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152
Namuna misol.
Ushbu
0
22
25
;
0
11
15
;
0
13
4
15
2
2
2
z
y
z
y
x
z
y
x
nochiziqli tenglamalar sistemasining yechimlarini oddiy iteratsiyalar, Zeydel,
Nyuton, Broyden usullari bilan 0,00001 aniqlikda toping.
Yechish.
Dastlab berilgan sistemani vektor fazoda vektor shakliga keltiramiz:
.
0
22
25
;
0
11
15
;
0
13
4
15
))
,
,
(
),
,
,
(
),
,
,
(
(
)
,
,
(
)
(
2
2
2
3
2
1
T
z
y
z
y
x
z
y
x
z
y
x
f
z
y
x
f
z
y
x
f
z
y
x
F
x
F
Maple dasturi yordamida ushbu sistemaning haqiqiy yechimini topamiz:
;
0
22
25
,
0
11
15
,
0
13
4
15
2
2
2
z
y
z
y
x
z
y
x
fsolve
Buni uch o‘lchovli grafikda chizib ham ko‘rish mumkin (3.18-rasm):
;
2
..
0
,
2
..
0
,
2
..
0
,
0
22
25
,
0
11
15
,
0
13
4
15
3
:
)
(
2
2
2
z
y
x
z
y
z
y
x
z
y
x
d
ot
implicitpl
plots
with
Ushbu sistemani oddiy iteratsiyalar, Zeydel, Nyuton, Broyden usullari bilan
0,00001 aniqlikda taqribiy yechamiz.
1)
Oddiy iteratsiyalar usuli
. Usul-
ning g‘oyasiga ko‘ra berilgan
F
(
x
)=0
sistemani
x
=
Ф
(
x
) ko‘rinishga keltira-
mizki, aniq yechim
D
={(
x
,
y
,
z
)
T
: 0
x
,
y
,
z
2} sohaga tegishli, boshlang‘ich
yaqinlashishni
x
0
= (1.0, 0.8, 0.9)
T
deb
olib, yechimni 10
-5
aniqlikda topish talab
etilsin.
Sistemaning tenglamalarini ushbu
25
22
25
1
)
,
,
(
;
15
11
15
1
15
1
)
,
,
(
;
0
15
13
15
4
15
1
)
,
,
(
2
3
2
2
2
1
y
z
y
x
z
z
x
z
y
x
y
z
y
z
y
x
x
3.18-rasm. Uchta sirtning bir nuqtada
kesishishini tasvirlovchi grafik.
ko‘rinishga keltiramiz. Bu yerda aniq yechim
D
= {(
x
,
y
,
z
)
T
: 0
x
,
y
,
z
2} sohaga
tegishli.
D
sohada olingan xususiy hosilalar:
153
;
0
1
x
;
3
.
0
15
2
1
y
y
;
3
.
0
15
4
1
z
;
3
.
0
15
2
2
x
x
;
0
2
y
;
1
.
0
15
1
2
z
;
0
3
x
;
2
.
0
25
2
3
y
y
.
0
3
z
Ko‘rinib turibdiki, barcha xususiy hosilalar qiymatlarining moduli 1 dan kichik, ya’ni
3
.
0
K
x
j
i
(
K
– maksimal chegaraviy qiymat), demak 3 ta noma’lumli 3 ta
nochiziqli tenglamalar sistemasi (
m
= 3) uchun
q
=
mK
= 3
0,3 = 0,9 < 1. Tanlangan
x
=
Ф
(
x
) bog‘lanish orqali quyidagi iteratsion formulalarni qurishimiz mumkin:
.
