MAT5107 : Combinatorial Enumeration
Mike Newman, winter 2017
16. Lagrange inverse function theorem
analytic functions
A complex function is said to be holomorphic at z0 if it is complex differentiable in some open
neighbourhood containing z0 . Note that this condition is stronger than real differentiability since it
says that
f (z) − f (z0 )
z→z0
z − z0
is independent of the path z → z0 in C. If we write z = x + iy, then f is complex-differentiable at
z0 if and only if
∂
∂
f (z) = −i f (z) ∂x
∂y
lim
z=z0
z=z0
This is the Cauchy-Riemann equation. Equivalently, it says that f (z) is conformal.
A complex function is said to be analytic at z0 if it can be expressed as a power series about z0 in
some open neighbourhood of z0 as
X
f (z) =
an (z − z0 )n
n≥0
For complex functions, being analytic and being holomorphic are the same thing. In particular, it
follows that f (z) is infinitely differentiable and that it is represented by its Taylor series.
For real functions, being analytic and being differentiable are not the same thing, primarily because
functions can be differentiable for all paths z → z0 in R while failing at many of the paths in C.
Notice that a real function can be thought of as a complex function (if we extend the domain); real
functions that are (real) differentiable but not analytic are not (complex) differentiable when we
extend the domain.
In a nutshell, this is why real analysis is technical and hard and ugly, while complex analysis is
elegant and simple and beautiful.
Example 16.1. The following functions are analytic at all points in the complex plane where they
are defined: polynomials, rational functions, sin, cos, log; the sum, difference, product and quotient
of analytic functions, compositions.
Problem 16.2. Verify, using the Cauchy-Riemann equation that any polynomial function of z is
∂
∂
∂
analytic at every point of C. Using the chain rule will help, as in ∂x
f = ∂z
f ∂x
z where z = x + iy.
Verify that f (z) = z is not analytic at any point of C.
Lagrange inverse function theorem
We consider the case where we have a formal power series u(z) that we wish to know the coefficients
of, where it satisfies some functional equation of the form u = zφ(u). It is not hard to see that if
φ(u) is a formal power series (in u), then this equation does uniquely determine define u(z). In other
words, u = u(z) is implicitly defined by the equation u = zφ(u) even if we have no explicit form for
∗
These notes are intended for students in mike’s MAT5107. For other uses please say “hi” to [email protected].
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u(z). We have used this fact, and even extracted coefficients by hand from equations such as these
(for instance trees).
Problem 16.3. Let φ(u) be a formal power series in u, and u be a formal power series in z. Assume
that u(z) = zφ(u(z)). Show that every coefficient un = [z n ] u(z) is uniquely determined.
We can extract a “formula” of sorts for the un in the previous problem, but in fact the process can
be automated, and even generalized.
Theorem 16.4 (LIFT). Let u(z) be a formal power series. Let φ(z) be a formal power series with
φ(0) 6= 0, and f (u) be a formal power series.
Suppose u(z) is implicitly defined by u(z) = zφ(u(z)). Then
[z n ] f (u(z)) =
1 n−1 0
[u
] f (u)φn (u)
n
Proof. To prove this it suffices to prove it for the case where φ and f are polynomials. This is because
the assertion of the theorem remains unchanged if we truncate φ and f at the n-th degree term.
We know that u = u(z) is uniquely defined as a formal power series, but we don’t have an expression
for it as a function. On the other hand we can directly derive an expression for z in terms of u, as
z = z(u) = u/φ(u). Notice that z(u) = u/φ(u) is an analytic function, and z(0) = 0, z 0 (0) 6= 0. So
it has an inverse which is analytic near the origin. This means that u(z), and hence f (u(z)), is an
analytic function in a neighbourhood of the origin (recall that f is assumed to be a polynomial).
We really don’t care what z = z(u) looks like as a function, but since it is an invertible analytic
function that guarantees that its inverse is also analytic. But it’s inverse is u = u(z) which is the
function we really do care about.
We use the Cauchy integral formula, integrating around a curve C = C(z) which circles the origin
within the region of analyticity. Since z = z(u) is at least locally invertible, we can change coordinates
from z to u (obtaining the curve C 0 in terms of u). Then we re-express the resulting integral and
apply the Cauchy integral formula backwards.
