Koshi masalasining ko’p qadamli ayirmali metodlar

Adamsning ekstrapolyasion metodlari. Yuqorida aytganimizdek, ^Adamsning m ^{-qadamli oshkor}

^yn ^{ y}n1 ^{ hb}mi ^fn1
i1
(1.13)

metodi uchun approksimasiyaning eng yuqori tartibi p  m . Noma’lum koeffisentlarni topish uchun(1.12) sistema bu holda ushbu ko’rinishga ega:



i1
^k  ¹ , k  0,1,...,m  1.

i b
^mi k  1
(1.14)

Har bir muayyan m uchun (1.12) sistemani yechib,
b_m1,b_m2 ,...,b_mn larni topamiz.

Agar
m 1 bo’lsa, u holda Adams metodi ushbu
^yn ^{ y}n1 ^{ hf}n1

Eyler metodiga aylanadi.

Adams mashhur ingliz artilleristi Boshfort iltimosiga ko’ra o’z metodlarini 1855 y. yaratgan edi. Bu metodlar keyinchalik unutilgan bo’lib, asrimizning boshida norvegiyalik matematik Shtyormer tomonidan qayta ochildi.

Osonlik bilan topish mumkinki,
m  2,3,4,5
bo’lganda mos ravishda

approksimasiya tartibi m ga teng bo’lgan quyidagi metodlarga ega bo’lamiz:

^yn ^{ y}n1

• 
  ^h ^{3 f}
  

2
n1 ^
fn2
^{, m  2;}

^yn ^{ y}n1
^{ h} ^{23 f}
12


n1
 16 f


n2

• 
  5 f
  

n3
^{, m  3;}

y_n  y_n1 
^h ^{55 f}
24


n1
 59 f


n2
 37 f


n3

• 
  9 f
  

n4
^{, m  4;}

y_n  y_n1 
h



720
^{1901 f}


n1
 2774 f


n2
 2616 f


n3
 1274 f


n4
 251 f


n5
^{, m  5.}

Amaliyotda Adams metodlari
m 1,2,...,10lar uchun ishlatiladi.

Mashq.Adams metodlari m  6,7,8,9,10, lar uchun chiqarilsin.
Adams metodlarini qurishda boshqacha yondashish ham mumkin. Faraz qilaylik,

y₀ , y₁ ,..., y_n1
(1.15)

taqribiy qiymatlar hisoblangan bo’lib, n  k  1 bo’lsin. Keyingi y_n ni hisoblash

uchun algebraik interpolyasiyalashdan foydalanamiz. Buning uchun ushbu
u^_ x_ning

x_n1k , x_nk ,..., x_n1, x_n ,..., x_n1q
k  q  1ta nuqtalaragi qiymatlaridan foydalanib, _k  q_
interpolyasion ko’phadini ko’ramiz.
(1.16)

tartibli Lagranj

k
L _ x_  _
^k q1
^{ xu'} ^x 

n1 j
,

k q
_{ }x  x
^' ^x 

bunda
^{j q} n1 j k q1

n1 j

^k q1 ^{ x   x  x}n1k ^{ x  x}nk ^{...x  x}n1q ^.

Tugunlar bir xil uzoqlikda joylashganligi almashtirishni bajaramiz. U holda
x_j  x_j1  h
uchun
x  x_n1  th

x  x  h_t  j _,
_{ x  h}k q1_*
_t _,

Bunda
n1 j k q1

k q1

k q1
^* _t _  _t  k _t  k  1_...t _t  1_..._t  q_,


*
k q1
^xn1 j
_  _1_q j h^{k q} _k 
j _!_ j  q!._

Bu holda (1.17) ko’phad quydagi ko’rinishga ega bo’ladi:

_{k 1}q j _*
^t 

^L  ^x
 th_  _ ^{k q1} u^_x _.
(1.18)

k q n1


jq
_t 
j _ j  q_!_k 
j _!
n1 j

Bu ko’phaddan foydalanib, quyidagi tenglikni yazamiz:
u^_ x_  L_{k q } x_n1  th_  r_{k q }x_n1  th_,
(1.19)

bunda
^r_{k q}  ^x_n1 ^{ th}
iterpolyaning qoldiq hadi. Agar
f _ x,u _
qaralayotgan sohada

_k  q  1_
tartibli uzluksiz xususiy hosilalarga ega bo’lsa, u holda qoldiq hadni

quyidagicha yozish mumkin:


^r_{k q}  ^x_n1 ^{ th} ^


_hk q1
*
k q1
^{k q2}


_ _.
(1.20)

 _t _u
Bu ifodani biz _x_n1 , x_{n 
}k  q  1_!
oraliqda ishlatamiz. Shuning uchun, agar


q  1
bo’lsa,

^xn1k ^{  x}n1q
Ushbu
va agar
q  0 bo’lsa,
^xn1k ^{  x}n
deb qaraymiz.

