[obm-l] Prove that 22/7 > pi.

[Luís Lopes <qed_texte@xxxxxxxxxxx>]

Sauda,c~oes,

Achei esta mensagem interessante.

Um abraço,
Luís


>From: "Nikolaos Dergiades" <ndergiades@.....>
>Reply-To: Hyacinthos@yahoogroups.com
>To: <Hyacinthos@yahoogroups.com>
>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


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