Multiplicative Functions: NumThy
Jonathan L.F. King
University of Florida, Gainesville FL 32611-2082, USA
squash@ufl.edu
Webpage http://www.math.ufl.edu/∼squash/
15 May, 2014 (at 10:01)
Let M ⊂ G be the set of multiplicative♥3 functions.
Basic functions. Define fncs δ, 1, Id ∈ M by:
Aside. In number_theory.ams.tex there is a
bit more on multiplicative functions also in mult.
convolution.ams.tex; run collect.ams.tex though.
See generating_func.latex for an application of the
Möbius fnc.
δ(1) := 1
and
δ(6=1) := 0 ;
1(n) := 1 and
Id(n) := n for all posints n.
Evidently δ() is a neutral element for convolution.
Also define, as n varies over the posints, these
MFs:
Basic
Given two arbitrary♥1 functions f ,g:Z+ →C, define their convolution (“Dirichlet convolution”) by
h
i
f ~ g (K) :=
X
f (a) · g(b) .
a·b=K
Each such sum is to be interpreted as over all
ordered pairs ( a, b)) of positive divisors of K. Easily, convolution is commutative?? and associative.
b be the set of functions f :Z →C such
Let G
+
b
that f (1) 6= 0. Inside is G ⊂ G, the set of good
functions, which have f (1) = 1. Say that a
good f is multiplicative if for all posints♥2 K
and Γ :
K ⊥ Γ =⇒ f (K·Γ ) = f (K) · f (Γ ) .
♥1
Indeed, they could map into a general ring. If this ring
is commutative then convolution will be commutative.
♥2
Use ≡N to mean “congruent mod N ”. Let n ⊥ k
mean that n and k are co-prime. Use k •| n for “k divides
n”. Its negation k r
| n means “k does not divide n.” Use
r
n |• k and n | k for “n is/is-not a multiple of k.” Finally, for
p a prime and E a natnum: Use double-verticals, p E •|| n,
to mean that E is the highest power of p which divides n.
Or write n ||• p E to emphasize that this is an assertion
about n. Use PoT for Power of Two and PoP for Power
of (a) Prime.
For N a posint, use Φ(N ) or ΦN for the set
{r ∈ [1 .. N ] | r ⊥ N }. The cardinality ϕ(N ) := |ΦN | is
the Euler phi function. (So ϕ(N ) is the cardinality of
the multiplicative group, ΦN , in the ZN ring.) Use EFT
for the Euler-Fermat Thm, which says: Suppose that integers b ⊥ N , with N positive. Then bϕ(N ) ≡N 1.
Webpage http://www.math.ufl.edu/∼squash/
τ (n) :=
X
σ(n) :=
X
d:d•|n
d:d•|n
1
and
d.
This τ is called the (number of ) divisor fnc, and
σ is called the sum of divisor fnc. So σ(4) =
1 + 2 + 4 = 7 and τ (4) = 1 + 1 + 1 = 3.
Each of δ, 1, Id, τ , σ is multiplicative. So is Euler ϕ; this will follow from (??).
b ~, δ)) is a commutative group
1.0: Theorem. ( G,
b are subgroups.
and M ⊂ G ⊂ G
♦
Proof. (The arguments showing G a group apply to show
b a group. So we only argue for G and M.)
G
Easily G is sealed under convolution. To show
the same for M, take fncs f, h ∈ M and posints
K ⊥ Γ . Given posints ( x, y)) with x · y equaling
the product KΓ , the UFThm says that we can
factor uniquely x = aα and y = bβ into posints so
that a, b •| K and α, β •| Γ . Thus
h
i
f ~ h (KΓ ) =
X
f (x)h(y)
x·y = KΓ
1.1:
=
X
f (aα)h(bβ) .
a·b = K
α·β = Γ
♥3
An f is totally (or completely ) multiplicative if
f (K · Γ ) = f (K) · f (Γ ) always holds, even if K 6⊥ Γ . This
class is less important than M. Indeed, the set of totally
multiplicative fncs is not sealed under convolution.
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Basic
Necessarily a ⊥ α and b ⊥ β, since K ⊥ Γ .
