arithmetic-geometric mean as a product

Recall that, given two real numbers $0<x\leq y$ , their arithmetic-geometric mean may be defined as $M(x,y)=\lim_{n\to\infty}g_{n}$ , where

	$\displaystyle g_{0}$	$\displaystyle=x$
	$\displaystyle a_{0}$	$\displaystyle=y$
	$\displaystyle g_{n+1}$	$\displaystyle=\sqrt{a_{n}g_{n}}$
	$\displaystyle a_{n+1}$	$\displaystyle={a_{n}+g_{n}\over 2}.$

In this entry, we will re-express this quantity as an infinite product. We begin by rewriting the recursion for $g_{n}$ :

$g_{n+1}=\sqrt{a_{n}g_{n}}=\sqrt{{a_{n}\over g_{n}}\cdot g_{n}^{2}}=g_{n}\sqrt{% a_{n}\over g_{n}}$

From this, it follows that

$g_{n}=g_{0}\prod_{m=0}^{n-1}h_{m}$

where $h_{n}=\sqrt{a_{n}/g_{n}}$ .

As it stands, this is not so interesting because no way has been given to determine the factors $h_{n}$ other than first computing $a_{n}$ and $g_{n}$ . We shall now correct this defect by deriving a recursion which may be used to compute the $h_{n}$ ’s directly:

	$\displaystyle h_{n+1}$	$\displaystyle=\sqrt{a_{n+1}\over g_{n+1}}$
		$\displaystyle=\sqrt{a_{n}+g_{n}\over 2\sqrt{a_{n}g_{n}}}$
		$\displaystyle=\sqrt{{1\over 2}\left(\sqrt{a_{n}\over g_{n}}+\sqrt{g_{n}\over a% _{n}}\right)}$
		$\displaystyle=\sqrt{{1\over 2}\left(h_{n}+{1\over h_{n}}\right)}$
		$\displaystyle=\sqrt{h_{n}^{2}+1\over 2h_{n}}$

Taking the limit $n\to\infty$ , we then have the formula

$M(x,y)=x\prod_{m=0}^{\infty}h_{n}$

where

$h_{0}={y\over x}$

and

$h_{n+1}=\sqrt{h_{n}^{2}+1\over 2h_{n}}.$

Title	arithmetic-geometric mean as a product
Canonical name	ArithmeticgeometricMeanAsAProduct
Date of creation	2013-03-22 17:09:59
Last modified on	2013-03-22 17:09:59
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Derivation
Classification	msc 26E60
Classification	msc 33E05