Bài tập toán cao cấp tập 3 part 7 docx

198 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
23.

n1
(−1)
n−1
(2n + 1)!!
2 · 5 ·8 ···(3n −1)
.(D
S. Hˆo
.
itu
.
tuyˆe
.
tdˆo
´
i)
24.

n1
(−1)
n+1


1 −cos
π
√
n

.(D
S. Hˆo
.
itu
.
c´o diˆe
`
ukiˆe
.
n)
25.

n1
(−1)
n
sin
π
n
n
.(D
S. Hˆo
.
itu
.

tuyˆe
.
tdˆo
´
i)
26.

n1
(−1)
n
√
n +2
.(D
S. Hˆo
.
itu
.
c´o diˆe
`
ukiˆe
.
n)
27.

n1
(−1)
n
n
√
n

.(D
S. Phˆan k`y)
28.

n1
(−1)
n+1
n −ln n
.(D
S. Hˆo
.
itu
.
c´o diˆe
`
ukiˆe
.
n)
29.

n1
(−1)
n−1
(n +1)a
2n
.
(D
S. Hˆo
.
itu

.
tuyˆe
.
tdˆo
´
i khi |a| > 1, hˆo
.
itu
.
c´o diˆe
`
ukiˆe
.
n khi |a| =1,
phˆan k`y khi |a| < 1)
30.

n1
(−1)
n
(n + 1)(
√
n +1− 1)
.(D
S. Hˆo
.
itu
.
tuyˆe
.

tdˆo
´
i)
31.

n1
(−1)
n+1

2+
1
n

n
5
n
.(DS. Hˆo
.
itu
.
tuyˆe
.
tdˆo
´
i)
32.

n1
(−1)
n

tg
π
3
n
.(DS. Hˆo
.
itu
.
tuyˆe
.
tdˆo
´
i)
Trong c´ac b`ai to´an sau d
ˆay, h˜ay t`ım sˆo
´
sˆo
´
ha
.
ng cu
’
a chuˆo
˜
id˜acho
cˆa
`
nlˆa
´
yd

ˆe
’
tˆo
’
ng cu
’
ach´ung v`a tˆo
’
ng cu
’
achuˆo
˜
itu
.
o
.
ng ´u
.
ng sai kh´ac nhau
mˆo
.
td
a
.
ilu
.
o
.
.
ng khˆong vu

.
o
.
.
t qu´a sˆo
´
δ cho tru
.
´o
.
c
33.

n1
(−1)
n−1
1
2n
2
, δ =0, 01. (DS. No =7)
13.3. Chuˆo
˜
il˜uy th`u
.
a 199
34.

n1
cos(nπ)
n!

, δ =0, 001. (D
S. No =5)
35.

n1
(−1)
n−1
√
n
2
+1
, δ =10
−6
.(DS. No =10
6
)
36.

n1
cos nπ
2
n
(n +1)
, δ =10
−6
.(DS. No = 15)
37.

n1
(−1)

n
2n
(4n + 1)5
n
, δ =0, 1?; δ =0, 01? (DS. No =2,No =3)
38.

n1
(−1)
n
n!
, δ =0,1; δ =0, 001? (D
S. No =4,No =6)
13.3 Chuˆo
˜
il˜uy th`u
.
a
13.3.1 C´ac di
.
nh ngh˜ıa co
.
ba
’
n
Chuˆo
˜
il˜uy th`u
.
ad

ˆo
´
iv´o
.
ibiˆe
´
n thu
.
.
c x l`a chuˆo
˜
ida
.
ng

n0
a
n
x
n
= a
0
+ a
1
x + a
2
x
2
+ ···+ a
n

x
n
+ (13.6)
hay

n0
a
n
(x −a)
n
= a
0
+ a
1
(x −a)+···+ a
n
(x −a)
n
+ (13.7)
trong d
´o c´ac hˆe
.
sˆo
´
a
0
,a
1
, ,a
n

, l`a nh˜u
.
ng h˘a
`
ng sˆo
´
.B˘a
`
ng ph´ep d
ˆo
’
i
biˆe
´
n x bo
.
’
i x − a t`u
.
(13.6) thu d
u
.
o
.
.
c (13.7). Do d
´odˆe
’
tiˆe
.

n tr`ınh b`ay
ta chı
’
cˆa
`
n x´et (13.6) l`a d
u
’
(t ´u
.
c l`a xem a = 0).
Chuˆo
˜
i (13.6) luˆon hˆo
.
itu
.
ta
.
id
iˆe
’
m x = 0, c`on (13.7) hˆo
.
itu
.
ta
.
i x = a.
Do d

´otˆa
.
pho
.
.
pd
iˆe
’
m m`a chuˆo
˜
il˜uy th`u
.
ahˆo
.
itu
.
luˆon luˆon = ∅.
D
ˆo
´
iv´o
.
ichuˆo
˜
il˜uy th `u
.
abˆa
´
t k `y (13.6) luˆon luˆon tˆo
`

nta
.
isˆo
´
thu
.
.
c
R :0 R  +∞ sao cho chuˆo
˜
id
´ohˆo
.
itu
.
tuyˆe
.
tdˆo
´
i khi |x| <Rv`a
phˆan k`y khi |x| >R.Sˆo
´
R d
´odu
.
o
.
.
cgo
.

il`ab´an k´ınh hˆo
.
itu
.
cu
’
a chuˆo
˜
i
200 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
(13.6) v`a khoa
’
ng I(R)=(−R, R)du
.
o
.
.
cgo
.
il`akhoa
’
ng hˆo

.
itu
.
cu
’
achuˆo
˜
i
l˜uy th `u
.
a (13.6).
B´an k´ınh hˆo
.
itu
.
R cu
’
a chuˆo
˜
il˜uy th`u
.
ac´othˆe
’
t´ınh thˆong qua c´ac
hˆe
.
sˆo
´
cu
’

an´obo
.
’
imˆo
.
t trong c´ac cˆong th´u
.
c
R = lim
n→∞
|a
n
|
|a
n+1
|
, (13.8)
ho˘a
.
c
R = lim
n→∞
1
n

|a
n
|
(13.9)
nˆe

´
u gi´o
.
iha
.
no
.
’
vˆe
´
pha
’
icu
’
a (13.8) v`a (13.9) tˆo
`
nta
.
i.
D
-
i
.
nh ngh˜ıa 13.3.1. Ngu
.
`o
.
i ta n´oi r˘a
`
ng h`am f(x) khai triˆe

’
nd
u
.
o
.
.
c
th`anh chuˆo
˜
il˜uy th`u
.
a

n0
a
n
x
n
trˆen khoa
’
ng ( −R, R)nˆe
´
u trˆen khoa
’
ng
d
´o c h u ˆo
˜
id˜anˆeuhˆo

