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CHIZIQSIZ TENGLAMALAR SISTEMANI YECHISH UCHUN NYUTON
USULI
Tayanch so‘z va iboralar: Chiziqsiz tenglamalar sistemasini yechishni ketma-ket
yaqinlashish usuli va Nyuton usullari
![Ilmiybaza.uz
V.1. Chiziqsiz tenglamalar sistemasini yechishni ketma-ket yaqinlashish usuli
Bir noma’lumli f(x)=0 tenglamalarni taqribiy yechish usullari.
1) Iteratsiya usuli. Avvalo iteratsiya funksiyasi g(x) tanlab olinadi:
f(x)
0
x
g(x) , q
max g'(x) ,
,
1
x
a b
.
Iteratsiya funksiyasi g(x) ko‘p hollarda f(x)=0 tenglamadan unda ishtirok
etayotgan birorta x ga nisbatan echib olishdan kelib chiqadi. Masalan,
a)
( )
cos( )
0,
cos( )
( ),
( )
sin( )
1,
[0,1].
f x
x
x
x
x
g x
g x
x
x
b)
3
3
2
( ):
5
4
0,
(4
)/5
( ),
( )
3
/5,
[0,1],
( )
1.
f x
x
x
x
x
g x g x
x
x
g x
Lekin, [0,1] dan tashqaridagi echimlar uchun bu iteratsiya funksiyasi
yaramaydi.
c)
3
3
3
5
( ):
5
4
0,
[2,3],
5
4
( ),
( )
1,.
3 5
4
f x
x
x
x
x
x
g x
g x
x
Boshqa hollarda iteratsiya funksiyasi quyidagicha tanlab olinishi mumkin.
a) agar f’(x)>0 bo‘lsa (aks holda f1=-f=0 deyish kerak)
2
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
'( ) ,
,
,
g x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
b) agar f’(x)<0 bo‘lsa f1=-f=0 desak,
(1)
(1)
(1)
1
1
1
1
2
( )
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
( ) ,
,
,
g x
x
lf x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
deb olish mumkin. Jumladan,
1
2
l
M
M
deb olish mumkin.
Demak, ikkita hol bo‘lishi mumkin ekan:
a)
(
1)
( )
( )
1
1
( )
( )
( ), 0
( )
,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
f
x
M x
x
f x
k
M
M
b)
(
1)
( )
( )
1
1
( )
( )
( ),
( )
0,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
M
f
x
x
x
f x
k
M
M
](/media/image/chiziqsiz-tenglamalar-sistemani-yechish-uchun-nyuton-usuli1098page_2.png "Ilmiybaza.uz
V.1. Chiziqsiz tenglamalar sistemasini yechishni ketma-ket yaqinlashish usuli
Bir noma’lumli f(x)=0 tenglamalarni taqribiy yechish usullari.
1) Iteratsiya usuli. Avvalo iteratsiya funksiyasi g(x) tanlab olinadi:
f(x)
0
x
g(x) , q
max g'(x) ,
,
1
x
a b
.
Iteratsiya funksiyasi g(x) ko‘p hollarda f(x)=0 tenglamadan unda ishtirok
etayotgan birorta x ga nisbatan echib olishdan kelib chiqadi. Masalan,
a)
( )
cos( )
0,
cos( )
( ),
( )
sin( )
1,
[0,1].
f x
x
x
x
x
g x
g x
x
x
b)
3
3
2
( ):
5
4
0,
(4
)/5
( ),
( )
3
/5,
[0,1],
( )
1.
f x
x
x
x
x
g x g x
x
x
g x
Lekin, [0,1] dan tashqaridagi echimlar uchun bu iteratsiya funksiyasi
yaramaydi.
c)
3
3
3
5
( ):
5
4
0,
[2,3],
5
4
( ),
( )
1,.
3 5
4
f x
x
x
x
x
x
g x
g x
x
Boshqa hollarda iteratsiya funksiyasi quyidagicha tanlab olinishi mumkin.
a) agar f’(x)>0 bo‘lsa (aks holda f1=-f=0 deyish kerak)
2
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
'( ) ,
,
,
g x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
b) agar f’(x)<0 bo‘lsa f1=-f=0 desak,
(1)
(1)
(1)
1
1
1
1
2
( )
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
( ) ,
,
,
g x
x
lf x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
deb olish mumkin. Jumladan,
1
2
l
M
M
deb olish mumkin.
