Arithmetic–geometric mean

File:Generalized means + agm.png

Plot of the arithmetic–geometric mean <math>\operatorname{agm}(1,x)</math> among several generalized means.

In mathematics, the arithmetic–geometric mean (AGM or agM<ref name="Cox" />) of two positive real numbers Template:Math and Template:Math is the mutual limit of a sequence of arithmetic means and a sequence of geometric means. The arithmetic–geometric mean is used in fast algorithms for exponential, trigonometric functions, and other special functions, as well as some mathematical constants, in particular, [[computing π|computing Template:Mvar]].

The AGM is defined as the limit of the interdependent sequences <math>a_i</math> and <math>g_i</math>. Assuming <math> x \geq y \geq 0</math>, we write:<math display=block>\begin{align}

a_0 &= x,\\
g_0 &= y\\
a_{n+1} &= \tfrac12(a_n + g_n),\\
g_{n+1} &= \sqrt{a_n g_n}\, .

\end{align}</math>These two sequences converge to the same number, the arithmetic–geometric mean of Template:Math and Template:Math; it is denoted by Template:Math, or sometimes by Template:Math or Template:Math.

The arithmetic–geometric mean can be extended to complex numbers and, when the branches of the square root are allowed to be taken inconsistently, it is a multivalued function.<ref name="Cox">Template:Cite journal</ref>

Example

To find the arithmetic–geometric mean of Template:Math and Template:Math, iterate as follows:<math display=block>\begin{array}{rcccl}

a_1 & = & \tfrac12(24 + 6) & = & 15\\
g_1 & = & \sqrt{24 \cdot 6} & = & 12\\
a_2 & = & \tfrac12(15 + 12) & = & 13.5\\
g_2 & = & \sqrt{15 \cdot 12} & = & 13.416\ 407\ 8649\dots\\
& & \vdots & &

\end{array}</math>The first five iterations give the following values:

Template:Math	Template:Math	Template:Math
0	24	6
1	Template:Underline5	Template:Underline2
2	Template:Underline.5	Template:Underline.416 407 864 998 738 178 455 042...
3	Template:Underline 203 932 499 369 089 227 521...	Template:Underline 139 030 990 984 877 207 090...
4	Template:Underline45 176 983 217 305...	Template:Underline06 053 858 316 334...
5	Template:Underline20...	Template:Underline06...

The number of digits in which Template:Math and Template:Math agree (underlined) approximately doubles with each iteration. The arithmetic–geometric mean of 24 and 6 is the common limit of these two sequences, which is approximately Template:Val.<ref>agm(24, 6) at Wolfram Alpha</ref>

History

The first algorithm based on this sequence pair appeared in the works of Lagrange. Its properties were further analyzed by Gauss.<ref name="Cox"/>

Properties

Both the geometric mean and arithmetic mean of two positive numbers Template:Mvar and Template:Mvar are between the two numbers. (They are strictly between when Template:Math.) The geometric mean of two positive numbers is never greater than the arithmetic mean.<ref>Template:Cite book</ref> So the geometric means are an increasing sequence Template:Math; the arithmetic means are a decreasing sequence Template:Math; and Template:Math for any Template:Mvar. These are strict inequalities if Template:Math.

Template:Math is thus a number between Template:Math and Template:Math; it is also between the geometric and arithmetic mean of Template:Math and Template:Math.

If Template:Math then Template:Math.

There is an integral-form expression for Template:Math:<ref>Template:Dlmf</ref><math display=block>\begin{align}

M(x,y) &= \frac{\pi}{2} \left( \int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}} \right)^{-1}\\

&=\pi\left(\int_0^\infty \frac{dt}{\sqrt{t(t+x^2)(t+y^2)}}\right)^{-1}\\

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

\end{align}</math>where Template:Math is the complete elliptic integral of the first kind:<math display="block">K(k) = \int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}} </math>Since the arithmetic–geometric process converges so quickly, it provides an efficient way to compute elliptic integrals, which are used, for example, in elliptic filter design.<ref name="Dimopoulos2011">Template:Cite book</ref>

The arithmetic–geometric mean is connected to the Jacobi theta function <math>\theta_3</math> by<ref>Template:Cite book pages 35, 40</ref><math display="block">M(1,x)=\theta_3^{-2}\left(\exp \left(-\pi \frac{M(1,x)}{M\left(1,\sqrt{1-x^2}\right)}\right)\right)=\left(\sum_{n\in\mathbb{Z}}\exp \left(-n^2 \pi \frac{M(1,x)}{M\left(1,\sqrt{1-x^2}\right)}\right)\right)^{-2},</math>which upon setting <math>x=1/\sqrt{2}</math> gives<math display="block">M(1,1/\sqrt{2})=\left(\sum_{n\in\mathbb{Z}}e^{-n^2\pi}\right)^{-2}.</math>

