ALGORITMLARNI LOYIHALASH
Oraliqni teng 2 ga bo’lish usuliga oid masalalar.
1-savol.
𝑥 3 + 𝑥 − 1 = 0 tenglamaning (0;1) oraliqdagi ildizini oraliqni teng ikkiga bo’lish usuli bo’yicha izlanayotgan
bo’lsa 2 qadamdan keyin qanday qoladi?
A. (0;0.25)
B. (0.25;0.5)
C. (0.5;0.75)
D. (0.75;1)
Oraliqni teng 2 ga bo’lish usuli xatolikka qarab berilgan oraliqni teng 2 ga bo’lib taqribiy yechimga
yaqinlashtirilib boriladi.
f(x) = 𝑥 3 + 𝑥 − 1 funksiyani tekshirib boramiz.
1-qadam:
Umumiy oraliq (0; 1) -> teng ikkitaga ajratsak (0; 0.5) va (0.5; 1)
Left = 0; right = 1; middle = (left+right)/2 = 0.5
Left: f(0) = 03 + 0 – 1 = -1 (manfiy)
Right: f(1) = 13 + 1 – 1 = 1 (musbat)
Middle: f(0.5) = 0.53 + 0.5 – 1 = -0.375 (manfiy)
0
0.5
1
manfiy
manfiy
musbat
Ishorasi almashadigan oraliqni tanlaymiz keyingi oraliq uchun.
(0; 0.5) -> (manfiy; manfiy)
=> xato
(0.5; 1) -> (manfiy; musbat) => to’g’ri (yechim shu oraliqda)
Demak ikkinchi qadamda (0.5; 1) oraliqdan izlaymiz.
2-qadam:
Umumiy oraliq (0.5; 1) -> teng ikkitaga ajratsak (0.5; 0.75) va (0.75; 1)
Left = 0.5; right = 1; middle = (left+right)/2 = 0.75
Left: f(0.5) = 0.53 + 0.5 – 1 = -0.375 (manfiy)
Right: f(1) = 13 + 1 – 1 = 1 (musbat)
Middle: f(0.75) = 0.753 + 0.75 – 1 = 0.171875 (musbat)
0.5
0.75
1
manfiy
musbat
musbat
Ishorasi almashadigan oraliqni tanlaymiz keyingi oraliq uchun.
(0.5; 0.75) -> (manfiy; musbat) => to’g’ri (yechim bor)
(0.75; 1) -> (musbat; musbat)
=> xato (yechim mavjud emas bu oraliqda)
Demak ikkinchi qadamda (0.5; 0.75) oraliqda yechim mavjudligini bildik. 2-qadam oxirida shu (0.5;
0.75) oraliq qolyapti.
To’g’ri javob (0.5; 0.75)
Xuddi shu usul bo’yicha
2-savol
𝑥 4 − 𝑥 − 1 = 0 tenglamaning (-1;0) oraliqdagi ildizini oraliqni teng ikkiga bo’lish usuli bo’yicha izlanayotgan
bo’lsa 2 qadamdan keyin qanday qoladi?
A. (-0.75;-0.5)
B. (-0.5;-0.25)
C. (-1;-0.75)
D. (-0.25;0)
f(x) = x4 – x – 1 uchun (-1; 0) oraliqda 2 ta qadamdan keyin …
1-qadam:
Umumiy oraliq (-1; 0). 2 ga ajratsak (-1; -0.5) va (-0.5; 0)
f(-1) -> musbat
f(-0.5) -> manfiy
f(0) -> manfiy
(-1; -0.5)
=>
(musbat; manfiy) yechim mavjud.
(-0.5; 0)
=>
(manfiy; manfiy) yechim mavjud emas
Demak (-1; -0.5) oraliqni olamiz.
2-qadam:
Umumiy oraliq (-1; -0.5). 2 ga ajratsak (-1; -0.75) va (-0.75; -0.5)
f(-1) -> musbat
f(-0.75) -> musbat
f(-0.5) -> manfiy
Demak:
(-1; -0.75)
=>
(musbat; musbat) bu oraliqda yechim mavjud emas.