^{xn 1}

^u  ^xn  ^{ u}  ^xn1  ^
_ u^_ x_dx  u _ x_{n1 }  h_u^_x_n1  th_dt

^xn1 ⁰

Formulada
^u ^{xn1  th} ^{ni (1.19) formulaning o’ng tomoni bilan almashtiramiz, u}

holda quyidagiga ega bo’lamiz:

1 1
^u  ^xn  ^{ u}  _n1  ^{ h}_ ^Lkq  ^{xn1  th}^{dt  h}_^rkq  ^{xn1  th} ^
0 0
(1.21)

k
 u _ x _  h _ b^qu^_x _  R^q ,

bunda
n1

^q

kj n1 j n,k
jq


_{jq }t  q_...t _t  1_..._t  k _
1

^bkj

^q

 _1<sub>


h</sub>k q2
^ _t 

0


1
j _ j  q_!_k 

^{k q2}

j _!
dt,

^Rn,k
^ _k  q  1_!^
*

0
k q1
_t _u
^{ x}n1

• 
  ^th^dt.
  

(1.22)

Bu yerda ^*
^t 
o’z ishorasini saqlaydi va
^{k q2}

^u  ^x
n1

• 
  ^th
  

uzluksiz bo’lganligi

k q1
uchun qoldiq hadni quyidagicha yozib olishimiz mumkin:

_R^q _
hk q2
1
_uk q2 _{  }*
_t _dt,
(1.23)

n,k
_k  q  1_!
_ k q1 0

bunda
^xn1k
   x_n1q ,
agar
q  1 bo’lsa va
x_{nk 1}    x_n , agar
q  0 bo’lsa. Hosil

qilingan (1.21) formuladan har xil ayirmali sxema va ular uchun qoldiq hadning

ifodasini ko’rsatish mumkin. Bu metodlar
q  1
bo’lganda interpolyasion deyiladi.

q  0
holda mos keladigan metod ekstrapolyasion deyiladi. Bunday

almashtirishning sababi quyidagidan iborat:
^Lmq ^{ x}
ko’rishda qatnashadigan,

(4.10) tugunlarni o’z ichiga olgan eng kichik oraliq _^x_n1k , x_n1q ^_ dir. Agar q  0
bo’lsa, qaralayotgan _x_n1 , x_{n }oraliq _x_n1k , x_{n1 } oraliqdan tasjqarida yotadi; shuning
uchun ham _x_n1 , x_{n } oraliqda ekstrapoyasiya qilinadi: agar q  1 bo’lsa,

^xn1k ^{, x}n1q ^_
oraliq _x_n1 , x_{n }
oraliqni o’z ichiga oladi va bu yerda asl ma’noda

interpolyasiya qilinadi.
Avvalo, ekstrtopolyasiya metodini ko’rib chiqamiz. (4.21) formulani quyidagicha yozib olamiz:


q  0

bo’lgan hol uchun

k
u _ x _  u _ x _  h_b^0u^_x
_  R^0 

n n1


k 1
kj n1 j n,k
j0

^{ u}  ^x  ^{ h}_^b0
u^_x
_ R^0 
(1.24)

n1
k , j1
j1
m
n j n,k

^{ u}  ^x
_  h_b u^_x _  R^o ,

n1
mj n j n,m1
j1

bunda
m  k  1 va b
_{ b}⁰
bo’lib,
q  0
bo’lganda u (1.22) formuladan

aniqlanadi:
mj k , j1


_j1 ¹ t _t  1_..._t  m  1_

^bmj
^ ^1
^ _t 
j  1_ j  1_!_m 
dt,
j !

(1.25)

0
j  1, 2,..., m.

Qoldiq had R
_{ R}⁰
esa
q  0 bo’lganda (1.23)dan quyidagicha hisoblanadi:

n,m n,m1


_hm1 R_n,m  _m! u
1
^m1 _ _t _t  1_..._t  m  1_dt.
0
(1.26)

Bu belgilashlarda (4.24) quyidagi ko’rinishga ega:

^u  ^xn  ^{ u}  ^xn1  ^{ h}_^bmju ^xn1  ^{ Rn,m.}
j1

Bu formula hisoblash uchun yaroqsizdir, chunki unda noma’lum had, izlanayotgan yechim hosilasining ushbu qiymatlari

^u ^xnm ^,u^xnm1 ^,...,u^xn1 
(1.27)


R_n,m qoldiq

(1.28)

^{va u}  ^xn1 
qatnashadi. Agar yechimning

^u  ^xnm ^,u ^xnm1 ^,...,u ^{x n1} 
aniq qiymatlari ma’lum bo’lsa, u holda (1.1) tenglamaga ko’ra (1.28) miqdorlarning aniq qiymatini topishimiz mumkin edi:

u^_x_{n j } 
Ammo bizga izlanayotgan
f _x_{n j} ,u _x_{n j }, j 1, 2,...,m.

y_nm , y_nm1,..., y_{n1 }n  m_
yechimning faqat taqribiy qiymatlari ma’lum va bular orqali
y_n^_1 taqribiy qiymatini topish mumkin:


^uxn j ^
hosilaning

^yn^ j ^{
f x}n j ^{, y}n j ^{ }
f_{n j} ,1,2,...,m.

