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:

:$$frac2pi=frac{sqrt2}2cdotfrac{sqrt{2+sqrt22cdotfrac{sqrt{2+sqrt{2+sqrt2}2cdots.

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=cosleft(frac{x}2 ight)cdotcosleft(frac{x}4 ight)cdotcosleft(frac{x}8 ight)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

:$$lim_{n ightarrow infty} prod_{i=1}^n {a_i over 2}=frac2piwhere $$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} oinfty}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)=2sin(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({yover 2} )cdot{1over {2^n sin({yover {2^n)=prod_{i=1}^{n-1} cosleft({yover {2^{i+1} ight).

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

:$$2sin({yover 2})}over {2^n sinleft(displaystyle{yover {2^n ight)=prod_{i=1}^{n-1} cosleft({yover {2^{i+1} ight).

Substituting "y"=π gives the identity

:$$ {2over {2^n sin({pi over {2^n)=prod_{i=2}^{n} cosleft({piover {2^i ight) .

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,

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

one derives that $$b_i=2cosleft({piover {2^{i+1} ight) satisfies the recursion $,$b_{i+1}=sqrt{2+b_i} with initial condition $$b_1= 2cosleft({pi over 4} ight)=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 ightarrow infty} {2over {2^n sin({pi over {2^n)={2over pi}

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

π