Koshi masalasining ko’p qadamli ayirmali metodlar

Adamsning interpolyasion metodlari. Yuqorida Adamsning interpolyasion metodi

^yn ^{ y}n1 ^{ hb}mi ^fni
i0
(1.36)

Formula bilan aniqlanib,
b_m0  0
ekanligini aytgan edik. Bu metod

approksimasiyaning tartibi (1.12) sistemadan, ya’ni
p  m  1
bo’lib,
^bmi
koeffisientlar
p  m  1
bo’lganda

m
_i^kb
 ¹ , k  1,2,...,m,b
 1  _b

i1
^mi k  1
m0 mi

m
i0

Sistemadan topiladi. Bunda metodni hosil qilamiz:
m 1
uchun approksimasiya tartibi ikki bo’lgan

y_n  y


n1

• 
  ^h  f 
  

₂ n
^fn1
, p  2.

Bu metod trapesiya metodi deb ham ataladi. Biz
m  2,3,4,5
bo’lganda mos

ravishda ushbu qilamiz:
p  m  1
tartibli approksimasiyaga ega bo’lgan ,etodlarni hosil

y_n  y


n1

• 
  ^h 5 f
  

12 ⁿ

• 
  8 f
  

n1 ^
fn2
, p  3;

y_n  y
n1 ^
h 9 f
24 ⁿ
 19 f


n1

• 
  5 f
  

n2 ^
fn3
, p  4;

y_n  y
n1 ^
h


720
251 f_n
 646 f


n1
 264 f


n2
 106 f


n3
 19 f


n4
, p  5;

y_n  y


n1

• 
  h
  

1440
475 f_n
 1427 f


n1
 798 f


n2
 482 f


n3
 173 f


n4
 27 f


n5
, p  6.

Yuqoridagi oshkormas metodlarda izlanayotgan
y_n chiziqli bo’lmagan

ko’rinishda. Shuning uchun ham bu tenglamalardan
y_n ni topish uchun iterasiya

metodini qo’llash kerak. Masalan, to’rtinchi tartibli Adams metodi uchun iterasion metod quyidagicha qo’llaniladi:

_ys1 _ ^3h
f x
, y^{s }  F ,

n ₈ n n n
(1.37)

F_n  y


n1
 ^h 19 f x
12


n1
^{, y}n1
 5 f x


n2
, y_n2  
f x


n3
, y_n3 ,

bu yerda s -iterasiya nomeri. Dastlabki yaqinlashish
0

y
n
sifatida Adamsning

uchinchi tartibli oshkor metodi yordamida topilgan yechimni olish mumkin, ya’ni

n
y^0  y


n1
 ^h 23 f x
12


n1
^{, y}n1
  16 f x


n2
^{, y}n2
  5 f x


n3
^{, y}n3
.

 M
Agar
^f ^
_y 
bo’lsa, u holda (1.34) iterasion metod yaqinlashuvchi bo’lishi

 

uchun
3hM _{ 1}
8
shart bajarilishi kerak, bu esa yetarlicha kichik h uchun doimo

bajariladi. Agar (1.34)da faqat bitta iterasiya olsak, ya’ni
s  0 bo’lsa, u holda

prediktor-korrektor (bashoratchi-tuzatuvchi) metodi deb ataluvchi metodga ega bo’lamiz.
Adams interpolyasion formulasini Lagraj interpolyasion ko’phadi hosil

k
qilishni ko’ramiz, buning uchun (1.21) formulada
q  1deb olamiz. U holda

ux_n
  ux


n1
  h _
b^1u^x


n1 j
1

 R
n,k
 ux


n1
  h_m b^* u^x


n j

*

 R

,
n,m
(1.38)

mj
j1 j0

kj
bu yerda
m  k  1,b^*
i 

 b
m1, j1
*

, R
n,m
n,m1 ^.

mj

*

 R
Endi (1.21) va (.23) formulalarda q 1,k  m 1 deb olib, quyidagi formulalarga ega bo’lamiz:
_{*  j} ¹ t  1tt  1...t  m  1dt

b_mj   1 _
0
t 
j  1m 
j j
, j  0,1...,m,
(1.39)

_{* h}m2
m2 _{ }

1
^{Rn,m } m  1!^u
ξ_t  1t...t  m  1dt.
0
(1.40)

Biz (1.24) formuladan (1.31) formulani qanday chiqargan bo’lsak, huddi shunga o’xshash mulohazalar yuritib, (1.36) formuladan quyidagi formulani chiqaramiz:

y  y  h_b_{n n1}
m

y
*
mj n
_{ j} . (1.41)

j0
Bu esa (1.3) formula bilan ustma-ust tushadi.

