O‘zbekiston respublikasi oliy va o‘rta maxsus ta’lim vazirligi samarqand davlat universiteti abdirashidov A., Babayarov A. I

 
Shu yechimni Nyuton usuli yordamida Maple dasturida taqribiy hisoblaymiz:
 
Avvalo Yakob matritsasini 
linalg
paketining 
jacobian
funksiyasi yordamida 
hisoblaymiz, keyin esa uning teskarisini 
linalg
paketining 
inverse
funksiyasidan foy-

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140 
dalanib hisoblaymiz. 
eval
funksiyasi ifodaning son qiymatini beradi. 
evalm
funksiyasi esa matritsa va vektorlar ustida amal bajarib, son natija beradi. Bosh-
lang‘ich vektorni 
xx:=[0.5;1.5]
va 
eps:=0.001
aniqlik darajasi deb, Nyuton usuli 
bo‘yicha taqribiy hisoblashlarni bajaramiz: 
> with(linalg): 
F:=(x,y)->[x*y-y^3-1,x^2*y^2+y^3-5];
FP:=jacobian(F(x,y),[x,y]); FPINV:=inverse(FP);
xx:=[0.5,1.5]; eps:=0.001; 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; 
Natijalar quyidagicha: 
:= 
F

(
)
,
x y
[
]
,
 
x y
y
3
1
 
x
2
y
2
y
3
5
:= 
FP






y

x
3
y
2
2
x y
2

2
x
2
y
3
y
2
:= 
FPINV













2
x
2
3
y
3
y
2
(
)

1
2
x y


x
3
y
2
3
y
3
(
)

1
2
x y

2
x
3
y
(
)

1
2
x y
1
3
y
2
(
)

1
2
x y
:= 
xx
[
]
,
0.5 1.5
:= 
eps
0.001
:= 
Err
1000
:= 
v
[
]
,
0.5 1.5
:= 
v1
[
]
,
0.1 10
11
0.1 10
11
:= 
j
0
:= 
v1
[
]
,
0.5 1.5
:= 
M






0.2962962963
0.2469135803
-0.08888888887 0.05925925927
:= 
v
[
]
,
1.836419753 1.240740741
:= 
Err
1.336419753
:= 
j
1
:= 
v1
[
]
,
1.836419753 1.240740741
:= 
M






0.4078488757
0.08736397583
-0.1775644435
0.03896485017
:= 
v
[
]
,
1.910372475 1.046712441
:= 
Err
0.194028300
:= 
j
2
:= 
v1
[
]
,
1.910372475 1.046712441
:= 
M






0.6353135777
0.08003024373
-0.2433868149
0.06085855173
:= 
v
[
]
,
1.992251965 1.002053670
:= 
Err
0.081879490
:= 
j
3
:= 
v1
[
]
,
1.992251965 1.002053670
:= 
M






0.7276965560
0.06768724860
-0.2654774883
0.06649093770
:= 
v
[
]
,
1.999976663 1.000005095
:= 
Err
0.007724698
:= 
j
4
:= 
v1
[
]
,
1.999976663 1.000005095
:= 
M






0.7333182897
0.06666959200
-0.2666635987
0.06666633793
:= 
v
[
]
,
2.000000000 1.000000000
:= 
Err
0.000023337
:= 
j
5
Iteratsion jarayonning 5-qadamida berilgan aniqlikdagi yechimga erishildi. 
 
4-Misol.
Quyidagi uch noma’lumli uchta nochiziqli tenglamalar sustemasini 
Nyuton usuli bilan yeching: 

---

141 
 
 
 





















.
0
2
3
.
0
,
,
0
3
2
.
0
,
;
0
2
1
.
0
,
2
3
2
2
2
1
z
xy
z
y
x
f
y
xz
y
y
x
f
x
yz
x
y
x
f
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:=<, 
, >; 
:= 
A










2
x0
1
2
z0
2
y0

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; 
:= 
A








-1
0
0
0
-1
0
0
0
-1
A1:=A^(-1); 
:= 
A1








-1
0
0
0
-1
0
0
0
-1
f:=; 
:= 
f








0.1
-0.2
0.3
X0:=: 
X:=Add(X0,(Multiply(A1,f)),1,-1); 
:= 
X








0.100000000000000004
-0.200000000000000010
0.299999999999999988
X0:=X; x0:=X[1];y0:=X[2];z0:=X[3]; 
A:=<,,>; 
:= 
A








-1.200000000
0.6000000000 -0.4000000000
-0.9000000000
-1.400000000
-0.3000000000
0.4000000000 -0.2000000000
-1.600000000
A1:=A^(-1); f:=; 
:= 
f








-0.1300000000
-0.0500000000
-0.0500000000
X:=Add(X0,(Multiply(A1,f)),1,-1); 
:= 
X








0.0224532224532224546
-0.174324324324324320
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:
:= 
X








0.0128241509376391898
-0.177800663726073254
0.244688047122264718
Sistemaning barcha haqiqiy yechimlarini topaylik: 
> 
solve
({0.1-
x
2
+2

x

y
-
x
=0,-0.2+
y
2
-3

x

z
-
y
=0,0,3-
z
2
-2

x

y
-
z
=0});
 

---

142 
5-Misol.
Quyidagi nochiziqli tenglamalar sistemasini yechishning iteratsion ja-
rayonini quring:











.
0
1
)
,
(
,
0
1
)
1
(
)
,
(
2
2
y
x
y
x
g
x
y
y
x
f
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: 
n
n
y
x
1
1
1



,
1
2
1
1





n
n
x
y
, 
n
=0,1,2,… 
Natijalar esa quyidagi jadvalda keltirilgan: 
n 
= 
0 
1
… 
17 
18 


n
n
y
x
2.0000 
1.7321 
1.5773 
1.2198 
… 
1.7166
1.3952 
1.7167 
1.3954
Buning uchun quyidagi baholash o‘rinli: 
4
18
18
10
*
8








y
y
x
x
. 

---

143 
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: 
1
2
1




n
n
y
x
,
1
1
1
1




n
n
x
y
. 
Natijalar esa quyidagi jadvalda keltirilgan: 
n 
= 
0 
1 
2 
3 
4 


n
n
y
x
-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: 






3
3
y
y
x
x
4
10
*
2


. 
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;