Giáo trình giải tích 2 part 8 ppt

A
g
Dg
g : U −→ R
n
C
1
U ⊂ R
n
A
A ⊂ U g det Dg =0 A
f : g(A) −→ R f ◦ g|det Dg| A

g(A)
f =

A
f ◦ g|det Dg|.
g ∈ C
1
A g(A)
f ◦ g|det Dg| A
g ∈ C
1
det Dg(a) =0 U
a
a
g =(Φ
n
◦ T
n

) ◦···◦(Φ
1
◦ T
1
),
T
i
(x)=a + σ
i
(x − a) σ
i
Φ
i
(x
1
, ··· ,x
n
)=(x
1
, ··· ,φ
i
(x), ··· ,x
n
)(i =1··· ,n)
det Dg(a) =0 i,
∂g
n
∂x
i
(a) =0.

B(x)=a + σ(x −a) σ n i
h = g ◦B
∂h
n
∂x
n
(a)=
∂g
n
∂x
i
(a) =0.
Φ(x)=(x
1
, ··· ,x
n−1
,h
n
(x)) Φ ∈ C
1
det DΦ(a)=
∂h
n
∂x
n
(a) =0
U a Φ Φ
−1
∈ C
1

g = h ◦ B
−1
= G ◦ Φ ◦ T
T = B
−1
= a + σ
−1
(x − a) G(x)=(h
1
(x), ··· ,h
n−1
(x),x
n
)
G
g(x):=T (x)=a + σ(x − a) σ
|det T | =1
U
a
g(x):=Φ
i
(x)=(x
1
, ··· ,φ
i
(x), ··· ,x
n
)
i = n
U

a
= S × [a
n
.b
n
],S R
n−1
Φ(U)=S × φ
n
(U) det DΦ=
∂φ
n
∂x
n
.

Φ(U)
f =

S
(

φ
n
(x
1
,···,x
n−1
,[a
n

,b
n
])
f(x)dx
n
)dx
1
···dx
n−1
.
=

S
(

b
n
a
n
f(x
1
, ··· ,φ
n
(x))|
∂φ
n
∂x
n
|dx
n

)dx
1
···dx
n−1
=

U
f ◦ Φ|det DΦ|.
T Φ Φ ◦ T

Φ◦T (C)
f =

T (C)
f ◦ Φ|det DΦ|
=

C
f ◦ Φ ◦ T |det(DΦ) ◦ T ||det DT|
=

C
f ◦ (Φ ◦ T )|det D(Φ ◦ T )|.
A A
P S ∈ P S ∩ A = ∅ g
S

g(A)
f =


S∈P
S∩A=∅

g(A∩S)
f =

S∈P
S∩A=∅

A∩S
f ◦ g|det Dg| =

A
f ◦ g|det Dg|. 
g : U −→ R
n
,U R
n
A A ⊂ U g
A f g(A) f ◦ g|det Dg|
A

g(A)
f =

A
f ◦ g|det g|.
µ{x :detDg(x)=0} =0
A = A ∪ ∂A ∂g(A) ⊂ g(∂A)
•

g : R
2
−→ R
2
g(r, ϕ)=(r cos ϕ, r sin ϕ)
|
D(x, y)
D(r, ϕ)
| = |det Dg(r, ϕ)| =





cos ϕ −r sin ϕ
sin ϕrcos ϕ





= r.
✲
✻
0 R
2π
r
ϕ
✲
g

✲
✻
0 R
x
y
ϕ
✟
✟
✟
✟
✟✯
r
A ⊂ R
2
g intA f A

g(A)
f(x, y)dxdy =

A
f(r cos ϕ, r sin ϕ)rdrdϕ.