Endi (1.39) formula yordamida
m  0,1,2,3,4,5 lar uchun (1.41) Adams

interpolyasion formulasi koeffisientlarini keltiramiz:

00

b
^*  1,
b^*  ¹ ,b^*  ¹ ,
10 ₂ 11 ₂

b^*  ⁵ ,b^*
 ² ,b^*   ¹ ,

20 ₁₂ 21
3 ²² 12

b^*  ³ ,b^*
 ¹⁹ ,b^*
  ⁵ ,b^*  ¹ ,

30 ₈ 31
₂₄ 32
24 ³³ 24

b^*  ²⁵¹,b^*
_ 323 _,b*
  ¹¹ ,b^*
_ 53
,b^*
  ¹⁹ ,

40 ₇₂₀ 41
360 ⁴²
₃₀ 43
360 ⁴⁴
720

b^* 
475 _,b*
 1427 _,b*
  ³⁹⁹ ,b^*
_ 241_{,b*
 } 173
,b^* 
27 _.

50 ₁₄₄₀ 51
1440 ⁵²
720 ⁵³
720 ⁵⁴
1440 ⁵⁵
1440

Adamsning ekstrapolyasion va interpolyasion metodlarini taqqoslaymiz. Buning uchun(1.31) formulani

m1

m
^yn ^{ y}n1 ^{ hb}* 1, j ^fn j
(1.42)

j0
Formula bilan solishtirish kerak, bu formula (1.41) formuladan m ni
m 1
bilan

almashtirish natijasida hosil bo’ladi, chunki bu formulalarni ko’rish uchun bir xil sondagi, ya’ni m ta nuqtalardan foydalaniladi. Jumladan, (1.31) formulada

^xnm ^{, x}nm1^,...,xn1^,
(1.42) formulada esa
^xnm1^{, x}nm2 ^,...,xn1^{, x}n
Tugunlardan foydalanilgan.
(1.43)

(1.44)

Ma’lumki, u^x funksiyani x_n1, x_n 
oraliqda (1.43) tugunlar yordamida

ko’rilgan interpolyasion ko’phad bilan yaqinlashtirishdan (1.44) tugunlar yordamida ko’rilgan ko’phad bilan yaqinlashtirish aniqroqdir. Shu ma’noda Adamsning interpolyasion metodi ekstrapolyasion metodiga nisbatan aniqroqdir. Buni yana ham yaxshiroq anglash uchun (1.31) va (1.42) formulalarning qoldiq hadlarni m 1,2,3,4 uchun (1.26) va (1.40) formulalar yordamida topamiz ((1.40) formulada

m ni
m 1 bilan almashtirish kerak):

^Rn,1
 _h2
2
uⁿ  ξ, R_n2
 5h³
12
u^m  ξ, R_n3
 ³ h³u
8
^IV  ξ, R_n4
 251 _h4u
720
^IV  ξ;

_ h² 2


ⁿ  
^*   ^h3
^m  
^*   ^h4
IV <sub> 
*  </sub> 19
5 IV _{ }

^Rn0
u ξ, R_n1
12 u ξ, R_n2
24 u ξ, R_n3
h u ξ.
720

Bulardan ko’rinadiki, kichikdir.
*

R
n,m1
sonli koeffisientlari
R_n,m nikiga nisbatan ancha

Endi Adams interpolyasion formulasining boshqa ko’rinishi, ya’ni
f x,u

chekli ayirmalarining qiymatlari qatnashadigan ko’rinishini keltiramiz, buning uchun Nyutonning ikkinchi interpolyasion formulasini (1.21) formulaga qo’yib, quyidagiga ega bo’lamiz:

y  ξ  ¹ ξ


 ¹ ² ξ


 ¹ ³ ξ


_ 19
⁴ ξ
 ³ ⁵ ξ 

n1
n ₂ n1 ₁₂
n2 ₂₄
n3
720
n4
160
n5

_ 863

m1
60480
^6 ^ξn6
 ... c^*
m1 ^ξnm1^,

bu yerda



0
 ^* 
¹  

  

    

1
ξ_i hf_i ,c_m1
m  1!^{ t}
1 t t
1 ... t m 1 dt.

(1.45) formulaning qoldiq hadini quyidagicha yozish mumkin:

R
*
n,m
m3 *

 h c
m2
_um3 _ξ.

Agar (1.33) formula bilan (1.45) formulani taqqoslasak, unda ko’rinib turibdiki, chekli ayirmalarning tartibi oshgan sari (1.45) formulada chekli ayirmalar oldidagi koeffisientlar absolyut qiymatlari bilan (1.33) formuladagiga nisbatan

tezroq kamayib boradi. Bunday holda esa, o’z navbatida, (1.45) yoyilmadagi hadlar absolyut qiymati bilan (1.33)dagiga nisbatan tezroq kamayadi.
Mashq. (1.45) formula isbotlansin.

