Giáo trình giải tích 2 part 1 potx

TRƯỜNG ĐẠI HỌC ĐÀ LẠT
KHOA TOÁN - TIN HỌC
Y  Z



TẠ LÊ LI






GIẢI TÍCH 2

(Giáo Trình)

Lưu hành nội bộ
Y Đà Lạt 2008 Z
R
n
R
n
5
R
n
R
n
R
n
R
n
R
n
X f
n
: X → R n ∈ N
(f
n
)
n∈N
x ∈ X (f
n
(x))
n∈N

D = {x ∈ X : (f
n
(x))
n∈N
}
(f
n
)
D  x → f(x) = lim
n→∞
f
n
(x) (f
n
)
f D
f
n
(x)=1−
1
n
|x| (n ∈ N) R R
f(x) = lim
n→∞
(1 −
1
n
|x|)=1, ∀x.
f
n

(x)=x
n
(n ∈ N) R (−1, 1]
f(x) = lim
n→∞
x
n
=

0 |x| < 1
1 x =1
f
n
f
(f
n
(x)) x ∈ D
(f
n
) f D
>0 N
n ≥ N ⇒|f
n
(x) − f(x)| <, ∀x ∈ D
M
n
=sup
x∈D
|f
n

(x) − f(x)|→0 n →∞
M
n
=sup|f
n
(x) − f(x)| =1
(f
n
) (g
n
) f g D (f
n
+ g
n
) (cf
n
)
f + g cf D
(f
n
) D
∀>0, ∃N : n, m ≥ N ⇒ sup
x∈D
|f
n
(x) − f
m
(x)| <
(f
n

) f D
∀>0, ∃N : n ≥ N ⇒ sup
x∈D
|f
n
(x) − f(x)| </2
m, n ≥ N
sup
x∈D
|f
n
(x) − f
m
(x)| < sup
x∈D
|f
n
(x) − f(x)|+sup
x∈D
|f
m
(x) − f(x)| <.
(f
n
) D x ∈ D
(f
n
(x)) f(x) ∈ R
m →∞  → 0 sup
x∈D

|f
n
(x) −f(x)|→0
n →∞ (f
n
) f D 
(f
n
) f D f
D lim
lim
n→∞
lim
x→x
0
f
n
(x) = lim
x→x
0
lim
n→∞
f
n
(x)
(f
n
) [a, b]
lim


lim
n→∞

b
a
f
n
(x)dx =

b
a
lim
n→∞
f
n
(x)dx
(f
n
) [a, b] (f

n
)
[a, b] (f
n
(c)) c ∈ [a, b] (f
n
)
f [a, b] lim
lim
n→∞

f

n
(x)=

lim
n→∞
f
n
(x)


x
0
∈ D >0
N |f
N
(x) − f(x)| </3, ∀x ∈ D.
f
N
x
0
δ>0 |f
N
(x) −f
N
(x
0
)| </3, ∀x, |x −x
0

| <δ.
|x − x
0
| <δ
|f(x)−f(x
0
)|≤|f(x)−f
N
(x)|+|f
N
(x)−f
N
(x
0
)|+|f
N
(x
0
)−f(x
0
)| </3+/3+/3=
f x
0
lim
x→x
0
f(x) = lim
x→x
0
lim

n→∞
f
n
(x)=f(x
0
) = lim
n→∞
lim
x→x
0
f
n
(x)
f
n
f
[a, b]






b
a
f
n
−

b

a
f





≤|b −a| sup
x∈[a,b]
|f
n
(x) − f(x)|→0, n →∞
lim
n→∞

b
a
f
n
=

b
a
f =

b
a
lim
n→∞
f

n
F
n
(x)=

x
c
f

n
(F
n
) F [a, b]
F (x)=

x
c
lim
n→∞
f

n
F
n
(x)=f
n
(x) − f
n
(c) f
n

= F
n
+ f
n
(c) [a, b]
f = F + lim
n→∞
f
n
(c)
f

(x)=F

(x)=

lim
n→∞

x
c
f

n


= ( lim
n→∞
f
n

)

(x)

X
∞

k=0
f
k
= f
0
+ f
1
+ ···+ f
n
+ ···
f
k
X
n S
n
= f
0
+ ···+ f
n
D = {x ∈ X : (S
n
(x))
n∈N

}
S(x)=
∞

k=0
f
k
(x) D
∞

k=0
f
k
D (S
n
)
n∈N
S D
M
n
=sup
x∈D
|S
n
(x) − S(x)| =sup
x∈D
|
∞

k=n+1

f
k
(x)|→0, n →∞
∞

k=0
x
k
=1+x + x
2
+ ···+ x
n
+ ···
D = {x ∈ R : |x| < 1}
S(x)=
1
1 − x
D
r
= {x : |x|≤r} 0 <r<1
S
n
(x)=
1 − x
n+1
1 − x
sup
|xleqr
|S
n

(x) − S(x)| =sup
|x|≤r





x
n+1
1 − x





≤
r
n+1
1 − r
→ 0, n →∞
D sup
|x|≤1
|S
n
(x) − S(x)| =+∞
∞

k=0
f
k

D
∀>0, ∃N : n, m ≥ N ⇒ sup
x∈D
|
m

k=n
f
k
(x)| <
∞

k=0
f
k
[a, b]
f
k
[a, b] k ∈ N
[a, b] lim

lim
x→x
0
∞

k=0
f
k
(x)=

∞

k=0
lim
x→x
0
f
k
(x)
f
k
[a, b]



b
a

∞

k=0
f
k
(x)

dx =
∞

k=0



b
a
f
k
(x)dx

f
k
[a, b]
∞

k=0
f

k
[a, b]
∞

k=0
f
k
[a, b]


