Hisoblash usullari fanidan laboratoriya ishlari

Urinma, vatarlar usullari iteratsiya usulining bir ko’rinishlari holos. Bular bir-biri bilan ni topish bilan farq qiladi. Iteratsiya usuliga chiqarilgan xossalar ularga ham tegishli bo’ladi.

Iteratsiya yoki ketma - ket yaqinlashish usuli metodik jihatdan eng sodda va qulay usullardan biridir, lekin hamma vaqt ildizga yaqinlashavermaydi. Iteratsiya usulini tenglamaga ko’llashdan oldin quyidagi teoremani bajarilishini tekshirib ko’rish kerak.

Teorema . Biror [a,b] oralikda differentsiallanuvchi bo’lsin. Agar shartni qanoatlantirsa iteratsiya usuli bilan topilgan ildiz haqiqiy ildizga yaqinlashadi.
Isbot.
Lagranj teoremasiga asosan

Bundan
(4)
(4) dan ko’rinadiki, bo’lganligi uchun n etarli darajada katta bo’lganda ildizni aniqligi shunchalik kichik bo’ladi.
I teratsiya usuli bilan yechishning algoritmining blok sxemasi va dasturi quyidagicha.

10 REM "Itaratsiya usuli"

20 DEF fnf (x) = 1 / (x + 1) ^ 2
30 INPUT a, e
40 x0 = a: n=0
50 x1 = fnf(x0)
60 n=n+1
70 IF ABS(x1 - x0) >= e THEN x0 = x1: GOTO 50
80 PRINT "x="; x1, "n="; n
90 END


Mustaqil yechish uchun misollar

№	Yechim bor oraliq	Aniqlik	Yechim
1	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,9216703
2	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	15,757300
3	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,3221853
4	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,7390851
5	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,3646556
6	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,6411858
7	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	2,094551
8	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,7861513
9	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,526534
10	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,6723832
11	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 2,596072
12	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,6071017
13	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,213412
14	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,557146
15	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,403928
16	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,933754
17	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,6823278
18	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	3,789278
19	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,2879097
20	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,246919
21	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,770917
22	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,512135
23	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,8652716
24	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,4655714
25	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,682328
26	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,8767263
27	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,1347284
28	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,6506561
29	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,52138
30	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,7468819
31	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,423622
32	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,4321718
33	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 0,3221853
34	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,841406
35	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 1,847708
36	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,9286265
37	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	- 1,179509
38	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,309907
39	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,855418
40	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	0,4777551
41	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,134728
42	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,26073
43	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,322185
44	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	3,004679
45	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	2,258259
46	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,283429
47	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,164619
48	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,221022
49	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,059253
50	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	1,416468

LABORATORIYA ISHI № 8

Gauss usuli.

Bu metod bir necha hisoblash sxemalariga ega. Shulardan biri Gaussning kompakt sxemasini ko’rib chiqamiz. Ushbu sistema berilgan bo’lsin:

EMBED Equation.3 (1)
Faraz qilaylik, EMBED Equation.3 (etakchi element) bo’lsin, aks holda tenglamalarning o’rnini almashtirib, EMBED Equation.3 oldidagi koeffitsenti noldan farqli bo’lgan tenglamani birinchi o’rniga ko’chiramiz. Sistemadagi birinchi tenglamalarni barcha koeffitsentlarini EMBED Equation.3 ga bo’lib,
EMBED Equation.3 (2)
ni hosil qilamiz, bu erda, EMBED Equation.3 .
(2) tenglamadan foydalanib, (1) sistemaning qolgan tenglamalarida EMBED Equation.3 ni yo’qotish mumkin. Buning uchun (2) tenglamani ketma-ket EMBED Equation.3 larga ko’paytirib, mos ravishda sistemaning ikkinchi, uchinchi va h.k. tenglamalaridan ayiramiz. Natijada, quyidagi sistema hosil bo’ladi:
EMBED Equation.3 (3)

