arithmetic-geometric mean as a product

Recall that, given two real numbers $0<x\\leq y$ , their arithmetic-geometricmean 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 givento determine the factors $h_{n}$ other than first computing $a_{n}$ and $g_{n}$ . We shall now correct this defect by deriving a recursion whichmay 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}}.