Demak, ikkita hol bo‘lishi mumkin ekan:
a)
(
1)
( )
( )
1
1
( )
( )
( ), 0
( )
,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
f
x
M x
x
f x
k
M
M
b)
(
1)
( )
( )
1
1
( )
( )
( ),
( )
0,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
M
f
x
x
x
f x
k
M
M
")
Ilmiybaza.uz
V.1. Chiziqsiz tenglamalar sistemasini yechishni ketma-ket yaqinlashish usuli
Bir noma’lumli f(x)=0 tenglamalarni taqribiy yechish usullari.
1) Iteratsiya usuli. Avvalo iteratsiya funksiyasi g(x) tanlab olinadi:
f(x)
0
x
g(x) , q
max g'(x) ,
,
1
x
a b
.
Iteratsiya funksiyasi g(x) ko‘p hollarda f(x)=0 tenglamadan unda ishtirok
etayotgan birorta x ga nisbatan echib olishdan kelib chiqadi. Masalan,
a)
( )
cos( )
0,
cos( )
( ),
( )
sin( )
1,
[0,1].
f x
x
x
x
x
g x
g x
x
x
b)
3
3
2
( ):
5
4
0,
(4
)/5
( ),
( )
3
/5,
[0,1],
( )
1.
f x
x
x
x
x
g x g x
x
x
g x
Lekin, [0,1] dan tashqaridagi echimlar uchun bu iteratsiya funksiyasi
yaramaydi.
c)
3
3
3
5
( ):
5
4
0,
[2,3],
5
4
( ),
( )
1,.
3 5
4
f x
x
x
x
x
x
g x
g x
x
Boshqa hollarda iteratsiya funksiyasi quyidagicha tanlab olinishi mumkin.
a) agar f’(x)>0 bo‘lsa (aks holda f1=-f=0 deyish kerak)
2
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
'( ) ,
,
,
g x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
b) agar f’(x)<0 bo‘lsa f1=-f=0 desak,
(1)
(1)
(1)
1
1
1
1
2
( )
( )
( ),
( )
1
( ), 1
( )
1
0
,
max
( ) ,
,
,
g x
x
lf x
x
lf x g x
lf
x
lf
x
l
M
f
x
a b
M
deb olish mumkin. Jumladan,
1
2
l
M
M
deb olish mumkin.
Demak, ikkita hol bo‘lishi mumkin ekan:
a)
(
1)
( )
( )
1
1
( )
( )
( ), 0
( )
,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
f
x
M x
x
f x
k
M
M
b)
(
1)
( )
( )
1
1
( )
( )
( ),
( )
0,
(
),
0,1,...
k
k
k
g x
x
lf x
x
f x
M
f
x
x
x
f x
k
M
M
Ilmiybaza.uz
s) Agar f(x)=0 tenglama x=g(x) ko‘rinishga keltirgandan keyin
aniqlansaki, max
( )
1
g x
q
, u holda
1
( )
( )
x
g
x
deb iteratsiyalarni
1 :
(
)
k
k
x
x
ko‘rinishda qurish kerak, chunki
( )
1/
( )
1
x
g x
bo‘ladi. Masalan,
2
2
( ),
( )
1/cos ( )
1,
( )
( ),( ( ))
1/(1
)
1
x
tg x tg x
x
x
x
arctg x
x
x
.
Iteratsiya funksiyasi g(x) tanlab olingach iteratsiyalar quyidagicha quriladi:
2,1 ,.....
,
,
),
(
(0)
1)
(
( )
a b k
x
g x
x
k
k
(0) ,
( )
x
q
х
k
л
1)
(
( )
)
(
1
k
k
k
x
x
q
q
x
.
2) Nyuton usuli
,)
('
( )
( ), ( )
0
( )
x
f
f x
x
g x g x
x
f x
( )
(
1)
(
),
1,2,....
k
k
x
g x
k
0
)
(''
)
(
(0)
(0)
x
f
f x
.
2
( )
(0))
1
,
k
x k
q
x
q
.)
('
min
,''
max
,
2
1
2
1
2
x
f
m
f
M
m
M
q
.
3) Uchinchi tartibli aniqlikdagi CHebыshev usuli (Nьyuton usulining rivoji)
2
(
1)
( )
3
( )
( )
( )
( )
0
( ), ( )
,
(
),
0,1,....
'( )
2
( )
k
k
f x
f
x f
x
f x
x
g x g x
x
x
g x
k
f
x
f
x
4) Yuqori tartibli Eyler usuli
(
1)
( )
(
1)
( )
( )
1
1
1
2
2
2
(
1)
( )
( )
2
1
1
2
1
2
1
2
1
2
(
),
(
),
(
),
(
( ))
( )(
( )
( )
,
(
),
(
).
