Funksiyaning hosilasi va differensiali

2⁰. Ko’phad uchun Teylor formulasi. Ushbu
P_n(x)  a₀  a₁(x  x₀)  a₂(x  x₀)²  a_n(x  x₀)ⁿ (6.24)
(bunda a₀ ,a₁,a₂ ,,a_n va x₀ o’zgarmas haqiqiy sonlar, nN ) ko’phadni qaraylik. Bu ko’phadni ketma–ket n marta differensiallab topamiz:
P_n^(x)  a₁  2a₂(x  x₀) 3a₃(x  x₀)²  na_n (x  x₀)^n1,
P_n^(x)  2a₂ 32a₃(x  x₀)  n(n1)a_n (x  x₀)^n2,
P_n(x)  32a₃  n (n 1) (n 2) a_n(x  x₀)^n3, (6.25)
………………………………
P_n^n(x)  n(n 1)(n  2)2a_n
Bu (6.24) va (6.25) tengliklarda x  x₀ deb olinsa, unda berilgan P_n (x) ko’phad va uning hosilalari P_n^k(x) (k 1, 2,,n) ning x₀ nuqtadagi qiymatlari topilsin:
P_n (x₀ )  a₀ ,
P_n(x₀ ) 1!a₁ ,
P_n(x₀ )  2!a₂ , …………… P_n^n(x₀)  n!a_n.
Undan
a₀  P_n (x₀ ),
Pⁿ(x⁰ ) a₁  ,
1!
Pⁿ(x⁰) a₂  ,
2!
……………
a_n  Pnn(x0)
n!
kelib chiqadi.
Shunday qilib, P_n (x) ko’phadning koeffitsientlari ko’phad va uning hosilalarining x₀ nuqtadagi qiymatlari orqali ifodalanadi. Koeffitsientlarning bu qiymatlarini (6.24) ga qo’ysak , unda
Pn(x0)(x Pn(x )
Pn(x)  Pn(x0)   x0)  n n! 0 (x  x0)n (6.26)
1!
bo’ladi.
(6.26) formula ko’phad uchun Teylor formulasi deb ataladi.
3⁰.Ixtiyori funksiya uchun Teylor formulasi. f (x) funksiyaa,b
intervalda aniqlangan bo’lib, u x₀ a,b nuqtada f (x₀), f (x₀)^,, f ^n(x₀) hosililarga ega bo’lsin. Funksiyaning nuqtadagi hosilalaridan foydalanib, quyidagi
f (1!x⁰)(x  x₀)  f 2(!x⁰)(x  x₀)²  f ^nn^(!x⁰)(x  x₀)ⁿ P_n( f ;x)  P(x₀)  f (x₀)  ko’phadni tuzaylik.
Agar qaralyotgan f (x) funksiya ndarajali ko’phad bo’lsa, unda yuqorida (2–bandda) aytilganga ko’ra f (x)  P_n ( f ;x)
ya’ni
f (1!x⁰)(x  x₀)  f 2(!x⁰)(x  x₀)²  f ^nn^(!x⁰)(x  x₀)ⁿ f (x)  f (x₀) 
bo’ladi.
Agar f (x) funksiya ko’phad bo’lmasa, Ravshanki,

f (x)≡P_n ( f ;x)
bo’lib, ular orasida farq yuzaga keladi. Biz uni R_n (x) orqali belgilaylik:
R_n (x)  f (x)  P_n ( f ;x)
Natijada ushbu f (x)  P_n ( f ;x)  R_n (x)
ya’ni
f (x⁰)(x  x₀)  f (x⁰)(x  x₀)²  f ^n_n^(_!x⁰)(x  x₀)ⁿ  R_n(x) (6.27) f (x)  f (x₀) 
_{1! 2!}
formulaga kelamiz. Bu (6.27) formula f (x) funksiya uchun Teylor formulasi deb ataladi. R_n (x) esa Teylor formulasining qoldiq hadi deyiladi.
f (x) funksiya P_n ( f;x) ko’phad bilan taqribiy f (x)  P_n ( f ;x) ifodalashda Teylor formulasidan keng foydalaniladi. Bunda qoldiq had R_n (x) ni baholash muhim. Bu masalani hal qilish uchun f (x) funksiyaga “og’irroq” shart qo’yishga to’g’ri keladi. f (x) funksiyaa,b intervalda aniqlangan bo’lib, u shu intervalda uzluksiz f (x), f (x),, f ^n(x ) hosilalarga ega bo’lsin degan edik. Endi a,b intervalda bu funksiyaning n 1tartibli
f ^n1(x) hosilasi ham mavjut bo’lsin deymiz. a,b intervalda argument x ning ixtiyori qiymatini tayinlab quyidagi

