Toshkent davlat transport universiteti
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Misol 2. ( )
arcsin f x x = funksiyani x ning darajalari bo‘yicha qatorga yoying.
Yechish. Eng avvalo IV formulada 1 2 m = − desak, quyidagi yoyilmani hosil qilamiz: 2 3 4 1 1 1 3 1 3 5
1 3 5 7 1 2 2 4 2 4 6
2 4 6 8 1
x x x x = −
+ − + − +
( 1 1)
−
x o’rniga 2
− ifodani qo’yib, quyidagini hosil qilamiz: 2 4 6 2 2 1 1 1 3 1 3 5 1 3 5
(2 1) 1 ... ...
2 2 4
2 4 6 2 4 6
2 1
n x x x x n x − = + + + + + + −
Darajali qatorlarni hadma-had integrallash qoidasidan foydalanib, 1 1 x − da quyidagiga ega bo’lamiz: 3 5 7 2 1 2 0 1 1 3 1 3 5
1 3 5 (2 1) arcsin ...
... 2 3
2 4 5 2 4 6 7
2 4 6 2 2 1 1
n dt x x x n x x x n n t +
− = = + + + + + + + − Bu qator ( )
− intervalda yaqinlashuvchi bo’ladi. Shu bilan birga qator 1
ham yaqinlashuvchi bo’ladi va ushbu qiymatlar uchun ham qatorning yig’indisi arcsin x ga teng bo’ladi. Bu tenglikda 1
ni hisoblash formulasini hosil qilamiz: 1 1
1 3 1 1 3 5 1
1 3 5 (2 1) 1 arcsin1
1 ...
... 2 2 3 2 4 5 2 4 6 7
2 4 6 2 2 1 n n n
− = = + + + + +
+
+
2 1
5 6
f x x x − = − + funksiyani x ning darajalari bo‘yicha qatorga yoying.
Yechish. Berilgan funksiyani eng sodda ratsional kasrlarga yoyib, quyidagiga ega bo’lamiz: 2 1
1 2 2 1 1 1 . 5 6 ( 2)(
3) 2 3 3 2 1 1 3 2 x x x x x x x x x x − − = = − = −
− + − − − − − −
Endi 2 1 1 1
x x x x = + +
+ + + − tenglikdan foydalanamiz: 2 2
1 1 3 3 3 3 1 3
n n n n x x x x x = = + + + +
+ − =
2 2 0 1 1 2 2 2 2 1 2 n n n n n x x x x x = = + + + +
+ − =
Natijada 1 1 0 0 0 2 1 2 1 ( ) 3 3 2 2 3 2 n n n n n n n n n n x x f x x + + = = = − = − =
yoyilmani hosil qilamiz. Loyiha-hisob ishlari topshiriqlari 8-masala. Funksiyani x ning darajalari bo’yicha qatorga yoying. 8.1. (
2 6 1 ln x x − − .
8.2. 2 2 3 x x − − . 8.3. ( ) 2 2
e − .
8.4.
( ) 2 3 x e − + . 8.5.
2 20 9 x x − − .
8.6. ( ) x x 3 sin 1 − . 8.7. 3 2 27 x x − .
8.8.
( ) 2 12 1 ln x x − − . 8.9. 4 5 16 x − .
8.10.
2 7 3 4 5
x − + . 8.11. x x x − 2 cos 2 2 .
8.12.
x x + 3 . 8.13.
( )
e x 1 − .
8.14.
( ) 2 8 5 3 ln x x − + . 8.15. 2 16 1 x − .
8.16.
x arctgx . 8.17. 3 2 27 x x − .
8.18.
2 12 7 x x − − . 8.19. 2
xe − .
8.20.
3 8
+ .
1 arcsin
− x x .
8.22. 2 4 1 x − . 8.23. ( ) 3 1
e + .
8.24.
( ) 2 3 ln 2 + + x x . 8.25. 2 9
x + .
8.26.
( )
e x − + 1 . 8.27. x xe 2 − .
8.28. 4 4 1 x − . 8.29. 2 3 4 1
x − − .
8.30. ( ) 2 ln 3
4 1
x − + . 9-masala. Aniq integralni 001
, 0 = aniqlikda hisoblang. 9.1.
1 , 0 0 6 2 dx e x .
9.2. ( ) 1 , 0 0 2 100 sin
dx x . 9.3. 1 0 2 cos
dx x .
