 5, 1999}
\centerline{Accepted: January 12, 1999}

\bigskip

\nonumsection Abstract

An asymptotic estimate is given for the coefficients of products of
large powers of generating functions.
This theorem and another local limit theorem which is useful for
conditioning are applied to various combinatorial enumeration problems
that involve multivariate Lagrange inversion.


\vfill\vfill\vfill

\noindent
1991 AMS Class. No.
\quad
Primary: 41A63
\quad
Secondary: 05A16, 05C05, 41A60

\eject


\section Introduction

If $f(0)\ne0$ has a (possibly formal) power series expansion at 0,
the equation\break
$w=xf(w)$ determines the power series $w(x)$.
Two forms of the Lagrange inversion formula are
$$
\eqalignno{
g_n = [x^n]\,g(w) 
&= [x^n]\,\Bigl(g(x)f(x)^n\{(1-xf'(x)/f(x)\}\Bigr)
&\equno\eqDiffLI\cr
&= (1/n)[x^n]\,\bigl(xg'(x)f(x)^n\bigr),
&\equno\eqPrimeLI\cr}
$$
where $[x^n]\,h(x)$ denotes the coefficient of the monomial $x^n$ in
the power series $h(x)$.
We obtained asymptotics for $g_n$ from \eqPrimeLI\ for some types of
formal power series~[\BRlifps].
When $f$ has a nonzero radius of convergence, various authors have
studied the asymptotics of $[x^n]\,g(w)$ using three basic
approaches:
\Item{$\bullet$}
{\it Exact Formula}.
Using \eqPrimeLI, obtain an exact formula $g_n$.
This is often either a simple expression or a summation with
alternating signs.
Obtain asymptotics from the exact formula.
This has been used only sporadically.
\Item{$\bullet$}
{\it Singularity Analysis}.
Determine the nature of the singularities of $w$ by looking at
$xf(w)-w=0$.
They are usually square root branch points due to the vanishing of
$\partial(xf(w)-w)/\partial w$.
Obtain asymptotics by what is essentially Darboux's Theorem.
For a systematic discussion of this approach, see Sprugnoli and
Verri~[\SprugVer].
\Item{$\bullet$}
{\it Contour Integration}.
Using the Cauchy Residue Theorem, one can estimate $g_n$ from 
\eqPrimeLI.
Since $f(x)^n$ leads to an integral with a simple dominant term it
suffices to use a circle.
For a systematic discussion of this approach, see Gardy~[\Gardy].
\itemskip

One can easily include other variables in \eqPrimeLI\ by simply
thinking of the coefficients of $f$, $g$, and $w$ as involving 
the new variables.
Furthermore, there are extensions of Lagrange inversion to several
functions and other variables can be included in these as well.

Recently Drmota~[\Drmota] treated a system of functional equations
using singularity analysis.
His results can be applied to multifunction Lagrange inversion when
$g(w_1,w_2,\ldots,w_d) = w_i$ for some $i$.
Not all cases of interest have this form, a prime example being map
enumeration.

The asymptotics of rooted convex polyhedra by faces and vertices (two
equations with no extra variables) were studied by us~[\BRpoly] using
singularity analysis and later by Bender and Wormald~[\BenWorm] using
an exact formula.
Rooted maps on general surfaces were dealt with in a similar manner
by us and Canfield~[\BCR].

In this paper, we are concerned with the asymptotic behavior of the
coefficients of large powers of multivariate generating functions and
their application to multivariate Lagrange inversion.
In Theorem~2 of~[\BRcllt] we studied coefficients of large powers of
a single multivariate function using a contour integration approach.
In Theorem~2.1 below, we extend this to products of powers of several
functions when the exponents tend to infinity in such a way that
their ratios are bounded.
When there is only one power, Theorem~2.1 is essentially contained
in Theorem~2 of~[\BRcllt], but we believe the conditions here are
more easily verified than those in [\BRcllt]. {}From a probabilistic viewpoint, our concern is with local limit
theorems (estimates of coefficients) rather than central limit
theorems (estimates for averages of coefficients).
One could certainly obtain a central limit theorem extending Theorem~1
of~[\BRcllt]; however,  more general central limit theorems have been
obtained by Hwang~[\Hwang] in the case of two variables.
Hwang also studies the rate of convergence (which we do not) and
points out that the central limit theorem we would derive would have
a convergence rate of $O(n^{-1/2})$.

\medskip

In the next section we state and prove Theorem~2.1, our theorem for
products of powers.
In Section~3 we explain how the theorem applies to Lagrange inversion
of a single function and discuss the problem of conditioning on some
of the indices.
This is useful when one wishes to study combinatorial objects
conditioned on things such as ``size'' or number of ``components.''
Section~4 illustrates applies these ideas to specific enumeration
problems.
Since neither conditioning nor Lagrange inversion applications were
discussed in [\BRcllt], the material in Sections~3 and~4 is new even
though Theorem~2.1 follows from Theorem~2 of~[\BRcllt] in this case.
In Section~5, we recall Lagrange inversion formulas for several
functions and show how the product of powers theorem can be applied
to these formulas.
We also prove a local limit theorem that is needed to continue
the discussion of conditioning.
Section~6 contains examples of specific applications.
Although Theorem 2.1 leads to Langrange inversion asymptotics for
many functions $g$; maps present a difficulty which we can resolve
only in the single variable situation.
This is explained in Section~6.
In the final section, we indicate some research problems suggested by
the limitations of our approach.

We thank Donatella Merlini and Renzo Sprugnoli for helpful
conversations.



\section A Limit Theorem for Products of Powers

Let $\integers$ denote the integers.
For a set $V$ of vectors, let ${\cal A}(V)$ be the additive abelian
group generated by $V$.
Bold face letters denote vectors,
$\vec x^{\vec n}=x_1^{n_1}x_2^{n_2}\cdots$, $|\vec x|$ denotes

the vector whose components are $|x_i|$, and $\|\vec x\|$ denotes the
length of $\vec x$.
As already noted $[\vec x^{\vec n}]\,h(\vec x)$ denotes the
coefficient of $\vec x^{\vec n}$ in the power series $h(\vec x)$.

Let $\vec m(f(\vec z))$ and $B(f(\vec z))$ be the vector and matrix 
given by
$$
\vec m(f(\vec z))_i
= {\partial(\log f)\over\partial(\log z_i)}
~~{\rm and}~~
B(f(\vec z))_{i,j} 
= {\partial^2(\log f)\over\partial(\log z_i)\,\partial(\log z_j)}.
$$
In all cases, the logarithms are real for real positive $\vec z$.
(This is possible since our functions are positive reals for such
$\vec z$.)
Note that
$$
\vec m(\vec f^\vec n) = {\textstyle\sum} n_i\vec m(f_i)
\quad{\rm and}\quad
B(\vec f^\vec n) = {\textstyle\sum} n_i B(f_i).
\eqlabel\eqmBpowers
$$

\break

\theorem\ThmPowers
Let $\vec u$ denote an $l$-dimensional vector over the complex
numbers, let $R$ be a compact subset of $(0,\infty)^l$ with nonempty
interior, and let $C$ be the set of complex vectors $\vec u$ with 
$|\vec u|\in R$.
Suppose $f_j(\vec u)$ $(1\le j\le d)$ and $h(\vec u)$ are such that
\Item{(a)}
$h$ and the $f_j$ are analytic in $C$ and strictly positive in $R$;
\Item{(b)}
in $R$, the $B(f_j)$ are positive semidefinite
and $\sum_{j=1}^d B(f_j)$ is nonsingular;
\Item{(c)}
in $C$, $|f_j(\vec u)|\le f_j(|\vec u|)$ with equality for all $j$
only in $R$.
\itemskip\noindent
Fix $\delta>0$ and let $n=\sum n_i$.
Then we have
$$
[\vec u^{\vec k}]\,
\bigl(h(\vec u)\vec f(\vec u)^\vec n\bigr)
=
{h(\vec r)\vec f(\vec r)^\vec n
     \bigl\{\exp(-\vec t B^{-1}\vec t'/2) +o(1)\bigr\}
  \over  \sqrt{\det(2\pi B)}\;\vec r^{\vec k}}
~~\hbox{as}~~n\to\infty
\eqlabel\eqPowers
$$
uniformly for $\vec n\in[n\delta,n/\delta]^d$ and $\,\vec r\in R$,
where $\,\vec i = \vec m(\vec f(\vec r)^\vec n)$, 
$B=B(\vec f(\vec r)^\vec n)$, $\vec t = \vec k-\vec i$, and 
$\,\vec t'$ denotes transpose.
\par
If, for all $i$, $f_i(\vec u)=\sum a_i(\vec k)\vec u^{\vec k}$ where 
$a_i(\vec k)\ge 0$ for all $\,\vec k$ and
$$
\Lambda(\vec f)
= {\cal A}\bigl\{\vec k-\vec j\mid a_i(\vec k)a_i(\vec j)\ne0
\hbox{~for some $i$}\bigr\}
= \integers^l,
\eqlabel\eqSpan
$$
then conditions (b) and (c) are satisfied.
Frequently $a_i(\vec 0)>0$ for all $i$, in which case
\eqSpan\ becomes 
${\cal A}\bigl\{\vec k\mid a_i(\vec k)>0
\hbox{~for some $i$}\bigr\}=\integers^l$.
\endthing