.
itu
.
v`a tˆo
’
ng cu
’
a n´o b˘a
`
ng f(x), t´u
.
cl`a
f(x)=

n0
a
n
x
n
,x∈ (−R, R).
D
i
.
nh ngh˜ıa 13.3.2. 1
+
Chuˆo
˜
il˜uy th`u
.
ada

.
ng
f(x
0
)+
f

(x
0
)
1!
(x −x
0
)+···+
f
(n)
(x
0
)
n!
(x − x
0
)
n
+
=

n0
f
(n)

(x
0
)
n!
(x −x
0
)
n
(13.10)
d
u
.
o
.
.
cgo
.
i l`a chuˆo
˜
i Taylor cu
’
a h`am f(x)v´o
.
i tˆam ta
.
id
iˆe
’
m x
0

(o
.
’
d
ˆay
0! = 1, f
(0)
(x
0
)=f(x
0
)).
2
+
C´ac hˆe
.
sˆo
´
cu
’
a chuˆo
˜
i Taylor
a
0
= f(x
0
),a
1
=

f

(x
0
)
1!
, ,a
n
=
f
(n)
(x
0
)
n!
(13.11)
d
u
.
o
.
.
cgo
.
il`ac´achˆe
.
sˆo
´
Taylor cu
’

a h`am f(x).
13.3. Chuˆo
˜
il˜uy th`u
.
a 201
3
+
Khi x
0
= 0, chuˆo
˜
i Taylor
f(0) +
f

(0)
1!
x + ···+
f
(n)
(0)
n!
x
n
+ ···=

n0
f
(n)

(0)
n!
x
n
(13.12)
d
u
.
o
.
.
cgo
.
il`achuˆo
˜
i Maclaurin.
13.3.2 D
-
iˆe
`
ukiˆe
.
n khai triˆe
’
nv`aphu
.
o
.
ng ph´ap khai
triˆe

’
n
D
-
i
.
nh l´y 13.3.1 (Tiˆeu chuˆa
’
n khai triˆe
’
n). H`am f(x) khai triˆe
’
ndu
.
o
.
.
c
th`anh chuˆo
˜
il˜uy th`u
.
a

n0
a
n
x
n
trˆen khoa

’
ng (−R, R) khi v`a chı
’
khi trˆen khoa
’
ng d´o h`am f(x) c´o da
.
o
h`am mo
.
icˆa
´
p v`a trong cˆong th´u
.
c Taylor
f(x)=f(0) +
f

(0)
1!
x + ···+
f
(n)
(0)
n!
x
n
+ R
n
(x)

phˆa
`
ndu
.
R
n
(x) → 0 khi n →∞∀x ∈ (−R, R).
Trong thu
.
.
c h`anh ngu
.
`o
.
i ta thu
.
`o
.
ng su
.
’
du
.
ng dˆa
´
uhiˆe
.
ud
u
’

nhu
.
sau.
D
-
i
.
nh l´y 13.3.2. D
ˆe
’
h`am f(x) khai triˆe
’
ndu
.
o
.
.
c th`anh chuˆo
˜
il˜uy th`u
.
a

n0
a
n
x
n
,x∈ (−R, R)
d

iˆe
`
ukiˆe
.
ndu
’
l`a trˆen khoa
’
ng d´o h`am f(x) c´o da
.
o h`am mo
.
icˆa
´
p v`a c´ac
d
a
.
oh`amd´obi
.
ch˘a
.
n, t´u
.
cl`a∃M>0:∀n =0, 1, 2, v`a ∀x ∈
(−R, R) th`ı
|f
(n)
(x)|  M.
Ta nˆeu ra d

ˆay hai phu
.
o
.
ng ph´ap khai triˆe
’
n h`am th`anh chuˆo
˜
il˜uy
th `u
.
a
202 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
1. Phu
.
o
.
ng ph´ap I (phu
.
o
.
ng ph´ap tru

.
.
ctiˆe
´
p) gˆo
`
m c´ac bu
.
´o
.
c sau:
a) T´ınh c´ac hˆe
.
sˆo
´
theo cˆong th´u
.
c (13.11)
b) Ch´u
.
ng to
’
r˘a
`
ng lim
n→∞
R
n
(x)=0.
Nhu

.
o
.
.
cd
iˆe
’
mcu
’
aphu
.
o
.
ng ph´ap n`ay l`a t´ınh to´an qu´a cˆo
`
ng kˆe
`
nh v`a
sau n˜u
.
a l`a viˆe
.
c kha
’
o s´at gi´o
.
iha
.
n R
n

(x) → 0(n →∞)la
.
i c`ang ph´u
.
c
ta
.
pho
.
n.
2. Phu
.
o
.
ng ph´ap II (phu
.
o
.
ng ph´ap gi´an tiˆe
´
p) l`a phu
.
o
.
ng ph´ap du
.
.
a
trˆen ba
’

ng c´ac khai triˆe
’
n “c´o s˘a
˜
n” (hay Khai triˆe
’
nba
’
ng)c`ung v´o
.
i c´ac
ph´ep t´ınh d
ˆo
´
iv´o
.
i chuˆo
˜
il˜uy th`u
.
a.
I. e
x
=1+x +
x
2
2!
+ ···+
x
n

n!
−···=

n0
x
n
n!
, x ∈ R.
II. sin x = x −
x
3
3!
+
x
5
5!
−···+(−1)
n
x
2n+1
(2n + 1)!
+ ···=
=

n0
(−1)
n
x
2n+1
(2n + 1)!

, x ∈ R.
III. cosx =1−
x
2
2!
+
x
4
4!
−···+(−1)
n
x
2n
(2n)!
+ ···=
=

n0
(−1)
n
x
2n
(2n)!
, x ∈ R.
IV.
(1 + x)
α
=1+αx +
α(α − 1)
2!

x
2
+ ···+
α(α −1)···(α −n +1)
n!
x
n
+
=1+

n1

α
n

x
n
, −1 <x<1,

α
0

=1,

α
n

=
α(α −1)···(α −n +1)
n!

,

α
n

= C
α
n
nˆe
´
u α ∈ N.
13.3. Chuˆo
˜
il˜uy th`u
.
a 203
Khi α = −1 ta c´o
1
1+x
=1−x + x
2
−···+(−1)
n
x
n
+
=

n0
(−1)

n
x
n
, −1 <x<1.
1
1 −x
=1+x + x
2
+ ···+ x
n
+ ···=

n0
x
n
, −1 <x<1.
V.
ln(1 + x)=x −
x
2
2
+
x
3
3
−···+(−1)
n−1
x
n
n

+ ; −1 <x<1
ln(1 −x)=−x −
x
2
2
−···−
x
n
n
− , −1 <x<1.
C
´
AC V
´
IDU
.
V´ı du
.
1. T`ım miˆe
`
nhˆo
.
itu
.
cu
’
a chuˆo
˜
il˜uy th`u
.

a

n1
(−1)
n−1
nx
n
.
Gia
’
i. 1
+
Tas˜e´apdu
.
ng cˆong th´u
.
c (13.8). V`ı a
n
=(−1)
n−1
n v`a
a
n+1
=(−1)
n
(n + 1) nˆen ta c´o
R = lim
n→∞
|a
n

|
|a
n+1
|
= lim
n→∞
n
n +1
=1.
Nhu
.
vˆa
.
y chuˆo
˜
ihˆo
.
itu
.
b´o
.
i −1 <x<1.
2
+
Ta c`on cˆa
`
n kha
’
o s´at su
.

