Download Lecture B - UCSD Math Department

Lecture B
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Notation: R denotes a ring, N denotes the set of sequences of natural numbers
with finite support, n is a generic element of N , 0 is the infinite zero sequence,
¯
¯
R[[X]] denotes the ring of formal power series over R where X = (X1 , X2 , . . . ). The
¯
¯
notation [X ¯n ]A extracts the coefficient of X ¯n in the formal power series A. Let
∫ ¯
¯
∂j and j denote the formal derivative and integral with respect to a variable Xj .
The reciprocal of a formal power series A when it exists is A1 and the compositional
inverse when it exists is A⊣ .
2 Introduction to combinatorial calculus
We introduce the combinatorial calculus of formal power series here, but with sparse emphasis on algebraic properties of rings of formal power series. We are more interested in
the combinatorial properties of the coefficients of a formal power series, and in interpreting
the combinatorial meaning of the usual operations from algebra and analysis and their
effect on coefficients. The main idea is to identify a sequence of elements of a ring R with
a formal power series which is treated as an algebraic object in the ring of formal power
series R[[X]]. Generally but not always, the ring R will be commutative with unity.
¯
Fibonacci numbers.
We start with a famous illustrative example: the Fibonacci numbers. The Fibonacci numbers are integers defined by the recurrence an = an−1 + an−2 for n ≥ 2 with a1 = a2 = 1
and a0 = 0. If we associate (an )n∈N with the series
A(X) =
∞
∑
an X n ,
n≥0
then summing the recurrence times X n on both sides from n = 2 leads to
A(X) =
X
.
1 − X − X2
Now of course for this to make sense, we have to make sense of this as a formal power
series in C[[X]] – in particular the reciprocal of 1 − X − X 2 must be defined. Assuming this
makes sense, we might now use the geometric series formula and then binomial theorem to
say
j ( )
∞
∞ ∑
∑
∑
j
2 2
A(X) = X
(X + X ) = X
X j+k .
k
j=0
j=0 k=0
Now we apply the coefficient operator: defining an = [X n ]A(X) we have for n ∈ N
)
n (
∑ (j ) ∑
k
n+1
an+1 = [X ]A(X) =
=
.
k
n
−
k
j+k=n
k=0
1
This is an explicit formula for Fibonacci numbers as well as a combinatorial identity for
the sum. Another approach is to write
X
1 ( 1
1 )
√
=
−
1 − X − X2
5 1 − Xφ 1 − Xφ
√
√
where φ = (1 + 5)/2 and φ = (1 − 5)/2. Now using the geometric series formula we see
for n ≥ 0:
1
an = √ (φn − φn ).
5
This in some sense is more explicit than the former expression for an , and proves the
identity
)
n (
∑
1
k
= √ (φn+1 − φn+1 ).
n−k
5
k=0
If we go a bit further, we note that the sum of the first n Fibonacci numbers is exactly the
coefficient of X n in the formal power series
A(X)
.
1−X
Consider the formal power series
∞
∑
(an+2 − 1)X n =
n=0
A(X) − X
.
X2
Now observe this is exactly A(X)/(1 − X). Therefore we immediately obtain
a0 + a1 + · · · + an = an+2 − 1.
The ideas in this example generalize greatly to many other algebraic operations in rings
of formal power series, and we develop many connections between algebraic identities and
combinatorial identities, as well as methods for extracting coefficients from formal power
series.
Derangements.
Let’s consider one more example, involving derangements of [n]. If an is the number of
derangements of [n], we saw that
an = n!
n
∑
(−1)k
k=0
k!
.
It makes sense to divide by n!, multiply by X n and then sum:
A(X) =
∞
∑
an
n=0
n!
n
X =
∞ ∑
n
∑
(−1)k
n=0 k=0
2
k!
X n.
Now the right hand side is recognizable as the product of two power series, similarly to the
end of the discussion of Fibonacci numbers, namely
∞
∑
(−1)k
k=0
k!
Xk ·
1
.
1−X
Since the first series is exp(−X) – again we have to make sure this is sensible notation –
(1 − X)A(X) = exp(−X).