(-0.75; -0.5)
=>
(musbat; manfiy) Yechim mavjud.
Javob: 2-qadamdan keyin (-0.75; -0.5) oraliq qolyapti.
Nyuton usuliga oid savollar
Nazariyacha.
Nyuton usulida:
a) Agar 𝑓(𝑎)𝑓 ′′ (𝑎) > 0 bo’lsa 𝑥0 = 𝑎
b) Agar 𝑓(𝑎)𝑓 ′′ (𝑎) < 0 bo’lsa 𝑥0 = 𝑏 deb olinadi
𝒇(𝒙𝒏 )
𝒇′ (𝒙𝒏 )
formula bilan hisoblanadi. Bunda ham yaqinlashtirish jarayoni |𝑥𝑛+1 − 𝑥𝑛 | ≤ 𝜀 shart bajarilmaguncha
davom etadi.
𝒙𝒏+𝟏 = 𝒙𝒏 −
3-savol
𝑥 4 − 𝑥 − 1 = 0 tenglamaning (-1;0) oraliqdagi ildizini Nyuton usuli bo’yicha izlaganda hisoblash formulasi
qanday bo’ladi?
A. 𝑥𝑛+1 =
B. 𝑥𝑛+1 =
C. 𝑥𝑛+1 =
D. 𝑥𝑛+1 =
4 +𝑥 +1
3𝑥𝑛
𝑛
; (shu javob sal yaqin)
3 −1
4𝑥𝑛
4
3𝑥𝑛 +2𝑥𝑛 +1
3 −1
4𝑥𝑛
4
3𝑥𝑛 +3𝑥𝑛 +1
3 −1
4𝑥𝑛
4
2𝑥𝑛 +𝑥𝑛 +1
3 −1
4𝑥𝑛
;
;
;
Bu savolda xatolik bor. Bu yilgi savollarda xatolar to’g’rilangan.
Umumiy formulasi: 𝒙𝒏+𝟏 = 𝒙𝒏 −
𝒇(𝒙𝒏 )
𝒇′ (𝒙)
𝑓(𝑥) = 𝑥 4 − 𝑥 − 1 , bundan hosila olsak 𝒇′ (𝒙) = 𝟒𝒙𝟑 − 𝟏
𝑓(𝑥𝑛 )
𝑥𝑛4 − 𝑥𝑛 − 1
𝑥𝑛 (4𝑥𝑛3 − 1) − (𝑥𝑛4 − 𝑥𝑛 − 1)
3𝑥𝑛4 + 1
𝑥𝑛+1 = 𝑥𝑛 − ′
= 𝑥𝑛 −
=
=
𝑓 (𝑥)
4𝑥𝑛3 − 1
4𝑥𝑛3 − 1
4𝑥𝑛3 − 1
4-savol
𝑥 3 + 𝑥 − 1 = 0 tenglamaning (0;1) oraliqdagi ildizini Nyuton usuli bo’yicha izlaganda hisoblash formulasi
qanday bo’ladi?
A. 𝑥𝑛+1 =
B. 𝑥𝑛+1 =
C. 𝑥𝑛+1 =
D. 𝑥𝑛+1 =
3 −𝑥 +1
2𝑥𝑛
𝑛
; (sal xatosi bor yaqin javob)
2 +1
3𝑥𝑛
3 −2𝑥 +1
2𝑥𝑛
𝑛
3 +1
3𝑥𝑛
3
3𝑥𝑛 −2𝑥𝑛 +1
2 +1
3𝑥𝑛
3
𝑥𝑛 +𝑥𝑛 −2
2 +1
3𝑥𝑛
;
;
;
Xuddi shu usulda ishlasak:
𝑓(𝑥) = 𝑥 3 + 𝑥 − 1 bundan hosila olsak 𝒇′ (𝒙) = 𝟑𝒙𝟐 + 𝟏
𝑓(𝑥𝑛 )
𝑥𝑛 3 + 𝑥𝑛 − 1
𝑥𝑛 (3𝑥𝑛 2 + 1) − (𝑥𝑛 3 + 𝑥𝑛 − 1)
2𝑥𝑛3 + 1
𝑥𝑛+1 = 𝑥𝑛 − ′
= 𝑥𝑛 −
=
=
𝑓 (𝑥)
3𝑥n 2 + 1
3𝑥𝑛 2 + 1
3𝑥𝑛 2 + 1
Vatarlar usuliga oid savollar tahlili
Bunga oid 4 ta nazariy (1 tasi tushishi mumkin) va kamida 1 ta masala bor.
