11

3

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`

?

11

3

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`

?

16

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}

9

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.

6

```
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}

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. – Xavier – 2018-04-18T16:33:03.997

@Xavier please see https://mathematica.stackexchange.com/help/merging-accounts

– Kuba – 2018-04-18T18:44:21.423