134
1
2
agar
,
8
2
agar
,
0
1
1
4
0
cos
cos
4
2
cos
4
2
sin
2
2
sin
5
2
2
sin
2
2
cos
5
2
cos
5
2
2
cos
2
8
10
2
2
5
2
2
2
2
2
2
2
2
2
2
0
2
2
2
0
2
0
2
0
0
2
0
2
2
0
0
0
n
k
k
n
k
k
k
k
x
k
k
xdx
k
k
x
k
k
x
x
k
k
v
xdx
k
dv
dx
du
x
u
xdx
k
x
xdx
l
k
x
f
l
a
x
dx
x
dx
x
f
l
a
k
l
k
l
Javob:
4-misol. Davri
2
T
bo’lgan quyidagi
1
0
agar
,
2
0
1
agar
,
1
x
x
x
x
f
funksiyani
1
;
1
kesmada Furye qatoriga yoying. (18-
rasm)
x
n
n
x
k
2
1
2
cos
1
2
1
8
4
5
1
2
2
135
18-rasm.
Yechish. Berilgan funksiya juft ham, toq ham emas,
1
l
. Demak,
0
a
,
k
a
va
k
b
koeffitsientlarning hammasini
topamiz.
2
5
2
2
2
1
1
0
2
0
1
1
0
0
1
0
x
x
x
dx
x
dx
dx
x
f
l
a
l
l
,
x
k
k
v
dx
du
dx
x
k
dv
x
u
dx
x
k
x
dx
x
k
a
k
sin
1
,
cos
,
2
cos
2
cos
1
0
0
1
1
0
1
0
0
1
sin
1
sin
2
sin
1
dx
x
k
k
x
k
k
x
x
k
k
n
k
n
k
k
k
x
k
k
k
2
agar
,
0
1
2
agar
,
2
1
1
1
cos
1
2
2
2
2
1
0
2
2
,
x
k
k
v
dx
du
dx
x
k
dv
x
u
dx
x
k
x
dx
x
k
b
k
cos
1
,
sin
,
2
sin
2
sin
1
0
0
1
x
y
0
-1
1
y=2-
x
y=1
136
1
0
1
0
0
1
cos
1
cos
2
cos
1
dx
x
k
k
x
k
k
x
x
k
k
k
x
k
k
k
k
k
k
k
1
sin
2
2
cos
1
1
1
1
1
0
2
2
.
Javob:
1
0
2
sin
1
1
2
1
2
cos
2
4
5
k
k
k
x
k
k
x
k
x
f
.
Mustaqil yechish uchun mashqlar
1. Davri
2
T
bo’lgan quyidagi:
1
0
agar
,
1
0
1
agar
,
2
1
3
x
x
x
x
f
funksiyani
1
;
1
kesmada Furye qatoriga yoying.
Javob:
1
2
2
2
sin
cos
3
8
5
k
k
x
k
k
x
k
x
f
.
2. Davri
2
T
bo’lgan
2
x
x
f
funksiyani
1
;
1
kesmada Furye qatoriga yoying.
Javob:
1
2
2
cos
1
4
3
1
k
k
x
k
k
x
f
.
3. Davri
2
T
bo’lgan
x
x
f
funksiyani
1
;
1
kesmada Furye qatoriga yoying.
Javob:
0
2
2
1
2
1
2
cos
4
2
1
k
k
x
k
x
f
.
4. Davri
2
T
bo’lgan
x
x
x
f
funksiyani
1
;
1
kesmada Furye qatoriga yoying.
137
Javob:
1
sin
1
2
1
k
k
x
k
x
f
.
5. Davri
4
T
bo’lgan
x
x
f
funksiyani
2
;
2
kesmada Furye qatoriga yoying.
Javob:
0
2
2
2
1
2
cos
1
2
1
8
1
k
x
k
k
x
f
.
4-§. Davriy bo’lmagan funksiyalarni Furye qatoriga
yoyish
Biror
l
;
0
kesmada uzluksiz va bo’lakli monoton
x
f
funksiyani
0
;
l
kesmada davom ettirib, bu
funksiyani
l
l ;
kesmada toq yoki juft holdagi funksiyaga
to’ldirib uni Furye qatoriga yoyish mumkin.
Davom ettirilgan toq funksiya uchun Furye qatori
faqat sinuslarni, juft funksiya uchun Furye qatori faqat
kosinuslarni o’z ichiga oladi.
1-misol.
x
x
f
2
funksiyani
;
0
kesmada toq
holda davom ettirib,
;
kesmada Furye qatoriga
yoying. (19-rasm).
Yechish. Bu funksiya
2
T
davrli bo’lib,
;
kesmada toq funksiya bo’ladi. U holda
0
0
a
,
0
k
a
,
0
k
b
bo’lib,
k
b
ni umumiy formuladan topish
kifoya. Bu yerda
l
.
138
19-rasm.
kx
k
v
dx
du
dx
kx
dv
x
u
dx
kx
x
b
k
cos
1
,
2
sin
,
2
sin
2
2
0
0
0
cos
cos
2
cos
2
cos
2
2
0
0
k
k
k
dx
kx
k
kx
k
x
1
2
agar
,
0
2
agar
,
4
1
1
2
1
1
2
n
k
n
k
k
k
k
k
k
.
Javob:
1
2
2
sin
4
k
k
kx
x
f
.
2-misol.
x
x
f
2
funksiyani
;
0
kesmada juft
holda davom ettirib,
;
kesmada Furye qatoriga
yoying. (20-rasm).
Yechish. Bu davriy bo’lmagan
x
x
f
2
funksiyani juft holda davom ettirib,
;
kesmada
2
T
davrli juft funksiyaga to’ldirsak bo’ladi. U holda
x
y
x
y
2
0
a
139
0
k
b
bo’lib,
0
a
,
k
a
larni umumiy formulalardan topish
kifoya. Bu yerda
l
.
20-rasm.
0
2
2
2
2
2
2
2
2
0
2
0
0
x
x
dx
x
a
,
kx
k
v
dx
du
dx
kx
dv
x
u
dx
kx
x
a
k
sin
1
,
2
cos
,
2
cos
2
2
0
0
0
0
cos
1
2
2
sin
2
sin
2
2
kx
k
k
dx
kx
k
kx
k
x
n
k
n
k
k
k
k
k
k
2
agar
,
0
1
2
agar
,
8
1
1
4
1
cos
4
2
2
2
.
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