(1.29)

Endi (1.27) formuladagi hosilalarni (1.29) tarqibiy qiymatlari bilan
^u  ^xn1 
ni esa

uning taqribiy qiymati
^yn1
bilan almashtiramiz va
^Rn,m
qoldiq hadni tashlaymiz,

natijada quyidagi taqribiy tenglikka ega bo’lamiz:

^u  ^x_n  ^{ y}_n1 ^{ h}_^b_mj ^f_{n j} ^.
j1
(1.30)

Bu tenglikning o’ng tomonini
y_n deb olamiz, u holda

^yn ^{ y}n1 ^{ hb}mj ^fn1
j1
(1.31)

kelib chiqadi. Shunday qilib, biz yana (1.13) tenglikka keldik. Bundan ko’ramizki,

^bmj
koeffisentlarni ikki xil usul bilan topishimiz mumkin: (1.14) sistemaning

yechimi yoki (1.25) integralning quymati sifatida.

Misol uchun keltiramiz:
m  5
bo’lganda
b_mj ning sonli qiymatini va
R_n,m ning ifodasini

_{b } 1901<sub>,b


 </sub> 2774 <sub>,b


</sub> 2616 <sub>,b


</sub> 1274 _,

51 ₇₂₀ 52
251
720 ⁵³
95 _{ }
720 ⁵⁴
720 ^

(1.32)

b₅₅ 

720
, R_n,5 

288
h⁶u ⁶  ξ ^




Mashq. (1.25) va (1.26) vformulalar yordamida m  6,7,8,9,10 uchun b_mj va R_n,m lar topilsin.
Biz yuqorida (1.18) Lagraj interpolyasion formulasidan foydalanib, natijada (1.25), (1.26) formulalarni chiqardik. Shunga o’xshash Nyutonning ikkinchi

interpolyasion formulasini qo’llab, (1.30) formula o’rniga
f x,u
funksiyaning

tugun nuqtalaridagi qiymatlari emas, balki chekli ayirmali qatnashadigan Adamsning ekstrapolyasion formulasini chiqarishimiz mumkin. Bu formula quyidagicha yoziladi:

y  y
 ξ  ¹ ξ  ⁵ ²ξ


 ³ ³ξ 

n n1
n ₂ n1 ₂
n2 ₂ n3

_ 251 <sub>4ξ


</sub> 95
⁵ξ
_ 19087 _{6ξ }
(1.33)

720
n4
288
n5
60480
n6

Bu yerda
_ 5275
17280
^7^ξn7
 ... c_m
^mξ_nm .

1


_i  hf_i ,c_m  _m!
^t ^{t 1}^...^{t  m 1}^dt,

k
ⁱ
esa
 _ x_
funksiyaning
x_k , x


k 1
...,x


k  j
nuqtalaridagi qiymatlari bo’yicha

tuzilgan i - tartibli chekli ayirmasidir. (1.33) formulaning qoldiq hadini quyidagicha yozish mumkin:
_{R  h}m2_{c u}^m2 _{ .}

Mashq. (1.33) formula isbotlansin.
n,m m1

Endi (1.33) formulaning
m  4
bo’lgandagi xususiy holini qaraymiz:

y  ξ  ¹ ξ


 ⁵ ² ξ


 ³ ³ ξ


 ²⁵¹ ⁴ ξ .

(1.34)

n1 n ₂ n1 ₁₂
n2 ₈
n3
720
n4

Bu yerda
n  4 deb olamiz, u holda

y  ξ  ¹ ξ ⁵ ²ξ  ³ ³ξ ²⁵¹ ⁴ξ.
(1.35)

4 4 ₂ 3 ₁₂
2 ₈ 1
720 ⁰

Hisoblashni (1.35) formula 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
		y1			^2 ξ0
				ξ1		^3 ξ0
x2	y2		ξ2		^2ξ1		^4ξ0
		y2		ξ2		^3ξ1
x3	y3		ξ3		^2ξ2
		y3		ξ3
x4	y4		ξ4
		y4

x5

y5

ξ5

(1.35) formulaning o’ng tomonidagi barcha miqdorlar aniq bo’lib, jadvalning pastki

qiya satrida joylashgan. Biz y₄ ni topamiz, demak, shu bilan y₅ ham aniqlanadi.
Topilgan y₅ ga ko’ra n ξ₅  hf x₅ , y₅ ni hisoblaymiz va chekli ayirmalar jadvalini

yana bir qiya satr bilan to’ldiramiz. Keyin (1.34)da davom ettiramiz.
n  6
deb olib, hisoblashni

4.