Related concepts

The reciprocal of the arithmetic–geometric mean of 1 and the square root of 2 is Gauss's constant.<math display=block>\frac{1}{M(1, \sqrt{2})} = G = 0.8346268\dots</math>In 1799, Gauss proved<ref group="note">By 1799, Gauss had two proofs of the theorem, but neither of them was rigorous from the modern point of view.</ref> that<math display="block">M(1,\sqrt{2})=\frac{\pi}{\varpi}</math>where <math>\varpi</math> is the lemniscate constant.

In 1941, <math>M(1,\sqrt{2})</math> (and hence <math>G</math>) was proved transcendental by Theodor Schneider.<ref group="note">In particular, he proved that the beta function <math>\Beta (a,b)</math> is transcendental for all <math>a,b\in\mathbb{Q}\setminus\mathbb{Z}</math> such that <math>a+b\notin \mathbb{Z}_0^-</math>. The fact that <math>M(1,\sqrt{2})</math> is transcendental follows from <math>M(1,\sqrt{2})=\tfrac{1}{2}\Beta \left(\tfrac{1}{2},\tfrac{3}{4}\right).</math></ref><ref>Template:Cite journal</ref><ref>Template:Cite journal</ref> The set <math>\{\pi,M(1,1/\sqrt{2})\}</math> is algebraically independent over <math>\mathbb{Q}</math>,<ref>G. V. Choodnovsky: Algebraic independence of constants connected with the functions of analysis, Notices of the AMS 22, 1975, p. A-486</ref><ref>G. V. Chudnovsky: Contributions to The Theory of Transcendental Numbers, American Mathematical Society, 1984, p. 6</ref> but the set <math>\{\pi,M(1,1/\sqrt{2}),M'(1,1/\sqrt{2})\}</math> (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over <math>\mathbb{Q}</math>. In fact,<ref>Template:Cite book p. 45</ref><math display="block">\pi=2\sqrt{2}\frac{M^3(1,1/\sqrt{2})}{M'(1,1/\sqrt{2})}.</math>The geometric–harmonic mean GH can be calculated using analogous sequences of geometric and harmonic means, and in fact Template:Math.<ref>Template:Cite journal</ref> The arithmetic–harmonic mean is equivalent to the geometric mean.

The arithmetic–geometric mean can be used to compute – among others – logarithms, complete and incomplete elliptic integrals of the first and second kind,<ref>Template:AS ref</ref> and Jacobi elliptic functions.<ref>Template:Cite book</ref>

Proof of existence

The inequality of arithmetic and geometric means implies that<math display=block>g_n \leq a_n</math>and thus<math display=block>g_{n + 1} = \sqrt{g_n \cdot a_n} \geq \sqrt{g_n \cdot g_n} = g_n</math>that is, the sequence Template:Math is nondecreasing and bounded above by the larger of Template:Math and Template:Math. By the monotone convergence theorem, the sequence is convergent, so there exists a Template:Math such that:<math display=block>\lim_{n\to \infty}g_n = g</math>However, we can also see that:<math display=block>a_n = \frac{g_{n + 1}^2}{g_n}</math> and so: <math display=block>\lim_{n\to \infty}a_n = \lim_{n\to \infty}\frac{g_{n + 1}^2}{g_{n}} = \frac{g^2}{g} = g</math>

Q.E.D.

Proof of the integral-form expression

This proof is given by Gauss.<ref name="Cox" /> Let

<math display=block>I(x,y) = \int_0^{\pi/2}\frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}} ,</math>

Changing the variable of integration to <math>\theta'</math>, where

<math display="block"> \sin\theta = \frac{2x\sin\theta'}{(x+y)+(x-y)\sin^2\theta'} \Rightarrow d(\sin\theta)=d\left(\frac{2x\sin\theta'}{(x+y)+(x-y)\sin^2\theta'}\right)\Rightarrow \cos\theta\ d\theta =2x \frac{(x+y)-(x-y)\sin^2\theta'}{((x+y)+(x-y)\sin^2\theta')^2}\ \cos\theta' d\theta' </math>

