Approximation diophantienne et nombres quadratiques

α d > 1
P
A > 0
p/q
α−p
q
>A
qd.
(a)α A
A p/q =α
(b)α d > 1k P1=kP
p/q
qd|P(α)−P(p/q)| ≥ 1.
(c)P(α)−P(p/q)
(d)
(e)
α
(a)a/b
|α−a/b|<1
2b2
α
α n > 1
k∈ {2,··· , n}
ϕk=qk−2/qk−1, ψk=rk+ϕk= [ak,···] + ϕk
(b)k≥2 1/ϕk+1 =ak+ϕk1/ϕk+ 1/rk=ψk−1
(c)ψk≤√5ψk−1≤√5
ϕk>√5−1
2.
k=n, n −1, n −2
α−pk
qk
≥1
√5q2
k
.
(d)α= (pnrn+1 +qn)/(qnrn+1 +qn−1)
ψk+1 ≤√5k=n, n −1, n −2
(e) (e)ϕnϕn+1 (√5−1)/2
(f)k=n−2, n −1, n
α−pk
qk
<1
√5q2
k

α α
(g) 1/√5
(a)α
n≥1
1
qn(an+1 + 2) <
α−pn
qn
≤1
q2
nan+1
.
(b)f:N∗→]0,+∞[α
α−p
q
< f(q)
p/q
(c)α
a0,··· , an,··· α c > 0
α−p
q
<c
q2
p, q
c > 0
(a)α, β a0,··· , an
α= [a0, a1,··· , an, β].
a, b, c, d
α=aβ +b
cβ +d.
(b)α
α
Aα2+bα +c= 0
A, B, C A 6= 0
[a0,··· , an,···]α pn/qn
rn= [an+1,···]
(c)rn
Anr2
n+Bnrn+Cn= 0
An=Ap2
n−1+Bpn−1qn−1+Cq2 n−1
Bn= 2Apn−1pn−2+B(pn−1qn−2+pn−2qn−1)+2Cqn−1qn−2
Cn=Ap2
n−2+Bpn−2qn−2+Cq2 n−2
(d)Cn=An−1B2
n−4AnCn=b2−4ac n ≥1
(e)δn−1pn−1=αqn−1+δn−1/qn−1δn−1
(f)pn−1
|An|<2|aα|+|a|+|b|
(f)An, Bn, Cnm
n rm=rm+n(a) (b)