Arithmetic–geometric mean

In mathematics, the arithmetic–geometric mean (AGM) of two positive real numbers x and y is defined as follows:

First compute the arithmetic mean of x and y and call it a₁. Next compute the geometric mean of x and y and call it g₁; this is the square root of the product xy:

$\begin{align} a_1 &= \frac{1}{2}(x + y)\\ g_1 &= \sqrt{xy} \end{align}$

Then iterate this operation with a₁ taking the place of x and g₁ taking the place of y. In this way, two sequences (a_n) and (g_n) are defined:

$\begin{align} a_{n+1} &= \frac{1}{2}(a_n + g_n)\\ g_{n+1} &= \sqrt{a_n g_n} \end{align}$

These two sequences converge to the same number, which is the arithmetic–geometric mean of x and y; it is denoted by M(x, y), or sometimes by agm(x, y).

This can be used for algorithmic purposes as in the AGM method.

Example

To find the arithmetic–geometric mean of a₀ = 24 and g₀ = 6, first calculate their arithmetic mean and geometric mean, thus:

$\begin{align} a_1 &= \frac{1}{2}(24 + 6) = 15\\ g_1 &= \sqrt{24 \times 6} = 12 \end{align}$

and then iterate as follows:

$\begin{align} a_2 &= \frac{1}{2}(15 + 12) = 13.5\\ g_2 &= \sqrt{15 \times 12} = 13.41640786500\dots\\ \dots \end{align}$

The first four iterations give the following values:

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

The arithmetic–geometric mean of 24 and 6 is the common limit of these two sequences, which is approximately 13.45817148173.

History

The first algorithm based on this sequence pair appeared in the works of Legendre. Its properties were further analyzed by Gauss.

Properties

The geometric mean of two positive numbers is never bigger than the arithmetic mean (see inequality of arithmetic and geometric means); as a consequence, (g_n) is an increasing sequence, (a_n) is a decreasing sequence, and g_n ≤ M(x, y) ≤ a_n. These are strict inequalities if x ≠ y.

M(x, y) is thus a number between the geometric and arithmetic mean of x and y; in particular it is between x and y.

If r ≥ 0, then M(rx,ry) = rM(x,y).

There is an integral-form expression for M(x,y):

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

where K(m) is the complete elliptic integral of the first kind:

$K(m) = \int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1 - m\sin^2(\theta)}}$

Indeed, since the arithmetic–geometric process converges so quickly, it provides an effective way to compute elliptic integrals via this formula. In engineering, it is used for instance in elliptic filter design.

Related concepts

The reciprocal of the arithmetic–geometric mean of 1 and the square root of 2 is called Gauss's constant, after Carl Friedrich Gauss.

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

The geometric–harmonic mean can be calculated by an analogous method, using sequences of geometric and harmonic means. The arithmetic–harmonic mean can be similarly defined, but takes the same value as the geometric mean.

The modified arithmetic–geometric mean was introduced and defined by Semjon Adlaj on p. 1094 of the September 2012 issue of the Notices of the AMS. It turned out to be useful for computing complete elliptic integrals of the second kind.

Proof of existence

From inequality of arithmetic and geometric means we can conclude that:

$g_i \leqslant a_i$

and thus

$g_{i + 1} = \sqrt{g_i \cdot a_i} \geqslant \sqrt{g_i \cdot g_i} = g_i$

that is, the sequence g_i is nondecreasing.

Furthermore, it is easy to see that it is also bounded above by the larger of x and y (which follows from the fact that both arithmetic and geometric means of two numbers both lie between them). Thus by the monotone convergence theorem the sequence is convergent, so there exists a g such that:

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

However, we can also see that:

$a_i = \frac{g_{i + 1}^2}{g_i}$

and so:

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

Q.E.D.

Proof of the integral-form expression

The following proof relies on the work of Carl Wilhelm Borchardt.

