•Rd
Rd
Rd•
x= (x1,···xd)∈Rdp∈]1,+∞[
kxkp= d
X
i=1 |xi|p!1/p
,kxk∞= sup
1≤i≤d|xi|.
k·k1k·k2k·k∞Rd
kxk∞≤ kxk2≤ kxk1≤dkxk∞kxk2≤√dkxk∞,
x∈Rdd= 1
R2{x= (x1, x2)∈R2|kxk ≤ 1}
k·k1,k·k2k·k∞k·kpRdp∈]1,+∞[
s t ≥0st ≤sp
p+tq
qq1/q = 1−1/p
t s 7→ st −sp
p−tq
q
x y ∈Rd\{0}α=kxkpβ=kykqi
|xiyi|
αβ ≤|xi|p
pαp+|yi|q
qβq,
d
X
i=1
xiyi≤ kxkpkykq
|xi+yi|p≤ |xi+yi|p−1|xi|+|xi+yi|p−1|yi| k·kp
k·k2|·|
A={(x, y)∈R2|0<|x−1|<1};B={(x, y)∈R2|0< x ≤1};
C={(x, y)∈R2| |x|<1,|y| ≤ 1};D={(x, y)∈R2|x∈Q, y ∈Q};
E={(x, y)∈R2|x6∈ Q, y 6∈ Q};F={(x, y)∈R2|x2+y2<4}.
AR2
A={(x, y)∈R2|x2+y2≤2}\{(x, y)∈R2|(x−1)2+y2≤1}.
A
Rd
Rd
A={(x, y)∈R2|x2+y2<1}\{(x, y)∈R2|x= 0,|y| ≤ 1},Z,Q.
C1C2RdC1∩C26=∅
C1∪C2
C⊂Rda, b ∈C
γ: [0,1] 7→ Rdγ([0,1]) ⊂C γ(0) = a γ(1) = b
C=⇒C=⇒C C
k·k RdK=B0(0,1)
K
• •
f1: (x, y)7→ r2x+ 3
y−2;f2: (x, y)7→ ln(x+y+ 1) ; f3: (x, y)7→ ln(1 + x)−ln(1 + y)
x2−y2;
f4: (x, y)7→ x
y
y
x;f5: (x, y)7→ 1
cos(x−y);f6: (x, y, z)7→ tan(px2+y2+z2).
R2
A={(x, y)∈R2|x2−sin(y)≤4}
B={(x, y)∈R2|x3−4ey>4}
C={(x, y)∈[0,1]2|cos(x)≥0}
K1K2Rdx∈K1y∈K2
d(K1, K2) = d(x, y)d
f:Rd→Rpf
f:Rd→R
∀M > 0∃R > 0|x|> R =⇒ |f(x)|> M
BRf−1(B)Rd
KRf−1(K)Rd
f f(x)→+∞
|x| → +∞f
Rdd≥2k · k T
T∈ L(Rd)
Rd→Rx7→ kT(x)kRd
S:= {x∈Rd| kxk= 1}Rd
|||T||| := max
kxk=1 kT(x)k
a∈ S kT(a)k=|||T|||
∀y∈Rd\{0} kT(y)k ≤ |||T|||kykT
sup
y6=0
kT(y)k
kyk≤ |||T||| a∈ S
lim
(x,y)→(0,0)
x2y
x+y; lim
(x,y,z)→(0,0,0)
xyz +z3
2x3+yz2; lim
(x,y)→(0,0) |x|+|y|
x2+y2;
lim
(x,y)→(0,0)
x4y
x2−y2; lim
(x,y,z)→(0,0,0)
xy +yz
x2+ 2y2+ 3z2; lim
(x,y,z)→(0,0,0)
xyz
x+y+z.
f:R2\ {(0,0)} → Rf(x, y) = x2y2
x2y2+ (x−y)2
lim
x→0lim
y→0f(x, y) = lim
y→0lim
x→0f(x, y) = 0
lim
(x,y)→(0,0) f(x, y)
R2
f1(x, y) = xy
x2+y2(x, y)6= (0,0) f1(0,0) = 0
f2(x, y) = x3+y3
x2+y2(x, y)6= (0,0) f2(0,0) = 0
(0,0)
(0,0) g:R2\{(0,0)} → Rg(x, y) =
xy ln(x2+y2)
fR2\ {(x, x)|x∈R}
f(x, y) = sin x−sin y
x−y.
fR2
f1
f2
URda∈U
f:U→Ra
d= 2
f:U→Ra f a
g:U→Rpa
g a g a
f1f2(0,0)
Γ = {(x, y)∈R2|x6+ 3x2y2+y4= 64}
ΓR2
Γ (2,0)
Γ
URda∈U v ∈Rdf:U→R
a v
lim
t→0, t6=0
f(a+tv)−f(a)
t.
f a v f0
v(a)
f a v
φ:R→Rφ(t) = f(a+tv)φ
f
f:R27→ Rf(x, y) = y2/x x 6= 0 f(0, y) = y y ∈R
f(0,0) v∈Rdf0
v(0,0) f
f a v = (v1, . . . , vd)
f0
v(a) =
d
X
i=1
∂f
∂xi
(a)vi=∇f(a)·v.
S(0,1) = {v∈Rd| |v|= 1}Rd
M= sup
v∈S(0,1)
f0
v(a).
M v
z= max(−x2−y2+
1800,0) A(20,20,1000)
f:R2→RC1s > 0
∀λ > 0,∀x∈R2, f (λx) = λsf(x).
f s −1x∈R2
sf(x) = x1
∂f
∂x1
(x) + x2
∂f
∂x2
(x).
hR2→R R2
∂1h ∂2h
f(x, y) = h(x−y, x +y), g(x, y) = h(x2+y2, xy).
fR2→R R21
g(r, θ) = f(rcos θ, r sin θ)∂g
∂r
∂g
∂θ f
f:R3→R R3∂f /∂z = 0
f
URda∈U f :U→R
C1a a C1a
a
h:R3→R3h(x, y, z) = (e2y+e2z, e2x+e2z, x −y)
h C1R3U
Ω = R2\ {(0,0)}f:R2→R
f(x, y) =
xy x2−y2
x2+y2(x, y)∈Ω
0 (x, y) = (0,0).
fΩ
f(0,0)
f∂2f
∂x∂y
∂2f
∂y∂x
(0,0)
(0,0)
IRa v Rd
f:I→R
a, b, c ∈I a < b < c
f(b)−f(a)
b−a≤f(c)−f(a)
c−a≤f(c)−f(b)
c−b.
x0∈I
x7→ f(x)−f(x0)
x−x0
I\{x0}
f x0∈˚
I f
˚
I
f:Rd→Rd≥2
f a v
f(0) = 0 u∈Rdkuk1= 1
f(u)≤
d
X
i=1
uif(ei).
fk·k1
φ: [−1,1] →Rφ(λ) = f(λu)φ
M > 0u∈Rdkuk1= 1
−λM ≤f(λu)≤Mλ ∀λ∈[−1,1].
fRd
1 / 11 100%
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!