Tengsizliklar-i. Isbotlashning klassik
Download 443.1 Kb. Pdf ko'rish
|
tengsizliklarni isbotlash
|
Amaliyot uchun masalalar. 1-masala. Tengsizliklarni isbotlang:
! ,
n n n n ⎛ ⎞
> ⎜ ⎟ ⎝ ⎠
N ∈ ; (1) 1 ! , : 2 n n n n N + ⎛ ⎞ < ∀ ∈
⎜ ⎟ ⎝ ⎠ 2
≥ ; (2) 2 ! , :
n n n N n > ∀ ∈ ≥ 3 n ≥ ; (3) ; (4) 1 ! 2 ,
: 3
n n N − > ∀ ∈
17 , 2 n n n n n e n N e ⎛ ⎞
⎛ ⎞ < < ∀ ∈
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
. (5)
(1) tengsizlikni isbotlaymiz. Induktsiya bazasi. 1
= da:
1 1 1! 3 ⎛ ⎞
> ⎜ ⎟ ⎝ ⎠
ga egamiz. Induktsiya bazasi isbotlandi. Induktiv o’tish. da
n k = ! 3 k k k ⎛ ⎞
> ⎜ ⎟ ⎝ ⎠
tengsizlik to’g’ri deb faraz qilamiz. da tengsizlik bajarilishini isbotlaymiz: 1
= +
( 1) ( 1) ( 1)! 3
k k + + ⎛ ⎞ + > ⎜ ⎟ ⎝ ⎠ . ( 1)! !( 1) ( 1 3 k k k k k k ⎛ ⎞
+ = + > + ⎜ ⎟
⎝ ⎠ )
ga egamiz. ( 1) 1 3
k + + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ songa ko’paytiramiz va bo’lamiz: 1 ( 1) ( 1) 1 3 ( 1) 3 ( 1) 3
k k k k k k k k + + + + ⋅ ⋅ + ⎛ ⎞ = ⎜ ⎟ + ⋅ ⎝ ⎠
1 1 1 3 1 1 3 3 (1 ) k k k k k k + + + + ⎛ ⎞ ⎛ ⎞ = > ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + . Bu yyerda quyidagi joriy hisoblashlarni bajaramiz: 2 1 1 ( 1) 1 ( 1) ... (
1) 1 (1 ) 1 ...
2! !
k k k k k k k k k k k k − − ⋅ ⋅ − + + = + +
⋅ + +
⋅ =
1 1 1 1 1 1 2 1 1 1 1 ...
1 1 ... 1 2! !
k k k k k < < − ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = + + − + + − − ⋅ ⋅ − < ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
18 2 1 2 1 1 1 1 1 2 3 2 1 2 3 ... 2 1 1 1 1 1 1 1 1 ... 1 1
... 2! 3! ! 2 2
2 k k k k − − = < =
⋅ ⋅ ⋅ ⋅ ⋅
= + + + + + < + + + + +
<
2 1 1 1 1 1 1 1 3 1 1 ... ... 1
3 (1 ) 3 1. 1 1 2 2
2 2 1 ( 2
k k k k k −
+ + +
= ⇒ + < ⇒ > − + 1 ) Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (1) tengsizlik bajariladi deb xulosa qilamiz. n (2) tengsizlikni isbotlaymiz. Induktsiya bazasi. 2
= da:
(2) tengsizlikning chap tomoni: 2! 2
= ; (2) tengsizlikning o’ng tomoni: 2 2 2 1 3 9 2, 25 2 2 4 + ⎛ ⎞ ⎛ ⎞ = = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . 2 2
demak, , 25
< induktsiya bazasi isbot bo’ldi. Induktiv o’tish. da
n k = 1 ! , 2 k k k + ⎛ ⎞ 2
< ≥ ⎜ ⎟ ⎝ ⎠ tengsizlik to’g’ri deb faraz qilamiz. da 1
= + 1 2) ( 1)!
, 3
k k + + ⎛ ⎞ + < ≥ ⎜ ⎟ ⎝ ⎠ 2 k tengsizlik bajarilishini isbotlash kerak. 1
1)! ! (
1) ( 1) 2 k k k k k k + ⎛ ⎞ + = ⋅ + < ⋅ + ⎜ ⎟ ⎝ ⎠ = ga egamiz. 1 2 2 k k + + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ songa ko’paytiramiz va bo’lamiz: 1 1
1 1 1 2 ( 1) ( 1) 2 2 2 ( 1) 2 2 2 ( 2) ( 2) k k k k k < k k k k k k k k k k + + + + + + + + ⋅ + ⋅ + ⋅ + ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⋅ +
+ ⎝ ⎠ ⎝ ⎠ 1 1 2 (
1) 1 ( 2) k k k k + + ⋅ + < +
tengsizlik bajarilishini isbotlaymiz.
