Giáo trình giải tích 2 part 5 docx


2
p : |
p
k
− x|≥δ |
p − kx
kδ
|≥1
|

2
|≤2M

|p−kx|≥kδ
r
p
(x) ≤ 2M
k

p=0

p − kx
kδ

2
r
p
(x)
≤
2M

kδ
2
kx(1 −x) ≤
M
2δ
2
k
.
>0 δ>0
|f(x) − B
k
(x)|≤|

1
| + |

2
| <+
M
2δ
2
k
k ≥ M/2δ
2
 sup
|x|≤1
|f(x) −B
k
(x)| < 2. 
f(x)=e

x
R
A K ⊂ R
n
∀f,g ∈A,α∈ R,f+ g, fg αf ∈A.
A
∀x, y ∈ K, x = y, ∃ϕ ∈A: ϕ(x) = ϕ(y).
R[x
1
, ··· ,x
n
] n R
n
a
0
+
k

p=1
(a
p
sin px + b
p
cos px),a
p
,b
p
∈ R,k ∈ N,
R
ϕ

1
, ··· ,ϕ
s
: K → R K
k

p
1
+···+p
s
=0
a
p
1
···p
s
ϕ
p
1
1
(x) ···ϕ
p
s
(x), a
p
1
···p
s
∈ R,k ∈ N.
K R

n
A⊂C(K)
K
K A
∀f ∈ C(K), ∃g
k
∈A:(g
k
)
k∈N
f.
A = {g : g A} A⊂C(K)
(h
k
) ⊂ A
h h ∈
A A = A
A
A (h
k
) ⊂ A h k
(g
k,i
) ⊂A h
k
i →∞

2
(g
k

= g
σ(k),i(k)
) ⊂A h h ∈ A
x, y ∈ K, α, β ∈ R h ∈
A h(x)=α, h(y)=β
h A ϕ ∈A ϕ(x) = ϕ(y) h(z)=
α +(β −α)
ϕ(z) −ϕ(x)
ϕ(y) −ϕ(x)
h
h
1
,h
2
∈ A max(h
1
,h
2
), min(h
1
,h
2
) ∈ A
max(h
1
,h
2
)=
h
1

+ h
2
+ |h
1
− h
2
|
2
min(h
1
,h
2
)=
h
1
+ h
2
−|h
1
− h
2
|
2
h ∈ A⇒|h|∈A.
h M>0
|h(x)| <M,∀x ∈ K (P
k
)
[−M,M]  t →|t| g
k

= P
k
◦ h (g
k
)
A |h|
f ∈ C(K)
∀>0, ∃g ∈
A : d(f(x),g(x)) <, ∀x ∈ K, i.e.
f(x) −<g(x) <f(x)+, ∀x ∈ K.
x, y ∈ K h
x,y
∈ A : h
x,y
(x)=f(x),h
x,y
(y)=f(y).
x y ∈ K h
x,y
(y)=f(y) U
y
y
h
x,y
(z) <f(z)+, ∀z ∈ U
y
∩ K
P
x
= {U

y
,y ∈ K} K K
U
y
1
, ··· ,U
y
p
K h
x
= min(h
x,y
1
, ··· ,h
x,y
p
) h
x
∈ A
h
x
(z) <f(z)+, ∀z ∈ K.
x ∈ K h
x
(x)=f(x) V
x
x
f(z) −<h
x
(z), ∀z ∈ V

x
∩ K.
P = {V
x
,x ∈ K} K
V
x
1
, ··· ,V
x
q
K g =max(h
x
1
, ··· ,h
x
q
) g ∈ A
f(z) −<g(z),z∈ K.
g 
R T
P
k
(x)=a
k,0
+
N
k

p=1

(a
k,p
sin(
2πpx
T
)+b
k,p
cos(
2πpx
T
)).
R T>0
C[0,T]

R
n
n
K
1
⊂ R
n
1
K
2
⊂ R
n
2
A
1
A

2
K
1
,K
2
A
1
A
2
f ∈ C(K
1
× K
2
)
k

i=1
g
i
(x)h
i
(y)
g
i
∈A
1
,h
i
∈A
2

,k ∈ N
K
1
× K
2

[a, b]
K R
n
f ∈ C[0, 1]
n
k f
B
k
(x
1
, ··· ,x
n
)=

0≤p
1
,···,p
n
≤k
C
p
1
k
···C

p
n
k
f(
p
1
k
, ··· ,
p
n
k
)x
p
1
1
···x
p
n
n
(1−x
1
)
k−p
1
···(1−x
n
)
k−p
n
.