Applying the coefficient operator,we find for n ≥ 1:
an+1 − (n + 1)an = (−1)n+1 .
Now this has a combinatorial interpretation: it says that the number of derangements of
[n + 1] and the number of permutations of [n + 1] with exactly one fixed point differ by 1
or by −1. Thus again an algebraic equation has provided combinatorial meaning.
2.1 Definitions
Let R be a ring with unity 1 ∈ R. Let N be the set of sequences of non-negative integers
with finite support each indexed by N. For n ∈ N , associate an element an ∈ R. Let
∏
¯
¯
ni
X = (X1 , X2 , . . . , Xn ) and define X ¯n to be the monomial ∞
i=1 Xi for indeterminates or
¯
¯ 0
variables Xi : i ∈ N. Here we define X¯ = 1 which is the multiplicative unity in R. Consider
¯
the set R[[X]] of objects
∑
¯
an X ¯n .
A(X) =
¯¯
¯
n
¯
∑
Define addition of A(X) and B(X) =
bn X ¯n as
¯¯
¯
¯
∑
A(X) + B(X) =
(an + bn )X ¯n
¯ ¯
¯
¯
¯
n≥0
¯
and multiplication
A(X) · B(X) =
¯
¯
∑( ∑
n
¯
m+ℓ=n
¯ ¯ ¯
)
am bℓ X ¯n .
¯ ¯ ¯
The set R[[X]] together with multiplication and addition is the ring of formal power series
¯
over R. In some texts, given a sequence a indexed by vectors in N , the notation a ↔ A(X)
¯
¯
¯
is used to denote that a is the sequence of coefficients in A(X) with an as the coefficient of
¯
¯
¯
¯
X ¯n . Traditionally, R[X] is the ring of polynomials over R, and topologically R[[X]] can be
¯
¯
¯
viewed as the completion of R[X] in the metric derived from RN as follows: let f : N → N
¯
be a bijection. For A, B ∈ R[[X]],
¯
−k
d(A, B) = 2
⇔ A ↔ a ∧ B ↔ b ∧ k = min{f (n) : an ̸= bn }.
¯
¯
¯ ¯ ¯ ¯¯
For us the most useful case is when R is an integral domain, for then so too are R[X]
¯
and R[[X]]. The multiplicative identity in R[[X]] is the formal power series in which the
¯
¯
3
coefficient of every monomial is zero except the monomial X¯0 appears with coefficient 1.
¯
The latter is called the constant term of the formal power series A, and denoted A(0). More
¯
generally, the coefficient operator is defined for each n by
¯
[X ¯n ]A(X) = an .
¯
¯
¯
We give a differential formula for extracting this coefficient below.
2.2 Reciprocal
The multiplicative inverse or reciprocal of a formal power series A(X), when it exists, is a
¯
formal power series B(X) such that A(X) · B(X) = B(X) · A(X) = 1. Not all elements
¯
¯
¯
¯
¯
of R[[X]] have reciprocals. One of the fundamental topics in combinatorial calculus is to
¯
determine inversion theorems – theorems which allow us to effectively compute coefficients
in inverses. Here is a necessary and sufficient condition for a formal power series to have
an inverse:
Theorem 1 Let R be a commutative ring with unity and let A ∈ R[[X]]. Then A has a
¯
reciprocal if and only if A(0) is invertible in R, and the reciprocal is unique.
¯
Proof ◃ Suppose A(0) is invertible with inverse a. We claim that the reciprocal is
¯
∞
∑
(1 − aA(X))k .