9.4. + 5 , 0 0 4 4 1 x dx .
9.5. 1 2 1 0,1
x e dx x − − .
9.6. ln 1
1 5 0,1 x dx x + . 9.7.
1 , 0 0 2 sin dx x .
9.8. + 5 , 1 0 3 3 27 x dx . 9.9. 25 , 0 0 cos dx x x .
9.10. 25 , 0 0 dx e x . 9.11. + 5 , 0 0 3 3 64 x dx .
9.12. ( ) 2 , 0 0 2 25 sin
dx x . 9.13. ( ) 2 2 1 0,1
ln 1 x dx x + .
9.14. + 1 0 4 4 16 x dx . 9.15. ( ) 5 , 0 0 2 4 cos dx x .
9.16. 0,2
1 0,1
x e dx x − − . 9.17. ln 1 0,4
2 0,1
x dx x + .
9.18. 4 , 0 0 2 2 5 sin dx x . 9.19. + 5 , 0 0 3 3 8 x dx .
9.20. ( ) 0,2 ln 1 2
0,1 x dx x + . 9.21.
( ) 5 , 0 0 2 4 sin dx x .
9.22. − 5 , 0 0 2 cos 1 dx x x . 9.23. ( ) + 1 , 0 0 2 1 ln
x x .
9.24. − 5 , 0 0 25 3 2 dx e x . 9.25. 5 , 0 0 sin xdx x .
9.26. + 5 , 2 0 4 4 625 x dx . 9.27. 2 , 0 0 sin dx x x .
9.28. 4 , 0 0 2 2 5 cos dx x . 9.29. 2 0,5
0,1 sin x dx x .
9.30. + 5 , 0 0 2 1 dx x x .
4-§. Loyiha hisob ishini bajarish namunasi 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. 2 1
25 10 24 n n n = − − Yechish. 2 10 10 1 1 25 10 24 (5 6)(5
4) 5 6 5 4
n n n n n = = − − − − + − +
tenglikdan foydalanib, qatorning qismiy yig‘indisini topamiz. 10 10 10 10 1 ... 1 9
4 14 9 19
14 24 (5 6)(5 4) n S n n = + + + + + = −
− + . 1 1 1 1 1 1 1 1 1 1 1 1 ... 9 4 14 9 19 14 24 5 11 5 1 5 6 5 4 n n n n = − − + − + − +
+ + − + − = − − − + 1 1 1 1 . 4 5 1 5 4
n = − + −
− − + Shuning
uchun, 1 1 1 3 1 . 4 5 1 5 4 4 lim lim(
) n n x n n S → → − + − − − + = = − ya’ni qator yaqinlashuvchi va yig‘indisi 3 4 − ga teng. 2-masala. Quyidagi qatorlarni (a) solishtirish alomatidan va (b) limit belgili solishtirish alomatidan foydalanib yaqinlashishga tekshiring. a) 3
ln n n n =
b) 1 1 cos n x n = −
Yechich. a) Ma’lumki, ihtiyoriy musbat haqiqiy 0 soni uchun etarlicha katta
n larda ln n n
3 3 3 2 ln 1 n n n n n − = tengsizlikni hosil qilamiz. ni shunday tanlaymizki, 3 1 2 − bo’lsin. Masalan, 1 4
= deb olishimiz mumkin. U holda 5 3 4 ln 1 n n n a b n n = = tengsizlikni o‘rinli va
5 5 5 4 4 4 1 1 1 1 ... ...
2 3
+ +
+
qator umumlashgan garmonik qator bo‘lib, 5 1 4 p = bo‘lgani uchun yaqinlashuvchi. Solishtirish alomatiga ko‘ra berilgan qator ham yaqinlashuvchi bo‘ladi. b) Bu qatorga limit belgili solishtirish alomatiga ko‘ra quyidagi yaqinlashuvchi bo‘lgan (umumlashgan garmonik qator 2 1
2 1 1 n n =
Birinchi ajoyib limitdan 0 sin
lim 1 → = foydalanib, quyidagiga ega bo‘lamiz: 2 2 2 2 2 2 2 2 2 2 1 cos 2sin sin
sin 2 2 2 lim
lim lim
lim 0. 1 1 2 2 2 2 4 n n n n x x x x x x x n n n n x x n n n n → → → → − = = = = Shunday qilib, limit belgili solishtirish alomatiga ko‘ra berilgan qator yaqinlashuvchi bo’ladi.
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