Did Archimedes could knew this?
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From: porky_pig_jr on 9 May 2010 16:36 On May 8, 11:22 am, Ludovicus <luir...(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... > Ludovicus Archimede doesn't never makes no mistakes.
From: Risto Kauppila on 10 May 2010 06:56 09.05.2010 23:36, porky_pig_jr(a)my-deja.com kirjoitti: > On May 8, 11:22 am, Ludovicus <luir...(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... >> Ludovicus > > Archimede doesn't never makes no mistakes. In my opinion, Archimedes was one of the greatest scientists in the whole history of science. His problem was that he had actually no predecessors. Rike
From: Ludovicus on 11 May 2010 06:01
On May 10, 8:56 pm, r...(a)trash.whim.org (Rob Johnson) wrote: > In article <d25e249b-e262-406b-af21-93ff40919...(a)r11g2000yqa.googlegroups..com>, > > Ludovicus <luir...(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. > Thanks. That's sufficient to show that Archimedes could have known the cited relation using only Euclidean theorems. |