.
hˆo
.
itu
.
cu
’
a chuˆo
˜
ita
.
i c´ac d
ˆa
`
um´ut cu
’
a
khoa
’
ng hˆo
.
itu
.
.
V´o
.
i x = −1 ta c´o

n1
(−1)

n−1
n(−1)
n
=

n1
(−1)
2n−1
n =

n1
(−n).
Do d
´o chuˆo
˜
id˜a cho phˆan k`y ta
.
idiˆe
’
m x = −1 (khˆong tho
’
a m˜an diˆe
`
u
kiˆe
.
ncˆa
`
n!)
204 Chu

.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
V´o
.
i x = 1 ta c´o

n1
(−1)
n−1
n ⇒ lim
n→∞
(−1)
n−1
n khˆong tˆo
`
nta
.
i
Do d
´o chuˆo
˜
i phˆan k`yta
.
idiˆe

’
m x = 1. Vˆa
.
ymiˆe
`
nhˆo
.
itu
.
cu
’
a chuˆo
˜
il`a
(−1, 1). 
V´ı d u
.
2. T`ım khoa
’
ng hˆo
.
itu
.
cu
’
a chuˆo
˜
i

n1

(−1)
n
(x − 2)
n
n
n
·
Gia
’
i. Trong tru
.
`o
.
ng ho
.
.
p n`ay ta su
.
’
du
.
ng cˆong th´u
.
c (13.9) v`a thu
d
u
.
o
.
.

c
R = lim
n→∞
1
n

|a
n
|
= lim
n→∞
1
n




(−1)
n
n
n



= lim
n→∞
n =+∞.
D
iˆe
`

ud´o c´o ngh˜ıa l`a chuˆo
˜
id˜achohˆo
.
itu
.
v´o
.
imo
.
i gi´a tri
.
x,t´u
.
cl`a
I(R)=(−∞, +∞).
V´ı d u
.
3. T`ım khoa
’
ng hˆo
.
itu
.
cu
’
a chuˆo
˜
i


n0
n!x
n
, 0! ≡ 1.
Gia
’
i.
´
Ap du
.
ng cˆong th´u
.
c (13.8) ta c´o
R = lim
n→∞
|a
n
|
|a
n+1
|
= lim
n→∞
n!
n!(n +1)
= lim
n→∞
1
n +1
=0.

Vˆa
.
y R =0.D
iˆe
`
ud´o c´o ngh˜ıa r˘a
`
ng chuˆo
˜
id˜a cho hˆo
.
itu
.
ta
.
idiˆe
’
m x =0.

V´ı du
.
4. Khai triˆe
’
n h`am
1
4 −x
th`anh chuˆo
˜
il˜uy th`u
.

ata
.
i lˆan cˆa
.
nd
iˆe
’
m
x
0
=2(c˜ung t´u
.
c l`a: theo c´ac l˜uy th`u
.
acu
’
ahiˆe
.
u x − 2 hay chuˆo
˜
il˜uy
th `u
.
av´o
.
i tˆam ta
.
id
iˆe
’

m x
0
= 2).
13.3. Chuˆo
˜
il˜uy th`u
.
a 205
Gia
’
i. Ta biˆe
´
ndˆo
’
i h`am d˜achodˆe
’
c´o thˆe
’
´ap du
.
ng khai triˆe
’
nba
’
ng:
1
1 −t
=

n0

t
n
, −1 <t<1.
Ta c´o
1
4 −x
=
1
2
·
1
1 −
x −2
2
Xem t =
x −2
2
ta c´o:
1
4 −x
=
1
2

1+

x −2
2

+


x −2
2

2
+ ···+

x − 2
2

n
+

⇒
1
4 −x
=

n0
1
2
n+1
(x −2)
n
.
Khai triˆe
’
n n`ay chı
’
d

´ung khi



x −2
2



< 1 ⇔|x − 2| < 2 ⇔−2 <x− 2 < 2 ⇔ 0 <x<4. 
Nhˆa
.
nx´et. Ba
.
nd
o
.
cc˜ung dˆe
˜
d`ang thu du
.
o
.
.
c khai triˆe
’
ntrˆend
ˆay
b˘a
`

ng phu
.
o
.
ng ph´ap tru
.
.
ctiˆe
´
p.
V´ı d u
.
5. Khai triˆe
’
n h`am f(x) = sin
πx
4
th`anh chuˆo
˜
i Taylor v´o
.
i tˆam
ta
.
id
iˆe
’
m x
0
=2.

Gia
’
i. Ta biˆe
´
nd
ˆo
’
i h`am d˜a cho nhu
.
sau
sin
π
4
x = sin
π
4
(x −2 + 2) = sin

π
4
(x − 2) +
π
2

= cos
π
4
(x −2).
Xem
π

4
(x −2) = t v`a ´ap du
.
ng khai triˆe
’
nba
’
ng I II ta c´o
sin
π
4
x =

n0
(−1)
n
(2n)!

π
4
(x −2)

2n
=

n0
(−1)
n
π
2n

4
2n
(2n)!
(x −2)
2n
,x∈ R. 
206 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
V´ı d u
.
6. Khai triˆe
’
n h`am f(x)=ln
1+x
1 −x
th`anh chuˆo
˜
i Maclaurin.
Gia
’
i. Ta c´o ln
1+x
1 −x

= ln(1 + x) − ln(1 −x). M˘a
.
t kh´ac
ln(1 + x)=x −
x
2
2
+ ···+(−1)
n−1
x
n
n
+ , −1 <x<1;
ln(1 −x)=−x −
x
2
2
−···−
x
n
n
− , −1 <x<1.
T`u
.
d
´o
ln(1 + x) − ln(1 − x)=2x +
2x
3
3

+
2x
5
5
+ ···+
2x
2n−1
2n −1
+
− 1 <x<1. 
B
`
AI T
ˆ
A
.
P
Su
.
’
du
.
ng c´ac khai triˆe
’
nba
’
ng d
ˆe
’
khai triˆe

’
n h`am th`anh chuˆo
˜
il˜uy
th `u
.
av´o
.
i tˆam ta
.
id
iˆe
’
m x
0
= 0 v`a chı
’
ra b´an k´ınh hˆo
.
itu
.
cu
’
a chuˆo
˜
i
1. f(x)=e
−2x
.(DS.


n0
(−1)
n
2
n
n!
x
n
, R =+∞)
2. f(x)=ln
1
1 −2x
.(D
S.

n1
2
n
n
x
n
, R =
1
2
)
3. f(x)=x ln

1+
x
3

3

.(D
S.

