qatorning
3
;
1
kesmada tekis yaqinlashuvchi ekanligini hosil qilamiz.
11.21-masala.
.
2
3
ln
2
x
x
x
f
funksiyani Makloren qatoriga
yoying.
Bu masalani yechish uchun
2
1
ln
2
ln
1
ln
2
ln
1
ln
2
1
ln
.
2
3
ln
2
x
x
x
x
x
x
x
x
x
f
deb olib
,
5
0
-
punktda keltirilgan
1
1
1
,
1
,
1
1
ln
n
n
n
x
n
x
x
tenglikdan foydalanamiz:
1
1
1
1
2
2
1
1
2
ln
2
1
ln
1
ln
2
ln
.
2
3
ln
n
n
n
n
n
n
x
n
n
x
x
x
x
x
1
1
.
2
1
1
1
2
ln
n
n
n
n
x
n
12.21-masala.
x
x
f
4
cos
funksiyani
4
0
x
nuqta atrofida
Teylor qatoriga yoying va bu qatorning yaqinlashish sohasini toping.
Avvalo
x
x
2
cos
1
2
1
cos
2
ekanini e`tiborga olib,
x
x
x
x
x
f
2
cos
2
cos
2
1
4
1
cos
cos
2
2
2
4
2
4
cos
1
2
cos
2
1
4
1
x
x
x
x
4
cos
8
1
2
cos
2
1
8
3
bo`lishini topamiz. So`ngra
4
t
x
almashtirishni bajaramiz:
.
4
cos
8
1
2
sin
2
1
8
3
4
cos
8
1
2
2
cos
2
1
8
3
4
t
t
t
t
t
f
x
f
Endi
x
sin
hamda
x
cos
larning 5
0
punktda
keltirilgan yoyilmalaridan
foydalanib, ushbu
230
,
!
1
2
2
1
2
sin
1
2
0
1
2
n
n
n
n
t
n
t
,
!
2
4
1
4
cos
2
0
2
n
n
n
n
t
n
t
tengliklarga ega bo`lamiz. Natijada
0
0
2
2
1
2
1
2
4
!
2
4
1
8
1
4
!
1
2
2
1
2
1
8
3
n
n
n
n
n
n
n
n
x
n
x
n
x
f
1
2
3
4
1
2
0
2
4
!
2
2
1
4
!
1
2
2
1
4
1
n
n
n
n
n
n
n
n
x
n
x
n
qatorni hosil qilamiz. Bu qatorning yaqinlashish sohasi
,
ekanligini
ko`rish qiyin emas.
13.21-masala. Quyidagi
...
5
4
1
4
3
1
3
2
1
2
1
1
qatorning yig`indisini toping.
Berilgan
1
1
1
1
n
n
n
n
qator uchun
2
*
1
1
0
1
1
1
1
n
n
n
n
n
n
taqqoslash amlomatiga ko`ra u absolut yaqinlshuvchi
Chekli yig`indiga
ega .
1
1
1
1
n
n
n
n
S
deb belgilaymiz.
Ushbu
1
1
1
n
n
n
n
x
x
S
(1)
yordamchi qatorni kiritamiz.
Bu qator
1
x
da absolut va tekis
yaqinlashadi. Abelning 2-teoremasiga ko`ra (5
0
-punktdagi 7-teorema va
uning natijasiga qarang)
x
S
S
x
0
1
lim
bo`ladi.
x
S
funksiyani
topish
uchun
(1)-tenglikni
2
marta
differensiallaymiz.
1
,
n
n
n
x
x
S
1
3
2
1
1
,
1
1
...
1
n
n
x
x
x
x
x
x
x
S
231
1
1
1
ln
c
x
x
S
va
0
0
S
2
1
1
ln
1
0
c
x
x
x
x
S
c
va
0
0
0
2
c
S
Deamk
,
x
x
x
x
S
1
ln
1
ekan
.
