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

1-misol. Tezkor tushish usili (gradi- yent usuli) yordamida ushbu x  x^2  2 yz  0,1; ^ y  y^2  3xz  0, 2;^  z  z^2  2xy  0,3. ^  tenglamalar sistemaning koordinata boshi atrofida yotgan ildizlarini taqribiy hisoblang. Yechish Berilganlarga ko‘ra

3.11-rasm. Gradiyent usuli algoritmining blok-sxemasi.

^ x  x²  2 yz  0,1 ^
_1  2x
2z
2 y _
⁰

 
f  ^ y  y²  3xz  0, 2^ ,
^{W }  ^{3z
1  2 y 3x}  ^{,
x0 } ⁰ ^.

 
^ z  z²  2xy  0,3 ^
_ ^{2 y
2x 1  2z} _
_0_

(3.41) va (3.42) formulalar bo‘yicha quyidagi birinchi yaqinlashishni olamiz:

_{ f} ⁰_{, f} ⁰ _
 0,1 _

   1^,


x^1  x^0 1 Ef ^0  _0, 2_ .

⁰ _{ f} ⁰_{, f} ⁰ _
 0,3 _

Xuddi shunday
_x²
- ikkinchi yaqinlashishni aniqlaymiz. Bu yerdan:

_{f 1  }0,13_{ ,
} 1, 2
W 
0,6 0, 4_
,
_{W f 1  } 0,181_{ .}

_0,05_
0,05_{
1 } 0,9 1, 4 0,3_{ 1
}0, 4 0, 2 1,6 _
0,002_
0,147_

WW ^f ^1  ^{ 0,181} ,
_{ } 0,13  0, 2748  0,05  0, 2098  0,05  0,1632 _{ 0,3719 .}

_{1 1 }0,002_
0,147_
x²
0, 2748²  0, 2098²  0,1632²
_ 0,1 _{ } 0,181_{ } 0,0327 _
 0,37119  .

 _0, 2_{ }0,002_  _0, 2007_

_ 0,3 _{
}0,147_{
} 0, 2453 _

Natijaning qanchalik to‘g‘ri va aniq ekanligini tekshirish uchun tafovut hisoblanadi.
135

Namunaviy misollar va ularning yechimlari

Nochiziqli tenglamalar sistemasini Maple dasturi yordamida taqribiy yechish.
1-misol. Quyidagi nochiziqli tenglamalar sistemaning yechimini  = 0,001 aniqlik bilan Nyuton usulida taqribiy hisoblang:
f₁x, y  2x³  y² 1  0;^



2
f x, y  xy³  y  4  0. ^
Yechish. Ushbu misolda berilgan tenglamalar sistemasi bitta musbat haqiqiy yechimga ega ekanligini quyidagi Maple dasturi plots paketining implication funk-

siyasidan foydalanib ^f^1(^x^,^y^)^^0va ^f^2 (^x^,^y^)^^0funksiyalarning chizil- gan grafiklaridan ko‘rish mumkin (3.12-rasm): plots[implicitplot]({2*x^3-y^2- 1=0,x*y^3-y-4=0},x=-2..2,y=-3..3); Bu usulga ko‘ra dastlabki yaqinlashish ^x0 ^^1,2;^y0 ^^1,7kabi bo‘lsin. U holda   6x0  2 y0 yoki 2  x0 , y0  y^3 3x y^2 1 0 0 0 1,2; 1,7  ^8,64^^3,40 97,910 . 4,91 9,40 (12) formulaga ko‘ra

3.12-rasm. 1-misolda berilgan nochiziqli tenglamalar sistemasi ildizining boshlang‘ich yaqinlashishni grafik usul bilan Maple dasturi yordamida aniqlash.

x₁  1,2 
 0,434
0,1956
8,64
 3,40
9,40
 0,434


 1,2  0,0349
 1,2349 ;^





y₁  1,7 


4,91


0,1956
 1,7  0,0390
 1,6610 . ^
_

Hisoblashlarni shu singari davom ettirib,
x₂  1,2343
y₂  1,6615
ni topamiz va

hisoblashlarni talab qilingan aniqlikkacha davom ettiramiz.
Berilgan tenglamalar sistemasining mavjud bitta haqiqiy yechimini Maple dasturi yordamida ham analitik usulda aniqlaylik:

• 
  solve({2*x^3-y^2-1=0,x*y^3-y-4=0},{x,y}); allvalues(%); evalf(%);
  

{x  1.234274484, y  1.661526467}

Natijalardan ko‘rinadiki, topilgan

x₂  1,2343

y₂  1,6615
- taqribiy yechimni

yetarlicha  aniqlikda topilgan deb hisoblash mumkin.
136

Endu bu masalani Maple tizimida sonli yechishni qaraymiz. Avvalo Yakob matritsasini linalg paketining jacobian funksiyasi yordamida hisoblaymiz, keyin esa uning teskarisini linalg paketining inverse funksiyasidan foydalanib hisoblaymiz. eval funksiyasi ifodaning son qiymatini beradi. evalm funksiyasi esa matritsa va vektorlar ustida amal bajarib, son natija beradi. Boshlang‘ich vektorni xx va eps aniqlik darajasi deb, Nyuton usuli bo‘yicha taqribiy hisoblashlarni bajaramiz:

• 
  with(linalg):
  

F:=(x,y)->[2*x^3-y^2-1,x*y^3-y-4];
FP:=jacobian(F(x,y),[x,y]); FPINV:=inverse(FP);
xx:=[1.2,1.7]; eps:=0.0001; Err:=1000; v:=xx; v1:=[1e10,1e10]; j:=0:
for i while Err>eps do v1:=eval(v);
M:=eval(eval(FPINV),[x=v[1],y=v[2]]):
v:=evalm(v-M&*F(v[1],v[2])); Err:=max(abs(v1[1]-v[1]),abs(v1[2]-v[2])); j:=j+1;
end do;

Hisob natijasi quyidagicha: F := ( x, y )  [ 2 x^3  y^2  1, x y^3  y  4 ] 6 x^2 2 y  ^F^P^:^=^ y^3 3 x y^2  1^  3 x y^2  1 y  ^2 ( 9 x^3 y^2  3 x^2  y^4 ) 9 x^3 y^2  3 x^2  y^4 ^ FPINV := ^^  y^3 3 x^2  ^ 2 ( 9 x^3 y^2  3 x^2  y^4 ) 9 x^3 y^2  3 x^2  y^4 ^ xx := [ 1.2, 1.7 ] eps := 0.0001 Err := 1000 v := [ 1.2, 1.7 ] v1 := [ 0.1 10 ^11, 0.1 10 ^11 ] v1 := [ 1.2, 1.7 ] M :=  0.09600350200 0.03470990077 -0.05015580660 0.08820398313^

v := [ 1.234876263, 1.660979681] Err := 0.039020319 j := 1 v1 := [ 1.234876263, 1.660979681] M :=  0.09258867450 0.03335772210 -0.04601453420 0.09187560387^ v := [ 1.234274675, 1.661526276] Err := 0.000601588 j := 2 v1 := [ 1.234274675, 1.661526276] M :=  0.09264916080 0.03338417877 -0.04608134315 0.09182868696^ v := [ 1.234274484, 1.661526467] Err := 0.191 10 ^-6 j := 3

Natija shuni ko‘rsatadiki, hisob jarayonining 3-qadamida berilgan aniqlikdagi yechimga erishildi.
2-misol. Quyidagi nochiziqli tenglamalar sistemasining musbat yechimini  = 0,001 aniqlik bilan Nyuton usulida taqribiy hisoblang:

1
f x, y  0,1x²  x  0,2 y²  0,3  0;^
  ^{2 }
f₂ x, y  0,2x  y  0,1xy  0,7  0._
137
Yechish. Boshlang‘ich yaqinlashishni tanlab olish uchun grafik usuldan, Maple dasturi plots paketining implication funksiyasidan foydalanib, (x,y) tekislikning biz-

ni qiziqtiradigan sohasida rini chizamiz (3.13-rasm):
f₁(x, y)  0 _va
f₂ (x, y)  0
egri chiziqlarning grafikla-

• 
  plots[implicitplot]({0.1*x^2+x+0.2*y^2-0.3=0,0.2*x^2+y-0.1*x*y-0.7=0},x=-
  

2..2,y=-2..2);

