teylor va makloren qatorlari

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1576320038.doc ( ) x f ¢ ( ) ( ) ( ) ( ) ( ) 22 0230400 213243...1... n n fxaaxxaxxnnaxx - ¢¢ =××+××-+××-++-×-+ ( ) 2 0 1 2 a x f × × = ¢ ¢ ( ) ( ) ( ) ( ) ( ) ï ï î ï ï í ì × = × × - - = × = × × × × = × = × × × = ¢ ¢ ¢ n n 0 n 4 4 0 iv 3 3 0 a ! n a 1 2 ... 2 n 1 n n x f .......... .......... .......... .......... .......... a ! 4 a 1 2 3 4 x f a ! 3 a 1 2 3 x f ( ) 0 0 x f a = ( ) ! 1 0 1 x f a ¢ = ( ) ! …

, 1 0 , 0 0 , 1 0 = = ¢ ¢ ¢ - = ¢ ¢ = ¢ = iv f f f f f ( ) ( ) ... ! 2 1 ... ! 4 ! 2 1 cos 2 4 2 + × - + - + - = n x x x x n n ( ) ( ) ( ) ( ) ( ) , 1 1 , 1 2 1 - - + - = ¢ ¢ + = ¢ k k x k k x f x k x f ( ) ( ) ( ) ( ) ,.., 1 2 1 3 - + - - = ¢ ¢ ¢ k x k k k x f ( ) ( ) ( ) ( ) ( ) ( ) ( ) n k n x 1 1 n k , ... …

; r) oraliqda darajali qatorga yoyilgan bo`lsin: f(x)=a0+a1(x-x0)+a2(x-x0)2+…+an(x-x0)n+… (1) (1) qatorning koeffisiyentlari va x0 nuqtadagi hosilalarini f(x) funktsiyaning qiymatlari orqali ifodalaymiz. u holda, qatorning birinchi hadi f(x0) =x0 (2) dan iborat bo`ladi. f(x) funktsiya x0 nuqtada aniqlangan va shu nuqtada istalgan tartibli hosilaga ega ekanligini e`tiborga olib, ni topamiz: f`(x)=a1+2a2(x-x0)+3a3(x-x0)2+…+nan(x-x0)n-1+… (3) bundan, x = x0 bo`lgan holda f`(x0)=a1 (4) ekanligi ko`rinadi. (3) ning ikkala tomonini differentsiallab, quyidagini hosil qilamiz: (5) x = x0 bo`lganda . (6) yuqoridagi jarayonni davom ettirsak, quyidagilar hosil bo`ladi: (7) (2), (4), (6) va (7) lardan (1)- qator koeffisiyentlarini topamiz: , , ,…, ,… (8) a0, a1, a2,… an lar teylor koeffitsiyentlaridan iborat. agar (8)- qatordagi a0,, a1,…an larning qiymatlari (1)- qatorga qo`yilsa, f(x) funktsiyaning x0 nuqtadagi teylor qatori hosil bo`ladi: (9) f(x) funktsiyaning x0 nuqtadagi integral ko`rinishdagi qoldiq hadli teylor formulasi quyidagidan iborat: rn (x) – qoldiq had. bunda, . 2. makloren qatori faraz qilaylik, …

i ketma–ket topamiz va x=0 nuqtada ularning qiy-matlarini aniqlaymiz: , , , ,… x=0 bo`lganda: , , , ,… bu qiymatlarni makloren qatoriga qo`ysak, quyidagi qator hosil bo`ladi: 2. f(x) = sinx funktsiyani makloren qatoriga yoyish. yechilishi: berilgan funktsiyaning hosilalarini topamiz: x =0 nuqtada ularning qiymatlarini topamiz va makloren qatoriga qo`yamiz: 3.f(x) = cos x funktsiyaning yoyilmasi. yechilishi: f(x) = cos x funktsiyaning hosilalarini topamiz: embed equation.3 … x = 0 nuqtada topilgan hosilalarning qiymatlarini aniqlaymiz: topilgan qiymatlarni makloren qatoriga qo`yamiz: 4.f(x) = (1+x)k – nyuton binomining yoyilmasi. yechilishi: berilgan nyuton binomidan ketma – ket hosilalar olamiz: ,… x = 0 nuqtada qiymatlarini topamiz: topilganlarni makloren qatoriga qo`yamiz: 5. ko`phadni (x - 1) ning darajasi bo`yicha qatorga yoyish. yechilishi: berilgan funktsiyaning hosilalarini topamiz: x=1 nuqtada ko`phad va uning hosilalari qiymatlarini topamiz: , , , , , topilgan qiymatlarni teylor qatoriga qo`yamiz: 6. funktsiyani x = 0 nuqtada teylor qatoriga yoyish. …

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1576320038.doc ( ) x f ¢ ( ) ( ) ( ) ( ) ( ) 22 0230400 213243...1... n n fxaaxxaxxnnaxx - ¢¢ =××+××-+××-++-×-+ ( ) 2 0 1 2 a x f × × = ¢ ¢ ( ) ( ) ( ) ( ) ( ) ï ï î ï ï í ì × = × × - - = × = × × × × = × = × × × = ¢ ¢ ¢ n n 0 n 4 4 0 iv 3 3 0 a ! n a 1 2 ... 2 n 1 n n x f .......... .......... .......... .......... .......... a ! 4 a 1 2 3 4 x f a …

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