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11. Nochiziqli tenglamalar sistemasini Maple dasturi
yordamida taqribiy yechish
1-misol.
Quyidagi
0
4
,
0
1
2
,
3
2
2
3
1
y
xy
y
x
f
y
x
y
x
f
nochiziqli tenglamalar sistemaning yechimini
= 0,001 aniqlik bilan
Nyuton usulida taqribiy hisoblang.
Yechish.
Ushbu misolda berilgan tenglamalar sistemasi bitta musbat
haqiqiy yechimga ega ekanligini quyidagi Maple dasturi
plots
paketining
implication
funksiyasidan
foydalanib
0
)
,
(
1
y
x
f
va
0
)
,
(
2
y
x
f
funksiyalarning chizilgan grafiklaridan ko‘rish mumkin (10-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
7
,
1
;
2
,
1
0
0
y
x
kabi bo‘lsin. U holda
1
3
2
6
,
2
0
0
3
0
0
2
0
0
0
y
x
y
y
x
y
x
yoki
910
,
97
40
,
9
91
,
4
40
,
3
64
,
8
7
,
1
;
2
,
1
.
(12) formulaga ko‘ra
10-rasm. 1-misolda berilgan nochiziqli
tenglamalar sistemasi ildizining boshlang‘ich
yaqinlashishni grafik usul bilan Maple dasturi
yordamida aniqlash.
37
6610
,
1
0390
,
0
7
,
1
1956
,
0
91
,
4
434
,
0
64
,
8
91
,
97
1
7
,
1
2349
,
1
0349
,
0
2
,
1
40
,
9
1956
,
0
40
,
3
434
,
0
91
,
97
1
2
,
1
1
1
y
x
Hisoblashlarni shu singari davom ettirib,
6615
,
1
2343
,
1
2
2
y
x
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
6615
,
1
2343
,
1
2
2
y
x
- taqribiy
yechimni yetarlicha
aniqlikda topilgan deb hisoblash mumkin.
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:
38
:=
F
(
)
,
x y
[
]
,
2
x
3
y
2
1
x y
3
y
4
:=
FP
6
x
2
2
y
y
3
3
x y
2
1
:=
FPINV
3
x y
2
1
2 (
)
9
x
3
y
2
3
x
2
y
4
y
9
x
3
y
2
3
x
2
y
4
y
3
2 (
)
9
x
3
y
2
3
x
2
y
4
3
x
2
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:
39
.
0
7
,
0
1
,
0
2
,
0
,
;
0
3
,
0
2
,
0
1
,
0
,
2
2
2
2
1
xy
y
x
y
x
f
y
x
x
y
x
f
Yechish.
Boshlang‘ich yaqinla-
shishni tanlab olish uchun grafik
usuldan, Maple dasturi
plots
paketining
implication
funksiyasidan foydalanib,
(
x
,
y
) tekislikning bizni qiziqtiradigan
sohasida
0
)
,
(
1
y
x
f
va
0
)
,
(
2
y
x
f
egri
chiziqlarning grafiklarini chizamiz (11-
rasm):
>
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
sistemasining biz izlayotgan musbat
yechimi 0<
x
<0,5; 0<
y
<1,0 kvadrat
ichida ekanligini ko‘ramiz.
11-rasm. 2-misolda berilgan
tenglamalar sistemasi ildizining
boshlang‘ich yaqinlashishini grafik
usul bilan Maple dasturi yordamida
aniqlash.
Boshlang‘ich yaqinlashishni
;
25
,
0
0
x
75
,
0
0
y
deb qabul qilamiz. U
holda qaralayotgan misolimiz uchun quyidagilarni yozib olamiz:
,
0
7
,
0
1
,
0
2
,
0
,
;
0
3
,
0
2
,
0
1
,
0
,
0
0
0
2
0
0
0
2
2
0
0
2
0
0
0
1
y
x
y
x
y
x
f
y
x
x
y
x
f
1
2
,
0
)
,
(
0
0
0
1
x
x
y
x
f
;
0
0
1
4
,
0
)
,
(
y
y
y
x
f
;
0
0
0
0
2
1
,
0
4
,
0
)
,
(
y
x
x
y
x
f
;
0
0
0
2
1
,
0
1
)
,
(
x
y
y
x
f
.
Tanlangan
)
,
(
0
0
0
y
x
X
larni (12) ning o‘ng tarafiga qo‘yib, dastlab
taqribiy
)
,
(
1
1
1
y
x
X
ni topamiz:
,
70654
,
0
97969
,
0
04258
,
0
75
,
0
;
19498
,
0
97969
,
0
05391
,
0
25
,
0
1
1
y
x
40
o‘z navbatida esa
)
,
(
2
2
2
y
x
X
= (0,19646, 0,70615) ni va
)
,
(
3
3
3
y
x
X
=
(0,19641, 0,70615) ni topamiz va hokazo. Iteratsiya jarayonini (13) shart
bajarilgunga 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);
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
x
2
x
0.2
y
2
0.3
0.2
x
2
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
FPINV
:=
1.
0.1000000000
x
0.1000000000
x
0.02000000000
x
2
1.
0.1600000000
x y
0.04000000000
y
2
0.4
y
0.1
x
0.02
x
2
1
0.16
x y
0.04
y
2
,
0.4000000000
x
0.1000000000
y
0.1000000000
x
0.02000000000
x
2
1.
0.1600000000
x y
0.04000000000
y
2
0.2000000000
x
1.
0.1000000000
x
0.02000000000
x
2
1.
0.1600000000
x y
0.04000000000
y
2
,
:=
xx
[
]
,
0.25 0.75
41
:=
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
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