n1
(−1)
n
x
3n+1
3
n
n
x
n
, R =
3
√
3)
4. f(x)=
3
√
1 −4x.(DS. 1 −
4
3
x +

n2
4
n

· 2 · 5 ···(3n − 4)
3
n
· n!
x
n
;
R =1)
5. f(x) = sin 5x.(D
S.

n0
(−1)
n
5
2n+1
(2n + 1)!
x
2n+1
; R =+∞)
6. f(x) = cos
x
3
3
.(D
S.

n0
(−1)
n

x
6n
3
2n
(2n)!
; R =+∞)
13.3. Chuˆo
˜
il˜uy th`u
.
a 207
B˘a
`
ng c´ach biˆe
´
ndˆo
’
i (trong tru
.
`o
.
ng ho
.
.
pcˆa
`
n thiˆe
´
t) sao cho c´o thˆe
’

´a p
du
.
ng c´ac khai triˆe
’
nba
’
ng d
ˆe
’
khai triˆe
’
n h`am f(x) th`anh chuˆo
˜
il˜uy th`u
.
a
v´o
.
i tˆam ta
.
id
iˆe
’
m x
0
. H˜ay chı
’
ra b´an k´ınh hˆo
.

itu
.
cu
’
a chuˆo
˜
i
7. f(x)=e
−
x
2
, x
0
= 10. (DS. e
−5

n0
(−1)
n
n!
(x −10)
n
2
n
, R =+∞)
8. f(x)=2
x
, x
0
= a.(DS. 2

a

n0
ln
n
2
n!
(x −a)
n
, R =+∞)
9. f(x)=2
x
3
−x
, x
0
= 0. (DS.

n0

ln
2
3

n
n!
x
n
, R =+∞)
10. f(x)=e

1−2x
3
, x
0
= 0. (DS.

n0
(−1)
n
2
n
e
n!
x
3n
, R =+∞)
11. f(x)=(2+x)e
x−1
, x
0
= −2. (DS.
1
e
3

n0
(x +2)
n+1
n!
, R =+∞)

12. f(x) = sin(a + x), x
0
=0.
(D
S. sin a

n0
(−1)
n
x
2n
2n!
+ cos a

n0
(−1)
n
x
2n+1
(2n + 1)!
,R=+∞)
13. f(x) = sin xcos 3x, x
0
=0.
(D
S.
1
2

n0

(−1)
n
(4x)
2n+1
(2n + 1)!
−
1
2

n0
(−1)
n
(2x)
2n+1
(2n + 1)!
,R=+∞)
14. f(x)=



cos x − 1
x
,x=0
0,x=0
; x
0
=0.
(D
S.


n0
(−1)
n
(x)
2n−1
(2n)!
, R =+∞)
15. f(x) = cos x, x
0
=
π
2
.(D
S.

n1
(−1)
n

x −
π
2

2n−1
(2n −1)!
, R =+∞)
16. f(x) = sin
2
x cos
2

x, x =0.
208 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
(DS.

n1
(−1)
n+1
· 2
4n−3
(2n)!
x
2n
, R =+∞)
17. f(x) = ln(x
2
+3x + 2), x
0
=0.
(D
S. ln 2 +

n1

(−1)
n−1
[1 + 2
−n
]
x
n
n
,R=+∞)
18. f(x) = ln(4 + 3x − x
2
), x
0
=2.
D
S. ln 6 +

n1
1
n

(−1)
n−1
3
−n
− 2
−n

(x −2)
n

)
19. f(x)=
1
x
2
− 2x − 3
, x
0
=0;x
0
=4.
(D
S. 1)

n0

(−1)
n+1
4
−
1
12

1
3

n

x
n

, |x| < 1;
2)

n0

(−1)
n
4
−
1
20
(−1)
n
5
n

(x −4)
n
, |x −4| < 1)
Chı
’
dˆa
˜
n. Biˆe
’
udiˆe
˜
n f(x)=
1
4


1
x −3
−
1
x +1

.
20. f(x)=
x
2
+ x +1
(x −1)( x +2)
, x
0
=0.
(D
S. 1 −

n0
2
n+1
− (−1)
n
2
n+1
x
n
, R =1)
21. f(x)=

1
x
2
+4x +7
, x
0
= −2.
(D
S.

n0
(−1)
n
(x +2)
2n
3
n+1
, R =
√
3)
Su
.
’
du
.
ng phu
.
o
.
ng ph´ap d

a
.
o h`am ho˘a
.
c t´ıch phˆan cu
’
achuˆo
˜
il˜uy th`u
.
a
d
ˆe
’
khai triˆe
’
n h`am f(x) th`anh chuˆo
˜
il˜uy th`u
.
av´o
.
i tˆam x
0
= 0 v`a chı
’
ra b´an k´ınh hˆo
.
itu
.

.
22. f(x) = arcctgx.(D
S.
π
2
x +
x
3
3
+ ···+
(−1)
n
x
2n−1
2n −1
+ , |x|  1)
13.3. Chuˆo
˜
il˜uy th`u
.
a 209
Chı
’
dˆa
˜
n. Khai triˆe
’
n arctgx v`a su
.
’

du
.
ng hˆe
.
th ´u
.
c arcctgx =
π
2
−
arctgx.
23. f(x)=
1
(1 − x)
3
.(DS.
1
2

n0
(n + 1)(n +2)x
n
, R =1)
24. f(x) = arc sin x.(D
S. x +

n1
(2n − 1)!!x
2n+1
(2n)!!(2n +1)

, R =1)
25. f(x)=
x

0
sin t
2
t
dt.(D
S.

n1
(−1)
n
x
4n+2
(2n + 1)!(4n +2)
, R =+∞)
26. f(x)=
x

0
ln(1 + t)
t
dt.(D
S.

n1
(−1)
n−1

x
n
n
2
, R =1)
27. f(x)=
x

0
1 − cos 2t
t
2
dt.(DS.

n1
(−1)
n
4
n
x
2n−1
(2n − 1)(2n)!
, R =+∞)
28. f(x)=
x

0
g(t)dt, g(x)=




ln(1 + x)
x
,x=0,
1,x=0.
(D
S.

n1
(−1)
n−1
x
n
n
2
,R=1)
29. f(x) = ln( x +
√
x
2
+ 1).
(D
S. x +

n1
(−1)
n
(2n − 1)!!
(2n)!!
x

2n+1
, R =1)
Chı
’
dˆa
˜
n. f(x)=
x

0
dt
√
1+t
2
.
´
Ap du
.
ng c´ac phu
.
o
.
ng ph´ap th´ıch ho
.
.
pd
ˆe
’
khai triˆe
’

n h`am f(x) th`anh
chuˆo
˜
il˜uy th `u
.
av´o
.
i tˆam ta
.
i x
0
v`a chı
’
ra b´an k´ınh hˆo
.
itu
.
cu
’
a chuˆo
˜
i
30. f(x)=e
x−1
, x
0
= 4. (DS. e
3

n0

(x −4)
n
n!
, R =+∞)
210 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
31. f(x)=e
3x
−2e
−x
, x
0
= 0. (DS.