<math display="block">\cos\theta = \frac{\sqrt{(x+y)^2-2(x^2+y^2)\sin^2\theta'+(x-y)^2\sin^4\theta'}}{(x+y)+(x-y)\sin^2\theta'}=\frac{\cos\theta'\sqrt{(x-y)^2\cos^2\theta'+4xy}}{(x+y)+(x-y)\sin^2\theta'}=\frac{\cos\theta'\sqrt{(x+y)^2\cos^2\theta'+4xy\sin^2\theta'}}{(x+y)+(x-y)\sin^2\theta'} ,</math>

<math display="block">\Rightarrow \cos\theta\ d\theta =\frac{\cos\theta'\sqrt{(x+y)^2\cos^2\theta'+4xy\sin^2\theta'}}{(x+y)+(x-y)\sin^2\theta'}\ d\theta =2x \frac{(x+y)-(x-y)\sin^2\theta'}{((x+y)+(x-y)\sin^2\theta')^2}\ \cos\theta' d\theta' ,</math>

<math display="block"> \Rightarrow d\theta = \frac{x((x+y)-(x-y)\sin^2\theta')}{((x+y)+(x-y)\sin^2\theta')} \frac{2 d\theta'}{\sqrt{(x+y)^2\cos^2\theta'+4xy\sin^2\theta'}}

\ ,</math>

<math display="block"> \sqrt{x^2\cos^2\theta+y^2\sin^2\theta} = \frac{\sqrtTemplate:X^2 ((x+y)^2-2(x^2+y^2)\sin^2\theta'+(x-y)^2\sin^4\theta')+4x^2y^2\sin^2\theta'}{((x+y)+(x-y)\sin^2\theta')}= \frac{x ((x+y)-(x-y)\sin^2\theta')}{((x+y)+(x-y)\sin^2\theta')}</math>

This yields <math display="block"> \frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}} = \frac{2 d\theta'}{\sqrt{(x+y)^2\cos^2\theta'+4xy\sin^2\theta'}} = \frac{d\theta'}{\sqrt{((\frac{x+y}{2})^2\cos^2\theta'+(\sqrt{xy})^2\sin^2\theta'}}

 ,</math>

gives

<math display="block">\begin{align} I(x,y) &= \int_0^{\pi/2}\frac{d\theta'}{\sqrt{((\frac{x+y}{2})^2\cos^2\theta'+(\sqrt{xy})^2\sin^2\theta'}}\\

      &= I\bigl(\tfrac{x+y}{2},\sqrt{xy}\bigr) .

\end{align}</math>

Thus, we have

<math display=block> \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} </math> The last equality comes from observing that <math>I(z,z) = \pi/(2z)</math>.

Finally, we obtain the desired result

<math display=block>M(x,y) = \pi/\bigl(2 I(x,y) \bigr) .</math>

Applications

The number π

According to the Gauss–Legendre algorithm,<ref>Template:Cite journal</ref>

<math display=block>\pi = \frac{4\,M(1,1/\sqrt{2})^2} {1 - \displaystyle\sum_{j=1}^\infty 2^{j+1} c_j^2} ,</math>

where

<math display=block>c_j = \frac{1}{2}\left(a_{j-1}-g_{j-1}\right) ,</math>

with <math>a_0=1</math> and <math>g_0=1/\sqrt{2}</math>, which can be computed without loss of precision using

<math display=block>c_j = \frac{c_{j-1}^2}{4a_j} .</math>

Complete elliptic integral K(sinα)

Taking <math>a_0 = 1</math> and <math>g_0 = \cos\alpha</math> yields the AGM

<math display=block>M(1,\cos\alpha) = \frac{\pi}{2K(\sin \alpha)} ,</math>

where Template:Math is a complete elliptic integral of the first kind:

<math display=block>K(k) = \int_0^{\pi/2}(1 - k^2 \sin^2\theta)^{-1/2} \, d\theta.</math>

That is to say that this quarter period may be efficiently computed through the AGM, <math display=block>K(k) = \frac{\pi}{2M(1,\sqrt{1-k^2})} .</math>

Other applications

Using this property of the AGM along with the ascending transformations of John Landen,<ref>Template:Cite journal</ref> Richard P. Brent<ref>Template:Cite journal</ref> suggested the first AGM algorithms for the fast evaluation of elementary transcendental functions (Template:Math, Template:Math, Template:Math). Subsequently, many authors went on to study the use of the AGM algorithms.<ref>Template:Cite book</ref>

See also

• Landen's transformation
• Gauss–Legendre algorithm
• Generalized mean

References

Sources

• Template:Cite journal
• Template:SpringerEOM
• Template:Mathworld