bu erda EMBED Equation.3 koeffitsentlar

EMBED Equation.3
formula yordamida hisoblanadi. Endi (3) sistema ustida ham shunga o’xshash almashtirishlar bajaramiz.
Buning uchun (3) sistemadagi birinchi tenglamaning barcha koeffitsentlarini etakchi element EMBED Equation.3 ga bo’lib,
EMBED Equation.3 (4)
ni hosil qilamiz, bu erda
EMBED Equation.3
(4) tenglama yordamida (3) sitemaning keyingi tenglamalarini yuqoridagidek EMBED Equation.3 ni yo’qotib,
EMBED Equation.3
sistemaga kelamiz, bu erda
EMBED Equation.3
Noma’lumlarni yo’qotish jarayonini davom ettirib va bu jarayonni EMBED Equation.3 - qadamgacha bajarish mumkin deb faraz qilib, EMBED Equation.3 - qadamda quyidagi sistemaga ega bo’lamiz:
EMBED Equation.3 (5)
bu erda
EMBED Equation.3 .
Faraz qilaylik, m mumkin bo’lgan oxirgi qadamning nomeri bo’lsin. Ikki hol bo’lishi mumkin: m=n yoki m. Agar m=n bo’lsa u vaqtda biz uchburchak matritsali va (1) sistemaga ekvivalent bo’lgan quyidagi
EMBED Equation.3 (6)
sistemaga ega bo’lamiz. Oxirgi sistemadan ketma-ket EMBED Equation.3 larni topish mumkin:
EMBED Equation.3 (7)
(6) uchburchak sistemaning koeffitsentlarini topish Gauss metodining to’g’ri yurishi, (7) sistemadan yechimni topish jarayoni teskari yurish deyiladi.
Faraz qilaylik, m bo’lsin va sistemaning m – va undan keyingi tenglamalari (5) ko’rinishga keltirilgan bo’lsin. Biz m – qadamni bajarilishi mumkin bo’lgan qadam deb hisoblagan edik, bu shuni bildiradiki (5) sistemaning ikkinchi tenglamasidan boshlab etakchi elementni ajratish mumkin emas, barcha EMBED Equation.3 EMBED Equation.3 nolga teng va (5) sistema quyidagi ko’rinishga ega
EMBED Equation.3
Agar bunda barcha ozod hadlar EMBED Equation.3 EMBED Equation.3 nolga teng bo’lsa, u holda biz faqat yagona birinchi tenglamaga ega bo’lamiz.
Barcha qadamdagi birinchi tenglamalarni birlashtirib, quyidagi sistemani hosil qilamiz:
EMBED Equation.3
Bu sistemadan biz EMBED Equation.3 noma’lumlarni EMBED Equation.3 noma’lumlar va ozod hadlar yordamida ifodalab olishimiz mumkin. Bu holda (1) sistema cheksiz ko’p yechimga ega bo’ladi. Agar m bo’lib, hech bo’lmaganda birorta EMBED Equation.3 EMBED Equation.3 bo’lsa, u holda (1) sistema yechimga ega bo’lmaydi.
Qo’lda hisoblanayotganda hatoga yo’l qo’ymaslik uchun, hisoblash jarayonini kontrol qilish ma’quldir. Buning uchun biz (1) matritsa satrlaridagi elementlar va ozod hadning yig’indisidan tuzilgan kontrol
EMBED Equation.3 (8)
yig’indidan foydalanamiz.
Agar EMBED Equation.3 larni (1) sistemaning ozod hadlari deb qabul qilsak, u holda almashtirilgan
EMBED Equation.3 (9)
sistemaning yechimi EMBED Equation.3 (1) sistemaning yechimi EMBED Equation.3 orqali quyidagicha ifodalanadi:
EMBED Equation.3 (10)
Haqiqatan ham, (10) ni (9) sistemaga qo’ysak, (1) sistema va (8) formulaga ko’ra
EMBED Equation.3
ayniyatga ega bo’lamiz.
Agar satr elementlar ustida bajarilgan amallarni har bir satrdagi kontrol yig’indi ustida ham bajarsak va hisoblashlar xatosiz bajarilgan bo’lsa, u holda kontrol yig’indilardan tuzilgan ustunning xar bir elementi mos ravishda almashtirilgan satrlar elementlarining yig’indisiga teng bo’ladi. Bu hol esa to’g’ri yurishni kontrol qilish uchun xizmat qiladi. Teskari yurishda esa, kontrol EMBED Equation.3 larni topish uchun bajariladi.
Tenglamalar sistemasi qo’lda echilganda hisoblashlarni 1-jadvalda ko’rsatilgan Gaussning kompakt sxemasi bo’yicha olib borish ma’quldir. Soddalik uchun jadvalda to’rtta noma’lumli to’rtta tenglamalar sistemasini yechish sxemasi keltirilgan.
Gauss metodi bilan n ta noma’lumli chiziqli algebraik tenglamalar sistemasini yechish uchun bajariladigan arifmetik amallarning miqdori quyidagilardan iborat. EMBED Equation.3 ta ko’paytirish va bo’lish EMBED Equation.3 ta qo’shish.
Yagona bo’lish jadvali.
1-jadval

EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	EMBED Equation.3	Ozod hadlar		Sxema qismlari
EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 1	EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3	EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3	EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3	EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3	EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3