\proof
Note that $B(\vec f^{\vec 1}) = \sum B(f_j)$.
Since
$$
B(\vec f^{\vec n}) 
= n\delta B(\vec f^{\vec 1}) + \sum (n_i-\delta n) B(f_i),
$$
$\vec n\in[n\delta,n/\delta]^d$, and the $B(f_i)$ are positive
semidefinite, it follows that
$$
\vec tB(\vec f^{\vec n})\vec t' 
\ge n\delta\vec t B(\vec f^{\vec 1})\vec t'.
$$
Since the domain of $\vec r$ is compact and $B(\vec f^{\vec 1})$ is
positive definite, it follows that $B(\vec f^\vec n)/n$ is positive
definite in a uniform sense; that is, there is a constant $C$ such
that  $\vec tB(\vec f(\vec r)^\vec n)\vec t'\ge nC\vec t\vec t'$ for
all $\vec r\in R$, all $\vec n\in[n\delta,n/\delta]^d$, and all 
$\vec t$.

The proof of \eqPowers\ follows that of Theorem~2 of~[\BRcllt] almost
exactly:

Expand the logarithm of $h(\vec z)\vec f(\vec z)^\vec n$ in a Taylor
series about $\vec r$, keeping quadratic terms and a third-order
error estimate.
Use the Cauchy Residue Theorem with the contours $|z_i|=r_i$ to
estimate the desired coefficient, with (b), (c), and the uniform
positive definiteness of $B(\vec f^\vec n)/n$ ensuring that
$$
\left|{h(\vec z)\vec f(\vec z)^\vec n \over
       h(\vec r)\vec f(\vec r)^\vec n}\right|
= O\bigl(\exp(-C\theta^2n)\bigr)
$$
uniformly for some $C>0$ and $\theta=\max(|\arg z_j|)$.
See [\BRcllt] for details.

We now prove the claims concerning \eqSpan.
Since $f_j$ has a power series with nonnegative coefficients:
\Item{(i)}
The first part of (c) holds.
\Item{(ii)}
By R\'enyi's number 2 on p.341 of [\Renyi], the first part of (b)
holds.
\Item{(iii)}
$\vec f^{\vec 1}$ has a power series with nonnegative coefficients
$a(\vec k)$ and
$$
\Lambda(\vec f^{\vec 1})
= {\cal A}\bigl\{\vec k-\vec j\mid a(\vec k)a(\vec j)\ne0\bigr\}
= \Lambda(\vec f) 
$$
\itemskip\noindent
Since
$\Lambda(\vec f) = \integers^l$, it follows from (iii) that
$|\vec f^{\vec 1}(\vec u)|=\vec f^{\vec 1}(|\vec u|)$ if and only if
$\vec u=|\vec u|$ and so the proof of (c) is complete.
The second half of (b) follows from Lemma~6 in~[\BRWcllt], with the
matrix $T$ in in that paper being the $1\times1$ matrix 
$\vec f^{\vec1}$ and 
${\cal A}^{(s)}_{i,j} = {\cal A}^{(1)}_{1,1}
= \Lambda(\vec f^{\vec 1})$.\qed
\betweenskip

For Theorem \ThmPowers\ to give more than an asymptotic upper bound,
the exponential in \eqPowers\ must not be $o(1)$.
In other words, we must have $|\vec t|=O(n^{1/2})$.
Thus the domain of useful $\vec k$ is asymptotically the same as the
domain of $\vec i$.
The latter depends on the problem and becomes evident only by
calculation; however, we can describe the typical situation.
Let $Z(\vec n)$ be the set of $\vec j$ such that
$[\vec u^{\vec j}]\,\bigl(h(\vec u)\vec f(\vec u)^\vec n\bigr)=0$.
It usually suffices to require that $\vec i$ be at least $\epsilon n$
from $Z(\vec n)$, where $\epsilon>0$ is an arbitrary constant.
In particular, all components of $\vec i$ will be at least 
$\epsilon n$.

Theorem \ThmPowers\ can be strengthened in at least two ways:
\Item{(a)}
The function $h$ can depend on $\vec n$ so long as its partials
through second order are uniformly $o(n)$.
\Item{(b)}
It may happen that the lattice $\Lambda(\vec f)$ in \eqSpan\ is a
proper sublattice of $\integers^l$ rather than all of $\integers^l$.
A theorem still exists, but it requires multisection as discussed
in~[\BRWcllt].
\itemskip
\noindent
We have omitted these from the theorem because they are relatively
rare and add complications.



\section Lagrange Inversion of One Function

How does Theorem \ThmPowers\ apply to Lagrange inversion of a single
function?
Since \eqDiffLI\ and \eqPrimeLI\ deal with formal power series over a
commutative ring of characteristic zero, we
are free to include extra variables $\vec y$ in the coefficients of
$g$, $f$ and $w$.
Thus, if
$$
w(x,\vec y) = xf(w(x,\vec y),\vec y)
~~{\rm with}~~
f(0,\vec y)\ne 0,
$$

we have $[x^n\vec y^\vec j]\,g(w(x,\vec y),\vec y)
  = [\vec y^\vec j]\,g_n$ where $g_n$ as in \eqDiffLI.
Apply Theorem~\ThmPowers\ with 
$$
d=1,~~
\vec n = (n),~~
\vec f = (f),~~
\vec u = (x,\vec y),~~
\vec k = (n,\vec j),
$$
and $h$ the remaining factors in \eqDiffLI\ or \eqPrimeLI\ after
$f^n$ is removed.
We start the indexing of $\vec k$, $\vec i$, and $\vec t$ at zero so
that $k_s= j_s$ for $s>0$.
For greatest accuracy in estimating the coefficient of 
$\vec u^\vec k$, one would normally set $\vec t=\vec 0$, that is,
$\vec i=\vec k$.
The equation for $\vec i$ is then
$$
n = n{r_0\over f(r_0,\vec r)}\,
    {\partial f(r_0,\vec r)\over\partial r_0}
~~{\rm and}~~
j_s = n{r_s\over f(r_0,\vec r)}\,
    {\partial f(r_0,\vec r)\over\partial r_s}
~~{\rm for}~~
s>0.
$$
Regarding the first of these as an equation in $n$, it has a
nontrivial solution if and only if
$$
1 - {r_0\over f(r_0,\vec r)}\,
    {\partial f(r_0,\vec r)\over\partial r_0}
= 0;
\eqlabel\eqDiffZero
$$
that is, the last factor in \eqDiffLI\ vanishes.
Hence $h$ fails to satisfy the $h>0$ condition in Theorem~\ThmPowers\
and so we must use \eqPrimeLI:
$$
[x^n\vec y^{\vec j}]\,g(w,\vec y) 
= [x^n\vec y^{\vec j}]\,\left(x{\partial g(x,\vec y)\over\partial x}
f(x,\vec y)^n\right).
$$


\bigskip
\noindent{\bf Conditioning}.
In addition to providing asymptotics, \eqPowers\ provides a local
limit theorem for $\vec j$ as $n\to\infty$.
One obtains a normal distribution by setting $\vec r=(r_0,\vec 1)$,
choosing $r_0$ so that $d\log(f(r_0,\vec 1)/d\log r_0=1$, and
conditioning on the zeroth component of $\vec i$ being $k_0=n$.
To condition, one drops the zeroth component of $\vec t$ and the
corresponding row and column of $B^{-1}$.
The latter corresponds to replacing the $l\times l$ ``covariance''
matrix $B$ with the inverse of the lower $(l-1)\times(l-1)$ block of
$B^{-1}$, say $C$. 
One can compute $C$ directly from the block matrix formula found on
pp.25--26 of~[\Press]:
$$
\left[\matrix{B_{1,1}&B_{1,2}\cr B_{1,2}'&B_{2,2}\cr}\right]^{-1}
\!\!= \left[\matrix{*&*\cr *&C^{-1}\cr}\right],
~~{\rm where}~~
C = B_{2,2} - B_{1,2}'(B_{1,1})^{-1}B_{1,2}.
\eqlabel\eqConditioning
$$
In this case $B_{1,1}$ is $1\times1$.

