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

${\displaystyle a_0=x}$

${\displaystyle g_0 = y}$

where

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

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

This sequence converges very quickly.

Properties

• ${\displaystyle \operatorname{agm}(x,y) \in [x,y]}$

• 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 ${\displaystyle N}$ correct decimal places, it must be the case that for the ${\displaystyle k}$th iteration of calculation

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

Proof of Existence

From the Arithmetic Geometric Mean Inequality

${\displaystyle a_n \geq g_n}$

This yields that

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

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

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

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

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

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

This implies that

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

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

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

This completes the proof.