Science and Education

b_n  0
^{n  1, 2,}





n
^f  ^x ^{~ a0 }  ^a
2 n1

cos nx

Aytaylik,
^f  ^x
funksiya ^{ , } da berilgan toq funksiya bo‘lsin:

^f ^x ^{  f}  ^x _{. U holda}

^f  ^x^{cos nx} _toq,
^f  ^x^{sin nx} _juft ^{n  1, 2,3,...}
funksiya bo‘ladi.

(1) formulalardan foydalanib, topamiz:
^f  ^x
funksiyaning Furye koeffitsiyentlarini

₁  _{1 } 0  _
^{an } _{ } ^f  ^x^{cos nxdx } _{  } ^f  ^x^{cos nxdx } _ ^f  ^x^{cos nxdx}_ ^
  0 
₁ 


^ ^ _ ^f  ^x^{cos nxdx  f}  ^x^{cos nxdx}_ ^{ 0}
0
^{n  0,1, 2,}

₁  _{1 } 0  _
^{bn } _{ } ^f  ^x^{sin nxdx } _{  } ^f  ^x^{sin nxdx } _ ^f  ^x^{sin nxdx}_ ^


_{2 }
 0 


^ _{ } ^f  ^x^{sin nxdx}_
^{n 1, 2,}

 0 

Demak, toq
^f  ^x
funksiyaning Furye koeffitsiyentlari

^{an  0,} ^{n  0,1, 2,}
₂ 

bo‘lib, Furye qatori
^{bn } _ ^f  ^x^{sin nxdx,} ^{n  1, 2,}


0

bo‘ladi.

2. 
   misol. Ushbu
   

^f  ^x <sup> x2



f</sup>  ^x ^~ _^{bn sin nx}
n1
^{  x   } juft funksiyaning Furye qatori topilsin.

Avvalo berilgan funksiyaning Furye koeffitsiyentlarini topamiz:

a₀ 


2
_ x²dx 
² ²,

 ₀ 3

a_n 


2
_ x² cos nxdx 

2. 
   sin nx ^
   

_x 2 _
⁴ _ x sin nxdx 


 ₀  n ₀ n ₀
^{4  x cos nx}  ^{1  } n ⁴

^  n ^ n

• 
  _{n } ^{cos nxdx} _ ^ ^1
  

^ _n2 ^. ^{n  1, 2,}

 ^{0 0} 

Demak,
^f  ^x ^{ x2}
funksiyaning Furye qatori

bo‘ladi.
2


^f  ^x ^{ x2 ~}
3



^{ 4}_^1
n1
n ^{cos nx} _n2

3. 
   misol. Ushbu
   

^f  ^x ^{ x}
^{  x  } 
toq funksiyaning Furye qatori topilsin.

Berilgan funksiyaning Furye koeffitsiyentlarini hisoblaymiz:

2 ^ 2 ^
x cos nx ^ 1 ^{
 2}^1n1

b_n  _{ } x sin nxdx  _{ }  _n  _{n } cos nxdx _  _n
0 _ ⁰ 0 _{ .}

Demak,
^{f  x  x} funksiyaning Furye qatori

bo‘ladi.



^f  ^x ^~ _
n1
^1n1 ^{2 sin nx} n

Faraz qilaylik, Ma’lumki, ushbu

^f  ^x
funksiya ^{ p , p}
 ^{p  0}

segmentda uzluksiz bo‘lsin.

almashtirish ^{ p , p}

t  ^ x p
oraliqni ^{ , } ga o‘tkazadi, ya’ni ^x o‘zgaruvchi ^{ p , p} da

o‘zgarganda ^t o‘zgaruvchi ^{ , } da o‘zgaradi. Endi
^f  ^x ^{ f  p t   } ^t ^.
 _ 

deymiz. Unda ^{ t }
funksiya ^{ , }
 
oraliqda berilgan uzluksiz funksiya bo‘ladi.

Bu funksiyaning Furye koeffitsiyentlari

1. 
    ¹
   

n _

2. 
    ¹
   

n _

_ ^ ^t ^{cos ntdt,}


_ ^ ^t ^{sin ntdt}

^{n  0 ,1, 2 ,}


^{n  1, 2 ,}

ni topib, Furye qatorini yozamiz:

2
^ ^t  ^{~ a0 } _ ^a
cos nt  b


^{sin nt} 

Modomiki,

n n
n1 _.

