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[parent] arithmetic-geometric mean as a product (Derivation)

Recall that, given two real numbers $0 < x \le y$ , 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:

$\displaystyle h_{n+1}$ $\displaystyle = \sqrt{a_{n+1} \over g_{n+1}}$    
  $\displaystyle = \sqrt{a_n + g_n \over 2 \sqrt{a_n g_n}}$    
  $\displaystyle = \sqrt{ {1 \over 2} \left( \sqrt{a_n \over g_n} + \sqrt{g_n \over a_n} \right)}$    
  $\displaystyle = \sqrt{ {1 \over 2} \left( h_n + {1 \over h_n} \right)}$    
  $\displaystyle = \sqrt{h_n^2 + 1 \over 2 h_n}$    

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




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Cross-references: formula, limit, defect, factors, infinite product, arithmetic-geometric mean, real numbers

This is version 3 of arithmetic-geometric mean as a product, born on 2007-05-27, modified 2007-05-28.
Object id is 9478, canonical name is ArithmeticGeometricMeanAsAProduct.
Accessed 825 times total.

Classification:
AMS MSC26E60 (Real functions :: Miscellaneous topics :: Means)
 33E05 (Special functions :: Other special functions :: Elliptic functions and integrals)

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