Z
f (u(z))
1
n
dz
[z ] f (u(z)) =
2πi C z n+1
Z
1
f (u)φn+1 (u) φ(u) − uφ0 (u)
=
du
2πi C 0
un+1
φ2 (u)
Z
Z
f (u)φn (u)
f (u)φn−1 (u)φ0 (u)
1
1
=
du
−
du
2πi C 0
un+1
2πi C 0
un
= [un ] f (u)φn (u) − [un−1 ] f (u)φn−1 (u)φ0 (u)
1
= [un−1 ] (f (u)φn (u))0 − [un−1 ] f (u)φn−1 (u)φ0 (u)
n
1
1
= [un−1 ] f 0 (u)φn (u) + [un−1 ] f (u)nφn−1 (u)φ0 (u) − [un−1 ] f (u)φn−1 (u)φ0 (u)
n
n
n−1 1 0
n
= [u
] f (u)φ (u)
n
Note that in order to express the result as a single coefficient extraction, we used the fact that for
any power series P (u) we have [un ] P (u) = [un−1 ] P 0 (u)/n, which follows directly from the definition
of the (formal power series) derivative.
In fact this proof is somewhat misleading, as it seems to depend heavily on analytic complex functions.
One can prove this without recourse to complex analysis but we won’t do this here.
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rooted labelled trees
Recall that if T is the set of rooted labelled trees (where size is measured by number of vertices) then
we found that T ∼
= {•}×Set(T ), which gave us the exponential generating function T = x exp(T (x)).
For u = T (x), φ(u) = exp(u) and f (u) = u, LIFT gives:
1
1
1 nn−1
n![xn ] u = n! [un−1 ] (eu )n = n! [un−1 ] enu = n!
= nn−1
n
n
n (n − 1)!
Every unrooted labelled tree corresponds to exactly n rooted labelled trees (the labelling effectively
kills the automorphisms). So the number of labelled trees (unrooted) is nn−2 .
As another example, let’s consider the average root degree of a rooted labelled tree.
Let fn,k be the number of rooted labelled trees on n vertices. We already saw (again using Set) the
bivariate generating function.
F (x, y) =
X
n,k
fn,k
xn k
y = x exp(yT (x))
n!
We analyze the random variable Xn , which measures the root degree of a random rooted labelled
tree. Recall that the factorial moments are given by the following (there “should” be a factor of
n! top and bottom to account for the fact that we want to extract coefficients from an exponential
generating function, but we have issued a pre-emptive retaliatory cancellation).
d
n
[x ] dy F (x, y) y=1
E[Xn ] =
n
[x ] F (x, 1)
d
d
[xn ] dy
y dy
F (x, y) y=1
E[(Xn )2 ] =
[xn ] F (x, 1)
So it is useful to calculate the following quantities.
[xn ] F (x, 1) = [xn ] x exp(T (x)) = [xn ] T (x)
d
n
[x ]
F (x, y) = [xn ] xT (x) exp(T (x)) = [xn ] T 2 (x)
dy
y=1
d d
[xn ]
y F (x, y) = [xn ] xT (x) exp(T (x)) + xT 2 (x) exp(T (x)) = [xn ] T 2 (x) + T 3 (x)
dy dy
y=1
We already worked out the first one, but in fact LIFT allows us to easily evaluate any of these.
We have u = T , and φ(u) = eu ; notice that we already used the implicit function to rewrite our
expression as a coefficient of a function of u alone, by eliminating the x on the right-hand side. This
time the function is not just f (u) = u, but f (u) = u2 or f (u) = u2 + u3 . Now using LIFT we obtain
[xn ] T (x) = [xn ] u =
1 n−1 u n
1
1 nn−1
[u
] (e ) = [un−1 ] enu =
n
n
n (n − 1)!
[xn ] T 2 (x) = [xn ] u2 =
1 n−1
2
2 nn−2
[u
] 2u(eu )n = [un−2 ] enu =
n
n
n (n − 2)!
[xn ] T 2 (x) + T 3 (x) = [xn ] u2 + u3 =
1 n−1
2
3
2 nn−2
3 nn−3
[u
] (2u + 3u2 )(eu )n = [un−2 ] enu + [un−3 ] enu =
+
n
n
n
n (n − 2)! n (n − 3)
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This gives us the moments.
