Multiplicative functions and their generating functions:
A multiplicative arithmetical function is a function f : Z+ → R that satisfies
f (ab) = f (a)f (b) when (a, b) = 1, and more generally
f (pe11 p2e2 · · · ) = f (pe11 )f (pe22 ) · · · .
When f is not nontrivial (not identically 0) then f (1) = 1.
Generating function of a non-trivial multiplicative function: Let f be a nontrivial multiplicative function and set
Fk (tk ) =
∞
X
f (pek )tek .
e=0
Then
f (pe11 pe22 · · · ) = f (pe11 )f (pe22 ) · · · = [te11 te22 · · · ]F1 (t1 )F2 (t2 ) · · · .
Therefore a generating function for f is Ff (t) = Ff (t1 , t2 , . . . ) = F1 (t1 )F2 (t2 ) · · · .
Any such product with constant terms 1 in the Fk is the generating function
of a multiplicative arithmetic function.
Products of generating functions:
If
F (t) = F (t1 , t2 , . . . ) =
X
f (pe11 pe22 · · · )te11 te22 · · · =
X
f (n)tn
n≥1
and
G(t) = G(t1 , t2 , . . . ) =
X
2
g(pe11 pe22 · · · )te11 te2 · · · =
X
g(n)tn
n≥1
then
F (t)G(t) =
X
f (pa11 pa22 · · · )g(pb11 pb22 · · · )ta11 +b1 ta22 +b2 · · · =
XX
f (d)g(n/d)tn .
n≥1 d|n
This implies that if a and b are multiplicative functions with generating
functions Fa (t) and Fb (t) then the multiplicative function c with generating
function Fa (t)Fb (t) is defined by
X
X
c(n) =
a(d)b(n/d) =
b(d)a(n/d).
d|n
d|n
1
Examples:
1. The unit function u(n) = 1 has generating function Fu (t) =
If f (n) is multiplicative then so is
X
X
g(n) =
f (n/d) =
f (d)
d|n
d|n
and
Fg (t) = Fu (t)Ff (t) =
Ff (t)
.
(1 − t1 )(1 − t2 ) · · ·
2. The identity function i(n) = n has generating function
f (n) is multiplicative then so is
X
X
n
h(n) =
df (n/d) =
f (d)
d
d|n
and
Fh (t) =
1
.
(1−t1 )(1−t2 )···
1
.
(1−p1 t1 )(1−p2 t2 )···
If
d|n
Ff (t)
.
(1 − p1 t1 )(1 − p2 t2 ) · · ·
3. The Möbius function µ(n) defined by
µ(pe11 · · · pekk ) = (−1)k χ(e1 = · · · = ek = 1)
has generating function
Fµ (t) = (1 − t1 )(1 − t2 ) · · · ,
hence is multiplicative. If f is a multiplicative function and g is defined by
X
g(n) =
f (d)
d|n
then we have seen by Example 1 above that
Fg (t) =
Ff (t)
Ff (t)
= Fu (t)Ff (t) =
.
(1 − t1 )(1 − t2 ) · · ·
Fµ (t)
This implies
Ff (t) = Fµ (t)Fg (t),
2
hence
f (n) =
X
µ(d)g(n/d) =
d|n
X
g(d)µ(n/d).
d|n
In particular,
f (pe ) = g(pe ) − g(pe−1 )
when p is prime and e ≥ 1.
4. The unit characteristic function ν(n) = χ(n = 1) has generating function
Fν (t) = 1. Given that Fν (t) = Fu (t)Fµ (t), we have
X
X
ν(n) =
µ(d) =
µ(n/d).
d|n
d|n
5. Euler’s (totient) function φ(n): This is defined as the number of natural
numbers ≤ n that are relatively prime to n. What we see in the textbook is
a proof of the inclusion-exclusion formula. Working through the details, let
p1 , . . . , pr be the primes which divide n. We want to count all the numbers
not divisible by any of these primes, i.e. throw away the numbers divisible at
least one of these. Setting Ai equal to the numbers in {1, 2, . . . , n} divisible by
pi , the numbers that are divisible by some pi are counted by n1 −n2 +n3 −· · ·
where nk is the sum of the sizes of the k-fold intersection intersection of sets.
Since
n
,
|Aa ∩ Ab ∩ · · · ∩ Ac | =
pa pb · · · pz
we have
z
z
z
nk = [z k ]n(1 + )(1 + ) · · · (1 + ).
p1
p2
pr
So
r
X
z
z
z
1
1
1
φ(n) =
(−1)k [z k ]n(1 + )(1 + ) · · · (1 + ) = n(1− )(1− ) · · · (1− ).
p1
p2
pr
p1
p2
pr
k=0
One can check that φ is multiplicative given this formula. Now define
X
g(n) =
φ(d).
d|n
This is multiplicative. It satisfies
g(pk ) =
k
X
φ(pi ) = 1 + (p − 1) + (p2 − p) + · · · + (pk − pk−1 ) = pk ,
i=0
3
hence g(n) = n for all n. Therefore
X
φ(d) = n.
d|n
To obtain a generating function for φ(n), note that
Fi (t) = Fg (t) = Fu (t)Fφ ,
hence
Fφ (t) = Fµ (t)Fi (t) =
(1 − t1 )(1 − t2 ) · · ·
.
