tài liệu toán giải tích 2

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)!
0 n →∞ f(x)=Tf(x) 
e
x
=1+x +
1
2!
x
2
+ ···+
1
n!
x
n
+ ···
cos x =1−
1
2!
x
2
+
1
4!
x
4
+ ···+
(−1)
n

(2n)!
x
2n
+ ···
sin x = x −
1
3!
x
3
+
1
5!
x
5
+ ···+
(−1)
n
(2n +1)!
x
2n+1
+ ···
1
1 − x
=1+x + x
+
···+ x
n
+ ··· , |x| < 1
ln(1 + x)=x −
1

2
x
2
+
1
3
x
3
+ ···+
(−1)
n+1
n
xan + ··· , |x| < 1
(1 + x)
α
=1+αx +
α(α − 1)
2!
x
2
+ ···+
α(α − 1) ···(α − n +1)
n!
x
n
+ ··· , |x| < 1
(x)=

x
0

e
−t
2
dt
e
x
x = −t
2
e
−t
2
=1−t
2
+
1
2!
t
4
+ ···+
(−1)
n
n!
t
2n
+ ···
erf(x)=x −
x
3
3
+

x
2
2!5
+ ···+
(−1)
n
n!(2n +1)
x
2n+1
+ ···=
∞

k=0
(−1)
k
k!(2k +1)
x
2k+1
x ∈ R
(x)=

x
0
sin t
t
dt sin x
Si(x)=

x
0

(1−
1
3!
t
2
+
1
5!
t
4
+···+
(−1)
n
(2n +1)!
t62n+···)dt =
∞

k=0
(−1)
k
(2k + 1)!(2k +1)
x
2n+1
ln 2
ln(1 + x)
ln(1 − x)=x +
1
2
x
2

+
1
3
x
3
+ ···+
x
n
n
+ ··· , |x| < 1
ln(1 + x) − ln(1 − x)
ln

1+x
1 − x

=2(x +
1
3
x
3
+ ···+
x
2n+1
2n +1
+ ···), |x| < 1
x =
1
3
ln 2 = 2(

1
3
+
1
3.3
3
+ ···+
1
(2n +1)3
2n+1
)+R
n
R
n
=

k>n
1
(2k +1)3
2k+1
<
1
3(2n +3)

k>n
1
9
k
=
1

3(2n +1)
(1/9)
n
1 − 1/9
= o(
1
9
n
)
sin cos
a
0
2
+
∞

k=1
(a
k
cos kx + b
k
sin kx)
f T ϕ(x)=f(
T
2π
x) 2π
2π
[−π, π]
<f,g>=


π
−π
f(x)g(x)dx f,g ∈ C[−π, π]
1, cos x, sin x, cos 2x, sin 2x, ··· , cos nx, sin nx, ···
0

π
−π
cos kx cos lxdx =0 k = l

π
−π
sin kxsin lxdx =0 k = l

π
−π
cos kx sin lxdx =0 ∀k, l

π
−π
dx =2π,

π
−π
cos
2
kxdx =

π
−π

sin
2
kxdx = πk=1, 2, ···
f
f(x)=
a
0
2
+
∞

k=1
(a
k
cos kx + b
k
sin kx),x∈ [−π, π]
f(x)coslx =
a
0
2
cos lx +
∞

k=1
(a
k
cos kx cos lx + b
k
sin kxcos lx)

f(x)sinlx =
a
0
2
sin lx +
∞

k=1
(a
k
cos kx sin lx + b
k
sin kxsin lx)
a
k
=
1
π

π
−π
f(x)coskxdx, k =0, 1, 2, ···
b
k
=
1
π

π
−π

f(x)sinkxdx, k =1, 2, ···
f
f [−π, π]
f
Ff(x)=
a
0
2
+
∞

k=1
(a
k
cos kx + b
k
sin kx)
a
k
,b
k
f
• f f(−x)=f(x) f(x)sinkx b
k
=0
Ff(x)=
1
2
a
0

