Toshkent davlat transport universiteti
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- Bu sahifa navigatsiya:
- 2.5. Integral alomati.
- Misol 13.
- Leybnits alomati.
- Yechish .
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Yechish. 2 1 , n n n a n + =
1 1 lim lim lim 1
1 n n n n n n n n l a e n n → → → + = = = + =
. Qator uzoqlashuvchi. 2.5. Integral alomati. Agar (n)
n a f = bo‘lib, ( ) f x funksiya 1
a da musbat, monoton kamayuvchi va uzluksiz bo‘lsa, u holda ( )
( )
a f x dx xosmas integral bir vaqtda yaqinlashuvchi yoki uzoqlashuvchi bo‘ladi.
1 1
n n = Dirixli qatorini yaqinlashishga tekshiring. Yechish. 1 ( ) p f x x = funksiya integral alomatining barcha shartlarini qanoatlantiradi. 1)
1 p = bo‘lsin. U holda 1 1 1 ( )
ln dx f x dx x x + + + = = = , ya’ni integral uzoqlashuvchi va demak qator ham uzoqlashuvchi. 2) 1
( ) ( ) 1 1 1 1 1 1 1 1 ( ) lim lim
lim 1 .
1 1
b p p p p b b x dx dx f x dx x b x x p p + + − − → → → = = = = − − − Shuning uchun 1
1 1 1 p dx x p + = − va demak qator yaqinlashuvchi. 1
1
+ = va qator uzoqlashuvchi bo‘ladi. 1 p = da Dirixli qatori garmonik qator bilan ustma-ust tushadi. Shuning uchun odatda Dirixli qatorini umumlashgan garmonik qator deb ham atashadi. Shunday qilib umumlashgan garmonik qator 1
1
3. Ishoralari o‘zgaruvchi qatorlar. Faraz qilaylik 1 2
... n a a a + + + +
( ) A qatorning hadlari ixtiyoriy ishorali sonlardan tashkil topgan bo‘lsin. Agar qator hadlarining absolyut qiymatlaridan tuzilgan 1 2 ... ...
n a a a + + + +
* ( ) A
qator yaqinlashuvchi bo‘lsa, u holda berilgan ( ) A qator ham yaqinlashuvchi bo‘ladi va absolyut yaqinlashuvchi qator deb ataladi. Agar ( )
qator
yaqinlashuvchi bo‘lib, * ( ) A qator uzoqlashuvchi bo‘lsa, u holda ( )
qator shartli yaqinlashuvchi qator deb ataladi. ( )
A qatorni absolut yaqinlashishga tekshirishda * (
A qator uchun musbat hadli qatorning ma’lum alomatlaridan foydalanish mumkin. Xususan, agar 1 lim 1 n n n a a → + yoki
lim 1
n n a → bo‘lsa, ( )
A qator absolut yaqinlashuvchi bo‘ladi. Umumiy holda, * ( ) A qatorning uzoqlashuvchiligidan ( )
qatorning uzoqlashuv- chiligi kelib chiqmaydi. Lekin, agar 1 lim 1 n n n a a → + yoki
lim 1
n n a → bo‘lsa, u holda na faqat * (
A qator, balki ( )
qator ham uzoqlashuvchi bo‘ladi. Leybnits alomati. Agar ishoralari almashinuvchi 1 1 2 3 4 ... ( 1) ...
( 0)
n n b b b b b b − − + − + + −
+
( ) B qator uchun 1)
1 2 3 ... ...
n b b b b va
2)
lim 0
n b → = .
shartlar o‘rinli bo‘lsa, u holda ( ) B qator yaqinlashuvchi bo‘ladi, uning yig‘indisi S musbat bo‘lib, birinchi hadidan oshib ketmaydi: 1
b . Natija. Agar ishoralari almashinuvchi ( ) B qator Leybnits alomati shartlarini qanoatlantirsa, u holda qatorni taqribiy hisoblashlardagi xatolik n n R S S = −
birinchi tashlab yuborilgan hadidan absolyut qiymati bo’yicha kichik bo’ladi: 1 | | .
n R b + Misol 18. Ushbu 2 2 2 2 cos cos 2 cos3
cos ...
... 1 2 3 n n + + + +
+
qatorni yaqinlashishga tekshiring, bu yerda −ixtiyoriy son. Yechish . Berilgan qator hadlarining absolyut qiymatlaridan qator tuzamiz:
2
2 2 cos cos 2 cos3
cos ...
... 1 2 3 n n + + + +
+
Hosil bo‘lgan musbat hadli qator – yaqinlashuvchi, chunki 2 2 cos 1 n n n tengsizlik o‘rinli bo‘lib, 2 2
1 1 1 1 ...
... 2 3 n + + + + + qator yaqinlashuvchi (Misol 10). Demak, berilgan qator absolut yaqinlashuvchi.