One can condition on a set of variables that includes $n$.
In this case, we set $r_i=1$ if we are {\it not\/} conditioning on
$j_i$.
The remaining components of the equation 
$(n,\vec j) = \vec m(\vec r)$, including the zeroth, are used to
solve for $r_0$ and the remaining $r_i$.
Again the indices of the variables being conditioned on are dropped
from $\vec t$ and $B^{-1}$.
Equation \eqConditioning\ still applies, but $B_{1,1}$ is no longer
$1\times1$ since it is indexed by all the variables on which we are
conditioning.
Since the asymptotics obtained from \eqPowers\ is uniform, so are the
asymptotics for limiting distributions, provided $(r_0,\vec r)\in R$
and all components of $\vec j$ lie in $[n\delta,n/\delta]$.
Of course $(r_0,\vec r)$ varies as $n\to\infty$ unless the
conditioned components grow at a rate proportional to $n$.

It is possible to condition on a set of variables that does 
{\it not\/} contain $n$.

This is more complex.
Rather than discuss it here, we treat the general case in the context
of multiple Lagrange inversion in Section~5.

Summing over variables, which is roughly the complement of
conditioning, is also discussed in Section~5.



\section Examples of Inversion of One Function

We now turn to examples of single function inversion.

\example\ExPartitions{Noncrossing Partitions}
A {\it noncrossing partition\/} of a set of integers is a set
partition such that there are no integers $a_1<b_1<a_2<b_2$ with
$a_i$ in one block and $b_i$ in another block.
Kreweras~[\Krew] showed that the number of noncrossing partitions
with $s_m$ blocks of size $m$ is
$$
a(n,\vec s) = {(n)_{k-1}\over s_1!\,s_2!\cdots},
\hbox{~~where $k = \sum s_m$ and $n=\sum ms_m$.}
$$
Asymptotic results can be obtained by summing this formula over
appropriate indices.
Alternatively, one can study the ordinary generating function
$A(x,\vec z)$ for noncrossing partitions with $z_m$ keeping track of
$s_m$ and $x$ keeping track of $n$ (the size of the set).
By the argument leading to (6.2) of Beissinger~[\Beiss],
$$
A(x,\vec z) = 1 +  \sum_m (xA(x,\vec z))^mz_m.
$$
With $w(x,\vec z) = xA(x,\vec z)$, we have
$$
w(x,\vec z) = xf(w,\vec z)
\hbox{~~where~} f(w,\vec z) = 1+\sum_m z_mw^m.
$$
By specializing $z_m$ to 0, 1, and a finite set of indeterminates we
can count various noncrossing partitions.
For example,
$$
\hbox{$z_1=y_1$,\quad $z_m=y_2$ for 
$m\in M\subseteq\{2,3,\ldots\}$,\quad and\quad $z_m=0$ otherwise,}
$$  
counts noncrossing partitions whose blocks are singletons or have
sizes in $M$, keeping track of the number of each type.
To verify \eqSpan, fix $m_0\in M$ and note that
$$
\eqalign{
   {\cal A}\bigl(\{(1,1,0),&\; (m,0,1) : m\in M\}\bigr)    \cr
&= {\cal A}\bigl(\{(1,1,0),\; (m-m',0,0),\; (m_0,0,1)   
         : m,m'\in M\}\bigr)                               \cr
&= {\cal A}\bigl(\{(1,1,0),\; (g,0,0),\; (m_0,0,1)\}\bigr),\cr}
$$
where $g=\gcd\{m-m': m,m'\in M\}$.
Hence \eqSpan\ holds if and only if $g=1$.
Thus
$$
f(x,\vec y) = 1 + y_1 x + y_2S_0,
~~{\rm where}~~ S_i = S_i(x) = \sum_{m\in M}m^ix^m,
$$
and so
$$
(n,k_1,k_2) = \vec m = (n/f)\bigl(
y_1 x + y_2S_1,\; y_1 x,\; y_2S_0\bigr).
$$
Since we want the zeroth component of $\vec m$ to be $n$,
$f = y_1x + y_2S_1$ and so
$$
y_2S_1 = 1 + y_2S_0
~~{\rm at}~(x,y_1,y_2) = \vec r.
$$
After some calculations,
$$
B
= {n\over f^2}\left[\matrix{
   fy_2(S_2-S_1) &      0       &      f         \cr
\vphantom{\Bigl(}
        0        & y_1x(f-y_1x) & -y_1xy_2S_0    \cr
        f        & -y_1xy_2S_0  & y_2S_0(1+y_1x) \cr}
  \right]
~~{\rm at}~(x,y_1,y_2) = \vec r.
$$
These equations can be used in the theorem to obtain asymptotics.

With $k_1$ the number of singleton blocks and $k_2$ the number of
other blocks, we can get a local limit theorem for the distribution
of $(k_1,k_2)$ as $n\to\infty$ when noncrossing partitions of
$\underline n$ are selected at random.
To do this, we set $r_1=r_2=1$ and use \eqConditioning\ to obtain the
covariance matrix.
It follows that the joint distribution of 
$(k_1-n\mu_1)\!\bigm/\!\!\Sqrt n$ and $(k_2-n\mu_2)\!\bigm/\!\!\Sqrt n$
is asymptotically normal with covariance matrix $C$ where
$$
S_i=S_i(r_0),\quad
S_1 = 1+ S_0 \hbox{~determines~}r_0,\quad
f = r_0 + S_1 = 1 + r_0 + S_0,
$$
$$
\mu_1 = r_0/f,\quad
\mu_2 = S_0/f,
$$
and
$$
C =
\left[\matrix{
   r_0S_1/f^2 &   -r_0S_0/f^2  \cr
  -r_0S_0/f^2 & (1+r_0)S_0/f^2 \cr}\right]
- \left[\matrix{ 0 \cr 1/f\cr}\right] {f\over S_2-S_1}
\left[\matrix{ 0 & 1/f\cr}\right].
$$
For example, when $M$ is the set of primes, $r_0=0.5580260$,
$$
\mu_1 = 0.263674,\quad
\mu_2 = 0.263815,\quad{\rm and}\quad
C = \left[\matrix{
\hfill 0.194150 & -0.069561 \cr -0.069561 &\hfill 0.067667}
\right].
\qed
$$

\example\ExPowers{Powers of an Inversion}
Suppose $w(x,\vec y) = xf(w,\vec y)$.
How do the coefficients of $[x^n]\,w^k$ behave as $k\to\infty$ with
$n$?
Meir and Moon~[\MMasymp] studied the case when $\vec y$ was absent
because $w(x)=xf(w(x))$ is associated with a variety of labeled and
unlabeled tree enumerations and $w^k$ counts forests with $k$
components.
The introduction of $\vec y$ allows us to keep track of additional
information (such as vertex degrees), but we can still follow Meir
and Moon's approach.
Furthermore, when $\vec y$ is absent, we obtain their result.
Since $g(w) = w^k$, Meir and Moon observed that Lagrange inversion gives
$$
[x^n]\,w(x,\vec y)^k
= (1/n)[x^n]\,\bigl(xkx^{k-1}f(x,\vec y)^n\bigr)
= (k/n)[x^{n-k}]\,f(x,\vec y)^n.
$$
One can now apply Theorem~\ThmPowers\ to obtain asymptotics.
The zeroth component of $\vec m$ gives the equation
$$
{n-k\over n}f(x,\vec y) = x{\partial f(x,\vec y)\over\partial x}.
\eqlabel\eqMMpowers
$$
It follows that ${n-k\over n}$ must be bounded away from 0 and so
the value of $k$ must be restricted to $1\le k\le\alpha n$ where
$\alpha<1$.
If \eqMMpowers\ has a solution $(r_0,\vec r)$ when $k=0$ and if the
power series for $f$ has nonnegative coefficients, letting $r_0$
decrease toward 0 produces a solution for the same $\vec r$ and all
larger values of $k$.
In particular, when $\vec r=\vec 1$, one obtains a local limit
theorem for the distribution of the variables counted by $\vec y$,
with means and covariance matrix proportional to $n$ and their values
depending on the value of $k/n$.\qed