+
∞

k=1
a
k
cos kx
• f f(−x)=−f(x) f(x)coskx a
k
=0
Ff(x)=
∞

k=1
b
k
sin kx
• F (af + bg)=aF f + bFg f, g a, b ∈ R
f(x), |x|≤π Ff(x)
x
4
π
∞

k=0
sin(2k +1)x
2k +1
x 2
∞


k=1
(−1)
k+1
sin kx
k
x
2
π
2
3
+4
∞

k=1
(−1)
k
cos kx
k
2
Ax
2
+ Bx + C A
π
2
3
+ C +4A
∞

k=1
(−1)

k
cos kx
k
2
+2B
∞

k=1
(−1)
k+1
sin kx
k
Ff(x)=f(x)
Ff(x) 2π
Ff(x) Ff(x) = f (x)
Ff(x)=f(x)
Ff(x)=f(x)
n f
F
n
f(x)=
a
0
2
+
n

k=1
(a
k

cos kx + b
k
sin kx)
F
n
f
F
n
f(x)=
a
0
2
+
n

k=1
(a
k
cos kx + b
k
sin kx)
=
1
2π

π
−π
f(u)du +
n


k=1
1
π

π
−π
f(u)(cos kucos kx +sinku sin kx)du
=
1
π

π
−π
f(u)

1
2
+
n

k=1
cos k(u − x)

du
g T

a+T
a
g(t)dt =


T
0
g(t)dt
t = u − x T =2π a = −π − x
F
n
f(x)=
1
π

π
−π
f(x + t)

1
2
+
n

k=1
cos kt

dt =

π
−π
f(x + t)D
n
(t)dt
D

n
(t)=
1
π

1
2
+
n

k=1
cos kt

2sin
t
2
cos kt =sin(k +
1
2
)t − sin(k −
1
2
)t
D
n
(t)=
1
π
sin
2n +1

2
t
2sin
t
2
D
n
2π

π
−π
D
n
(t)dt =1
g [a, b]
lim
λ→+∞

b
a
g(t)cosλtdt = lim
λ→+∞

b
a
g(t)sinλtdt =0
g
lim
λ→+∞


b
a
g(t)cosλtdt =
g(t)sinλt
λ




b
a
−
1
λ

b
a
g

(t)sinλtdt
g

→ 0 λ → +∞
g
g

g >0
s

π

−π
|g −s| <

b
a
g(t)cosλtdt =

b
a
(g(t) − s(t)) cos λtdt +

b
a
s(t)cosλtdt
s |cos λx|≤1
lim
λ→+∞






b
a
g(t)cosλtdt






≤

b
a
|g(t) − s(t)|dt < 
lim
λ→+∞

b
a
g(t)cosλtdt =0 
f
[a, b]
a = a
0
<a
1
< ···<a
s
= b f (a
i−1
,a
i
)
lim
x→a
+
i
f(x)=f(a

+
i
) lim
x→a
−
i
f(x)=f(a
−
i
) i =0, ··· ,s
f x
f

+
(x) = lim
t→0
+
f(x + t) −f(x6+)
t
,f

−
(x) = lim
t→0
+
f(x −t) −f(x
−
)
t
,

f(x)=|x| 0 f

+
(0) = 1,f

−
(0) = −1
f(x)= x 0
f(0
+
)=1,f(0
−
)=−1 f

(0
+
)=f

−
(0) = 0
f 2π [−π, π] f

+
(x),f

−
(x)
F
n
f(x) f x

Ff(x)=
1
2
(f(x
+
)+f(x
−
))
f x Ff(x)=f(x)
A
f
(x)=
1
2
(f(x
+
)+f(x
−
))
D
n
F
n
f(x) −A
f
(x)=

π
−π
(f(x + t) − A

f
(x))D
n
(t)dt
=2

π
0

f(x + t)+f(x −t)
2
− A
f
(x)

