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

```Sauda,c~oes,

Achei esta mensagem interessante.```
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Um abraço,
Luís

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```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
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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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