Viète formula

From Wikipedia, the free encyclopedia

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:

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

The expression on the right hand side has to be understood as a limit expression (as $n \rightarrow \infty$ )

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

where a_n is the nested quadratic radical given by the recursion $a_n=\sqrt{2+a_{n-1}}$ with initial condition $a_1=\sqrt{2}$ .

[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

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

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.