Your American History Reference Guide!
- Proof of Viète formula

HistoryMania Information Site on Proof of Viète formula American History American History Search        American History Browse welcome to our free resource site for all enthusiasts!

Proof of Viète formula

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

\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}.

Proof

Using an iterated application of the double-angle formula

\, \sin(2x)=2\sin(x)\cos(x)

for sine (see the "double-angle formulas" section in the trigonometric identity article) 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 intial 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).

Last updated: 05-21-2005 11:54:14
The contents of this article are licensed from Wikipedia.org under the
GNU Free Documentation License. How to see transparent copy
Search | Browse | Contact | Legal info