Sauda,c~oes, Achei esta mensagem interessante.
Um abraço, Luís
From: "Nikolaos Dergiades" <[EMAIL PROTECTED]> Reply-To: [EMAIL PROTECTED] To: <[EMAIL PROTECTED]> Subject: RE: [EMHL] Prove that 22/7 > pi Date: Sun, 19 Mar 2006 22:33:33 +0200 Dear friends, M. T. ZED wrote: > Prove that 22/7 > pi. > Help me please. We have 4/(x^2+1) = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - x^4(1-x)^4/(x^2+1) or in the interval ( 0, 1) 4/(x^2+1) < x^6 - 4x^5 + 5x^4 - 4x^2 + 4 and by integration from 0 to 1 we get pi < 22/7. Does anybody knows a geometric or a simpler proof? Best regards Nikos Dergiades
Escreva x^4(1-x)^4 = x^4(x^4 - 4x^3 + 6x^2 - 4x + 1) . Agora some e subtraia termos para obter múltiplos de x^2 +1: x^8 - 4x^7 + 6x^6 - 4x^5 + x^4 = x^8 + 6x^6 + x^4 - 4x^5(x^2 + 1) = (x^6-4x^5)(x^2 + 1) + 5x^6 + x^4 = (x^6 - 4x^5 + 5x^4)(x^2+1) - 4x^4 = (x^6 - 4x^5 + 5x^4 - 4x^2)(x^2 + 1) + 4x^2 = (x^6 - 4x^5 + 5x^4 - 4x^2 + 4)(x^2 + 1) - 4 Transpondo e dividindo por x^2 + 1, vem: 4/(x^2+1) = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - x^4(1-x)^4/(x^2+1) qed ========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html =========================================================================