(B
k
) f
U R f : U → R a ∈ U
f

(a)
lim
x→a
f(x) −f(a)
x − a
= lim
h→0
f(a + h) −f(a)
h
= f

(a)
f(a + h)=f(a)+f

(a)h + o(h)
f(x) T (x)=f(a)+f

(a)(x −a) x a
U R
n
f : U → R
m
a ∈ U A : R
n

→ R
m
f(a + h) −f(a) −Ah
h
→ 0, h → 0.
A
f a Df(a) f

(a)
f a
f(a + h)=f(a)+Df(a)h + o(h),
o(h) ϕ(h) lim
h→0
ϕ(h)
h
=0.
f a f a
T
T (x)=f(a)+Df(a)(x −a)
f a
f a
(a, f(a)) f
G
f
= {(x, y) ∈ R
n
× R
m
: y = f(x),x∈ U} ,
T

T
a
= {(x, y) ∈ R
n
× R
m
: y = T(x)=f(a)+Df(a)(x −a),x∈ R
n
}.
d((x, f(x)); T
a
) ≤ d(f(x),T(x)) = o(x −a) x → a
f a Df(a)
f a
A, B
lim
h→0
A(h) − B(h)
h
=0.
x ∈ R
n
\ 0
A(x) − B(x)
x
= lim
t→0
A(tx) −B(tx)
tx
=0

A(x)=B(x), ∀x ∈ R
n
A = B
f Df(a)
lim
x→a
(f(x) − f(a)) = lim
x→a
(f(x) − f(a) −Df(a)(x − a)) + lim
x→a
Df(a)(x − a)=0
f a 
0
T DT(a)=T,∀a
R → R
m
h → <A,h> A ∈ R
m
A ∈ R
m
f

(a) = lim
h→0
f(a + h) − f(a)
h
.
n>1
y
h

y ∈ R
m
,h∈ R
n
R
n
−→ R
m
m ×n R
n
R
m
Jf(a) Df(a)
j e
j
∈ R
n
Jf(a)e
j
= j Jf(a).
a
Df(a)(te
j
)=f(a + te
j
) − f(a)+o(t).
j f a D
j
f(a)
∂f

∂x
j
(a)
D
j
f(a)=
∂f
∂x
j
(a) = lim
t→0
f(a + te
j
) − f(a)
t
.
∂f
∂x
j
a =(a
1
, ··· ,a
n
) x
k
= a
k
k = j
x
j

→ f(a
1
, ··· ,x
j
, ··· ,a
n
) a
j
e ∈ R
n
\ 0 e f a
D
e
f(a)=
∂f
∂e
(a) = lim
t→0
f(a + te) −f(a)
t
.
f e a
f(x, y)=x
y
∂f
∂x
(x, y)=yx
y−1
,
∂f

∂y
(x, y)=x
y
ln y (x, y > 0).
f(x, y)=

|xy|
∂f
∂x
(0, 0) = lim
t→0
f(t, 0) −f(0, 0)
t
=0,
∂f
∂y
(0, 0) = 0.
f(x
1
, ··· ,x
n
)=(f
1
(x
1
, ··· ,x
n
), ··· ,f
m
(x

1
, ··· ,x
n
)).
f a ∈ U Df(a)
f a
Jf(a)
f a
∂f
i
∂x
j
(a), (i =1, ··· ,m; j =
1, ··· ,m)
Jf(a)=






∂f
1
∂x
1
(a) ···
∂f
1
∂x
n

(a)
··· ··· ···
∂f
m
∂x
1
(a) ···
∂f
m
∂x
n
(a)






.
Df(a):R
n
→ R
m
dx =



dx
1
dx

n



→ dy =



dy
1
dy
m



= Jf(a)dx













df

1
=
∂f
1
∂x
1
(a)dx
1
+ ···+
∂f
1
∂x
n
(a)dx
n
df
m
=
∂f
m
∂x
1
(a)dx
1
+ ···+
∂f
m
∂x
n
(a)dx

n
f : R
2
−→ R
3
f(x, y)=(x
2
+ y
2
,x+ y,xy) (x, y) ∈ R
2
,
Jf(x, y)=



2x 2y
11
yx



.
f(x, y)=x
2
+ y
2
(x
0
,y

0
) T (x, y)=x
2
0
+ y
2
0
+2x
0
(x − x
0
)+2y
0
(y −y
0
)
z = x
2
+ y
2
R
3
(x
0
,y
0
,z
0
) T
z −z

0
=2x
0
(x − x
0
)+2y
0
(y −y
0
).
dz =2x
0
dx +2y
0
dy
f a f a
f a f a
f(x, y)=

|xy|
∂f
∂x
(0, 0) =
∂f
∂y
(0, 0) = 0
Df(0, 0) f
f (0, 0)






f(h, k) −f(0, 0) −

∂f
∂x
(0, 0)
∂f
∂y
(0, 0)


h
k






√
h
2
+ k
2
→ 0 , (h, k) → (0, 0).

