Respublikasi oliy va o‘rta maxsus ta’lim vazirligi samarqand davlat universiteti

x = Ф(x) bog‘lanish orqali quyidagi iteratsion formulalarni qurishimiz mumkin:

x

_ n 1

n

  ¹
15
1
y ² 
4
15 ^zn
1
 ¹³  0;
15
11

y

^ n 1

  x² 

n
15
15 ^{zn } 15 ^;

^z  ¹ y ²  ²² .

_ n 1
25 ⁿ 25

Bu hisoblashlarni
^xn1 ^{ x}n
_  10^5
shart bajarilgunga qadar davom ettirsak,

quyidagi jadval natijalariga kelamiz:

n	xn	yn	zn	^xn1 ^^xn 
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 x₇ yechim yetarlicha aniqlikda. Bunda yaqinlashish tezligini quyidagicha baholash o‘rinli:

_{* q}7

0.9⁷

* 8

x₇  x
_  x₁  x_0
  _{1  q}  0.07333  _{1  0.9}  0.351 , chunki
x₇  x
^{ 7.6 10} .


2) Zeydel usuli. Iteratsion jarayonni x₀ = (1.0, 0.8, 0.9)^T boshlang‘ich yaqinlash- ish bilan quyidagi formulalarda amalga oshiramiz (hisob natijalari quyidagi jadvalda keltirilgan):
153

x

_ n1
  ¹
15
y²  ⁴ z

n
15 ⁿ
 ¹³  0;
15



y

^ n1



1

  ¹
15

2

x
n1

• 
  ¹ z
  

15 ⁿ
_ 11_;
15

_^zn1 ^ ₂₅
2

y
n1
_ 22 _.
25

n	xn	yn	zn	^xn1 ^^xn 
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 x₅ yechim yetarlicha aniqlikda. Bu yerda ham

_{* q}5
0.9⁵
* 10

x₅  x
_  x₁  x_0
  _{1  q}  0.08214  _{1  0.9}  0.486 , chunki
x₅  x
 10 .


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
F(x)  F(x, y, z)  ( f₁(x, y, z), f₂ (x, y, z), f₃ (x, y, z)) 
 15x  y²  4z  13  0; x²  15y  z  11  0; y²  25z  22  0^T .
belgilashlarni yuqorida qabul qilgan edik, bu yerda
^ f₁(x, y, z)  15x  y²  4z 13  0,


_ f₂

f

^_ 3
(x, y, z)  x² 15y  z 11  0, (x, y, z)  y²  25z  22  0.

Boshlang‘ich yaqinlashishni x₀ = (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:


_15 2 y J (x, y, z)  ^2x 15
^_ 0 2 y
 4 _


1 ^.
 25^_

Ushbu x₀ = (1.0, 0.8, 0,9)^T boshlang‘ich yaqinlashish uchun
F(x₀ )  (0.96, 1.1, 0.14)^T

154

va
_15


J (x₀ )  ^ 2
^_ 0

1.6
15

1.6

 4 _


1 ^.
 25^_

Endi
J (x₀ )u₀  F(x₀ ) ^{tenglamaning yechimi quyidagicha:}

_ 0.07293361_
u₀  ^ 0.0830388^
_1.07293361_
va x₁  x₀  u₀  ^0.71696122^.

   
^_ 0.00028552^_ ^_0.90028552_^
Bu hisoblash jarayonini xuddi shu tartibda k = 1, 2, … lar uchun davom ettirsak,

^{ x}k 1 ^
 ^x_k 
^u_k 

 _{yk 1} _  _yk  _  _vk  _,
     

bu yerda

^u_k 
^ ^zk 1 ^ ^ ^zk ^ ^^wk ^
_ 15
2 y_k
 4 _

^ v_k ^  J (x_k , y_k , z_k )^1Fx_k , y_k , z_k ; J (x_k )  ^2x_k 15 1 ^.
   

^^wk ^
^_ 0
2 y_k
 25^_

Nyuton usuli iteratsiyalarining yaqinlashishini quyidagi jadvaldan ko‘rish mumkin:

n	xn	yn	zn	^xn1 ^^xn 
0	1.00000000	0.80000000	0.90000000
1	1.07293361	0,71696122	0.90028552	0.073
2	1.07257038	0.71667586	0.90054497	3.610^-4
3	1.07257037	0.71667585	0.90054497	8,0510^-9

Bu yerda topilgan x₃ yechim yetarlicha aniqlikda. Bu yerda ham

_{* q}3
0.9³

* 9

x₃  x
_  x₁  x_0
  _{1  q}  0.073  _{1  0.9}  0.533 , chunki
x₃  x
 10 .


Demak, ushbu misolni yechishda Nyuton usulining yaqinlashish tezligi oddiy it- eratsiyalar va Zeydel usullariga nisbatan ancha yuqori ekan.
4) Broyden usuli. Bu yerda ham Nyuton usulidagi kabi berilgan sistema uchun ushbu
F(x)  F(x, y, z)  ( f₁(x, y, z), f₂ (x, y, z), f₃ (x, y, z)) 

 15x  y²  4z  13  0; x²  15y  z  11  0; y²  25z  22 
0^T .

belgilashlarni yuqorida qabul qilgan edik, bu yerda

155

^ f₁(x, y, z)  15x  y²  4z 13  0,


_ f₂

f

^_ 3
(x, y, z)  x² 15y  z 11  0, (x, y, z)  y²  25z  22  0.

Boshlang‘ich yaqinlashishni x₀ = (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:


_15 2 y J (x, y, z)  ^2x 15
^_ 0 2 y
 4 _


1 ^.
 25^_

Ushbu x₀ = (1.0, 0.8, 0,9)^T boshlang‘ich yaqinlashish uchun
_15

1.6

 4 _


F(x₀ )  (0.96, 1.1, 0.14)^T va A₀ = J (x₀ )  ^ 2
^_ 0
15
1.6
1 ^.


 25^_

Endi tenglamaning yechimini quyidagicha izlaymiz:
_1.07293361_