Download 11. Dirichlet generating functions

MAT5107 : Combinatorial Enumeration
Mike Newman, winter 2017
11. Dirichlet generating functions
Dirichlet generating functions
We have seen ordinary generating functions and exponential generating functions; each corresponds
naturally to a particular method of multiplying sequences. There are other multiplications possible,
depending on the combinatorial constructions or applications.
P
Given a sequence (an )n≥1 we define its Dirichlet generating function to be A(x) = n≥1 nanx .
Notice that the indeterminate x appears in the exponent. Such sequences are useful in number theory
(where typically an s is used as the indeterminate instead of x).
For instance, the sequence 1, 1, 1, . . . corresponds to the Dirichlet generating function
1
1
1
ζ(x) = 1 + x + x + x + · · ·
2
3
4
What is the multiplication for Dirichlet generating function? More precisely, how do we compute the
cn in the following?



X bn
X cn
X an

=

nx
nx
nx
n≥1
n≥1
n≥1
The answer is that we need to multiply a term from each bracket in all possible ways so as to obtain
n−x . In other words we need indices i and j such that the product ij = n.
(
α ∈ A , ω(α) = r
−→
(α, β) ∈ A × B , ω(α, β) = ω(α)ω(β) = ij
β ∈ B , ω(β) = s
Multiplication of Dirichlet generating functions corresponds to a multiplicative weight. In contrast,
ordinary and exponential generating functions both combined weight additively, with the multiplication rule for exponential generating functions additionally naturally adding a factor corresponding
to relabelling. So the cn in the product of Dirichlet generating functions is computed as follows.



X an
X an
X cn
XX
X
1


=
=
a
b
←→
c
=
ad bn/d
n
d
n/d
nx
nx
nx
nx
n≥1
n≥1
n≥1
n≥1 d|n
d|n
As an example, let’s compute ζ 2 (x).



