Buxoro davlat universiteti qo’lyozma huquqida udk abdullayev Behzod Rajabovich

Teorema 3.2.2. (3.23) tenglik tasdiqlaydiki,

_   n ,k 


^pˆ k ,


N

• 
  ^{ C}  ^Yk ,
  

_ (  ) _

^U ^  ^ _{ }
 1 _Y
^ 

k 1 1
^k m  ^{n  k  2} m k ,

Haqiqatdan ham
U _ _
funksiya (3.15) vaznli kubatur formula uchun ekstremal

funksiya bo’ladi va
U _ _  L^m _ S _ , bunda
^Yk ,
( ) - ortonormallangan sferik

2
^{garmonika k - tartibli ko’rinishdagi va} bo’lmagan sferik garmonikalar soni:
 (n, k ) - k – tartibli chiziqli bog’liq

3.3. Davriy bo’lmagan funksiyalar uchun optimal interpolyatsion formulalar qurish.

Quyidagi formula berilgan bo’lsin


_ ^f  ^{cos }  ^{sin  d }
0

va
^ N '

_N '
 _ C _{ '}
 ^' 1
f _cos  _{ ' }

(3.24)

yoki
_ cos^  d cos 
0
^{  C}  '
 '1
cos^ 
 ' ^,
^{  0,1,..., m  1}  ^{, (3.25)}

1
_ f (t ) dt
 1

va

N '
^{  C}  '
 '1
^f  ^t_{ '} 

(3.26)

1
t ^ dt


N '
 _ C t ^

, (

 0,1,..., m  1) .



 1

Belgilash kiritamiz:

 '  '

 '1

N '
_ C _{ '} f (cos  _{ '} )  L ( f ; 0,  )
 '1
yoki
N '
_ C _{ '} f (t_{ '} )  L ( f ;  1,1) .
 '1

2
Agar f (t )  L^{(m )} (1,1), unda

1 ₁ 1

 1
_ ^{f (t ) d t  L}  ^{f ;  1,1} ^
 1
( m  1)! ^{ F}m
 ^u 
^f ( m )  ^u  ^{du ,}
(3.27)

bu yerda

F_m (u ) 

va
¹ N '

 '

m
_ (t  u ) ^{m 1} dt  _ C k
_u  '1

(t  u )

_{ x} m 1 _,
k_m ( x )  _
 0 ,
x  0 uchun
x  0 uchun

(3.13) kubatur formulada , 0  
 2
, ^ bo’yicha davriy bo’lganligi uchun

to’g’ri to’rtburchaklar formulasini qo’llaymiz, unda
_x (  ) _
2 _,
N

C₁  C ₂  ...  C _N
^{ 2} , 0    
N
davriy bo’lmagan hol uchun yuqori darajali

kvadratur formulani qo’llaymiz.
Yuqoridagi fikrlarimizga asosan kubatur formula quyidagi ko’rinishda keladi.

2  1

_2 N N₁
2

2
_{ } f (t ,  )dtd 
 _{ } L ( f (t
, ² ); y , y
⁾ , (3.28)

0  1 ^N
 1 j 1

2
2
 ' _N
2
j j 1

bu yerda


L ( f

(t_{ '}

, ^22 ); y N ₂
2


j ^{, y} j 1 ⁾

• 
  kvadratur formula:
  

^y j 1 ^
y _j  h  ,
N ₁
j  1, 2, ..., N ₁ .

Shunday qilib (3.28) ko’rinishdagi kubatur formulani xatolik funksionali normasini baholaymiz: ma’lumki

||


(3.29) dan quyidagi kelib chiqadi

| 
(3.29)

^{2  1 2 } N ₁

 | _{ }  (t , ) d t d   _{ } C _1 
(t , )d  


1

0  1 0
₁ 1

^{2 } N _1
2 N ₁ N ₂
2

 _{ } C _1 
₀ ₁ 1
2  _ 1
(t_1
, )d  
N ₂

1
N_{1
 } C ₁ 


₁ 1 ₂ 1

(t_1
, ² ) | 
N ₂

 _{  } 
(t , ) d t  _ C _1  (t_ , ) _ d  

⁰  ^{ 1}
_{N 1} 2 
₁ 1 _

• 
  _ C
  

₁ 1
^1 ^{ }
0

^{2  1} N ₁

 _ | _  (t , ) d t  _ C _1 
(t , ) | d  


1

0  1


_{N 1} 2 
₁ 1

 _
₁ 1
C _1 _ 
0
. (3.30)

endi (3.30) formulaning o’ng tomonidagi har bir qo’shiluvchini baholaymiz. (3.27) ni hisobga olsak quyidagicha bo’ladi,
1)


1
_ 
 1

1
 _ 
 1


N ₁
(t , ) d t  _ L ( 
j 1

(t _j , ); y _j , y _{j 1} ) 

_{N1 } ^y j 1 _

 _{  } 
j 1 _{ y j}
(t , ) d t  L ( 
(t _j , ); y _j , y _{j 1} ) _ 
_

_{N 1 } 1 _h

 _{  } 
j 1 _{  1}
( y _j 
(  1), ) d  
2

 L [ 

( y _j 
^h (  1) , );  1;1 ]^ 


2 <sub>


N 1</sub> 1 _h

2
 _ F ( )  ^{(m )} ( y 


(  1),  ) d 


. (3.31)

_ m j
j 1 _{ 1}

Koshi – Bunyakovskiy tengsizligini qo’llasak (3.31) ni o’ng tomoniga quyidagini olamiz.

