Toshkent davlat transport universiteti
-masala. 6.1.-6.24 masalalarda funksional qatorning yaqinlashish sohasini aniqlang. 6.1
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6-masala. 6.1.-6.24 masalalarda funksional qatorning yaqinlashish sohasini aniqlang. 6.1. 3 5 7 1 2 1 ... ( 1)
... n n x x x x x − − − + − + + − +
6.2. 4 6 8 2 2 ... ...
n x x x x + + + + + +
6.3. 2 3 1 ln (1 ) ln (1 ) ... ln (1 ) ... n x x x + + + + + + + +
6.4. 2 3 ... ...
x x x nx e e e e − − − − + + + +
+ 6.5. 1 1
1 1 1 1 1 ... ( 1) ... 2 3 n x x x n − + + + − + − + −
+ 6.6. 1 ln 2
1 1 ( 1) n x n n + = − 6.7. 2 ...
... 4 16 4 n n x x x xtg x tg x tg + + + +
6.8. 2 2 4 2 ...
... x x nx x x nx e e e + + + + 6.9. 2 3
4 8 (2 ) ... ...
10 100
100 10
n x x x x + + + + +
6.10. 3 5
1 2( 2) 4( 2) 8( 2) ... 2 (
2) ...
n n x x x x − − + − + − + +
− +
6.11. 2 3 1 9 27 (3 ) 3 ... ( 1) ... 3 5 2 1
n x x x x n − − + − + −
+ −
6.12. 6 9 3 3 2 1 ( 5) ( 5) ( 5) ( 5) ... ...
2 3 3 3
3 n n x x x x n − − − − − + + + + + 6.13. 1 1
1) 3 1 n n n n x n + = + + 6.14. 3 1
1) n n n x n = +
6.15. 3 6 3 3 2 1 1 ... ... 3 3 3 n n x x x − − + + + + + 6.16. 4 8
4 2 1 ... ...
2 5 16
8 16 (3 1) 16 n n x x x x n − + + + +
+ − 6.17. 1 2
2 1 3 (3 2)
n n x n − − = − 6.18. 2 1
4 ln n n n x n n − = 6.19. ! 1
n n n x = 6.20. 1
= 6.21. 3( 1)
1 ( 1) n n n x n − − = + 6.22. 6 9
3 2 1 ... ...
3 3 3 n n x x x x − + + + +
+ 6.23. 8 12
4 1 ... ... 3 4
5 16 (2 1) 4 n n x x x x n − + + + +
+ − 6.24. 2 1 2 1 5 (3 2)
n n x n − = − 6.25-6.30 masalalarda hadma-had differensiallash va
integrallash qoidalaridan foydalanib qatorning yig‘indisini toping. 6.25. 2 3 ... ...
2 3
x x x x n + + + + + 6.26. 3 5 2 1 1 ... ( 1) ...
3 5 2 1 n n x x x x n − − − + − + − + −
6.27. 2 1 1 2 3 ... ( 1) ( 1) ... n n x x n x − − + − + −
+ +
6.28. 2 1 1 2 2 3
3 4 ...
( 1) ... n x x n n x − + + + + +
+ 6.29. 2 1
2 3 ... ... 2 4 8 2
n x x n x − + + + +
+ 6.30. 2 3
1 1 ... ( 1) ... 2 3
3 9 4 27
( 1)3
n n n x x x x n + + − + − + − + +
7-masala. Darajali qatorning yaqinlashish radiusini, yaqinlashishi oralag’ini toping, hamda oraliq chekkalarida yaqinlashishga tekshiring. 7.1. (
= − 2 ln 1 n n n n x .
7.2. ( ) ( ) ( ) = + + − 1 1 3 1 1 n n n n n x . 7.3. ( ) ( ) = + − 1 2 1 2 n n n n x .
7.4. =1 2 n n n n x . 7.5. ( ) = + − 1 1 2 3 2 n n n n x .
7.6. ( ) ( ) = + + 1 3 1 8 2
n n n x . 7.7. = + + 1 1 1 2 3 n n n n x .
7.8. ( ) ( )
= + − 1 ! 1 3 1 n n n n x n . 7.9. ( ) ( ) = − + − 1 1 2 5 1
n n n n x .
7.10.
( ) = + − 0 2 2 1 1 1 n n n x n .
7.11. ( )
( ) = + + − 1 1 3 2 4 1
n n n n x .
7.12.
( ) n n n n 0 2 x 1 n 1 = − + . 7.13.
( ) ( ) = + + 1 3 1 2 1
n n n x .
7.14. ( ) = − − 1 1 3 5 1 n n n n x . 7.15. ( ) n n n n 2 x 1 3 n ln n = − .
7.16. ( ) n n 2 n 2 x 1 n ln n = − . 7.17. ( ) ( ) = + − 1 1 ln 2 2 n n n n x .
7.18. ( ) = + + 1 1 1 3
n n n x . 7.19. ( ) ( ) n n n n 0 5 x 1 n 2 ! n = − + .
7.20. n 3 n 0 n! x n 1 = + . 7.21. ( )
n 3 n 0 x n 1 = − + .
7.22. ( )
3 n 1 n n n 0 n 1 x e + = − . 7.23.
2 n n 2 ln n x n = .
7.24. n 3 n 0 x n 2 n 3 = + + . 7.25.
( ) ( ) = + + 1 2 1 2 1
n n n x .
