33
n
n
1
0
x
c
..
x
c
c
P(x)
+
+
+
=
(1)
ko‘phad uchun ushbu
n
,
0,1,
k
),
f(x
)
P(x
k
k
=
=
tengliklar bajarilsin. Bu
tengliklarni ochib yozsak
)
,
0
(
c
m
n
m
=
larga
nisbatan
1)
(n
+
noma’lumli
1)
(n
+
ta tenglamalar sistemasi hosil bo‘ladi.
=
+
+
+
+
=
+
+
+
+
=
+
+
+
+
)
(
....
.....
..........
..........
..........
..........
..........
)
(
....
)
(
....
2
2
1
0
1
1
2
1
2
1
1
0
0
0
2
0
2
0
1
0
n
n
n
n
n
n
n
n
n
n
x
f
x
c
x
c
x
с
с
x
f
x
c
x
c
x
с
с
x
f
x
c
x
c
x
с
с
(2)
bu sistemaning determenanti Vandermond determenantidir:
).
.x
,
.x
W(x
n
1
0
Masala
mazmunidan
ravshanki,
k
x
nuqtalar
bir-biridan farqli, demak bu determinant noldan farqlidir. Shuning uchun ham (2)
sistema va shu bilan birga qo‘yilgan interpolyatsiya masalasi yagona yechimga ega. Bu sistemani yechib,
m
c
larni topib (1) ga qo‘ysa,
)
(
x
P
ko‘phad aniqlanadi. Biz
P(x)
ning oshkor ko‘rinishini topish uchun boshqacha
yo‘l tutamiz, fundamental ko‘phadlar deb atluvchi
(x)
Q
nj
larni, ya’ni
=
=
=
,
lg
'
,
1
,
lg
'
,
0
)
(
anda
bo
j
i
anda
bo
j
i
x
Q
j
i
i
nj
shartlarni qanoatlantiradigan n-darajali ko‘phadlarni ko‘ramiz. U holda
(3)
izlanayotgan interpolyatsion ko‘phad bo‘ladi. Haqiqatdan ham barcha
n
..,
0,1,2,
i
=
lar uchun
=
=
=
=
=
n
j
n
j
j
j
i
j
i
nj
j
i
n
x
f
x
f
x
Q
x
f
x
L
0
0
)
(
)
(
)
(
)
(
)
(
va ikkinchi tomondan
(x)
L
n
-
n
darajali ko‘phaddir.
Endi
(x)
Q
nj
ning oshkor ko‘rinishini topamiz,
i
j
bo‘lganda
0
)
(x
Q
i
nj
=
shuning uchun ham
(x)
Q
nj
ko‘phad
i
j
bo‘lganda
x
-
x
i
ga bo‘linadi. Shunday qilib,
-
n
darajali ko‘phadning n ta
bo‘luvchilari bizga ma’lum, bundan esa
kelib chiqadi. Noma’lum ko‘paytuvchi
C
ni esa
1
)
x
-
(x
С
)
(x
Q
j
i
i
j
nj
=
=
shartdan topamiz, natijada:
−
=
j
i
j
i
nj
)
x
-
x
(
С
(x)
Q
i
x
x
bu ifodani (3 )ga qo‘yib kerakli ko‘phadni aniqlaymiz:
=
−
−
=
n
j
j
i
i
j
i
j
n
x
x
x
x
x
f
x
L
0
)
(
)
(
(4)
bu ko‘phad Lagranj interpolyatsion ko‘phadi deyiladi.
Bu formulaning xususiy hollarini ko‘raylik:
1
n
=
bo‘lganda Lagranj ko‘phadi ikki nuqtadan o‘tuvchi
to‘gʻri chiziq formulasini beradi:
)
(
)
(
)
(
1
0
1
0
0
1
0
1
1
x
f
x
x
x
x
x
f
x
x
x
x
x
L
=
−
−
+
−
−
=
Agar
2
n
=
bo‘lsa, u vaqtda kvadratik interpolyatsion ko‘phadga ega bo‘lamiz. Bu ko‘phad
uchta
nuqtadan o‘tuvchi va vertikal o‘qqa ega bo‘lgan parabolani aniqlaydi
)
(
)
)(
(
)
)(
(
)
(
)
)(
(
)
)(
(
)
(
)
)(
(
)
)(
(
)
(
2
1
2
0
2
1
0
1
2
1
0
1
2
0
0
2
0
1
0
2
1
2
x
f
x
x
x
x
x
x
x
x
x
f
x
x
x
x
x
x
x
x
x
f
x
x
x
x
x
x
x
x
x
L
−
−
−
−
+
−
−
−
−
+
−
−
−
−
=
Endi Lagranj interpolyatsion formulasining boshqa ko‘rinishini keltiramiz. Buning uchun
=
+
−
=
n
i
i
n
x
x
x
0
1
)
(
)
(
=
=
n
j
nj
j
n
x
Q
x
f
x
L
0
)
(
)
(
)
(
=
j
i
i
nj
)
x
-
(x
С
(x)
Q