|
To’rni tuzib olish:
I - variant
1
1
i
i
q
a
h
,
n
i
,
1
(1)
1
)
1
(
1
1
1
1
1
q
q
a
q
a
a
b
h
n
n
i
i
n
i
i
, q>1 – o’suvchi geometrik progressiya
q
q
a
a
b
n
1
)
1
(
1
, q<1 – kamayuvchi geometrik progressiya (2.11)
1)
1
)
1
(
1
q
q
a
a
b
n
, (2.12)
q
a
q
a
b
n
ln
)
)
1
)(
(
1
ln(
1
, q>1. (2.13)
2)
q
q
a
a
b
n
1
)
1
(
1
, (2.14)
q
a
q
a
b
n
ln
)
)
1
)(
(
1
ln(
1
, q<1. (2.15)
1
a
va
q
- oldindan beriladi.
Misol.
Faraz qilaylik, ushbular
5
,
1
,
05
,
0
1
q
a
).
1
(
1
,
0
a
b
b
a
uchun quyidagilarni hisoblaylik:
379687
.
0
5
.
1
05
.
0
,
253125
.
0
5
.
1
05
.
0
16875
.
0
5
.
1
05
.
0
,
1125
.
0
5
.
1
05
.
0
075
.
0
5
.
1
05
.
0
,
05
.
0
5
6
4
5
3
4
2
3
2
1
1
h
h
h
h
h
h
a
q>1 va (2.13) formula bo’yicha
n
6
039062
.
1
6
1
i
i
h
Misol.
Faraz qilaylik, ushbular
l
a
b
b
a
q
a
,
1
,
0
,
8
.
0
,
3
.
0
1
uchun (2.15) formula bo’yicha quyidagilarni hisoblaylik
27
5
2222
.
0
2030
.
1
,
2222
.
0
8
.
0
ln
;
2030
.
1
333333
.
0
ln
3
.
0
2
.
0
1
1
ln
n
Haqiqatan ham
.
00848
.
1
12288
.
0
)
8
.
0
(
3
.
0
1536
.
0
)
8
.
0
(
3
.
0
,
192
.
0
)
8
.
0
(
3
.
0
24
.
0
8
.
0
3
.
0
,
3
.
0
)
8
.
0
(
5
1
4
5
3
4
2
3
2
0
1
1
i
i
h
h
h
h
h
a
h
II - variant
Quyidagi holni ham qarash mumkin::
1
1
i
i
q
a
h
,
n
i
,
1
,
b
x
a
x
n
,
0
,
n
i
h
x
x
i
i
i
,
1
,
1
,
q
q
a
a
b
n
1
)
1
(
1
,
h
a
1
.
a)
n
q
q
a
b
h
1
)
1
)(
(
, q<1 – kamayuvchi geometrik progressiya,
n
va
q
oldindan
beriladi.
в)
1
)
1
)(
(
n
q
q
a
b
h
, q>1 – o’suvchi geometrik progressiya.
Shunday qilib, to’rlarning quyidagi modullarini qarash mumkin:
1)
Teng o’lchovli to’r
h
n
a
b
.
2)
Kvaziteng o’lchovli to’r (
2
,
1
h
h
…).
3)
O’suvchi geomerik progressiya bo’yicha notekislik o’lchovli
)
(
,
1
*
i
h
q
.
4)
Kamayuvchi geomerik progressiya bo’yicha notekislik o’lchovli
)
(
,
1
i
h
q
.
5)
Yuqoridagi 3) va 4) usullar o’rta arifmetigi
n
i
h
h
h
i
i
i
,
1
,
2
*
.
2.2.
O’zgarmas koeffisiyentli bir o’lchovli ko’chirish tenglamasini notekis
to'r yordamida sonli yechish.
Masalaning qo’yilishi.
Ko’chirish tenglamasini quyidagi shaklida o’rganamiz:
),
,
(
t
x
f
qu
x
u
p
t
u
,
0
l
x
,
0
T
t
(2.16)
uning boshlang’ich sharti:
)
(
)
0
,
(
0
x
u
x
u
(2.17)
va chegaraviy shartlari:
1. P>0,
2
)
,
(
t
l
u
, p>0, chap chegarada shart berilmagan.
2. P<0,
1
)
,
0
(
t
u
, p<0, o’ng chegarada shart berilmagan. (2.18)
Dastlabki berilganlar:
28
1) P>0
.
1
,
1
,
)
,
(
:
yechim
aniq
)
(
,
)
(
);
1
(
)
,
(
;
1
,
1
,
1
)
,
(
)
(
)
(
2
0
)
(
B
A
Ae
t
x
u
Ae
t
Ae
x
u
B
q
p
ABe
t
x
f
T
l
t
x
p
t
x
B
t
l
B
Bx
t
x
B
2) P<0
.