7.26. ( ) = + − 1 3 1 2 3 1
n n n x . 7.27. ( ) ( ) ( ) = + + − 1 1 ln 1 2 n n n n x .
7.28.
( ) =1 ! 3 n n n x . 7.29. ( ) ( )
= + − 1 ! 2 1 1 1 n n n n n x n .
7.30.
( ) ( ) = − + 1 1 5 3 2
n n n x .
3-§. Funksional qatorlarni darajali qatorlarga yoyish 1. Teylor qatori. ( )
| | x a r −
atrofida cheksiz differensiallanuvchi bo‘lsin. Agar bu funksiyani shu atrofda x a − ning darajalari bo‘yicha darajali qatorga yoyish mumkin bo‘lsa, u holda bu qator (Teylor qatori) quyidagi ko‘rinishga ega bo‘ladi: ( ) 2
( ) ( )
( ) ( )(
) ( ) ( ) . 2! !
n f a f a f x f a f a x a x a x a n = + − + − + + − + (1)
( )
1 ( )
( ) ( )
( ) ( ) ! k n k n k f a R x f x f a x a k = = − + − Teylor qatorining qoldiq hadi bo‘lsin. (1) tenglik o‘rinli bo‘lishi uchun
−
tengsizlikni qanoatlantiruvchi barcha
x larda lim
( ) 0
n R x → = bo‘lishi zarur va yetarlidir. Qoldiq hadni baholash uchun quyidagi formuladan foydalanish mumkin:
( 1) ( ) ( )
( ) ,
0 1 ( 1)! n n n x a R x f a x a n + + − = + − + .
(2) 0 a = da Makloren qatorini hosil qilamiz: ( )
2 (0)
(0) ( )
(0) (0)
. 2! ! n n f f f x f f x x x n = + + + + +
2 ( )
cos f x x = funksiyani x ning darajalari bo‘yicha qatorga yoying. Yechish. Funksiyani 1
2 ( )
cos ;
x =
( ) 2cos sin
sin 2 cos 2
; 2
x x x x = − = − = +
( ) 2cos 2 2cos 2
2 ; 2 f x x x = − = +
2 2 ( )
2 sin 2 2 cos 2
3 ; 2 f x x x
= = +
3 3 ( )
2 cos 2 2 cos 2
4 ; 2 IV f x x x = = +
..........................
( ) 1 ( ) 2 cos 2
; 2
n f x x n − = +
( 1) ( ) 2 cos 2 ( 1) . 2
n f x x n + = + +
0 a = deb, quyidagilarni hosilqilamiz: 3 5 ( ) 1 ( 1) (0) 1,
(0) 0, (0) 2, (0)
0, (0)
2 , (0)
0, (0)
2 , ..., (0)
2 cos
, ( ) 2 cos 2 ( 1) , 2 2
V VI n n n n f f f f f f n f f f x x n − + = = = − = = = = − = = + + bu yerda 0 1
. Qoldiq hadni (2) formulaga asosan topamiz: 1 1
( 1) 1 (2 ) 2 ( )
cos 2 ( 1) . ( 1)!
2 ( 1)!
2 n n n n x n x R x x x n n n + + + + = = + + + + x ning har qanday qiymatida 1 (2 ) lim 0 ( 1)! n n x n + → = + bo‘lib, cos ( 1) 2 x n + + – chegaralangan miqdor bo‘lganligidan lim
( ) 0
n R x → = ekanligi kelib chiqadi. Bu esa 2 ( ) cos f x x = funksiyani Makloren qatorining yig‘indisi ko‘rinishida ifodalash mumkinligini anglatadi: 3 5 2 1 2 2 4 6 2 2 2 2 2 cos 1 ( 1)
. 2! 4! 6! (2 )!
n n n x x x x x n − = − + − + + − +
2. Darajali qatorga yoyishda qo’llaniladigan usullar. Ba’zi funksiyalarning darajali qatorga yoyilmalarini keltiramiz. I. 2
1 1! 2! 3! ! ( ). n x x x x x n e x = + +
+ + + + −
II.
3 5 2 1 sin
( 1) 1! 3! 5! (2 1)! ( ).
n x x x x x n x + = − + − + − + + − III.
2 4 2 cos 1 ( 1) 2! 4! (2 )! ( ).
n x x x x n x = −
+ − + − + − IV. (
2 3 ( 1) ( 1)( 2) 1 1 1! 2! 3! m m m m m m m x x x x − − − + = + + + + +
( 1) (m n 1) ! ,
m m x n − − + + +
bu yoyilma quyidagi hollarda o’rinli bo’ladi: 0
1
− ; 1 0
− da, 1 1
− ; 1
1
− .
V. ( ) 2 3 1 ln 1 ( 1)
2 3 ( 1 1). n n x x x x x n x − + = − + − + − + − Ko’p hollarda yuqoridagi yoyilmalardan, hamda geometric progressiya yig’indisi formulasidan foydalanib berilgan funksiyani osongina darajali qatorga yoyish mumkin, bunda qatorning qoldiq hadini tekshirish zarurati yo’qoladi. Ba’zida darajali qatorlarni hadma-had differensiallash va integrallash qoidalaridan foydalanish maqsadga muvofiq bo’ladi. Rasional funksiyalarni darajali qatorlarga yoyishda ularni eng sodda rasional kasrlarga yoyish tavsiya etiladi.
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