𝑓(𝑥) = 0 tenglamaning (a; b) oraliqdagi ildizini topishning vatarlar usuli.
1) Agar 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) > 0 bo’lsa 𝑥0 = 𝑎 hamda b nuqta qo’zga’lmas bo’ladi. xn+1 esa quyidagicha: 𝑥𝑛+1 =
𝑓(𝑥 )(𝑏−𝑥 )
𝑛
𝑛
𝑥𝑛 − 𝑓(𝑏)−𝑓(𝑥
)
𝑛
2) Agar 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) < 0 bo’lsa 𝑥0 = b hamda a nuqta qo’zga’lmas bo’ladi. xn+1 esa quyidagicha: 𝑥𝑛+1 =
𝑓(𝑥 )(a−𝑥 )
𝑛
𝑛
𝑥𝑛 − 𝑓(a)−𝑓(𝑥
)
𝑛
Eslab qoling:
0 dan katta bo’lsa x0 = a,
b esa qo’zgalmas
0 dan kichik bo’lsa x0 = b,
a esa qo’zg’almas
5-savol
𝑓(𝑥) = 0 tenglamaning (a; b) oraliqdagi ildizini topishda 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) > 0 shart bajarilsa vatarlar usuli
formulasi qanday ko’rinishda bo’ladi?
𝑓(𝑥𝑛 )(𝑏−𝑥𝑛 )
A. 𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑏)−𝑓(𝑥
; 𝑥0 = 𝑎
)
𝑛
B.
C.
D.
𝑓(𝑥𝑛 )(𝑎−𝑥𝑛 )
𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑎)−𝑓(𝑥
; 𝑥0 = 𝑏
𝑛)
𝑓(𝑥 )
𝑥𝑛+1 = 𝑥𝑛 − 𝑓′ (𝑥𝑛 ) ; 𝑥0 = 𝑏
𝑛
𝑓(𝑥 )
𝑥𝑛+1 = 𝑥𝑛 − 𝑓′ (𝑥𝑛 ) ;
𝑛
𝑥0 = 𝑎
6-savol
𝑓(𝑥) = 0 tenglamaning (a;b) oraliqdagi ildizini topishda 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) < 0 shart bajarilsa vatarlar usuli
formulasi qanday ko’rinishda bo’ladi?
𝑓(𝑥𝑛 )(𝑎−𝑥𝑛 )
A. 𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑎)−𝑓(𝑥
; 𝑥0 = 𝑏
)
𝑛
B.
C.
D.
𝑓(𝑥𝑛 )(𝑏−𝑥𝑛 )
𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑏)−𝑓(𝑥
; 𝑥0 = 𝑎
𝑛)
𝑓(𝑥 )
𝑥𝑛+1 = 𝑥𝑛 − 𝑓′ (𝑥𝑛 ) ; 𝑥0 = 𝑏
𝑛
𝑓(𝑥 )
𝑥𝑛+1 = 𝑥𝑛 + 𝑓′ (𝑥𝑛 ) ;
𝑛
𝑥0 = 𝑎
7-savol
𝑓(𝑥) = 0 tenglamaning (a;b) oraliqdagi ildizini vatarlar usulida izlanganda “a” nuqta qo’zg’almas bo’lishi
uchun qanday shart bajarilishi kerak?
A. 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) < 0
B. 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) > 0
C. 𝑓(𝑎)𝑓(𝑏) < 0
D. 𝑓 ′ (𝑎)𝑓 ′ (𝑏) < 0
8-savol
𝑓(𝑥) = 0 tenglamaning (a;b) oraliqdagi ildizini vatarlar usulida izlanganda “b” nuqta qo’zg’almas bo’lishi
uchun qanday shart bajarilishi kerak?
