Giáo trình giải tích 2 part 2 pdf

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 kx sin 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 kx cos lx)
f(x)sinlx =
a
0
2

sin lx +
∞

k=1
(a
k
cos kx sin lx + b
k
sin kx sin 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 + bF g 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 ku cos 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