n0
3
n
− 2(−1)
n
n!
x
n
, R =+∞)

32. f(x)=
1
2
3x−2
, x
0
= −1. (DS. 32

n0
(ln 8)
n
n!
(x +1)
n
, R =+∞)
33. f(x)=2
x
e
x−1
, x
0
= 1. (DS. 2

n0
(ln 2 + 1)
n
n!
(x −1)
n
, R =+∞)

34. f(x) = sin 3x, x
0
=
π
4
.
(D
S.
√
2
2

n0
(−1)
n
3
2n
(2n)!

x −
π
4

2n
−
√
2
2

n0

(−1)
n
3
2n+1
(2n + 1)!

x −
π
4

2n+1
,R=+∞)
35. f(x) = cos
x
2
, x
0
= −
π
3
.
(D
S.
√
3

n0
(−1)
n


x +
π
3

2n
(2n)!2
2n+1
+

n0
(−1)
n
(2n + 1)!(2
2n+1

x +
π
3

x +
π
3

2n+1
,R=+∞)
36. f(x) = cos
2
x, x
0
=

π
4
.
(D
S.
1
2
+

n1
(−1)
n
4
n−1

x −
π
4

2n−1
(2n −1)!
,R=+∞)
37. f(x) = sin x cos
2
x, x
0
=0.
(D
S.


n0
(−1)
n
4(2n)!
(1+3
2n+1
)x
2n+1
,R=+∞)
38. f(x) = cos x cos 2x, x
0
=0.
(D
S.
1
2

n0
(−1)
n

1
(2n)!
+
3
2n
(2n)!

x
2n

,R=+∞)
13.4. Chuˆo
˜
i Fourier 211
39. f(x)=x ln x, x
0
=1.
(D
S.

n1
(−1)
n−1
(x −1)
n+1
n
+

n1
(−1)
n−1
(x −1)
n
n
,R=1)
40. f(x) = ln(2 x + 3), x
0
=4.
(D
S. ln 11 +


n1
(−1)
n

2
11

n+1
1
n +1
(x −4)
n
,R=
11
2
)
41. f(x) = ln(3 −4x), x
0
= −2.
(D
S. ln 11 −

n1

4
11

n
(x +2)

n
n
,x∈

−19
4
,
3
4

)
42. f(x) = arctg
x +3
x −3
, x
0
=0.
(D
S. −
π
4
+

n0
(−1)
n+1
3
2n+1
x
2n+1

2n +1
,R=3)
Chı
’
dˆa
˜
n. f

(x)=−
3
x
2
+9
=

n0
(−1)
n+1
x
2n
3
2n+1
.T`u
.
d
´o
x

0
f(t)dt = f(x) − f(0) =


n0
(−1)
n+1
3
2n+1
x

0
t
2n
dt.
13.4 Chuˆo
˜
i Fourier
13.4.1 C´ac di
.
nh ngh˜ıa co
.
ba
’
n
Hˆe
.
h`am
1
2
, cos
πx


, sin
πx

, ,cos
nπx

, sin
nπx

, (13.13)
d
u
.
o
.
.
cgo
.
il`ahˆe
.
lu
.
o
.
.
ng gi´ac co
.
so
.
’

.D
´o l `a h ˆe
.
tru
.
.
c giao trˆen d
oa
.
n[−, ]
(t ´u
.
c l`a t´ıch phˆan theo d
oa
.
nd´ocu
’
a t´ıch hai h`am kh´ac nhau bˆa
´
tk`ycu
’
a
hˆe
.
b˘a
`
ng 0).
212 Chu
.
o

.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
D
-
i
.
nh ngh˜ıa 13.4.1. 1) Chuˆo
˜
i h`am da
.
ng
a
0
2
+

n1

a
n
cos
nπx

+ b
n
sin

nπx


, (13.14)
trong d
´o >0; a
n
, b
n
l`a c´ac h˘a
`
ng sˆo
´
(go
.
il`ahˆe
.
sˆo
´
cu
’
a chuˆo
˜
i (13.14))
d
u
.
o
.
.

cgo
.
il`achuˆo
˜
ilu
.
o
.
.
ng gi´ac.
2) Chuˆo
˜
ilu
.
o
.
.
ng gi´ac (13.14) d
u
.
o
.
.
cgo
.
i l`a chuˆo
˜
i Fourier cu
’
a h`am

f(x) theo hˆe
.
lu
.
o
.
.
ng gi´ac co
.
so
.
’
(13.13) (go
.
it˘a
´
t l`a chuˆo
˜
i Fourier) nˆe
´
u
c´ac hˆe
.
sˆo
´
cu
’
an´od
u
.

o
.
.
c t´ınh theo cˆong th´u
.
c
a
n
=
1



−
f(x) cos
nπx

dx, n =0, 1,
(13.15)
b
n
=
1



−
f(x) sin
nπx


dx, n =1, 2,
C´ac hˆe
.
sˆo
´
a
n
, b
n
du
.
o
.
.
c t´ınh theo cˆong th´u
.
c (13.15) d
u
.
o
.
.
cgo
.
il`ahˆe
.
sˆo
´
Fourier cu
’

a h`am f theo hˆe
.
lu
.
o
.
.
ng gi´ac co
.
so
.
’
.
Gia
’
su
.
’
h`am f(x) tho
’
a m˜an d
iˆe
`
ukiˆe
.
n


−
|f(x)|dx < +∞. Khi d´o

luˆon luˆon x´ac d
i
.
nh du
.
o
.
.
cc´achˆe
.
sˆo
´
a
n
, b
n
theo (13.15) v`a lˆa
.
pchuˆo
˜
i
Fourier d
ˆo
´
iv´o
.
i h`am f(x) v`a viˆe
´
t
f(x) ∼

a
0
2
+

n1

a
n
cos
nπx

+ b
n
sin
nπx


, (13.16)
o
.
’
d
ˆay dˆa
´
u“∼”du
.
o
.
.

cd`ung khi d
˘a
’
ng th´u
.
cchu
.
ad
u
.
o
.
.
cch´u
.
ng minh (v`a
d
u
.
o
.
.
cgo
.
il`adˆa
´
utu
.
o
.

ng ´u
.
ng).
13.4.2 Dˆa
´
uhiˆe
.
udu
’
vˆe
`
su
.
.
hˆo
.
itu
.
cu
’
achuˆo
˜
i Fourier
Dˆa
´
utu
.
o
.
ng ´u