1
2
ln
2
lim
0
1
x
S
S
x
14.21-masala.
1
1
2
.
1
2
2
n
n
n
n
x
qatorning yig`indisini toping.
Bu qatorning
yaqinlashish sohasi
1
;
1
kesmadan iborat bo`lib, bu
kesmaning ichki nuqtalarida qatorni hadlab, differensiallash mumkin:
1
1
2
5
3
1
2
2
1
1
2
,
1
...
2
.
1
2
2
)
(
n
n
n
n
n
n
x
x
x
x
x
x
x
S
n
x
x
S
n
n
x
x
S
1
;
1
x
1
2
1
2
1
ln
2
1
1
c
x
c
dx
x
x
x
S
va
0
0
0
1
c
S
2
2
2
1
ln
2
1
c
dx
x
c
dx
x
S
x
S
((bo`laklab
integrallash
usulidan foydalanamiz))
2
2
1
1
ln
2
1
1
ln
2
c
x
x
x
x
x
va
0
0
S
0
2
c
.
Demak,
.
1
ln
2
1
1
ln
2
1
1
2
2
2
1
1
2
x
x
x
x
x
n
n
x
n
n
ekan. Tenglik
1
;
1
intervalda o`rinli.
15.21-masala.
.
4
5
2
0
2
n
n
x
n
n
qatorning yig`indisini toping.
Berilgan qator
1
;
1
intervalda absolut
va tekis yaqinlashadi va
shu intervaldagi
oraliqda bu qatorni hadlab integrallash mumkin:
0
0
3
0
2
2
2
4
1
4
1
4
5
n
n
n
n
n
n
x
n
n
x
S
x
x
n
n
x
n
n
x
S
0
0
1
1
4
1
4
1
4
1
1
n
n
n
n
c
x
S
x
c
x
n
x
c
x
n
dx
x
S
x
0
x
da
0
0
0
0
1
1
c
S
S
Demak,
x
S
x
dx
x
S
x
1
4
va
0
1
1
1
n
n
dx
x
S
x
n
x
S
0
2
3
2
1
1
...
n
n
c
x
x
x
x
x
x
0
x
da
.
0
0
0
2
1
c
S
232
2
1
1
1
1
1
1
x
x
x
x
S
x
x
dx
x
S
Bu tenglik va
x
S
x
dx
x
S
x
1
4
dan
4
4
2
3
'
4
4
1
4
1
1
2
1
4
1
x
x
x
x
x
x
x
x
S
x
x
S
x
.
1
2
2
1
2
2
3
2
3
3
x
x
x
x
S
x
x
x
Shunday qilib,
0
3
2
2
2
.
1
;
1
,
1
2
2
4
5
n
n
x
x
x
x
x
n
n
16.21-masala. Integral ostidagi funksiyani darajali qatorga
yoyish usuli yordamida
1
0
2
dx
e
x
integralni 0,001 aniqlikda hisoblang.
Agar 5
0
-punktda
x
e
uchun keltirilgan yoyilmadan foydalansak,
0
2
!
1
2
n
n
n
x
n
x
e
ekanligini, bu yerdan esa
1
0
1
0
0
1
0
0
0
1
0
1
2
2
0
2
.
1
2
!
1
1
2
!
1
!
1
!
1
2
n
n
n
n
n
n
n
n
n
n
n
x
n
n
n
n
x
dx
n
x
dx
n
x
dx
e
bo`lishini topamiz. Bu hosil bo`lgan qator Leybnis qatori bo`lib, uning m-
hadidan keyingi qoldig`i
1
1
2
!
1
m
n
n
m
n
n
r
uchun
3
2
!
1
1
m
m
r
m
bo`lishi bizga ma`lum.
001
,
0
m
r
bajarilishi
uchun oxirgi tengsizlikdan
4
m
bo`lishi kifoyaligini aniqlaymiz. Demak,
1
0
.
747
,
0
216
1
42
1
5
1
3
1
1
2
dx
e
x
