Transcende ntni chi sla ce chisla yaki ne zadovolnyayut zhodne algebrayichne rivnyannya z racionalnimi koeficiyentami Zmist 1 Vlastivosti 2 Prikladi 3 Istoriya 4 Shema dovedennya togo sho chislo UNIQ postMath 0000000E QINU ye transcendentnim 5 Div takozh 6 LiteraturaVlastivosti RedaguvatiMnozhina transcendentnih chisel kontinualna Kozhne transcendentne dijsne chislo ye irracionalnim ale zvorotne nepravilno Napriklad chislo 2 displaystyle sqrt 2 nbsp irracionalne ale ne transcendentne vono ye korenem mnogochlena x 2 2 0 displaystyle x 2 2 0 nbsp Mira irracionalnosti majzhe bud yakogo v sensi miri Lebega transcendentnogo chisla dorivnyuye 2 Prikladi RedaguvatiOsnova naturalnih logarifmiv chislo e Chislo p displaystyle pi nbsp Desyatkovij logarifm bud yakogo cilogo chisla okrim chisel 10 n displaystyle 10 n nbsp Sinus kosinus tangens bud yakogo nenulovogo algebrayichnogo chisla a displaystyle a nbsp zgidno z teoremoyu Lindenmanna Veyershtrassa Istoriya RedaguvatiVpershe ponyattya transcendentnogo chisla vviv Zhozef Liuvill v 1844 koli za dopomogoyu diofantovih nablizhen doviv teoremu pro te sho algebrayichne chislo nemozhlivo dovoli dobre nabliziti racionalnim drobom U 1873 Sharl Ermit doviv transcendentnist chisla e displaystyle mathrm e nbsp osnovi naturalnih logarifmiv U 1882 Ferdinand fon Lindeman doviv teoremu pro transcendentnist stepenya chisla e displaystyle mathrm e nbsp z nenulovim algebrayichnim pokaznikom tim samim dovivshi transcendentnist chisla p displaystyle pi nbsp i nerozv yaznist zavdannya kvadraturi kruga Nekonstruktivne dovedennya isnuvannya transcendentnih chisel majzhe trivialnij naslidok teoriyi mnozhin Kantora U 1900 na II Mizhnarodnomu Kongresi matematikiv David Gilbert sered sformulovanih nim problem sformulyuvav somu problemu Yaksho a 0 displaystyle a not 0 nbsp a displaystyle a nbsp algebrayichne chislo i b displaystyle b nbsp algebrayichne ale irracionalne chi pravilno sho a b displaystyle a b nbsp transcendentne chislo Zokrema chi ye transcendentnim chislo 2 2 displaystyle 2 sqrt 2 nbsp Cya problema bula virishena v 1934 A O Gelfondom yakij doviv sho vsi taki chisla ye transcendentnimi Shema dovedennya togo sho chislo e displaystyle mathrm e ye transcendentnim RedaguvatiPershe dovedennya togo sho chislo e displaystyle e nbsp osnova naturalnogo logarifma ye transcendentnim datuyetsya 1873 rokom Nadali sliduvatimemo strategiyi Davida Gilberta yakij sprostiv originalne dovedennya zaproponovane Sharlem Ermitom Ideya polyagaye v zastosuvanni metodu vid suprotivnogo Pripustimo sho e displaystyle mathrm e nbsp algebrayichne chislo Todi isnuye skinchennij nabir cilih koeficiyentiv c 0 c 1 c n displaystyle c 0 c 1 c n nbsp sho zadovolnyayut rivnyannya c 0 c 1 e c 2 e 2 c n e n 0 c 0 c n 0 displaystyle c 0 c 1 mathrm e c 2 mathrm e 2 cdots c n mathrm e n 0 qquad c 0 c n neq 0 nbsp Dlya dodatnogo cilogo chisla k displaystyle k nbsp rozglyanemo nastupnij mnogochlen f k x x k x 1 x n k 1 displaystyle f k x x k left x 1 cdots x n right k 1 nbsp i pomnozhimo obidvi chastini rivnyannya vishevkazanogo rivnyannya na 0 f k e x d x displaystyle int 0 infty f k mathrm e x operatorname d x nbsp takim chinom otrimayemo c 0 0 f k e x d x c 1 e 0 f k e x d x c n e n 0 f k e x d x 0 displaystyle c 0 left int 0 infty f k mathrm e x operatorname d x right c 1 mathrm e left int 0 infty f k mathrm e x operatorname d x right cdots c n mathrm e n left int 0 infty f k mathrm e x operatorname d x right 0 nbsp Ce rivnyannya mozhna zapisati v nastupnij formi P Q 0 displaystyle P Q 0 nbsp de P c 0 0 f k e x d x c 1 e 1 f k e x d x c 2 e 2 2 f k e x d x c n e n n f k e x d x Q c 1 e 0 1 f k e x d x c 2 e 2 0 2 f k e x d x c n e n 0 n f k e x d x displaystyle begin aligned P amp c 0 left int 0 infty f k mathrm e x operatorname d x right c 1 mathrm e left int 1 infty f k mathrm e x operatorname d x right c 2 e 2 left int 2 infty f k mathrm e x operatorname d x right cdots c n mathrm e n left int n infty f k mathrm e x operatorname d x right Q amp c 1 mathrm e left int 0 1 f k mathrm e x operatorname d x right c 2 mathrm e 2 left int 0 2 f k mathrm e x operatorname d x right cdots c n mathrm e n left int 0 n f k mathrm e x operatorname d x right end aligned nbsp Lema 1 Isnuye take k displaystyle k nbsp dlya yakogo viraz