D
n
(t)dt
=2

π
0
g(t)sin(n +
1
2
)tdt
g(t)=
f(x + t) −f(x
+
)+f(x −t) −f(x

−
)
t
t
2π sin
t
2
f

+
(x),f

−
(x) lim
t→0
+
g(t)=
1
π
(f

+
(x) − f

−
(x)) g
0 n →∞
F
n
f(x) → A

f
(x) n →∞ 
x =
4
π
∞

k=0
sin(2k +1)π
2k +1
0 < |x| <π
x =0, −π, π
1
2
( (x
+
)+ (x
−
)) = 0
x = π/2
∞

k=0
(−1)
k
2k +1
=
π
4
1 −

x
2
π
2
=
2
3
−
4
π
2
∞

k=1
(−1)
k
cos kx
k
2
|x|≤π
x = ±π
x = π
∞

k=1
1
k
2
=
π

2
6
x =0
∞

k=1
(−1)
k
k
2
= −
π
2
12
∞

k=1
1
(2k −1)
2
=
1
2

∞

k=1
1
k
2

−
∞

k=1
(−1)
k
k
2

=
π
2
8
f
2
[π, π]
a
2
0
2
+
∞

k=1
(a
2
k
+ b
2
k

) ≤
1
π

π
−π
f
2
(x)dx

π
−π
(f(x)−F
n
f(x))F
n
f(x)dx =0

π
−π
(F
n
f(x))
2
dx = π

a
2
0
2

+
n

k=1
(a
2
k
+ b
2
k
)


π
−π
f
2
(x)dx =

π
−π
(f(x) −F
n
f(x)+F
n
f(x))
2
dx
=


π
−π
(f(x) −F
n
f(x))
2
dx +

π
−π
(F
n
f(x))
2
dx +2

π
−π
(f(x) −F
n
f(x))F
n
f(x)dx
=

−π
6π(f(x) − F
n
f(x))
2

dx + π(
a
2
0
2
+
n

k=1
(a
2
k
+ b
2
k
))
a
2
0
2
+
n

k=1
(a
2
k
+ b
2
k

) ≤

π
−π
f
2
(x)dx
n → +∞

f 2π f

[−π, π]
Ff f R
F
n
f(x) f(x)
a

k
,b

k
f

a
k
=
1
π


π
−π
f(x)coskxdx =
1
π

f(x)
sin kx
k
|
π
−π
−
1
k

π
−π
f

(x)sinkxdx

= −
1
k
b

k
b
k

=
1
π

π
−π
f(x)sinkxdx =
1
π

−f(x)
cos kx
k
|
π
−π
+
1
k

π
−π
f

(x)coskxdx

=
1
k
a


k
|a
k
cos kx + b
k
sin kx|≤|a
k
| + |b
k
|≤
1
2
(b

2
k
+
1
k
2
)+
1
2
(a

2
k
+
1

k
2
)
∞

k=0
(a

2
k
+ b

2
k
)
∞

k=1
1
k
2
Ff

• f(x) T x =
T
2π
X
f(x)=f(
T
2π

X) 2π X
X
a
0
2
+
∞

k=1
( a
k
cos kX + b
k
sin kX )
a
k
=
1
π

π
−π
f(
T
2π
X)coskXdX, b
k
=
1
π


π
−π
f(
T
2π
X)sinkXdX
X =
2π
T
x
a
0
2
+
∞

k=1
( a
k
cos
2kπ
T
x + b
k
sin
2kπ
T
x )
f

a
k
=
2
T

T/2
−T/2
f(t)cos
2kπ
T
tdt, k =0, 1, 2, ···
b
k
=
2
T

T/2
−T/2
f(t)sin
2kπ
T
tdt, k =1, 2 , ···
• f [a, b]
f
˜
f R T ≥ b − a
˜
f(x + kT)=f(x),x∈ [a, b],k ∈ Z

˜
f
• cos sin f [0,l]
f(x) cos f
(−l, l] f(x)=f(−x) x ∈ (−l, 0)
f(x) sin f
(−l, l] f(x)=−f(−x) x ∈ (−l, 0)
[−π, π] 2π
f(x)= x, x ∈ [−π, π] Ff(x)=
4
π
∞

k=0
sin(2k +1)x
2k +1
✲
x
✻
y
✲
✲
rr
✲
✲
rr
✲
✲
rr
✲

✲
rr
✲
✲
rr
−ππ
f(x)=x, x ∈ [−π,π] Ff(x)=2
∞

k=1
(−1)
k+1
sin kx
k
✲
x
✻
y




✒
r




✒
r





✒
r




✒
r




✒
r
−ππ
f(x)=x
2
,x∈ [−π,π] Ff(x)=
π
2
3
+4
∞

k=1
(−1)