\example\ExTreeDegS{Plane Trees by Vertex Degree}
A planted plane tree is a rooted plane tree in which the root has
degree 1.
If $x$ counts nonroot vertices and $y_k$ counts nonroot vertices of
degree $k$, then the generating function satisfies
$$
w(x,\vec y) = x\sum_{k\ge 0}z_{k+1}w^k. 
\eqlabel\eqPlanted
$$
Goulden and Jackson~[\GouldJack,~Sec.2.7.7] obtain the formula
$$
[x^n\vec y^{\vec k}]\,w(x,\vec y) = {(n-1)!\over\prod k_i!},
$$
provided $\sum k_i = n$ and $\sum ik_i=2n-1$, and is zero otherwise.
where the last factor is a multinomial coefficient
If one wishes to keep track of only a few degrees, say those in a
finite set $D$, summing this formula could be impractical.
In Exercise~2.7.2 of~[\GouldJack], Goulden and Jackson obtain formulas
when $D$ is a singleton or a pair of degree.
The former is an alternating sum and the latter an alternating double
sum.

Specializing \eqPlanted\ by setting $y_k=1$ for $k\notin D$, we can
apply the theorem with $h=x$ and

$$
f 
= \sum_{k\notin D}x^{k-1} +  \sum_{k\in D}y_kx^{k-1}
= {1\over1-x}  +  \sum_{k\in D}(y_k-1)x^{k-1}
\eqlabel\eqDegf
$$
Since $f$ has positive coefficients, we now verify \eqSpan.
If $k$ is the $j$th element in $D$, let $\vec e_k$ be the unit vector
whose $j$th component is 1 and let $i\notin D$ be fixed.
Since $\gcd\{i-j\mid i,j\notin D\}=1$ when $D$ is finite,
$$
\eqalign{
{\cal A}\Bigl\{\vec k-\vec j \Bigm| a(\vec k)a(\vec j)\ne0\Bigr\}
&= {\cal A}\Bigl\{(i-j,\vec 0),\;\; (k-j,\vec e_k) \Bigm|
     i,j\notin D,\; k\in D\Bigr\} \cr
& = {\cal A}\Bigl\{(1,\vec 0),\;\; (k-j,\vec e_k) \Bigm|
     j\notin D,\; k\in D\Bigr\} \cr
& = \integers^{1+|D|}. \cr}
$$
One easily computes that the component of $\vec m$ associated with
$x$ is
$$
m_0={n\over f}\biggl({x\over(1-x)^2}
+ \sum_{k\in D} (k-1)(y_k-1)x^{k-1}\biggr),
\eqlabel\eqDegmzero
$$
and that associated with $y_k$ is 
$$
m_k=ny_kx^{k-1}/f.
\eqlabel\eqDegmk
$$
After some calculation, and using the fact that $m_0=n$ for
Lagrange inversion, we find that
$$
\eqalign{
b_{0,0} 
&= {n\over f}\biggl({x(1+x)\over(1-x)^3}
+ \sum_{k\in D}(k-1)^2(y_k-1)x^{k-1}\biggr) - 1, \cr
b_{0,k}
&= (k-2)m_k/f, \cr
b_{k,k}
&= m_k(1-m_k/n), \cr
b_{k,j}
&= -m_km_j/n,\quad k\ne j.\cr}
$$
We can use the theorem to obtain either asymptotics or a local limit
theorem.

To obtain asymptotics, we want $\vec m$ to give the number of
vertices of each type so that then $\vec t=\vec 0$ in \eqPowers,
which will give the greatest accuracy.
The values of $r_0$ and $r_k$ are given by setting $x=r_0$ and
$y_k=r_k$ and then combining \eqDegf, \eqDegmzero, and~\eqDegmk:
With $\mu_k=m_k/n$, the fraction of vertices of degree $k$, we have
$$
{1\over1-r_0}  +  \sum_{k\in D}(\mu_k-r_0^{k-1})
= {r_0\over(1-r_0)^2} + \sum_{k\in D} (k-1)(\mu_k-r_0^{k-1}),
$$
which can be solved numerically for $r_0$ once $D$ and the fractions
$\mu_k$ are given.
Then $r_k=\mu_k/r_0^{k-1}$.
Using these values in the formulas for $b_{i,j}$ and then in
\eqPowers\ with $\vec t=\vec 0$ gives the asymptotics.

The local limit theorem is easily obtained since we simply set
$y_k=r_k=1$ for $k\in D$ and $x=r_0$.
This leads to
$$
r_0     = 1/2,           \quad
f       = 2,             \quad
\mu_k   = 2^{-k},        \quad
b_{0,0} = 2n.
$$
Using \eqConditioning, we obtain
$$
{c_{k,k}\over n} = \mu_k - \mu_k^2\left(1+{(k-2)^2\over2}\right)
\quad{\rm and}\quad
{c_{k,j}\over n} = -\mu_k\,\mu_j\left(1+{(k-2)(j-2)\over2}\right).
$$

We could equally well have looked at out-degrees in simply generated
families of trees.
In that case, \eqDegf\ becomes
$$
f = \sum_{k\ge1} f_kx^k + \sum_{k\in D}f_k(y_k-1)x^k,
$$
and the analysis proceeds as above.
In particular, when $D$ is a singleton set, we recover Theorem~1(i)
of Meir and Moon~[\MMdegree].\qed


\example\ExMaps{3-Connected Rooted Maps}
The asymptotics for 3-connected rooted maps by number of edges were
determined by Tutte~[\Tutte].
We use Mullin and Schellenberg's parameterization~[\MullSchell].
They found that the generating function with $x^my^n$ counting
3-connected rooted planar maps with $m+1$ vertices and $n+1$ faces is
$$
p(x,y) = \left({1\over1+x}+{1\over1+y}-1\right)xy
- {rs\over(r+s+1)^3},
$$
where
$r=x(s+1)^2$ and $s=y(r+1)^2$.
Setting $x=y$ and $r=s$, we obtain the generating function by number
of edges since $E=V+F-2$ by Euler's relation.
For asymptotic purposes, we can ignore the first part of $p(x,y)$ and
look at 
$$
[x^n]\,\left({-r^2\over(1+2r)^3}\right) ~~{\rm where}~~
r = x(1+r)^2.
$$
We have
$$
\eqalign{
[x^n]\,\left({-r^2\over(1+2r)^3}\right)
&= n^{-1}[x^n]\,\left(x\left({-x^2\over(1+2x)^3}\right)'
       (x+1)^{2n}\right)\cr
&= n^{-1}[x^n]\,\left({2x^2(x-1)\over(1+2x)^4} (x+1)^{2n}\right).}
\eqlabel\eqMapsZero
$$
The fraction in the last line is $h(x)$.
Since condition \eqDiffZero\ becomes $1-{2r_0\over1+r_0}=0$,
$h(r_0)=0$ and Theorem~\ThmPowers\ fails to apply.
Fortunately, we can rewrite the last line of \eqMapsZero\ in the form
of \eqDiffLI\ and then use \eqPrimeLI:
$$
\eqalign{
n^{-1}[x^n]\,\left({2x^2(x-1)\over(1+2x)^4} (x+1)^{2n}\right)
&= n^{-1}[x^n]\,\left({-2x^2(x+1)\over(1+2x)^4} (x+1)^{2n}
  \left(1-{2x\over1+x}\right)\right)\cr
&= n^{-2}[x^n]\,\left(x\left({-2x^2(1+x)\over(1+2x)^4}\right)'
   (x+1)^{2n}\right) \cr
&= n^{-2}[x^n]\,\left({2x^2(2x^2+x-2)\over(1+2x)^5}
   (x+1)^{2n}\right). \cr}
$$
Noting that $r_0=1$ and putting all this in the theorem we find that
the number of rooted 3-connected maps with $n$ edges is asymptotic to
$2^{2n+1}\bigm/3^5\Sqrt{n\pi}\,n^2$.\qed