¹ N _1
  _{
 1} j 1
1

^{| F}_m ^  ^|2 ^{d } _ 2 ^


1
^ 1

 ¹ _h 
_ ^y j 1


N ₁



( m ) 2





j


 _{ }
|  ^{(m ) } y 
_  1_ , |² d ^_
2
_ 2 h
₂ m 1
_{M } 
_ | 
y _j
 ^{t ,}  ^|
dt _ 
_

  1
 _
j 1

m  ¹
_h 2
^_ ^N1


^y j 1

1
_ 2

1
 ^N1  ²
1
_M  ¹  ²

 M _
 |  ^{(m )} _t , _ |² dt _{ } 1² _ 
^{m } 
|  ^{(m )} _t , _ |² dt _ ,
(3.32)

bu yerda
_ j 1 _{y j}

_{ } j 1 _
^{N 1} 1 

₁  ¹ 

 
^{M } m 1 ^
2 m  1 !
_ | F_m
 _ |² d _ .

Shunday qilib

  1





2
_ t , _  L^{(m )} _ S _ , bundan kelib chiqadiki

 ¹ 
_ |  ^{(m )} _ t , _ |² dt _

 C || 

| L^{(m )} ( S ) ||

, С₁ – o’zgarmas. (3.33)

_ 1 2
 1 
(3.33) va (3.32) dan foydalanib quyidagini olamiz


1
_ 
 1

^{ M C ||  | L} m ^  ^S  ^{|| 1 M} ' ^{||  | L} m ^  ^S  ^||
, MC₁= M ^' . (3.34)

_N m ^{1 2} _N m ²
1 1

2. 
   (3.30) ning o’ng tomonidagi ikkinchi qo’shiluvchi uchun olingan natijalardan foydalansak quyidagicha bo’ladi.
   

2 
_ 
0
1  ^ 2 

2



N


_m ^
₂  k 1
_k 2 m ^
d || 
| L^{(m )} ( S ) || , d – o’zgarmas. (3.35)

(3.34) va (3.35) ni (3.30) ga qo’ysak quyidagi bahoni olamiz:
2 ^{ } 1 ^

2
 _N |,   
 ^{ m}

N

M ^ || 
/ L^{(m )} ( S ) ||_ d  

0 ^{1 
N1} 1  ^ 2 

2


 _ | C _1 |
₁ 1
_m ^

N
₂  k 1
_k 2 m ^
d || 
| L^{(m )} ( S ) ||

^ ^2
d  ^
1


^{2  2 N1} 

|| 
| L^{(m )} ( S ) || _

2
_N m
M ^ 
_m ^
_k 2 m ^
_ | C _1
|_ ,
(3.36)

N
_ 1 2
 k 1
 _{1 }

yoki (3.36) dan quyidagi kelib chiqadi

2 M ^'
|| L^{(m )*} ( S ) ||  



d  _



1

2  ^{2 N1}


| C | ^.


^{N 2} _N m _N m  _k 2 m  ^1
_{1 2}  k 1  ₁ 1
Ma’lumki kvadratur formulaning koeffisiyentlari musbat bo’lsa (masalan Gaus tipidagi kvadratur formulalar) quyidagicha bo’ladi.

N ₁


_ C 1
₁ 1
N ₁


 _ C 1 .
₁ 1

Shunday qilib quyidagi lemmani isbotladik.

Lemma 3.3.1.

L^{(m )} ( S ), (2 m  3)
fazoda (3.28) ko’rinishdagi kubatur

2
formulaning xatolik funksionali normasi uchun quyidagi baho o’rinli:

2 M ^'
|| L^{(m )*} ( S ) ||  



d  _



1

2  ^{2 N1}


| C |,

(3.37)


^{N 2} _N m _N m 
_k 2 m  ^1

bu yerda
_{1 2}  k 1

 ₁

_{d } 1
1


_{ 2 } m 1 _,
  0


uchun


_M ^' _1 _
( m  1)! _1
o’zgarmaslar.
| F ( ) |² d  _ ,

m

F_m ( )  _
 0 ,
  0


uchun
, ^d va ^d₁

Quyidagi o’rinli bo’lsin
N₁ = N₂ va N=N₁ × N_{2 .} (2.38)
(3.38), va (3.37) dan foydalanib quyidagi bahoni olamiz



m
1 ^  ^


1 _
^{2  2} N 1 

2

k
|| _N
L^m* ( S ) || 
_ 2 M ^'  d

 ^ 2 m  ^{ | C 1}

|_ . (3.39)

_N ² _
 k 1
 _{1 }

Baxvalov teoremasini keltiramiz.