19 1 1 1 2 ( 1) 2 1 2 ( 2) 2 1 1 1 1 k k k k k k k k 1
+ +
+ ⋅ +
= = ⋅
+ + ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ + + ⎝ ⎠ ⎝ ⎠
1 2 0 1 1 (
1) 1 1 (1 ) 1 ... 2 1 1 2! 1 ( 1)
k k k k k k k k + > + + ⋅
⎛ ⎞ + = + + ⋅ + + ⎜ ⎟ + + + + ⎝ ⎠ > . 1 1 1 1 2 1 (1 ) 2 2 1 1 2 1 2 k k k k k + + + ⎛ ⎞ ⇒ + < ⇒ ⋅
< ⋅ = ⎜ ⎟ + + ⎝ ⎠ . 1 1 2 2 1 2 2 k k k k + + + + ⎛ ⎞ ⎛ ⎞ < ⋅ =
⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (2) tengsizlik bajariladi deb xulosa qilamiz. 2
≥ (3) tengsizlikni isbotlaymiz. Induktsiya bazasi. 3
= da:
(3) tengsizlikning chap tomoni: 3! 6
36 = =
; (3) tengsizlikning o’ng tomoni: 3 2
2 = 7 . 36 27 < demak, induktsiya bazasi isbot bo’ldi.
da
n k = 2 ! ,
k k k > 3 ≥ tengsizlik to’g’ri deb faraz qilamiz. da 1
= + 1 2 ( 1)!
, k k k k + + > ≥ 3
tengsizlik bajarilishini isbotlash kerak. 1 1 1 2 2 2 2 1 1 2 2 ( 1) ( 1) ( 1)! ! ( 1) ( 1) ( 1) ( 1) ( 1) k k k k k k k k k k k k k k k k k + + + + ⋅ + + = ⋅ + >
⋅ + ⋅ = + ⋅ + + 1 2 2 2 k k ⋅ > ⋅
20 (6) masalada tengsizlik isbotlangan edi. Bu tengsizlikdan kelib chiqadi: 1 ( 1) , n n n n n + > + ∀ ≥ 3
1 2 2 ( 1) ,
n n n n n 3 + > + ∀ ≥
. U holda n k = da 1 1 1 2 2 2 2 1 2 2 1 ( 1) ( 1) 1 ( 1) ( 1) (1 ) ( 1) ( 1) k k k k k k k k k k k + + > + ⋅ + > + ⋅ = + ⋅ + > + + ⋅ 1 1 2 2
k + . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (3) tengsizlik bajariladi deb xulosa qilamiz. 3
≥ (4) tengsizlikni isbotlaymiz. Induktsiya bazasi. 3
= da:
(4) tengsizlikning chap tomoni: 3! 6
36 = =
; (4) tengsizlikning o’ng tomoni: 3 1 2
− = . demak, induktsiya bazasi isbot bo’ldi. 6 4 >
n k = da bajariladi deb faraz qilamiz: . da
1 ! 2 ,
3 k k k − > ∀ ≥ 3
1
= +
1 1 ( 1)! 2 , k k + −
+ > ≥ tengsizlik bajarilishini isbotlash kerak. 1 1
( 1)!
! ( 1) 2
( 1) 2
2 , 3 2 k k k k k k k k k − > + + = ⋅ + > ⋅ + = ⋅ > ≥ .
Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (4) tengsizlik bajariladi deb xulosa qilamiz. 3
≥ (5) tengsizlikni isbotlaymiz. Induktsiya bazasi. 1
= da:
1 1 1 1 1! 2 e e ⎛ ⎞
⎛ ⎞ < < ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠ bo’ldi.
21 Induktiv o’tish. n k = da (5) tengsizlik to’g’ri deb faraz qilamiz: ! 2
k k k k e e ⎛ ⎞
⎛ ⎞ < < ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠ .
da 1
= + 1
1 1 ( 1)! 2
k k k k e e + + + + ⎛ ⎞ ⎛ ⎞ < +
⎜ ⎟
⎟ ⎝ ⎠ ⎝ ⎠ tengsizlik bajarilishini isbotlash kerak. Bu tengsizlikning chap tomonini isbotlaymiz. 1 1 ( 1) 1 ( 1)!
! ( 1) (
1) 1
k k k k k k k e k k k k e e k e + + ⎛ ⎞ + ⋅⎜ ⎟
+ ⎛ ⎞
⎛ ⎞ ⎝ ⎠ + = ⋅ + >
+ ⋅ = ⋅ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ =
1 1 1 1 1 ( 1) 1 1 ( 1) 1 1 k k k k k k e k k k e k e k e e k k + + + + + + ⋅ ⋅ + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = ⋅ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ 1 k e + > . (5) tengsizlikning chap tomonini isbotlaymiz. 1 1
1 1 2 ( 1)!
! ( 1) 2 2 2 1 2 k k k k k k k k k k k e e e k + + + ⎛ ⎞
⎜ ⎟ + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ⎠ + = ⋅ + <
= ⋅ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
×
1 1 1 2 1 ( 1) 1 2 k k k k e k k + + < ≤ + ⎛ ⎞ ⎛ ⎞ × ⋅ < ⎜ ⎟ ⎜ ⎟ + + ⎝ ⎠ ⎝ ⎠ . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (5) tengsizlik bajariladi deb hulosa qilamiz. n Eslatib o’tamiz, (2) tengnsizlik va 1 1
e n ⎛ ⎞ + < ⎜ ⎟ ⎝ ⎠ tengsizlikdan foydalanib, da 1
>
1 1 1 1 2 ! 2 2 2 2 n n n n n n n n n n n k e e e e n e + ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ ⎠ ⎝ ⎠ < = ⋅
⋅ = ⋅
⋅ < ⋅ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎛ ⎞
⋅⎜ ⎟ ⎝ ⎠
2 n n .
Download 443.1 Kb. Do'stlaringiz bilan baham: 
|