\section Lagrange Inversion of Several Functions

Suppose we have
$$
w_i(\vec x,\vec y) = x_if_i(\vec w(\vec x,\vec y),\vec y)
~~{\rm for}~~ 1\le i\le d
\eqlabel\eqMLfunctions
$$
and want $[\vec x^\vec n\vec y^\vec j]\,g(\vec w,\vec y)$.
The two forms of Lagrange inversion for several equations that
parallel \eqDiffLI\ and \eqPrimeLI, respectively, are~[\BRliplus]
$$
\eqalignno{
[\vec x^\vec n\vec y^\vec j]&\,g(\vec w(\vec x,\vec y),\vec y)\cr
&= [\vec x^\vec n\vec y^\vec j]\,\left\{
     g(\vec x,\vec y)\; \vec f(\vec x,\vec y)^{\vec n}\;
   \left\|\delta_{i,j} - 
     {x_i\over f_j(\vec x,\vec y)}\,
     {\partial f_j(\vec x,\vec y)\over\partial x_i}\right\|
   \right\}
&\equno\eqDetLI \cr
&= {1\over \prod n_i}\;[\vec x^\vec n\vec y^\vec j]
  \sum_{\graph T}\vec x^\vec 1
  {\partial(g,f_1^{n_1},\ldots,f_d^{n_d})^{\vphantom{\bigr)}}
    \over\partial\graph T},
&\equno\eqOurLI \cr}
$$
where $\vec 1$ is the vector of all ones,
$\delta_{i,j}$ is the Kronecker delta,
\Bitem
$\|D\|$ is the determinant of the $d\times d$ matrix $D$,
\Bitem
the vector $(g,f_1^{n_1},\ldots,f_d^{n_d})$ has indices
$0,\ldots,d$,
\Bitem
$\graph T$ runs over all trees with vertex set indexed on
$0,\ldots,d$ and edges directed toward 0,
and
\Bitem
for a directed graph $\graph D$ whose vertex set $V$ is
the indices of $\vec h$ and $\vec x$
$$
{\partial \vec h\over\partial\graph D}
= \prod_{j\in V}\left\{
    \biggl(\prod_{(i,j)\in E}{\partial\over\partial x_i}\biggr)
    h_j(\vec x)\right\},
$$
where $E$ is the edge set of $D$.

\noindent
To apply the theorem, we let $\vec u=(\vec x,\vec y)$,
$\vec k=\vec i = (\vec n,\vec j)$, and we let $h$ be what is left in
\eqDetLI\ or \eqOurLI\ after removing $\vec f^\vec n$.
By Theorem~\ThmPowers,

$$
[\vec x^\vec n\vec y^\vec j]\,g(\vec w(\vec x,\vec y),\vec y)
\sim
{h(\vec r)\vec f(\vec r)^\vec n
\over  \sqrt{\det(2\pi B)}\;\vec r^{\vec k}},
\eqlabel\eqMLasymp
$$
where $B=B(\vec f(\vec r)^\vec n)$ and 
$\vec k = \vec m\bigl(\vec f^\vec n\bigr)$.
The last equation can be written
$$
\vec n L(\vec r) = \vec 0
\quad{\rm where}\quad
L_{i,j}(\vec x) = \delta_{i,j}
  - {x_j\over f_i}{\partial f_i\over\partial x_j},
\eqlabel\eqLmatrix
$$
where $\delta_{i,j}$ is the Kronecker delta.
Thinking of $\vec n$ as unknown, we see the condition for this set
of equations to have a nontrivial solution is precisely that the
determinant in \eqDetLI\ vanish, and so $h$ will violate condition
(a) in Theorem~\ThmPowers.
Hence we use \eqOurLI\ rather than \eqDetLI.
This is the multiple inversion case of what happened with~\eqDiffLI.

\medskip

We now turn our attention to conditioning.
Since the discussion is somewhat involved, you may wish to read
Example~6.1 beforehand.

To discuss conditioning, we first need an appropriate local limit
theorem.
It will be simpler not to distinguish between the variables $\vec x$
and $\vec y$.
This can be done by supplementing \eqMLfunctions\ with the additional
equations $w_i=y_i$, which means $f_i=1$.
In this way, we eliminate references to $\vec y$ and incorporate 
$\vec j$ in $\vec n$.
Since the new $f_i=1$, \eqMLasymp\ and \eqLmatrix\ are still valid
and $L_{\rm new} = L_{\rm old}\oplus I$, where $I$ is an identity
matrix indexed for the added Lagrange equations $w_i=y_i$.


\theorem\ThmLLT
Let $\vec r$ be the solution to 
$\vec i L(\vec r) = \vec 0$.
Under the assumptions of Theorem~\ThmPowers, with the extra variables
$\vec y$ eliminated as described above, we have
$$
[\vec x^\vec k]\,\bigl(h(\vec x)\vec f(\vec x)^\vec k\bigr)
=
{h(\vec r)\vec f(\vec r)^\vec k
     \bigl\{\exp\bigl(-\vec t LB^{-1}L'\vec t'/2\bigr) +o(1)\bigr\}
  \over  \sqrt{\det(2\pi B)}\;\vec r^{\vec k}}
\eqlabel\eqLLT
$$
uniformly as $\|\vec k\|\to\infty$, where $L=L(\vec r)$,
$B=B(\vec f(\vec r)^\vec k)$, and $\vec t=\vec k-\vec i$.
Since $\vec iL=\vec 0$, we can replace $\vec t$ by $\vec k$.
\Endthing

\noindent
Equation \eqLLT\ does not give a distribution becausee $\det(L)=0$
implies that $LB^{-1}L'$ is singular.
It turns out that conditioning leads to a nonsingular matrix.


\proof
Let $n=\|\vec k\|$.
Let $\vec r^*$ be the solution to $\vec kL=\vec 0$.
Let $\vec s$ and $\vec s^*$ be the the componentwise logarithms of
$\vec r$ and $\vec r^*$, respectively.
Let $B$ denote $B(\vec f(\vec r)^\vec k)$.
Note that $B = O(n)$ and $\det(B)$ is of order $n^{\dim(B)}$, where
the latter follows from (a)~$B$ is positive definite, (b)~$B(f_i)$ is
positive semidefinite, and (c)~$k_i/n$ is bounded away from 0.
It follows that $B^{-1}=O(1/n)$.
These results also hold for $B(\vec f(\vec r^*)^\vec k)$.

For now, we assume that $\|\vec tL\| \le n^{3/5}$.
By Taylor series with remainder,
$$
\vec k L(\vec r^*) 
= \vec k L(\vec r) + (\vec s^*-\vec s)B
+ O\left(\|\vec s^*-\vec s\|^2n\right).
\eqlabel\eqkLTaylor
$$
Since $\vec k L(\vec r^*)=\vec0$, it follows from the above that
$$
\|\vec s^*-\vec s\| = O(n^{3/5})B^{-1} = O(n^{-2/5})
$$
and so, from \eqkLTaylor,
$$
\eqalignno{
\vec s^*-\vec s
&= -\vec k L(\vec r^*)B^{-1} + O(n^{-4/5}n)B^{-1} \cr
&= -\vec t L(\vec r^*)B^{-1} + O(n^{-4/5})
= O(n^{-2/5}).
&\equno\eqDeltas\cr}
$$
Since the components of $\vec r$ are bounded away from 0 and
$\infty$, we also have $\vec r^*-\vec r = O(n^{-2/5})$.
We now consider the expansion of the logarithm of the ratio
$$
\left.
{h(\vec r)\vec f(\vec r)^\vec k/\vec r^\vec k
   \over \sqrt{\det\bigl(2\pi B(\vec f(\vec r)^\vec k)\bigr)}}
\right/
{h(\vec r^*)\vec f(\vec r^*)^\vec k/(\vec r^*)^\vec k
   \over \sqrt{\det\bigl(2\pi B(\vec f(\vec r^*)^\vec k)\bigr)}}
\eqlabel\eqAllRatio
$$
in a Taylor series about $\vec s^*$.
Note that 
$$
h(\vec r^*) = h(\vec r) + O(\|\vec r^*-\vec r\|)
= h(\vec r)(1+O(n^{-2/5})
$$
since $h$ is bounded away from 0.
Since
$$
B
= B(\vec f(\vec r^*)^\vec k) + O(n\|\vec r-\vec r^*\|)
= B(\vec f(\vec r^*)^\vec k) + O(n^{3/5}).
$$
and $B^{-1}=O(1/n)$, it follows that
$$
B(\vec f(\vec r)^\vec k)^{-1}B(\vec f(\vec r^*)^\vec k)
= I + O(B^{-1}n\|\vec r-\vec r^*\|)
= I + O(n^{-2/5}).
\eqlabel\eqBratio
$$
Finally, with $\Delta\vec s=\vec s-\vec s^*$,
$$
\eqalign{
\log\left(
{\vec f(\vec r  )^\vec k(\vec r^*)^\vec k  \over
\vec f(\vec r^*)^\vec k \vec r   ^\vec k} \right)
&= -\vec k L(\vec f(\vec r^*))(\Delta\vec s)'
+ {\textstyle{1\over2}}
   \Delta\vec s B(\vec f(\vec r^*)^\vec k)(\Delta\vec s)'
+ O\left(n\|\Delta\vec s\|^3\right) \cr
&= {\textstyle{1\over2}}
   \Delta\vec s B(\vec f(\vec r^*)^\vec k)(\Delta\vec s)'
+ O(n^{-1/5}), \cr}
$$
since $\vec kL(\vec r^*)=\vec0$.
Combining this with \eqDeltas,
$$
\log\left(
{\vec f(\vec r  )^\vec k(\vec r^*)^\vec k  \over
\vec f(\vec r^*)^\vec k \vec r   ^\vec k} \right)
= {\textstyle{1\over2}}
   \vec t L(\vec r)B^{-1}B(\vec f(\vec r^*)^\vec k)B^{-1}
         L(\vec r)'\vec t' + O(n^{-1/5}).
\eqlabel\eqLogRatio
$$
{}From \eqBratio,
$$
B^{-1}B(\vec f(\vec r^*)^\vec k)B^{-1}
= B^{-1}(I+O(n^{-2/5}))
= B^{-1}+O(n^{-7/5})$$
and $B(\vec f(\vec r^*)^\vec k)^{-1}=B^{-1}+O(n^{-7/5})$.
Combining the various equations and \eqPowers\ with 
$\vec i=\vec k$ and the fact that $\vec tL(\vec r)=O(n^{3/5})$, we
obtain \eqLLT.

