# arithmetic-geometric mean as a product

Recall that, given two real numbers $0, 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 ArithmeticgeometricMeanAsAProduct 2013-03-22 17:09:59 2013-03-22 17:09:59 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Derivation msc 26E60 msc 33E05