8-Ma’ruza: Nyutonning interpolyatsion formulalari Reja

_{8-Ma’ruza: Nyutonning interpolyatsion formulalari}
Reja:
1. Nyutonning 1-interpolyatsion formulasi.
2.Nyutonning 2-interpolyatsion formulasi.
3. Sonli differensiallashda Nyuton interpolyatsion formulasidan foydalanish.

Bu yerda [a,b] kesmada kiritilgan teng qadamli, ya’ni yonma-yon turgan tugun nuqtalarining orasidagi masofa h o‘zgarmas bo‘lgan, _n to‘rda qiymatlari berilgan f(x) funktsiya uchun interpolyatsiyalash ko‘phadini qurish masalasini qaraymiz. Bu ko‘phadni Lagranj interpolyatsiyalash ko‘phadi sifatida ham qurish mumkinligi aniq. Ammo bu yerda qurish jihatidan Lagranj interpolyatsiyalash ko‘phadidan soddaroq bo‘lgan Nyuton interpolyatsiyalash ko‘phadlarini qurish usulini beramiz.

Avvalo, chekli ayirmalar tushunchasini kiritamiz. Agar teng h qadamli _n to‘rda f(x) funktsiyaning qiymatlari
f(x_i)=y_i(i=0,1,2,…, n) (5.9)
berilgan bo‘lsa
y_i=y_i+1- y_i (i=0,1,2,…, n-1)
ayirmalar 1-tartibli chekli ayirmalar,
²y_i=y_i+1-y_i (i=0,1,2,…, n-2)
ayirmalar 2-tartibli chekli ayirmalar va hokazo
^m(y_i)=^m-1y_i+1-^m+1y_i (i=0,1,2,…,n-m), (mn)
ayirmalar m-tartibli chekli ayirmalar deb yuritiladi. CHekli ayirmalarning taorifidan ko‘rinadiki, _n to‘rda berilgan funktsiyaning y, ²y, …., ⁿy chekli ayirmalari mavjud bo‘lib, n-dan yuqori tartibli chekli ayirmalari yo‘qdir.
Yuqoridagi formulalar asosida 5-tartibli chekli ayirmalar jadvalini tuzamiz:
5.3-jadval

x	y	y	^2y	^3y	^4y	^5y
x0	y0	y0=y1-y0	^2y0= y1-y0	^3y0= ^2y1-^2y0	^4y0= ^3y1-^3y0	^5y0= ^4y1-^4y0
x1= x0+h	y1	y1=y2-y1	^2y1= y2-y1	^3y1= ^2y2-^2y1	^4y1= ^3y2-^3y1
x2= x0+2h	y2	y2= y3-y2	^2y2= y3-y2	^3y2= ^2y3-^2y2
x3= x0+3h	y3	y3= y4-y3	^2y3= y4-y3
x4= x0+4h	y4	y4= y5-y4
x5= x0+5h	y5
…	…

Masalan ushbu jadval uchun

5.4-jadval

X	0.1	0.2	0.3	0.4
Y	0.25	0.37	0.40	0.48

3-tartibli chekli ayirmalarni quyidagi jadvalda yaqqol ko‘rish mumkin:
5.5-jadval

i	X	yi	yi	y^2i	y^3i
0 1 2 3	0.1 0.2 0.3 0.4	0.25 0.37 0.40 0.48	0.12 0.03 0.08	-0.09 0.05	0.14

Teng qadamli _n to‘rda berilgan funktsiyaning interpolyatsiyalash ko‘phadini

P_n=a₀+a₁(x-x₀)+a₂(x-x₀)(x-x₁)+a₂(x-x₀)(x-x₁)(x-x₂)+…+a_n(x-x₀)(x-x₁)…(x-x_n-1) (5.10)

ko‘rinishda izlaylik. U holda (5.10) da (5.9) ga asosan koeffitsentlarni quyidagicha aniqlaymiz.

5.6-jadval

X	Koeffitsentlarni aniqlash	Koef fitsentlar
x=x0	y0 =a0	a0=y0
x=x1	y1=a0+a1h,
x=x2	y2=a0+a1(x2-x0)+a2(x2-x0)(x2- x1) y2=y0+ 2h+a22hh, y1+y1=u0+2 y0+2a2h^2, y0+y0+y1=y0+2y0+2a2h^2, y1-y0= 2a2h^2, ^2y0= 2a2h^2
…	…	…
x=xn	yn=a0+a1(xn-x0)+a2(xn –x0)( xn – x1)+...+an(xn-x0)( xn – x1)...(xn-xn-1) yn=y0+ 2h+ 2hh+ 6hhh+…+123…n an hh...h

(bu ishlarni to‘liqroq bajarishini o‘quvchiga havola qilamiz). Topilganlarni (5.10) ga qo‘ysak,
(5.11)
ni olamiz. Buni Nyutonning birinchi – interpolyatsiyalash ko‘phadi deb yuritiladi.
Agar deb olsak,