25
22
25
1
;
15
11
15
1
15
1
;
0
15
13
15
4
15
1
2
1
2
1
2
1
n
n
n
n
n
n
n
n
y
z
z
x
y
z
y
x
Bu hisoblashlarni
5
1
10
n
n
x
x
shart bajarilgunga qadar davom ettirsak,
quyidagi jadval natijalariga kelamiz:
n
x
n
y
n
z
n
n
n
x
x
1
0
1.000000000
0.800000000
0.900000000
1
1.064000000
0.726666667
0.905600000
0.07333
2
1.072957037
0.718233600
0.901121778
0.00896
3
1.072575174
0.716658998
0.900634380
9.0
10
-5
4
1.072595827
0.716681125
0.900544005
2.6
10
-5
5
1.072569612
0.716672146
0.900545273
8.2
10
-5
6
1.072570809
0.716675980
0.900544759
3.8
10
-6
7
1.072570305
0.716675775
0.900544978
5.0
10
-7
Bu yerda topilgan
x
7
yechim yetarlicha aniqlikda. Bunda yaqinlashish tezligini
quyidagicha baholash o‘rinli:
351
.
0
9
.
0
1
9
.
0
07333
.
0
1
7
7
0
1
*
7
q
q
x
x
x
x
, chunki
8
*
7
10
6
.
7
x
x
.
2)
Zeydel usuli
. Iteratsion jarayonni
x
0
= (1.0, 0.8, 0.9)
T
boshlang‘ich yaqinlash-
ish bilan quyidagi formulalarda amalga oshiramiz (hisob natijalari quyidagi jadvalda
keltirilgan):
154
.
25
22
25
1
;
15
11
15
1
15
1
;
0
15
13
15
4
15
1
2
1
1
2
1
1
2
1
n
n
n
n
n
n
n
n
y
z
z
x
y
z
y
x
n
x
n
y
n
z
n
n
n
x
x
1
0
1.000000000
0.800000000
0.900000000
1
1.064000000
0.717860267
0.900612934
0.08214
2
1.072475225
0.716693988
0.900546011
0.00848
3
1.072568918
0.716676128
0.900544987
9,4
10
-5
4
1.072570352
0.716675855
0.900544971
1,4
10
-6
5
1.072570374
0.716675851
0.900544971
2,2
10
-8
Bu yerda topilgan
x
5
yechim yetarlicha aniqlikda. Bu yerda ham
486
.
0
9
.
0
1
9
.
0
08214
.
0
1
5
5
0
1
*
5
q
q
x
x
x
x
, chunki
10
*
5
10
x
x
.
Demak, ushbu misolni yechishda Zeydel usulining yaqinlashish tezligi yuqo-
riroq ekan. Ammo bu ijobiy hol ba’zi nochiziqli tenglamalar sistemasini Zeydel usuli
bilan yechishda kuzatilmasligi mumkin.
3)
Nyuton usuli
. Berilgan sistema uchun ushbu
.
0
22
25
;
0
11
15
;
0
13
4
15
))
,
,
(
),
,
,
(
),
,
,
(
(
)
,
,
(
)
(
2
2
2
3
2
1
T
z
y
z
y
x
z
y
x
z
y
x
f
z
y
x
f
z
y
x
f
z
y
x
F
x
F
belgilashlarni yuqorida qabul qilgan edik, bu yerda
.
0
22
25
)
,
,
(
,
0
11
15
)
,
,
(
,
0
13
4
15
)
,
,
(
2
3
2
2
2
1
z
y
z
y
x
f
z
y
x
z
y
x
f
z
y
x
z
y
x
f
Boshlang‘ich yaqinlashishni
x
0
= (1.0, 0.8, 0,9)
T
deb olib, yechimni 10
-5
aniq-
likda topamiz.
Usul qoidalariga ko‘ra
J
(
x
) – Yakob matritsasini quyidagicha yozamiz:
.
25
2
0
1
15
2
4
2
15
)
,
,
(
y
x
y
z
y
x
J
Ushbu
x
0
= (1.0, 0.8, 0,9)
T
boshlang‘ich yaqinlashish uchun
T
)
14
.
0
,
1
.
1
,
96
.
0
(
)
(
0
x
F
155
va
.
25
6
.
1
0
1
15
2
4
6
.
1
15
)
(
0
x
J
Endi
)
(
)
(
0
0
0
x
F
u
x
J
tenglamaning yechimi quyidagicha:
00028552
.
0
0830388
.
0
07293361
.
0
0
u
va
.
90028552
.
0
71696122
.
0
07293361
.
1
0
0
1
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