(pp. 1-12)

1. A formula of Vieta. We start from simple trigonometry. Write

$\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}$

$ = {2^2}\sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}$

(1.1)$ = {2^3}\sin \frac{x}{8}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}$

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$ = {2^n}\sin \frac{x}{{{2^n}}}\prod\limits_{k = 1}^n {\cos \frac{x}{{{2^k}}}.} $

From elementary calculus we know that, for$x \ne 0,$

$1 = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{x}{{{2^n}}}}}{{\frac{x}{{{2^n}}}}} = \frac{1}{x}\mathop {\lim }\limits_{n \to \infty } {2^n}\sin \frac{x}{{{2^n}}},$

and hence

(1.2)$\mathop {\lim }\limits_{n \to \infty } {2^n}\sin \frac{x}{{{2^n}}} = x.$

Combining (1.2) with (1.1), we get

(1.3)$\frac{{\sin x}}{x} = \prod\limits_{k = 1}^\infty {\cos \frac{x}{{{2^k}}}} .$

A special case of (1.3) is of particular interest. Setting$x = \frac{\pi }{2},$we obtain

(1.4)$\frac{2}{\pi } = \prod\limits_{n = 1}^\infty {\cos \frac{\pi }{{{2^{n + 1}}}}} $

$ = \frac{{\sqrt 2 }}{2}\frac{{\sqrt {2 + \sqrt 2 } }}{2}\frac{{\sqrt {2 + \sqrt {2 + \sqrt 2 } } }}{2}$...,

a classical formula due to Vieta.

2. Another look at Vieta’s formula. So far everything has been straightforward and familiar.

Now let us take a look a t (1.3) from a different point of view.

It is known that every real number$t,0 \le t \le 1$, can be written*uniquely*in the form

(2.1)$t = \frac{{{\varepsilon _1}}}{2} + \frac{{{\varepsilon _2}}}{{{2^2}}} + ...,$

where each ε is either 0 or 1....