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

:$lim\_\{n\; ightarrow\; infty\}\; prod\_\{i=1\}^n\; \{a\_i\; over\; 2\}=frac2pi$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

:$lim\_\{mathbf\{n\}\; oinfty\}2^\{mathbf\{n\}+1\}sqrt\{2-underbrace\{sqrt\{2+sqrt\{2+sqrt\{2+cdots+sqrt\{2\}\_\{mathbf\{n\};=;pi$

(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