Exercice Problème 1. Fonctions arithmétiques.

cotan cos
sin x
e2iπx
θ θ 6≡ 0 mod 2π i Z+1
Z−1Z=eiθ
(X−i)5(i+ cotan(πx)) + (X+i)5(i−cotan(πx))
4
X
k=0
cotan((x+k)π
5),
4
Y
k=0
cotan((x+k)π
5)
n D(n)nN
C(n)
C(n) = (d1, d2)∈N2d1d2=n
N∗
F+∗
∗ F
∀(f, g)∈ F2,∀n∈N∗:
(f+g)(n) = f(n) + g(n)
(f∗g)(n) = X
(d1,d2)∈C(n)
f(d1)g(d2) = X
d∈D(n)
f(d)g(n
d)
f
∀(p, q)∈N∗2, p ∧q= 1 ⇒f(pq) = f(p)f(q)
n
I(n) = n
e0(n) = (1n= 1
0n > 1
e(n)=1
d(n)nN
σ(n)nN
φ(n)k∈J1, nKn
φ(1) = 1
β(n) =
n
X
k=1
k∧n
µ(n) = 
1n= 1
0n
(−1)sn s
1
β(6) (σ∗µ)(12)
e∗e I ∗e
p φ(p)σ(p)p d(p)
(µ∗e)(pm)m
∗
e0∗
n∈N∗
T(n) = (d1, d2, d3)∈N3n=d1d2d3
T(n)∗
(F,+,∗)e0

m n
P:(D(m)×D(n)→D(mn)
(a, b)7→ ab
f g f ∗g
I e0e d σ µ
f
N(f)
N(f) = min {k∈N∗f(k)6= 0}
f g f ∗g
N(f∗g) = N(f)N(g)
µ∗e=e0
f g
f=g∗e⇔g=f∗µ
n d δ n n =dδ
F={k∈J1, dKk∧d= 1}∆ = {s∈J1, nKs∧n=δ}
k7→ δk F ∆
F
a∈J1, nK
n∧x=a x ∈J1, nK
I=e∗φ φ =I∗µ
β∗e=I∗I
σ∗φ=I∗I
n φ(n) + σ(n) = nd(n)n
∀x∈Re P(x)≥0
X2+X+ 1, X2−X+ 1, X2+ 2X+1
2
m θ ]0, π[
sin 2θ≥0,sin 3θ≥0,· · · ,sin(m+ 1)θ≥0
C=X2+c1X+c2
u1u2Im u1>0φ∈]−π, π]u1
r
m∈N∗
Dm= (Xm+1 −um+1
1)(Xm+1 −um+1 2)
m φ
BmBmC=Dm
Bm=b0+b1X+· · · +b2mX2mk∈ {0,· · · , m}bk
b2m−k
m BmDm
C
B D BC =D