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

4-Misol. Quyidagi uch noma’lumli uchta nochiziqli tenglamalar sustemasini Nyuton usuli bilan yeching:

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f₁x, y  0.1 x²  2 yz  x  0; ^


f₂ x, y  0.2  y²  3xz  y  0,^


f₃ x, y  0.3  z²  2xy  z  0. ^
Yechish. Misolni Maple dasturida yechamiz. Yakob matritsasini tuzamiz:

>restart; with(LinearAlgebra): f1:=0.1-x0^2+2*y0*z0-x0; f1 := 0.1  x0^2  2 y0 z0  x0 f2:=-0.2+y0^2-3*x0*z0-y0; f2 := 0.2  y0 ^2  3 x0 z0  y0 f3:=0.3-z0^2-2*x0*y0-z0; f3 := 0.3  z0^2  2 x0 y0  z0 f1x:=diff(f1,x0); f1y:=diff(f1,y0); f1z:=diff(f1,z0); f2x:=diff(f2,x0); f2y:=diff(f2,y0); f2z:=diff(f2,z0); f3x:=diff(f3,x0); f3y:=diff(f3,y0); f3z:=diff(f3,z0); A:=<, , >; 2 x0  1 2 z0 2 y0  A := ^3 z0 2 y0  1 3 x0 ^ ^2 y0 2 x0 2 z0  1^   # Ildizga yaqin bo‘lgan boshlang‘ich yaqinlashishni tanlaymiz: x0:=0: y0:=0: z0:=0: A:=A; -1 0 0 A := ^0 -1 0^    0 0 -1 A1:=A^(-1); -1 0 0 A1 := ^0 -1 0^    0 0 -1

f:=;  0.1  f := -0.2 ^ 0.3 ^ X0:=: X:=Add(X0,(Multiply(A1,f)),1,-1);  0.100000000000000004 X := -0.200000000000000010 ^ 0.299999999999999988^ X0:=X; x0:=X[1];y0:=X[2];z0:=X[3]; A:=<,,>;  -1.200000000 0.6000000000 -0.4000000000 A := -0.9000000000 -1.400000000 -0.3000000000^ ^ 0.4000000000 -0.2000000000 -1.600000000^ A1:=A^(-1); f:=; -0.1300000000 f := -0.0500000000 ^-0.0500000000^ X:=Add(X0,(Multiply(A1,f)),1,-1); 0.0224532224532224546 X := ^-0.174324324324324320    0.246153846153846140 i:=2: while (Norm(f))>0.0001 do X0:=X; x0:=X[1];y0:=X[2];z0:=X[3]; A:=<,,>; A1:=A^(-1); f:=; X:=Add(X0,(Multiply(A1,f)),1,-1); i:=i+1; end do: X:=X; # Natija: 0.0128241509376391898 X := ^-0.177800663726073254    0.244688047122264718

Sistemaning barcha haqiqiy yechimlarini topaylik:

• 
  
  
  solve({0.1-x²+2xy-x=0,-0.2+y²-3xz-y=0,0,3-z²-2xy-z=0});
  

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5-Misol. Quyidagi nochiziqli tenglamalar sistemasini yechishning iteratsion ja- rayonini quring:

_ f (x, y)  y(x 1) 1  0,


^g(x, y)  x²  y² 1  0.
Yechish. Dastlab berilgan nochiqli tenglamalar sistemasining yechimi mavjud- ligini Maple dasturi yordamida grafik usulda aniqlaylik (3.15-rasm):

• 
  plots[implicitplot]({y*(x-1)-1=0,x^2-y^2-1=0},x=-3..3,y=-3..3);
  

3.15-rasm. 5-misolda berilgan tenglamalar sistemasi ildizining boshlang‘ich yaqin- lashishini grafik usul bilan Maple dasturi yordamida aniqlash.

3.14-rasmdagi grafikdan ko‘rinadiki, bu tenglamalar sistemasi 2 ta haqiqiy yechimga ega.

Berilgan nochiziqli tenglamalar sistemasining 1-chorakdagi aniq yechimi anali- tik usulda Maple dasturi yordamida quyidagicha topiladi:

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

{ x  1.716672747, y  1.395336994}

Birinchi chorakda yotgan yechimni taqribiy usul bilan topish uchun quyidagi it- eratsion jarayondan foydalaniladi:

^xn1
 1 ¹ _,
y_n
y_n1  

2

x
n1
^1, n=0,1,2,…

Natijalar esa quyidagi jadvalda keltirilgan:

n =	0	1	…	17	18
xn  yn 	2.0000 1.7321	1.5773 1.2198	…	1.7166 1.3952	1.7167 1.3954

.
Buning uchun quyidagi baholash o‘rinli:


x  x₁₈ y  y

^  8*10^4

18 ^_
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Hisoblashlarning Maple dasturi quyidagicha bo‘lib, yuqoridagi jadval natijalarini tasdiqlaydi:
> with(linalg): G:=(x,y)->[1+1/y,sqrt(x^2-1)]; v:=evalf(G(v[1],v[2])); eps:=0.0001; Err:=1000; v:=[2,1.7]; j:=0;
for i while Err>eps do
v1:=evalf(v): v:=evalf(G(v[1],v[2])); Err:=max(abs(v1[1]-v[1]),abs(v1[2]-v[2])): j:=j+1; end do;
Uchinchi chorakda yotgan yechimni taqribiy usul bilan topish uchun quyidagi iteratsion jarayondan foydalaniladi:

x_n1  
, ^yn1 ^
1
^xn1
1 ^.

