The regular octagon below will serve to represent a general 2^{n}-gon.

Since the polygon has unit perimeter, each side is of length 2^{−n}.

The area of the colored right triangle is ½ × 2^{−(n+1)} × h.

The 2^{n}-gon consists of 2^{n+1} such triangles; therefore its area is ½h.

We also have tan x = 2^{−(n+1)}/h, from which h = 2^{−(n+1)}/tan x.

Further, x = /2^{n}.

Hence the area of the 2^{n}-gon is 2^{−(n+2)}/tan(/2^{n}).

The problem is therefore reduced to finding the tangent of /2^{n}, which we can determine by repeated application of appropriate half angle formulae.

Although there is a half angle formula that gives tan (½a) directly in terms of tan a, this is not suitable for repeated application. More fruitful is to recursively determine cos(/2^{n−1}), and thereby derive tan(/2^{n}).

We shall use the following two identities. Since our angles all lie in the first quadrant (0 to /2), we will always take the positive square root.

We begin with the standard result that cos(/4) = /2.

Then, using (1), cos(/8) = [(1 + /2)/2] = ½.

Similarly, cos(/16) = ½.

From the algebraic manipulation, it's clear that each application of (1) will generate one more nested radical sign, and an additional prefix of "2 +". (This can be proved more formally using mathematical induction, if required.)

Therefore cos(/2^{n}) = ½, where there are n−1 twos under the nested radical signs.

Substituting the expression for cos(/2^{n−1}) into (2), we have:

tan(/2^{n}) = , where there are n−1 twos under the nested radical signs.

Therefore the area of the 2^{n}-gon | = 2^{−(n+2)}/tan(/2^{n}) |

= , where there are n−1 twos under both sets of nested radical signs. |

As n increases, we should expect the area of the regular 2^{n}-gon to approach that of a circle with unit perimeter, that is 1/(4).

The table below, which shows the approximate area, A_{n}, of the regular 2^{n}-gon for various values of n, and of the circle, illustrates this.

n | 2^{n} | A_{n} |
---|---|---|

2 | 4 | 0.0625 |

3 | 8 | 0.0754442 |

4 | 16 | 0.0785522 |

5 | 32 | 0.0793216 |

6 | 64 | 0.0795135 |

7 | 128 | 0.0795615 |

8 | 256 | 0.0795735 |

9 | 512 | 0.0795765 |

10 | 1024 | 0.0795772 |

Circle | 0.0795775 |

As n tends to infinity, A_{n} tends to 1/(4).

Furthermore, since cos(/2^{n}) tends to 1 as n tends to infinity, it must be the case that a_{n} = tends to 2 as n, the number of twos under the nested radical signs, tends to infinity.

An independent, rigorous, proof of this result begins by showing that the limit of sequence {a_{n}}, as n tends to infinity, exists. To do this we show that {a_{n}} is monotonic increasing and bounded above.

Firstly, we prove by mathematical induction that a_{n} < 2, for all n.

The basis is straightforward: a_{1} = < 2.

For the induction step, note that, if a_{k} < 2, then a_{k+1} = (2 + a_{k}) < 2.

Therefore a_{n} < 2, for all n.

Now consider a_{n+1} = (2 + a_{n}).

Then a_{n+1}^{2} = 2 + a_{n}.

So (a_{n+1} + 1)(a_{n+1} − 1) = a_{n} + 1.

Since for all n, a_{n} < 2, a_{n+1} − 1 < 1, and so a_{n+1} + 1 > a_{n} + 1.

Therefore a_{n} is monotonic increasing; in fact, it is *strictly* monotonic increasing.

Putting these two results together, by the Monotonic Convergence Theorem, {a_{n}} has a limit.

Having proved that the limit exists, we can calculate its value, L, from the quadratic equation L^{2} = 2 + L.

The only positive solution is L = 2, and therefore this is the limit of {a_{n}}.

Having devised and solved this puzzle, I realised that, in the limit, the solution affords a formula for . Of course, such a result must already be known, and indeed a little searching on the web turned up the closely related Viète's formula: number 64 in this list of pi formulas. This was the first ever exact formula for , and was developed in 1593.

Source: Original (but anticipated by Viète by 410 years!)