# 算术-几何平均数

$a_1 = \frac{x+y}{2}$
$g_1 = \sqrt{xy}.$

$a_{n+1} = \frac{a_n + g_n}{2}$
$g_{n+1} = \sqrt{a_n g_n}.$

## 例子

$a_1=\frac{24+6}{2}=15,$
$g_1=\sqrt{24 \times 6}=12,$

$a_2=\frac{15+12}{2}=13.5,$
$g_2=\sqrt{15 \times 12}=13.41640786500\dots$ etc.

n an gn
0 24 6
1 15 12
2 13.5 13.41640786500...
3 13.45820393250... 13.45813903099...
4 13.45817148175... 13.45817148171...

24和6的算术-几何平均数是两个数列的公共极限，大约为13.45817148173。

## 性质

M(x, y)是一个介于xy的算术平均数和几何平均数之间的数。

M(x,y)还可以写为如下形式：

$\Mu(x,y) = \frac{\pi}{4} \cdot \frac{x + y}{K \left( \frac{x - y}{x + y} \right) }$

1和$\sqrt{2}$的算术-几何平均数的倒数，称为高斯常数

$\frac{1}{\Mu(1, \sqrt{2})} = G = 0.8346268\dots$

## 存在性的证明

$g_n \leqslant a_n$

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

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

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

$\lim_{n\to \infty}a_n = \lim_{n\to \infty}\frac{g_{n + 1}^2}{g_{n}} = \frac{g^2}{g} = g$

## 关于积分表达式的证明

$I(x,y) = \int_0^{\pi/2}\frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}},$

$\sin\theta = \frac{2x\sin\theta'}{(x+y)+(x-y)\sin^2\theta'},$

\begin{align} I(x,y) &= \int_0^{\pi/2}\frac{d\theta'}{\sqrt{\bigl(\frac12(x+y)\bigr)^2\cos^2\theta'+\bigl(\sqrt{xy}\bigr)^2\sin^2\theta'}}\\ &= I\bigl(\tfrac12(x+y),\sqrt{xy}\bigr). \end{align}

\begin{align} I(x,y) &= I(a_1, g_1) = I(a_2, g_2) = \cdots\\ &= I\bigl(M(x,y),M(x,y)\bigr) = \pi/\bigr(2M(x,y)\bigl). \end{align}

$M(x,y) = \pi/\bigl(2 I(x,y) \bigr).$

## 参考文献

### 引用

1. ^ David A. Cox. The Arithmetic-Geometric Mean of Gauss. (编) J.L. Berggren, Jonathan M. Borwein, Peter Borwein. Pi: A Source Book. Springer. 2004: 481. ISBN 978-0-387-20571-7. first published in L'Enseignement Mathématique, t. 30 (1984), p. 275-330