Viète's Product

The hook

The last line of this question is one of the most beautiful formulas in mathematics, and one of the oldest of its kind. It was published in 1593 by the French lawyer and mathematician François Viète, and it was the first time anyone had written \(\pi\) exactly as the result of an infinite process. Before Viète, \(\pi\) could only be approximated; after him, it could be defined by a formula that, carried on for ever, gives every digit.

The formula looks mysterious: where could nested square roots of \(2\) come from? The question reveals that it comes from repeatedly using the double-angle formula, and it gets a second, less famous formula, for \(\ln 2\), from the same machinery along the way.

Before you start

\(\sinh 2y = 2\sinh y\cosh y\), exactly like \(\sin 2y = 2\sin y\cos y\). That one identity, used over and over, drives the whole question. The question works with hyperbolic functions until the last part, where we will switch to ordinary trigonometry.

Working it through

Stage 1 Halving repeatedly

The case \(n = 1\) is just the double-angle formula with \(y = x/2\): \[\sinh x = 2\cosh\left(\frac{x}{2}\right)\sinh\left(\frac{x}{2}\right).\]

Apply the double-angle formula to it: \(\sinh(x/2^n) = 2\cosh(x/2^{n+1})\sinh(x/2^{n+1})\). Substituting, \[\sinh x = 2^{n+1}\cosh\left(\frac{x}{2}\right)\cdots\cosh\left(\frac{x}{2^n}\right)\cosh\left(\frac{x}{2^{n+1}}\right)\sinh\left(\frac{x}{2^{n+1}}\right),\] which is the formula for \(n + 1\). By induction it holds for all \(n\).■

For the deduction, divide both sides by \(x\), and then multiply both sides by \(\dfrac{x/2^n}{\sinh(x/2^n)}\). On the right, \(2^n \sinh(x/2^n) \cdot \dfrac{x/2^n}{x\,\sinh(x/2^n)} = 1\), leaving just the product of cosh terms.■

Stage 2 The limit

From the series, for \(y \neq 0\), \[\frac{\sinh y}{y} = 1 + \frac{y^2}{3!} + \frac{y^4}{5!} + \cdots,\] which tends to \(1\) as \(y \to 0\), so its reciprocal \(y/\sinh y\) also tends to \(1\). (If you want more care: every factorial in the tail \(\frac{y^2}{3!} + \frac{y^4}{5!} + \cdots\) is at least \(6\), so for \(0 < |y| < 1\) the tail is at most \(\frac{1}{6}\bigl(y^2 + y^4 + \cdots\bigr) = \frac{y^2}{6(1-y^2)}\), which tends to \(0\).)

Now let \(n \to \infty\) in part (i), with \(x\) fixed. Then \(y = x/2^n \to 0\), so the left-hand side tends to \(\dfrac{\sinh x}{x} \times 1\). The right-hand side is the partial product, which therefore tends to \(\dfrac{\sinh x}{x}\). That is what the infinite product means.■

Stage 3 A formula for \(\ln 2\)

With \(x = \ln 2\), \(e^{x/2} = \sqrt2\), so \[\cosh\left(\frac{x}{2}\right) = \frac{2^{1/2} + 2^{-1/2}}{2} = \frac{2 + 1}{2\sqrt2} = \frac{3}{2\sqrt2}.\] Similarly \(e^{x/4} = 2^{1/4}\), so \(\cosh(x/4) = \dfrac{2^{1/4} + 2^{-1/4}}{2}\), and multiplying top and bottom by \(2^{1/4}\) gives \(\dfrac{2^{1/2} + 1}{2 \times 2^{1/4}}\).■