P k displaystyle tfrac P k nbsp ye cilim nenulovim chislom Dovedennya Kozhen dodanok v P displaystyle P nbsp ye dobutkom cilogo chisla na sumu faktorialiv ce viplivaye z rivnosti 0 x j e x d x j displaystyle int 0 infty x j mathrm e x operatorname d x j nbsp yaka ye spravedlivoyu dlya bud yakogo cilogo dodatnogo j displaystyle j nbsp div Gamma funkciya Vin ne dorivnyuye nulyu oskilki dlya bud yakogo a displaystyle a nbsp takogo sho 0 lt a n displaystyle 0 lt a leq n nbsp pidintegralnij viraz v c a e a a f k e x d x displaystyle c a mathrm e a int a infty f k mathrm e x operatorname d x nbsp ye dobutkom e x displaystyle mathrm e x nbsp na sumu dodankiv v yakih najmenshij stepin pri x displaystyle x nbsp dorivnyuye k 1 displaystyle k 1 nbsp pislya zamini v integrali x displaystyle x nbsp na x a displaystyle x a nbsp Otrimayemo sumu integraliv viglyadu 0 x j e x d x displaystyle int 0 infty x j mathrm e x operatorname d x nbsp de k 1 j displaystyle k 1 leq j nbsp i tomu vona ye cilim chislom sho dilitsya na k 1 displaystyle k 1 nbsp Pislya dilennya na k displaystyle k nbsp otrimayemo 0 za modulem k 1 displaystyle k 1 nbsp Prote mozhna zapisati 0 f k e x d x 0 1 n n k 1 e x x k d x displaystyle int 0 infty f k mathrm e x operatorname d x int 0 infty left left 1 n n right k 1 mathrm e x x k cdots right operatorname d x nbsp i todi pri dilenni pershogo dodanku na k displaystyle k nbsp otrimayemo 1 k c 0 0 f k e x d x c 0 1 n n k 1 0 mod k 1 displaystyle frac 1 k c 0 int 0 infty f k mathrm e x operatorname d x equiv c 0 1 n n k 1 not equiv 0 pmod k 1 nbsp Tomu pri dilenni kozhnogo integrala v P displaystyle P nbsp na k 1 displaystyle k 1 nbsp lishe pershij ne bude dilitisya nacilo na k 1 displaystyle k 1 nbsp i lishe todi koli k 1 displaystyle k 1 nbsp ye prostim chislom i k 1 gt n displaystyle k 1 gt n nbsp k 1 gt c 0 displaystyle k 1 gt c 0 nbsp Z cogo viplivaye sho viraz P k displaystyle tfrac P k nbsp ne dilitsya nacilo na k 1 displaystyle k 1 nbsp i tomu ne mozhe dorivnyuvati nulyu Lema 2 Q k lt 1 displaystyle left tfrac Q k right lt 1 nbsp dlya dostatno velikih k displaystyle k nbsp Dovedennya Zauvazhimo shof k e x x k x 1 x 2 x n k 1 e x x x 1 x n k x 1 x n e x u x k v x displaystyle begin aligned f k mathrm e x amp x k x 1 x 2 cdots x n k 1 mathrm e x amp left x x 1 cdots x n right k cdot left x 1 cdots x n mathrm e x right amp u x k cdot v x end aligned nbsp de u x v x displaystyle u x v x nbsp neperervni dlya vsih x displaystyle x nbsp i tomu ye obmezhenimi na promizhku 0 n displaystyle 0 n nbsp Ce oznachaye sho isnuyut konstanti G H gt 0 displaystyle G H gt 0 nbsp taki sho f k e x u x k v x lt G k H displaystyle left f k mathrm e x right leq u x k cdot v x lt G k H quad nbsp dlya 0 x n displaystyle quad 0 leq x leq n nbsp Tomu kozhen z integraliv v Q displaystyle Q nbsp ye obmezhenim i v najgirshomu vipadku 0 n f k e x d x 0 n f k e x d x 0 n G k H d x n G k H displaystyle left int 0 n f k mathrm e x operatorname d x right leq int 0 n left f k mathrm e x right operatorname d x leq int 0 n G k H operatorname d x nG k H nbsp Todi mozhna obmezhiti i Q displaystyle Q nbsp Q lt G k n H c 1 e c 2 e 2 c n e n G k M displaystyle Q lt G k cdot nH left c 1 mathrm e c 2 mathrm e 2 cdots c n mathrm e n right G k cdot M nbsp de M displaystyle M nbsp ye nezalezhnoyu vid k displaystyle k nbsp konstantoyu Z cogo viplivaye sho Q k lt M G k k 0 displaystyle left frac Q k right lt M cdot frac G k k to 0 quad nbsp de k displaystyle k to infty nbsp sho zavershuye dovedennya lemi Viberemo k displaystyle k nbsp sho zadovolnyaye umovi oboh lem Otrimayemo nastupne cile chislo P k displaystyle left tfrac P k right nbsp sho ne dorivnyuye nulyu dodane do neskinchenno maloyi velichini Q k displaystyle left tfrac Q k right nbsp dorivnyuye nulyu sho nemozhlivo Tomu nashe pripushennya sho e displaystyle mathrm e nbsp ye algebrayichnim chislom hibne otzhe e displaystyle mathrm e nbsp transcendentne chislo Div takozh RedaguvatiTeoriya transcendentnih chiselLiteratura RedaguvatiGelfond A O Transcendentnye i algebraicheskie chisla M 1952 Feldman N Algebraicheskie i transcendentnye chisla Arhivovano 19 veresnya 2004 u Wayback Machine ros Kvant 7 1983 Otrimano z https uk wikipedia org w index php title Transcendentne chislo amp oldid 40228933
