Viète's formula

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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 now considered as a result of one of Leonhard Euler's formula - branded more than one century after. 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).

Upon 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).

[edit] 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

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

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.