\medskip

We now assume that $\|\vec tL\| \ge n^{3/5}$.
In this case, the exponential in \eqLLT\ is $o(1)$.
Thus, it suffices to prove that \eqAllRatio\ tends to $\infty$ with
$n$.
The compactness of $R$ and the nonsingularity of $B$ assure us that
$h(\vec r)/h(\vec r^*)$ and the ratios of the square roots in
\eqAllRatio\ are bounded away from 0 and $\infty$.
Thus it suffices to consider the ratio $F(\vec s)/F(\vec s^*)$ where
$$
F(\vec u) = \vec f(\vec x)^\vec k/\vec x^\vec k,
~~ u_i=\log x_i,
~~{\rm and}~~\vec x\in R.
$$
We may assume that the region $R$, viewed in $\vec s$ coordinates
is convex:
Other than the compactness requirement, the main restriction on $R$
was that the various power series converge.
If $\sum a_\vec n\exp(\vec n\cdot\vec s)$ and
$\sum a_\vec n\exp(\vec n\cdot\vec s^*)$ converge absolutely, then so
does
$\sum a_\vec n\exp\bigl(
       \vec n\cdot(\lambda\vec s+(1-\lambda)\vec s^*)\bigr)$
because, for series with nonnegative terms,
$$
\eqalign{
\sum A_\vec n^\lambda B_\vec n^{1-\lambda}
&\le \sum \max(A_\vec n,B_n)^\lambda 
          \max(A_\vec n,B_n)^{1-\lambda}\cr
&=   \sum \max(A_n,B_n)
\le \sum A_n + \sum B_n.\cr}
$$

By standard calculus, the gradient and matrix of second derivatives
of $F$ are
\hbox{$\nabla F=\vec kL(\vec x)$} and $B(\vec f(\vec x)^\vec k)$.
Since $B$ is positive definite, $F$ is convex and so has just one
minimum, which is given by the solution $\vec s^*$ to
$\nabla F(\vec u)=\vec0$.
Think of $\vec k$ as fixed and $\vec i$ as variable.
(Thus $\vec s$ is also variable.)
Using \eqLogRatio\ for $\|\vec tL\|= n^{3/5}$, we see that 
$F(\vec s^*)=o(F(\vec s))$ uniformly for such values.
{}From the convexity of $F$, it follows that
$F(\vec s^*)=o(F(\vec s))$ for $\|\vec tL\| \ge n^{3/5}$.\qed


\bigskip
\noindent{\bf Conditioning and Summing}.
Let $\cal C$ be the indices of variables on which we are conditioning
and $\cal N$ the remaining indices.
One uses the equation $\vec iL=\vec 0$ to determine $r_j$ for
$j\in\cal C$ and $i_j$ for $j\in\cal N$.
In addition, one has the equations $f_j(\vec r)/r_j=1$ for 
$j\in\cal N$.
The latter equations guarantee that $i_j$ will be chosen to be at the
peak of the distribution when $j\in\cal N$.
One extracts the diagonal submatrix of $LB^{-1}L'$ that is indexed by
$\cal N$.
If we condition on {\it all\/} the original $\vec x$, then $r_j=1$
for $j\in\cal N$ (since $f_j=1$) and the relevant portion of $L$ is the
identity matrix so we are just extracting a diagonal submatrix of
$B^{-1}$; all of which is as described in Section~3 for the single
Lagrange equation.
In particular, \eqConditioning\ applies with $B_{1,1}$ indexed by
$\cal C$ and $B_{2,2}$ and $C$ indexed by $\cal N$.

The theorem can also be used for summing over variables.
Suppose we have a formula as in \eqLLT\ and  want to sum over
certain components of $\vec t$.
Let the set of indices be $\cal S$
The corresponding $\vec i$ components must be chosen so that we are
the peak; that is, $f_j(\vec r)/r_j=1$ for $j\in\cal S$.
Partition $M=LB^{-1}L'$ into a $2\times2$ block matrix where the
subscript 1 refers to elements of $\cal S$.
In summing \eqLLT, the tail will be negligible because of the
convexity discussed in the last paragraph of the theorem's proof.
Thus we are faced with the problem of summing $\exp(-z/2)$ over 
$\vec t_1$ where
$$
\eqalign{
z
&= (\vec t_1,\vec t_2)
   \left[\matrix{M_{1,1}&M_{1,2}\cr M_{2,1}&M_{2,2}\cr}\right]
   (\vec t_1,\vec t_2)'\cr
&= (\vec t_1+\vec t_2T) M_{1,1} (\vec t_1+\vec t_2T)'
+ \vec t_2(M_{2,2}-TM_{1,1}T')\vec t_2',\cr}
$$
and $T$ satisfies $TM_{1,1}=M_{2,1}$.
(We have used the fact that $M$ is symmetric.)
In order to be able to carry out the summation, $M_{1,1}$ must be
nonsingular.
In that case, we obtain a factor of 
$\bigl(\det(2\pi M_{1,1}^{-1})\bigr)^{1/2}$ from the summation and
the exponential becomes
$$
\exp\bigl(-\vec t_2(M_{2,2}-TM_{1,1}T')\vec t_2'/2\Bigr),
$$
which is still singular if $M$ is singular.
If $M$ is not singular, the above matrix is the inverse of the lower
righthand block of $M^{-1}$.

(In this case, an alternative proof can be found in Section 3.4 of
Press~[\Press].)
Conditioning, before or after summing, will normally make the matrix
nonsingular.

Before conditioning and/or summing, one may wish to make a linear
change of variables.
For example, if $m_k$ is the number of vertices of type $k$, one
might introduce the coordinate $n=\sum m_k$, the total number of
vertices, and condition on it.
After changing coordinates, the rules for selecting $\vec r$ and
$\vec i$ differ somewhat, but the underlying principles are the same:
If $\vec n =\vec mA$ gives the old coordinates $\vec n$ in terms of
the new coordinates $\vec m$, then
$$
{\vec f(\vec r)^\vec n  \over \vec r   ^\vec n}
=
{\vec f^*(\vec r)^\vec m\over(\vec r^*)^\vec m},
$$
where $\log\vec f^*=(\log\vec f)A$ and
$\log\vec r^*=(\log\vec r)A$.
The earlier rules now apply with $\vec f^*(\vec r)$, $\vec r^*$, and 
$\vec iA^{-1}$ in place of $\vec f(\vec r)$, $\vec r$, and $\vec i$.
(This is illustrated in the next example.)