1 1
1 1 1 ... ( 1) ...
2 3 4 n n − − + − + + − +
qator yaqinlashuvchi, chunki Leybnits alomatining barcha shartlari bu qator uchun o‘rinli bo‘ladi. Bu qator shartli yaqinlashuvchi, chunki 1 1
1 1 ... ... 2 3 4 n + + + + + +
qator uzoqlashuvchi (garmonik qator). Loyiha-hisob ishlari topshiriqlari 1-masala. Berilgan har bir qator uchun:
a) qatorning dastlabki n ta hadining yig‘indisi ( )
n S ni toping;
b) ta’rifdan foydalanib qatorni yaqinlashishini isbotlang; v) qatorning yig‘indisi ( )
ni toping. 1.1. 1
1 1 ... ... 2 4
3 5 4 6
( 1) (
3) n n + + + + + + + 1.2.
1 1 1 1 ...
... 3 7
5 9 7 11
(2 1) (2
5) n n + + + + + + +
1.3. 2 1 1 1 1 ... ... 10 40 88 9 3 2 n n + + + + + + −
1.4. 2 6 6 6 6 ... ... 15 54 111 9 12 6 n n + + + + + + −
1.5. 4 4 4 4 ...
... 1 5
5 9 9 13
(4 3) (4
1) n n + + + + + − +
1.6. 24 24 24 24 ...
... 2 4
1 7 4 10
(3 5) (3
1) n n + + + + + − − +
1.7. 2 2 2 2 2 ... ... 15 35 63 4 8 3 n n + + + + + + +
1.8. 2 1 1 1 1 ... ... 4 10 18 3
n + + + + + + − 1.9.
8 8 8 8 ...
... 1 7
3 11 7 15
(4 5) (4
3) n n + + + + + − − +
1.10. 9 9 9 9 9 ... ...
1 8 2 11
5 14 8 17
(3 4) (3
5) n n + + + + +
+ −
− +
1.11. 2 6 6 6 6 ... ...
7 91 247 36 24 5 n n + + + + + − −
1.12. = − + 1 2 8 6 9 6
n n
1.13.
= − + 1 2 2 3 9 3
n n
1.14. = − − 1 2 45 28 49 14
n n
1.15.
= − − 1 2 12 7 49 7
n n
1.16. = − − 1 2 33 56 49 14
n n
1.17.
= − − 1 2 35 12 36 12
n n
1.18. = − + 1 2 15 8 16 8
n n
1.19.
= − + 1 2 12 7 49 7
n n
1.20. = − − 1 2 2 3 9 3
n n
1.21.
= − + 1 2 10 21 49 7
n n
1.22. = − + 1 2 35 12 36 12
n n
1.23.
= − − 1 2 24 70 49 14
n n
1.24. = − + 1 2 6 5 25 5
n n
1.25.
= − − 1 2 10 21 49 7
n n
1.26. = − − 1 2 40 42 49 14
n n
1.27.
= − + 1 2 6 35 49 7
n n
1.28. = − + 1 2 3 4 4 4
n n
1.29.
= − − 1 2 13 84 49 14
n n
1.30. = − − 1 2 48 14 49 14
n n
. 2-masala. Solishtirish alomatlaridan foydalanib quyidagi qatorlarni yaqinlashishga tekshiring: 2.1.
2 1 sin 2 n n n n =
2.2.
2 1 ln n n =
2.3. 1 1 3 1 5 n n n =
+
2.4.
1 1 (5n 3)7 n n = −
2.5. 1 sin 3 n n n =
2.6. 1 sin
3 2
n = − 2.7.
1 1 100 99 n n = −
2.8.
2 1 ( 1)( 2)
n n = − + 2.9.
2 1 1 20 n n n = +
2.10.
1 1 (2 1)( 3)
n n = − + 2.11.
5 4 1 ln n n n =
2.12.
3 1 ln n n n =
2.13. ( ) 1 2 2 1 n n n = − −
2.14.
1 1 ( 3 1 3 1) 2 n n n n = + − − 2.15.
3 5 1 cos 3
n n =
2.16.
4 3 1 ln n n n =
2.17. 4 3 1 1
tg n n =
2.18.
1 1 2 5 n n n − = + 2.19.
2 3 1 sin n n n =
2.20.
2 1 1 3 n n n n = + + 2.21.
= + 1 3 1 n n n
2.22. = + 1 3 1 1 n n
2.23. = − − 2 3 1 1 ) 1 ( 1
n arctg n
2.24. =1 3 4 1 n n arctg n
2.25. = − 1 1 1
n e n
2.26. 1 1 5 3 2 n n n − = + 2.27.
1 2 sin 5 n n =
2.28.
=1 3 n n tg
2.29. 1 2 1 2
n n = −
2.30.
3 3 1 sin 2
n n = +
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