2 nn−2
nn−1
2(n − 1)
2
÷
=
=2−
n (n − 2)!
n!
n
n
2 nn−2
nn−1
3 nn−3
2(n − 1) 3(n − 1)(n − 2)
11
6
E[(Xn )2 ] =
÷
+
=
+
=5−
+ 2
n (n − 2)! n (n − 3)!
n!
n
n2
n
n
E[Xn ] =
The average and variance is then
2
µ = E[Xn ] = 2 −
n
2(n − 1) 3(n − 1)(n − 2)
2(n − 1) 2 (n − 1)(n + 2)
3
2
2
σ =
−
+
=
=1− + 2
n
n2
n
n2
n n
exercises
1.
In the proof of Theorem 16.4, we insisted on φ(0) 6= 0, which is to say that φ has a nonzero
constant term. What happens if φ(0) = 0, so that the constant term is zero? The proof of
Theorem 16.4 fails, but what can we say about u(z) when u(z) = zφ(u(z)) and φ(0) = 0?
2.
Recall that the Catalan numbers have ordinary generating function C(x) where C = 1 + xC 2 .
a) Explain why we can’t (directly) apply Lagrange inversion to the equation C = 1 + xC 2 ;
in other words, why taking u = C does not (directly) work.
b) Set u = C(x) − 1, and show that u = x(u + 1)2 . Show that Lagrange inversion does
apply here and use it to derive the coefficient in u(x) and hence the coefficients of C(x).
Of course you know the final answer already. . .
c) Set u = xC(x), and show that u = x/(1 − u). Show that Lagrange inversion does apply
here and use it to derive the coefficient in u(x) and hence the coefficients of C(x). Of
course you know the final answer already. . .
3.
Consider the expression u5 − u + x = 0, where we imagine that u is a power series in x.
a) Show that this gives u = x/(1 − u4 ), and explain why Lagrange inversion applies here.
b) Using Lagrange inversion, find an expression for u as a formal power series in x.
c) Substitute the constant a for the variable x, and observe that you have a series solution
to the quintic polynomial u5 − u + a = 0.
d) Substitute u = xT into u5 − u + x = 0, and show that T = 1/(1 − x4 T 4 ). Explain why
T is the generating function for a certain class of trees and describe this class precisely.
You might notice that this only gives one root to u5 − u + a = 0; as a → 0 the five roots
converge to ±1, ±i, 0; this expression gives the root that converges to 0 as a → 0. Solving a
general quintic can be reduced to solving the Bring-Girard form u5 − u + a, so this gives (with
some preprocessing!) a power series solution to (one root of) an arbitrary quintic, where the
coefficients count certain trees.
4.
Let T be the set of rooted labelled plane trees where size is measured by number of vertices.
a) Recall (or rederive!) a functional equation for the corresponding exponential generating
function T .
b) Check that this equation is amenable to applying Theorem 16.4, and use Theorem 16.4
to find the number of such trees on n vertices. (Note that we already found this number
without Theorem 16.4.)
c) Using Theorem 16.4, find the average root degree of such a tree, and the variance.
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5.
Let T be the set of rooted labelled plane trees where each vertex has either 0 or d children,
and size is measured by number of vertices.
a) For d = 2, show that T = x(1 + T 2 ). Solve this directly using the quadratic formula,
and notice that we know the power series of the resulting closed-form expression for T .
Hence find tn .
b) For d = 2, find tn using Theorem 16.4. Of course this should be the same answer as
before.
c) For arbitrary d, give the corresponding functional equation for T . Using Theorem 16.4
find tn .
d) For arbitrary d, notice that many of the tn are zero. Explain.
e) Without really calculating anything, give the average root degree of such a tree, and the
variance (notice that n = 1 is a special case).
f) Using Theorem 16.4, find the average root degree of such a tree, and the variance.
Compare with the “obvious” answer.
6.
Let T be the set of rooted labelled plane trees where the number of children is a multiple of
d, and size is measured by number of vertices.
a) Recall (or rederive!) a functional equation for the exponential generating function T (x).
b) Using Theorem 16.4, find tn .
c) Using Theorem 16.4, find the average root degree of such a tree, and the variance.
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