(1 − p1 t1 )(1 − p2 t2 ) · · ·
6. Möbius Inversion: Let f : R → R be given and define
X
g(x) =
f (x/n),
n≤x
summing over positive integers. Then
X
X
X
X
X
µ(n)g(x/n) =
µ(n)
f (x/mn) =
µ(n)
f (x/mn) =
n≤x
n≤x
X
f (x/l)
l≤x
n≤x
m≤x/n
X
µ(l/m) =
X
mn≤x
f (x/l)ν(l) = f (x).
l≤x
m|l
Conversely, if we define
f (x) =
X
µ(n)g(x/n)
n≤x
then
X
X X
XX
X
X
f (x/n) =
µ(n)g(x/kn) =
µ(n)g(x/kn) =
g(x/l)
µ(l/m) =
n≤x
n≤x k≤x/n
n≤x kn≤x
X
l≤x
m|l
g(x/l)ν(l) = g(x).
l≤x
When a multiplicative function is used to define the other this way then the
second function is also multiplicative, and we obtain
X
g(n) =
f (d)
d|n
4
if and only if
f (n) =
X
µ(d)g(n/d).
d|n
We already derived this by the method of generating functions above.
7. Applying Möbius inversion to the functions
X
τ (n) =
1,
d|n
σ(n) =
X
d,
d|n
n=
X
φ(d),
d|n
we obtain
1=
X
n=
X
d|n
d|n
φ(n) =
n
µ(d)τ ( ),
d
n
µ(d)σ( ),
d
X
d|n
n
µ(d) .
d
The identities above also follow from Fu = Fτ Fµ , Fi = Fσ Fµ , Fφ = Fi Fµ .
8. Summary of generating functions:
µ(n): Fµ = (1 − t1 )(1 − t2 ) · · ·
P
ν(n) = χ(n = 1) = d|n µ(n): Fν = 1
u(n) = 1: Fu =
1
(1−t1 )(1−t2 )···
1
i(n) = n: Fi = (1−p1 t1 )(1−p
2 t2 )···
P
P
1
τ (n) = d|n 1 = d|n u(d): Fτ = Fu2 = (1−t1 )2 (1−t
2
2 ) ···
P
P
1
σ(n) = d|n d = d|n i(d): Fσ = Fu Fi = (1−t1 )(1−t2 )···(1−p
1 t1 )(1−p2 t2 )···
φ(n): Fφ =
(1−t1 )(1−t2 )···
(1−p1 t1 )(1−p2 t2 )···
= Fµ Fi .
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9. The Riemann zeta-function. Take any generating function F (t) = F (t1 , t2 , . . . ) =
F1 (t1 )F2 (t2 ) · · · for a multiplicative function f . Making the substitution
ti 7→ p1s where s is a complex number yields an infinite product. For exami
ple, recall that we have
X
1
=
te1 te2 te3 · · ·.
Fu (t) =
(1 − t1 )(1 − t2 ) · · · e ,e ,e ,···≥0 1 2 3
1
2
3
Hence
∞
X 1
1
=
.
e1 e2 e3
s
p
n
p
p
·
·
·
1
2
3
n=1
,···≥0
X
Fu (s) = Fi (1/ps1 , 1/ps2 , . . . ) =
e1 ,e2 ,e3
This is called the Riemann zeta-function ζ(s). In particular,
∞
ζ(2) =
X 1
1
π2
=
=
.
1 − (1/p2 ) n=1 n2
6
Y
p
We will derive this evaluation this shortly.
P
n
More generally, if Ff (t) = ∞
n=1 f (n)t then
Ff (s) =
Ff (1/ps1 , 1/ps2 , . . . )
=
∞
X
f (n)
n=1
Examples:
1. Fµ (s) =
1
.
Fu (s)
This implies
∞
X
µ(n)
n=1
In particular,
=
ns
∞
X
µ(n)
n=1
n2
=
1
.
ζ(s)
6
.
π2
2. Fτ (s) = Fu (s)2 . This implies
∞
X
µ(n)
n=1
ns
6
= ζ(s)2 .
ns
.
In particular,
∞
X
µ(n)
n=1
3. Fi (s) =
P∞
n
n=1 ns
n2
=
π4
.
36
= ζ(s − 1).
4. Fσ (s) = Fi (s)Fu (s). This implies
∞
X
σ(n)
ns
n=1
= ζ(s − 1)ζ(s).
5. Fφ (s) = Fµ (s)Fi (s). This implies
∞
X
φ(n)
n=1
ns
=
ζ(s − 1)
.
ζ(s)
6. For arbitrary functions f : Z+ → R and g : Z+ → R we have
∞
∞
X
f (n) X g(n)
n=1
ns
n=1
ns
=
∞
X f (a)g(b) X
1 X
=
f (d)g(n/d)
s
s
(ab)
n
n=1
a,b≥1
d|n
assuming the expressions converge.
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