B(X) = a
¯
¯
k=0
One checks quickly by definition of multiplication that A(X)B(X) = 1. The only question
¯
¯
is to show B ∈ R[[X]], and the only way that can fail is if the coefficient of some X ¯n in
¯
¯
B is not a finite sum of elements of R. Note that C(X) = 1 − aA(X) has a zero constant
¯
¯
term and therefore the lowest power of Xj in any monomial in (1 − aA(X))k for k ∈ N is
¯
at least k. This means that the coefficient of X ¯n in B(X) is the same as the coefficient of
¯
¯
X ¯n in
∑
¯
a
C(X)k
¯
k∈K
which is a finite sum of elements of R. Now suppose A(0) is not invertible. Then for
¯
any power series B(X) ∈ R[[X]], by definition of multiplication the coefficient of X 0 is
¯
¯
A(0)B(0), and this is not 1 since A(0) is not invertible. So in this case A(X) has no
¯
¯
¯
¯
inverse. Uniqueness is clear.
The reciprocal of a formal power series A is usually denoted 1/A. For instance, the geometric series identity familiar from calculus for |x| < 1, namely
∞
∑
xk =
k=0
4
1
1−x
can now be proved more generally in the ring R[[X]] of formal power series in one variable
X: by the rules of multiplication:
(1 − X)(1 + X + X 2 + . . . ) = 1 + 0X + 0X 2 + · · · = 1
so 1 + X + X 2 + . . . is indeed the reciprocal of 1 − X. Most important in our work will be
methods for computing coefficients of reciprocal series and so-called inversion theorems.
2.3 Infinite sums and products
We have defined the sum and product of two formal power series, and these definitions
extend to sums and products of finitely many formal power series. Now if A1 , A2 , . . . are
formal power series and the coefficient of X ¯n is zero in all but finitely many of the power
¯
series, we can define
∞
∑
Ak
k=1
to be the formal power series in which the coefficient of X ¯n is the sum of the coefficients
¯
of X ¯n in the Ak . We call the family {Ak : k ∈ N} summable. If {Ak : k ∈ N} is summable,
¯
then we can define the infinite product
∏
(1 + Ak )
k∈N
as follows: the coefficient of X ¯n in this product is determined by taking the product over
∑
¯
k ∈ N of the coefficients of monomial X ¯nk in Ak such that ∞
k=1 nk = n. For instance, the
¯
¯
Euler products discussed for partitions are now valid formal power series:
∞
∏
(1 + Xi ) ∈ R[[X]]
¯
n=1
is well-defined since {Xk : k ∈ N} is summable, and if Xk is replaced with X k for k ∈ N then
we recover the formal power series whose coefficients are partition numbers (see Lecture
A).
2.4 Formal derivative and integral
For this section, R is a commutative ring with unity 1 ∈ R and zero characteristic. Let n
¯
and fix j ∈ N. For a monomial X ¯n , denote ∂j X ¯n = 0 if nj = 0 and
¯
¯
∏
n −1
∂j X ¯n = nj Xj j
Xini .
¯
i̸=j
The formal derivative of A ∈ R[[X]] is
¯
∂j A =
∑
n
¯
5
an ∂j X ¯n .
¯
¯
Similarly, let
∏
1
nj +1
n
X
=
X
Xini
¯
j¯
nj + 1 j
i̸=j
∫
and, provided R has characteristic zero, define the formal integral
∑ ∫
∫
A
=
an j X ¯n .
j
n
¯
¯
Formal integration and differentiation follow the usual rules of calculus, for instance ∂j (A +
B) = ∂j A + ∂j B and ∂j (A · B) = B∂j A + A∂j B and similarly for the quotient rule. The
proofs are straight from the definitions of addition and multiplication of power series, and
also extend to infinite sums and products of a summable set of power series. One of the
nice facts in R[[X]] is that ∂j and ∂k are commuting operators (unlike in analysis). A useful
¯
result which allows the extraction of coefficients is Taylor’s Theorem:
Theorem 2 If R has characteristic zero and ∂j A = ∂j B and A(0) = B(0) then A = B.
∑
¯
¯
Furthermore, let A(X) = n an X ¯n where n ∈ N and let ∂ ¯n denote the partial derivative
¯
¯
¯ ¯¯
operator with nj derivatives of
Xj for j ∈ N.
[X ¯n ]A(X) = B(0)
¯
¯
where
B(X) =
∞
∏
1 n
∂ ¯ A(X).
n
!¯
j
j=1
2.5 Compositional inverse
Another important operation on formal power series is composition. Let A ∈ R[[X]] and
¯
suppose that B = (Bk )k∈N is a set of formal power series such that {an B ¯n } is summable.
