Kompleks sonlar nazariyasi

§. Yig’indi va ko’paytmalarni kompleks sonlar yordamida hisoblash

1-m i s o l. Ayniyatni isbotlang:

1  C ²  C ⁴  ...  C¹  C ³  ..  2^n1.

n n n n

Yechish. Nyuton binomini qo’lab quyidagi tengliklarni hosil qilamiz:

1  C¹  C ²  C ³  ...  C^n1  Cⁿ  1  1ⁿ  2ⁿ ,

n n n n n

1  C¹  C ²  C³  ...(1)ⁿ Cⁿ  (1  1)ⁿ  0

n n n n

1  C¹  C ²  C ³  ...(1)ⁿ Cⁿ

 (1 1)ⁿ

 0 .

n n n n

Bu tengliklarni hadlab qo’shib, undan keyin ayirib, kerakli ayniyatni hosil qilamiz. ■

2-m i s o l. Ayniyatni isbotlang:

C⁰  C ³  C⁶  ....  ^{1 } 2ⁿ  2cos^{рn } .




n n n ^{ }


3
 ³ 

Yechish. Quyidagi tenglikni qaraymiz:

1  xⁿ  C ⁰  C¹ x  C ² x ²  C ³ x³  ...  C^n1x^n1  Cⁿ xⁿ .

n n n n n n

Bu tenglikga ketma-ket x = 1,

 ,  ²

larni qo’yamiz, bu yerda

 ²    1  0 . Natijada quyidagi tengliklar hosil bo’ladi:

2ⁿ  C ⁰  C¹  C ²  C ³  ...,

n n n n

1   ⁿ

 C ⁰  C ¹  C ² ²  C ³ ³  ...,

n n n n

^1   ^{2 }n

 C ⁰  C¹ ²  C ³ ⁴  C ³ ⁶  ...

n n n n

Lekin k son 3 ga bo’linmaganda 1   ^

_{ } 2

 0 , k son 3 ga bo’linganda

esa

1   ^

  ^2  3

bo’ladi. Shuning uchun yuqoridagi tengliklarni hadlab

qo’shib,

2ⁿ  1   ⁿ  1   ² n

 3C ⁰  C ³  C ⁶  ...

tenglikni hosil qilamiz.

n n n

  cos ^2

3

• 
  i sin ^2
  

3

deb olish mumkin bo’lganligi uchun

1     ²  cos ^4

3

1   ²    cos ^2

3

• 
  isin ^4
  

3

• 
  isin ^2
  

3

 cos ^

3

 cos ^

3

• 
  isin ^ ,
  

3

• 
  isin ^ .
  

3

Shuning uchun

2ⁿ  1   ⁿ  ^1   ^{2 }n

 2ⁿ  2 cos ^n . Bu yerdan

3

C ⁰  C ³  C ⁶  ...  ^{1 } 2ⁿ  2 cos ^{n } . ■




n n n ^{ }


3
 ³ 

3-m i s o l. Tenglikni isbotlang:

1²  2²  ...  n²  C ²

 2C ²  C ²  ...C ² .


  1  ²

n1



n n1 2

2 2

C


Yechish. ²

1 2

  bo’lganligi uchun

2 2

2C_  

  .

Shuning uchun

² _ k ²  _ k .

2_C

n n



n
 2

 2

 2

Bundan yuqoridagi ayniyat kelib chiqadi. ■

4-m i s o l. Yig’indini hisoblang:   1  C ²  C ⁴  C ⁶  ...

n n n

Yechish. 1  iⁿ

 1  C¹i  C ²i ²  C ³i³  ifodani qaraymiz. Bundan

n n n

1  iⁿ

 ^1  C ²  C ⁴  ...^ i^C¹  C ³  C ⁵  ^.

n n n n n

Lekin 1  i 

2^cos ^  i sin ^{ } . Shuning uchun

4

4
 

 

n

  1  C ²  C ⁴  C ⁶  ... 

n _.

n n n

2² cos

4

Bu yerdan n = 4m bo’lganda   1^m 2^2m , n = 4m+1 bo’lganda

  1^m 2^2m , n = 4m+3 bo’lganda    1^m1 2^2m1 , n = 4m+2 bo’lganda

  0 bo’lishi kelib chiqadi. ■

5-m i s o l. Ayniyatni isbotlang:

1 ¹

2

C¹ 

¹ C ²  ... 


n
3

1

n 1

Cⁿ 

₂n1 _1 n 1 ^.

n

n

n1
Yechish. Quyidagi ko’phadni qaraymiz:

n1
Bundan

1  xn1  1  C¹

2

x  C
n1

x²  ...C ^n1 x^n1 .

