Fwd: NEW FORMULA FOR THE NUMBER PI (?)

Does anyone know anything about this?
Charles Lucy lucy@lucytune.com

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Begin forwarded message:

> From: "candido otero ramos" <oteropera@hotmail.com>
> Date: 20 March 2007 12:04:28 GMT+01:00
> To: info@lucytune.com
> Subject: NEW FORMULA FOR THE NUMBER PI
>
>
> New formula for the number Pi
>
> The method is to circumscribe a polygon around the circumference
> [of a circle] and calculate the perimeter of this polygon.
> A polygon with infinitely many sides results in the number Pi.
> The method is that of Archimedes for polygons who are
> outside the circumference of those (in other words).
>
>
> The formula is the following iterative one:
>
>
> 2 * A[n] * B[n]
> A[n+1] = ------------------
> 1 + B[n]
>
>
> 1 + B[n]
> B[n+1] = sqrt ( ----------- )
> 2
>
>
>
> The initial values are A[0] = 4 and B[0] = sqrt( 1 / 2 )
>
>
> The value 4 of A[0] equals 4 * tan[Pi/4]
> considered as a circumference of radius 1 of the square
> around the unit circle.
> the value sqrt(1/2) of B[0] equals cos(Pi/4)
>
>
> The value of A[n] ( lim(A[n]) for n--> infinity ) equals Pi.
>
>
> Proof
>
> we consider A[n] the perimeter of the polygon of 2^q sides
> and A[n + 1] the polygon of 2^(q + 1) sides , the proof of
> the relation between A[n] and A[n + 1] is the following:
>
>
>
> sin [ pi/n ]
> A[n] = 2^q * tan[ pi/n ] = 2^q * ------------------
> cos [ pi/n ]
>
>
>
> If we divide by the inverse cosine of Pi/n
>
>
> sin [ pi/n ]
> 2^q * ----------------
> cos [ pi/n ]
> ----------------------------
> 1
> ------------------
> cos [ pi/n ]
>
>
>
> the cosines cancel.The sine of Pi/n is equal to
>
>
> 2 * sin [ pi/(2*n) ] * cos [ pi/(2*n) ]
>
>
> with this we divide between
>
>
> 2^q * 2 * sin [ pi/(2*n) ] * cos [ pi/(2*n) ]
> ---------------------------------------------------
>
> cos [ pi/(2*n) ] * cos [ pi/(2*n) ]
>
>
> the cosine cancels and it remains
>
>
> 2^q * 2 * sin [ pi/(2*n) ]
> ------------------------------
> cos [ pi/(2*n) ]
>
>
> therefore A[n] = 2^q * tan[ pi/n ]
>
> and A[n + 1] = 2^(q + 1) * tan[ pi/(2*n) ]
>
>
> therefore the final formula is
>
>
> sin [ pi/n ]
> 2^q * -------------------
> cos [ pi/n ]
> -------------------------------------------------------------------
> 1
> ----------------------- * cos [ pi/(2*n) ] * cos [ pi/(2*n) ]
>
> cos [ pi/n ]
>
>
> which is equal to
>
>
> sin [ pi/n ]
> 2^q * -------------------
> cos [ pi/n ]
> ----------------------------------------------------
> 1 1 + cos [ pi/n ]
> ------------------------ * ------------------------
> cos [ pi/n ] 2
>
>
> this leaves us finally with an identity that we did write in the
> beginning
>
>
> 2 * A[n] * B[n]
> A[n+1] = ------------------
> 1 + B[n]
>
>
> for which A[n] > A[n + 1] > PI
> and consequentially the computation of A[n] for n-->infinity
> is greater of equal to Pi. A[infinity] == PI.
>
>
> The computation of B[n] is the computation of the successive
> cosines for lengths of the sides in the preceding steps.
>
>
>
> para cualquier duda o consulta contactar con la direccion
> MSN MESSENGER oteropera@hotmail.com
>
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>

--- In tuning-math@yahoogroups.com, Charles Lucy <lucy@...> wrote:
>
> Does anyone know anything about this?

It's an old idea, which Vieta used to make an infinite product formula
for pi which was the first such formula ever discovered, before even
the Leibniz series.