.
ng “∼” trong hˆe
.
th ´u
.
c (13.16) c´o thˆe
’
thay b˘a
`
ng dˆa
´
ud
˘a
’
ng
th ´u
.
cnˆe
´
u h`am f(x) tho
’
a m˜an dˆa
´
uhiˆe
.
ud
u
’
sau dˆay vˆe
`

khai triˆe
’
n h`am
th`anh chuˆo
˜
i Fourier.
13.4. Chuˆo
˜
i Fourier 213
D
-
i
.
nh l´y Dirichlet. Gia
’
su
.
’
f(x) l`a h`am tuˆa
`
n ho`an chu k`y T (go
.
il`a
h`am T -tuˆa
`
n ho`an), f(x) v`a f

(x) ho˘a
.
cliˆen tu

.
c kh˘a
´
pno
.
i ho˘a
.
cliˆen tu
.
c
t`u
.
ng kh´uc (t´u
.
c l`a chı
’
c´o mˆo
.
tsˆo
´
h˜u
.
uha
.
nd
iˆe
’
m gi´an doa
.
n loa

.
i I trong
mˆo
˜
i chu k`y).
Khi d
´o chuˆo
˜
i Fourier cu
’
a h`am f(x) hˆo
.
itu
.
v´o
.
imo
.
i x d
ˆe
´
ntˆo
’
ng S(x)
v`a
1) Ta
.
imo
.
id

iˆe
’
mliˆen tu
.
ccu
’
a h`am f(x), chuˆo
˜
ihˆo
.
itu
.
dˆe
´
nch´ınh
h`am f(x) t´u
.
cl`aS(x)=f(x).
2) Ta
.
imo
.
id
iˆe
’
m gi´an doa
.
n, chuˆo
˜
ihˆo

.
itu
.
dˆe
´
nnu
.
’
atˆo
’
ng c´ac gi´o
.
i
ha
.
nmˆo
.
tph´ıa bˆen tr´ai v`a bˆen pha
’
icu
’
a h`am, t ´u
.
cl`a
S(x
0
)=
f(x
0
+0)+f(x

0
− 0)
2
,x
0
l`a diˆe
’
m gi´an doa
.
n.
3) Nˆe
´
u f(x) liˆen tu
.
c kh˘a
´
pno
.
i th`ı chuˆo
˜
i Fourier cu
’
an´ohˆo
.
itu
.
tuyˆe
.
t
d

ˆo
´
iv`adˆe
`
u.
C´o hai tru
.
`o
.
ng ho
.
.
pd
˘a
.
cbiˆe
.
t sau dˆay:
1) Nˆe
´
u f(x) l`a h`am ch˘a
˜
n, t´u
.
cl`af(x)=f(−x) ∀x th`ı b
n
=0
∀n  1 v`a chuˆo
˜
i Fourier cu

’
a n´o chı
’
ch´u
.
a c´ac h`am cosin:
f(x)=
a
0
2
+

n1
a
n
cos
nπx

,a
n
=
2



0
f(x) cos
nπx

dx n =0, 1, 2,

2) Nˆe
´
u f(x) l`a h`am le
’
,t´u
.
cl`af(x)=−f(−x) ∀x th`ı a
n
=0
∀n =0, 1, v`a chuˆo
˜
i Fourier cu
’
a n´o chı
’
ch´u
.
a c´ac h`am sin:
f(x)=

n1
b
n
sin
nπx

,b
n
=
2




0
f(x) sin
nπx

dx n =1, 2,
Nˆe
´
u h`am f(x)chı
’
x´ac d
i
.
nh trˆen doa
.
n[a, b] ⊂ [−, ] v`a khˆong x´ac
d
i
.
nh trˆen [−, ] \ [a, b] th`ı c´o thˆe
’
du
.
.
ng h`am phu
.
F (x) sao cho
F (x)=




f(x),x∈ [a, b]
g(x),x∈ [−, ] \ [a, b],
214 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
trong d´o g(x) l`a h`am kha
’
t´ıch t`uy ´y.
Sau d
´o gia
’
thiˆe
´
t F(x +2)=F (x) ∀x ∈ R v`a thu du
.
o
.
.
c h`am tuˆa
`
n

ho`an F (x)c´ochuk`y2. Ph´ep du
.
.
ng h`am F (x)nhu
.
vˆa
.
yd
u
.
o
.
.
cgo
.
il`a
ph´ep th´ac trıˆe
’
n tuˆa
`
n ho`an f(x).
H`am f(x)d
u
.
o
.
.
c cho trˆen [0,]c´othˆe
’
th´ac triˆe

’
nt`uy ´y sang khoa
’
ng
kˆe
`
[−, 0] v`a do vˆa
.
yn´od
u
.
o
.
.
cbiˆe
’
udiˆe
˜
nbo
.
’
i c´ac chuˆo
˜
i Fourier kh´ac
nhau (m`a thu
.
`o
.
ng g˘a
.

p l`a chuˆo
˜
i Fourier chı
’
ch´u
.
a ho˘a
.
c c´ac h`am cosin
ho˘a
.
c c´ac h`am sin).
Ta x´et hai tru
.
`o
.
ng ho
.
.
pd
˘a
.
cbiˆe
.
t sau dˆay:
1) Chuˆo
˜
i Fourier theo c´ac h`am cosin thu d
u
.

o
.
.
c khi th´ac triˆe
’
nch˘a
˜
n
h`am d
˜a cho trˆen doa
.
n[0,] sang khoa
’
ng kˆe
`
[−, 0]. Trong tru
.
`o
.
ng ho
.
.
p
n`ay d
ˆo
`
thi
.
cu
’

a h`am F(x)dˆo
´
ix´u
.
ng qua tru
.
c tung.
2) Chuˆo
˜
i Fourier theo c´ac h`am sin thu d
u
.
o
.
.
c khi th´ac triˆe
’
nle
’
h`am
d
˜a cho sang khoa
’
ng kˆe
`
[−, 0). Trong tru
.
`o
.
ng ho

.
.
p n`ay d
ˆo
`
thi
.
cu
’
a h`am
F (x)l`ad
ˆo
´
ix´u
.
ng qua gˆo
´
cto
.
ad
ˆo
.
.
Gia
’
su
.
’
h`am f(x)liˆen tu
.

c trˆen d
oa
.
n[−, ]v`af(−)=f()v`aa
0
,
a
n
, b
n
, n ∈ N l`a c´ac hˆe
.
sˆo
´
Fourier cu
’
a h`am f(x). Khi d´o ta c´o d˘a
’
ng
th ´u
.
c
1



−
f
2
(x)dx =

a
2
0
2
+

n1
(a
2
n
+ b
2
n
). (13.17)
D
˘a
’
ng th ´u
.
c (13.17) d
u
.
o
.
.
cgo
.
il`ad
˘a
’

ng th´u
.
c Parseval.
C
´
AC V
´
IDU
.
V´ı d u
.
1. Khai triˆe
’
n h`an f(x) = signx, −π<x<πth`anh chuˆo
˜
i
Fourier v`a su
.
’
du
.
ng khai triˆe
’
nˆa
´
yd
ˆe
’
t`ım tˆo
’

ng cu
’
a chuˆo
˜
i Leibnitz

n1
(−1)
n
2n +1
·
13.4. Chuˆo
˜
i Fourier 215
Gia
’
i. V`ı h`am f(x)le
’
nˆen a
n
=0,n =0, 1, v`a
b
n
=
2
π
π

0
signx sin nxdx =

2
π

−
cos nx
n



π
0

=
2
nπ
([1 −cos nπ]) =





4
π(2m −1)
nˆe
´
u n =2m − 1
0nˆe
´
u n =2m
,m∈ N.