k
cos kx
k
2
✲
x
✻
y
rr
−ππ
[0, 2π] 2π
f(x), 0 ≤ x<2π Ff(x)
x π −2
∞

k=1
sin kx
k
x
2
4
3
π
2
+4
∞

k=1
cos kx
k

2
− 4π
∞

k=1
sin kx
k
Ax
2
+ Bx + C A
4
3
π
2
+ Bπ + C +4A
∞

k=1
cos kx
k
2
− (4πA − 2B)
∞

k=1
sin kx
k
Ff(x)=x, 0 <x<2π
✲
x





✒
r




✒
r




✒
r




✒
r




✒
r

02π
Ff(x)=x
2
, 0 <x<2π
✲
x
✕
r
✕
r
✕
r
✕
r
✕
r
02π
f(x)
f(x)=x, x ∈ [−π,π] f (x)=x, x ∈ [0, 2π]
2π
f(x)=x, x ∈ [0,π]
f(x) cos f
f(x)=|x|,x∈ [−π, π] f
|x| =
π
2
−
4
π
∞


k=1
cos(2k +1)x
(2k +1)
2
, −π ≤ x ≤ π
✲
x
✻
y


❅
❅
❅


❅
❅
❅


❅
❅
❅


❅
❅
❅



❅
❅
❅
−ππ
f(x) sin f
f(x)=x, x ∈ [−π,π] f
x =2
∞

k=1
(−1)
k+1
sin kx
k
, −π<x<π
✲
x
✻
y


✒



r 

✒




r 

✒



r 

✒



r 

✒



r
−ππ
∞

k=1
sin kx
k
=
π −x

2
0 <x<2π
∞

k=1
cos kx
k
2
=
3x
2
− 6πx +2π62
12
0 <x<2π
∞

k=1
(−1)
k+1
sin kx
k
=
x
2
|x| <π
∞

k=1
(−1)
k+1

cos kx
k
2
=
π
2
− 3x
2
12
|x| <π
∞

k=0
sin(2k +1)x
2k +1
=
π
4
0 <x<π
∞

k=0
cos(2k +1)x
(2k +1)
2
=
π
2
− 2πx
8

0 <x<2π
∞

k=1
sin 2kx
2k
=
π −2x
4
0 <x<π
∞

k=1
cos 2kx
(2k)
2
=
6x
2
− 6πx + π
2
24
0 <x<2π
x
∞

k=1
1
k
2

=
π
2
6
,
∞

k=1
(−1)
k+1
k
2
=
π
2
12
,
∞

k=0
(−1)
k
2k +1
=
π
4
R
n
R
n

R
n
R
n
= {x =(x
1
, ··· ,x
n
):x
i
∈ R,i =1, ··· ,n}
x + y =(x
1
, ··· ,x
n
)+(y
1
, ··· ,y
n
)=(x
1
+ y
1
, ··· ,x
n
+ y
n
)
αx = α(x
1

, ··· ,x
n
)=(αx
1
, ··· ,αx
n
),α∈ R.
R
n
n R
e
1
=(1, 0, ··· , 0), ··· ,e
n
=(0, ··· , 0, 1).
x =(x
1
, ··· ,x
n
)=
n

i=1
x
i
e
i
0=(0, ··· , 0)
R
n

x =(x
1
, ··· ,x
n
),y =(y
1
, ··· ,y
n
) ∈ R
n
<x,y>= x
1
y
1
+ ···+ x
n
y
n
.
x =
√
<x,x> =(x
2
1
+ ···+ x
2
n
)
1
2

.
d(x, y)= x −y = {(x
1
− y
1
)
2
+ ···+(x
n
− y
n
)
2
}
1
2
.
x, y, z ∈ R
n
α, β ∈ R
(S1) <αx+ βy,z > = α<x,y>+β<x,z>.
(S2) <x,y> = <y,x>.
(S3) <x,x> ≥ 0, <x,x>=0 x =0.
(N1) x≥0, x =0 x =0.
(N2) αx = |α|x.
(N3) x + y≤x + y.
(M1) d(x, y) ≥ 0, d(x, y)=0 x = y.
(M2) d(x, y)=d(y, x).
(M3) d(x, y) ≤ d(x, z)+d(z, y).
(N3)

| <x,y>|≤xy.
tx + y
2
= x
2
t
2
+2<x,y>t+ y
2
≥ 0 ∀t ∈ R
∆=<x,y>
2
−x
2
y
2
≥ 0
x + y
2
= x
2
+ y
2
+2<x,y>≤x
2
+ y
2
+2xy =(x+ y)
2
(N3)