¯¯
¯
Then the composition of A with B is
¯
∑
an B ¯n .
A ◦ B = A(B) =
¯¯
¯
¯
n
¯
In the special case of the ring R[[X]] of single variable formal power series, we define the
compositional inverse of A ∈ R[[X]] to be a power series B ∈ R[[X]] such that A ◦ B(X) =
X. This entity exists and is unique when A(0) = 0 and ∂A(X) is invertible, and is denoted
¯
A⊣ – as opposed to the reciprocal which is A1 . The Lagrange inversion formula is a way to
express the coefficients of the compositional inverse of a formal power series, and we prove
this later.
2.6 Explicit Examples
For simplicity we start off with power series in the ring C[[X]] of one-variable formal power
series over the complex numbers, although in applications we will often need two or more
variables.
6
Reciprocal and geometric series. Recall the geometric series formula
∑
1
=
Xk
1−X
k=0
∞
which comes from the definition of reciprocal given before.
Compositional inverse. The compositional inverse of A(X) = X + X 2 + X 3 + · · · =
X/(1−X) is A⊣ (X) = X −X 2 +X 3 −· · · = X/(1+X) and there is no reciprocal. However,
1/(1 − X) has no compositional inverse.
Exponential and logarithmic power series. Define the exponential and logarithmic
power series respectively as
exp(X) =
∞
∑
Xk
k=0
log(1 − X) = −
k!
∞
∑
Xk
k=1
k
.
The definition of formal derivative implies ∂ exp(X) = exp(X) and ∂ log(1 − X) = −1/(1 −
X), and we also have the rule (exp(X))k = exp(kX) for k ∈ N and exp(kX) = 1/ exp(−kX),
in accordance with the corresponding power series from calculus. The composition log(exp(X))
makes sense and by the chain rule, ∂ log(exp(X)) = 1 and log(exp(0)) = 0. However X also
satisfies these conditions, and therefore since C has characteristic zero, Taylor’s Theorem
shows log(exp(X)) = X. Similarly, exp(log(1 − X)) = 1 − X, but note that log X is not
a formal power series. Another way to derive exp(X) is from a differential equation: one
can show that if A ∈ C[[X]] and ∂A(X) = A(X) then A is proportional to exp(X): the
differential equation shows (n + 1)an+1 = an for n ≥ 1 which has solution an = a0 /n!.
The Binomial Series. This is one of the most important series combinatorially. For
indeterminates X and Y , defined the binomial series in C[X, Y ] as
∞ ( )
∑
Y
(1 + X) =
X k.
k
k=0
Y
If n is a positive integer, then
n ( )
∑
n
(X + Y ) =
X k Y n−k
k
k=0
n
is a finite sum. The usual rules (1 + X)Y (1 + X)Z = (1 + X)Y +Z , ((1 + X)Y )Z = (1 + X)Y Z
and (1 + X)−Y = 1/(1 + X)Y are satisfied by this formal power series.
Taylor’s Theorem. Let f : C → C be a function that is analytic at the origin. Then the
power series provided by a Taylor expansion of f at the origin can be viewed as a formal
power series. The methods of complex analysis extend to formal power series, provided
7
that each operation does indeed produce a formal power series i.e. the coefficients must be
finite sums of complex numbers. So for instance
sin(X) =
∞
∑
(−1)k X 2k+1
k=0
(2k + 1)!
defines a formal power series, and it satisfies the same rules as the ordinary power series for
X, for instance the formal power series we get from arcsin(X) is indeed the compositional
inverse of sin(X). However, log(X) is not a formal power series since log(z) is not analytic
at z = 0.
2.7 Identities
Using operations on formal power series, we can obtain combinatorial identities from algebraic identities. Generally we may pass from an algebraic identity to a combinatorial
identity by comparing coefficients, and we may take a combinatorial identity and sum each
side multiplied by X ¯n to get an algebraic identity, taking care to ensure that coefficients
¯
are finitely generated.
