How do I program the approximation to π devised by François Viète, which is given by
2 * 2/Sqrt[2] * 2/Sqrt[2 + Sqrt[2]] * 2/Sqrt[2 + Sqrt[2 + Sqrt[2]]] * ...
using some of these functions: Nest, For, Module, Product, Do?
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$\begingroup$ I appreciate all of your answers, but these commands are not in the list I posted... Right now, I have come to this: PieApprox[n_] := 2/Nest[Sqrt[# + 2] &, Sqrt[2], n]. So I was thinking if someone could help me plug in the Product fonction to finish it, because I can't figure out how to do it. $\endgroup$ – Xavier Apr 18 '18 at 16:33
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$\begingroup$ @Xavier please see mathematica.stackexchange.com/help/merging-accounts $\endgroup$ – Kuba♦ Apr 18 '18 at 18:44
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You could use
VietePiApprox[n_] := (Times @@ NestList[Sqrt[2 + #] &, Sqrt[2], n])/ 2^(n + 1)
SetAttributes[VietePiApprox, Listable]
which approximates Pi as follows.
N[VietePiApprox[Range[5]] - 2/\[Pi]]
{0.0166617, 0.00410909, 0.0010238, 0.000255735, 0.0000639204}
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Well, FoldList also can finish this job:
2/Times @@ (1/2 FoldList[Sqrt[2 + #] &, ConstantArray[Sqrt[2.], 10]])
By the way, by putting the one half just before NestList, FoldList can save one's effort for counting the number of twos in the denominators.
answered Apr 18 '18 at 9:37
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Clear[VietePiApprox];
VietePiApprox[n_] := Product[FunctionExpand[Cos[Pi/2^(i + 1)]], {i, 1, n}];
Table[VietePiApprox[i], {i, 1, 10}]
% - 2.0/Pi
{0.070487, 0.0166617, 0.00410909, 0.0010238, 0.000255735, \ 0.0000639204, 0.0000159792, 3.99476*10^-6, 9.98687*10^-7, 2.49671*10^-7}