Set,

$\begin{align} a_0 \equiv x\\ g_0 \equiv y \end{align}$

So $\scriptstyle \lim_{n \to \infty}a_n \;=\; \lim_{n\to \infty}g_n \;=\; g \;=\; M(a_0, g_0)$

$\begin{align} a_1 &= \frac{1}{2}(a_0 + g_0)\qquad(1) \\ g_1 &= \sqrt{a_0g_0} \end{align}$

It is obvious that $\scriptstyle M(a_0,\, g_0) \;=\; M(a_1,\, g_1) \;=\; \ldots \;=\; M(a_n,\, g_n)$. As mentioned above, if $\scriptstyle r \;\ge\; 0$ then $\scriptstyle M(ra_0,\, rg_0) \;=\; rM(a_0,\, g_0)$. Therefore following expression holds:

$g = a_0M\left(1, \frac{g_0}{a_0}\right) = a_1M\left(1, \frac{g_1}{a_1}\right)\qquad(2)$

Next we define 4 new variables:

$\begin{align} x &= \frac{g_0}{a_0}, \quad x_1=\frac{g_1}{a_1} \\ y &= \frac{1}{M\left(1, \frac{g_0}{a_0}\right)}, \quad y_1 = \frac{1}{M\left(1, \frac{g_1}{a_1}\right)}\qquad(3) \end{align}$

Furthermore, from (1) we can deduce the following relation between $\scriptstyle x$ and $\scriptstyle x_1$:

$x_1 = \frac{2\sqrt{x}}{1 + x}$

$\Rightarrow{dx_1 \over dx} = \frac{1 - x}{(1 + x)^2\sqrt{x}} = \frac{\left(x_1 - x_1^3\right)(1 + x)^2}{2\left(x - x^3\right)}\qquad(4)$

From (2) and (3) we can deduce that

$\begin{align} y &= y_1\frac{a_0}{a_1} = y_1\frac{2a_0}{a_0 + g_0} = \frac{2y_1}{1 + x}\\ {dy \over dx} &= -\frac{2}{(1 + x)^2}y_1 + \frac{2}{1 + x}{dy_1 \over dx_1}{dx_1 \over dx} \end{align}$

If we substitute (4) to the last expression and multiply it by $\scriptstyle x \,-\, x^3$ we'll get

$\left(x - x^3\right){dy \over dx} = \frac{2x(x - 1)}{1 + x}y_1 + (1 + x)\left(x_1 - x_1^3\right){dy_1 \over dx_1}$

Taking derivative on both sides we will get the next expression:

$\frac{d}{dx}{\left[\left(x - x^3\right){dy \over dx}\right]} = 2y_1{d \over dx}\left[\frac{x(x - 1)}{1 + x}\right] + \frac{2x(x - 1)}{1 + x}{dy_1 \over dx_1}{dx_1 \over dx} + \left(x_1 - x_1^3\right){dy_1 \over dx_1} + (1 + x) {d \over dx_1}{\left[\left(x_1 - x_1^3\right){dy_1 \over dx_1}\right]}{dx_1 \over dx}$

After some elementary rearrangement we get:

${d \over dx}\left[\left(x - x^3\right){dy \over dx}\right] - xy = \frac{1 - x}{(1 + x)\sqrt{x}}\left\{{d \over dx_1}\left[\left(x_1 - x_1^3\right){dy_1 \over dx_1}\right] - x_1y_1\right\}$

Using the same considerations we can deduce that:

${d \over dx_1}\left[\left(x_1 - x_1^3\right){dy_1 \over dx_1}\right] - x_1y_1 = \frac{1 - x_1}{(1 + x_1)\sqrt{x_1}}\left\{{d \over dx_2}\left[\left(x_2 - x_2^3\right){dy_2 \over dx_2}\right] - x_2y_2\right\}$

where $\scriptstyle x_2 \;=\; \frac{g_2}{a_2}$.

We can continue the process. Assuming,

$\begin{align} H^*(y) &\equiv {d \over dx}\left[\left(x - x^3\right){dy \over dx}\right] - xy \\ H^*(y) &= \frac{1 - x}{(1 + x)\sqrt{x}}\frac{1 - x_1}{(1 + x_1)\sqrt{x_1}} \ldots \frac{1 - x_n}{(1 + x_n)\sqrt{x_n}}H^*(y_n) \end{align}$

We know that $\scriptstyle \lim_{n \to \infty}x_n \;=\; \lim_{n \to \infty}\frac{g_n}{a_n} \;=\; 1$, so $\scriptstyle 1 \,-\, x_n \;\rightarrow\; 0$ as $\scriptstyle n \;\rightarrow\; \infty$. Therefore:

$H^*(y) = 0$

So we found a differential equation for y:

${d \over dx}\left[\left(x - x^3\right){dy \over dx}\right] - xy = 0$

which is equivalent to:

$\left(x - x^3\right){d^2y \over dx^2} + \left(1 - 3x^2\right){dy \over dx} - xy = 0$

One of the solutions to the above equation is the complete elliptic integral of the first kind $\scriptstyle K(x)$.

