Arithmetic–geometric mean

Example

To find the arithmetic–geometric mean of and , iterate as follows:\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}The first five iterations give the following values:

The number of digits in which a**n and g**n 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 .

History

The first algorithm based on this sequence pair appeared in the works of Joseph-Louis Lagrange. Its properties were further analyzed by Carl Friedrich Gauss.

Properties

Both the geometric mean and arithmetic mean of two positive numbers x and y are between the two numbers. (They are strictly between when x ≠ y.) The geometric mean of two positive numbers is never greater than the arithmetic mean. So the geometric means are an increasing sequence g ≤ g ≤ g ≤ ...; the arithmetic means are a decreasing sequence a ≥ a ≥ a ≥ ...; and gn ≤ M(x, y) ≤ an for any n. These are strict inequalities if x ≠ y.

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

If r ≥ 0 then .

There is an integral-form expression for M(x, y):\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}where K(k) is the complete elliptic integral of the first kind:K(k) = \int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}} 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.

The arithmetic–geometric mean is connected to the Jacobi theta function \theta_3 byM(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},which upon setting x=1/\sqrt{2} givesM(1,1/\sqrt{2})=\left(\sum_{n\in\mathbb{Z}}e^{-n^2\pi}\right)^{-2}.

Related concepts

The reciprocal of the arithmetic–geometric mean of 1 and the square root of 2 is Gauss's constant.\frac{1}{M(1, \sqrt{2})} = G = 0.8346268\dotsIn 1799, Gauss provedBy 1799, Gauss had two proofs of the theorem, but neither of them was rigorous from the modern point of view. thatM(1,\sqrt{2})=\frac{\pi}{\varpi}where \varpi is the lemniscate constant.

In 1941, M(1,\sqrt{2}) (and hence G) was proved transcendental by Theodor Schneider.In particular, he proved that the beta function \Beta (a,b) is transcendental for all a,b\in\mathbb{Q}\setminus\mathbb{Z} such that a+b\notin \mathbb{Z}_0^-. The fact that M(1,\sqrt{2}) is transcendental follows from M(1,\sqrt{2})=\tfrac{1}{2}\Beta \left(\tfrac{1}{2},\tfrac{3}{4}\right). The set {\pi,M(1,1/\sqrt{2})} is algebraically independent over \mathbb{Q}, but the set {\pi,M(1,1/\sqrt{2}),M'(1,1/\sqrt{2})} (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over \mathbb{Q}. In fact,\pi=2\sqrt{2}\frac{M^3(1,1/\sqrt{2})}{M'(1,1/\sqrt{2})}.The geometric–harmonic mean GH can be calculated using analogous sequences of geometric and harmonic means, and in fact .{{cite journal 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, and Jacobi elliptic functions.

Proof of existence

The inequality of arithmetic and geometric means implies thatg_n \leq a_nand thusg_{n + 1} = \sqrt{g_n \cdot a_n} \geq \sqrt{g_n \cdot g_n} = g_nthat is, the sequence gn is nondecreasing and bounded above by the larger of x and y. By the monotone convergence theorem, the sequence is convergent, so there exists a g such that:\lim_{n\to \infty}g_n = gHowever, we can also see that:a_n = \frac{g_{n + 1}^2}{g_n} 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

This proof is given by Gauss. Let

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

Changing the variable of integration to \theta', where

\begin{align} \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' \end{align}

\begin{align} \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'}, \end{align}

\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' ,

\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'}} \ ,

\begin{align} \sqrt{x^2\cos^2\theta+y^2\sin^2\theta} &= \frac{\sqrt}{((x+y)+(x-y)\sin^2\theta')}\ &= \frac{x ((x+y)-(x-y)\sin^2\theta')}{((x+y)+(x-y)\sin^2\theta')} \end{align}

This yields \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'}},

gives

\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}

Thus, we have

\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} The last equality comes from observing that I(z,z) = \pi/(2z).

Finally, we obtain the desired result

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

Applications

The number ''π''

According to the Gauss–Legendre algorithm,

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

where

c_j = \frac{1}{2}\left(a_{j-1}-g_{j-1}\right) ,

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

c_j = \frac{c_{j-1}^2}{4a_j} .

Complete elliptic integral ''K''(sin''α'')

Taking a_0 = 1 and g_0 = \cos\alpha yields the AGM

M(1,\cos\alpha) = \frac{\pi}{2K(\sin \alpha)} ,

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

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

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

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 (e**x, cos x, sin x). Subsequently, many authors went on to study the use of the AGM algorithms.

References

Notes

Citations

Sources

References

1. Cox, David. (January 1984). "The Arithmetic-Geometric Mean of Gauss". L'Enseignement Mathématique.
2. [https://www.wolframalpha.com/input/?i=agm%2824%2C+6%29 agm(24, 6)] at [[Wolfram Alpha]]
3. Bullen, P. S.. (2003). "Handbook of Means and Their Inequalities". Springer Netherlands.
4. "Elliptic Integrals".
5. (2011). "Analog Electronic Filters: Theory, Design and Synthesis". Springer.
6. (1987). "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity". Wiley-Interscience.
7. Schneider, Theodor. (1941). "Zur Theorie der Abelschen Funktionen und Integrale". Journal für die reine und angewandte Mathematik.
8. Todd, John. (1975). "The Lemniscate Constants". [[Communications of the ACM]].
9. G. V. Choodnovsky: ''Algebraic independence of constants connected with the functions of analysis'', Notices of the AMS 22, 1975, p. A-486
10. G. V. Chudnovsky: ''Contributions to The Theory of Transcendental Numbers'', American Mathematical Society, 1984, p. 6
11. (1987). "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity". Wiley-Interscience.
12. {{AS ref. 17. 598–599
13. King, Louis V.. (1924). "On the Direct Numerical Calculation of Elliptic Functions and Integrals". Cambridge University Press.
14. Salamin, Eugene. (1976 <!--). "Computation of π using arithmetic–geometric mean". [[Mathematics of Computation]].
15. Landen, John. (1775). "An investigation of a general theorem for finding the length of any arc of any conic hyperbola, by means of two elliptic arcs, with some other new and useful theorems deduced therefrom". [[Philosophical Transactions of the Royal Society]].
16. Brent, Richard P.. (1976). "Fast Multiple-Precision Evaluation of Elementary Functions". [[Journal of the ACM]].
17. (1987). "Pi and the AGM". Wiley.