Viète's formula

This article is not about Viète's formulas for symmetric polynomials.

In mathematics, the Viète formula, named after François Viète, is the following infinite product type representation of the mathematical constant π:

$\frac2\pi= \frac{\sqrt2}2\cdot \frac{\sqrt{2+\sqrt2}}2\cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots.$

The above formula is a result of one of Leonhard Euler's formula. Euler discovered that:

$\frac{\sin(x)}x=\cos\left(\frac{x}2\right)\cdot\cos\left(\frac{x}4\right)\cdot\cos\left(\frac{x}8\right)\cdots$

Substituting x=π/2 will produce the formula for 2/π, that is represented in an elegant manner by Viète.

The expression on the right hand side has to be understood as a limit expression

$\lim_{n \rightarrow \infty} \prod_{i=1}^n {a_i \over 2}=\frac2\pi$

where $a_n=\sqrt{2+a_{n-1}}$ with initial condition $a_1=\sqrt{2}.$

(Wells 1986, p. 50; Beckmann 1989, p. 95). However, this expression was not rigorously proved to converge until Rudio (1892).

Do some simplification, a pretty formula for π is given by

$\lim_{\mathbf{n}\to\infty}2^{\mathbf{n} +1}\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_{\mathbf{n}}}\;=\;\pi$

(J. Munkhammar, pers. comm., April 27, 2000).

Proof

Using an iterated application of the double-angle formula

$\, \sin(2x)=2\sin(x)\cos(x)$

for sine one first proves the identity

${{\sin(2^n x)}\over {2^n \sin(x)}}=\prod_{i=0}^{n-1} \cos(2^i x)$

valid for all positive integers n. Letting x=y/2ⁿ and dividing both sides by cos(y/2) yields

${{\sin( y)}\over {\cos({y\over 2} )}}\cdot{1\over {2^n \sin({y\over {2^n}})}}=\prod_{i=1}^{n-1} \cos\left({y\over {2^{i+1}}}\right).$

Using the double-angle formula sin y=2sin(y/2)cos(y/2) again gives

Failed to parse (unknown function\displaystyle): {{2\sin({y\over 2})}\over {2^n \sin(\displaystyle{y\over {2^n}})}}=\prod_{i=1}^{n-1} \cos\left({y\over {2^{i+1}}}\right).

Substituting y=π gives the identity

${2\over {2^n \sin({\pi \over {2^n}})}}=\prod_{i=2}^{n} \cos\left({\pi\over {2^i}} \right) \ .$

It remains to match the factors on the right-hand side of this identity with the terms a_n. Using the half-angle formula for cosine,

$2\cos(x/2)=\sqrt{2+2\cos x},$

one derives that $b_i=2\cos\left({\pi\over {2^{i+1}}}\right)$ satisfies the recursion $\,b_{i+1}=\sqrt{2+b_i}$ with initial condition $b_1= 2\cos\left({\pi \over 4}\right)=\sqrt{2}=a_1$ . Thus a_n=b_n for all positive integers n.

The Viète formula now follows by taking the limit n → ∞. Note here that

$\lim_{n \rightarrow \infty} {2\over {2^n \sin({\pi \over {2^n}})}}={2\over \pi}$

as a consequence of the fact that $\lim_{x\rightarrow 0} \,{x\over {\sin x}}=1$ (this follows from l'Hôpital's rule). π

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Viète's formula

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