( )
( )
(
( ))
i
k
k
k
k
k
r
i
k
k
k
r
x
g x
x
g
x
x
O h
xg g
x
g x g
x
g x
x
g x
x
O h
x
g x
g
x
g g
x
Jumladan,
2
(
1)
( )
( )
2
1
1
2
( ( )
( )
( )
( )
( )
,
(
),
(
).
2 ( )
( ( )
k
k
k
r
xg g x
g
x
g x
g
x
g x
x
g x
x
O h
x
g x
g g x
V.2. Tenglamani tekshirish. Ushbu tenglamalarni qaraymiz:
3
3
3
1( ):
6
12
0,
2( ):
6
12
0, 3( ):
5
4
0
f
x
x
x
f
x
x
x
f
x
x
x
Ravshanki,
![Ilmiybaza.uz
3
1( ):
6
12
0
f
x
x
x
tenglama uchun, f1(1)=-7<0, f1(2)=8>0,
1
[1,2]
,
3
2( ):
6
12
0
f
x
x
x
tenglama uchun, f2(-3)=3>0, f2(-4)=-28<0,
2
[ 4, 3]
,
3
3( ):
5
4
0
f
x
x
x
tenglama uchun, f3(0)=-4<0, f3(1)=2>0, 3
[0,1]
Yana,
2
1( )
3
6
0,
[1,2],
1
max
1( )
18, 1 1/
1.
f
x
x
x
M
f
x
l
M
,
2
2 ( )
3(
2)
0,
[ 4, 3],
1
max
2 ( )
48, 1 1/
1.
f
x
x
x
M
f
x
l
M
2
3 ( )
3
5
0,
[0,1],
1
max
3 ( )
8, 1 1/
1.
f
x
x
x
M
f
x
l
M
Iteratsiya funksiyalari kiritamiz:
1( ):
1( )/
1
1( )/18
g
x
x
f
x
M
x
f
x
,
2( ):
2( )/
1
2( )/ 48
g
x
x
f
x
M
x
f
x
,
3
2
3( ): (4
)/5, 3 ( )
3
/5,
3 ( )
1,
[0,1]
g
x
x
g
x
x
g
x
x
.
Iteratsiyalarni quyidagicha quramiz:
1
1
1
1
:
1(
), 2
:
2(
), 3
:
3(
),
: 0.. , : 10.
k
k
k
k
k
k
x
g
x
x
g
x
x
g
x
k
n n
](/media/image/chiziqsiz-tenglamalar-sistemani-yechish-uchun-nyuton-usuli1098page_4.png "Ilmiybaza.uz
3
1( ):
6
12
0
f
x
x
x
tenglama uchun, f1(1)=-7<0, f1(2)=8>0,
1
[1,2]
,
3
2( ):
6
12
0
f
x
x
x
tenglama uchun, f2(-3)=3>0, f2(-4)=-28<0,
2
[ 4, 3]
,
3
3( ):
5
4
0
f
x
x
x
tenglama uchun, f3(0)=-4<0, f3(1)=2>0, 3
[0,1]
Yana,
2
1( )
3
6
0,
[1,2],
1
max
1( )
18, 1 1/
1.
f
x
x
x
M
f
x
l
M
,
2
2 ( )
3(
2)
0,
[ 4, 3],
1
max
2 ( )
48, 1 1/
1.
f
x
x
x
M
f
x
l
M
2
3 ( )
3
5
0,
[0,1],
1
max
3 ( )
8, 1 1/
1.
f
x
x
x
M
f
x
l
M
Iteratsiya funksiyalari kiritamiz:
1( ):
1( )/
1
1( )/18
g
x
x
f
x
M
x
f
x
,
2( ):
2( )/
1
2( )/ 48
g
x
x
f
x
M
x
f
x
,
3
2
3( ): (4
)/5, 3 ( )
3
/5,
3 ( )
1,
[0,1]
g
x
x
g
x
x
g
x
x
.
Iteratsiyalarni quyidagicha quramiz:
1
1
1
1
:
1(
), 2
:
2(
), 3
:
3(
),
: 0.. , : 10.
k
k
k
k
k
k
x
g
x
x
g
x
x
g
x
k
n n
")
Ilmiybaza.uz
3
1( ):
6
12
0
f
x
x
x
tenglama uchun, f1(1)=-7<0, f1(2)=8>0,
1
[1,2]
,
3
2( ):
6
12
0
f
x
x
x
tenglama uchun, f2(-3)=3>0, f2(-4)=-28<0,
2
[ 4, 3]
,
3
3( ):
5
4
0
f
x
x
x
tenglama uchun, f3(0)=-4<0, f3(1)=2>0, 3
[0,1]
Yana,
2
1( )
3
6
0,
[1,2],
1
max
1( )
18, 1 1/
1.