X 1
X 1
XX
X d(n)
1

=
1
·
1
ζ 2 (x) = 
=
nx
nx
nx
nx
n≥1
n≥1
n≥1 d|n
n≥1
where d(n) is the number of divisors of n. Note that this sequence is not monotonic, far from it! This
is in contrast to our usual generating functions, where the coefficients have (mostly) increased with
n. Or to put it another way, for OGF or EGF we often (but not always) have that i + j = n implies
that an is larger than ai or aj . With Dirichlet generating functions we often (but not always) have
that ij = n implies that an is larger than ai or aj . A more correct (but still heuristic) observation
might be that the sequences from OGF or EGF are ordered according to their indices under the total
ordering of “≤” on N, while sequences from DGF are ordered according to their indices under the
partial ordering of “|” on N \ {0}. So in this sense ζ 2 is monotonic.
∗
These notes are intended for students in mike’s MAT5107. For other uses please say “hi” to [email protected].
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multiplicativity
A function f (n) defined on the positive integers is said to be multiplicative if whenever gcd(a, b) =
1 then f (ab) = f (a)f (b). Note that we require nothing when the factors are not coprime. Both d(n)
and µ(n) defined above are multiplicative, as you are invited to show in the exercises.
Prime powers play a special role for multiplicative functions.
Lemma 11.1. If f (n) is a multiplicative function then it is determined by its values on prime
powers.
This has an important consequence. If f (n) is multiplicative then it behaves very nicely as the
coefficients of a DGF.
Lemma 11.2. Let f be a multiplicative function. Then we have the following equivalent form for
its Dirichlet generating function.
X f (n)
Y f (p) f (p2 ) f (p3 )
1 + x + 2x + 3x + · · ·
=
nx
p
p
p
primes p
n≥1
To show this, extract [n−x ] from both sides, and note that by uniqueness of factoring there is only
one choice of term in each factor of the infinite product on the right-hand side. Note that equality of
generating functions means equality of every coefficient. In particular, in expanding the product on
the right-hand side we only ever take products for which all but finitely many of the terms is 1. In
fact this is not quite true: we never actually “expand” anything, we simply ask what the coefficient of
n−x is on each side, for all n ∈ N. We saw this before without commenting on it, when we considered
the generating function for MSet and PSet as a product before “simplifying” it to something of
the form exp (· · · ). The terms in the infinite product that correspond to x∞ never arise, since the
ordinary generating function is defined exactly by the coefficient of xn for n ∈ N.
Back to the business at hand, the function 1 (whose value is 1 for all n) is multiplicative (a short
exercise!), so we get
Y
Y
X 1
1
1
1
=
1 + x + 2x + · · · =
ζ(x) =
x
n
p
p
1 − p1x
p
p
n≥1
We also know that the function µ(n) is multiplicative, so
Y
X µ(n) Y µ(p) µ(p2 ) µ(p3 )
1
=
1 + x + 2x + 3x + · · · =
1− x
nx
p
p
p
p
p
p
n≥1
We conclude that these two DGFs are reciprocals of each other since their product is the DGF 1.
Möbius inversion
Given an =
X
br , we want to invert, and solve for the bn . How do we do it? We first identify
r|n
the finite sum as the coefficient in a product of DGFs. Then we multiply by the reciprocal of the
convoluting function. Then we extract coefficients, using the multiplication rule backwards.
X
X
an =
br =
1 · br
⇐⇒
A(x) = ζ(x)B(x)
r|n
r|n
1
A(x) = B(x)
ζ(x)
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⇐⇒
X
r|n
µ(n/r)ar = bn
This is the proof of Lemma 10.3.
Compare this with the following analogue, which is based on an EGF convolution:
n n
X
X
n
n−k n
an =
bk ⇐⇒ bn =
ak
(−1)
k
k
k=0
k=0
cyclotomic polynomials
We offer a short example of another application of Möbius inversion, as a way of studying primitive
n-th roots of unity.
The n-th roots of unity are exactly the roots of the polynomial 1 − xn . These form a group with
respect to multiplication. Among these roots, there are some that multiplicatively generate the group
of all n-th roots: these are the primitive roots.
Definition 11.3. α = e2πir/n is a primitive n-th root of unity if α, α2 , . . . , αn = 1 are all distinct.
We have the following characterization of primitive roots.
Lemma 11.4. The primitive n-th roots of unity are exactly the complex numbers = e2πir/n for
gcd(r, n) = 1.
Furthermore, every n-th root of unity is a primitive d-th root of unity for exactly one d, and d | n.
We are interested in the polynomial whose roots are exactly the primitive n-th roots of unity: this is
the cyclotomic polynomial Φn (x). Certainly Φn (x) | 1 − xn . In fact, Lemma 11.4 gives the following.
Y
Lemma 11.5.
Φd (x) = 1 − xn
d|n
We can find Φn (x) using Möbius inversion. But first we need to transform the product to a sum
Y
Φd (x) = 1 − xn
d|n
X
log Φd (x) = log(1 − xn )
d|n
log Φn (x) =
X
µ(n/d) log(1 − xd )
d|n
Y
Φn (x) =
(1 − xd )µ(n/d)
d|n
Recall that µ(k) ∈ (0, ±1). So for instance we have
Φ18 (x) = (1 − x18/1 )µ(1) (1 − x18/2 )µ(2) (1 − x18/3 )µ(3) (1 − x18/6 )µ(6) (1 − x18/9 )µ(9) (1 − x18/18 )µ(18)
=
(1 − x18 )(1 − x3 )
(1 − x9 )(1 − x6 )
= 1 − x3 + x6
Note that the unsimplified form of Φ18 (x) does not at first glance appear to be a polynomial. The
formula for Φn (x) given by Möbius inversion is obviously a rational function. The fact that it is
a polynomial follows from the proof that it is a polynomial whose roots are exactly the primitive
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n-th roots of unity (in fact, among all such polynomials it is the one with constant term 1). So as a
“corollary” of the fact that Φn (x) is a polynomial we have the following.
Y
Y
n/r (1 − x ) (1 − xn/r )
r|n
µ(r)=−1
r|n
µ(r)=1
As an analogous situation one can give a combinatorial proof that the number of k-sets of a n-set is
given by the following, which is hence an integer.
n
n!
=
k!(n − k)!
k
If one takes the factorials as “definition” of the binomial coefficient it is not immediately obvious that
it is an integer. A better perspective is to define the binomial coefficient as the number of k-subsets
of an n-set. This is an integer since it is the size of a set (the set of k-subsets of an n-set), so we
obtain as a “corollary” that k!(n − k)! | n!.
exercises
1.
State the multiplication rule for products of three (or better yet t) DGFs.
2.
Verify that d(n) and µ(n) really are multiplicative. Also verify that the constant function
1(n) = 1 is multiplicative. Can you think of other multiplicative functions?
3.
Prove that if f is a multiplicative function then f (1) = 1. This explains the leading term in
Lemma 11.2.
4.
Prove Lemma 11.1. (hint: factor n)
5.
Prove that if we arbitrarily assign a positive integer value to f on each prime power, then
there is a unique extension to a function f : N \ {0} → N \ {0}. This is something of a
converse to Lemma 11.1, and can be seen as analogous to defining a function on a basis of a
vector space and extending by linearity.
6.
Both Möbius inversion and the EGF analogue mentioned above use the sequence 1, 1, 1, . . .
(and it’s reciprocal) to establish an equivalence between identities. What would the OGF
analogue be, using the sequence 1, 1, 1, . . .? (Of course, one can use other sequences as well,
provided they have a reciprocal.)
P
7. Prove that if f is multiplicative then n≥1 fn(n)
x has a DGF-reciprocal.
(hint: start with Exercise 11.3, but note that Lemma 1.4 does not directly apply.)
8.
Prove Lemma 11.4.
9.
Prove Lemma 11.5.
10.
Compute Φpk (x) where p is a prime number. Draw the roots in the complex plane for Φ4 (x)
and Φ9 (x). Can you give a nice geometric description?
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