Exercices de révision mathématiques : Analyse avancée

e−(ln n)αα > 0
ln 1 + (−1)n
nαα > 0
√2π
C
n
X
k=2
ln k
k=1 2ln2n+C+o(1)
f:R+→R+
f(1) + ··· +f(n)∼Zn
0
f.
Pn
k=0 k!P+∞ k=n1
k!

ϕ a > 1
+∞
X
n=1
ϕ(n)
an−1.
X
d|n
ϕ(d) = n.
f[0,1] [0,1] f
R+
R
fR+R
a b R+∗
∀x∈R+,|f(x)| ≤ ax +b.
∀x∈[0, π/2],2
πx≤sin x≤x,
∀(a1,··· , an)∈(R+)n,(a1···an)1/n ≤1
n(a1+··· +an)
AR R f
R R
∀x∈R, f(x) = sup {d(x), d ∈A, d ≤f}.
fR R f
pR R
f0
B(u, v)B
E F u v
IRE F
MRn(R)
∀t∈R, N(t) = M(t)−1.
NRN0(t)t
f I Rf
I f0I

Ck
f C∞R R x0∈R
f x0f0(x0)=0 f00(x0)≥0
f0(x0) = 0 f00(x0)>0f
x0
f Ckk∈N∗x0f
∀j∈ {0,··· , k}, f(j)(x0)=0.
C∞R
]−1,1[
Ck
C1R+
+∞f0(x) +∞
α∈R+∗
t∈]0,1[7−→ ln(t)
(1 −t)α, t ∈R+∗7−→ ln(t) (sin(1/t))α
u v L2(I)uv L1(I).
u Lpv Lqp > 11
p+1 q= 1
R+R
f
f(x)→0x→+∞
fR+R+
xf(x) 0 +∞
Zx
0
sin t
tdt
+∞t7→ sin t
tR

x > 0 0+
Z+∞
xtdt.
+∞Zx
2
dt
ln t.
+∞
In=Zn
01 + t
nn
e−2tt, Jn=Z1
0
e−nt ln(t)
√1−t2t.
s]1,+∞[
Z+∞
0
ts−1
et−1t= Γ(s)ζ(s).
s]0,1[
Z+∞
0
ts−1
1 + tt=
N
X
n=−N
(−1)n
s−n.
[1,+∞[u= 1/t
C1
ΓC∞R+∗
x≥0
f(x) = Z+∞
0
1−cos(t)
t2e−tx dt
f00(x)x > 0f(x)x≥0
fR+R
`+∞x > 0Lf(x) = R+∞
0f(t)e−xt dt
xLf(x)−→
x→0`.
x+∞

[a, b]C
C1
C1
C1Ck
ζ C∞]1,+∞[
+∞1+
x > 1|h|<1−x
ζ(x+h) =
+∞
X
k=0
ζ(k)(x)
k!hk.
ζ2
ζ2(x) =
+∞
X
n=1
(−1)n−1
nx,
C∞R+∗
xR
f(x) =
+∞
X
n=0
e2nix
n!.
f C∞R
f0
C∞