∞

k=0
f
k



(x)=
∞

k=0
f

k
(x)
|f
k
(x)|≤a
k
, ∀x ∈ D
∞

k=0
a
k
∞

k=0
f
k
D
(f
k
) 0
∞


k=0
ϕ
k
D
∞

k=0
f
k
ϕ
k
D
(f
n
)
∞

k=0
ϕ
k
D
∞

k=0
f
k
ϕ
k
|f

k
(x)|≤a
k
m

k=n
|f(x)|≤
m

k=n
a
k
∞

k=0
f
k

∞

k=0
a
k
x
k
x
0
∞

k=0

a
k
(x − x
0
)
k
z = x − x
0
x
0
S(x)=
∞

k=0
a
k
(x −x
0
)
k
R, 0 ≤ R ≤ +∞
R>0
S(x) |x − x
0
| <R |x − x
0
| >R
S D
r
= {x : |x − x

0
|≤r} 0 <r<R
R
S
1
R
= lim sup
k→∞
k

|a
k
|
x
0
0 z = x −x
0
|z|≤r<R ρ : r<ρ<R lim sup k
0
|a
k
|
1
k
<
1
ρ
, ∀k>k
0
. |a

k
z
k
| <

r
ρ

k
S(z)
D
r
S(z) |z| <R
|z| >R ρ : R<ρ<|z| lim sup k
|a
k
|
1
k
>
1
ρ
|a
k
z
k
| >

|z|
ρ


k
k a
k
z
k
→ 0
∞

k=0
a
k
z
k

1
R
= lim
k→∞
|a
k+1
|
|a
k
|
∞

k=0
k!x
k

R = lim
k→∞
|a
n
|
|a
n+1
|
= lim
n→∞
k!
(k +1)!
=0
∞

k=0
x
k
k!
∞
|x−x
0
| = R
∞

k=0
x
k
,
∞


k=1
x
k
k
,
∞

k=1
x
k
k
2
1
|x| =1
∞

k=0
x
k
x = ±1
∞

k=1
x
k
k
2
|x| =1
∞


k=1
x
k
k
x =1 x = −1
∞

k=0
a
k
(x − x
0
)
k
R>0
S(x)=
∞

k=0
a
k
(x −x
0
)
k
(x
0
−R, x
0

+ R)

∞

k=0
a
k
(x − x
0
)
k


=
∞

k=1
ka
k
(x − x
0
)
k−1


∞

k=0
a
k

(x − x
0
)
k

dx =
∞

k=0
a
k
k +1
(x − x
0
)
k+1
+ C

∞

k=0
(−1)
k
x
k
=
1
1+x
|x| < 1
∞


k=1
(−1)
k
kx
k−1
= −
1
(1 + x)
2
|x| < 1
∞

k=0
(−1)
k
x
k+1
k +1
=ln(1+x) |x| < 1
1
1+x
2
=
1
1 − (−x
2
)
=1−x
2

+ x
4
− x
6
+ ···=
∞

k=0
(−1)
k
x
2k
, |x| < 1
arctan x = x −
x
3
3
+
x
5
5
−
x
7
7
+ ···=
∞

k=0
(−1)

k
x
2k+1
2k +1
, |x| < 1
f
k
(x)=x
k
ϕ
k
(x)=a
k
∞

k=0
a
k
S S(x)=
∞

k=0
a
k
x
k
|x| < 1
lim
x→1
−

S(x)=S
ln 2 = 1 −
1
2
+
1
3
−
1
4
+
1
5
−···+
(−1)
n+1
n +1
+ R
n
π
4
=1−
1
3
+
1
5
−
1
7

+
1
9
−···+
(−1)
n
2n +1
+ R
n
R
n
O(
1
n
)
f x
0
f(x)=
∞

k=0
a
k
(x − x
0
)
k
a
k
=

f
(k)
(x
0
)
k!
k =0, 1, 2, ···
n ∈ N x x
0

∞

k=0
a
k
(x − x
0
)
k

(n)
=
∞

k=n
k(k −1) ···(k −n +1)a
k
(x − x
0
)

k−n
x = x
0

f x
0
f x
0
Tf(x)=
∞

k=0
a
k
(x − x
0
)
k
, a
k
=
f
(k)
(x
0
)
k!
Tf(x)=f(x)
Tf(x) f(x)=
∞


k=0
sin 2
k
x
k!
Tf(x) Tf(x) = f(x) f(x)=e
−
1
x
2
x =0 f (0) = 0
f
(k)
(0) = 0, ∀k Tf(x) ≡ 0 = f(x)
Tf(x)=f(x), |x − x
0
| <R f D = {x :
|x − x
0
| <R}
f C |f
(k)
(x)|≤C, ∀x ∈
(x
0
− R, x
0
+ R) f
x ∈ (x

0
−R, x
0
+ R) θ ∈ (0, 1)
|f(x) −T
n
(x)| = |R
n
(x)| =





f
(n+1)
(x
0
+ θR)
(n +1)!
(x − x
0
)
n+1





≤

CR
n+1
(n +1)!