Arithmetic-geometric mean

The arithmetic-geometric mean,denoted $\operatorname{agm}(x,y)$, of two non-negative numbers $x, y \in \R$ is defined as the limit of two sequences $(a_n)_{n=0}^{\infty},(g_n)_{n=0}^{\infty}$, with

$a_0=x$

$g_0 = y$

where

$a_{n+1} = \dfrac{a_n + g_n}{2}$

$g_{n+1} = \sqrt{a_n g_n}$

This sequence converges very quickly.

Properties

• $\operatorname{agm}(x,y) \in [x,y]$

• Because the arithmetic mean approaches from above, and the geometric mean approaches from below, the two can be used to estimate an error value during calculation. For $N$ correct decimal places, it must be the case that for the $k$th iteration of calculation

$|a_k - g_k| < 10^{-N}$

Proof of Existence

From the Arithmetic Geometric Mean Inequality

$a_n \geq g_n$

This yields that

$(g_n)_{n=0}^{\infty}$ is a non-decreasing sequence.

$g_{n+1} = \sqrt{g_n a_n} \geq \sqrt{g_n g_n} = g_n, \forall n\in\N$

Because the mean of two numbers lies between the two numbers, there exists a lower bound for $g_n$. Then by the Monotone Convergence Theorem, there exists a limit $g\in\R$ such that

$\lim_{n \rightarrow \infty} g_n = g$

Using the definition of the sequence $(g_n)_{n=0}^{\infty}$

$g_{n+1} = \sqrt{a_n g_n}$

This implies that

$a_n = \dfrac{g_{n+1}^2}{g_n}$

Taking the limit as $n \rightarrow \infty$, and using the Algebra of limits

$\lim_{n\rightarrow \infty} a_n = \lim{n\rightarrow \infty} \dfrac{g_{n+1}^2}{g_n} = \dfrac{g^2}{g} = g$

This completes the proof.