(5.11) ni
(5.12)
ko‘rinishda yozib olish mumkin. Bu Nyutonning birinchi interpolyatsiyalash ko‘phadining yakuniy ko‘rinishi bo‘lib, hisoblash uchun ancha qulaydir.
Agar interpolyatsiyalash ko‘phadini
P_n(x)=a₀+a₁(x-x_n)+a₂(x-x_n)(x-x_n-1)+….+a_n(x-x_n)…(x-x₁)
ko‘rinishda izlasak, yuqoridagi qilingan o‘xshash mulohazalar asosida
(5.13)
ni olamiz. Buni Nyutonning ikkinchi – interpolyatsiyalash ko‘phadi deb yuritiladi.
Agar (5.13) da desak,
(5.14)
ko‘phadni olamiz. Bu Nyutonning ikkinchi interpolyatsiyalash ko‘phadining yakuniy ko‘rinishidir.
BERILGAN JADVALGA ASOSAN Nyuton INTERPOLYaTSIYa
FORMULASIDAN FOYDALANIB,ARGUMENT
KIYMATIGA MOS FUNKTSIYa KIYMATINI ANIKLASH

X:	2.000	3.000	4.000	5.000	6.000
Y:	1.583	1.436	1.372	1.238	1.084

Y( 3.500)= 1.4079
Y( 4.100)= 1.3625


(*----5.4 – Dastur--------*)
{ * Nyuton interpolyatsiya ko‘phadining qiymatini aniqlash * }
uses crt;
var
i,j,n:integer;
s,’,s1,t,x1:real;
x:array[0..7] of real;
y:array[0..7,0..7] of real;
begin
clrscr;
writeln(‘ Nyuton interpolyatsiya ko‘phadining qiymatini aniqlash ‘);
write(‘(x,y)-juftliklar soni N= ‘);read(n);
writeln(‘(x,y)-juftliklarni kriting ‘);
for i:=0 to n do
begin
{gotoxy((i)*10,4);}
write(‘x(‘,i,’)=’);read(x[i]);
{gotoxy((i)*10,4);}
write(‘y(‘,i,’)=’);read(y[0,i]);
end;
writeln(‘ berilgan argument qiymati:’);
write(‘x=’);read(x1);
t:=(x1-x[0])/(x[2]-x[1]);
for i:=1 to n do
for j:=0 to n-1 do y[i,j]:=y[i-1,j+1]-y[i-1,j];
s:=y[0,0];
s1:=1;’:=1;
for i:=1 to n do begin
for j:=1 to i do begin
s1:=s1*(t-(j-1));
‘:=‘*j;
end;
s:=s+y[i,0]*s1/’;
end;
readln;
writeln(‘ Ko‘phadning qiymati: ‘);
write(‘y(‘,x1:2:3,’)=’,s:4:4);
readln;
end.
Nyuton interpolyatsiya ko‘phadining qiymatini aniqlash
(x,y)-juftliklar soni N=3
(x,y)-juftliklarni kriting
x(0)=0.1
y(0)=0.25
x(1)=0.2
y(1)=0.37
x(2)=0.3
y(2)=0.4
x(3)=0.4
y(3)=0.48
berilgan argument qiymati:
x=0.212
Ko‘phadning qiymati:
y(0.212)=0.3774


5.4. Sonli differentsiyalashda Nyuton interpolyatsiyalash
formulasidan foydalanish

Berilgan jadval bo‘yicha hosila topish amalini bajarish uchun bu Nyutoning 1- interpolyatsiya ko‘phadidan foydalanamiz:

bunda t=(x-x0)/h . Murakkab funktsiyani differentsiallash qoidasiga asosan:
da deb olsak


……………………..

Bu holda birinchi va ikkinchi tartibli hosilalarni hisoblash formulalarini quyidagicha yozamiz:


5.3-masala. Berilgan quyidagi jadval asosida jadval asosida x=0.15 bo‘lgandagi birinchi va ikkinchi tartibli hosilalarni hisoblang.
5.7-jadval

X	0	0.1	0.2	0.3	0.4
Y	2	2.1152	2.2614	2.4399	2.6518

Yechish. Bu jadval asosida chekli ayirmalarni tuzamiz:
5.8-jadval

X	Y	y	^2y	^3y
0 0.1 0.2 0.3 0.4	2 2.1152 2.2614 2.4399 2.6518	0.1152 0.1462 0.1785 0.2119	0.0310 0.0323 0.0334	0.013 0.011

x=0.15 va h=0.1, y₀=2.1152, y₀ =0.1462, ²y₀=0.0323, ³y₀=0.011 ga asosan:

t=0.5 bo‘lsa, yuqoridagi formulalar asosida hosilalarni hisoblaymiz:

O‘z-o‘zini tekshirish uchun savollar

1. Nyuton interpolyatsiyalash ko‘phadini tanlash qoidasi va uning ahamiyati.
2. CHekli ayirmalar asosida Nyuton interpolyatsiyalash ko‘phadining koeffitsentlarini topish.
3. Ikkinchi va uchunchi tartibli Nyuton ko‘phadini yozing.
4. Lagranj va Nyuton interpolyatsiyalash ko‘phadini tanlash koidalarining farqi
5. Sonli differentsiyalashda Nyuton interpolyatsiyalash formulasidan foydalanish.