A. 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) > 0
B. 𝑓(𝑎)𝑓(𝑏) < 0
C. 𝑓 ′ (𝑥)𝑓 ′′ (𝑥) < 0
D. 𝑓 ′ (𝑎)𝑓 ′ (𝑏) < 0
9-savol
ex − 10x − 2 = 0 funksiyani [-1;0] oraliqdagi ildizini topishda 𝑓′ (𝑥)𝑓′′ (𝑥) < 0 shart bajarilsa vatarlar usuli bilan i=2
qadamlar orasidagi xatoligini aniqlang.
A. x2-x1=0.004 (shunga yaqinroq)
B. x2-x1=0.0014
C. x2-x1=0.00244
D. x2-x1=0.003444
Berilgan:
f(x) = ex − 10x − 2 ;
a = -1;
b=0
Yuqoridagi shartga ko’ra demak bizda x0 = b bo’lar ekan.
𝑓(𝑥 )(a−𝑥 )
𝑛
𝑛
𝑥𝑛+1 = 𝑥𝑛 − 𝑓(a)−𝑓(𝑥
shu formula bo’yicha ishlaymiz.
)
𝑛
1-qadam:
𝑥1 = 𝑥0 −
𝑓(𝑥0 )(a − 𝑥0 )
𝑓(0) ∗ (−1 − 0)
𝑓(0)
= 0 −
=
=
𝑓(a) − 𝑓(𝑥0 )
𝑓(−1) − 𝑓(0)
𝑓(−1) − 𝑓(0)
𝑒 0 − 10 ∗ 0 − 2
1-2
-1
= -1
=
=
≈ -0.10674774
(𝑒 − 10 ∗ (−1) − 2) − (𝑒 0 − 10 ∗ 0 − 2)
(e-1 + 8) - (1-2)
e-1 + 9
2-qadam:
𝑓(𝑥1 ) = e-0.10674774 − 10 ∗ (−0.10674774) − 2 ≈ -0.03377 ≈ 0
𝑥2 = 𝑥1 −
𝑓(𝑥1 ) ∗ (−1 − (−0.10674774))
𝑓(𝑥1 )(a − 𝑥1 )
= −0.10674774 −
≈. . ..
𝑓(a) − 𝑓(𝑥1 )
𝑓(−1) − 𝑓(𝑥1 )
Bunaqa savol tushmasa kerak. Keyin |𝑥1 − 𝑥2 | ni hisoblash kerak hali. Xullas ishlanishi shunaqa. Shuni
osonroq kalkulyatorsiz hisoblasa bo’ladigan variantdagisi tushishi mumkin.
Aniq integrallarni taqribiy hisoblashlar
(To’g’ri to’rtburchaklar, Trapetsiya, Simpson usullari)
Bularning xatoligi ham mos ravishda 𝑂(ℎ), 𝑂(ℎ2 ), 𝑂(ℎ4 ) bo’ladi.
To’g’ri to’rtburchaklar usulida xatoligi 𝑂(ℎ), formulasida h (belgisi TT)
Trapetsiya usulida xatoligi 𝑂 (ℎ2 ), formulasida
Simpson usulida xatoligi 𝑂 (ℎ4 ) formulasida
h
3
ℎ
2
(belgisi T)
(belgisi С)
10-savol
1
2
∫0 𝑒 −𝑥 𝑑𝑥 aniq integralni to’g’ri to’rtburchaklar formulasi bo’yicha h=0.01 qadam bilan hisoblansa xatolik
tartibi qanday bo’ladi?
A. O(0,01)
B. O(0,0001)
C. O(0,0000001)
D. O(0,001)
To’g’ri to’rtburchaklar usulida xatoligi 𝑂(ℎ)
11-savol
𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥 aniq integralni trapetsiyalar formulasi bo’yicha h=0.01 qadam bilan hisoblansa xatolik tartibi
qanday bo’ladi?
