Pn=1
n
n
k=1
(k+n)1
n
un=1
n
n
(n+ 1)(n+ 2)...(n+n)
Sn=
n
k=0
kCk
nf(x) =
n
k=0
xkCk
nf′(x)
SnCk
nCk+1
n+1
Sn
(u)u0>0u1>0nN
un+2 =√unun+1 unn
un+3 =3
√unun+1un+2
lim
n→+∞
n
2n
√a−n
1
aa > 0
n∈N∗xn−nx + 1 = 0
un]0,1[ (un)
un
u0>0n un+1 =u2
n+ 3
2un+ 2
(un)un+1 +un∼1
n
E=Cp([0,1],C)k0≤k≤p f E
Nk(f) = |f(0)|+|f′(0)|+... +|f(k−1)(0)|+||f(k)||∞
NkE
NjNk
E=Mn(R)
KR2O
O K
K, L R2K
K∩L=∅K L
(−1)n
nα+ (−1)nα > 0
un= (n4+n2)1/4−P(n)1/3P
Pun
(un)n∈NnN
vn=un
1 + un
, wn=un
1 + u2
n
unvn
unwn
Un= cos n2πln n−1
n
p∈ {2,3,4}Un= sin π(1 + np)1
p
p
N
u0= 1 nN∗un+1 =un
1 + 2un
1 + 3un
1
un+1 −1
unun
a, b u u0>0
∀n∈Nun+1 =n+a
n+bwn=
ln nb−aun(a, b)
un
lim nun= 0
un=(−1)n
√n+ (−1)nun= ln 1 + (−1)n
√n
t∈Rt̸= 0 un=e−it ln n
un+1 −un(un/n)
(un)
un2u2n1
n1+it
fR R
∀x∈R, f(2x) = 2f(x).
fR R
∀(x, y)∈R2, f(x)−f(y) = y+2x
x+2y
f(t)dt
f:x∈R∗7→ ex2−1
xf(0) = 0
f f′(0)
f, f 3, f 5f
f f
c∞R R f−1
x7→ (f−1(x))5
f c2]−1,1[ R
lim
x→0
1
xf′(x)−f(x)−f(0)
x
fRlim
x→+∞f′(x) = +∞
lim
x→+∞
f(x)
x= +∞
f∈C2(R,R)f f′′ Rf′
R
f:x7→ ln |ln |x||
F c1[0,1] R
1
0
xF (x)F′(x)dx ≥ −1
21
0
F2(x)dx
f[0,1] RF F (x) = 1
xx
0
f(t)dt
1
0
F2(t)dt ≤41
0
f2(t)dt
+∞
0
ln t
etdt
f]0,1]
]0,1] 1
0f(t)dt
limx→0xf(x) = 0
x7→ x
1
et
tdt
f:x7→ 1
0
t−1
ln(t)txdt
a < b b
a
dx
(x−a)(b−x)
F(x) = +∞
−∞
e−t2e−ixtdt
F
I=π
2
0
ln(sin x)dx J =π
2
0
ln(cos x)dx
I=π
0
ln a−cos t
b−cos tdt a > 1b > 1
F(u) = π
0
ln(u−cos t)dt
J(t) = 1
ππ
0
cos(tcos θ)dθ L(x) = +∞
0
J(t)e−xtdt
J c1RJ
+∞
φ ψ J(t) =+∞φ(t) + o(ψ(t))
J
a0, b0, a1, b1, a2, b2
∀t∈R,(a0+b0t)J(t)+(a1+b1t)J′(t)+(a2+b2t)J′′(t) = 0.
L x ≥0L(0)
x > 0+∞
0
tJ(t)dt +∞
0
tJ′(t)dt
L
f:x7→ e−x2F:x7→ +∞
x
f(t)dt F
+∞F(k)k∈ {1,2, ..., 40}
f(x)/F (x)F(x)∼+∞
f(x)
2x
1
0
ln(1 + tn)dt π2
12n
f(a) = +∞
0
xln x
(1 + x2)adx
f(2) f(3)
a∈RJ(a) = +∞
0
sin ax
ex−1dx
J(a) =
+∞
n=1
a
a2+n2
2π eax [0,2π[J(a)
an=+∞
−∞
1
(1 + t+t2)ndt
f(t) = 1
t+t2+ 1
ananun=+∞
0
1
(1 + u2)ndu
unun+1
+∞
0
x
1 + x4|sin x|3/2dx
fR R F:x7→ x
0
sin(x−t)f(t)dt
Ry” + y=f
n∈N∗π/2
0
sin2(nx)
sin2xdx eix
lim
x→+∞+∞
0
sin(xt2)
1 + t2dt
n∈N∗x > 0fn(x) = sin nx
nx +x√x
fnf
]0,+∞[
Un=+∞
0
fn(t)dt (Un)
f0: [a, b]→R[a, b]⊂]−1,1[
∀n∈N,∀x∈[a, b], fn+1(x) = x
a
fn(t)dt.
fn
S(x) =
+∞
n=1
1
xn2+n
S c1R∗
+
0++∞S(x)
x f(x) =
+∞
n=1
(−1)n−1n
x2+n2
f f(0) = ln 2
f(x) = +∞
0
cos xt
et+ 1dt
a f(x)∼ax2x→+∞
f:x7→
+∞
n=0
ein2x
2nfRc∞
p∈Nap=f(p)(0)
p!Gp:x7→
p
k=0
akxk
1 / 22 100%
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans l'interface ou les textes ? Ou savez-vous comment améliorer l'interface utilisateur de StudyLib ? N'hésitez pas à envoyer vos suggestions. C'est très important pour nous!