$K(x) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1 - x^2\sin^2(\theta)}}$

The second solution is $\scriptstyle K\left(\sqrt{1 \,-\, x^2}\right)$, assuming $\scriptstyle 0 \;<\; x \;<\; 1$.

So, we can write the complete expression for y:

$y = \alpha K(x) + \beta K\left(\sqrt{1 - x^2}\right)$

Using the definition of y and x as

$\begin{align} y &= \frac{1}{M\left(1, \frac{g_0}{a_0}\right)} = \frac{a_0}{M(a_0, g_0)} \\ x &= \frac{g_0}{a_0} \end{align}$

we conclude that

$\frac{a_0}{M(a_0, g_0)} = \alpha K\left(\frac{g_0}{a_0}\right) + \beta K\left(\sqrt{1 - \frac{g_0^2}{a_0^2}}\right) \qquad(5)$

Finally we need to find the values of α and β. It is easy to see that $\scriptstyle M(a_0,\, a_0) \;=\; a_0$. Substituting this to the last equation we get:

$\frac{a_0}{a_0} = \alpha K(1) + \beta K(0)$

The values of K(x) at x = 0, 1 are: $\scriptstyle K(0) \;=\; \frac{\pi}{2}$, $\scriptstyle K(1) \;=\; \infty$, so α must be equal to 0. Therefore

$1 = \beta \frac{\pi}{2} \Rightarrow \beta = \frac{2}{\pi}$

Returning to (5) and subtitutiong for β, we get an expression for $\scriptstyle M(a_0,\, g_0)$:

$\frac{a_0}{M(a_0, g_0)} = \frac{2}{\pi} K\left(\sqrt{1 - \frac{g_0^2}{a_0^2}}\right)$

hence

$M(a_0, g_0) = \frac{a_0\pi}{2 K\left(\sqrt{1 - \frac{g_0^2}{a_0^2}}\right)}$

The expression in the properties section stated that

$M(a_0, g_0) = \frac{\pi}{4} (a_0 + g_0) \; / \; K\left(\frac{a_0 - g_0}{a_0 + g_0}\right)$

To prove it we can use the expression mentioned above:

$g = M(a_0, g_0) = M(a_1, g_1)$

So,

$g = \frac{\frac{\pi}{2}(a_0 + g_0)}{2 K\left(\sqrt{1 - \frac{a_0g_0}{\frac{1}{4}(a_0 + g_0)^2}}\right)} = \frac{\pi(a_0 + g_0)}{4 K\sqrt{\left(\frac{a_0 - g_0}{a_0 + g_0}\right)^2}} = \frac{\pi}{4} (a_0 + g_0) \; / \; K\left(\frac{a_0 - g_0}{a_0 + g_0}\right)$

which completes the proof.

See also

• Generalized mean
• Inequality of arithmetic and geometric means
• Gauss–Legendre algorithm

References

• Adlaj, Semjon (September 2012). "An eloquent formula for the perimeter of an ellipse". Notices of the AMS 59 (8): 1094–1099. 
• Jonathan Borwein, Peter Borwein, Pi and the AGM. A study in analytic number theory and computational complexity. Reprint of the 1987 original. Canadian Mathematical Society Series of Monographs and Advanced Texts, 4. A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1998. xvi+414 pp. ISBN 0-471-31515-X MR1641658
• Zoltán Daróczy, Zsolt Páles, Gauss-composition of means and the solution of the Matkowski–Suto problem. Publ. Math. Debrecen 61/1-2 (2002), 157–218.
• M. Hazewinkel (2001), "Arithmetic–geometric mean process", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4 
• Weisstein, Eric W., "Arithmetic–Geometric mean", MathWorld.