Viète formula

From Freepedia

This article is not about Viète's formulas for symmetric polynomials.

In mathematics, the Viète formula is the following infinite product type representation of the mathematical constant π:

<math>\frac2\pi=

\frac{\sqrt2}2 \frac{\sqrt{2+\sqrt2}}2 \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots</math>

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

<math>\lim_{n \rightarrow \infty} \prod_{i=1}^n {a_i \over 2}

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

Proof

Using an iterated application of the double-angle formula

<math>\, \sin(2x)=2\sin(x)\cos(x) </math>

for sine (see the "double-angle formulas" section in the trigonometric identity article) one first proves the identity

<math> {{\sin(2^n x)}\over {2^n \sin(x)}}=\prod_{i=0}^{n-1} \cos(2^i x)

</math>

valid for all positive integers n. Letting x=y/2ⁿ and dividing both sides by cos(y/2) yields

<math> {{\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).

</math>

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

<math> {{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).

</math>

Substituting y=π gives the identity

<math> {2\over {2^n \sin({\pi \over {2^n}})}}=\prod_{i=2}^{n} \cos\left({\pi\over {2^i}} \right) \ .

</math>

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,

<math>2\cos(x/2)=\sqrt{2+2\cos x},</math>

one derives that <math> b_i=2\cos\left({\pi\over {2^{i+1}}}\right) </math> satisfies the recursion <math> \,b_{i+1}=\sqrt{2+b_i} </math> with initial condition <math> b_1= 2\cos\left({\pi \over 4}\right)=\sqrt{2}=a_1 </math>. Thus a_n=b_n for all positive integers n.

The Viète formula now follows by taking the limit n → ∞. Note here that

<math> \lim_{n \rightarrow \infty} {2\over {2^n \sin({\pi \over {2^n}})}}={2\over \pi} </math>

as a consequence of the fact that <math> \lim_{x\rightarrow 0} \,{x\over {\sin x}}=1 </math> (this follows from l'Hôpital's rule).