|hk|
√

h
2
+ k
2
→ 0 (h, k) → (0, 0)
f : U → R
m
U ⊂ R
n
∂f
∂x
i
,i=1,··· ,n,
U f x ∈ U
m =1 h =(h
1
, ··· ,h
n
) 0
f(x + h) −f(x)=
n

j=1
(f(x + v
j
) − f(x + v
j−1
)), v
j
=(h

1
, ··· ,h
j
, 0, ··· , 0).
j g
j
(h
j
)=f(x + v
j
)
f(x + v
j
) − f(x + v
j−1
)=
∂f
∂x
j
(c
j
)h
j
, c
j
= v
j−1
+ θ
j
h

j
e
j
, 0 <θ
j
< 1.
x
lim
h→0
1
h
|f(x + h) −f(x) −

j
∂f
∂x
j
(x)h
j
| = lim
h→0
1
h
|

j
(
∂f
∂x
j

(c
j
) −
∂f
∂x
j
(x))h
j
| =0,
f x 
f,g x f + g x
D(f + g)(x)=Df(x)+Dg(x)
f,g x m =1 fg x
D(fg)(x)=Df(x)g(x)+f(x)Dg(x)
f,g x g(x) =0
f
g
x
D(
f
g
)(x)=
Df(x)g(x) −f(x)Dg(x)
g(x)
2
f : U −→ V g : V −→ W U, V, W R
n
, R
m
, R

p
f x g y = f(x) g ◦ f x
Dg ◦ f(x)=Dg(f(x))Df(x)
f(x + h)=f(x)+Df(x)h + ϕ
1
(h) ϕ
1
(h)=o(h)
g(f(x)+k)=g(f (x)) + Dg(f (x))k + ϕ
2
(k) ϕ
2
(k)=o(k)
g ◦f(x + h)=g(f(x)+Df(x)h + ϕ
1
(h)
  
k
)
= g(f(x)) + Dg(f(x))Df(x)h + Dg(f(x))ϕ
1
(h)+ϕ
2
(Df(x)h + ϕ
1
(h))
Dg(f(x))ϕ
1
(h)≤Dg(f(x))ϕ
1

(h) = o(h),
ϕ
2
(Df(x)h + ϕ
1
(h)) = o(Df(x)h + ϕ
1
(h))=o(h).
g ◦f x D(g ◦ f)(x)=D(g(fx))Df(x). 
Jh(x)=Jg(f(x))Jf(x)
f(x)=(f
1
(x
1
, ··· ,x
n
), ··· ,f
m
(x
1
, ··· ,x
n
))
g(y)=(g
1
(y
1
, ··· ,y
m
), ··· ,g

p
(y
1
, ··· ,y
m
))
y = f(x)
h(x)=g ◦f(x)=(h
1
(x
1
, ··· ,x
n
), ··· ,h
p
(x
1
, ··· ,x
n
))






∂h
1
∂x
1

···
∂h
1
∂x
n
··· ··· ···
∂h
p
∂x
1
···
∂h
p
∂x
n






=






∂g
1

∂y
1
···
∂g
1
∂y
m
··· ··· ···
∂g
p
∂y
1
···
∂g
p
∂y
m












∂f

1
∂x
1
···
∂f
1
∂x
n
··· ··· ···
∂f
m
∂x
1
···
∂f
m
∂x
n






∂h
i
∂x
j
=
∂g

i
∂y
1
∂f
1
∂x
j
+
∂g
i
∂y
2
∂f
2
∂x
j
+ ···
∂g
i
∂y
m
∂f
m
∂x
j
=
m

k=1
∂g

i
∂y
k
∂f
k
∂x
j
f(x, y) x, y x = r cos ϕ, y = r sin ϕ
h(r, ϕ)=f(r cos ϕ, r sin ϕ)
∂h
∂r
=
∂f
∂x
cos ϕ +
∂f
∂y
sin ϕ,
∂h
∂ϕ
=
∂f
∂x
(−r sin ϕ)+
∂f
∂y
r cos ϕ.
f : R
n
−→ R f x

∇f(x)= f(x)=(
∂f
∂x
1
(x), ··· ,
∂f
∂x
n
(x)).
c ∈ R M
c
= {x ∈ R
n
: f(x)=c} = f
−1
(c)
f n =2
γ :(−1, 1) −→ R
n
γ R
n
t
γ

(t)=
dγ(t)
dt
= lim
∆t→0
γ(t +∆t) −γ(t)

∆t
.
γ

(t) γ t
γ

(t) γ γ(t)
γ M
c
γ(t) ∈ M
c
, ∀t
(f ◦ γ)

(t)=f

(γ(t))γ

(t)=< f(γ(t)),γ

(t) >=0.
f(x) M
c
x
M
c
a =(a
1
, ··· ,a

n
)
< f(a),x−a>=0 D
1
f(a)(x
1
− a
1
)+···+ D
n
f(a)(x
n
− a
n
)=0.
v ∈ R
n
f(a + tv)=f(a)+ < f(a),v >t+ o(t).
< f(a),v > f a v
| < f(a),v >|≤ f(a)v =
v = λ f(a) ± f(a)
f
f