Endi (1.45) formulaning
m  3 bo’lgandagi xususiy holini qaraymiz:

y  ξ  ¹ ξ  ¹ ² ξ


 ¹ ³ ξ


_ 19
⁴ ξ .
(1.46)

n1 n ₂ n1 ₁₂
n2 ₂₄ n3
720
n1

Bu yerda
n  5 deb olamiz, u holda quyidagi hosil bo’ladi:

y  ξ  ¹ ξ 
¹ ³ ξ
1 _{3 ξ } 19
⁴ ξ.
(1.47)

4 5 ₂
⁴ 12
³ 24
² 720 ¹

Hisoblashni (1.46), (1.47) formulalar bilan bajarish uchun quyidagi jadvaldan foydalangan ma’qul:

x	y	y	ξ hf	ξ	^2 ξ	^3ξ	^4ξ
x0	y0		ξ0
		y0		ξ0
x1	y1		ξ1		^2 ξ0
		y1		ξ1		^3 ξ0
x2	y2		ξ2		^2ξ1		^4ξ0
		y2		ξ2		^3ξ1
x3	y3		ξ3		^2ξ2		^4ξ1
		y3		ξ3
x4	y4		ξ4		^2ξ3	^3ξ2
		y4		ξ4
x5	y5		ξ5

Hisoblashni (1.45) formula yordamida olib borganda

ξ_n ,ξ_n1,² ξ_n2 ,...,^m1 ξ_nm1
noma’lum ayrimalarning dastlabki yaqinlashishlari Adamsning ekstrapolyasion

n
metodi yordamida hisoblanadi. Darhaqiqat, y^0 ni (1.33) formula yordamida

hisoblash kerak. Bu esa
ξ^0 ni topishga imkon beradi, natijada qolgan

n
ayirmalarning dastlabki yaqinlashishini toppish mumkin bo’ladi.
Dastlabki yaqinlashishlarni topishning boshqacha usulini ham ko’rsatish mumkin. Buni (1.47) formula misolida ko’ramiz. Bu formulaning o’ng tomonida va jadvalning pog’onali siniq chizig’ining pastida

ξ₅ ,ξ₄ ,² ξ₃ ,³ ξ₂ ,⁴ ξ₁
(1.48)

Ayirmalar joylashgan bo’lib, ularning qiymatlari noma’lum. Bularni iterasiya metodi bilan toppish uchun ularning dastlabki yaqinlashishini ko’rsatish kerak. Agar h qadam to’g’ri tanlangan bo’lsa, u holda oxirgi ma’noli raqamning bir necha birligi chegarasida
⁴ ξ^0  ⁴ ξ
1 0

Bo’ladi, shuning uchun
⁴ ξ₁ning dastlabki yaqinlashishi sifatida deb olishimiz

mumkin. Bu esa (1.48) ayirmalarning dastlabki yaqinlashishlarini quyidagi formulalar yordamida topishga imkon beradi:
³ ξ^0  ³ ξ ⁴ ξ^0,
2 1 1
²ξ^0  ²ξ  ³ξ^0,
3 2 2
ξ^0  ξ  ²ξ^0,
4 3 3
ξ^0  ξ  ξ^0.

4
5 4 4

Endi (1.49) dastlabki yaqinlashishlarni (1.47) formulaga qo’yib,
y^1 ni va

5
ni topamiz. Bundan keyin
_y114 _{ y}4
 y

ξ^1  hf x , y^1 
5 5 5

ni hisoblaymiz. Agar
ξ^1  ξ^0 tenglik bajarilsa, u holda y  y^1 deb olib, y ni

5 5 5 5 5

hisoblashni tugatamiz. Agar
_ξ1 _{ ξ}0
bo’lsa, u holda
ξ^1 ga ko’ra (1.49)

5 5 5
ayirmalarning yangi

_ξ1_,ξ1_,2 _ξ1_{, }3 _ξ1_,4 _ξ1
(1.50)

5 4 3 2 1

5
qiymatini ketma-ket quyidagi formulalar yordamida hisoblaymiz:

4
ξ^1 
_ξ1 _
ξ₄ ,

²ξ^1  ξ^1  ξ,
3 4 3
³ξ^1  ³ξ^1  ²ξ,
2 3 2
⁴ξ^1  ³ξ^1  ³ξ.
1 2 1

Topilgan (1.50) qiymatlarni (1.47) formulaga qo’yib,
y^2ni, demak,
y^2 I topamiz.

4

5
Agar
y^2  y^1bo’lsa, u holda y₅  y^2 eb olamiz. Aks holda iterasiyani davom

5 5 5
ettiramiz. Taiiyki, iterasiyani ko’p davom ettirishning foydasi yo’q. Qadam shunday

tanlanishi kerakki, bitta yoki ikkita iterasiya yetarli bo’lsin.
y₅ topilgandan keyin

shu usul bilan
y₆ va hakozolar topiladi.

5.