x
2
+y
2
≤R
2
e
−x

2
−y
2
dxdy =

2π
0

R
0
e
−r
2
rdrdϕ = π(1 − e
−R
2
).
R → +∞

∞
−∞
e
−x
2
dx =
√
π
•
g : R
3

−→ R
3
(r, ϕ, z) → (r cos ϕ, r sin ϕ, z).
|det g(r, ϕ, z)| = r A ⊂ R
2
g intA f
A

g(A)
f(x, y, z)dxdydz =

A
f(r cos ϕ, r sin ϕ, z)rdrdϕdz
A = {x
2
+ y
2
≤ 1, 0 ≤ z ≤ 1}

A
z
1+x
2
+ y
2
dxdydz =

1
0


2π
0

1
0
z
1+r
2
rdrdϕdz =

1
0
zdz

2π
0
dϕ

1
0
r
1+r
2
dr =
π
2
ln 2.
•
g : R
3

−→ R
3
, (ρ, ϕ, θ) → (ρ cos ϕ sin θ, ρsin ϕ sin θ, ρcos θ)
|det Dg(ρ, ϕ, θ)| = ρ
2
sin θ
s
M
0
✓
✓
✓
✓✼
θ
ϕ
ρ
✁
✁
✁
✁☛
x
✲
y
✻
s
z

g(A)
f(x, y, z)dxdydz =


A
f(ρ cos ϕ sin θ, ρ sin ϕ sin θ, ρ cos θ)ρ
2
sin θdρdϕdθ,
A g intA f

x
2
+y
2
+z
2
≤R
2

x
2
+ y
2
+ z
2
dxdydz =

R
0

2π
0

π

0
ρ
3
sin θdρdϕdθ
=

R
0
ρ
3
dρ

2π
0
dϕ

π
0
sin θdθ =
R
4
4
2π.2.
A R
n
λ ∈ R λ : R
n
→ R
n
,λ(x)=λx

(y
1
, ··· ,y
n
)=(λx
1
, ··· ,λx
n
)
v(λ(A)) =

λ(A)
dy =

A
λ
n
dx = λ
n
v(A)
v
1
, ··· ,v
n
∈ R
n
A = {y ∈ R
n
: y = x
1

v
1
+ ···+ x
n
v
n
, 0 ≤ x
1
, ··· ,x
n
≤ 1}
v(A)=|det(v
1
, ··· ,v
n
)| = |det(v
ij
)
1≤i,j≤n
|
v
i
=(v
i1
, ··· ,v
in
)
1+x + x
2
+ x

3
+ ··· =
1
1 − x
1
x
+
1
x
2
+
1
x
3
+ ··· =
1
x(1 − 1/x)
=
1
x − 1
···+
1
x
3
+
1
x
2
+
1

x
+1+x + x
2
+ x
3
+ ··· =0 x =0, 1
∞

k=1
x
k
k
2
∞

k=0
k!x
k
∞

k=0
k
k +1
(x −1)
k
∞

k=2
(x +2)
k

ln k
∞

k=1
1
2
k
k
x
k
∞

k=1
(−1)
k
k
x
2k+1
∞

k=1
1
k

x
2

2k
∞


k=0
x
k
∞

k=0
(k +1)x
k
∞

k=0
x
k+1
k +1
∞

k=0
(k
2
+2k − 2)x
k
∞

k=1
(−1)
k
x
k
k
∞


k=0
x
2k+1
2k +1
∞

k=0
2k
2
+1
k!
x
k
f (a, b) f c ∈ (a, b)
f(x)=e
−
1
x
2
(x =0),f(0) = 0 f
f
0
f(x)=sin
3
x f(x)=
x
(1 − x)(1 − x
2
)

f(x)=
1
(1 − x)
n
f(x)=
√
a + x f(x)=ln

1+x
1 − x
(x)=
1
√
2

x
0
e
−
t
2
2
dt
(1)
(x)=

x
0
sin t
t

dt
(1)
f T

a+T
a
f(x)dx =

T
0
f(x)dx, ∀a
f [−π, π] a
k
,b
k
a

k
,b

k
f f

a

k
= kb
k
,b


k
= −ka
k
f(x)=0 −π ≤ x ≤ 0 f (x)=1 0 <x≤ π
f(x)=|x|, −π ≤ x ≤ π
π
4
=1−
1
3
+
1
5
−
1
7
+ ··· va
π
2
8
=
1
1
2
+
1
3
2
+
1

