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Teore ma Bolca no Ko shi teore ma pro promi zhne zna chennya nepere rvnoyi fu nkciyi dilitsya na dvi chastini persha z yakih ye teoremoyu pro prohodzhennya neperervnoyu funkciyeyu cherez nul druga uzagalnyuye pershu i stverdzhuye sho yaksho neperervna funkciya prijmaye dva znachennya vona takozh prijme znachennya na vidrizku mizh nimi Zmist 1 Persha teorema Bolcano Koshi 1 1 Dovedennya 2 Druga teorema Bolcano Koshi 2 1 Dovedennya 3 Vikoristannya teoremi na praktici 4 Div takozh 5 DzherelaPersha teorema Bolcano Koshi RedaguvatiNehaj funkciya f x viznachena ta neperervna v zamknenomu promizhku a b ta na kincyah cogo promizhku prijmaye znachennya riznih znakiv Todi mizh a ta b neodminno znajdetsya tochka c v yakij funkciya obertayetsya v nul f c 0 displaystyle f c 0 nbsp a lt c lt b displaystyle a lt c lt b nbsp Dovedennya Redaguvati Dovedennya ciyeyi teoremi zrobimo metodom poslidovnogo dilennya vidrizku metod Bolcano Nehaj dlya viznachenosti f a lt 0 ta f b gt 0 Rozdilimo vidrizok a b navpil tochkoyu a b 2 displaystyle frac a b 2 nbsp Yaksho v danij tochci funkciya dorivnyuye nulyu todi teorema dovedena Yaksho c a b 2 0 displaystyle c frac a b 2 neq 0 nbsp todi na kincyah odnogo z vidrizkiv a a b 2 displaystyle left a frac a b 2 right nbsp a b 2 b displaystyle left frac a b 2 b right nbsp funkciya bude prijmati znachennya riznih znakiv Poznachivshi cej promizhok cherez a 1 b 1 displaystyle left a 1 b 1 right nbsp mayemo f a 1 lt 0 displaystyle f a 1 lt 0 nbsp f b 1 gt 0 displaystyle f b 1 gt 0 nbsp Rozdilimo navpil vidrizok a 1 b 1 displaystyle left a 1 b 1 right nbsp ta znovu vidkinemo vipadok z rivnistyu funkciyi nulevi u comu vipadku teorema dovedena Poznachimo cherez a 2 b 2 displaystyle left a 2 b 2 right nbsp tu z polovin vidrizku dlya yakoyi f a 2 lt 0 displaystyle f a 2 lt 0 nbsp f b 2 gt 0 displaystyle f b 2 gt 0 nbsp Prodovzhimo cej proces pobudovi vidrizkiv Pri comu pislya deyakoyi kincevoyi kilkosti iteracij mi abo natknemosya na rivnist funkciyi nulevi i dovedennya teoremi zakinchitsya abo otrimayemo neskinchennu poslidovnist vkladenih odin v odnogo promizhkiv Zupinimos na comu ostannomu vipadku Todi dlya n go vidrizku a n b n displaystyle left a n b n right nbsp n 1 2 3 budemo mati f a n lt 0 displaystyle f a n lt 0 nbsp f b n gt 0 displaystyle f b n gt 0 nbsp Posilannya 1 Prichomu dovzhina jogo dorivnyuye b n a n b a 2 n displaystyle b n a n frac b a 2 n nbsp Posilannya 2 Pobudovana poslidovnist vidrizkiv zadovolnyaye lemu pro vkladeni vidrizki tomu sho vidpovidno do 2 lim b n a n 0 displaystyle lim b n a n 0 nbsp Tomu isnuye tochka s iz promizhku a b dlya yakoyi lim a n lim b n c displaystyle lim a n lim b n c nbsp Pokazhemo sho same cya tochka zadovolnyaye vimogam teoremi Perejshovshi do granici v nerivnostyah 1 ta vikoristovuyuchi pri comu neperervnist funkciyi zokrema v tochci x c otrimayemo sho odnochasno f c lim f a n 0 displaystyle f c lim f a n leq 0 nbsp ta f c lim f b n 0 displaystyle f c lim f b n geq 0 nbsp Tak sho dijsno f c 0 Teorema dovedena Druga teorema Bolcano Koshi RedaguvatiNehaj funkciya f x viznachena ta neperervna na deyakomu promizhku X zamknutomu abo ni skinchennomu abo neskinchennomu Yaksho v dvoh tochkah x a ta x b a lt b cogo promizhku funkciya prijmaye nerivni znachennya f a A displaystyle f a A nbsp ta f b B displaystyle f b B nbsp to yake b ne bulo chislo S sho lezhit mizh A ta B znajdetsya taka tochka x c mizh a ta b sho f c C Dovedennya Redaguvati Budemo vvazhati sho A lt B todi A lt C lt B Rozglyanemo v promizhku a b dopomizhnu funkciyu f x f x C displaystyle varphi x f x C nbsp Cya funkciya neperervna v promizhku a b ta na kincyah jogo maye rizni znaki f a f a C A C lt 0 displaystyle varphi a f a C A C lt 0 nbsp f b f b C B C gt 0 displaystyle varphi b f b C B C gt 0 nbsp Todi zgidno z pershoyu teoremoyu Bolcano Koshi mizh a ta b znajdetsya tochka x c dlya yakoyi f x 0 displaystyle varphi x 0 nbsp tobto f x C 0 displaystyle f x C 0 nbsp abo f c C displaystyle f c C nbsp Sho i treba bulo dovesti Vikoristannya teoremi na praktici RedaguvatiZa dopomogoyu ciyeyi teoremi mozhna viznachiti nayavnist koreniv u rivnyanni Napriklad dlya vsih ochevidnij korin x 4 u rivnyanni 2 x 4 x displaystyle 2 x 4x nbsp ale skladno pomititi isnuvannya she odnogo korenya cogo rivnyannya Funkciya f x 2 x 4 x displaystyle f x 2 x 4x nbsp pri x 0 maye znachennya f 0 1 gt 0 a pri x 1 2 znachennya f 1 2 2 2 lt 0 displaystyle f left frac 1 2 right sqrt 2 2 lt 0 nbsp vidpovidno funkciya oskilki vona neperervna obertayetsya na 0 v deyakij tochci mizh 0 ta 1 2 Div takozh RedaguvatiTeorema DarbuDzherela RedaguvatiGrigorij Mihajlovich Fihtengolc Kurs diferencialnogo ta integralnogo chislennya 2023 1100 s ukr Fihtengolc G M Kurs differencialnogo i integralnogo ischisleniya Moskva Nauka 1962 T 1 607 s ros Otrimano z https uk wikipedia org w index php title Teorema Bolcano Koshi amp oldid 39882708