B. Abdurahmonov Toshkent–2008 2 B. Abdurahmonov. Matematik induksiya metodi
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Matematik induksiya metodi @aniq fan
- Bu sahifa navigatsiya:
- 4.3-masala.
- Isbotlash.
- 4.6-masala.
- 1-teorema.
- 4.7-masala.
- 1-teorema .
- Isbotlaymiz.
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2 2
2 3 2 1 1 6 12 8 1 1 1 2 2 6 12 9 k k k k k k k k k k + + < + + + ⎛ ⎞ ⎛ ⎞ = + < + ⎜ ⎟ ⎜ ⎟ + + + + + ⎝ ⎠ ⎝ ⎠
. 29
2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.2.2.1) tengsizlik ixtiyoriy n natural son uchun bajariladi.
1 2 1 2 (1 ) (1 ) ... (1
) 1 ...
, ,
n x x x x x x n N + ⋅ + ⋅ ⋅ + ≥ + +
+ + ∀ ∈
(4.3) Bu yerda , 1,...,
i x i n = , – (-1) dan katta bo’lgan bir xil ishorali son. 1-teorema. 1
= da
1 1 1 1 x x + = +
ga ega bo’lamiz. 1-teorema isbotlandi. 2-teorema. (4.3) tengsizlikning n=k da bajarilishi berilgan: 1 2 1 2 (1 ) (1 ) ... (1
) 1 ...
k k x x x x x x + ⋅ + ⋅ ⋅ + ≥ + +
+ + . (4.3) tengsizlikning n = k + 1 da bajarilishini isbotlash lozim: 1 2
1 2 1 (1 ) (1
) ... (1 ) (1
) 1 ...
k k k k x x x x x x x x + + + ⋅ +
⋅ ⋅ + ⋅ +
≥ + + + +
+ .
1 2
2 1 1 1 2 (1 ) (1 ) ... (1
) (1 ) (1
... ) (1
) 1 ... k k k k k x x x x x x x x x x x + + + ⋅ +
⋅ ⋅ + ⋅ +
≥ + + + + ⋅ + = ≥ + + + +
1 1 1 1 1 1 1 1 ...
... 1 ... 0 k k k k k k k x x x x x x x x x x + + + + + + + + + + + ≥ +
+ + + ≥ =
. 2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.3) tengsizlik ixtiyoriy n natural son uchun bajariladi. 4.4-masala. Tengsizlikni isbotlang:
1 3 2
1 ...
, 2 4
2 2 1 n n N n n − ⋅ ⋅ ⋅ < ∀ ∈
+ . (4.4)
2 1 1 1 1 1 2 4 3 2 1 1
2 = = < = ⋅ + . 1-teorema isbotlandi. 2-teorema. n = k da berilgan: 1 3
2 1 1 ... . 2 4 2 2 1 k k k − ⋅ ⋅ ⋅ < +
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1 3 2 1 2( 1) 1 1 1 ... 2 4
2 2( 1) 2( 1) 1
2 3
k k k k k − + − ⋅ ⋅ ⋅ ⋅
= +
+ tengsizlikni isbotlash lozim.
2 3 1 3 2 1 2( 1) 1 1 2 1 ...
2 4 2 2( 1) 2( 1) 2 1 2 3 1 2 1 k k k k k k k k k k + − + − + ⋅ ⋅ ⋅ ⋅ < ⋅ = + + + + < +
2 2 1 2 3 1 1 4 8 3 1 . 2 2 2 3 2 3 2 3 4 8 4 (2 2) 1
k k k k k k k k k + ⋅
+ + + = ⋅ = ⋅ < + + + + + + <
2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.4) tengsizlik ixtiyoriy n natural son uchun bajariladi.
Tengsizliklarni isbotlang:
1
1 1 ... , : 2 2 3
n N n n + + + + > ∀ ∈ ≥ ; (4.5.1)
1 1 1 1 ... 2 , : 2 2 3 n n N n n + + + + < ∀ ∈
≥ . (4.5.2) (4.5.1) tengsizlikni isbotlaymiz. 1-teorema . n = 2 da: 1 2 1 1 1
1 2. 2 2 2 + + + = > = ega bo’lamiz. 1-teorema isbotlandi. 2-teorema . n = k d 1 1
1 ...
2 3
k + + + + > tengsizlik berilgan. 1 1 1 1 1 ... 1 2 3 1 k k k + + + + + > + + tengsizlikni isbotlash lozim. Isbotlash.
1 1 1 1 1 1 ... 2 3 1 1
k k k + + + + + > + = + +
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2 2 1 1 1. 1 1 k k k k k k + +
+ = > = + + + 2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.5) tengsizlik ixtiyoriy 2
≥ natural son uchun bajariladi. (4.5.2) tengsizlikni isbotlaymiz. 2-teorema . n = 2 da: 1 2 1 2 2
1 2 2
2 2 2 + + + = < = . 1-teorema isbotlandi. 2-teorema . (4.5.2) tengsizlik n = k da bajarilishi berilgan: . 2
2 1 ... 3 1 2 1 1 ≥ < + + + +
k k
1 1
1 1 ... 2 1 2 3 1
k k + + + + +
+ +
Isbotlash.
< + + + + + + < 1 1 1 ...