5
2
+ ···
x := 0, ±π f(x)
t ∈ R \ Z f
t
(x)=costx, |x|≤π
f
t
(x)
tπ =
1
tπ
+
∞

k=1
2t
π(t
2
− k
2
)
π
2
sin
t
π
= lim
n→+∞

n

k=−n
1
(t − k)
2
sin tπ
tπ
= lim
n→+∞
(1 −
t
2
1
)(1 −
t
2
2
) ···(1 −
t
2
n
2
)
f(x)=e
x
,x∈ [0, 2π] f(x)=0 x ∈ [0,l] f(x)=1 x ∈ (l, 2l)
f(x)=x, x ∈ (−2, 2)
cos
f(x)=1 x ∈ [0,π/2] f(x)=0 x ∈ (π/2,π]

f(x)=x(π −x),x∈ [0,π]
sin
x =2
∞

k=1
(−1)
k+1
k
sin kx, −π<x<pi
x
2
,x
3
,x
4
−π<x<π
R
n
|d(x, y) − d(y, z)|≤d(x, z).
m, M, a, b > 0 x =(x
1
, ···,x
n
) ∈ R
n
m max
1≤i≤n
|x
i

|≤x≤M max
1≤i≤n
|x
i
|.
ax≤
n

i=1
|x
i
|≤bx.
x, y ∈ R
n
| <x,y>| = xy, x + y = x + y.
f,g [a, b]






b
a
fg






≤


b
a
f
2

1/2


b
a
g
2

1/2
.
T : R
n
→ R
m
M
T (h)≤Mh, ∀h ∈ R
n
.
R
n
{x : x≤1}, {x : x =1}, {x =(x
1

, ···,x
n
):x
i
∈ Q,i=1, ···,n}
a ∈ R x
1
= a, x
k
= x
2
k−1
− x
k−1
+1 a
(x
k
) (x
k
) (x
k
)
(x
k
)
lim
k→∞
x
k+1
x

k
= L>0 lim
k→∞
k
√
x
k
= L
∀>0, ∃A, B > 0,N ∈ N : k ≥ N ⇒ A(L − )
k
<x
k
<
B(L + )
k
.
x
k
=
k
k
k!
lim
k→∞
k
(k!)
1/k
= e
lim
k→∞

x
k
= M (x
k
)
lim
k→∞
1
k
(x
1
+ ···+ x
k
)=M
0 <x
k
<y
k
x
k+1
=(x
k
y
k
)
1
2
y
k+1
=

1
2
(x
k
+y
k
) x
k
<y
k
(x
k
), (y
k
)
(x
k
) R
n
x
k+1
−x
k
 <
1
k
2
+ k
, ∀k (x
k

)
x
k
=1+
1
2
+ ···+
1
k
(k ∈ N)
R [a, b] (a, b) (a, b]
B(a, r) R
n
A R [0, 1] A
A [0, 1]
U
k
(k ∈ N) R
n

k∈N
U
k

k∈N
U
k

k∈N
(R

n
\ U
k
)

k∈N
(R
n
\ U
k
)
X ⊂ Y
X ⊂ Y
X ∪ Y = X ∪ Y
∂(X ∪Y ) ⊂ ∂X∪∂Y ∂(X ∩Y ) ⊂ ∂X∩∂Y ∂(X ×Y )=∂X×
Y ∪X × ∂Y.
X R
n
X
Z R {x ∈ R
n
: x < 1}
X x ∈ X d
x
> 0
d(x, y) ≥ d
x
, ∀y ∈ X.
X K X ∩ K = ∅ d>0
d(x, y) ≥ d, ∀x ∈ K, ∀y ∈ X.