1 x^n1 1

_C1 _C 2 _Cn

 C ⁰ x  ⁿ x ²  ⁿ x³  ... ^ⁿ x^n1

(*).

n 1

ⁿ 2 2

n 1

Bu tenglikka x = 1 ni qo’yib, izlanayotgan ayniyatni hosil qilamiz. ■ 6-m i s o l. Ayniyatni isbotlang:

Cⁿ  Cⁿ

• 
  Cⁿ
  

 ...  Cⁿ

_{ C}n1 _.

n n1

n 2

n

n 1

Yechish. Tenglikning chap tomonidagi ifoda

S  1  xn  1  xn1  1  xn2  ...  1  xn

ko’phaddagi xⁿ oldidagi koeffisiyentdan iborat. Bu ko’phadni quyidagicha almashtiramiz:

 ⁿ 

   ²

_{    n} 1  x^{ 1}  1

S  1  x

1  1  x  1  x  ...  1  x

 1  x 

x

C
 ¹ 1  x^{n 1}  1  xⁿ . x

kvadrat qavs ichidagi ko’phadda bo’ladi. ■

_xn1

oldidagi koeffisiyent

n1 n 1

ga teng

7-m i s o l.

C ⁰C ^p  C¹C ^p1  ...  C ^pC ⁰

 C ^p

tenglik o’rinli bo’lishini

ko’rsating.

n m n m

n m mn

Yechish. Quyidagi ko’phadlarni qaraymiz:

1  xⁿ 

C^s x^s , 1  x^m 

m

C^t x^t .

n
 _n

s0

 _m

t0

U holda: 1  xⁿ 1  x^m 

n

_Cs _xs

m

_Ct _xt

 1  x^nm 

_C p _x p

 _n

s0

 _m

t 0

^ mn

tenglikdan talab qilinayotgan tenglik kelib chiqadi. ■ 8-m i s o l. Ayniyatni isbotlang:

C ⁰ 2  C¹ 2  C ² 2  ...  C ⁿ 2  Cⁿ .

n n n n 2n

Yechish. 1  xⁿ 1  xⁿ  1  x²ⁿ ko’paytmani qaraymiz. Bu ko’paytmani

n n 2 n

quyidagicha yozish mumkin:

_{ C}s _xs _{ C}t _xt

 _ C^

x^ .

^{Bu yerdan} C^

n

s0

 _ C^sC^{t .}

n

t0

2n

k 0

2n n n

st

^Natijada, Cⁿ

 _ C^sC^t  _

_Cs _Cns _{ }

C ^s 2 . ■

n
2n n n

stn

n n

s0

n


n
s0

9-m i s o l. Tenglikni isbotlang:

m1


2m
2^2m cos^2m  _ 2C^
cos2m     C^m .

2m
Yechish.

 0

_{cos } cos  i sin   cos  i sin   _tenglikda

2

cos  isin  z

deb olamiz. U holda

cos  isin  z ^1,

^ z  z ^{1 }2m

₁ 2m
 2m

cos ^2m  ^





2 _

m1

 C^ z z .

^22m  0 2m




2m

_ C

2m
2m

Bundan

2^2m cos ^2m

 _ C ^

_z 2m  _{ C}m _

^ z ^{2m } . Ikkinchi

2m
 0  m1

yig’indida

m  k  m  k ¹ 

deb olamiz. U holda bu yig’indi quyidagi

ko’rinishga keladi:

0
_{ C} 2m ' _z 2m '  _

m1<sub>C 

z</sub> 2m  _.

2 m

 'm1`

^ 2m

 0`

Shunday qilib,

2^2m cos ^2m

_ m1<sub>C 

z</sub> 2m  _{ z} 2m  _{ C}m _.

^ 2m 2m

 0

Lekin

_Z 2m  _{ Z} 2m 

 2cos2m  k . Shuning uchun

m1

2m

2m
2^2m cos^2m   _ 2C ^ cos 2m     C^m . ■

 0

10-m i s o l. Tenglikni isbotlang:

_sin n _sin n  1

sin  sin2  ...  sin n 

2 2

sin ^

2

  2k ,

k  Z .

Yechish.

A  cos  cos2  ...  cosn

yig’indini kiritamiz. U holda

isbot qilinayotgan tenglikning chap tomonini B orqali belgilab

A  Bi  cos  isin   cos2  isin2   ...  cosn  isinn 

ni hosil qilamiz.

Bu geometrik progressiya hadlarining yig’indisidan iborat. Quyidagi

belgilashni kiritamiz:   cos ^

2

• 
  i sin ^ . U holda
  

2

_ 2n2 _{ } 2

A  Bi   ² ⁴  ...   ²ⁿ 

 ²  1 ^.

