Systèmes dynamiques Examen Corrigé 2006-2007

mnmn
Ik= [9·10k,10k+1[nlog10 m∈JkJk= [log10 9+
k, k + 1[= k+ [log10 9,1[ k∈N=k+J0mn≥1mn
nlog10 m J0Z
T:R/Z→R/ZT x =x+ log10 m
mnTn0∈J0
log10 m p, q mq= 10p
m10
T
R/Z
lim rn/n = 1 −log10 9
m m
f0>0f:R→R
f−id 1f
f x ∈R/Z
ρ(f) = 0
x∈R/Zlim fn(x) = 0 φ
R/ZSnφ(x)/n
φ(0) x
f ν(φ) = δ(φ)
f d {0}
f
gα
f
f C2f C0
f f
χ6= 1
lim
n→∞
1
NnX
x∈Pn
χ(x) = 0.

χ= 1
P
 > 0 supx∈T2|φ(x)−P(x)| ≤
/3
1
NnX
x∈Pn
φ(x)−1
NnX
x∈Pn
χ(x)
≤/3.
n
1
NnPx∈Pnχ(x)−Rχ(x)dx
≤/3
n
1
NnPx∈Pnφ(x)−Rφ(x)dx
≤RT2|φ−χ|dx ≤
/3
ˆ
f(k)k∈Z2f∈L2(T2, m)
f(·+z) = f(·)L2ˆ f(k) = e2πizk ˆ f(k)
k∈Z2ˆ f(k)6= 0 z∈Pnχk(z) = 1
En
X=T2/PnPn
E
C
En¯x∈T2/Pn
¯x6= 0T2/Pnx /∈Pn¯x¯x=x+Pn
χ=η◦Tn/η χ(x)6= 1
η∈Γη(Tnx) = η(x)
Tnx=xmod Z2x∈Pn
EC0C0(T2/Pn)
T2Pn
χ PnFn
L2(m)χ
T2Pn
χ
f(x) = Pz∈Pnχ(x+z)f
Pnf(x) = χ(x)Pz∈Pnχ(z)χFn
Pz∈Pnχ(z) = 0
c
kp/q −ξk>1
kP(p/q)−P(ξ)k ≤ C· kp/q −ξk
C= maxξ−1≤z≤ξ+1 |P0(z)|P(ξ) = 0 q2P(p/q)
p/q P

χ=χkFnln∈Z2
Anln−ln=k A
lnAnln−ln=k
ln= (An−id)−1k λ := (3 + √5)/2>1
A1/λ
(An−id)−1C
n An−id
cλnAn−id λn
ln
l
n Anl=l+k
l l A
Anl
Anl=l+k
χ n χ /∈ Fn
h(T) = h(T, ξ)Wn−1
k=0 T−kξ kn
H(Wn−1
k=0 T−kξ)≤log knh(T, ξ)≤log k
(1/n)H(Wn−1
k=0 T−kξ)
n n log k≤H(Wn−1
k=0 T−kξ)
H(Wn−1
k=0 T−kξ) = nlog k
Wn−1
k=0 T−kξ µ 1/k
µ x ∈X
(ωi) = φ(x)Tix∈CiCiξ
ξ φ
Wn−1
k=0 T−kξ