\section Examples of Inversion of Several Functions

We now turn to examples of inversion of more than one function.
Because of the intimate connection between Lagrange inversion and
tree enumeration, it is not surprising that Lagrange inversion is
often ideal for studying tree enumeration questions.
For a single function, this can be seen in the examples of the
Section~4.
Good~[\Good] may have been the first to point this out for inversion
of several functions.
In this connection, see also Goulden and Jackson~[\GouldJack].
As the examples in this section illustrate, Lagrange inversion of
several functions leads to more complicated calculations than occur
for a single function.
Thus one should reduce the inversion problem to a single function
when possible.


\example\ExColorTree{Plane Rooted Colored Trees}
As usual, the set of colors is finite.
We consider situations in which local conditions determine the
possible colors of a vertex.
Similar methods apply to rooted labeled colored trees (no longer
planar).
Examples of situations that can be dealt with in this manner are:
\Bitem
A vertex must have a different color from its children (proper
coloring).
\Bitem
The children of a vertex must have distinct colors.
\Bitem
A vertex with grandchildren must have the same color as a grandchild.
\itemskip
\noindent
The more complex the conditions and the greater the number of colors,
the greater the number of equations that must be inverted, and so the
more complicated the calculations.

We begin by considering trees with green and red vertices.
Our local condition will be that a nonleaf green vertex must have
exactly one red child and two green children, while a nonleaf red
vertex must have exactly one child of each color with the left one
being red.
Associate the subscript 1 with green and 2 with red.
Let $t_i$ keep track of number of vertices of color $i$ and let $w_i$
be the generating function for trees by root color.
The conditions translate to
$$
\eqalign{
w_1(\vec x) &= x_1f_1(\vec w),
~~{\rm where}~~
f_1(\vec w) = 1 + 3w_1(\vec x)^2 w_2(\vec x),\cr
%
w_2(\vec x) &= x_2f_2(\vec w),
~~{\rm where}~~
f_2(\vec w) = 1 + w_1(\vec x) w_2(\vec x).\cr}
$$
Lagrange inversion gives us $c(\vec k)$, the number of trees with
$k_1$ green and $k_2$ red vertices:
$$
c(\vec k) = [\vec x^\vec k]\,(w_1+w_2).
$$
There are 3 directed trees in the sum over $\graph T$.
Their edge sets are 
$$
E_1 = \{(1,0),\;(2,1)\},~~
E_2 = \{(2,0),\;(1,2)\},~~
{\rm and}~~
E_3 = \{(1,0),\;(2,0)\}.
\eqlabel\eqThreeTrees
$$
Thus
$$
\eqalign{
c(\vec k) 
&= (k_1k_2)^{-1}[\vec x^\vec k]\left\{x_1x_2\left(
  {\partial(x_1+x_2)\over\partial x_1}
   {\partial f_1(\vec x)^{k_1}\over\partial x_2}f_2(\vec x)^{k_2}
\right.\right.\cr
&\qquad\left.\left.
+ {\partial(x_1+x_2)\over\partial x_2}
   {\partial f_2(\vec x)^{k_2}\over\partial x_1}f_1(\vec x)^{k_1}
+ {\partial^2(x_1+x_2)\over\partial x_1\partial x_2}
   f_1(\vec x)^{k_1} f_2(\vec x)^{k_2}\right)\right\}\cr
&= (k_1k_2)^{-1}[\vec x^\vec k]\left\{x_1x_2\left(
   {3k_1x_1^2\over f_1(\vec x)}f_1(\vec x)^{k_1}f_2(\vec x)^{k_2}
+  {k_2x_2\over f_2(\vec x)}f_1(\vec x)^{k_1}f_2(\vec x)^{k_2}
\right)\right\}.\cr}
$$
We apply Theorem~\ThmPowers\ with $\vec n=(k_1,k_2)$ and
$h=3x_1^3x_2/f_1(\vec x)+x_1x_2^2/f_2(\vec x)$.
We set $\vec k=\vec i$ and $\nu=\vec 0$ in \eqPowers.
The equation $\vec i=\vec m(\vec f(\vec r)^\vec i)$ is
$$
\eqalign{
k_1 &= {6r_1^2r_2\over1+3r_1^2r_2}k_1 + {r_1r_2\over1+r_1r_2}k_2\cr
k_2 &= {3r_1^2r_2\over1+3r_1^2r_2}k_1 + {r_1r_2\over1+r_1r_2}k_2\cr}
\eqlabel\eqCTmvector
$$
and \eqPowers\ becomes
$$
c(\vec k) \sim
\left({3r_1^3r_2\over k_2(1+3r_1^2r_2)}
    + {r_1r_2^2\over k_1(1+r_1r_2)}\right)
{(1+3r_1^2r_2)^{k_1}(1+r_1r_2)^{k_2}\over
   \sqrt{\det(2\pi B)}\; r_1^{k_1}r_2^{k_2}},
\eqlabel\eqCTasymp
$$
where $B=k_1B(f_1)+k_2B(f_2)$.
To verify \eqSpan, note that
$$
\Lambda(\vec f) = {\cal A}\{(2,1),\;(1,1)\} = \integers^2.
$$
It remains to compute $\vec r$, verify that $h>0$, compute $B$, and,
perhaps, simplify \eqCTasymp, details of which we omit.
One can check that, with $1<\rho=k_1/k_2<2$,
$$
r_1 = {(\rho-1)^2\over3(2-\rho)}
~~{\rm and}~~
r_2 = {3(2-\rho)^2\over(\rho-1)^3}
\eqlabel\eqCTrho
$$
$r_1$ and $r_2$ are positive and \eqCTmvector\ is satisfied.
Hence the theorem applies if, for some $\epsilon>0$, we have
$1+\epsilon\le k_1/k_2\le2-\epsilon$ as $\|\vec k\|\to\infty$.

Let $n=k_1+k_2$.
After some calculations
$$
\eqalign{
B(f_1)
&= {3r_1^2r_2\over(1+3r_1^2r_2)^2}
   \left[\matrix{4&2\cr2&1\cr}\right]
= {\rho-1\over\rho^2} \left[\matrix{4&2\cr2&1\cr}\right] \cr
B(f_2)
&= {r_1r_2\over(1+r_1r_2)^2}
   \left[\matrix{1&1\cr1&1\cr}\right]
= (2-\rho)(\rho-1) \left[\matrix{1&1\cr1&1\cr}\right] \cr
B(\vec f^\vec n)
&= k_1B(f_1) + k_2B(f_2)
= {n\rho\over\rho+1}B(f_1) + {n\over\rho+1}B(f_2) \cr
&= {n(\rho-1)\over\rho(\rho+1)}
   \left[\matrix{
   4 + 2\rho - \rho^2 & 2 + 2\rho - \rho^2 \cr
   2 + 2\rho - \rho^2 & 1 + 2\rho - \rho^2 \cr}\right]. \cr}
$$
One can eliminate $\vec r$ from \eqCTasymp\ by using \eqCTrho,
and $\vec k$ by using
$\vec k = n\left({\rho\over\rho+1},\,{1\over\rho+1}\right)$.

\medskip

We now obtain a local limit theorem for the number of red vertices in
trees having a fixed number of vertices.

To do so, we use \eqLLT, change coordinates from $\vec k$ to
$(n,k_2)$, choose $i_1/i_2$ so that we are at a peak, and condition on
$n$.
Since the change of coordinates is given by
$$
(k_1,k_2) = (n,k_2)A
~~{\rm where}~~
A = \left[\matrix{1&0\cr-1&1\cr}\right],
\eqlabel\eqCTcoords
$$
$\vec tLB^{-1}L'\vec t'=\vec u(ALB^{-1}L'A')\vec u'$, where
$\vec u=(n-i_1-i_2,\;k_2-i_2)$.
To condition on $u_1=0$, we delete the first row and column of
$ALB^{-1}L'A'$, which leaves a single number, say $1/n\sigma^2$.
Thus the number of trees satisfies
$$
c(n,k_2)
= {h(\vec r) (f_1/r_1)^n
     \bigl\{\exp(-(k_2-i_2)^2/2n\sigma^2) + o(1)\bigr\}
  \over \sqrt{\det(2\pi B)}}\,
\left({f_2r_1\over f_1r_2}\right)^{k_2}.
$$
To be a peak as a function of $k_2$, the last factor must be 1;
that is $f_1/r_1=f_2/r_2$.
With Maple's help we found that $\rho=1.73473$.
Using this in the preceding formula, we obtain
$$
c(n,k_2)
= {C_1C_2^n \bigl\{\exp(-(k_2-n\mu)^2/2n\sigma^2) + o(1)\bigr\}
  \over 2\pi n^2 C_3},
$$
where $C_1=1.00829$, $C_2=2.55726$, $C_3=0.105061$, $\mu=0.36567$,
and $\sigma=0.110205$.
(The extra factor of $n$ in the denominator is due to $h(\vec r)$.)
Of course, since the local limit theorem states that $k_2$ is normally
distributed with mean $n\mu$ and variance $n\sigma^2$, it does not
require the various $C_i$ values.