Pascal’s Triangle Identity and Binomial Sums
This arises by factoring an algebraic quantity: recall the identity states for n, k ∈ N,
( ) (
) (
)
n
n−1
n−1
=
+
.
k
k−1
k
( )
We have (1 + X)n = (1 + X)(1 + X)n−1 for n ∈ N and the coefficient of X n is nk by the
binomial theorem. On the other
(n−1)hand, by the definition of multiplication, with a0 = 1 = a1 ,
a2 = a3 = · · · = 0 and bk = k ,
n−1
(1 + X)(1 + X)
=
k
∞ (∑
∑
k=0
)
k
al bk−l X =
l=0
)
∞ ((
∑
n−1
k=0
(
)
n−1 ) k
X
+
k−1
k
and comparing coefficients gives the result. For many identities involving binomial sums,
for instance
( )
n ( )2
∑
2n
n
=
n
k
k=0
using the basic rule of multiplication of formal power series is effective. The left hand side
is [X n ](1 + X)2n . Now
)
n ∑
k ( )(
∑
n
n
(1 + X) · (1 + X) =
Xk
ℓ
k
−
ℓ
k=0 ℓ=0
n
n
8
and reading off the coefficient with k = n gives the identity. Another trick is to interchange
the order of summation in multiple sums. For instance, while it is not possible to explicitly
compute the inner (harmonic) sum in
∞ ∑
k
∑
k=1 ℓ=1
1
2k+1 ℓ
if we interchange the sums we get
∞ ∑
∞
∑
ℓ=1 k=ℓ
∞
∑
1
=
.
k+1
ℓℓ
2 ℓ
2
ℓ=1
1
This sum is computed by letting X = 1/2 in the following computation:
∫ ∑
∞
X ℓ−1 = − log(1 − X).
ℓ=0
We conclude that the given sum is log 2.
Evaluating sums.
We may use operations on formal power series to evaluate explicit sums. Let d ≥ 1. To
determine
n
∑
kd
k=1
we notice for any polynomial P ,
P (X∂)A =
∑
P (k)ak X k .
k≥0
Applying this to the geometric series
A(X) =
n
∑
Xk =
k=0
1 − X n+1
1−X
with P (X) = X d we obtain
∞
∑
k d X k = P (X∂X)
k=0
1 − X n+1
.
1−X
Setting X = 1, we get n(n + 1)/2 for d = 1 and n(n + 1)(2n + 1)/6 for d = 2, and in
principle we can compute the sum for any value of d. As a second example, consider the
sum
∞
∑
1
.
(k + d − 1)(d)
k=1
9
This can be determined via formal integration: if A(X) is the geometric power series, and
P (X) = X d , then
∞
∫ d∑
X k.
k=0
For d = 2 we get
∫2
∫
1
= − log(1 − X) = X + (1 − X) log(1 − X).
1−X
Putting X = 1 we get that the sum is 1. Of course there is an easier way: it is a telescoping
sum in disguise
∞
∞ (
∑
∑
1
1 )
1
=
−
= 1.
k(k + 1) k=1 k k + 1
k=1
For d = 3 we get
−
X 3 2 1
1
+ X − log(1 − X) + X log(1 − X) − X 2 log(1 − X).
2
4
2
2
Again putting X = 1 we get that the sum is 14 . We should note that making a substitution
into a formal power series converts it to a power series where issues of convergence arise.
The substitution step is only valid within the radius of convergence of the power series.
Formal power series for partial sums.
The nth Harmonic sum is
1 1
1
+ + ··· + .
2 3
n
In the exercises you will be asked to show via formal integration that:
Hn = 1 +
∞
∑
Hn X n =
n=1
−1
log(1 − X).
1−X
This relies on the basic observation that if A(X) is a formal power series for the sequence
(an )n∈N , then
A(X)
1−X
is a formal power series for the partial sums of (an )n∈N .
Bivariate series.