f
x
x
x
M
f
x
l
M
,
2
2 ( )
3(
2)
0,
[ 4, 3],
1
max
2 ( )
48, 1 1/
1.
f
x
x
x
M
f
x
l
M
2
3 ( )
3
5
0,
[0,1],
1
max
3 ( )
8, 1 1/
1.
f
x
x
x
M
f
x
l
M
Iteratsiya funksiyalari kiritamiz:
1( ):
1( )/
1
1( )/18
g
x
x
f
x
M
x
f
x
,
2( ):
2( )/
1
2( )/ 48
g
x
x
f
x
M
x
f
x
,
3
2
3( ): (4
)/5, 3 ( )
3
/5,
3 ( )
1,
[0,1]
g
x
x
g
x
x
g
x
x
.
Iteratsiyalarni quyidagicha quramiz:
1
1
1
1
:
1(
), 2
:
2(
), 3
:
3(
),
: 0.. , : 10.
k
k
k
k
k
k
x
g
x
x
g
x
x
g
x
k
n n
Ilmiybaza.uz
V.3. Masalaning Mathcad dasturida yechilishi.
V.3.1. Polyroots komandasi bilan echish (ko‘phadli tenglama uchun o‘rinli).
1)
2)
3)
Ko‘rinib turibdiki, ikkita qo‘shma kompleks ildizlar ham bor ekan
V.3.2. Nochiziq tenglama uchun iteratsiya usulini Mathcad da tashkil etamiz.
v1
12
6
0
1
x1
polyroots v1
(
)
x1
0.73514
2.76067i
0.73514
2.76067i
1.47028
v1
12
6
0
1
x2
polyroots v1
(
)
x2
3.135
1.567
1.171i
1.567
1.171i
v1
4
5
0
1
x3
polyroots v1
(
)
x3
0.362
2.322i
0.362
2.322i
0.724
x
0
f1 x
( )
x3
6 x
12
x1
root f1 x
( ) x
(
)
x1
1.4703
r1
f1 x1
(
)
r1
0
f2 x
( )
x3
6 x
12
x2
root f2 x
( ) x
(
)
x2
3.1349
r2
f2 x2
(
)
r2
0.0006
f3 x
( )
x3
5 x
4
x3
root f3 x
( ) x
(
)
x3
0.7241
r3
f3 x3
(
)
r3
0
x
2
f1 x
( )
x3
6 x
12
x1
root f1 x
( ) x
(
)
x1
1.4703
r1
f1 x1
(
)
r1
0
f2 x
( )
x3
6 x
12
x2
root f2 x
( ) x
(
)
x2
3.1349
r2
f2 x2
(
)
r2
0.0008
f3 x
( )
x3
5 x
4
x3
root f3 x
( ) x
(
)
x3
0.7241
r3
f3 x3
(
)
r3
0.0001
Ilmiybaza.uz
Turli xil boshlang‘ich o‘artlarda natija bir xil.
V.3.3. Nochiziq tenglamalar uchun iteratsiya usulini tashkil etish.
V.3.4. Nochiziq tenglamalar uchun Nьyuton iteratsiya usulini tashkil etish.
g1 x
( )
x
f1 x
( )
18
g2 x
( )
x
f2 x
( )
48
g3 x
( )
4
x3
5
x10
2
x20
3
x30
1
k
0 10
x3k 1
g3 x3k
x1k 1
g1 x1k
x2k 1
g2 x2k
x1T
0
1
2
3
4
5
6
7
8
0
2
1.5556
1.4946
1.4776
1.4725
1.471
1.4705
1.4703
1.4703
x2T
0
1
2
3
4
5
6
7
0
-3
-3.0625
-3.0969
-3.1152
-3.1248
-3.1297
-3.1323
-3.1336
x3T
0
1
2
3
4
5
6
7
8
0
1
0.6
0.7568
0.7133
0.7274
0.723
0.7244
0.724
0.7241
p1 x
( )
x
f1 x
( )
d
d
p2 x
( )
x
f2 x
( )
d
d
p3 x
( )
x
f3 x
( )
d
d
z10
1
z20
2
z30
5
z1k 1
z1k
f1 z1k
p1 z1k
z2k 1
z2k
f2 z2k
p2 z2k
z3k 1
z3k
f3 z3k
p3 z3k
z1T
0
1
2
3
4
5
6
7
8
0
1
1.5556
1.4728
1.4703
1.4703
1.4703
1.4703
1.4703
1.4703
z2T
0
1
2
3
4
5
6
7
0
-2
-4.6667
-3.628
-3.2104
-3.1371
-3.1349
-3.1349
-3.1349