Since f and h are each multiplicative fncs,
h
i
f ~ h (KΓ ) =
X
As f (1·1) equals 1, we can write
∗:
−h(KΓ ) =
=
X
X
f (a)h(b) ·
And
this
last
h
equals
X
f (aα)·h(b)h(β) ,
f (α)h(β) .
α·β=Γ
a·b=K
a·b = K,
α·β = Γ ,
( b,β))6=( K,Γ )
a·b=K
α·β=Γ
1.2:
h
f (a)f (α)h(b)h(β)
Prof. JLF King
i
f ~ h (K)
since, by (??), our h is multiplicative on all the
( b, β)) pairs on RhS(∗). Adding to each side gives
times
i
h(K)h(Γ ) − h(KΓ ) =
f ~ h (Γ ).
X
f (aα)·h(b)h(β) .
a·b = K,
α·β = Γ
Each good f has a convolution inverse.
We construct an h ∈ G so that f ~ h = δ. For
each posint K, then,
X
δ(K) = f (1) · h(K) +
(We’ve again used that f (1·1) = 1.) And since f is
multiplicative, this RhS equals
f (a) · h(b) .
X
a·b=K
( a,b))6=( 1,K))
X
f (a)h(b) ·
a·b=K
f (α)h(β)
α·β=Γ
def
Since f (1)=1 is not zero, there is a unique value
that we can assign h(K) so as to make the
displayed-eqn hold. Finally note that h(1) will
indeed be assigned the value 1, so h is good. This
shows that G is a group.
M is sealed under convolution-inverse.
The last step –to showing that (M, ~, δ)) is a
group– is to show that when the above f is multiplicative, then the corresponding h is too.
An inductive proof that h(KΓ ) = h(K)h(Γ )
proceeds by considering a pair ( K, Γ)) so that
1.3:
h(bβ) = h(b)h(β) for each proper divisor pair ( b, β)) •| ( K, Γ)).
That is, b •| K and β •| Γ with at least one of
these being proper.
Now WLOG ( K, Γ)) 6= ( 1, 1)), so δ(KΓ ) = 0, i.e
[f ~ h](KΓ ) is zero. Hence
X
0 =
f (aα)h(bβ)
a·b = K
α·β = Γ
=
X
f (aα)h(bβ) + f (1·1) · h(KΓ ) .
a·b = K,
α·β = Γ ,
( b,β))6=( K,Γ )
== [f ~ h](K) · [f ~ h](Γ ) = δ(K)δ(Γ ) .
Putting it all together,
∗∗:
h(K)h(Γ ) − h(KΓ ) = δ(K·Γ ) ,
since δ() is a MF. Finally, RhS(∗∗) is zero,
since K·Γ =
6 1.
2: Lemma.
Suppose that f and g are MFs.
Then f · g (pointwise product) is a MF. If g is nowhere zero, then ptwise quotient f /g is a MF.
Proof. Routine.
♦
Analogy. Recall that CRThm allows us to convert
polynomial congruences f (x) ≡ 0 modulo a composite M to just examining f (x) ≡ 0 modulo p L ;
we only need consider powers of primes.
In analogy, the MF theory allows us to convert
proving an identity h() = g(), between MFncs,
on all of Z+ , to just verifying it on each primepower p L .
Filename: Problems/NumberTheory/multiplicative_fncs.latex
Prof. JLF King
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Applications of Möbius Inversion
Applications
ii : µ ~ 1 = δ.
Define the Möbius
fnc µ to
be the convolution~
1
, so µ is characterized
inverse of 1. I.e µ := 1
by µ ~ 1 = δ. By Thm ??, it is enough to know
µ()’s values on the powers of primes.
iii : ϕ ~ 1 = Id.
3: Lemma. For each prime p and exponent n in
[2 .. ∞):
µ(p) = 1
µ(p n ) = 0 .
and
Proof. Well 0 = δ(p) = µ(p) + µ(1) = µ(p) + 1.
For a higher power of p:
0 = δ(p n ) = µ(1) + µ(p) + S + µ(p n ) ,
j
S, by induction,
is
where S := n−1
j=2 µ(p ). This
h
i
n
zero. Thus µ(p ) = − µ(1) + µ(p) , which is
zero.