Do d
´ov´o
.
i −π<x<πta c´o
signx =
4
π

m1
sin(2m −1)x
2m −1
·
Nˆe
´
ud
˘a
.
t x =
π
2
th`ı t`u
.
chuˆo
˜
i Fourier thu d
u
.
o
.
.

c, ta c´o
1=
4
π

m1
(−1)
m+1
2m −1
⇒

m0
(−1)
m
2m +1
=
π
4
· 
V´ı d u
.
2. T`ım khai triˆe
’
n Fourier cu
’
a h`am f(x)=x
3
+ x
2
− x +1,

x ∈ [−1,1].
Gia
’
i. Nhu
.
vˆa
.
y ta pha
’
i khai triˆe
’
n h`am d
˜a cho theo hˆe
.
co
.
so
.
’
{1, cos πx, sin πx,cos 2πx, sin 2πx, }.
Ta c´o
a
0
=
1

1
(x
3
+ x

2
− x − 1)dx = −
4
3
,
216 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
a
n
=
1

−1
(x
3
+ x
2
− x − 1) cos nπxdx
=
1
nπ

(x

2
− 1) sin nπx


1
−1
−2
1

−1
x sin nπxdx

=
2
π
2
n
2

x cos nπx


1
−1
−
1

−1
cos nπxdx


=
4
π
2
n
2
(−1)
n
,n=1, 2,
b
n
=
1

−1
(x
3
+ x
2
− x − 1) sin nπxdx =
12
π
3
n
3
(−1)
n
,n=1, 2,
V`ı h`am f(x) kha
’

vi trˆen khoa
’
ng ( −1, 1) nˆen ta c´o
x
3
+ x
2
− x − 1=−
2
3
+
4
π
2

n1

(−1)
n
n
2
cos nπx +
3(−1)
n
πn
3
sin nπx

,
∀x ∈ (−1, 1).

V´ı d u
.
3. Khai triˆe
’
n h`am f(x)=x, x ∈ [0,]:
1) theo c´ac h`am cosin;
2) theo c´ac h`am sin.
Gia
’
i. 1) D
ˆe
’
khai triˆe
’
n h`am f(x) th`anh chuˆo
˜
i Fourier theo c´ac h`am
cosin, ta thu
.
.
chiˆe
.
n ph´ep th´ac triˆe
’
nch˘a
˜
n h`am f(x)=x, x ∈ [0, 1] sang
d
oa
.

nkˆe
`
[−, 0] v`a thu du
.
o
.
.
c h`am f
∗
(x)=|x|, x ∈ [−, ]. T`u
.
d
´o th´ac
triˆe
’
n2-tuˆa
`
n ho`an h`am f
∗
(x) ra to`an tru
.
csˆo
´
.Hiˆe
’
n nhiˆen h`am f
∗
(x)
liˆen tu
.

c ∀x ∈ R. Ta c´o:
a
0
=
2



0
xdx = ,
a
n
=
2



0
x cos
nπx

dx =



0nˆe
´
u n ch˘a
˜
n,

−
4
n
2
π
2
nˆe
´
u n le
’
.
13.4. Chuˆo
˜
i Fourier 217
T`u
.
d
´o ta thu du
.
o
.
.
c
x =

2
−
4
π
2


n1
cos(2n − 1)x
(2n −1)
2
, 0  x  .
2) D
ˆe
’
khai triˆe
’
n h`am f(x) theo c´ac h`am sin, ta thu
.
.
chiˆe
.
nph´ep
th´ac triˆe
’
nle
’
h`am f(x)=x, x ∈ [0,] sang d
oa
.
nkˆe
`
[−, 0] v`a thu du
.
o
.

.
c
h`am f
∗
(x)=x, x ∈ [−, ]. T`u
.
d
´o th´ac triˆe
’
n2-tuˆa
`
n ho`an h`am f
∗
(x)
ra to`an tru
.
csˆo
´
. H`am d
˜adu
.
o
.
.
c th´ac triˆe
’
n tho
’
am˜and
i

.
nh l´y Dirichlet.
Do vˆa
.
y ta c´o: a
0
=0,a
n
=0,∀n =1, 2,
b
n
=
2



0
x sin
nπx

dx =
2


−
x
nπ
cos
nπx






0
+

nπ


0
cos
nπx

dx

=
2
nπ
(−1)
n+1
.
Do d
´o
x =
2
π

n1
(−1)

n+1
n
sin
nπx

, 0  x  ,
V´ı du
.
4. Khai triˆe
’
n h`am
f(x)=



0nˆe
´
u x 
π
2
|x|−
π
2
nˆe
´
u
π
2
< |x| <π
v`a f(x +2π)=f(x), x ∈ R th`anh chuˆo

˜
i Fourier.
Gia
’
i. H`am f(x) l`a h`am ch˘a
˜
n nˆen n´o khai triˆe
’
nd
u
.
o
.
.
c th`anh chuˆo
˜
i
Fourier theo c´ac h`am cosin. Ta c´o
a
0
=
2
π
π

0
f(x)dx =
2
π
π


π/2

x −
π
2

dx =
π
4
,
a
n
=
2
π
π

0
f(x) cos nxdx =
2
π
π

π/2

x −
π
2


cos nxdx
=
2
π

cos nπ
n
2
−
cos
nπ
2
n
2

,n∈ N.
218 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
T`u
.
d
´o suy ra r˘a
`

ng v´o
.
i x ∈ (−π,π)
f(x)=
π
8
+
2
π

n1

cos nπ
n
2
−
cos
nπ
2
n
2

cos nx
=
π
8
+
2
π


n1
(−1)
n
n
2
cos nx −
1
2π

n1
(−1)
n
n
2
cos 2nx,
trong d´otad˜asu
.
’
du
.
ng c´ac hˆe
.
th ´u
.
c
cos
nπ
2
n
2

cos nx =0 v´o
.
i n =2m − 1
cos
nπ
2
n
2
cos nx =
(−1)
m
4m
2
cos 2mx, n =2m. 
13.4. Chuˆo
˜
i Fourier 219
B
`
AI T
ˆ
A
.
P
Khai triˆe
’
n h`am th`anh chuˆo
˜
i Fourier trˆen d
oa

.
n (khoa
’
ng) d˜a c h o v `a
trong mˆo
.
tsˆo
´
tru
.
`o
.
ng ho
.
.
p h˜ay su
.
’
du
.
ng khai triˆe
’
nthud
u
.
o
.
.
cd
ˆe