(N3) (M3) 
| <x,y>| = xy x, y
max
1≤i≤n
|x
i
|≤x≤
√
n max
1≤i≤n
|x
i
|.
R
n
X ⊂ R
n
x : N −→ X x(k)=x
k
=(x
k,1
, ··· ,x
k,n
)
(x
k
)
k∈N
(x
k

)
(x
k
) a ∈ R
n
lim
k→∞
x
k
= a x
k
→ a
∀>0, ∃N : k ≥ N =⇒ d(x
k
,a) <.
(x
k
) a =(a
1
, ··· ,a
n
) ∈ R
n
lim
k→∞
x
k
= a lim
k→∞
x

k,i
= a
i
,i=1, ··· ,n.
lim
k→∞
x
k
x
k
=

1
k
p
,
k
100
e
k
,
ln k
k
p
,
k
√
2,
k
√

k
p
,
1
k
√
k!

(p>0).
(x
k
)
∀>0, ∃N : k, l ≥ N =⇒ d(x
k
,x
l
) <.
R
n
R

R
n
R
n
a ∈ R
n
r>0
a r B(a, r)={x ∈ R
n

: d(x, a) <r}.
a r B(a, r)={x ∈ R
n
: d(x, a) ≤ r}.
1, 2, 3
X ⊂ R
n
a ∈ R
n
a
X ∃r>0: B(a, r) ⊂ X
a
X ∀r>0: B(a, r) ∩ X = ∅,B(a, r) ∩ (R
n
\ X) = ∅
[α, β] R x α<x<β
α, β
Q R
X ⊂ R
n
X
∀a ∈ X, ∃r>0:B(a, r) ⊂ X
X
o
X
X X
X X =
o
X
R

∅ R
n
U
i
,i ∈ I x ∈ U =

i∈I
U
i
i
0
∈ I,x ∈ U
i
0
B(x, r) ⊂ U
i
0
⊂ U x
U U 

i∈N
(−
1
i
,
1
i
)
X ⊂ R
n

R
n
\ X
Z [a, b]
Q
∅ R
n
a ∈ R
n
X ∀r>0,B(a, r) ∩ X
a
X
X = X∪ X X
R [a, b) {1/k :
k ∈ N} Q
X ⊂ R
n
X X =
X X
(x
k
) X x x ∈ X
⇒ x X ∀r>0,B(x, r)∩X =
∅ ∀r>0,B(x, r) ⊂ R
n
\ X x ∈ (R
n
\ X)=R
n
\ X

x ∈ X
⇒
⇒ (x
k
) ⊂ X x
k
→ x {x
k
}
k
0
,x = x
k
0
x ∈ X {x
k
} x
X x ∈ X
⇒ R
n
\ X x ∈ R
n
\ X
∀r>0,B(x, r) ∩ X = ∅ x X
x ∈ X 
K ⊂ R
n
K
K R>0: K ⊂ B(0,R)
[a, b] R

B(a, r)
[a
1
,b
1
] ×···×[a
n
,b
n
] R
n
P = {U
i
,i ∈ I} I
K R
n
i ∈ I U
i
R
n
K ⊂

i∈I
U
i
.
(a −
1
k
,b +

1
k
),k∈ N [a, b]
(a, a +1),a∈ R R
K R
n
K
K
(x
k
) K (x
σ(k)
) x x ∈ K
K
P = {U
i
,i∈ I} K {U
i
1
, ··· ,U
i
s
} K
⇔ ⇔
⇒ (x
k
) ⊂ K R>0 x
k
 <R
(x

k,i
)
k∈N
, (i =1, ···n)
R (x
k,1
) (x
σ
1
(k),1
) a
1
(x
k
) (x
σ(k)
) σ : N → N