P
Before developing further results, note that:
For each good f , necessarily f ~0 = δ,
since δ is the neutral element for ~. We now need
a little tool.
4: Bijection Lemma. Fixing a posint N , let g(`)
mean Gcd{`, N }. Then there is a bijection
a: [1 .. N ] ,
G
note
Φ(d)×{d} === ( x, d))
d:d •| N
d •| N
and
x∈Φ(d)
realized by the mapping
b:
` 7→ ( x` , d`) , where
: `/g(`) and
x` =
: N/g(`)
d` =
The inverse bijection is
c:
iv : ϕ ~ τ = σ and ϕ ~ σ = Id~2 .
Proof of ??. Take cardinalities in (??a), which
gives Id = ϕ ~ 1. As for (??), note that
ϕ ~ τ = ϕ ~ 1 ~ 1 = Id ~ 1 ,
which is σ, by definition.
6: Prop’n. The fnc ϕ/Id is multiplicative. And
a,b: limsup ϕ(K)
= 1 and
K
liminf
K→∞
K→∞
ϕ(K)
K
= 0.
Finally, ϕ(K) → ∞ as K%∞ .
c:
Proof. Exercise.
Convolution powers. Using binomial coefficients, one can show that
!
L+n−1
1 (p ) =
,
L
for each prime p. Furthermore
!
n
~n L
L
µ (p ) = [ 1] ·
.
L
~n
L
These are routinely proved by induction on n.
Applications of Möbius Inversion
An N th -root ω of unity is primitive if, for each k
in [1 .. N ), this ω is not a k th -root of unity. Use VN
for the set of primitive N th -roots. Define the “N th
cyclotomic polynomial ” by
7a:
CN (z) :=
Y
[z − ω] .
ω∈VN
Letting FN (z) := z N − 1, note that
( x, d)) 7→ x ·
N
d
.
We now establish a few useful relations.
5: Basic Lemma. Among MFs 1, τ , σ, Id, µ and ϕ,
the following relations hold.
i : 1 ~ 1 = τ . Also Id ~ 1 = σ, so Id = σ ~ µ.
7b:
FN (x) =
Y
d•|N
Cd (x) .
Used recursively, we compute these cyclo-polys:
7c:
C1 (z) = z − 1;
C2 (z) = z + 1;
Examples:
C3 (z) = z 2 + z + 1;
C4 (z) = z
2
Filename: Problems/NumberTheory/multiplicative_fncs.latex
+ 1;
C5 (z) = z 4 + z 3 + z 2 + z + 1;
C6 (z) =
z2 − z + 1 .
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Applications of Möbius Inversion
8: Prop’n. The sum of all the N th -roots of unity
equals δ(N ). And their product is [ 1]N +1 .
The sum of all primitive N th -roots equals µ(N ).
Their product is 1, except when N = 2 where the
product is 1.
♦
Pf for all roots. Let S be the sum♥4 , and P the
product, of the set, A, of all N th -roots. Thus
xN − S·xN −1 + · · · + [ 1]N ·P =
Y
[x − Z] = xN − 1.
Z∈A
Pf for primitive roots. Let S(n) := (Vn ). Summing over the (positive) divisors a •| N gives
P
X
S(a)
=
h
Sum of all
N th roots
i
=
Prof. JLF King
Reversal. Consider a Z ∈ VN . Algebra gives
x·[ x1 − Z] = 1·[xZ − 1] = 1·Z·[x − Z1 ] .
←−
Now CN (x) equals xK · CN ( x1 ) which equals
Y
x · [ x1 − Z] = [ 1]K ·
Z∈VN
i Y
Z ·
Z∈VN
[x − Z1 ] .
Z∈VN
Since C2 (x)=[x+1] is palindromic, WLOG N >3.
Thus
K :=
hQ
i ϕ(N ) is even. Also, (??) says that
Z∈VN Z = 1. Thus
Y
←−
note
CN (x) = 1 · 1 · [x − Z1 ] === CN (x) .
Z∈VN
δ(N ) .
a·b=N
h Y
Lastly, (??) follows from (??).
