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Viète formula - Wikipedia, the free encyclopedia

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 an 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 an. 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 an=bn 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).en:Viète's formula

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