f (t) f (t) ² f ^n(t)(x  t)ⁿ (6.28) F(t)  f (x)  f (t)  1! (x  t)  2! (x  t)  n!
yordamchi funksiyani tuzamiz va uni x₀ , x a,b (yoki x, x₀  a,b ) segmentda qaraymiz. F(t) funksiyaning (6.28) ifodasidan uning x₀ ,x segmentda uzluksiz bo’lishini ko’rish qiyin emas. Bu funksiya x₀ , x intervalda hosilaga ham ega. Haqiqatan ham,
f (t)   f (t) 2 f ^n(t) (x  t)
F(t)   f (t)  ^_ 1! (x  t)  f (x)_{ } 2! (x t) 1! _ 
_  _  
- _ f nn1!(t) (x  t)n  (fnn_(1t)!) (x  t)n1_   f nn1!(t) (x  t)n
Demak,
f n1(t) ⁿ
F(t)   (x  t) (6.29)
n!
Endi x₀ ,x segmentda uzluksiz va x₀ , x intervalda chekli (nolga teng bo’lmagan ) hosilaga ega bo’lgan biror  (t) funksiyani olaylik.
F(t) va  (t) funksiyalarga x₀ ,x segmentda Koshi teoremasini qo’llanib topamiz.
F(x)  F(x0)  F(c) (6.30)  (x)  (x₀) (c)
bunda,
x₀  c  x (c  x₀ (x  x₀ ),0 1)
Yuqoridagi (6.28) funksiya uchun
F(x)  0, F(x₀ )  R_n (x)
tengliklarga egamiz. Endi (6.29) tenglikdan t  0 da
f n1(c)
F(c)   (x  c) n!
bo’lishini etiborga olsak, Unda (6.30) tenglikdan
 (x)  (x⁰ ) f ^n1(c) (x c)ⁿ (6.31)
R_n (x)  
(c) n!
c  x₀ (x  x₀ ) formula kelib chiqadi.
Shunday qilib, Teylor formulasining qoldiq hadi uchun (6.31) formula topildi. Bu holda f (x) funksiyaning Teylor formulasi quyidagi
f (x ) f (x ) f n(x )
f (x)  f (x₀)  1! 0 (x  x0)  2! 0 (x  x0)2  n! 0 (x  x0)n 
 (x)  (x⁰) f ^n1(c) (x c)ⁿ ,
 
(c) n!
c  x₀  (x  x₀ ), 0 1 (6.32)
ko’rinishda yoziladi.
Terlor formulasidan kengroq foydalanish maqsadida, uning qoldiq hadining turli ko’rinishlarini keltiramiz.
1). Koshi ko’rinishidagi qoldiq hadli Teylor formulasi. Yuqorida qaralgan  (t) funksiya sifatida  (t)  x t funksiyani olaylik. Ravshanki, bu funksiya x₀ , x a,b segmentda uzluksiz, x₀ , x intervalda esa chekli (t)  1 hosilaga ega. Bu funksiya uchun  (x)  0,  (x₀ )  x  x₀ bo’ladi. Natijada (6.31) formula quyidagi
R_n(x)  f n1(c) (x  c)n(x  x0)  n!

 f n1(c) x  x₀ (x  x₀)n(x  x₀ f n1(c) n1(1)n, 0 1
)  (x  x0) n! n!
ko’rinishni oladi. Qoldiq hadning bu ifodasini (6.32)ga qo’yib topamiz.
f (x0) (x f (x ) f n(x )
f (x)  f(x0)n1(c)1!(x  x)xn0)1(12!)n0. (x  x0)2  n! 0 (x  x0)n  (6.33) f 0 n!
Bu (6.33) formula f (x) funksiyaning Koshi ko’rinishidagi qoldiq hadli Teylor formulasi deb ataladi.
2). Lagranj ko’rinishidagi qoldiq hadli Teylor formulasi. Endi  (t) xt^n1 funksiyani olaylik. Bu funksiya ham x₀ , x a,b segmentda uzluksiz, x₀ , x intervalda esa chekli (t)  (n1)(xt)ⁿ hosilaga ega. Bu funksiya uchun
 (x)  0,  (x₀)  (x  x₀)^n1,
(c)  (n1)(x c)ⁿ, (c  x₀  (x  x₀), 0 1)
bo’ladi. U holda yuqoridagi (6.31) formula ushbu

f n1(c) n  (x  x0 )n1 f n|(c)(x  x0 )n1 Rn (x)  n! (x  c)  (n 1)(x  c)n  (n 1)!
ko’rinishni oladi. Qoldiq hadning bu ifodasini (6.32) qo’yib topamiz:
f (x)  f (x₀ )  f (1!x₀ ) (x _ x0 ) _ f (x₀ ) (x _ x0 ) _{2 }_ f ^nn^(!x₀ ) (x _ x0 )_{n }
2!
 n1(c) (x  x₀ )n1. (6.34) f
n 1!
Bu formula f (x) funksiyaning Lagranj ko’rinishidagi qoldiq hadli
Teylor formulasi deb ataladi.
Teylor formulasi qoldiq hadining bu ko’rinishi sodda bo’lib, u (6.34) formuladagi navbatda keladigan hadni eslatadi. Faqat bunda funksiyaning n 1tartibli hosilasining x₀ nuqtadagi qiymati o’rnida bu hosilaning c (c  x₀ (x  x₀ )) nuqtadagi qiymati olinadi.