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From: Ludovicus on 8 May 2010 11:22
Lim (Arc - cord) = (Tangent - arc) / 2

Because in that case he could have improved the calcul of Pi
through: Lim Arc = (2*cord + tangent) / 3
As he stablished that 3 + 10 / 71 < pi < 3 + 1 / 7
he could have taken C = (20 / 71 + 1 / 7) / 3 and then:
pi aprox. = 3 + C = 3.141515...
Ludovicus
From: facemelter1729 on 8 May 2010 11:29
speek american, pus-nozzle.

"Ludovicus" <luiroto(a)yahoo.com> wrote in message
news:d25e249b-e262-406b-af21-


From: Rob Johnson on 10 May 2010 20:56
In article <d25e249b-e262-406b-af21-93ff409198a1(a)r11g2000yqa.googlegroups.com>,
Ludovicus <luiroto(a)yahoo.com> wrote:
>Lim (Arc - cord) = (Tangent - arc) / 2
>
>Because in that case he could have improved the calcul of Pi
>through: Lim Arc = (2*cord + tangent) / 3
>As he stablished that 3 + 10 / 71 < pi < 3 + 1 / 7
>he could have taken C = (20 / 71 + 1 / 7) / 3 and then:
> pi aprox. = 3 + C = 3.141515...

There is little reason that he couldn't have known that

tan(x) - x
lim ---------- = 2 [1]
x->0 x - sin(x)

In <http://www.whim.org/nebula/math/sintan.html>, it is shown not
only that

tan(x) sin(x)
lim ------ = lim ------ = 1 [2]
x->0 x x->0 x

but also that

1 - cos(x) 1
lim ---------- = - [3]
x->0 x^2 2

and that

x - sin(x) 1
lim ---------- = - [4]
x->0 x^3 6

Multiply [3] by tan(x)/x to get

tan(x) - sin(x) 1
lim --------------- = - [5]
x->0 x^3 2

Subtract [4] from [5] to get that

tan(x) - x 1
lim ---------- = - [6]
x->0 x^3 3

Divide [6] by [4] to get [1].

I don't know if the idea of telescoping infinite sums was around
in Archimedes' time, but the rest of the argument above and in the
page cited is pretty basic.

Rob Johnson <rob(a)trash.whim.org>
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