X, K X ∩K = ∅ d>0
(F
k
) R
n
k ∈ N F
k
F
k
⊃ F
k+1
(F
k
)=sup{d(x, y):x, y ∈ F
k
}→0
k →∞

k∈N
F
k
I
k
=(0,
1
k
),k ∈ N

k
I

k
= ∅
{(x, y) ∈ R
2
: |x|≤1} {(x, y) ∈ R
2
: x
4
+ y
4
=1}
{x ∈ R
n
: x≤r} {x ∈ R
n
:1≤x≤2} {x ∈ R
n
: x =1}
Z [0, 1]
K R
n
R
n
\ K
K R
n
R
n
\ K
L

i
,i ∈ I, L
i

L
j
= ∅ i, j

i∈I
L
i
C
C
C = {(x, y) ∈ R
2
:0<y≤ x
2
,x =0}∪{(0, 0)} C
C (0, 0)
C
C = {(x, y) ∈ R
2
: y =sin
1
x
,x =0}∪{(0,y):|y|≤1} C
C
C
C F
0

=[0, 1]
F
1
=[0,
1
3
] ∪ [
2
3
, 1] F
0
F
2
=[0,
1
9
] ∪ [=
2
9
,
1
3
] ∪ [
2
3
,
7
9
] ∪ [
8

9
, 1] F
1
F
k
F
k−1
F
k
2
k
[
k
3
k
,
k +1
3
k
] C =

k
F
k
x ∈ C x =

∞
k=0
a
k

3
k
a
k
∈{0, 2}
C C (C)=∅
C C
1
3
+2.
1
9
+ ···+2
k−1
/3
k
+ ···=1
(0, 0) f(x, y)=
x
2
y
x
4
+ y
2
n → +∞
n
2
+n ∼ n
2

,
√
n +1−
√
n ∼
√
n
2
, (−1)
n
n
2
= O(n
2
),n
2
+2 = o(n
3
), sin n = O(1)
x → 0
(1 + x)
α
=1+αx + o(x) (1 + x)
α
∼ 1+αx
e
x
=1+x + o(x) e
x
∼ 1+x

ln(1 + x)=x + o(x)ln(1+x) ∼ x
sin x = x + o(x)sinx ∼ x
cos x =1−
1
2
x
2
+ o(x)cosx ∼ 1 −
1
2
x
2
x →∞ o, O
x
α
,x
β
,a
x
,b
x
, log
c
x, log
d
x (α, β, a, b, c, d > 0)
f(x, y)=
xy(x + y)
x
2

+ y
2
(x, y) =(0, 0) f(0, 0) = 0.
f(x, y)=
e
xy
− 1
2xy
xy =0 f(x, y)=0 xy =0
f(x, y)=
sin xy
x
x =0 f ((0,y)=y
f(x, y)=
xy
x
2
+ y
2
,f(0, 0) = 0)
f : R
2
→ R
(0, 0)
f(x, y)=
x
2
y
x
4

+ y
2
(x, y) =(0, 0) f(0, 0) = 0
f : R → R f(x + y)=f(x)+f(y), ∀x, y ∈ R f
0 f
g : R → R g(x + y)=g(x)g(y), ∀x, y ∈ R g
0 g
f :[0, 1] −→ [0, 1] f(x)=
1
q
x =
p
q
f(x)=0
x f
f : R
n
→ R
m
f R
n
f
−1
(V ) V ⊂ R
m
f
−1
(F ) F ⊂ R
m
U ⊂ R {(x, y) ∈ R

2
: x ∈ U}
f : R
n
→ R {x ∈ R
n
:0≤ f(x) ≤ 1}
f : R → R U ⊂ R f(U)
{A ∈ Mat(n, n) : det A =0}
(n, n) n R
f : R → R
{x : f(x)=0}{x : f (x) > 1}{f(x):x ≥ 0}{f (x):0≤ x ≤ 1}
f : R
2
→ R {f(x, y):x
2
+ y
2
=1}
X ⊂ R
n
d(x, X)= inf
y∈X
d(x, y)
R
n
 x → d(x, X)
|d(x, X) −d(x

,X)|≤d(x, x


)