The same calculation gives, for every \(k \geq 1\), \[\cosh\left(\frac{x}{2^k}\right) = \frac{1 + 2^{1/2^{k-1}}}{2} \times 2^{-1/2^k}.\] (For \(k = 1\) this is \(\frac{1 + 2}{2} \times 2^{-1/2} = \frac{3}{2\sqrt2}\), as before.) Multiplying over \(k = 1, 2, 3, \ldots\), the powers of \(2\) combine to \(2^{-(1/2 + 1/4 + 1/8 + \cdots)} = 2^{-1}\), and the first fraction is \(\frac{1+2}{2} = \frac32\). So the infinite product is \[\frac32 \times \frac12 \times \frac{1 + 2^{1/2}}{2} \times \frac{1 + 2^{1/4}}{2} \times \frac{1 + 2^{1/8}}{2} \cdots.\] On the other side, \(\sinh(\ln 2) = \frac{2 - \frac12}{2} = \frac34\), so \(\dfrac{\sinh x}{x} = \dfrac{3}{4\ln 2}\). The factor \(\frac34\) cancels from both sides, leaving \[\frac{1}{\ln 2} = \frac{1 + 2^{1/2}}{2} \times \frac{1 + 2^{1/4}}{2} \times \frac{1 + 2^{1/8}}{2} \cdots. \qed\]

Stage 4 Viète’s formula

It does. The induction used only the double-angle formula, and \(\sin 2y = 2\sin y\cos y\). The limit used only \(\frac{\sin y}{y} \to 1\) as \(y \to 0\), which follows from \(\sin y = y - \frac{y^3}{3!} + \cdots\) in the same way. So for \(x \neq 0\) (and \(x\) not a multiple of \(\pi\), so that no factor is zero along the way), \[\frac{\sin x}{x} = \cos\left(\frac{x}{2}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{8}\right)\cdots.\]

Take \(x = \frac{\pi}{2}\). Then \(\frac{\sin x}{x} = \frac{1}{\pi/2} = \frac{2}{\pi}\), and the factors are \(\cos\frac\pi4, \cos\frac\pi8, \cos\frac{\pi}{16}, \ldots\). The first is \(\cos\frac\pi4 = \frac{\sqrt2}{2}\). If a factor is \(\cos\theta = \frac{c}{2}\), the half-angle formula makes the next one \[\cos\frac{\theta}{2} = \sqrt{\frac{1 + c/2}{2}} = \frac{\sqrt{2 + c}}{2}.\] So the numerators go \(\sqrt2\), \(\sqrt{2 + \sqrt2}\), \(\sqrt{2 + \sqrt{2 + \sqrt2}}\), …, each obtained by adding \(2\) to the previous one and taking the square root. That is Viète’s formula.■

Looking back

• Telescoping a product. The identity in part (i) is a product that collapses, just as some sums telescope.

• Finite identity first, limit second. As in Section 2.4, every step was exact for finite \(n\), and the only limit needed was the very simple \(\frac{\sin y}{y} \to 1\).

• Hyperbolic and circular functions are twins. The same proof worked for both, because both satisfy the same double-angle formula and the same small-angle limit. This is no coincidence: \(\cosh(\mathrm{i}x) = \cos x\), as you will see in Section 5.7.

What’s really going on

Infinite products

Viète’s product was the first infinite product in mathematics. Others followed: John Wallis found \(\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots\) in 1655, and Euler’s product for \(\frac{\sin x}{x}\) appeared in Section 2.4. At university you will learn when an infinite product converges. Roughly, \(\prod(1 + a_n)\) with small positive \(a_n\) behaves like \(\sum a_n\), because \(\ln(1 + a_n) \approx a_n\). Here \(1 - \cos(\pi/2^n) \approx \frac12(\pi/2^n)^2\), which shrinks by a factor of \(4\) each time, so the product converges quickly.

How quickly?

After \(n\) factors the partial product equals \(\dfrac{1}{2^n\sin(\pi/2^{n+1})}\) exactly, and its error is about \(\frac{\pi^2}{6 \cdot 4^{n+1}}\) times \(\frac2\pi\). Each new factor makes the answer about \(4\) times more accurate, gaining about \(0.6\) of a decimal digit. Ten factors give \(2/\pi\) correct to six decimal places. That is good, but nothing like as fast as the methods of Section 2.6, which double the number of correct digits at each step.

The thinking skill

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Further reading

• Petr Beckmann, A History of \(\pi\) (St Martin’s Press). Archimedes, Viète, Wallis and the rest, told with gusto.

• Jonathan Borwein & Peter Borwein, Pi and the AGM (Wiley). The serious mathematics of computing \(\pi\), including the method of Section 2.6. Chapter 1 is accessible with this chapter behind you.

• Eli Maor, Trigonometric Delights (Princeton University Press). Contains a chapter on Viète and his product.