1
,
1
,
)
,
(
:
yechim
aniq
)
(
,
)
(
);
1
(
)
,
(
;
1
,
1
,
1
)
,
(
)
(
1
0
)
(
B
A
Ae
t
x
u
Ae
t
Ae
x
u
B
q
p
ABe
t
x
f
T
l
t
x
p
t
x
B
Bt
Bx
t
x
B
Oshkor sxemalar va uning turlari bo’yicha hisob natijalari.
Har ikkala hol
uchun ham yugiruvchi hisob sxemasidan foydalanami.
1) p>0
Bu ho’lda o’ng ayirmali sxemadan foydalaniladi:
;
1
,
0
,
1
,
0
,
0
1
1
1
1
1
1
1
1
j
j
n
i
f
qy
h
y
y
p
y
y
j
i
j
i
i
j
i
j
i
j
j
i
j
i
(2.16′)
,
,
0
),
(
0
0
n
i
x
u
y
i
i
; (2.17′)
;
1
,
0
,
0
1
2
1
j
j
y
j
j
N
. (2.18′)
(2.16′) tenglamadan kelib chiqadiki,
.
*
yerda
bu
1
,
0
,
0
,
1
,
1
1
1
1
0
1
1
1
1
1
1
1
1
p
h
R
j
j
n
i
q
R
f
y
R
y
y
i
j
j
i
j
j
i
j
i
j
j
i
j
i
j
i
j
i
2) p<0
Bu ho’lda chap ayirmali sxemadan foydalaniladi:
;
1
,
0
,
,
1
,
0
1
1
1
1
1
1
1
1
j
j
n
i
f
qy
h
y
y
p
y
y
j
i
j
i
i
j
i
j
i
j
j
i
j
i
;
(2.16″)
,
,
1
),
(
0
0
n
i
x
u
y
i
i
; (2.17″)
.
1
,
0
,
0
1
1
1
0
j
j
y
j
j
(2.18″)
(2.15″) tenglamadan kelib chiqadiki,
.
*
yerda
bu
1
,
0
,
,
1
,
1
1
1
1
0
1
1
1
1
1
1
1
1
p
h
R
j
j
n
i
q
R
f
y
R
y
y
i
j
j
i
j
j
i
j
i
j
j
i
j
i
j
i
j
i
Dastur matni 1-ilovada keltirilgan.
29
1-jadval. O’zgarmas koeffisiyentli bir o’lchovli ko’chirish tenglamasini oshkor (o’ng
ayirmali) sxemaning yugiruvchi hisob sxemasi bo’yicha sonli hisob
natijalari
p>0 va 50-qatlam uchun
N Taqribiy yechim Aniq yechim
Xatolik
0
1.37301170
1.35914091 0.01387078
1
1.41878826
1.40520915 0.01357911
2
1.46606506
1.45283887 0.01322618
3
1.51488985
1.50208301 0.01280684
4
1.56531173
1.55299629 0.01231544
5
1.61738112
1.60563527 0.01174585
6
1.67114985
1.66005846 0.01109139
7
1.72667123
1.71632633 0.01034490
8
1.78400003
1.77450141 0.00949863
9
1.84319260
1.83464833 0.00854427
10
1.90430684
1.89683395 0.00747290
11
1.96740228
1.96112735 0.00627493
12
2.03254007
2.02759998 0.00494008
13
2.09978305
2.09632572 0.00345734
14
2.16919578
2.16738091 0.00181487
15
2.24084454
2.24084454 0.00000000
2.1-rasm. Yechim ustivir ekan.
2-jadval. O’zgarmas koeffisiyentli bir o’lchovli ko’chirish tenglamasini oshkor (chap
ayirmali) sxemaning yugiruvchi hisob sxemasi bo’yicha sonli hisob
natijalari
p<0 va 50-qatlam uchun
N Taqribiy yechim Aniq yechim
Xatolik
30
0
0.03678794
0.03678794 0.00000000
1
0.03444494
0.03558189 0.00113696
2
0.03220334
0.03441538 0.00221204
3
0.03005929
0.03328711 0.00322782
4
0.02800907
0.03219583 0.00418676
5
0.02604910
0.03114032 0.00509122
6
0.02417592
0.03011942 0.00594350
7
0.02238620
0.02913199 0.00674579
8
0.02067672
0.02817693 0.00750021
9
0.01904439
0.02725318 0.00820879
10
0.01748622
0.02635971 0.00887349
11
0.01599934
0.02549554 0.00949620
12
0.01458096
0.02465970 0.01007874
13
0.01322842
0.02385126 0.01062284
14
0.01193914
0.02306932 0.01113018
15
0.01071063
0.02231302 0.01160239
2.2-rasm. Yechim noustivir ekan.
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