IOWords, S ~ 1 = δ. Thus S = δ ~ µ = µ.
As for the product, non-real primroots come in
conjugate pairs. Each conjugate pair multiplies
P
to 1. So (VN ) is 1 –unless [ 1] is an N th primroot, which happens exactly when N = 2.
10: Theorem.
Each CN can be described by
inclusion-exclusion as:
CN (z) =
Y
note
[z a − 1]µ(b) ===
a·b=N
Defn. For a poly with non-zero constant term, e.g
−
g(x) := 9x3 +8x−6, its reversal ←
g has the coeffs
←
−
3
in reverse order. So g (x) = 6x + 8x2 + 9.
9: Lemma. Each CN is a monic intpoly, of degree
note
|VN | === ϕ(N ). When N ∈ [2 .. ∞), moreover,
←−
CN is palindromic in that CN = CN .
Finally, for N > 2 and K := ϕ(N ),
K
9a: CN (x) = x − µ(N )x
K−1
+ · · · + [ µ(N )]x + 1 .
Y
C1 (z a )µ(b)
a·b=N
note Y
11:
C1 (z a )µ(b) .
===
a·b=N , with
b square-free
Below: P is prime, Γ and N are posints and K a
natnum.
??i: Now suppose P ⊥ Γ. Then
CPK+1 Γ (z) = CΓ (z P
K+1
.
In particular, setting Γ := 1,
Proof. Equation (??) gives
K
CN = FN
K
) CΓ (z P ) .
Y
d•|N
d6=N
Cd .
By induction on N , each of the Cd is a monic
intpoly. And easily: If a monic intpoly divides
another, then the quotient is a monic intpoly.
[z P ]P − 1
K
= CP (z P ) .
CPK+1 (z) =
K
P
[z ] − 1
??ii: Consider an N ⊥ Γ. Then
†N :
CN Γ (z) =
♥4
For the sum, an alternative
PN −1 proof is to fix a primitive
N root ω. Now SN := k=0 ω k is the sum of all N th
roots. So S1 = 1 = δ(1). For N ∈ [2 .. ∞), note ω 6= 1 so
N
−1
SN = ωω−1
= 0 = δ(N ).
Y
CΓ (z a )µ(b)
a·b=N
th
This generalizes (??).
Filename: Problems/NumberTheory/multiplicative_fncs.latex
Prof. JLF King
Applications of Möbius Inversion
Pf of (??).
Set L := K+1 and apply (??) to
L
N := P Γ. Omitting all pairs a·b=N with b |• P 2
means that a has form either αP L or αP K . So
CN =
h Y
i h
[FαP L ]µ(β) ·
αP L ·β=N
=
Y
Y
[FαP K ]µ(Pβ)
i
αP K ·Pβ=N
Y
[FαP L ]µ(β)
[FαP K ]µ(β) .
α·β=Γ
α·β=Γ
K
L
I.e CN (z) = CΓ (z P )/CΓ (z P ).
Aside. Letting R := P L , we can restate the above
conclusion as
‡:
Y
CRΓ (z) =
µ(j)
CΓ (z i )
.
i·j=R
Pf of (??). Whoa! This proof, while correct, should be
rewritten
Eqn (†1 ) is a tautology. So, given M ⊥Γ, we’ll
establish (†M ) by induction on the number of
PoPs in M . Write M = N R with R a PoP coprime to N . Then CM Γ (z) equals
by († )
Y
N
CN RΓ (z) =====
=
CRΓ (z a )µ(b)
a·b=N
by (‡)
Y h Y
=
===
=
CΓ [z a ]i
µ(j) iµ(b)
a·b=N i·j=R
Y
=
µ(bj)
CΓ (z ai )
.
a·b=N
i·j=R
This, since b ⊥ j, because N ⊥ R. Letting x := ai
and y := bj, the co-primeness N ⊥ R again gives
CM Γ (z) =
CΓ (z x )µ(y) .
Y
x·y=N R
Filename: Problems/NumberTheory/multiplicative_fncs.latex
As of: Monday 26May2003.
Typeset: 15May2014 at 10:01.
Filename: Problems/NumberTheory/multiplicative_fncs.latex
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