\medskip

We now turn our attention to the last problem raised at the start of
this example, namely, a vertex with grandchildren must have the same
color as a grandchild.
We want to count trees according to the number of vertices of each
color.
This problem has a new feature: with each vertex we must keep track of
its color and the set of colors of its children, with the empty set
arising when a vertex has no children.
As a result, we consider generating functions $T_{c,S}(\vec x)$ for
trees where $c$ is the root color and $S$ is the set of children's
colors.
This leads to functional equations of the form
$$
T_{c,S} = x_c\, f_{c,S}(\vec T),
$$
which are inappropriate for Lagrange inversion.
Consider instead
$$
T_{c,S} = x_{c,S}\, f_{c,S}(\vec T).
$$
We may think of $x_{c,S}$ as keeping track of $k_{c,S}$, the number
of vertices of color $c$ whose children's colors are $S$.
After applying \eqLLT\ to obtain asymptotics, we change coordinates
to $k_c=\sum_T k_{c,T}$ and $k_{c,S}$ with $S\ne\emptyset$,
then we sum over all values of $k_{c,S}$ (with $S\ne\emptyset$) to
obtain a result in terms of just the $k_c$.
The change of coordinates is done as in \eqCTcoords.
The method for summation is described at the end of Section~5.
After changing coordinates, the condition for setting the components
of $\vec r$ becomes ``$f_{c,S}(\vec r)/r_{c,S}$ must be independent
of $S$.''
We omit the details.\qed


\example\ExTreeDegM{Plane Trees by Vertex Degree, Continued}
Counting planted plane trees by vertex degree was treated in
Example~\ExTreeDegS.
We now consider a multiple Lagrange equation approach so that one can
compare the two approaches.

Let the sets $S_i$ be a finite partition of a subset of
$\{1,2,\ldots\}$ with $1\in S_1$.
(We need 1 to have leaves.)

Let $y_i$ keep track of the number of vertices having degree in
$S_i$.

To follow the idea of Example~\ExTreeDegS, we introduce one further
variable $x$ keeping track of all vertices except the root and write
the single equation
$$
T(x,\vec y) = x f(T(x,\vec y),\vec y)
~~{\rm where}~~
f(w,\vec y) = \sum_iy_i\biggl(\sum_{k\in S_i} w^{k-1}\biggr),
$$
which is dealt with as follows:
The equations
$$
n = {nx \partial f(x,\vec y)/\partial x\over f(x,\vec y)}
~~{\rm and}~~
k_i = {ny_i \partial f(x,\vec y)/\partial y_i\over f(x,\vec y)}
\eqlabel\eqDegreeCond
$$
must be solved for $x=r_0$ and $\vec y=\vec r$ if we have values of
$n$ and $\vec k$ in mind.
If we want to study the distribution of $\vec k$ conditioned on $n$,
we must set $\vec r=\vec 1$ and solve the first equation from
\eqDegreeCond,
$$
1 = {r_0\partial f(r_0,\vec 1)/\partial r_0\over f(r_0,\vec 1)}
\eqlabel\eqDegreeOne
$$
for $r_0$.
The remaining equations in \eqDegreeCond\ determine $\vec k/n$, the
value of $\vec k$ that gives the peak of the normal.
We can then proceed with conditioning as in the previous example.

If we do not introduce $x$, it is natural to introduce $T_i(\vec y)$,
the enumerator for trees where the degree of the root's son is in
$S_i$.
We have
$$
T_i(\vec y) = y_if_i(T(\vec y)),
~~{\rm where}~~
f_i(T) = \sum_{k\in S_i} T^{k-1}
~~{\rm and}~~
T(\vec y) = \sum_i T_i(\vec y).
$$
This leads to the equations
$$
k_i = \sum_j k_j r_i f_j'(r)/f_j(r),
~~{\rm where}~~
r=\sum_j r_j.
$$
Hence $r_i$ is proportional to $k_i$ and so $r_i=r k_i/n$, which is 
chosen so that
$$
1 = \sum_j r_j f_j'(r)/f_j(r).
\eqlabel\eqDegreeMulti
$$
If we want to get a distribution by conditioning on $\sum k_i=n$,
then, as in the previous example, we want $r_j/f_j(r)$ to be
independent of $j$.
This together with \eqDegreeMulti\ gives us $\dim\vec r$ equations in
the same number of unknowns; however, that does not seem to be as
easily solved as \eqDegreeOne.\qed



\example\ExMapsVF{3-Connected Rooted Maps, Continued}
Continuing Example~\ExMaps, we now consider enumeration of
3-connected rooted maps by vertices and faces.
As noted in that example, we want
$$
[\vec x^\vec n]\,\left({-w_1w_2\over(1+w_1+w_2)^3}\right)
~~{\rm where}~~
w_1 = x_1(1+w_2)^2 ~~{\rm and}~~ w_2 = x_2(1+w_1)^2.
$$
We must sum over 3 trees, namely those in \eqThreeTrees.
The tree with $E_3$ produces a term that is smaller by a factor on
the order of $n$ than those for $E_1$ and $E_2$.
The sum over the trees with $E_1$ and $E_2$ lead to a term which
vanishes when we set $\vec x$ to the $\vec r$ according to
Theorem~\ThmPowers.
Paralleling Example~\ExMaps, we could write this sum $S$ as
$(S/D)\times D$, where $D$ is the determinant in \eqDetLI, and then
use the fact that \eqDetLI\ equals \eqOurLI\ to get a more tractable
formula.
This approach requires that $S/D$ be well behaved as we approach
$\vec r$.
In Example~\ExMaps, this was the case since we had cancellation
between $S$ and $D$.
In this case there is no such cancellation and, based on Maple
calculations, the limit of $S/D$ as we approach $\vec r$ depends on
how $\vec r$ is approached.

Consequently, we are unable to proceed.\qed


\eject

\section Unsolved Problems

Examples~\ExMaps\ and \ExMapsVF\ raise the issue that the function
$g(\vec w)$ can cause problems.
We were able to avoid them in the first example but not in the
second.
If one attempts to enumerate all rooted maps a similar problem
arises.
There is also a new problem in that case:
One can enumerate all maps on general
surfaces~[\ArqPlane,\thinspace\BCR] and the vanishing determinant
problem causes difficulty on the projective plane even in the one
variable case because of a branch point.
Can the approach in this paper be extended to such situations or
must one use singularity analysis?
If the latter, is there a useful general formulation that will deal
with problems of this sort?

Suppose additional variables appear in $g$ but not in $h$.
In particular, suppose $h=h(w)$ and $g=g(w,y)$ and $g$ does not
have a singularity at $w=r$.
If $g(r,y)$ has finite radius of convergence and has nice
singularities there, then the arguments in~[\BRWcllt] may be usable.
If $g(r,y)$ is entire, this approach breaks down.
For example, if $w=xf(w)$ is the functional equation for the
exponential generating function of a regular family of labelled 
trees, then $(1-w(x))^{-y}$ is the enumerator for functional digraphs
such that removal of cyclic edges results in a forest of trees
enumerated by $w$ and $y$ keeps track of the number of cycles.
When $w(r)<1$, $(1-w(r))^{-y}$ is a Hayman admissible
function~[\Hayman].
We are not aware of any multivariate singularity analysis that
combines algebraic singularities and Hayman-admissibility.
We discussed multivariate Hayman admissibility in~[\BRhayman].

We believe it should be possible to extend Drmota's functional
equation results~[\Drmota] to remove the limitation that $g(\vec w)$
be $w_i$ for some $i$..
Also, one should be able to eliminate the conditioning on $n$ in his
Theorem 1, thereby obtaining a result like our Theorem~\ThmLLT,
probably with his $I-F_\vec y$ and $F_{\vec y,\vec y}$ playing a role
akin to our $L$ and $B$.
We have not attempted to develop these ideas and currently have no
plans to do so.



\nonumsection References

\frenchspacing

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\bye

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