A. O(0,0001)
B. O(0,000001)
C. O(0,001)
D. O(0,00001)
Trapetsiya usulida xatoligi 𝑂 (ℎ2 )
12-savol
1
∫0 𝑠𝑖𝑛𝑥 2 𝑑𝑥 qiymatini Simpson formulasi bo’yicha O(10-4) aniqlikda hisoblash uchun h – qadamni qanday
olish kerak?
A. h=0,1
B. h=0,01
C. h=0,001
D. h=0,05
Simpson usulida xatoligi 𝑂 (ℎ4 )
13-savol
2
∫1 𝑐𝑜𝑠𝑥 2 𝑑𝑥 qiymatini Simpson formulasi ko’ra h=0,1 qadam bilan hisoblasak xatolik qanday tartibda bo’ladi?
A. O(10-4)
B. O(10-3)
C. O(10-2)
D. O(10-1)
Simpson usulida xatoligi 𝑂 (ℎ4 )
14-savol
To’g’ri to’rtburchak formulasi to'g'ri keltirilgan javobni ko'rsating.
A. 𝐽ℎ𝑇𝑇 (𝑓) = ℎ ∑n-1
i=0 𝑓𝑖+0.5
B.
h
C. 𝐽ℎ𝑇 (𝑓) = 2 {𝑓0 + ⋯ + 𝑓𝑛−1 + 𝑓𝑛 }
h
h
D. 𝐽ℎ𝐶 (𝑓) = 3 ∑m-1
i=0 [𝑓2𝑖 + 4𝑓2𝑖+1 + 𝑓2𝑖+2 ] = 3 [𝑓0 + 4(𝑓1 + . . . + 𝑓2𝑚−1 ) + 2(𝑓2 +. . . +𝑓2𝑚−2 )]
To’g’ri to’rtburchaklar formulasida faqat h ishtirok etadi, belgisi TT
n-1
𝐽ℎ𝑇𝑇 (𝑓) = ℎ ∑ 𝑓𝑖+0.5
i=0
15-savol
Trapetsiya formulasi to'g'ri keltirilgan javobni ko'rsating.
ℎ
A. 𝐽ℎ𝑇 (𝑓) = {𝑓0 + ⋯ + 𝑓𝑛−1 + 𝑓𝑛 }
2
B.
C. 𝐽ℎ𝑇𝑇 (𝑓) = ℎ ∑n-1
i=0 𝑓𝑖+0.5
h
h
D. 𝐽ℎ𝐶 (𝑓) = 3 ∑m-1
i=0 [𝑓2𝑖 + 4𝑓2𝑖+1 + 𝑓2𝑖+2 ] = 3 [𝑓0 + 4(𝑓1 + . . . + 𝑓2𝑚−1 ) + 2(𝑓2 +. . . +𝑓2𝑚−2 )]
Trapetsiya formulasida h/2 ishtirok etadi, belgisi T
h
𝐽ℎ𝑇 (𝑓) = {𝑓0 + ⋯ + 𝑓𝑛−1 + 𝑓𝑛 }
2
16-savol
Simpson formulasi to'g'ri keltirilgan javobni ko'rsating.
A.
B.
h
C. 𝐽ℎ𝑇 (𝑓) = 2 {𝑓0 + ⋯ + 𝑓𝑛−1 + 𝑓𝑛 }
D. 𝐽ℎ𝑇𝑇 (𝑓) = ℎ ∑n-1
i=0 𝑓𝑖+0.5
Simpson formulasida h/3 ishtirok etadi, belgisi С
m-1
𝐽ℎ𝐶 (𝑓) =
h
h
∑[𝑓2𝑖 + 4𝑓2𝑖+1 + 𝑓2𝑖+2 ] = [𝑓0 + 4(𝑓1 + . . . + 𝑓2𝑚−1 ) + 2(𝑓2 +. . . +𝑓2𝑚−2 )]
3
3
i=0
Eslab qoling yana bir bor!!!
To’g’ri to’rtburchaklar usulida xatoligi 𝑂(ℎ), formulasida h (belgisi TT)
Trapetsiya usulida xatoligi 𝑂 (ℎ2 ), formulasida
Simpson usulida xatoligi 𝑂 (ℎ4 ) formulasida
h
3
ℎ
2
(belgisi T)
(belgisi С)