Natijalar esa quyidagi jadvalda keltirilgan:

n =	0	1	2	3	4
xn  yn 	-1.1 -0.5	-1.1076 -0.4721	-1.1059 -0.4745	-1.1069 -0.4749	-1.1070 -0.4746

Buning uchun quyidagi baholash o‘rinli:


x  x₃ y  y
^{  2*104} .

₃ _

Hisoblashlarning Maple dasturi quyidagicha bo‘lib, jadval natijalarini tasdiqlaydi:
> with(linalg): G:=(x,y)->[-sqrt(y^2+1),1/(x-1)]; v:=evalf(G(v[1],v[2])); eps:=0.0001; Err:=1000; v:=[-1.1,-0.5]; j:=0;
for i while Err>eps do v1:=evalf(v): v:=evalf(G(v1[1],v[2])); Err:=max(abs(v1[1]-v[1]),abs(v1[2]-v[2])): j:=j+1; end do;


Nochiziqli tenglamalar sistemasini Mathcad dasturi yordamida taqribiy yechish.
Quyida Mathcad hisoblash tizimida tuzilgan 3 ta dasturiy modul keltirilgan bo‘lib, ular ushbu
F(x)=0
nochiziqli tenglamalar sistemasini taqribiy yechadi. Bu modullarning sarlavhalari quyidagicha:
NSys_Z(x,F,ε) – Zeydel algoritmini amalga oshiruvchi dastur; NSys_N(x,F,ε) – Nyuton algoritmini amalga oshiruvchi dastur; NSys_B(x,F,ε) – Broyden algoritmini amalga oshiruvchi dastur,
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bu yerda x – boshlang‘ich yaqinlashishning ustun-vektori; F – tenglamalar sistemasi chap qismining vektor funksiyasi nomi; J – Yakob matritsasi nomi; ε – yechimni topishning aniqligi. Bu dasturlardan NSys_B(x,F,ε) dastur Der(x,F,ε) – yordamchi modulga murojaat qiladi, bu Yakob matritsasining x nuqtadagi qiymatini hisoblab be- radi.
Dasturlarning ishlash jarayonini ko‘rsatish maqsadida quyidagi misolni keltiril- gan modullarda yechish ko‘rsatilgan:

• 
  Nochiziqli tenglamalar sistemasini yechish dasturiga murojaat:
  

Nochiziqli tenglamalar sistemasini yechishning Zeydel usuli dasturi:

Nochiziqli tenglamalar sistemasini yechishning Nyuton usuli dasturi:

144

Nochiziqli tenglamalar sistemasini yechishning Broyden usuli dasturi:

1-misol. Quyidagi nochiziqli tenglamalar sistemasi berilgan:
^_tg (x x  1)  x ²
_ 1 2 2

1

2
^_x ²  0,7x ²  1
Bu tenglamalar sistemasini oddiy iteratsiya usulining Zeydel takomillashgan var- ianti bilan  = 0.001 va  = 0.0001 aniqliklarda yechish talab qilinadi.
Yechish. Berilgan sistemani standart shaklda yozib olamiz:

_{ f}
(x , x
)  tg (x x
1)  x

²  0

_ 1 1 2 1 2 2
2 2

1

2
^_ f₂ (x₁, x₂ )  x  0,7x 1 0
Bu funksiyalarning aniqlanish sohalari:
D_f1 = {–∞ < x₁ < ∞; –∞ < x₂ < ∞};
D_f2 = {–1 ≤ x₁ ≤ 1; –1,195 ≤ x₂ ≤ 1,195};
Bu funksiyalarning qiymatlar sohalari:
D_o = {–1 ≤ x₁ ≤ 1; –1,195 ≤ x₂ ≤ 1,195};
Вu funksiyalarning grafiklarini Mathcad dasturida chizamiz (3.16-rasm). Grafi- klardan ko‘rinadiki, misolda berilgan sistema 4 ta haqiqiy yechimga ega:
D_I = {0 < x₁ < 0,1; –1,3 < x₂ < –1,1};
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D_II = {0,7 < x₁ < 0,9; 0,6 < x₂ < 0,8};
D_III = {–0,1 < x₁ < 0; 1,1 < x₂ < 1,3};
D_IV = {–0,9 < x₁ < –0,7; 0,6 < x₂ < 0,8};

3.16-rasm. 1-misolda berilgan tenglamalar sistemasi ildizining boshlang‘ich yaqin- lashishini grafik usul bilan Mathcad dasturi yordamida aniqlash.

Sistemaning barcha yechimlari uchun iteratsion jarayonning yaqinlashish formu- lasini chiqaramiz.

^x  x  ¹ (x ²  0,7  x ²  1)
 ^{1 1} ₃ ^{1 2}
I-yechim uchun: ^

x

 


^ ₂ tg(x₁  x₂  1)
Xususiy hosilalarni aniqlaymiz:

_{2  }
^x1 2
x_{2 ;}
tg(x₁  x₂  1)
_{2  }
^x2 2
x_{1 ;}
tg(x₁  x₂  1)

Birinchi yechim uchun boshlang‘ich yaqinlashishni aniqlanish sohasidan unga yaqin bo‘lgan х₁=0,1; х₂=–1,2 nuqtani olib, yaqinlashish shartini tekshiramiz:

    
 

 0,59  1.

Ko‘rinadiki, yaqinlashish sharti bajarilayapti. Demak, hosil qilingan ekvivalent sistemadan birinchi yechimni aniqlashtirish uchun foydalanish mumkin.
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2. 
   yechim uchun:
   

^_x₁ 