Let k ∈ N. Suppose we wish to determine
∞ ( )
∑
n
Bk (Y ) =
Y n.
k
n=0
10
If we sum Bk (Y )X k we get
∞ ( )
∞ ∑
∞ ∑
∞ ( )
∞
∑
∑
∑
n
n
1
k n
k n
A(X, Y ) =
X Y =
X Y =
Y n (1 + X)n =
.
k
k
1
−
Y
(1
+
X)
n=0
n=0
n=0
k=0
k=0
To determine Bk (Y ) we require [X k ]A(X, Y ). This is straightforward:
[X k ]A(X, Y ) = [X k ]
1
1
1
1 ( Y )k
Yk
= [X k ]
·
=
.
=
XY
1 − Y (1 + X)
1 − Y 1 − 1−Y
1−Y 1−Y
(1 − Y )k+1
In particular we see independent of k that
∞
∑
2
−n
n=0
( )
n
= 2.
k
Of course there is a more direct approach: notice using formal derivatives
∞
Yk k∑ n Yk k 1
Yk
.
Bk (Y ) =
∂
∂
=
Y =
k!
k! 1 − Y
(1 − Y )k+1
n=0
Some other examples of bivariate series will be encountered as formal power series of Stirling
numbers of the first and second kind.
Inverse pairs.
Suppose we are given an identity
bn =
n
∑
a k ck .
k=0
A useful thing to do is to determine a sequence dk such that
an =
n
∑
bk dk .
k=0
The sequences an and bn are referred to as inverse pairs. This may occasionally be done via
formal power series. We give an example, which might be referred to as binomial inversion:
let (an )n∈N∪{0} be a sequence of complex numbers and for n ∈ N ∪ {0} let (bn )n∈N be defined
by:
n ( )
∑
n
bn =
ak .
k
k=0
Then we claim for all n ∈ N ∪ {0},
an =
n
∑
(−1)
k=0
11
n−k
( )
n
bk .
k
To see this, we have
∞
∑
bn
n=0
n!
n
X =
∞ ∑
n
∑
ak
n=0 k=0
1
X n.
k! (n − k)!
·
Now this is the product of the two series
exp(X) =
Therefore
∞
∑
ak
k=0
k!
k
∞
∑
1 n
X
n!
n=0
X = exp(−X)
and
∞
∑
ak
k=0
k!
X k.
∞
∑
bn
n=0
∞ ∑
n
∑
(−1)k bk n
=
X .
n! n=0 k=0 k!(n − k)!
n
Comparing coefficients of X gives the result. In other words, we managed to find a formal
power series relation B(X) = C(X)A(X) where A(X) and B(X) have the an and bn as
coefficients, or some simple function of an and bn , and then writing A(X) = B(X)/C(X)
gives the inverse relation.
Recurrence equations.
By a recurrence equation we mean a sequence (an )n∈N defined by an = f (an−1 , an−2 , . . . , a0 )
for some function f together with some initial conditions on the sequence. For instance,
the Fibonacci equation an = an−1 + an−2 for n ≥ 2 and a0 = 0 and a1 = 1 we solved
by creating a formal power series for each side and then determining a closed formula for
that power series and reading off coefficients. In general this approach leads to functional
and partial differential equations for the unknown formal power series A(X), however often
these are difficult or impossible to solve explicitly. This also generalizes to rings R[[X]]
however the technical complications often become quite a barrier. We will nevertheless
discuss particular well-known cases where these can be overcome. Here we give one more
example: let (an )n∈N satisfy a1 = 1 and a2n = 2an and a2n+1 = an + an+1 for n ∈ N.
Summing the recurrence,
A(X) = (1 + X + X 2 )A(X 2 )
∑
n−1
where A(X) = ∞
. Repeating this functional equation gives (noting summabiln=1 an X
ity):
∞
∏
k−1
k
A(X) =
(1 + X 2 + X 2 ).
k=1
We may interpret this Euler product has coefficients representing the number of partitions
of an integer into powers of two with some restrictions on the multiplicities of consecutive
powers of two. The special topic of linear recurrences with constant coefficients can be
effectively solved using formal power series.
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