Английская Википедия:Arithmetic–geometric mean

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Файл:Generalized means + agm.png

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

In mathematics, the arithmetic–geometric mean of two positive real numbers Шаблон:Math and Шаблон:Math is the mutual limit of a sequence of arithmetic means and a sequence of geometric means:

Begin the sequences with x and y: <math display=block>\begin{align}

a_0 &= x,\\
g_0 &= y.

\end{align}</math>

Then define the two interdependent sequences Шаблон:Math and Шаблон:Math as

<math display=block>\begin{align}

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 Шаблон:Math and Шаблон:Math; it is denoted by Шаблон:Math, or sometimes by Шаблон:Math or Шаблон:Math.

The arithmetic–geometric mean is used in fast algorithms for exponential and trigonometric functions, as well as some mathematical constants, in particular, [[computing π|computing Шаблон:Mvar]].

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, in general, a multivalued function.

Example

To find the arithmetic–geometric mean of Шаблон:Math and Шаблон: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:

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

The number of digits in which Шаблон:Math and Шаблон: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 Шаблон:Val.

History

The first algorithm based on this sequence pair appeared in the works of Lagrange. 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, for Шаблон:Math, Шаблон:Math is an increasing sequence, Шаблон:Math is a decreasing sequence, and Шаблон:Math. These are strict inequalities if Шаблон:Math.

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

If Шаблон:Math, then Шаблон:Math.

There is an integral-form expression for Шаблон:Math:

<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 Шаблон: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>

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

The arithmetic–geometric mean is connected to the Jacobi theta function <math>\theta_3</math> by

<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 called Gauss's constant, after Carl Friedrich Gauss.

<math display=block>\frac{1}{M(1, \sqrt{2})} = G = 0.8346268\dots</math>

In 1799, Gauss proved 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 proven transcendental by Theodor Schneider. The set <math>\{\pi,M(1,1/\sqrt{2})\}</math> is algebraically independent over <math>\mathbb{Q}</math>, 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,

<math display="block">\pi=2\sqrt{2}\frac{M^3(1,1/\sqrt{2})}{M'(1,1/\sqrt{2})}.</math>

The geometric–harmonic mean can be calculated by an analogous method, using sequences of geometric and harmonic means. One finds that Шаблон:Math. The arithmetic–harmonic mean can be similarly defined, but takes the same value as the geometric mean (see section "Calculation" there).

The arithmetic–geometric mean can be used to compute – among others – logarithms, complete and incomplete elliptic integrals of the first and second kind, and Jacobi elliptic functions.

Proof of existence

From the inequality of arithmetic and geometric means we can conclude 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 Шаблон:Math is nondecreasing.

Furthermore, it is easy to see that it is also bounded above by the larger of Шаблон:Math and Шаблон:Math (which follows from the fact that both the arithmetic and geometric means of two numbers lie between them). Thus, by the monotone convergence theorem, the sequence is convergent, so there exists a Шаблон: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. 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'} ,</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'} ,</math>

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

d\theta'\ ,</math>

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

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

= \frac{2\cos\theta' d\theta'}{\sqrt{(x+y)^2-2(x^2+y^2)\sin^2\theta'+(x-y)^2\sin^4\theta'}}

 ,</math>

gives

<math display=block> \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} </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

For example, according to the Gauss–Legendre algorithm:

<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 Шаблон: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, Richard P. Brent suggested the first AGM algorithms for the fast evaluation of elementary transcendental functions (Шаблон:Math, Шаблон:Math, Шаблон:Math). Subsequently, many authors went on to study the use of the AGM algorithms.

See also

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

References

Sources

• Шаблон:Cite journal
• Шаблон:SpringerEOM
• Шаблон:Mathworld