ekan, unda
t  ^ x p

_ ^  _x ^ _~ a_{0  }^{ }

cos n ^ x  b sin n ^{ }

x ,

a
 _p  ₂  n
p ⁿ p ^

 
bo‘lib, uning koeffitsiyentlari
^n1  

1 ^{p }  ^ 

a_n 
_{p }  _{ p} x _cos n _p xdx ,
^{n  0,1, 2}

 p ^{ }
1 ^{p }  ^ 

b_n 
_{p }  _{ p} x _sin n _p xdx .
^{n  1, 2}

• 
  p
  

bo‘ladi. Natijada ^{ p , p}
quyidagicha
 
da berilgan


^f  ^x

funksiyaning Furye qatorini

a₀ ^{ } n x n x <sup>


f</sup>  ^x ^{~ } _ ^{an cos  bn sin} _
2 p p
^n1  
bo‘lishini topamiz, bunda

1 ^p n x

a_n 
_ ^f  ^x^{cos dx}

• 
  
  p
  
  p
  p
  

^{n  0,1, 2}

1 ^p n

b_n 
_ ^f  ^x^sin

• 
  
  p
  p
  

xdx
p
^{n  1, 2}

4. 
   misol. Ushbu
   

^f  ^x ^{ ex}
^{1  x  1}
funksiyaning Furye qatori topilsin.

Yuqoridagi formulalardan foydalanib, koeffitsiyentilarini topamiz:
^f  ^x ^{ ex}
funksiyaning Furye

^{1 1} n sin n x  cos n x ¹

a₀  _ e^xdx  e  e^1,
1
a_n  _ e^x cos n xdx 
1
1 n² ²
e^x 
1


 ¹ e cos n  e^1
1 n² ²
cos n _  _1_n
e  e^1
1 n² ²
^{n  1, 2 ,}

b_n 

1
_ e^x cos n xdx 

1
sin n x  n cos n x _x

1

1
1 n² ^{2 e }

_ 1
1 n² ²
_en cos n  n e^1 cos n _ 

Demak,
^n ^1n
^ 1 n² ²
_ n ^{e  e1}

_e ¹  e  _1_ ¹ _
1 n² ²
^{n  1, 2,}

funksiyaning Furye qatori

^f  ^x ^{ ex}
^{1  x  1}

e  e^{1
 } ^1n
^1n1 ^

bo‘ladi.
e^x ~
 _e  e^1 _
2 _{n1 }1 n² ²
cos n 


1 n² ^{2
n sin n x}
_

Aytaylik,
^f  ^x _funksiya ^{a, b}
da berilgan bo’lsin. ^{a, b}
segment ^ak
nuqtalar

yordamida bo‘laklarga ajratilgan.
(a₀  a,
a_n  b) _.

Agar har bir ^{ak , ak 1 }
^{k  0,1, 2,}
_da ^f  ^x
funksiya differensiallanuvchi

bo‘lib,
x  a_k
nuqtalarda chekli o‘ng
^{f }^{ak  0} ^{k  0,1, 2, , n 1} _,

va chap


^{f }^{ak  0} ^{k  0,1, 2,}

hosilalarga ega bo‘lsa, deyiladi.
^f  ^x
funksiya ^{a, b}
da bo‘lakli-differensiallanuvchi

Endi Furye qatorining yaqinlashuvchi bo‘lishi haqidagi teoremani isbotsiz keltiramiz.

Teorema. ^2 davrli
^f  ^x
funksiya ^{ , } oraliqda bo‘lakli-differensiallanuvchi

bo‘lsa, u holda bu funksiyaning Furye qatori

^f  ^x ^{~ a0 } _ ^{a cos kx  b
sin kx}

2
k k
k 1
^{ , } da yaqinlashuvchi bo‘lib, uning yig‘indisi
^f  ^{x  0} ^{ f}  ^{x  0}
2

ga teng bo‘ladi.

5. 
   misol. Ushbu
   

^f  ^x ^{ cos ax}

^{  x   , a  n  Z} 

funksiyaning Furye qatori

topilsin va u yaqinlashishga tekshirilsin.
Bu funksiyaning Furye koeffitsiyentlarini topamiz. Qaralayotgan funksiya juft bo‘lgani uchun

bo‘lib,
b_n  0

^{n  1, 2,3,}

_{  }    
₂  
a_n  cos ax cos nxdx  _cos a  n x  cos a  n x_ dx 
0 0
_ ^{sin a} _1ⁿ  ¹ _ ¹ 


bo‘ladi. Demak,
^_ a  n a  n ^_



n1
^{sin a  1  n}  ^{1 1}  ^



n a




^f  ^x ^~ _{
 a} ^ _^1
^ a  ^

• 
  _{n }cos nx_{ .}
  

unda
Agar

^{f  x  cos ax} funksiya teoremaning shartlarini bajarishini e’tiborga olsak,

^{sin a  1  n}  ^{1 1}  ^

cos ax  _{
 a} ^ _^1
^ a  ^

• 
  _{n }cos nx_
  

bo‘lishini topamiz.

 ^{n1
 n a } 