Bundan berilgan tenglamalar sistemasi- ning biz izlayotgan musbat yechimi 0<x<0,5; 0<y<1,0 kvadrat ichida ekanligini ko‘ramiz. Boshlang‘ich yaqinlashishni ^x0 ^^0,25;^y0 ^^0,75deb qabul qilamiz. U holda qara- layotgan misol uchun quyidagilarni yozamiz: f1x0 , y0   0,1x^2  x  0,2 y^2  0,3  0; ^ 0 0 0  f2 x0 , y0   0,2x^2  y  0,1x y  0,7  0, 0 0 0 0  ^^f^1^(^x^0^,^y^0^) 0,2x  1 ; ^^f^1^(^x^,^y^0^) 0,4 y ; x ^0 y ^0

3.13-rasm. 2-misolda berilgan tenglamalar sistemasi ildizining boshlang‘ich yaqinlashishini grafik usul bilan aniqlash.

f₂ (x₀ , y₀ ) _{ 0,4x}
 0,1y _;
f₂ (x₀ , y₀ ) _{ 1  0,1x .}

x ^{0 0} y ⁰

Tanlangan
X ₀  (x₀ , y₀ )
larni (3.12) ning o‘ng tarafiga qo‘yib, dastlab taqribiy

X₁  (x₁, y₁ )
ni topamiz:


x  0,25  ^0,05391  0,19498;^

1


y₁  0,75 
0,97969
0,04258

0,97969



 0,70654,^
_

o‘z navbatida esa
X ₂  (x₂ , y₂ )
= (0,19646, 0,70615) ni va
X ₃  (x₃ , y₃ ) ₌

(0,19641, 0,70615) ni topamiz va hokazo. Iteratsiya jarayonini (3.13) shart bajaril- gunga qadar davom ettiramiz. Bu hisoblashlar berilgan sistemaning yechimi (x,y) = (0,1964; 0,7062) ekanligini ko‘rsatadi.
Bu topilgan yechimning qanchalik to‘g‘riligini Maple dasturi yordamida aniqlashtiramiz:

• 
  solve({0.1*x^2+x+0.2*y^2-0.3=0,0.2*x^2+y-0.1*x*y-0.7=0},{x,y});
  

{ x  .1964115055, y  .7061541848} _.

Endi Nyuton usuli bilan misolning taqribiy yechimini topamiz:

• 
  with(linalg):
  

F:=(x,y)->[0.1*x^2+x+0.2*y^2-0.3,0.2*x^2+y-0.1*x*y-0.7];
FP:=jacobian(F(x,y),[x,y]); FPINV:=inverse(FP);

138

xx:=[0.25,0.75]; eps:=0.0001; Err:=1000; v:=xx; v1:=[1e10,1e10]; j:=0:
for i while Err>eps do
v1:=eval(v); M:=eval(eval(FPINV),[x=v[1],y=v[2]]):
v:=evalm(v-M&*F(v[1],v[2])); Err:=max(abs(v1[1]-v[1]),abs(v1[2]-v[2])); j:=j+1; end do;
F := ( x, y )  [ 0.1 x²  x  0.2 y²  0.3, 0.2 x²  y  0.1 x y  0.7 ]

FP :=  0.2 x  1 0.4 y  ^0.4 x  0.1 y 1  0.1 x^ xx := [ 0.25, 0.75 ] eps := 0.0001 Err := 1000 v := [ 0.25, 0.75 ] v1 := [ 0.1 10 ^11, 0.1 10 ^11 ] v1 := [ 0.25, 0.75 ] M :=  0.9594095941 -0.2952029520 ^-0.02460024600 1.033210332 ^^  v := [ 0.1969557196, 0.7064883149] Err := 0.0530442804 j := 1	v1 := [ 0.1969557196, 0.7064883149] M :=  0.9642769266 -0.2779750296 ^-0.008000478288 1.022397604 ^ v := [ 0.1964115443, 0.7061542263] Err := 0.0005441753 j := 2 v1 := [ 0.1964115443, 0.7061542263] M :=  0.9643276107 -0.2778427597 ^-0.007819206552 1.022287533 ^ v := [ 0.1964115055, 0.7061541848] Err := 0.415 10 ^-7 j := 3