Oxirgi kasrning surat va maxrajida  ning shunday darajalarini qavsdan tashqariga chiqaramizki, qavs ichida  ning qarama-qarshi ko’rsatkichli

darajalarining ayirmasi qolsin (buning mumkin bo’lishi uchun biz cos  i sin 

ni emas

cos ^

2

• 
  i sin ^
  

2

ni belgilardik):

A  Bi 
_ n2 _ n _{ }  n _



    ^1


^cos^{n 1}  isin ^{n 1} ^2isin ^{n n}

^{ n1  n  n  } 2

2 ^ 2

sin _

n 1

n 1 _

 _1 

_ ^ ^2 _cos

  isin

 _.

 

2isin

2

sin^{  2 2 }

2

_sin n _sin n  1

Bu yerdan B 

2 2

sin ^

2

ni va bir vaqtda

_sin n _cos n  1

_{A } 2 2

sin ^

ni hosil qilamiz. ■

2

Xuddi shunday

a₁ cos b₁  a₂ cos b₂  ...  a_n cos b_n
va a₁ sin b₁  a₂ sin b₂  ...  a_n sin b_n

yig’indilarni ham hisoblash mumkin, agar

b₁ , b₂ ,...,b_n

argumentlar arifmetik

progressiyani, etsa.

a₁ , a₂ ,..., a_n

koeffisiyentlar esa gometrik progressiyani tashkil

M A S H Q L A R

46*. Tengliklarni isbotlang:

• 
  C
  

C

a)
1 3

2n 2n

 ...  C^n1  2^2n2

, agar n  juft bo’lsa;

2n

2n

2 n
b) 1  C ²  ...  C ^2n1  2^2n2 , agar n  toq bo’lsa.

47*. Tengliklarni isbotlang:

a) C¹  C ⁴  C ⁷      ^{1 } 2ⁿ  2cos ^{n  2 } ;




n n n ^{ }


3
 ³ 

b) C ²  C ⁵  C ⁸      ^{1 } 2ⁿ  2cos ^{n  4 } .

n n n ^{ }


3
 ³ 

48*. Quyidagi yig’indini hisoblang:

C¹  C ³  C ⁵  C ⁷  ... .

49*. Ayniyatni isbotlang:

n n n n

 _{0 } 2_C1

_ 3_C 2 _ 4_C 3

_ n1_Cn

  1^n1  1

C_n  ⁿ 

2

n _ n

3 4

 .... 

n _

n  1

n  1 ^.

50*. Ayniyatni isbotlang:

C ⁰  C¹  C ²  ...(1)^ C ^

  1^ C ^ .

n n n

n n1

51*.

_C 0_C

 C¹C^{ 1}  ...  C^n Cⁿ  _

2n!

tenglik o’rinli

n

n

n

n

n

n

n -  !n   !
bo’lishini ko’rsating.

52*. Ayniyatni isbotlang:

  C

  C

2n1

2n

2n

2n

2n

2n

b) C
a) C ⁰ 2  C¹ 2  C ² 2  ...  C ² n 2   1ⁿ Cⁿ ;

0

2n1

2 ₁

2n1

2 ₂

2n1

2  ...  C ^2n1 2  0 .

53*. Quyidagi tengliklarni isbotlang:

a) 2^2m sin^2m  m1 1^m 2C^

cos2m     C^m ;



 0

m


2m1
b) 2^2m cos ^2m1  _C^
2m
cos2m  2
2m
 1 ;

 0

c) 2^2m sin ^2m1   m  1^m C^

sin2m  2

 1 .



 0

54*. Tengliklarni isbotlang:

2m1

1. 
   cos ^
   

n

• 
  cos ^3
  

n

• 
  cos ^5
  

n

 ...  cos ^{2n  1}

n

 0 ;

2. 
   sin ^
   

n

• 
  sin ^3
  

n

• 
  sin ^5
  

n

 ...  sin ^{2n  1}

n

 0 .

55*. Yig’indilarni toping:

1. 
   cos x  C¹ cos 2x  ...  Cⁿ cosn  1x ;
   

n n

2. 
   sin x  C¹ sin 2x  ...  Cⁿ sinn  1x .
   

n

56*. Yig’indilarni toping:

57. Isbotlang:

n

sin ² x  sin ² 3x  ...  sin ² 2n 1x .

1. 
   cos ² x  cos ² 2x  ...  cos ²nx  ⁿ  ^{cosn  1x sin nx} ;
   

2 2sin x

2. 
   sin ² x  sin ² 2x  ...  sin ²nx  ⁿ  ^{cosn  1x sin nx} .
   

2

58*. Yig’indilarni toping:

2sinx

1. 
   cos x  2 cos 2x  3cos 3x  ...  n cos nx ;
   
2. 
   sin x  2sin 2x  3sin 3x  ...  n sin nx .
   

3.