’
t´ınh
tˆo
’
ng cu
’
a chuˆo
˜
isˆo
´
:
1. f(x)=
x
2
, x ∈ (0,2π). (D
S.
π
2
−

n1
sin nx
n
)
2. f(x)=



6nˆe
´

u0<x<1
3x nˆe
´
u2<x<4.
(D
S.
15
2
+
12
π
2

cos
πx
2
+
1
9
cos
3πx
2
+
1
25
cos
5πx
2
+


−
6
π

sin
πx
2
+
1
2
sin
2πx
2
+
1
3
sin
3πx
2
+

)
3. f(x)=e
−x
, x ∈ (−π,π).
(D
S. e
−x
=
e

π
− e
−π
π

1
2
+

n1
(−1)
n
1+n
2
(cos nx + n sin nx)

)
4. f(x)=x
2
, x ∈ [−π, π). T´ınh tˆo
’
ng c´ac chuˆo
˜
isˆo
´

n1
(−1)
n−1
n

2
v`a

n1
1
n
2
.
(D
S. x
2
=
π
2
3
− 4

n1
(−1)
n−1
cos nx
n
2
. Thay x
0
= 0, ta thu du
.
o
.
.

c

n1
(−1)
n−1
n
2
=
π
2
12
. Thay x = π thu d
u
.
o
.
.
c

n1
1
n
2
=
π
2
6
)
5. f(x)=x
2

, x ∈ (0,π), f(x)=f(x + pi).
(D
S.
π
2
3
+

n1

1
n
2
cos 2nx −
π
n
sin 2nx

)
220 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
6. f(x) = cos
x

2
, x ∈ (0,2π], f(x)=f(x +2π).
(D
S. cos
x
2
=
8
π

n1
n sin nx
(2n −1)(2n +1)
)
7. f(x)=



0nˆe
´
u − 3 <x 2
x nˆe
´
u0<x<3.
T´ınh tˆo
’
ng cu
’
achuˆo
˜

i

n1
1
(2n −1)
2
.
(D
S.
3
4
−
6
π
2

n1
1
(2n −1)
2
cos
(2n −1)nx
3
−
3
π

n1
(−1)
n

n
sin
nπx
3
,
tˆo
’
ng cu
’
a chuˆo
˜
ib˘a
`
ng
π
2
8
v´o
.
i x
0
=0)
8. f(x)=x sin x, x ∈ [−π,π].
(D
S. 1 −
cos x
2
+2

n2

(−1)
n
cos nx
n
2
−1
)
H˜ay khai triˆe
’
n c´ac h`am sau d
ˆay th`anh chuˆo
˜
i Fourier theo c´ac h`am
sin ho˘a
.
c h`am cosin
9. f(x)=x cos x, x ∈ (0,π). T´ınh tˆo
’
ng chuˆo
˜
i

n0
4n
2
+1
(4n
2
− 1)
2

.
(D
S. 1) x cos x = −
2
π
+
π
2
cos x −
4
π

n1
4n
2
+1
(4n
2
− 1)
2
cos 2nx.
2) x cos x = −
sin x
x
+2

n2
(−1)
n
n

n
2
− 1
sin nx, 0  x<π.
3)

n0
4n
2
+1
(4n
2
− 1)
=
π
2
8
+
1
2
(thˆe
´
x = 0 v`ao 1))
10. f(x)=x(π − x), 0  x<π, f(x)=f(x + π).
(D
S.
π
2
6
−


n1
cos
2nx
n
2
)
13.4. Chuˆo
˜
i Fourier 221
11. f(x) = cos x, x ∈

0,
π
2

.T´ınh S =

n1
(−1)
n
(2n −1)(2n +1)
.
(D
S.
4
π

1
2

−

n1
(−1)
n
cos 2nx
4n
2
− 1

; S =
1
2
−
π
4
khi x
0
=0)
12. f(x)=



1nˆe
´
u0 x  1
0nˆe
´
u1<x π.
T´ınh

σ
1
=

n1
sin n
n
; σ
2
=

n1
(−1)
n
sin n
n
(D
S.
2
π

1
2
+

n1
sin n
n
cos nx


,σ
1
=
π − 1
2
;
khi x
0
=0; σ
2
= −
1
2
v´o
.
i x = π)
13. f(x)=|cos x|.T´ınh
S
1
=

n1
(−1)
n
1
4n
2
− 1
v`a S
2

=

n1
1
4n
2
− 1
.
(D
S.
2
π
+
4
π

n1
(−1)
n+1
4n
2
−1
cos 2nx;
S
1
=
π − 2
4
khi x
0

=0,S
2
=
1
2
khi x
0
=
π
2
)
14. f(x)=|sin x|, x ∈ [−π,π]. T´ınh S =

n1
1
(2n−1)(2n+1)
.
(D
S.
2
π
−
4
π

n1
cos 2nx
(2n −1)(2n +1)
,S=
1

2
khi x
0
=0)
15. f(x) = sign(sin x). T´ınh S =

n1
(−1)
n−1
2n −1
.
(D
S.
4
π

n1
sin(2n −1)x
2n −1
,

n1
(−1)
n+1
2n −1
=
π
4
v´o
.

i x
0
=
π
2
)
222 Chu
.
o
.
ng 13. L´y thuyˆe
´
tchuˆo
˜
i
16. f(x)=x −[x]={x} - phˆa
`
n thˆa
.
p phˆan cu
’
asˆo
´
x.
(D
S. {x} =
1
2
−
1

π

n1
sin 2nπx
n
)
17. f(x)=



−x nˆe
´
u − π  x  0
x
2
π
nˆe
´
u0<x π.
(D
S.
5π
12
+

n1

3(−1)
n
− 1

πn
2
cos nx −
4
π
2
(2n −1)
3
sin(2n −1)x

)
18. f(x)=



0 khi − 2  x  0
1
2
x khi 0 <x 2.
(D
S.
1
4
+

n1

−
2
π

2
(2n −1)
3
cos
nπx
2
+
(−1)
n+1
nπ
sin
nπx
2

)
19. f(x)=x, x ∈ [3, 5]. (D
S. 4 +

n1
2
nπ
(−1)
n+1
sin nπx)
Chı
’
dˆa
˜
n. Ta hiˆe
’

u khai triˆe
’
n Fourier cu
’
a h`am f(x)=x trˆen khoa
’
ng
(3, 5) l`a khai triˆe
’
n Fourier cu
’
a h`am tuˆa
`
n ho`an v´o
.
ichuk`y2 =5−3=2
tr `ung v´o
.
i h`am f(x) trˆen khoa
’
ng (3, 5).
20. f(x) = sin 2x + cos5x, x ∈

−
π
2
,
π
2


.
(D
S. −
2
5π
+ sin 2x +
20
π

n1
(−1)
n
4n
2
− 25
cos 2nx)
21. f(x)=x sin 2x, x ∈

−
π
4
,
π
4

.
(D
S.
1
π

+
8
π

n1
(−1)
n+1
n
4n
2
− 1
cos 4nx)