proof of vite formula

In mathematics, the Vite 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 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}

Proof

Using an iterated application of the double-angle formula

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

for the sine (see paragraph "double-angle formulas" under trigonometric identity) one first proves the identity

=\prod_{i=0}^{n-1} \cos(2^i x)

valid for n > 0. Substitute here x by

{y \over {2^n}}

and divide both sides by

\cos\left({y\over 2}\right)

. This first yields

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

and then

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

valid for n > 0, using the double-angle formula

\sin y=2\sin{y\over 2}\cos{y\over 2}

again. Now specialize this relation for

\, y=\pi

to obtain the identity (valid for n > 0)

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

This identity may be regarded as some kind of "finite" (nth-order) variant of the Vite formula. It remains to match the terms a_n with the factors on the right side of the last identity. Using the half-angle formula for the cosine