3 1 2 1 1 2 k k k
1 1 1 1 2 + + ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + < k k k k ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + ⋅ + = 1 1 2 1 2
k k k . 2 1 1 2 2 < + + + k k k tengsizlikning 1 2
2 4 2 2 1 4 2 4 + < + ⇔ + < + + k k k k k k
tengsizlikka teng kuchli ekanligini isbotlash lozim. Oxirgi tengsizlik quyidagi tengsizlik va tenglikdan kelib chiqadi: N k k k k k k ∈ ∀ + = + + < + , 1 2 1 4 2 4 4 2 4 . 2-teorema isbotlandi. . Matematik induksiya prinsipiga ko’ra (4.5.2) tengsizlik ixtiyoriy 2 ≥ n natural son uchun bajariladi. Eslatma:
1 1 1 1 1 1 1 ... ...
2 3
n n n n n n n + + + + > + + + = = .
Tengsizlikni isbotlang
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1 ( 1) , : 3 n n n n n N n + > + ∀ ∈
≥ . (4.6) 1-teorema. n = 3 da: 4 3
81 64 4 . = > = 1-teorema isbotlandi. 2-teorema. n = k da 1 (
k k k k + > + . tengsizlikning bajarilishi berilgan. 2 1
1) ( 2) k k k k + + + > + tengsizlikning bajarilishini isbotlash lozim. Isbotlash. Agar
1 ( 1) k k k k + > + bo’lsa, u holda 1 (
1 k k k k + + > tengsizlik bajariladi. Oxirgi tengsizlikni musbat 2 ) 1 ( + + k k songa ko’paytirib, quyidagi tenglik va tengsizlik zanjirini hosil qilamiz: 1 2 2 1 2 2 1 1 ( 1) (
1) (( 1) ) 2 1 ( 1) k k k k k k k k k k k k k k k k + + + + + + ⎛ ⎞ + ⋅ +
+ + + + > = = = ⎜ ⎟ ⎝ ⎠ 1 1 1 1 2 0 ( 2) k k k k k k + + ⎛ ⎞ = + + > > > + ⎜ ⎟ ⎝ ⎠ . 2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.5.2) tengsizlik ixtiyoriy 3
≥
natural son uchun bajariladi. 4.7-masala. Tengsizlikni isbotlang:
! , 3 n n n n N ⎛ ⎞
> ∀ ∈
⎜ ⎟ ⎝ ⎠
; (4.7.1)
: 2
! , 2 n N n n n n ⎛ ⎞ ∀ ∈ ≥ ⎜ ⎟ ⎝ ⎠ + < ; (4.7.2)
: 3 2 ! , n N n n n n ∀ ∈
≥ > ; (4.7.3) : 3 1 ! 2
, n N n n n ∀ ∈
≥ − > ; (4.7.4)
! ,
n n n n n e n N e ⎛ ⎞
⎛ ⎞ ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠ < < ∀ ∈
. (4.7.5) (4.7.1) tengsizlikni isbotlaymiz. 1-teorema . n = 1 da 1 3
! 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ > . Tengsizlikka ega bo’lamiz. 33
2-teorema. n = k da (4.7.1) tengsizlik berilgan: k k k ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ > 3 ! . Bu tengsizlikning n = k+1 bajarilishini isbotlaymiz: ) 1 ( 3 ) 1 ( ! ) 1 ( + + > + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ k k k .
= +
⎠ ⎞ ⎜ ⎝ ⎛ > + = + ) 1 ( 3 ) 1 ( ! ! ) 1 ( k k k k k k
) 1 ( 3 1 + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + k k songa kopaytiramiz va bo’lamiz
>
⎠ ⎞ ⎜ ⎝ ⎛ +
⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⋅ + + ⋅ ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = + + + +
k k k k k k k k k k k k 1 1 3 3 1 3 ) 1 ( ) 1 ( 3 3 1 1 ) 1 ( ) 1 ( 1 Oraliq hisoblashlar:
=
+ − ⋅ ⋅ − + + ⋅ − + + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + k k k k k k k k k k k k k k 1 ! ) 1 ( ... ) 1 ( ...
1 ! 2 ) 1 ( 1 1 1 2
− −
⋅ − − + + − + + = < < ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛
1 1 1 1 ...
2 1 1 1 ! 1 ... 1 1 ! 2 1 1 1
k k k k k
2 1 2 1 1 1 1 1 2 3
12 3... 2 2 1 1 1 1 1 1 1 1 ...
1 1 ...
2! 3! ! 2 2 2
k k k − − = < =
⋅ ⋅ ⋅ ⋅ ⋅
< + + + + +
< + + + + +
<
= + + + + + + + < − ... 2 1 2 1 ...
2 1 2 1 1 1 1 2
k
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1 1 1 3 3 1 1 3 2 1 1 1 1 > ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +
⇒ < ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⇒ = − + = k k k k .
1 1 3 k k + + ⎛ ⎞ > ⎜ ⎟ ⎝ ⎠ . 2-teorema isbotlandi. Matematik induksiya prinsipiga ko’ra (4.7.1) tengsizlik ixtiyoriy
natural son uchun bajariladi. (4.7.2) tengsizlikni isbotlang.
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