yagonalik teoremasi

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1525843724_71453.doc ( ) ( ) 0 0 0 0 () () k kk k k kk k fzazz gzbzz ¥ = ¥ = =- =- å å 00 () zzd î ) ( d e ì ()(),() fzgzzd =î ()()(). fzgzzd ºî 0 z 12 ,,...,,1,2,3,4,... n zzzen î= 0 z e z î " ()(),(1,2,3,...) nn fzgzn == 0 z } 0 , : { 0 > 1 1 d ( ) 0 uzd d ì ( ) 0 uzd d ì ( ) 0 uz d ( ) (),1,2,3,... n fzn = ( ) 0 uz d ( ) ( ) ,1,2,3,... n fzn = ( ) 0 uz d ( ) 0 uz d ( ) 0 uz d g ( ) ( ) 0 uz d gì g ( ) ( ) ( ) 11 . nn nn fzdzfzdzfzdz ¥¥ == ggg éù == êú ëû …
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’plamda bir-biriga teng bo’lsa, u holda f(z) va g(z) funksiyalar d sohada aynan bir-biriga teng bo’ladi: isbot. modomiki, nuqta e to’plamning limit nuqtasi ekan, unda e to’plamga tegishli turli nuqtalardan tuzilgan va ga intiluvchi da f(z)=g(z) bo’lgani uchun bo’ladi. endi f(z) va g(z) funksiyalarni nuqtaning atrofida (bunda esa nuqtadan gacha bo’lgan masofa) teylor qatoriga yoyamiz: bo’lganligi sababli k ning biror qiymatidan boshlab keyingi lar doiraga tegishli bo’ladi. shuning uchun bo’lib, (1) dan (2) bo’lishi kelib chiqadi. bu tenglikda da limitga o’tib (3) bo’lishini topamiz. bu (3) tenglikni e’tiborga olib (2) ni har ikkala tomonini ga bo’lsak, unda (4) hosil bo’ladi. keyingi tenglikda da limitga o’tib (5) bo’lishini topamiz. bu (5) tenglikni e’tiborga olib, (4) ning har ikkala tomonini ga bo’lsak, unda hosil bo’ladi. so’ng da limitga o’tib, bo’lishini topamiz. bu jarayoni davom ettira borib, bo’lishini topamiz. shunday qilib lar uchun bo’ladi. demak, doirada f(z)=g(z) bo’ladi. d sohada ixtiyoriy nuqtani olib, …
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rtga ko’ra (6) qator da tekis yaqinlashuvchi. demak, qator da ham tekis yaqinlashuvchi bo’ladi. funksiya d sohada golomorf bo’lgani uchun u (6) qatorning har bir hadi da ham golomorf bo’ladi. binobarin, da funksiya uzluksiz. unda qator yig’indisi f(z) funksiya ham da uzluksiz bo’ladi. endi da yotuvchi yopiq silliq chiziqni olaylik . (7) qatorni chiziq bo’yicha hadlab integrallab, topamiz: (8) koshi teoremasiga ko’ra (9) bo’ladi. (8) va (9) dan bo’lishi kelib chiqadi. morera teoremasidan foydalanib f(z) funksiyani da va, demak, nuqtada golomorf bo’lishini topamiz. qaralayotgan nuqta d sohaning ixtiyoriy nuqtasi bo’lganligidan f(z) funksiyani d sohada golomorf bo’lishi kelib chiqadi. natija: yuqorida keltirilgan veyershtrass teoremasining sharti bajarilganda qatorni istalgan marta hadlab differensiallash mumkin bo’lib, bo’ladi. golomorf funksiyaning nollari. faraz qilaylik, biror funksiyaning kengaytirilgan kompleks tekislikda da, berilgan bo’lib, bo’lsin. agar bo’lsa, kompleks son funksiyaning noli deyiladi. aytaylik, funksiya nuqtada golomorf bo’lsin. bu funksiyani nuqta atrofida darajali qatorga yoyamiz: (10) agar nuqta funksiyaning …
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erda funksiya uchun bo’lib, funksiya nuqtada golomorf bo’ladi. aksincha, agar funksiya quyidagicha ifodalanib, funksiya nuqtada golomorf bo’lsa, u holda nuqta funksiyaning karrali noli bo’ladi. teorema. faraz qilaylik, funksiya z=a nuqtada golomorf bo’lib, shu z=a nuqta funksiyaning noli bo’lsin: f(a)=0. u holda yo f(z) funksiya a nuqtaning biror atrofida aynan nolga teng embed equation.3 yoki a nuqtaning shunday atrofi topiladiki, bu atrofda f(z) funksiyaning z=a nuqtadan boshqa noli bo’lmaydi. isbot. shartga ko’ra, f(z) funksiya z=a nuqtada golomorf. unda funksiya z=a nuqta atrofida qatorga yoyiladi: . (14) aytaylik, (14) da barcha lar nolga teng bo’lsin: . ravshanki, bu holda funksiya z=a nuqta atrofida f(z)=0 bo’ladi. endi (14) da bo’lib, bo’lsin. bu holda z=a nuqta f(z) funksiyaning m karrali noli bo’lib, u quyidagicha ifodalanadi. bu erda g(z) funksiya z=a nuqtada golomorf va . ayni paytda g(z) funksiya z=a nuqtada uzluksiz ham bo’ladi. unda bo’lganligi sababli z=a nuqtaning shunday atrofi topiladiki, bu atrofda bo’ladi. …
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1525843724_71453.doc ( ) ( ) 0 0 0 0 () () k kk k k kk k fzazz gzbzz ¥ = ¥ = =- =- å å 00 () zzd î ) ( d e ì ()(),() fzgzzd =î ()()(). fzgzzd ºî 0 z 12 ,,...,,1,2,3,4,... n zzzen î= 0 z e z î " ()(),(1,2,3,...) nn fzgzn == 0 z } 0 , : { 0 > 1 1 d ( ) 0 uzd d ì ( ) 0 uzd d ì ( ) 0 uz d ( ) (),1,2,3,... n fzn = ( ) 0 uz d ( ) ( ) ,1,2,3,... n fzn = ( ) 0 uz d ( ) 0 uz d ( ) 0 …

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